I got a 133.5 on the 10b. Big improvement over my 10a score (114).

[quote="cognos599"][quote="rainbowpenguin"]What did you guys get on the amc 10/12 b?

I got a 132 on the 10b.[/quote]

um yeah, 121.5 on 10A and 126 on 10B, btw rainbowpenguin, r u mark??!?!?!?! if so I LOVE YOU and hate you at the same time.[/quote]

hahaha, no, I wish I was mark though- he's amazing at math (6th grade- omg).

Why do you hate me cognos? Or did you mean you love/hate mark? (Did he beat you in MathCounts or something?)

just someone in West Lafayette

who r u?

144 on the 10B

Failure on the 12A (well since i thought I could get 130+)

no he was talking about the indiana mathleague. oh and btw thank you rainbowpenguin, i will find out who u r.

[quote="rainbowpenguin"]hahaha, cognos, I found out who you are :)

only one person fits "my team got a 176, we got pwn by Honey Creek and I got a 37" So, that narrows it to one person and you're pretty amazing dude- you tied Justin (National MathCounts Written Winner), but oh yeah, you lost to  Mark by one point :)

I found out who you are, but there's no way you could find out who I am.

I took the IML though (39 in 7th grade and 40 in 8th grade, but I'm 2 years older than you, so thank god, I still can beat you in the amc). I gave you a huge hint. now u can find me.

p.s. good luck at state- I'm rooting for you, mark, and justin to go to nationals![/quote]

no one matches that discription. i checked the mathleague website and the only person who is 2 years older than me and got a  39 in 7th grade and a 40 in 8th grade is Andrew Huh and he is from honey creek not from west lafayette. i am really puzzled. i do not have the intellect to figure out who you are. if you don't want people to know who you are, you can pm me ur name and i won't tell ne one.

haha well allowing for a few typos, rainbowpenguin could also be milner, though he got a 38 in 7th grade. He's the only one whom I think would actually take the time to get on AoPS anyway..

132 on the 10 B - two guesses, both wrong, and another careless mistake. I was aiming for a perfect.

are we allowed to discuss this now anyway?

sorry, undefined and cognos, but I never said that I always lived in IN, but checking every state's IML scores would just be long and boring- hint... I used to live in a State touching IN (that narrows it down to like 4 states), but I'll also say Alex Zhai lives in the same state (that totally gave it away).

undefined, are you also in 8th grade?

good luck you guys in MathCounts State! Isn't next weekend, like March 8th?

[quote="cognos599"]you did say that you live in west lafayette though i think. i don't recognized anyone that is in 10th grade and got a perfect in 8th and a 39 in 7th and lives in Illinois.  :oops: , i don't believe that you really know anything about me other than my name school and the reason you know that i tied with justin and got 1 pt less than mark is that i have posted that before. did u play a trick on me? if so, well done.  :idea:[/quote]

you're right, I really don't know anything about you other than what is on the Internet, but yeah, all that you said is the paragraph above is right, check again- you'd notice one person who fits those stuff. Next year, we'll be competing rivals :) West Side vs. HHS

When I was in 8th grade, I had all the speed and talent needed to place top 4 at State and go to Nationals, but unfortunately, after doing really well at regionals (I placed 3rd like u), I wasn't as focused and worried as I should have been. Hopefully, you won't make my mistake- be focused, score good on both sprint and target, and place top four (hahaha, I wish doing that was as easy as saying that)

Is state at Purdue? (I might come if it is)

I'm a disgrace to people from Indiana and Indians.  I got a 73.5 on the AMC 10.  Oh well, there's still two more years to (not make it to AIME) through the AMC 10.

[quote="cognos599"]ya we are allowed to discuss it. omg rainbowpenguin r u daniel milner?[/quote]
OMG...If it [b]was[/b] Milner, and you didn't know for this long, that would be just funny.

Milner's not [i]that[/i] amazing.. Catlin was, however :P

Michael Catlin actually gave his second place nationals (team) trophy to the school...why would anyone do that?

there's a very small chance that anyone else in the future will ever get as close as Catlin did to being a national champion (in west lafayette, at least), which ensures that his trophy (with his name on it) stays at the school for a very long time.. :D

The non-greedy hypothesis would be that he wanted to let the school show off for having educated such a bright young student. My name, however, is supposedly on a plaque in the Jr. High office, though. dunno. maybe not.

10 years from now ppl will look at that plaque and say "who the heck is soo-hyun?"

if anyone even looks at it. what was it for?

No they'll already have started saying that.. no one really looks at plaques at the school anyway, do they?

It was for being the first middle schooler from west side to make the USAMO.. which is kinda weird since it was only because of the lowered index.. meh.

They were [i]given[/i] a plaque for that?

[quote="mathcountsfreak"][quote="cognos599"]150
141
132
123
114
105
96
87
78
69[/quote]
What? Michael is confused.[/quote]

oops, did i actually post this, i was just typing it and i dunno i guess it just submitted itself.

[quote="undefined117"]--seconded.

Dudeeeee those look like depressing practice AMC scores..[/quote]

i took 150 and kept subtracting 9.

[quote="undefined117"]yeah.. it'd be like replacing every two right answers with two blanks.

the AIME is getting my all tense for no reason :([/quote]

i have a feeling ill get a 2 like last year.  :(

[quote="undefined117"]goal is a 9, but i'll probably get the same score as last year, too -_-[/quote]

or lower

[quote="undefined117"]you wish :D[/quote]

you've gotten all rusty though, sooey.

[quote="MustItAlwaysBeMe"]What if you make it to IMO, but aren't in the US...lol

I know, it won't happen.[/quote]

idk, would you be on the JMO team? japan gets pwned every year no offense.

They do?

I honestly would expect them to do pretty well.  :D

Nah, I won't be on the team in Japan cause I haven't taken the Japanese equivalent of the AMC/AIME.

and as for being rusty, I'm paying for being lazy. argh. Just to make myself feel better, I'll say I've been concentrating on other stuff. [i]and of courseeeee they were educational[/i] how could you suspect otherwise?

[quote="the future"]Anyone want to make a suggestion of AIME techniques/revive the problem marathon with AIME level questions.[/quote]

um shawn, you are the one who should be giving people AIME suggestions. n e way, ill start a new topic for only AIME problems.

ohhhh no I made 6 careless mistakes on the last 15 AIME questions I attempted (1993 and 2001 I)

dunno.. I might do another one today, and if I don't decrease my mistakes to [b]0[/b] I'll qualify as a failure.


algebra
difference of squares
special factorizations


Recently, I've been getting the type of e-mail spam that requests you to help some random monarch with some transaction at 15%...etc.

The most recent is gijado.
This username should be banned.

I got continuously spammed by that person in my private message inbox.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=707044]this[/url] for more information. A transcript of the pm is availbale there as well as on my blog.

Yeah, me too.  I'm getting continually spammed by this one person (not gijado) in my private message box.  I'll put up the pms if anyone wants to see them.

ccy tell me who the person is on pm please.

I just received an email notifying me of a PM from a certain "gijado".  When I went to my inbox, there was no new pm.  Is that normal?
Thanks.

gijado got banned before you checked your inbox. The email also got deleted.

[quote="Valentin Vornicu"]One more thing guys: please make the difference next time between PM (private message) and Email. It's really helpful for us.[/quote]
Well , in my case it was a PM, and i have received an email ( as usual) to notify me of that.
Thank you.


geometry
incenter
circumcircle
geometry unsolved

Let $C,B,E$ be three points on a straight line $l$ in that order. Suppose that $A$ and $D$ are two points on the same side of $l$ such that

(i)  $\angle ACE = \angle CDE = 90^{o}$ and

(ii) $CA = CB = CD$.

Let $F$ be the point of intersection of the segment $AB$ and the circumcircle of $\triangle ADC$. Prove that $F$ is the incentre of $\triangle CDE$.

Since $CA=CB=CD$, so $C$ is the centre of the circle passing through $A$, $B$ and $D$. Thus $\angle{BCD}=2\angle{BAD}$. Because $F$ is on the circumcircle of $\triangle{ACD}$, so $\angle{DAF}=\angle{DCF}$, hence $\angle{BCD}=2\angle{DCF}$, thus $\angle{BCF}=\angle{DCF}$.

Because $AC \perp CB$ and $AC=CB$, so $\angle{CDF}=\angle{CAF}=\angle{CAB}=45^o$. Since $\angle{CDE}=90^o$, so $\angle{CDF}=\angle{EDF}$, thus $F$ is the intersection of two of the angle bisectors of $\triangle{CDE}$, so $F$ is the incentre.


The source is a 1987 chinese olympiad.

A3.  In a tournament each player plays every other player. Every game results in win or loss. A prize is awarded to every player X who fulfils the following condition: for each player Z who beats X there must be a player Y such that X beats Y and Y beats Z. Only one prize is awarded. Show that the recipient must have won every game.

Not sure if this counts as a proof:

[hide]Call the players Ai. Call the winner A0. Anyone who is unbeaten automatically fulfills requirements to be a winner.

Suppose A1 beat A0. There is only one winner, thus there must exist A2 that beat both A1 and A0. However, A2 is not the winner, thus there must exist A3 that beat both A2 and A1. But A3 is not the winner, thus there must exist A4... ad infinitum.

There are only a finite number of players, therefore, no one beat A0.[/hide]

I dont think that your proof is rigerous because you assume that the player who beats A_3, A_4, A5, etc. is not beaten by any of the players whom you have already mentioned.  Hmmmm....im not sure.

Anyways heres my solution:

[hide]I will first demonstrate by induction that in any tournament, in which every player plays all other players, there must be at least one player worthy of a prize.  In any one person tournament there is clearly a prize winner because there is only one person and he beats all the other players in the tournament.  Now, let us consider a tournament of n players and assume that any tournament of smaller size has at least one prize winner.  First consider one player in the tournament.  I will refer to this player as Bob.  We can now divide the rest of the players into two groups: those that lose to Bob (which I will refer to as group L) and those that defeat Bob (which I will refer to as group W).  We know that there are at most n-1 players in group W and there is a nonzero number of players in group W (unless Bob is a prize winner, in which case we are done).  So there is at least one player in W who would win a prize if both Bob and the players in group L were eliminated from the tournament.  This player beats Bob in the tournament (because he is in group W) and thus he also defeats everyone in group L by one degree of separation (through Bob).  Thus this player has beaten everyone in the tournament directly or through another player and he/she deserves a prize.

Now consider a completely separate tournament in which Bob also plays and earns an award.  The other players in the tournament can also be divided into groups L and W depending on whether they win or lose against Bob.  Assume, for the sake of contradiction, that group W consists of a nonzero number of players and also that there are no other prize winners in the tournament.  We demonstrated above that there must be at least one prize winner in any round robin tournament.  As a direct application of this fact, we know that there is at least one player in group W that would earn a prize if all players not in group W were eliminated.  Now we add the players not in group W back in as we did above and we can see that the prize winner in group W is a prize winner of the whole tournament.  This shows that whenever one player earns a prize either there are no players in group W (the prize winner defeats every other player) or there is another prize winner.[/hide]

I like that proof! Very clever. Yeah, I was unable to rigorize mine.


induction geometry integration rectangle calculus function symmetry

[quote="JRav"]This is a problem a friend tried to solve.  Needless to say, he was not successful so I'd like to see if anyone here can tackle it:

Find the self-inductance of a rectangle.  Let's assume current $ I$, and sides of length $ a$ and $ b$.

Hoping for a solution, or at least an attempt at one.  Thanks. :)[/quote]

The self-inductance, $ L \equal{}\frac{\Phi}{I}$.

\[ \Phi \equal{}\mathbf{B}\cdot\mathbf{A}\]

For a rectangular loop of area $ a \times b$,

\[ \mathbf{B} \equal{}\frac{\mu_0 I}{4\pi}\left(\frac{2b}{z^2 \plus{}({a\over 2})^2} \plus{} \frac{2a}{z^2 \plus{}({b\over 2})^2}\right) \hat{e}_z\]

At $ z \equal{} 0$, the center of the loop, this reduces to
\[ \mathbf{B} \equal{} \frac{2\mu_0 I}{\pi}\left(\frac{b}{a^2} \plus{}\frac{a}{b^2}\right)\hat{e}_z\]

The area is given by $ \mathbf{A} \equal{} ab \,\hat{e_z}$, and thus the flux is:
\[ \Phi \equal{}\mathbf{B}\cdot\mathbf{A} \equal{}\frac{2\mu_0 I}{\pi}\left(\frac{a^2}{b} \plus{}\frac{b^2}{a}\right)\]

Finally,
\[ \boxed{L \equal{}\frac{\Phi}{I} \equal{} \frac{2\mu_0}{\pi}\left(\frac{a^2}{b} \plus{}\frac{b^2}{a}\right)}\]

[quote="JRav"]Thanks for the solution.  Pretty simple actually.[/quote]

No sweat  :)

Not so easy...

(i)[quote="Dr. No"] \[ \mathbf{B} =\frac{\mu_0 I}{4\pi}\left(\frac{2b}{z^2 +({a\over 2})^2} + \frac{2a}{z^2 +({b\over 2})^2}\right) \hat{e}_z\] [/quote]

For which point?

(ii)[quote="Dr. No"]\[ \Phi =\mathbf{B}\cdot\mathbf{A} =\frac{2\mu_0 I}{\pi}\left(\frac{a^2}{b} +\frac{b^2}{a}\right)\] [/quote].

As $ \; \vert {B}$ is not constant in the retangle,

  \[ \Phi = \int _A{\mathbf{B}\cdot\mathbf{dA}}\]

I already specified for $ z \equal{} 0$, i.e. at the center of the loop (assuming the loop is in the x-y plane).

$ \mathbf{B}$ is pointed in the z-direction. I calculated $ \mathbf{B}$ from the Biot-Savart law. Evaluated it at $ z \equal{} 0$, and multiplied by the area of the loop. The B-field is independent of the x and y, which are what one integrates over to find the flux $ \Phi \equal{}\iint_{S} \mathbf{B}\cdot\mathrm{d}\mathbf{A}$.

[quote="Immanuel Bonfils"]Sorry, but it's all wrong.

Seems you calculated the field in a generic point of a perpendicular through the rectangle center (that is where I asked; for which point?" and not the particular one for z=0). That last one, you used as constant for all the points of the rectangle, and that is not true.
And obviously B  depends on x and y ![/quote]
I invite you to show the calculation which proves that. I can show what I did, and I will if need be, but not just now as the calculation is somewhat tedious and I don't have the time to spend. Nevertheless, unless I've made some gross misinterpretation of the geometry, the B-field depends only upon $ z$. Thus the B-field is independent of integration over $ x$ and $ y$.

Actually, in the center

\[{ \mathbf{B} = \frac{4\mu_0 I}{\pi\sqrt{a^2+b^2}}}\hat{e}_z\]

Let's make it interesting:

$ \vec{B}\equal{}B$ is constant throughout the rectangle.  Dimensions $ a$ and $ b$, current $ I$.  Calculate $ L$ for an arbitrary point in $ \mathbb{R}^{3}$.

If we really want to go crazy, let's say $ I\equal{}I(t)$ where $ I(t)$ is some function, whatever it happens to be (maybe oscillating, maybe exponential, maybe sawtooth, whatever).  Find $ L$ for an arbitary point in $ \mathbb{R}^{4}$.

[quote="Immanuel Bonfils"]Actually, in the center
\[{ \mathbf{B} = \frac {4\mu_0 I}{\pi\sqrt {a^2 + b^2}}}\hat{e}_z
\]
[/quote]

Okay, but what I had in mind is for you to show the calculation.

With that in mind,
\[ \mathrm{d}\mathbf{B} = \frac {\mu_{0}I}{4\pi}\left(\frac {2\mathrm{d}x}{z^2 + ({a\over 2})^2} + \frac {2\mathrm{d}y}{z^2 + ({b\over 2})^2}\right)\hat{e}_z
\]

\[ \mathbf{B} = \frac {\mu_0I}{4\pi}\int_{ - {b\over 2}}^{{b\over 2}}\frac {2\mathrm{d}x}{z^2 + ({a\over 2})^2} + \int_{ - {a\over 2}}^{{a\over 2}}\frac {2\mathrm{d}y}{z^2 + ({b\over 2})^2}\quad\hat{e}_z
\]

\[ \mathbf{B} = \frac {\mu_0I}{4\pi}\left(\frac {2b}{z^2 + ({a\over 2})^2} + \frac {2a}{z^2 + ({b\over 2})^2}\right)\hat{e}_z
\]

[quote="JRav"]Let's make it interesting:

$ \vec{B} \equal{} B$ is constant throughout the rectangle.  Dimensions $ a$ and $ b$, current $ I$.  Calculate $ L$ for an arbitrary point in $ \mathbb{R}^{3}$.

If we really want to go crazy, let's say $ I \equal{} I(t)$ where $ I(t)$ is some function, whatever it happens to be (maybe oscillating, maybe exponential, maybe sawtooth, whatever).  Find $ L$ for an arbitary point in $ \mathbb{R}^{4}$.[/quote]
The problem with that is $ L$ is independent of $ I$.

[quote="Immanuel Bonfils"]Dr. No

In the diagram you  took e [i]I[/i] $ \vert {dl}$ at the center of each of the rectangle side, and  after that, in the text, integrated over the whole side?!

JRaw

I guess [i]L[/i], means inductance, but it isn't  defined in a point...[/quote]

Immanuel,

Despite my kindergarten art work, you do have to integrate over the entire length of the wire.

OK

Let the origin of the Cartesian Plane be in the rectangle center and the sides  of length b an a parallel to the x and y axis, respectively.

(i)[u] In the center (origin)[/u] :

Due to the horizontal sides, parallel to x axis we have
\[ B_b = {{\mu_oI} \over { \pi a}}{b \over{\sqrt{a^2+b^2}}}\]
For the vertical ones:
\[ B_a ={{\mu_oI} \over { \pi b}}{a \over{\sqrt{a^2+b^2}}}\]

So, in the center, \[ B = 2  {{\mu_oI} \over { \pi ab}}\sqrt{a^2+b^2}\]

(ii)[u]Now let P(0,a/4) be another point of the rectangle.[/u]
\[ B{}_b{}_1 = {{4\mu_oI} \over {3 \pi a}}{b \over{\sqrt{9a^2+4b^2}}}\]
\[ B{}_b{}_2 = {{4\mu_oI} \over { \pi a}}{b \over{\sqrt{a^2+4b^2}}}\]
\[ B_a ={{\mu_oIa} \over {2 \pi b}}\left({1 \over \sqrt{a^2+4b^2}}+{3 \over\sqrt{9a^2+4b^2}}\right)\]

So, on P,  \[ B = {{\mu_oI} \over {3 \pi ab}}( 3\sqrt{a^2+4b^2}+\sqrt{9a^2+4b^2})\]

Is B constant in the rectangles surface?

Immanuel,

What I'm asking is how are you making the calculations for the B-field due to each of the wires? I can add so presumably if I accept your field values I will get your answer for the total field. But what I want to know is how you calculated the B-field in each wire? Notice that my solution doesn't have any radicals in it. How did you wind up with radicals in yours? I'm not trying to start an argument here, I would really like to know. In problems of this sort I don't always get the geometry right -- if I've done so here, I'd like to figure out why.

Let's take the  field at the origin, due to the lower horizontal side , AB, as a sample.
$ \; I\vec{dl}\; \equal{} I dx \hat{x}  \;$  at point Q(x , A/2),  and  $ \; \vec{r}\equal{}O \minus{} Q\;$ so $ r^2 \equal{} {a^2 \over 4} \plus{}x^2$
So $ d\vec{B}\;\equal{}{{\mu_o I dx} \over {4 \pi r^2}}sin \theta \;  \hat{z}\;$.

From the triangle QMO (M central point of AB) we get
$ \; x \equal{} {a \over 2} \cot \theta\; \rightarrow \; dx \equal{}  \minus{} {a \over 2} \; \csc^2 \theta\;d\theta\;$ and  $ \;r \equal{} {a \over 2}\; csc \theta \;$.

In the expression for  the intensity of $ \;d \vec{B}\;$ we have:
$ \;dB \equal{} {{\mu_o I} \over {2 \pi a}}\;sin \theta\; d\theta$

Integration from A to B leads to

$ \; B_b \equal{} {{\mu_o I} \over {2 \pi a}}\;(cos \alpha \minus{} cos \beta) \; , \alpha\; and \; \beta \;$ represents  $ \;\theta\;$ in the extremes A and B.

By symmetry, $ \;\minus{}cos \beta \equal{}  cos \alpha\;$ that comes from the triangle AMO as
\[ \cos \alpha \equal{} {b \over \sqrt{a^2\plus{}b^2}}\]
So
\[ B_b \equal{} {{\mu_o I} \over { \pi a}}\;{b \over \sqrt{a^2\plus{}b^2}}\]


Every triangle can be split into two parts....for a formal solution:
Let's call the triangle $ \triangle {ABC}$.  WLOG, draw a median from $ A$ to $ BC$ to point $ M$.  Since $ MB \equal{} MC$, we get that $ [\triangle{AMB}] \equal{} [\triangle{AMC}]$ since they also use the same altitude from $ A$ to $ BC$ (or an extension of it).  And so, we have split our triangle into two congruent parts.  Because we have not assumed anything about $ < A$ (except, of course, that it lies in the range $ 0^{\circ} < A < 180^{\circ}$), we are done.

Oh wow..he meant actually congruent.
This is also easily done..you look at the two cases for which these two inner triangles can be congruent.  On one, we get an isosceles triangle (where two of the equal sides are two sides of the triangle).  The other possibility would lead you to getting that the two base angles are equal, once again asserting that the triangle is isosceles.  Because we have covered all of the possibilities and the results are only isosceles triangles, we are done.

[quote="lukezli"]i think he was talking about the two smaller triangles being congruent...[/quote]

The proof does get that the two smaller triangles are congruent, it just refers to the fact that the larger triangle is isosceles.

[quote="saszs"]explain how you would proove that not every triangle can be split into 2 congruent parts.

show how you got that proof for the not so good problem solvers.[/quote]

suppose for absurd that it is possible.obviously these two triangles must share $ 2$ common points(otherwise we would not partition the triangle, we would have a part left out). let the triangle be ABC.let D be the third point.if D does not lie on one of the sides ,it would be impossible to make such partition.therefore D must lie in one of the sides,say AB.you´ll see that for they to be congruent you must have an non-acute angle .but not all triangles are non-acute.absurd. :wink:

If two parts are congruent, they must be similar. Let a triangle have angles $ \alpha$, $ \beta$ and $ \gamma$, at vertices $ A$, $ B$, and $ C$. Construct a point $ M$ on $ AB$ and draw the segment $ CM$.  Basically we need to show that there exists a triangle such that no such $ M$ results in $ ACM$ congruent to $ BCM$.

Let $ \angle ACM \equal{} \delta$ and $ \angle BCM \equal{} \epsilon$, such that $ \gamma \equal{} \delta \plus{} \epsilon$. Let the third angles of $ ACM$ and $ BCM$ be $ \mu$ and $ \nu$.

$ ACM$ and $ BCM$ congruent means they must also be similar and have the same angles.

In one case, we can have $ \alpha \equal{} \beta$, resulting in an isoceles triangle.

Otherwise, we are obviously not interested in the case $ M\equal{}A$, so we cannot have $ \nu \equal{} \alpha$. Thus, $ \epsilon  \equal{} \alpha$. Similarly, we cannot have $ \mu \equal{} \beta$, so $ \delta \equal{} \beta$. Thus, $ \gamma \equal{} \delta \plus{} \epsilon \equal{} \alpha \plus{} \beta$, and we have $ \pi \equal{} \alpha \plus{} \beta \plus{} \gamma \equal{} 2 (\alpha \plus{} \beta)$. Or, $ \gamma \equal{} \alpha \plus{} \beta \equal{} \pi /2$, a right triangle.

But in a right triangle, it is easy to see the two parts aren't congruent unless it is also isoceles.

I know that not every triangle can be split into 2 equal parts. I just couldn't proove it so thanks for the help.


geometry
analytic geometry
rectangle
graphing lines
slope

For a convex polygon whose vertices are lattice points on a plane, prove the following contexts.

(1) Double of the area is an integer.

(2)The tangent of internal angle is a rational number except in case of right angle.

Isn't number one basically

[hide]Shoelace method?  All vertices are lattice points, so the final area we get before halving is an integer.[/hide]

#2

Let's say their coordinates are $ A \equal{} (x_1, y_1), B \equal{} (x_2, y_2), C \equal{} (x_3, y_3)$.

$ \angle ABC \equal{} \arctan{\frac{y_2 \minus{} y_1}{x_2 \minus{} x_1}} \plus{} \arctan{\frac{y_3 \minus{} y_2}{x_3 \minus{} x_2}}$.

The tangent of that, by the sum of tangents formula, is clearly rational.

What's shoelace Method?

Japanese high school students don't study arctangent.

(1) Lemma: double the area of any triangle with vertices that are integer coordinates is an integer
Proof: take a rectangle with edges parallel to the coordinate axes that is the smallest possible. this has integer area. the triangles that you add are either integers or half integers, subtracting this finishes the lemma.

Now triangluate the polygon and we're done

(2) let the slopes of the two edges by an angle be m_1, m_2, then tangent of the angle is $ \frac{m_1\minus{}m_2}{1\plus{}m_1m_2}$ which is rational since the slopes of the edges must be rational.

That's correct answer. Just let me say one thing.

(2) $ \left|\frac {m_1 \minus{} m_ 2}{1 \plus{} m_1m_2}\right|$ is more better, I think.

To Japanese high school students:

[url=http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf]What's shoelace Method?[/url]

Yes, you often use the formula of triangle ABC:  $ \boxed{\triangle{OAB} \equal{} \frac {1}{2}|ad \minus{} bc|}$ with $ O(0, \ 0), A(a,\ b ),\ B(c,\ d)$.

Part 1 is merely a special case of Pick's Theorem.

[b](1)[/b] Pick a vertex, and draw in all diagonals of the polygon with this vertex as an endpoint. This divides the polygon into n-2 triangles, each with lattice-point vertices. So now it suffices to prove the case of a triangle.


Mr. $A$ owns a home worth $\$$10,000.  He sells it to Mr. $B$ at a 10 % profit based on the worth of the house.  Mr. $B$ sells the house back to Mr. $A$ at a 10 % loss.  Then:

(A)	 $A$ comes out even
(B)	 $A$ makes $\$$1100 on the deal
(C)	 $A$ makes $\$$1000 on the deal
(D)	 $A$ loses $\$$900 on the deal
(E)	 $A$ loses $\$$1000 on the deal

The answer is A I think because after Mr.A sells the house, the houe's price doesn't change, so 10%-10% is 0, making the profit neutral.

depends on the real estate market... :lol:

True, but there is no realty market mentioned in the problem, so it won't be applied to it  :ninja:


This is probably an classic. Homever I'm interested in a nice (short !  :) ) solution

Find all continuous functions [tex]f : R \rightarrow R[/tex] such that [tex]f ( x \cdot y )
= f ( x ) \cdot f ( y )[/tex]

[quote="lightrhee"]This is a well known result. This is one of the Cauchy's equation.
f(x) is $x^c$ for any real constant c, or 0.[/quote]

this is not necessarily true, the general continuous solutions are:

$f(x) = |x|^c$

$f(x) = \mathrm{sgn}(x)\cdot |x|^c$ or

$f(x) = 0$


logarithms
number theory
prime numbers
number theory proposed

Prove that for all natural numbers $ n$ there exist a set $ M_n$ of natural numbers consisting of exactly $ n$ elements so that:
a) arithmetic mean of elements of any non-empty subset of $ M_n$ is integer,
b) geometric mean of elements of any non-empty subset of $ M_n$ is integer.

a lot of examples will do i think. for instance:

if $ x_1,x_2,....x_n$ are these elements of $ M_n$ then it will work if we set them $ x_i \equal{} (n!\cdot a_i)^{n!}$ where $ a_1,...a_n$ are arbitrary natural numbers.

if we set them in following way we also get that each two elements of $ M_n$ are coprime to each other. $ x_i \equal{} p_i^{n!}$ where $ p_i$ are prime numbers such that $ p_i \equal{} 1 (\mod n!)$ (there are infinitely many by dirichlet's theorem)

(infact it's not necesary to write $ n!$it would suffice to write $ \prod_{i \equal{} 1}^{\infty}q_i^{[\log_{q_i}n]}$ (here $ q_i$ means i-th prime number) but $ n!$ kinda looks better)


topology
real analysis
real analysis theorems

How do we show that the following two definitions of "compact" are the same?:

If $ X$ is a topological space, then $ X$ is compact if every open cover of $ X$ has a finite subcover.

If $ M$ is a metric space, then $ M$ is compact if every sequence in $ M$ has a convergent subsequence in $ M$.

These definitions are not the same on a general topological space; for them to be the same, we must have a certain countability condition that metric spaces satisfy.

Let $ (1)$ be the open cover definition of compactness, and let $ (2)$ be the  convergent subsequence (Bolzano-Weierstrass style) definition of compactness.

Let me now state $ (1'):$ the [b]finite intersection property.[/b]

A space $ X$ has the FIP iff we can say this: given a collection of closed subsets of $ X$ such that every finite subcollection has a nonempty intersection, then the whole collection has a nonempty intersection.

We have that $ (1)\iff(1').$ Showing that is just an exercise in complements and contrapositives.

We now claim that $ (1')\implies (2).$

Let $ (x_n)$ be a sequence in $ X.$ For each $ k,$ define the closed set $ F_k$ to be the closure of $ \{x_k,x_{k\plus{}1},x_{k\plus{}2},\dots\}.$ Then the intersection of any finite number of them must still contain members of the original sequence and is thus nonempty. By $ (1'),$ the intersection of all of the $ F_k$ is nonempty. Let $ x\in\bigcap_{k\equal{}1}^{\infty}F_k.$

Then for each $ k,\ x$ is in the closure of $ \{x_k,x_{k\plus{}1},x_{k\plus{}2},\dots\}.$ For any $ \epsilon>0,$ we can find infinitely many members of the original sequence that lie within $ B(x: \epsilon).$ Hence, we can abstract a subsequence that converges to $ x.$

I'll leave proving $ (2)\implies(1)$ for someone else.


algebra
polynomial
modular arithmetic
function
limit
algebra open

Find all integer polynomials $ P(x)$ such that $ P(x)^2 | P(P(x) \plus{} x)$ or show that none exist.

waaaait....
$ P(P(x)\plus{}x)\equiv P(x)(P'(x)\plus{}1) \pmod{P(x)^2}$
so we need $ P(x)|P'(x)\plus{}1$??

Was not aware of that lemma (and I'll think about it!).

Let me ask a second question then:  let $ Q(x) \equiv \frac{P(P(x) \plus{} x)}{P(x)}$.  What is the maximum value of $ \text{deg} \left( \gcd(P(x), Q(x)) \right)$ as a function of $ n \equal{} \text{deg} P$?

Isn't it always $ 0$? Because $ \forall x P(x) \equal{} 0 \rightarrow P(P(x) \plus{} x) \equal{} 0$. So, we have that $ \forall x P(x) \equal{} 0 \rightarrow Q(x) \neq 0$ and vice versa. Therefore, $ \deg(\gcd(P(x),Q(x))) \equal{} \deg 1 \equal{} 0$. Is this wrong? Maybe I don't understand the meaning of $ \gcd$..

It's not non zero. I said that it is not zero for any $ x$ that is the root of $ P(x)$.

yes exactly, why is it non zero in that case?

Counterexample:  let $ P(x) \equal{} x^2 \plus{} x \equal{} x(x \plus{} 1)$.  Then $ Q(x) \equal{} (x \plus{} 1)(x \plus{} 2)$ and $ \gcd(P(x), Q(x)) \equal{} x \plus{} 1$.  :)  I am looking for a generalization here.

Hm.. I thought this might happen. OK, so this is the deal. Denote $ \forall x P(x)\equal{}Q(x)\equal{}0 \rightarrow x \in S$ and we have:

$ \deg(\gcd(P(x),\,Q(x))) \equal{} \sum_{x \in S} \min\left(d(P,x),\,d(Q,x)\right)$

where

$ d(F,r) \equal{} \min_{p \in \mathbb{N}}\left(\Big| \lim_{x \rightarrow r} \frac{(x\minus{}r)^p}{F(x)}\Big| < \infty\right)$

I wrote this already, but since I thought that $ S\equal{}\{\}$, I got  $ 0$ as an answer. :) Hope I got it right this time. BUT, I must say that the expression required cannot be a function of $ n$. Using the counterexample t0rajir0u wrote and the following one, this is easily concluded:

$ P(x)\equal{}x^2$

$ P(P(x)\plus{}x)\equal{}(x^2\plus{}x)^2\equal{}x^2(x\plus{}1)^2$

$ Q(x)\equal{}(x\plus{}1)^2$

so

$ \deg(\gcd(P(x),Q(x))) \equal{} 0$

But, what can we say about the maximum of the expression? Well, we know that the expression cannot exceed $ \min\deg(P(x),Q(x))$ so we could say that for $ \deg P(x) < 2$ the expression has a max of $ 0$ and for $ \deg P(x) \geq 2$, $ \deg P(x)$ is definitely the upper bound.

....I think.... :)

We can do better:  by Albanian Eagle's observation (which is now obvious to me), our upper bound is $ \deg P' \equal{} n \minus{} 1$.

Upon examination of the case $ n \equal{} 3$, it appears that we can do no better than a linear common factor.

hmmm. According to my observations the answer might just as well be $ \left[\frac{n}{2}\right]$ (I have proved we can reach this bound but can't improve it higher...) :wink:
btw the linear factor thing is wrong see for example:
$ x(x\plus{}1)(x^2\minus{}x\minus{}1)$


function
induction
algebra solved
algebra

Find all functions f defined on the nonnegative reals with values in the same set such that f([f(x)])+{f(x)}=x for all x>=0. Here [] and {} are the integer and fractional part.

This is kind of long, so I wonder if ne1 will have the patience to check it out... :D :

Put x=0 to get f(0) in N and f(f(0))=0. We now put x=n from N. We have f([f(n)])+{f(n)}=n, so |f([f(n)])| = n, where |x| is the upper integer part of x, which means that [f([f(n)])] is n if f(n) is in N and n-1 otherwise (*).

We now take x=[f(n)], with n in N. We get f([f([f(n)])])+{f([f(n)])}=[f(n)] (**). We assume f(n) not in N. This means that [f([f(n)])]=n-1, so from (**) we get f(n-1)-(n-1)=[f(n)]-f([f(n)]) (***).

We write [f(n)]+{f(n)}=f(n), we subtract the initial relation from this to get
[f(n)]-f([f(n)])=f(n)-n (****). From (***) and (****) we get f(n-1)=f(n)-1, so f(n-1) is also not in N, so we can repeat the process: f(0)=f(n)-n, so f(0) isn't in N, which is a contradiction.

We have f(n) is in N, so {f(n)}=0, so f(f(n))=n for all n in N. This means that f takes any value in N, so for any x>=0 we have [f(x)]=t=f(n) for some n, which means that f([f(x])=f(f(n))=n, so {f(x)}=x-n. Since we have [f(x)]+{f(x)}=f(x), this means that f(x)=f(n)-n+x, so f(n)=[f(x)]=f(n)-n+[x], so [x]=n, so we have this:
[f(x)]=f(n)=>[x]=n. Conversely, if [x]=n_1 and [f(x)]=f(n_2) => [x]=n_2, we get n_1=n_2, so we have the result:

f([x])=[f(x)] (1)

From here we find f([f(x)]) = f(f([x])) = [x] = x-{f(x)}, so we have found

{f(x)}={x} (2)

I think any function satisfying (1) and (2) checks out.

Observe that f([f(x)])=[f(f(x))] (from (1)) and {f(x)}={f(f(x))} (from (2)), so, by adding them we get f(f(x))=x.

To find the functions we do this:
We divide N in pairs (ai, bi) s.t. f(ai)=bi and f(bi)=ai. Then f is defined in any nonnegative x by using  rules (1) and (2).

Your solution is very nice. I also found a quite long solution to this hard problem-I think you agree with me when I say hard. But I proved by induction that f(n) is i\an integer for n natural number. This can be deduced by replacing x with [f(x)]. Congratulation, grobber!

The problem is nice, but the solution is rather unusual. I thought the solution was f(x)=x until I reached the last few lines of the proof. :D

I think this is the most interesting part of this problem:it has lots of solutions. For another interesting problem, look at the following post: Constant-Non-constant-constant-...


Hey Guys,

In Calculus, today, I received a fairly difficult project -- probably not as hard as you would think, but I do need help. We are asked to create one picture on our Graphing Calculators. I have done about 11 so far, but none are as I like them.

This is what I put into the Y Values:

Y1= Ö(81-X )
Y2= -Ö(81-X2)
Y3= Ö(4-(X+3)2)+3.7
Y4= Ö(4-(X-3)2)+3.7
Y5= -Ö(1-X2)
Y6= -Ö(16-X2)-2
Y7= -Ö(1-(X+3)2)+3
Y8= -Ö(1-(X-3)2)+3
Y9= -.7abs X+12

*Note: The Ö symbol stands for the square root symbol. The big 2s at the ends of the Xs stands for "squared". I couldn't find a way to make them smaller.

Here are my window values:

Xmin= -12
Xmax= 12
Xscl= 1
Ymin= -10
Ymax= 15
Yscl= 1

This should produce a smiley face. Can you show me some of your graphics and formulas you have used? I would greatly appreciate it!

Thanks,

Jamond

it might be easier to make a smiley thing if you switch to polar.

I have already made the smiley face; I just need the some formulas or equations for another pic.

Thanks,


So yesterday my band went to see a football game in Rhode Island and it was raining. Our band director gave us jackets to put on... I put it on the wrong way and couldn't take it off... I was running out of air....... and was  :wallbash_red: this close  to dieing...but I ripped open the jacket and took a DEEP breath.. so I survived...

may this be a learning lesson that tomorrow isn't promissed. :mad:

you told Mrs Page about the "almost died" thing......  :mad:  tell me what she said...


Prove that for a whole number $ n>2,  (n!)^2 > n^n$.

[hide="Solution"]
Proof by induction:

1) Base case: $ n\equal{}3$

$ (3!)^2>3^3$, indeed.

2) $ n\equal{}k$
Assume that $ (k!)^2>k^k$.

3) $ n\equal{}k\plus{}1$
We have to prove that $ ((k\plus{}1)!)^2>(k\plus{}1)^{k\plus{}1}$

$ ((k\plus{}1)!)^2\equal{}(k!)^2\cdot(k\plus{}1)^2$

Thus, now we have to prove that $ (k!)^2>(k\plus{}1)^{k\minus{}1}$.

Here, we claim that $ k^k\ge(k\plus{}1)^{k\minus{}1}$ for any positive integer $ k$.

Above claim is true by simple induction.

Thus, $ (k!)^2>k^k>(k\plus{}1)^{k\minus{}1}$.

Hence, we have $ ((k\plus{}1)!)^2>(k\plus{}1)^{k\plus{}1}$, and the induction is complete.[/hide]

Alternative solution:
For $ n>2$,

$ (n!)^2\equal{}(n \cdot 1)((n\minus{}1) \cdot 2)...(2 \cdot (n\minus{}1))(1 \cdot n)>n^n$


I've been working on my Caley contest of year 1999 and I upon a question that I did not understand. Can someone explain to be what distinct positive integers are with examples? (and no it's not 2,4,6 ... because in the question it said an example of this number was abc = 16).

Thank you~

"Distinct" means "not the same."  So $1, 2, 3$ is a list of distinct positive integers, while $1, 2, 1$ is not.  If this isn't what you are asking, you should give more information.  (Like, for instance, what the actual question you are working on is.)


Find all triples $a,b,c \in N$ such that ${1 \over a}+{2 \over b}-{3 \over c}=1$

$\frac{1}{a} + \frac{2}{b} = 1 + \frac{3}{c}$
It's obvious that right side is bigger than 1. Now, suppose that $a, b \geq 3$. Then:
$\frac{1}{a} + \frac{2}{b} \leq \frac{1}{3} + \frac{2}{3} = 1$
which is impossible.
So $a \leq 2$ or $b \leq 2$. We are left with the 4 cases, which are quite easy to investigate...

I love this method of solution when $f(x)$ is increasing and the $f(x)'$ is deacresing...

Davron


linear algebra
matrix
linear algebra unsolved

what can you say about matrices A in M(n,R) such that
there exist P orthogonal matrix and $^{t}PAP=^{t}A$

If $n=2$ all matrices $A$ satisfy this property.

Let $A\in\mathcal{M}_{n}(\mathbb{C})$ s.t. $trace(A)=0$; then $A$ is unitarly similar to a matrix with 0 diagonal (not easy exercise).
Thus if $A\in\mathcal{M}_{n}(\mathbb{R})$ then $A,A^{T}$ are simultaneously unitarly similar to $aI_{n}+N, aI_{n}+N^{*}$ where $a\in\mathbb{R}$, $N$ is nilpotent and $N^{*}$ is the complex adjoint. It remains to find the nilpotent matrices $N$ s.t. $N,N^{*}$ are unitarly similar.
Remark 1: Let $A,B\in\mathcal{M}_{n}(\mathbb{R})$; then $A,B$ are unitarly similar iff they are orthogonally similar.
Remark 2: if $n=2$, $A,B$ are unitarly similar iff $trace(A)=trace(B),trace(A^{2})=trace(B^{2}), trace(AA^{*})=trace(BB^{*})$ that is always satisfied if $A\in\mathcal{M}_{2}(\mathbb{R})$ and $B=A^{T}$.

My last post is obviously wrong (lines 3,4,5,6)!!  Yet it's true that if $A\in\mathcal{M}_{n}(\mathbb{R})$ then $A$ is unitarly similar to a triangular matrix $U$ (Schur). In the same basis $A^{T}$ becomes $U^{*}$; now the matrices are triangular but complex.

I hope that there isn't any mistake in the following:
Proposition: Here $n=3$; if $A\in\mathcal{M}_{3}(\mathbb{R})$ then $A,A^{T}$ are orthogonally  similar iff $trace(A^{T}A^{2}(A^{T})^{2}A)=trace(A(A^{T})^{2}A^{2}A^{T})$.

Remark 1: one finds only one algebraic condition; for example if $A$ is randomly choosen, except the element $A_{1,1}$, then $A_{1,1}$ is a root of an equation of degree $2$ which have, except in few cases, 2 real roots.

Proof: 1) $A,A^{T}$ are orthogonally similar iff they are unitarily similar.
2) 2 complex matrices $A,B$ of dimension $3$ are unitarily similar iff these 9 relations are satisfied: $trace(A)=trace(B),trace(A^{2})=trace(B^{2}),trace(A^{*}A)=trace(B^{*}B),trace(A^{3})=trace(B^{3}), trace(A^{*}A^{2})=trace(B^{*}B^{2}),trace((A^{*})^{2}A^{2})=trace((B^{*})^{2}B^{2}),trace(A^{*}AA^{*}A)=trace(B^{*}BB^{*}B), trace(A^{*}A^{2}A^{*}A)=trace(B^{*}B^{2}B^{*}B),trace(A^{T}A^{2}(A^{T})^{2}A)=trace(A(A^{T})^{2}A^{2}A^{T})$.
(cf Horn and Johnson,Matrix analysis p. 76 for a reference).
The 8 first relations are always satisfied if $A$ is a real $(3,3)$ matrix and $B=A^{T}$; it remains the last one.

Remark 2: if one writes this relation with the $A_{i,j}$ then it runs over more than 2 Maple-pages!
I think that it's not the result that the prof. hoped.

Little correction:
The last relation is in fact (before the change $B\rightarrow{A^{T}}$): $trace(A^{T}A^{2}(A^{T})^{2}A)=trace(B^{T}B^{2}(B^{T})^{2}B)$.


Εξωτερικά των πλευρών τριγώνου $ ABC$ κατασκευάζουμε κανονικά $ n$ - γωνα. Να βρεθούν οι τιμές του $ n$ έτσι ώστε τα κεντρα των $ n$γωνων να σχηματίζουν ισόπλευρο τρίγωνο.......

Άντε ρε παιδιά!!! Κανένας????

[quote="Γιώργος"]Εάν σε κάθε πλευρά ενός τριγώνου $ ABC$ κατασκευάσουμε ισόπλευρα τρίγωνα όλα εξωτερικά, τα κέντρα αυτών των τριγώνων σχηματίζουν νέο ισόπλευρο τρίγωνο. :)[/quote]
To ίδιο και με τετράγωνα αλλά το θέμα είναι πώς θα αποδείξουμε ότι είναι μόνο αυτά (αν είναι μόνο αυτά...)

[quote="Γιώργος"]Εάν σε κάθε πλευρά ενός τριγώνου $ ABC$ κατασκευάσουμε ισόπλευρα τρίγωνα όλα εξωτερικά, τα κέντρα αυτών των τριγώνων σχηματίζουν νέο ισόπλευρο τρίγωνο. :)[/quote]

Θεώρημα Toricceli για την ιστορία...

Εγώ παιδιά έχω λύση με μιγαδικούς αλλά με ενδιαφέρει μήπως μπορεί κανένας σας να δώσει και γεωμετρική. Α και πληροφοριακά, είναι απο shortlist ΒΜΟ......

Α και επειδή μπορεί να έγινε παρανόηση τα $ n$ γώνα κατασκευάζονται με πλευρές τις πλευρές του τριγώνου


Putnam Harvard college MIT search Stanford geometry

I'd like some input on the following colleges, specifically on the math programs.. and of course on the school overall.

From visiting college message boards and researching, i've been impressed with [b] [url=http://www.hmc.edu]Harvey Mudd College[/url][/b]. The college has had an impressive history participating in the Putnam, but is it worth the money? It's a very expensive college, and very far away from my state (Florida)

I have also looked at the following for their math programs:
University of Chicago
Rose-Hulman Institute
UC at Berkeley..

any insight would be greatly appreciated

U.Chicago is fabulous, inlcuding the math dept. But only for people who want a well-rounded education. People who are heavily into math/science to the exclusion of humanities might be happier elsewhere.

I really fell for U. Chicago when I visited -- I think Paul Salley (erm, or maybe Paul Sally ... not sure) may be the best math teacher in the country.  They have a very good math program, but they will really make you work for everything.

texas137 - Hate to contradict you, but the 750 thing you mentioned (actually it is 750 Math & 700 Verbal) is not 50% of tuition, but $5,000 a year.  Still a large sum, but not quite as big as 50%.  There is a scholarship that is 50% of tuition, though: the NOVA Scholar award.  I think, however, that this year is the last year they are offering it to incoming freshman (it might be next year - the wording is a bit ambiguous).

All this is not to disparage Harvey Mudd College in any way, though.  I'm going to be attending HMC in the fall myself, and, so far as I can do so not having actually attended as a student yet, I'd like to recommend it.  The professors there are [i]incredible[/i], and the atmosphere is great.  I think I'll refrain from any more specific praise, as I have only visited thus far and my words might not carry as much weight.  I'll leave the describing job to current students.

The Putnam is not the only indicator of quality math education, though.  If that were the case, all of the good students would be going to Harvard and MIT, which is not the case (although a lot of good students do go there).  One thing that really impressed me about Harvey Mudd when I was doing research was the fact that, being a small, undergraduate institution, it offers students the great opportunity of working closely with professors in research mathematics - even, often, as freshmen or sophomores.

So if you like the idea of a smaller school that still offers many of the attractions (quality professors, rigor, etc.) of tech schools, as well as a strong humanities program (they call themselves a "liberal arts college of math and science"), I'd definitely suggest looking into HMC.  If you are really interested, see if you can contact a current student or a professor there; they would be able to give you better descriptions than I.  Good luck on the college search!

[quote="Setofallsets"]texas137 - Hate to contradict you, but the 750 thing you mentioned (actually it is 750 Math & 700 Verbal) is not 50% of tuition, but $5,000 a year.  Still a large sum, but not quite as big as 50%.  There is a scholarship that is 50% of tuition, though: the NOVA Scholar award.  I think, however, that this year is the last year they are offering it to incoming freshman (it might be next year - the wording is a bit ambiguous). [/quote]
I *was* referring to the Nova Scholar Award. Both that and the
$5,000 one are listed in the catalogue I got from them in the fall, but you're right, Nova isn't mentioned on the website. The catalogue says the requirements are SAT 1450, Top 10%, and selection by a committee. I'm sorry to hear it's being phased out. It's on my son's list for fall 2005 (although the end of Nova Scholarships moves it down quite a bit).

Good luck there. Report back after you've started.

Same question, different list of schools. Does anyone have any input on how the following schools would compare for a math/CS major?
MIT
Stanford
Caltech

As for MIT/Stanford/Caltech, one thing to keep in mind is the possibility of the student changing his/her mind and wanting to study something outside of tech-y stuff.  If that's a reasonable possibility, I think you have to go with Harvard/Stanford/Princeton rather than MIT/Caltech.  While MIT may have a few other notable areas of study, the diversity and excellence of those outside math/science is not what it is at H/S/P.

Another obvious note Stanford and Caltech are in California.  It's nice out here.

But really, you can hardly go wrong with any of them if you really are dedicated to math/CS.  I'd recommend school visits while in session and see which student body fits your child the best.

Setofallsets does bring up an interesting point to research: how likely is it that you will work with a professor if research is your game?  I don't know much about Caltech, but most of my friends who went to MIT were in research groups pretty quickly if they wanted to be.  My (brief) experience at Stanford grad school made me view Stanford as being somewhat more geared towards the grad students.  But that was 10 years ago, and in engineering.  A lot could have changed, but this undergrad vs grad emphasis is something you'll want to look at closely (and why I picked Princeton over Harvard and Stanford for undergrad).

For Math or CS, MIT, Stanford, and Caltech will all be similar to some degree. One difference will be in the freshman experience. Each of them have a different way of handling freshman (from Pass/Fail option to the degree of guidance and support).

But the biggest difference between the three schools is their culture. Each one has a very different and distinct culture, that you can only get a good feel for if you visit. Also the sizes of the schools are very different - with Caltech being the smallest and Stanford being the biggest. This affects a lot of things - both socially and academically.

Caltech is very small, overwhelmingly male, and somewhat insular. It has a predominant "science and math culture". The few people that I know who went to Caltech really loved it - but I have heard stories of people who just burned out on the science and math, or didn't feel they had enough options to study outside of those areas. Considering their small size, I imagine they have very good guidance and support from their teachers and administrators though.

MIT is a lot larger student body-wise, and is in a more urban area. While the majority of the students there do major in science of math, there are also a lot of students who major in other things - like psychology, economics, politics, etc. Though Caltech and MIT are both science and math schools, they have very different cultures. MIT has a heavily fraternity-driven social scene. They also have a lot of resources, academically (libraries, conferences, speakers, lab equipment) and socially (Boston's sports, concerts, foods, and other colleges).

Stanford is the largest of the schools (and when you count it's professional schools and graduate schools, it's much larger than the other two). But that also means more diversity - academically, socially, and culturally. They have more classes to offer each semester, both in math and in other subjects, a more diverse student body (including many more women and minorities than MIT or Caltech), and a much more vibrant sports program. This has it's pros and cons, depending on who you are. In addition to being strong academically, Stanford is much more of a "jock" school than people realize. They have a lot of major sports programs and some very gifted atheletes.

I agree with Richard though about taking visits and seeing which school fits best for you. And also about the undergrad vs. grad emphasis, which can be hard to determine just from a visit. But be sure not just to look at the relative emphasis that the school puts on the two; but also within the individual departments. For example, Princeton as a whole puts a great deal of emphasis on it's undergraduate students - but in the Math department the undergraduates are outnumbered by far by the grad students, and the emphasis is given accordingly.

Also consider housing, food, and transportation. That's often a REALLY overlooked aspect of college life. Of course you're there to learn - but having a nice place to live, good food to eat, and the ability to get around are nice things to have!

[quote="rrusczyk"]As for MIT/Stanford/Caltech, one thing to keep in mind is the possibility of the student changing his/her mind and wanting to study something outside of tech-y stuff.[/quote]

Not a likely possibility at my house. I think we can safely eliminate colleges that are heavily into producing Renaissance men and women, no matter how wonderful they might be.

[quote="rrusczyk"] Another obvious note Stanford and Caltech are in California.  It's nice out here.[/quote]

It certainly is! And both campuses are beautiful.

[quote="rrusczyk"] I'd recommend school visits while in session and see which student body fits your child the best. [/quote]

Actually, we already did that. It seemed to confuse things by raising more issues. It's clear he'll have to go again next year before it's time to make any decisions.

[quote="rrusczyk"]undergrad vs grad emphasis is something you'll want to look at closely [/quote]

Point well taken. My own college experience was at a top-20 research institution where undergrad teaching was the lowest possible priority. It was generally left to grad students who might not be fluent in English. That was a million years ago, but I bet it hasn't changed much.


If the point whose coordinates are (-5,3) us reflected about the line y=-2, what are the coordinates of its images?

$ (\minus{}5,3)$ is 5 units away from the line $ y\equal{}\minus{}2$, so the y-coordinate would be $ 3\minus{}10\equal{}\minus{}7$, so the point would be $ (\minus{}5,\minus{}7)$


Find all n (positive integers) such that

n^2 | 2^n + 1.

Isn't it from some IMO? I think the only answer is 3, but it's a tough problem.

90 imo a3

http://www.kalva.demon.co.uk/imo/isoln/isoln903.html

here's my solution
Clearly, n must be odd. Now assume that 3^klln; that is, 3k
is the highest power of 3 dividing n. Then 3^(2k) | n^2 | (2^n + 1) which implies 2^n = −1(mod 32k) ) 3^(2k−1) | n which implies  2k − 1  \leq  k
 k  \leq  1, showing
that n has at most one factor of 3. We observe that n = 3 is a solution.
Suppose that n has a prime factor greater than 3; let p be the least such
prime. Then p | (2^n+1) ) .2^n = −1 (mod p). Let d be the order of 2 modulo
p. Since 2^(2n) = 1(mod p), d | 2n. If d is odd, then d | n which implies 2^n = 1 (mod p), contradiction,
so d is even, say d = 2d1. Then 2d1 | 2n which implies d1 | n. Also,
d | (p − 1), or
2d1 | (p−1) ) d1  \leq  (p−1)/2 < p. But d1 | n, so d1 = 1 or d1 = 3. If d1 = 1,
then d = 2, and 2^2 = 1 (mod p), contradiction. If d1 = 3, then d = 6, and
2^6 = 1 (mod p), or p | 63 whichimplies p = 7. However, the order of 2 modulo 7 is 3,
which is odd, again contradiction. Therefore, no such p can exist, and the
only solution is n = 3.

I proposed the following problem

    Problem: Let be given a positive integer a such that a has only prime divisor like 2 or 3.Find all n such that
            n^2 | a^n +1

Regards


conics
parabola
function
calculus
derivative
algebra unsolved
algebra

Let $a \in \mathbb{R}$ be a parameter.

(1) Prove that the curves of $y = x^2 + (a + 2)x - 2a + 1$ pass through a fixed point; also, the vertices of these parabolas all lie on the curve of a certain parabola.

(2) If the function $x^2 + (a + 2)x - 2a + 1 = 0$ has two distinct real roots, find the value range of the larger root.

(1) It is clear that all parabolas pass through $(2,9)$, and the vertices lie on the parabola $y = -x^2+4x+5$.

(2) The larger root is $ \frac{-a-2+\sqrt{a^2+12a}}{2}$, so $a \in (-\infty , -12)$ or $(0,\infty )$. Now, as $a$ gets smaller, the root gets bigger. On the other hand, the root is negative in $(0,\frac{1}{2})$, and the function is concave, so the lower bound is $-1$.

note that , irrespective of the value of a , the parabola passes through $(2,9)$
note that , the vertex is $(-\frac{a+2}{2},1-2a-[\frac{a+2}{2}]^2)$ and these points lie on the parabola
$y=5+4x-x^2$.
now , we come to part (b)
note that for having 2 distinct real roots , we must have $D=a(a+12)>0$ , i.e. $a>0$ , or $a<-12$
now , note that , the larger root is $\frac{-(a+2)+\sqrt{a^2+12a}}{2}$
if $a>0$ , then the range of this root is clearly $(-1,\infty)$ .
[for finding this , we calculate the derivative , check its sign for a>0 ,find limits etc. (everything for a>0)]
 we can similarly find it for $a<-12$ using simple calculus.

For $a<-12$, it is clear that the range is (5,infinity), but then for $a>0$, isn't it that the range is only $(-1,2)$ by using limits and derivatives?

Cause like for the $a>0$ case we have $a<\sqrt{a^2+12a}<a+6$, so it's clearly bounded between $(-1,2)$


Can you construct 36 degree angle, provided that you can't use anything but a compass! It is equivalent to divide 180 degree into five equal angles, which is the actual problem problem given by my friend.
 :)

First idea I got: construct a regular pentagon ABCDE, draw a diagonal connecting A and C, and voila, the angle BAC (or BCA) is 36 degrees.

Of course, there can be an quicker method, but this idea is easy to find...

You have only got a compass. It was also my idea first...

That's GREAT! Now I would request you to stop working with 36 alone, and find a way to construct it using a compass by means of dividing 180 into five equal parts!
Have a look at the Subject(Title) please!
 :)

[quote="hsiljak"]Well, I should start reading the problems completely before answering  :lol:

[size=59]"Sto je brzo, to je i kuso":D[/size]

EDIT: Compass only? Well, that should be enough:

1. Construct the pentagon ABCDE with compass only:

[url]http://www.geocities.com/robinhuiscool/Pentagon.html[/url]

[url]http://forumgeom.fau.edu/FG2008volume8/FG200819.pdf[/url] (it's useful to be a passionate Forum Geometricorum reader  :lol: )

Now, the angle is there...[/quote]

[size=59]''Ispeci pa reci''[/size]

Does anybody know how to solve this?

What exactly do you mean?  Are you looking for a technique to extend the above construction to make it possible to divide an arbitrary angle into five equal parts by compass?  This is known to be impossible; see for example http://www.jstor.org/sici?sici=0025-5572(198503)2%3A69%3A447%3C9%3ADAAIEP%3E2.0.CO%3B2-J .

No! You just divide $ 180$ into five! That's very much for me! :lol:

Actually, Akashnil provided me with a brilliant solution......dividing $ 180$ into $ \frac{1}{5}th$....I am waiting to see if there is any other solutions. The solution deals with only compass and scale(for joining two points only....no measurement!) constructions.


geometry
geometric transformation
reflection

AAAAAAAAAAAAAAAAAAAAAAAAH!!!!! how did you do?  :?

from the scores my school showed, its looking like a high cutoff this year.... so, what'd you get? and whats your breakdown?

me:
critical reading: 80
math: 80
writing: 73 (you know, if that took that durn sentence error part off, i would have killed this part....)

total: 233

Here are my scores:

Math: 73 (I hate stupid mistakes so much)
Critical Reading: 62
Writing: 58

Total: 193

I don't think this is really that bad for an 8th grader, even though I should have gotten a 80 in the math part.

PSATs aren't here for me:(

Hopefully I'll get my scores on Monday.

77 Verbal / 80 Math / 76 Writing

=233. One point improvement over last year.

Writing was a bit dissappointing, compared to my 80 last year.

lol. im sorry but i totally have to say this... theres a reason you didnt ace that math, 77+78+76?... thats uh... 231.... b'aint it?

Yeah, I got an 80 on math. I seem to have been thinking of my score from last year. lol

Critical Reading: 80
Math: 80
Writing Skills: 80

Total: 240

Overall, I got two questions wrong (on the writing section). I must say that I am rather pleased. I was pretty confident coming out of the test, but I didn't expect this. I credit AoPS for turning math from my weakest area on the test (63 last year) to a perfect score.

I suppose this means I can look forward to being a National Merit Semifinalist. Hopefully I can pull a scholarship out of it.

The fact that mine aren't going to arrive until January just makes me want to burn down my school.

Oh wow, pascal012.  Perfect score?  (Well not perfect if you got two problems in correct, but perfect since you got 80/80/80.)  I thought I did well until I found out what my friend got (combined score of 226), but you did even better!  I got:

[list]
Critical Reading - 62
Math - 76
Writing Skills - 71
[/list]

I'm so ticked off about my math score.  The [i]one[/i] problem I missed was one of the easiest ones!

[quote]
[b]12.[/b] Typing at a rate of [i]w[/i] words in 20 seconds, a typist can type how many words, in terms of [i]w[/i], in 15 [u]minutes[/u]?
(A) 300[i]w[/i]
(B) 45[i]w[/i]
(C) 5[i]w[/i]
(D) (4/3)[i]w[/i]
(E) (3/4)[i]w[/i]
[/quote]

I mean, come on!  It's so easy!  I don't understand how I missed it!  Anyways, on my critical reading score, I wasn't expecting much -- I'm not too great at analysis, although I love to read -- except that I expected a bit better, especially for my writing score.  *cries*

Fortunately, my percentile ranking improved 12 from 86 last year to 98 this year, which in fact made me incredibly happy.  ^_^

-iamunknown73

[quote]Math is very top heavy (it always is). One of my friends missed a question, and lost 4 pts from math for it too. Whereas, on the verbal, you get still get 80 for missing like 3...[/quote]

I happen to have missed 2 on the verbal. I in fact got a 77.

Heh, oops.  :D
I assumed that from the way most of my practice tests were scored. Anyways I missed 7 and got 65. :(

Still, missed questions are much are punishing for math.

Eh...I like it this way. (the 4 pt deduction) In fact, math needs to be harder so that the number of perfect scores are about equal with verbal and writing.

I think the scoring system was a lot meaner overall this year. :-)

PSATs,..
79 Verbal
78 Writing
80 Math
Total : 237
It's not a perfect 240, but considering I'm a sophomore.. pretty good, I guess.
Good thing there wasn't an essay section, or I would have gotten a 50 or something..
My school took it on a test date that didn't release the answers or something,
so the only portion of my score report worth looking at is the top third of the page.

[quote="h_s_potter2002"]Did you guys get the official reults by the time you posted, becuase I didn't get my test booklet and all that stuff untill yesterday, even though my teacher just told me my score back in Febuary.[/quote]

February just started...you must mean last year.  How did your teacher know it?  I don't think your teacher is supposed to grade your test.

I meant to say December, sorry  :blush:   But I don't know how my teacher knew it.  I know she didn't grade it or anything like that.  And this is my test prep teacher btw, which is the class where you prepare for the PSATs.

O, for some reason I thought you were saying that she gave you your score earlier than December.  Schools get the results back in December so they had them then but your school just decided to wait to hand you the scores.

73/76/76 soph

Grr... one math question... grr... I did overall pretty good, but need to get the perfect math and writing for next year... hopefully boost my reading up to 75+.

tetrahedr0n

I happen to have [color=red]missed 2 on the verbal. I in fact got a 77[/color].
_________________

I'm in Gr.10

Math - 80
Writing Skills - 80
Critical Reading - 73
I got [color=red]3 wrong on the verbal but got a 73[/color].
_________________

I'm in grade 10.
Critical reading: 65, down from a 74 last year
Math: 80
Writing/Grammar: 80

Total 225.

...

Wow I got my PSATs in January. I recalled my scores and compared them to your scores, and all I have to say is: man do I feel stupid or what?

Sophmore:
Verbal:70
Math:71
Writing:78

Junior:
Verbal: 68
Math: 70-ish
Writing: 60-ish

All I remember is my total this year was 208. I hope I can still qualify for National Merit. >_<

Has anyone had a huge difference in their PSAT scores and their old SAT scores? I have over 100 points difference and I was wondering if any of you guys have this too.

[quote="Forsaken"]Has anyone had a huge difference in their PSAT scores and their old SAT scores? I have over 100 points difference and I was wondering if any of you guys have this too.[/quote]
My old SAT score was 60 pts lower than the PSAT CR + Math. Not huge, but pretty big, I guess. Maybe it's because I missed a couple on the analogies section.

Soph:
Math 80
Writing 80
Verbal 62..... what the heck...

total 222(hehe i liek that numvber :D )

[quote="MithsApprentice"][quote="iamunknown73"]Math - 76

The [i]one[/i] problem I missed... [/quote]

You missed only one problem and got a 76? You're kidding right?[/quote]

I got one wrong and got 76 on math too. Why do they subtract 4 points?!

[quote="henryyangrui"]I got one wrong and got 76 on math too. Why do they subtract 4 points?![/quote]

same here. i think it's a percentile thing, but who really knows. it's really stupid tho because on the writing section i know someone who missed 2 or 3 questions and still got an 80. now that's not fair.

The SAT/PSAT math sections are stupid and too easy so the curve is ridiculous, causing 4 point deductions for missing 1 question... now if they made them at least AMC difficulty, I'm sure we'd be seeing 80(0)'s while missing 3-4 questions ;)

213, sophomore
80 Math
67 Writing
66 Critical Reading

I did not do so well on the language parts because I decided to guess on a lot of them, causing me to lose 5-7 raw points on each of the two sections. for I believed the score did not matter for a sophomore. I looked at my score reports and I *would* have gotten

227, sophomore
80 Math
74 Writing
73 Critical Reading

had I not guessed so wildly.

I feel dumb for blowing off the PSAT's :blush: . What are you aiming for next year?
Me: 235ish.

The scaled scores are all supposed to be statistical, with a defined median and scores ranging up or down from that median with so many score points relating to so many standard deviations from the median. The SAT I scoring scale is supposed to be median 500, each standard deviation 100, so presumably the PSAT score is related to the same idea, with different numbers.


search
number theory unsolved
number theory

Determine all pairs (x,y) of integers with $y$l$x^{2}+1$and$x$l$y^{3}+1$
i think this was posted before.
how can i search?

click on the search button or [url=http://www.mathlinks.ro/Forum/search.php]here[/url]


In the adjoining figure triangle $ ABC$ is inscribed in a circle. Point $ D$ lies on $ \stackrel{\frown}{AC}$ with $ \stackrel{\frown}{DC} \equal{} 30^\circ$, and point $ G$ lies on $ \stackrel{\frown}{BA}$ with $ \stackrel{\frown}{BG}\, > \, \stackrel{\frown}{GA}$. Side $ AB$ and side $ AC$ each have length equal to the length of chord $ DG$, and $ \angle CAB \equal{} 30^\circ$. Chord $ DG$ intersects sides $ AC$ and $ AB$ at $ E$ and $ F$, respectively. The ratio of the area of $ \triangle AFE$ to the area of $ \triangle ABC$ is
[asy]
size(200);
defaultpen(linewidth(.8pt));

pair C = origin;
pair A = 2.5*dir(75);
pair B = A + 2.5*dir(-75);
path circ =circumcircle(A,B,C);
pair D = waypoint(circ,(7/12));
pair G = waypoint(circ,(1/6));
pair E = intersectionpoint(D--G,A--C);
pair F = intersectionpoint(A--B,D--G);

label("$A$",A,N);
label("$B$",B,SE);
label("$C$",C,SW);
label("$D$",D,SW);
label("$G$",G,NE);
label("$E$",E,NW);
label("$F$",F,W);
label("$30^\circ$",A,12S+E,fontsize(6pt));

draw(A--B--C--cycle);
draw(circ);
draw(Arc(A,0.25,-75,-105));
draw(D--G);[/asy]$ \textbf{(A)}\ \frac {2 \minus{} \sqrt {3}}{3}\qquad \textbf{(B)}\ \frac {2\sqrt {3} \minus{} 3}{3}\qquad \textbf{(C)}\ 7\sqrt {3} \minus{} 12\qquad \textbf{(D)}\ 3\sqrt {3} \minus{} 5\qquad$
$ \textbf{(E)}\ \frac {9 \minus{} 5\sqrt {3}}{3}$

I thought this was very hard. Here is my solution:
Let O be the center of the circle. Since $ \angle DGC \equal{} 15 \equal{} \angle CAO$ and $ DG \equal{} AC$, O lies on CG.
Any chord subtending an arc of $ \theta$ has a length of $ r \cdot \sqrt {2(1 \minus{} cos \theta)}$, where r is the radius of the circle. Now, $ \triangle EDC \sim \triangle AEG$, so $ \frac {AE}{EC} \equal{} \frac {AG}{DC} \equal{} 1$, since $ \angle AOG \equal{} \angle DOC \equal{} 30$. So $ \frac {AE}{AC} \equal{} \frac {1}{2}$. Also, $ \frac {AF}{FB} \equal{} \frac {AG}{DB} \equal{} \sqrt {\frac {1 \minus{} cos 30}{1 \minus{} cos 90}} \equal{} \sqrt {1 \minus{} \sqrt {3}/2} \equal{} x$. After a ton of algebra, we get $ \frac {AF}{AB} \equal{} \frac {x}{x \plus{} 1} \equal{} 2 \minus{} \sqrt {3}$. Thus, the final answer is $ \frac {AE}{AC} \cdot \frac {AF}{AB} \equal{} \frac {2 \minus{} \sqrt {3}}{2}$, which is not an answer. Can someone show me an easier way?

[hide="You were on the right track but made some mistakes."]
Angle chase shows that $ \triangle DFB$ is equilateral, that $ AD\perp EO$, and $ \triangle DCE$ is 30-60-90. Thus we can assign lengths as follows:
    [asy]defaultpen(linewidth(.8pt));

    pair C = origin;
    pair A = 2.5*dir(75);
    pair B = A + 2.5*dir(-75);
    path circ =circumcircle(A,B,C);
    pair D = waypoint(circ,(7/12));
    pair G = waypoint(circ,(1/6));
    pair E = intersectionpoint(D--G,A--C);
    pair F = intersectionpoint(A--B,D--G);
dot(circumcenter(A,B,C));
    label("$A$",A,N);
    label("$B$",B,SE);
    label("$C$",C,SW);
    label("$D$",D,SW);
    label("$G$",G,NE);
    label("$E$",E,NW);
    label("$F$",F,W);
MP("$O$",circumcenter(A,B,C));
    label("$30^\circ$",A,12S+E,fontsize(6pt));
label("$1$",(E+F)/2,red);
label("$1$",(F+G)/2,red);
label("$1$",(A+F)/2,E,red);
label("$\sqrt{3}+1$",(B+F)/2,red);
label("$\sqrt{3}$",(A+E)/2,W,red);
label("$\sqrt{3}$",(D+E)/2,red);
label("$2$",(C+E)/2,red);

    draw(A--B--C--cycle);
    draw(circ);
    draw(Arc(A,0.25,-75,-105));
    draw(D--G);[/asy]
Thus we directly compute an answer of \[ \frac{\sqrt{3}}{(2\plus{}\sqrt{3})^2}\equal{}\sqrt{3}(2\minus{}\sqrt{3})^2\equal{}7\sqrt{3}\minus{}12,{\bf (C)}\]
[/hide]

[hide="stuff I found without getting a solution"]Hmm I drew in DB, found that DFB was an equilateral triangle and was similar to AFG.  From there I got that EF = FG and that <AFE = 30.  Let K be the intersection between DB and AC.  Triangle AKB is a 30-60-90 triangle.  I guess you could use that to find the area of triangle ACB.  Also CG is the diameter of the circle.. but I don't see how that helps.  D;  This is a nice problem by the way.[/hide]

[quote=mathwizarddude]This is what I got - I don't see the flaw:
If we let the radius be 1, the $ BD\equal{}\sqrt2$ since $ \angle BOD\equal{}90$. And $ BF\equal{}BD\equal{}\sqrt2$ since triangle DFB is equilateral. Also $ AB\equal{}\sqrt3$ because of the 30-60-90 triangle. So the area ratio is $ (1/2)(\sqrt3\minus{}\sqrt2)/\sqrt3$, which does not seem to be correct.[/quote]

To find AB you obtain a 15-75-90 right triangle, and if the radius is 1 then $ AB\equal{}(\sqrt6+\sqrt2)/4$ , and you can procede from there.


[quote="Euclid_great"]Can i have IMO 2000-2007 solutions?[/quote]

[url=http://www.mathlinks.ro/resources.php?c=1&cid=16]Here[/url] you can find the problems of IMO 1959-2008. Many problems have already solutions. You are asked to contribute further solutions or post ones for yet unsolved problems.


You need at first that they are coprime.
Then it's still wrong, e.g. $a=2,b=1,m=2,n=1$.

Either change all $+$ to $-$ or see http://www.mathlinks.ro/Forum/viewtopic.php?t=67642 .

Then how to prove that \[ (a^n - b^n, a^m - b^m) = a^{(m, n)} - b^{(m, n)}  \]


Find all positive integers $n$ such that $4^{n}+2007$ is a perfect square.

[quote="silouan"]the problem posed in the contest was when $4^{n}+2007=k^{2}$
The answer is tha there is not exist such $n$ (just take mod 4)[/quote]

...you mean mod 5, right?

trying the first problem, I find that bounding seems to work... you show that $n<31$, then $n$ can't be in $22..31$, etc.
Can someone check this or post their own?

[quote="chess64"]4 works fine... 3 is not a quadratic residue[/quote]

And if $n$ is odd?

[quote="boxedexe"]If $n$ is odd, then $n^{2}\equiv 1\pmod 4$.  If $n$ is even, $n^{2}\equiv 0\pmod 4$  Since $4^{n}+2007\equiv 3\pmod 4$, $4^{n}+2007$ cannot be a square. :wink:[/quote]

O sorry I was thinking of the $n^{4}+2007$ problem.

First note that if n is even,then 4^n ends with a 6 but there are no such perfect squares that ends with 3. Thus,n must be odd if there exists such n's.
Let 4^n+2007=m^2. Then we have (m-2^n)(m+2^n)=2007.
Hence,(m-2^n,m+2^n)=(1,2007),(3,669) or (9,223).
If m-2^n=1,then 2m=2008 which implies that m^2=1008016 but there are no such n such that 4^n ends with 9.
Similarly,there are no solutions for the other 2 pairs as well.
Therefore there exists no such n such that 4^n+2007 is a perfect square.  :lol:

One can take $mod 4$ this kills it immediatly.
Otherwise, we can work in an other way like this :
Suppose there existe $k$ such that : $4^{n}+2007=k^{2}$ and since $2007=3^{2}.223$ where $223$ is prime.
Then we have either : $k-2^{n}=1$ and $k+2^{n}=2007$ this give us $2^{n}=1003$ this is obviously impossible since $1003$ is not a 2-th power. In the same way we check the other cases and we find a contradiction.

[color=#f00]Claim[/color] : There is no such $n$ that satisfies the condition $4^n +  2007 = a^2$, $a \in \mathbb{Z}$
Taking $\pmod 4$ yields, that $4^n + 2007= 3 \pmod 4$. To be a square it would need to be either $0 \pmod 4$ or $1 \pmod 4$. Hence there is no such $n$.  $\square$


inequalities
algebra
Sums and Products
Real Roots
China

Let $\alpha_1, \alpha_2, \dots, \alpha_n$, and $\beta_1, \beta_2, \ldots, \beta_n$, where $n \geq 4$, be 2 sets of real numbers such that
    \[\sum_{i=1}^{n} \alpha_i^2 < 1 \qquad \text{and} \qquad \sum_{i=1}^{n} \beta_i^2 < 1.\]
    Define
    \begin{align*}
        A^2 &= 1 - \sum_{i=1}^{n} \alpha_i^2,\\
        B^2 &= 1 - \sum_{i=1}^{n} \beta_i^2,\\
        W &= \frac{1}{2} (1 - \sum_{i=1}^{n} \alpha_i \beta_i)^2.
    \end{align*}
    Find all real numbers $\lambda$ such that the polynomial \[x^n + \lambda (x^{n-1} + \cdots + x^3 + Wx^2 + ABx + 1) = 0,\] only has real roots.

:P ,motivated from bulgaria final round pro.
 idea is common to get equalities in classical inequalities,try which. :?:

My Solution, Please Check this solution!

We will show for infinitely many solution for $\lambda$ is provide the conditions. Let's begin.

From equations in the problem we have $$A^2+B^2+\sum_{i=1}^{n}\alpha_i^2+\sum_{i=1}^{n}\beta_i^2=2\ge 2AB+\sum_{i=1}^{n}\alpha_i^2+\sum_{i=1}^{n}\beta_i^2\ge 2AB+2\left( \sum_{i=1}^{n}\alpha_i\beta_i\right)$$

$$=2\left( AB+\sum_{i=1}^{n}\alpha_i\beta_i\right)\le 2 \Longleftrightarrow AB+\sum_{i=1}^{n}\alpha_i\beta_i\le 1 = \sqrt{\left( 1 - \sum_{i=1}^{n}\alpha_i^2\right)\left( 1 - \sum_{i=1}^{n}\beta_i^2\right)}+\sum_{i=1}^{n}\alpha_i\beta_i\le 1$$

$$1+\left( \sum_{i=1}^{n}\alpha_i^2+\sum_{i=1}^{n}\beta_i^2\right)+ \sum_{i=1}^{n}\sum_{i=1}^{n}\alpha_i^2\beta_i^2 \le 1+ \left( \sum_{i=1}^{n}\alpha_i\beta_i\right)^2-2\left( \sum_{i\not=j}^{n}\alpha_i \beta_j\right)$$

İts easy to see,

$$\left(\sum_{i=1}^{n}\alpha_i^2\right)\left(\sum_{i=1}^{n}\beta_i^2\right)+2\sum_{i\not=j}\alpha_i\beta_j\le \left( \sum_{i=1}^{n}a_ib_i \right)^2 +\sum_{i=1}^{n}\alpha^2+\sum_{i=1}^{n}\beta_i^2 \Rightarrow \left( \sum_{i=1}^{n}\alpha_i\right)\left( \sum_{i=1}^{n}\beta_i\right)\le \sum_{i=1}^{n}\alpha^2+\sum_{i=1}^{n}\beta_i^2<2$$ Cause we have $$\left( \sum_{i=1}^{n}a_ib_i \right)^2\ge 2\sum_{i\not=j}\alpha_i\beta_j$$

For simplicity we put, $n=2k$ and $\lambda = -n$, let $x_i$ be root of equation. Put prime factors of $n$ to, $x_1,x_2,\dots$ then if number of prime factors of $n$ smaller then number of roots, we put ${-1,1}$ to this roots and we have all real roots. And We ar :-D e done!


analytic geometry
graphing lines
slope
logarithms
calculus
derivative
function

sorry. Can you find $ \frac{d}{dx}\ln y ?$

It's incorrect. $ \frac {d}{dx}\ln y \equal{} \frac {d}{dy}\ln y\ \frac {dy}{dx} \equal{} \frac {1}{y}\ y'.$

im not used to doing derivative in respect to x of y's.

i get screwed up

O.K. Generarly speaking, $ \frac{d}{dx}f(g(x))\equal{}f'(g(x))g'(x)$ holds.

The way I explain implicit differentiation:

We have the equation $ 3x^2\plus{}5\ln y\equal{}12$. We assume that there is a differentiable function $ Y(x)$ so that $ (x,Y(x))$ satisfies the equation: $ 3x^2\plus{}5\ln(Y(x))\equal{}12$.
Differentiating by the chain rule, $ 6x\plus{}5\frac{Y'(x)}{Y(x)}\equal{}0$, and $ Y'(x)\equal{}\minus{}\frac65 x\cdot Y(x)$ wherever this expression is defined.
That much is elementary calculus; [i]any[/i] differentiable $ Y$ satisfying the original equation must have its derivative satisfy the new equation. There is also a theorem of multivariable calculus (the Implicit Function Theorem) which states that such a $ Y$ exists, at least on some interval around the original point.

The same problem is posted right [url=http://www.mathhelpforum.com/math-help/calculus/25201-slope-line-tanget-graph.html]here[/url].

(The poster didn't even change the form to post.)

[quote="ProblemSets"]The slope of the line tangent to the graph of 3x^2 + 5ln(y) = 12 at (2,1) is...

a) -12/5
b) 12/5
c) 5/12
d) 12
e) -7[/quote]

$ 3x^2 + 5lny = 12$<=>$ y=e^{\frac{12-3x^2}5}$
=>$ y'=-\frac{6x}5.e^{\frac{12-3x^2}5}$
=>The slope of the line tangent to the graph at (2,1) is $ y'(2)=-\frac{12}5$.


trigonometry
inequalities
geometry solved
geometry

In a circle, a chord AB is perpendicular to diameter CD. Let them intersect at N. P is a point on the arc ACB. Let PN intersect the circle at P and Q and PD intersect AB at R. Prove that $RD\geq NQ$.

[color=red][Moderator edit: Made some corrections.][/color]

I remember this being posted before.

Anyway, after replacing $NQ$ with $NC\cdot\frac{ND}{NP}$, we're left with showing that $RD>NC\cdot\frac{ND}{NP}$, or $\frac{RD}{ND}\ge\frac{NC}{NP}$. We have $\frac{RD}{ND}=\frac{CD}{PD}$, so we want $\frac{CD}{PD}\ge\frac{CN}{PN}$. This is really easy to show (you can use a quick application of the sine law, for example).

[quote="indybar"]In a circle, a chord AB is perpendicular to diameter CD. Let them intersect at N. P is a point on the arc ACB. Let PN intersect the circle at P and Q and PD intersect AB at R. Prove that $RD\geq NQ$.[/quote]

Indeed this has been posted on http://www.mathlinks.ro/Forum/viewtopic.php?t=15104 , but let me use the occasion to give a slightly extended version of the problem:

[color=blue][b]Problem.[/b] Let k be a circle with center O, and let A and B be two points on the circle k. Further, let D be a point on the circle k such that $OD\perp AB$ (thus, the point D is the midpoint of one of the two arcs intercepted from the circle k by the chord AB). Let the lines OD and AB intersect at a point N. Let P be an arbitrary point on the circle k, and let the line PN intersect the circle k at a point Q (apart from the point P). Finally, denote by R the point of intersection of the lines PD and AB. Prove that $RD\geq NQ$.[/color]

[i]Solution.[/i] (Well, this is more or less the solution Pierre posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=15104 post #2, but rewritten using directed angles in order to make it arrangement-independent.)

We will use directed angles modulo 180°.

Let H be the foot of the perpendicular from the point N to the line DQ. Then, < NHQ = 90°. On the other hand, $OD\perp AB$ yields < DNR = 90°. Thus, < DNR = < NHQ.

The point O is the center of the circle k, which passes through the points D, P and Q. Thus, by the central angle theorem, < ODP = 90° - < PQD. In other words, < (OD; PD) = 90° - < (PQ; DQ). But since $OD\perp AB$, we have < (AB; OD) = 90°, and thus

  < (AB; PD) = < (AB; OD) + < (OD; PD) = 90° + (90° - < (PQ; DQ)) = 180° - < (PQ; DQ)
                     = - < (PQ; DQ) = < (DQ; PQ).

In other words, < NRD = < HQN. Together with < DNR = < NHQ, this yields that the triangles NRD and HQN are similar. Thus, $\frac{RD}{QN}=\frac{DN}{NH}$. But we have $DN\geq NH$, since the triangle NHD is right-angled, and the hypotenuse of a right-angled triangle is always greater or equal to any catet (equality holds in the degenerate case when the hypotenuse and the catet fall together). Now, of course, from $\frac{RD}{QN}=\frac{DN}{NH}$ and $DN\geq NH$, we conclude that $RD\geq QN$; in other words, $RD\geq NQ$. And the problem is solved.

In the inequality $RD\geq NQ$, equality holds if and only if the point P is the midpoint of one of the two arcs intercepted on the circle k by the chord AB. The proof of this is left to the reader (just note that in the above, I showed that equality holds if and only if the triangle NHD is degenerate in the sense that its hypotenuse DN and its catet NH fall together, what is equivalent to < NDH = 90°, or, equivalently, < ODQ = 90°, but this is impossible unless the points D and Q coincide, what forces the point P to be the midpoint of one of the two arcs intercepted on the circle k by the chord AB).

  darij


Give an example for the following cases of a metric space $ (X,d)$ where $ X\subset C$ and $ d$ is the inherited metric then

$ 1.$ $ X$ has exactly $ 3$ limit points

$ 2.$ there exists a sequence $ \{a_n\} \subset X$ such that $ \{a_n\}$ is Cauchy, but $ \{a_n\}$ does not converge.

Yes, $ \mathbb{C}$ denotes as complex number

Is that two separate questions?  They don't seem related.  For 1) you can take the union of three bounded and monotonic sequences in $ \mathbb{R}$ and for 2) you can take a sequence of rational numbers that converges to an irrational number.  So if you needed to satisfy both conditions you would just take the union of three bounded monotonic sequences, at least one of which converges to an irrational number.


[quote="SplashD"]Don't know if it's been posted before.

If $a,b,c$ are the sides of a triangle, prove:
$a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\le abc$[/quote]

L-R$= (a+b-c)(a-c-b)(a+c-b)$


http://en.wikipedia.org/wiki/Moving_sofa_problem
http://mathworld.wolfram.com/MovingSofaProblem.html

This is an unsolved problem.

What is the solution to the one-dimensional version of the problem, i.e. find the sofa with maximum length that can be moved around a right-angled hallway of unit width.

At some point the segment will be perpendicular to the bisector of the angle, so...

[quote="LeFromage"]What is the solution to the one-dimensional version of the problem, i.e. find the sofa with maximum length that can be moved around a right-angled hallway of unit width.[/quote]

you mean this one...   :D

http://mathworld.wolfram.com/MovingLadderProblem.html


If 1 skip equals 5 hops and 1 jump equals 2 skips, how many hops are in a hop, skip, and a jump?

$ s \equal{} 5h$

$ j \equal{} 2s$

$ 2s \equal{} 10h$

$ j \equal{} 2s \equal{} 10h$

$ 10h \plus{} 5h \plus{} h \equal{} \boxed{16h}$


Is anybody on AOPS studying Cambridge AICE curriculum in India (11th and 12th grade)? If so, how do you like it?

I do know that a few international schools in Bangalore offer AICE. There must be some in other cities as well.

Thanks.


just wondering of any of you read books on mental calculation, tricks, techniques and so on

I went through Henry Sticker's "How to Calculate Quickly" cover to cover, and learned a lot of neat tricks:

http://www.amazon.com/How-Calculate-Quickly-Course-Arithmetic/dp/048620295X/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1218664605&sr=8-1

It's got a lot of exercises and practical tricks.


[quote="4everwise"]Find the number of integers between 45 and 317 inclusive and exclusive.[/quote]
[hide]
2 through 4 is 2, 3, 4 digits. 4-2=2

317-45=272
272+1 = 273
[/hide]

[hide]
sum of numbers between 45 and 317 inclu. =

sum of numbers between 1 and 317 inclu. - sum of numbers between 1 amd 44 inclusive.

(317x318)/2 - (44x45)/2 =

50403-990=49413
[/hide]

Edit: Oh! just the number. ;)  I need to pay attention.

[hide]

317-45=272
272+1=273
[/hide]

I'm going to craw under my rock in shame now...

[hide]
317-45=272, since you have to count 45, the answer is 273[/hide]

[hide]
To find the number of numbers between 317 and 45,
you need a simple subtraction problem.  You can write
 $317-45$ which is $272$.  Since you have to find the
 answer inclusivley you would need to count the number
$45$ in your answer.  So $272+1=\boxed{273}$[/hide]

[hide]317-44=273 :idea: our answer[/hide]

[hide]271 and 273.

317-45-1=271.
you subtract the 1 to get rid of the number 317.

317-45+1=273.
in this case you must add 1 to include the number 45.

maybe some of you have forgotten the meanings of inclusive and exclusive. :roll: [/hide]

notice that i am the [i]only[/i] person who has answered both questions correctly. :P


inequalities
number theory proposed
number theory

In an old script found in ruins of Perspolis is written:
[code]
This script has been finished in a year whose 13th power is
258145266804692077858261512663
You should know that if you are skilled in Arithmetics you will know the year this script is finished easily.[/code]
Find the year the script is finished. Give a reason for your answer.

The year is $ 183$, for sure....
But I have no reasons for it... :(

Anyway...is it a problem of inequality??
I am very eager to see a solution then... :maybe:

easy one.
put the number$ A$.
we have $ 17^2 < 300$ so $ 170^{13} < 17*3^6*10^{25} < A$ and its easy to prove$ A < 200$
$ a^{13} \equal{} A$ so we have $ (a,9) \equal{} 3$ and$ (a,5) \equal{} 1$ and$ a \equal{} 8k \minus{} 1$ so $ a \equal{} 183$

Hi,see here for another problem like this. :P
 http://www.artofproblemsolving.com/community/c146h150612p849619


Let $ S \equal{} \{1,2,...,100\}$ be. If $ A \equal{} \{(a,b,c): a,b,c \in S , a < b$ and $ a < c\}$ then find $ S(A)$ (the number of elements of $ A$)

First let $ (a,b,c)\equal{}(p,q,r)$ so we can reconstruct $ (p,q,r)$ become $ (p,r,q)$ if $ q\ne r$ so there are $ C^{100}_3 . 2\equal{} 323400$ elements of $ A$ in this case
If $ b\equal{}c$ then there are $ C^{100}_2\equal{}4950$ elements of $ A$ in this case
so $ S(A)\equal{}323400\plus{}4950\equal{}328350$


Each of the twenty dots on the graph below represents one of
Sarah's classmates.  Classmates who are friends are connected with
a line segment.  For her birthday party, Sarah is inviting only
the following:  all of her friends and all of those classmates who
are friends with at least one of her friends.  How many classmates
will not be invited to Sarah's party?
[asy]/* AMC8 2003 #18 Problem */
pair a=(102,256), b=(68,131), c=(162,101), d=(134,150);
pair e=(269,105), f=(359,104), g=(303,12), h=(579,211);
pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501);
pair m=(282,411), n=(147,451), o=(103,437), p=(31,373);
pair q=(419,175), r=(462,209), s=(477,288), t=(443,358);
pair oval=(282,303);
draw(l--m--n--cycle);
draw(p--oval);
draw(o--oval);
draw(b--d--oval);
draw(c--d--e--oval);
draw(e--f--g--h--i--j--oval);
draw(k--oval);
draw(q--oval);
draw(s--oval);
draw(r--s--t--oval);
dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h);
dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p);
dot(q); dot(r); dot(s); dot(t);
filldraw(yscale(.5)*Circle((282,606),80),white,black);
label(scale(0.75)*"Sarah", oval);[/asy]


How to get the table in attachment?

How about the following:

[color=red]\begin{tabular}{@{.}}
\end{tabular}
  [/color]

If I will not use [color=red]@{}[/color], How to write the code?

thank you very much!

How about
[code]\renewcommand{\arraystretch}{1.5}
\begin{tabular}{rp{5cm}}
Problem 1 (10 points). & \\ \cline{2-2}
Problem 2 (10 points). & \\ \cline{2-2}
Total (70 points). & \\ \cline{2-2}
\end{tabular} [/code]
This gives something like
\[ \begin{tabular}{rp{5cm}}
Problem 1 (10 points). & \\[1ex] \cline{2\minus{}2}
Problem 2 (10 points). & \\[1ex] \cline{2\minus{}2}
Total (70 points). & \\[1ex] \cline{2\minus{}2}
\end{tabular}\]
I can't use \renewcommand{\arraystretch}{1.5} on the forum so have simulated it by using [1ex].
Obvously you can change the 5cm to any length you like.

the key to the problem is the final  line

[quote="stevem"]The 'final line' of what? Not sure if you are still asking a question.[/quote]

             Problem 1 (10 points). .....................
             Problem 2 (10 points). .....................
[color=red]Total (70 points): ........................................[/color]      (the final line)

[color=darkred]I am sorry!
My english is terrible[/color]

If you want the final line to be longer add another column:
[code]\begin{tabular}{rp{5cm}p{2cm}}
Problem 1 (10 points). & \\ \cline{2-2}
Problem 2 (10 points). & \\ \cline{2-2}
Total (70 points). & & \\ \cline{2-3}
\end{tabular}[/code]
\[ \begin{document}\begin{tabular}{rp{5cm}p{2cm}}
Problem 1 (10 points). & \\[1ex] \cline{2-2}
Problem 2 (10 points). & \\[1ex] \cline{2-2}
Total (70 points). & & \\[1ex] \cline{2-3}
\end{tabular}\]

text-align:right
the column is aligned to the right

actually, I  want  to use [color=red]@{.}[/color] to write the code
[color=red]\begin{tabular}{@{.}}
\end{tabular}[/color]

I am sorry!
My english is terrible

You can change alignment by using \multicolumn for a single column
[code]\begin{tabular}{rp{5cm}p{2cm}}
Problem 1 (10 points). & \\ \cline{2-2}
Problem 2 (10 points). & \\ \cline{2-2}
\multicolumn{1}{l}{Total (70 points):} & & \\ \cline{2-3}
\end{tabular}[/code]
\[ \begin{document}\begin{tabular}{rp{5cm}}
Problem 1 (10 points). & \\[1ex] \cline{2-2}
Problem 2 (10 points). & \\[1ex] \cline{2-2}
\multicolumn{1}{l}{Total (70 points): } & \\[1ex] \cline{2-2}
\end{tabular} \]

Note: \multicolumn{1}{l} is \multicolumn{[i]number one[/i]}{[i]letter l[/i]}

\rule{}{} will be better than \cline
in my opinion

Good point.

Using \cline
\[ \begin{tabular}{rp{5cm}p{2cm}}
Problem 1 (10 points). & \\[1ex]  \cline{2\minus{}2}
Problem 2 (10 points). & \\[1ex]  \cline{2\minus{}2}
\multicolumn{1}{l}{Total (70 points): } & & \\[1ex]  \cline{2\minus{}3}
\end{tabular}
\]
Using \rule
\[ \begin{tabular}{rp{5cm}p{2cm}}
Problem 1 (10 points). & \rule{5cm}{0.1ex} \\[1ex]
Problem 2 (10 points). & \rule{5cm}{0.1ex}\\[1ex]
\multicolumn{1}{l}{Total (70 points): } & \rule{8cm}{0.1ex} \\
\end{tabular}
\]
Plus \rule can be adjusted up and down with an optional parameter eg \rule[-0.5ex]{5cm}{0.1ex} moves the line down 0.5ex

How about the following code?

\renewcommand{\arraystretch}{1.5}
  \begin{tabular}{r}
 Problem\hspace{2mm}1 (10 points). \rule{2.5cm}{1pt}\\
 Problem\hspace{2mm}2 (10 points). \rule{2.5cm}{1pt}\\
Total (70 points): \rule{5cm}{1pt}
 \end{tabular}

\[ \begin{tabular}{r}
Problem\hspace{2mm}1 (10 points). \rule{2.5cm}{1pt}\\[1ex]
Problem\hspace{2mm}2 (10 points). \rule{2.5cm}{1pt}\\[1ex]
Total (70 points): \rule{5cm}{1pt}
\end{tabular}
\]
Looks odd with the last line not left-aligned with the others, using a colon instead of a dot for only one line, and putting in extra spacing between Problem and the number. But if that is what you want then go ahead.

I try this:

\renewcommand{\arraystretch}{1.5}
  \begin{tabular}{r@{.\hspace{2mm}}l}
Problem\hspace{2mm}1 (10 points)&\rule{2.5cm}{1pt} \\
Problem\hspace{2mm}2 (10 points)&\rule{2.5cm}{1pt} \\
 \multicolumn{2}{r}{Total (70 points): \rule{5cm}{1pt}}
 \end{tabular}

[color=red]but...[/color]

You can show the tables here except that \renewcommand won't work so simulate it with \\ [1ex]
\begin{tabular}{r@{.\hspace{2mm}}l}
Problem\hspace{2mm}1 (10 points)&\rule{2.5cm}{1pt} \\[1ex]
Problem\hspace{2mm}2 (10 points)&\rule{2.5cm}{1pt} \\[1ex]
\multicolumn{2}{r}{Total (70 points): \rule{5cm}{1pt}}
\end{tabular}
Wouldn't it be better to align the horizontal lines?
\begin{tabular}{ll}
Problem 1 (10 points)&\rule{2.5cm}{1pt} \\[1ex]
Problem 2 (10 points)&\rule{2.5cm}{1pt} \\[1ex]
Total (70 points) & \rule{5cm}{1pt}}
\end{tabular}
or perhaps
\begin{tabular}{lll}
Problem 1 & (10 points) & \rule{2.5cm}{1pt} \\[1ex]
Problem 2 & (10 points) &\rule{2.5cm}{1pt} \\[1ex]
Total & (70 points)  & \rule{5cm}{1pt}}
\end{tabular}

Thank you very, very much!!!!!

 \begin{tabular}{[color=red]p{2cm}[/color]r@{.\hspace{2mm}}l}
[color=red]&[/color]Problem\hspace{2mm}1 (10 points)&\rule{2.5cm}{1pt} \\
[color=red]&[/color]Problem\hspace{2mm}2 (10 points)&\rule{2.5cm}{1pt} \\
 \multicolumn{[color=red]3[/color]}{r}{Total (70 points):\hspace{2mm}\rule{5cm}{1pt}}
 \end{tabular}

[color=red]or [/color]

  \begin{tabular}{r@{.\hspace{2mm}}l}
[color=red]\hspace{2cm}[/color]Problem\hspace{2mm}1 (10 points)&\rule{2.5cm}{1pt} \\
[color=red]\hspace{2cm}[/color]Problem\hspace{2mm}2 (10 points)&\rule{2.5cm}{1pt} \\
 \multicolumn{[color=red]2[/color]}{r}{Total (70 points):\hspace{2mm}\rule{5cm}{1pt}}
 \end{tabular

Not sure what/if you are asking. If you click on the tables I gave you can see the code (remove the [1ex] of course).


Let a,b,c>0 such that   $ a^4\plus{}b^4\plus{}c^4\equal{}3$                                Prove that
$ \sum\frac{a}{b\plus{}c^3}>\equal{}\frac{3}{2}$


Determine all pairs of integers $(a, b)$ such that \[\frac{a^{2}}{2ab^{2}-b^{3}+1}\] is a positive integer.


Many people seem to use it without proof, and I have not been able to acquire one.
can anyone help? :?:


quadratics
number theory unsolved
number theory

For which prime numbers p is 2 is quadratic residue mod p?

For $p \equiv \pm 1 \mod 8$ (see also the formulary for such themes: http://www.mathlinks.ro/Forum/viewtopic.php?t=76610 ).
See http://www.mathlinks.ro/Forum/viewtopic.php?t=23681 for a method to prove it.


number theory
prime numbers
number theory proposed

Let (2,a,b,c) a set of 4 numbers where a,b,c are prime

By combination we compute 7 numbers x(i) for i=0 to 6

x(0)= 2a+bc
x(1)= 2b+ac
x(2)= 2c+ab
x(3)= 2+abc
x(4)= a+2bc
x(5)= b+2ac
x(6)= c+2ab

All the x(i) are simultaneously prime if

(2,a,b,c)= (2,3,5,7)

or (2,3,5,11)
or (2,3,5,197)
or (2,3,5,1033)
or (2,3,7,11)
or (2,3,7,71)
or (2,3,7,709)
or (2,3,7,1489)
and so on

Example

When the set (2,a,b,c)=(2,3,7,11)

x(0)= 83
x(1)= 47
x(2)= 43
x(3)= 233
x(4)= 157
x(5)= 73
x(6)= 53

I called the combination 37 because we have to find 3 prime numbers to produce 7 prime numbers

Can we find some rules to find the triplet (a,b,c)?

2,3,11,73
2,3,11,397
2,3,11,1847

Is there any explanation why the increment of one digit?

73 (2digits)
397(3digits)
1847(4digits)

Can we predict that the next one will have 5 digits?
No! the next one is 4 digit-number 1867
Another 4 digit-number 2897

Any help to find some rule or some  explanation?

Thank you for any.

I have found that the sum of the seven numbers x(i) is always composite : divisible by 5 or 7 or 13.


A bin has 8 black balls and 7 white balls.  3 of the balls are drawn at random.  What is the probability of drawing 2 of one color and the 1 of the other color?

Case 1: 2 black 1 white

There are $ 3$ ways to order these, and $ 8*7*7$ to pick, so there are $ 1176$ ways total.

Case 2: 2 white 1 black

Still 3 ways to order, $ 7*6*8$ to pick, so $ 1008$ total.

THere are $ 15*14*13$ total ways, so the probability that we get 2 of one color and 1 of another color is $ \frac{1176\plus{}1008}{2730}\equal{}\boxed{\frac{4}{5}}$.

EDIT: beaten

[hide]Another way to think about it:

How can we NOT have two of one color and one of the other? If all three are the same color.

Probability of all three Black: $\frac{8*7*6}{15*14*13}$

Probability of all three White: $\frac{7*6*5}{15*14*13}$

Sum of these two fractions = 1/5.  So probability of the two-and-one scenario is $1 - \frac{1}{5}$, or $\frac{4}{5}$.[/hide]


I hope this is the right place..  :D

Here is the problem:
Given a finite alphabet $\Sigma$
$L=\(\{xu^{2}\)y\textbar x,u,y\(\epsilon\Sigma^{* }\) u\(\neq\varepsilon\) \}$
1) If $|\Sigma|=2$ prove that $L$ is regular
2) if $|\Sigma|>2$ prove that $L$ is not regular

a) is easy but I can't seem to get b) right... can it be done using the Pumping lemma?

I think its more complicated than just using the pumping lemma. By the way how do you prove 1). I mean for proving that a language is not regular you use the pumping lemma or something simillar. But to prove its regular what do u do.. constuct the finite automata or ... ?
10x

Sure, constructing an automata is not very hard when you figure out what $L$ is for a two-letter alphabet. More exactly, for $|\Sigma|=2$ is $L=\Sigma^{*}$ minus 7 words.

For the second part of the problem i figured out something.
if somebody can prove that there exists a string $w$ so that $|w|=\infty$ and $w\notin L$ for  $|\Sigma|>2$  i think i can solve it. So.. can somebody prove this :) ?

Yes, I've been trying to find a similiar string..... :) Although not quite with these properties... no luck though.


If $ x,y,z$ are real numbers such that $ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}2$, then $ x\plus{}y\plus{}z\leq xyz\plus{}2$

Not hard problem:
$ 2\equal{}x^{2}\plus{}y^{2}\plus{}z^{2}\geq x^{2}\plus{}2yz\Rightarrow yz\leq 1$.
$ x\plus{}y\plus{}z\leq xyz\plus{}2\Longleftrightarrow x(1\minus{}yz)\plus{}1\cdot (y\plus{}z)\leq\sqrt{(x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}2yz)(1\plus{}1\minus{}2yz\plus{}y^{2}z^{2})}$.Denote $ t\equal{}zy\Rightarrow$
$ \sqrt{(x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}2yz)(1\plus{}1\minus{}2yz\plus{}y^{2}z^{2})}\equal{}\sqrt{2(1\plus{}t)(2\minus{}2t\plus{}t^{2})}\leq 2\Longleftrightarrow (1\plus{}t)(2\minus{}2t\plus{}t^{2})\leq 2\Longleftrightarrow$
$ 2\minus{}2t\plus{}t^{2}\plus{}2t\minus{}2t^{2}\plus{}t^{3}\leq2\Rightarrow t^{3}\leq t^{2}$,or $ t\leq 1$. :)


function
number theory open
number theory

I don't find form to solve it, You could help me? The expression:

$2 \cdot \prod_{i=1}^{n \to\infty} \frac{p_{2i-1} \cdot p_{2i+1}}{(p_{2i})^2}$

$p\in\mathbb{P}$

And, $p_i$ is i-nth prime.

There is way to know if converge or diverge the value of that?

Thanks

[quote="elchuy"]
$2 \cdot \prod_{i=1}^{i \to\infty} \frac{p_{2i-1} \cdot p_{2i+1}}{(p_{2i})^2}$

$p\in\mathbb{P}$ [color=red]what is P? Is this the set of the prime numbers?[/color]

And, $p_i$ is i-th prime.

There is way to know if converge or diverge the value of that?
[/quote]

$\prod_{i=1}^{\infty} \frac{p_{2i-1} \cdot p_{2i+1}}{(p_{2i})^2}= \frac{p_1p_3}{p^2_2}\cdot\frac{p_3p_5}{p^2_4}\cdot\frac{p_5p_7}{p^2_6}\cdots$

I used the first 1000 primes and it seems that it tends to $\sqrt 2$

Could we use the relation $p_n \sim n\cdot lnn$ ?

[quote="pontios"]Could we use the relation $p_n \sim n\cdot lnn$ ?[/quote]
That relation is good!! but, only work for big primes?? it's demonstrated?

Well, here a graphic created by Maple:

[img]http://img509.imageshack.us/img509/4962/limite1ux.jpg[/img]

Seems that it converges :roll:

Although that does not prove the convergence since the twin primes do that the function jumps.

Thank you! :oops:

[quote="elchuy"][quote="pontios"]Could we use the relation $p_n \sim n\cdot lnn$ ?[/quote]
That relation is good!! but, only work for big primes?? it's demonstrated?
[/quote]

Well, it's a consequence of the Prime Number Theorem :
http://en.wikipedia.org/wiki/Prime_number_theorem .

:)


Each edge and each diagonal of the convex $n$-gon (where $n\geq 3$) is colored in red or blue. Prove that the vertices of the $n$-gon can be labeled as $A_1,A_2,...,A_n$ in such a way that one of the following two conditions is satisfied:

1) all segments $A_1A_2,A_2A_3,...,A_{n-1}A_n,A_nA_1$ are of the same color;

2) there exists a number $k$, $1<k<n$ such that the segments $A_1A_2,A_2A_3,...,A_{k-1}A_k$ are blue , and the segments $A_kA_{k+1},...,A_{n-1}A_n,A_nA_1$ are red.