although my teacher and textbook do not speak of this i have found reference to "pentagonal bipyramidal" valence structures with 7 domains.  My questions are
do these really exist and when(examples)?
what is the hybridization involved?
are there structures with even more domains?

$IF_{7}$ is an example with pentagonal bipyramidal. Look here for more information:

http://dbhs.wvusd.k12.ca.us/webdocs/Bonding/Molecular-Polarity.html

yeah but that doesn't answer my other ?s it just gives an example

The hybridization is most likely $d^{3}sp^{3}$ since octahedral is $d^{2}sp^{3}$.

When there are 2 lone pairs and 5 bonding pairs about central atom, the pentagon arrangement occurs.

Since the atoms having more effective pairs than 4 usually are placing electrons in their d orbitals, I would imagine that there could be even 8 effective pairs (18 electrons).

It should be $sp^{3}d^{3}$  By the way does there even exist any $d^{3}sp^{3}$ hybridised species?

[quote="gregx007"]Ehh, keep in mind that the orbital hybridization model is an approximation that is pretty good for s and p, but starts to break down while mixing in d.  Textbooks like to write about sp3d2, etc., but from what I know, when you do calculations you pretty much die unless you switch to MO theory.

I think there is a quote that goes something along the lines of: "Linus Pauling invented the orbital hybridization model and set American chemistry behind twenty years."[/quote]

What is MO theory?


It's a well-known physics formula that force equals mass times acceleration.  Jen wants to throw a softball with the same force as Jack throws a baseball.  If the softball has a mass of $ 200$ g and the baseball has a mass of $ 150$ g, what is the ratio of acceleration of Jen's ball to Jack's.  Answer as a fraction in lowest terms.

Let $ x$ equal the acceleration of the softball and $ y$ the acceleration of the baseball.

$ 200x\equal{}150y \implies \frac{x}{y}\equal{}\boxed{\frac{3}{4}}$.

[quote="gaussintraining"]Let $ x$ equal the acceleration of the softball and $ y$ the acceleration of the baseball.

$ 200x \equal{} 150y \implies \frac {x}{y} \equal{} \boxed{\frac {3}{4}}$.[/quote]

i couldnt have said it better myself, so i didnt :P

As the problem says, f = ma, where f is force, m is mass, and a is acceleration.

Jen's equation: f = 200a

Jack's equation: f = 150a

(Jack)/(Jen) = 3/4.

_-_-_-_-_-_-_-_-_-_-
~pi man!!  :juggle: :icecream:  :heli:

Suppose that $a$ varies inversely with $b^2$. If $a=9$ when $b=2$, find the value of $a$ when $b=3$.

1) Don't revive.

2) Your question is not related to the original question.

3) The answer would be 4.


inequalities
number theory solved
number theory

Do there exist infinitely many integer solutions to \[ x^2 +y^2 +z^2 =x^3+y^3+z^3 ? \]

evidently (x,y,z) = (a(2a 2 + 1),-a(2a 2 + 1),(2a 2 + 1)) have the required properties
q.e.d
cheers

Galois, how did you find this family of solutions ?

Is there a trick to find such a family ?

well,basically try to reduce the complexity of the problem.suppose if we take y=-x,we get an equation like
2x 2 = (z-1)z 2 .now if we can find z  s.t
(z-1)/2 is a perfect square ,we are done.so set z=2a 2  + 1

So this trick doesn't always work. I think it does not work in this case :

Are there infinitely many solutions in $\mathbb{Z}^3$ to \[ x^2 + y^2 + z^2 = x^5 + y^5 = z^5 ? \]

But I saw a problem which can be solved using the same trick :

Prove that there are infinitely many solutions in $\mathbb{Z}^3$ to $x^2 + y^5 = z^3.$

An infinite family of solutions to x^2 + y^5 = z^3 is (0, t^3, t^5), where t is an arbitrary integer.

Cheers,
Thanos

As far as Arne's first problem is concerned, here is a solution:

The condition x^5 + y^5 = z^5 implies that xyz = 0 (this really trivial fact has a rather mundane proof but this margin is, I am afraid, too small to contain it). Suppose:

--> z = 0. Then, (x,y,z) = (0,0,0).

--> x = 0. Then, y = z and the other condition yields (0,0,0).

--> y = 0. Same treatment as above.

Hence, there is only one solution, namely (0, 0, 0).

i am sure we can quote the FLT in this case, but Arne's problem is most likely \[ x^2 + y^2 + z^2 = x^5 + y^5 + z^5 \]

:D hmm thanos - unfortunately we don't have any margins here :D:D

For the Canadian problem there are lots of different solutions.

1) Note that $10^2 + 3^5 = 7^3$ hence an infinite family of solutions is
$(a, b, c) = (10k^{15}, 3k^6, 7k^{10})$ for $k$ an integer.

2) Note that $3^2 + (-1)^5 = 2^3$. Continue in the same way as in 1).

3) Treating x, y, z as powers of two, we get
$(a, b, c) = (2^{10k}, 2^{4k}, 2^{7k})$ for $k$ a positive integer.

4) ...

Now remains the problem : Are there infinitely many integer solution to
$x^2 + y^2 + z^2 = x^5 + y^5 + z^5$
for which $xyz$ is different from 0 ?

for the last problem, i dont think there are infinitely many integer solutions.  one solution is x=y=z=1
if x,y,z are positive and one of them is greater than 1, then x^5>=x^2, y^5>=y^2, z^5>=z^2 with strict inequality of x^5+y^5+z^5>xx+yy+zz

similar argument for negative numbers
hence the only case is when x>=y>0>z and x>0>y>=z


We see that from minutes 0-10, she was going 6kmph.. From minute 10-20, she was going 7.5kmph.... From minute 20-25 she was going 6kmph..... From minute 25-30 she was going 13.2 kmph... From here, we see that her maximum speed occurs at 13.2 kmph, which rounded to the nearest who number is 13..

Looking at the graph, it is obvious that the greatest speed was achieved between minute 10 and minute 20, it being the steepest. We see that Lisa traveled 3.5-1=2.5 km in ten minutes. From here we can set up an equation to solve the distance per 60 minutes, or in other words, km per hour.
We have: $ \dfrac{10}{60}\equal{}\dfrac{2.5}{x}$, and solving for x, we have $ x\equal{}\fbox{15}$.


Let F be a field of characteristic zero and let V be a finite dimensional vector space over F. If $ a_{1},...,a_{m}$are finitely many non-zero vectors from V prove that there is a linear functional f on V satisfying $ f(a_{i})\ne 0$ when $ 1\le i\le m$

Let $ V^{*}$ be the dual space of $ V$ - that is, the set of all linear functionals on $ V.$ $ V^{*}$ is also finite dimensional and has the same dimension as $ V.$ Let $ S_{i}\equal{}\{f\in V^{*}: f(a_{i}) \equal{} 0\}.$ Each $ S_{i}$ is a proper subspace of $ V^{*}$ - in fact, a subspace of codimension one.

What we are asked to show is that $ \bigcup_{i \equal{} 1}^{m}S_{i}\ne V^{*}.$

Here's another way to phrase it: the problem is equivalent to the statement that a vector space over a field of characteristic zero cannot be a finite union of proper subspaces.

(Note that this is clearly false over a finite field.)

[color=green][Edit: I finished this post by adding the proof below before I saw perfect_radio's post.][/color]

We can then prove this statement by induction on $ m,$ the number of subspaces.

Suppose $ \bigcup_{i \equal{} 1}^{m}S_{i}\ne V^{*}.$ Choose $ u\not\in\bigcup_{i \equal{} 1}^{m}S_{i}.$

Now find some $ v\not\in S_{m\plus{}1}.$ Consider the line $ \{u\plus{}tv: t\in\mathbb{F}\}.$

We note that this line intersects each $ S_{i}$ at most once. That's because if $ u\plus{}tv$ and $ u\plus{}sv$ both belong to $ S_{i},$ we could find a linear combination of these that would equal $ u,$ which would force $ u\in S_{i},$ which is known to be false for $ 1\le i\le m.$ And if $ u\plus{}tv\in S_{m\plus{}1}$ for any $ t\ne 0,$ then we could make a linear combination of $ u$ and $ u\plus{}tv$ that equaled $ v,$ which would force $ v\in S_{m\plus{}1},$ which is known to be false.

Therefore, the line $ \{u\plus{}tv: t\in\mathbb{F}\}$ intersects each $ S_{i}$ at most once, so it has a finite number of intersections with $ \bigcup_{i \equal{} 1}^{m\plus{}1}S_{i}.$ Since the field $ \mathbb{F}$ is infinite, there must be a $ t$ such that $ u\plus{}tv$ is not in thus union.

Note that we did not explicitly use that the field had characteristic zero. We used that the field was infinite. All fields of characteristic zero are infinite, but there are also infinite fields of finite characteristic. We could change the hypothesis of the original problem to the field being infinite.

That's really an [b]interesting[/b] way to rephrase the problem :) (grobber gave a solution a long, long time ago).

Here's what I'd have done:
Let $ v_{1},\ldots, v_{n}$ be a basis of $ V$, so that each $ a_{k}$ can be written as $ \sum_{i \equal{} 1}^{n}c_{k,i}v_{i}$.
Assign some non-zero number to $ v_{1}$. What should we assign to $ v_{2}$? More precisely, what should we not? Some $ a_{k}$'s can be linear combinations of $ v_{1}, v_{2}$, but that excludes only finitely many values. So we can define $ f\left( v_{2}\right)$ (as we have an infinite field at our disposal). Continuing in the same manner we can define $ f\left( v_{j}\right)$ for all $ j$, so that's our linear functional.
[EDIT: Really minor correction]

Thanks perfect radio, I was thinking along those lines but I thought that was too messy and there might be a slicker proof using just properties of hyperspaces and linear functionals. I phrased the problem the way Kent did, which is probably why I drew a blank. Are there any other ways to prove it? How does the result fare in infinite dimensional spaces?

I doubt that there's anything interesting over infinite dimensional vector spaces. Given finitely many $ a_{i}$ we can limit ourselves to the span of those vectors, which is a finite dimensional space. And if you allow infinitely many $ a_{i},$ then every argument that either perfect_radio or I used goes flying out the window, and I suspect the result would be false.

Equivalent formulation:

If we regard $ R^n$as a linear space with an inner product, this says that given non-zero vectors $ a_1 ,..,a_m$ in $ R^n$, there is some vector v in $ R^n$ for which

$ v \cdot a_k  \ne 0$ $ k \equal{} 1,...,m$

Relating this formulation to the one Kent suggested is the real feat.

[quote="Kent Merryfield"]Let $ V^{*}$ be the dual space of $ V$ - that is, the set of all linear functionals on $ V.$ $ V^{*}$ is also finite dimensional and has the same dimension as $ V.$ Let $ S_{i} \equal{} \{f\in V^{*}: f(a_{i}) \equal{} 0\}.$ Each $ S_{i}$ is a proper subspace of $ V^{*}$ - in fact, a subspace of codimension one.

What we are asked to show is that $ \bigcup_{i \equal{} 1}^{m}S_{i}\ne V^{*}.$

[b]Here's another way to phrase it: the problem is equivalent to the statement that a vector space over a field of characteristic zero cannot be a finite union of proper subspaces.[/b][/quote]

Kent can you prove this? I'm having a hard time showing it.

Here's the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=224519]finite-dimensional version[/url].
In an infinite-dimensional space $ V$, let $ V_1,V_2,\dots,V_n$ be the proper subspaces, and choose $ x_i\not\in V_i$. Intersect everything with the span of the $ V_i$, and use the finite-dimensional version there.

As for "equivalent"- don't get too hung up on that word.

Thanks jmerry. I won't lose any more sleep over the equivalence  :surrender:

I do not understand what you mean by "intersect everything with the span of the $ {V_i }$" though. In my thinking, you would have to intersect the union of the $ V_i$by some finite dimensional subspace W of V. Then $ V \cap W \equal{} \bigcup\limits_{i \equal{} 1}^n {V_i } \cap W$, which imples that the finite dimensional subspace $ V \cap W$ is the union of finitely many proper subspaces. Contradiction.


summer program
MathZoom
AMC
AIME
USA(J)MO
USAMO

Hi DrKevin (or other mathzoomers),

I'm interested in going to a math summer camp and I found out about MathZoom. I don't know whether I should go or not. Can you please explain a typical day at the camp? I also have a question about the "Groups", (a.k.a Black Group, Red Group, etc.) What exactly are these groups and how does it affect your stay? Also, what's the difference between a day camper and a residential camper? Also, how is MathZoom overall? (This is for mathzoomers, please.)

Thank you
rts2007

P.S. I received a score of 5 on this year's AIME, and I know I can get higher than that next year. Will that effect my "group?"

I wonder why no one has responded yet..... :huh:

A typical day at Math Zoom Summer Program is like this:

7:30-8:30: Breakfast
9:00-12:15: Lecture and problem session
12:15-1:15: Lunch
1:15-4:30: Lecture and problem session
4:30-6:00: Sports and recreation
6:00-7:00: Dinner
7:00-8:30: Evening Seminars or group activities.
8:30-9:00: Supervised self study
10:00pm: lights out

Day campers are those who live close by and don't need live in the dorms.
The groups are decided by the staff, based on your experiences.  The Black Group is for USAMO and advanced AIME level.  The Blue Group is the intermediate level (AMC 10/12 and beginner AIME).  The Red Group is the beginner level.

Since your AIME score is 5, I can put you in the Black Group.  The materials will be challenging.   This is flexible too; if it is not suitable for you (too hard), you can move to the Blue Group without a problem.  The goal is to put you in a group that can benefit you the most.

Hope to see you at the camp!  The deadline is approaching...


invariant
geometry
3D geometry
combinatorics proposed
combinatorics

Let $n \geq 3$ be a natural number. A cube $W$ of edge length $n$ is composed of $n^3$ unit cubes $W_1, \ldots, W_{n^3}$. Each unit cube gets assigned a number. Show that you can choose the numbers such that

a.) all numbers are distinct

b.) the sum of these following numbers in any row equals 0: consider all the cubes in a row which are parallel to any edge of $W$'s parallel layers of unity cubes.

First fill assign numbers $a_{ij}$ to each cube s.t. the sum is zero on each row (this is, of course, possible, since we can put $a_{ij}=0,\ \forall i,j$). Then, whenever we have $a_{ij}=a_{kl}$, we can consider eight cubes, one of which is $a_{ij}$ and none of which is $a_{kl}$ which are the vertices of a parallelepiped with sides parallel to those of the large cube. Now change $a_{i,j}$ to $a_{ij}+x$, and then change the numbers assigned to the other seven considered cubes accordingly (to preserve the property that the sum on any row is zero).

If we make $x$ large enough, the eight chosen cubes will have values outside a certain interval containing the values of all the other cubes, so the number of pairs of equal numbers in the cube has decreased by at least one. This means that by this precedure, we will eventually reach a stage when all the numbers are different.


[color=cyan]What is the remainder when 15^17 + 17^15 is divided by 32?[/color]

SOlution:

[hide]By Fermat's little theorem, we get that

15^17 :equiv: 15(mod17)

This means the numerator is congruent to 15 (mod17)

ANd since the denominator is also congruent to 15(mod17), the remainder is just 1.[/hide]

But why are you taking this mod 17? You should be working mod 32.

Ok, hopefully, correct solution:

[hide]15^17 :equiv: 15(mod32).

17^15 :equiv: 17(mod32)

SO the remainder is 0.[/hide]

Wait, so how did you derive those two congruences?

Um, basically, 15^odd is congruent to 15(mod32), and 15^even is congruent to 1(mod32)

17^odd is congruent to 17(mod32), and 17^even is congruent to
1(mod32)

Try it out..
it cycles every 2, so...

But how can you actually show that (without just looking at the pattern)?

You can apply multiplication to a modulus. That is, ab (mod m) :equiv: a (mod m) * b (mod m).

Well, sorta like this, since 15,17,32 are relatively prime, multiplying is nice.

It's easy to see that 15^1  :equiv: 15(mod32)
and 15^2 :equiv: 1(mod32)
now, 15^3 :equiv: 15*15^2 :equiv: 15*1 :equiv: 15(mod32)

Then, the same argument for the rest of the powers of 15, and for the powers of 17 as well

Ah, ok, I see now.

[quote]Well, sorta like this, since 15,17,32 are relatively prime, multiplying is nice. [/quote]

But of course, it doesn't matter whether they are relatively prime or not, since that stuff only factors into division in mods, right?

Um, perhaps. Im not too familiar with division and multiplication in mods, and too lazy to test it out myself.

Perhaps Simon would care to give us the short answer....

I kno in division, u need relative primality. Not sure in multiplication tho, probably don't...

With multiplication, you're always fine. For division, if things aren't relatively prime, you can end up in trouble.

if in division, i think i remember Joel telling me that u need to divide everything by the common factor,

meaing something like 6(mod32) /2 = 3(mod16),
or something like that....


Acute triangle ABC is inscribed in a circle. Altitude AM is extended to P on the circle and altitude CN is extended to Q on the circle. If $ AC: PQ=7: 2$, find $ \sin B$.

$ AM$ intersect $ CN$ at $ H$

Since $ BHP$ is a isosceles triangle, $ M$ is the midpoint of $ HP$.
$ AHC$ and $ QHP$ are similar triangles so $ \frac{CH}{HP}=\frac{AC}{QP}=\frac{7}{2}$

We have

$ (\sin B)^{2}=(\sin MHC)^{2}=1-(\cos MHC)^{2}=1-(\frac{MH}{HC})^{2}=1-(\frac{HP}{2HC})^{2}=1-(\frac{2}{14})^{2}=\frac{48}{49}$

Then

$ \sin B=\frac{4\sqrt{3}}{7}$


Umm..I'm sure there's is any easy way to do this other than doing those grids...umm...any help would be great.

Particle Man is at the origin in the three-dimensional space.  How many ways can Particle Man take a series of 12 unit-lengh steps, each step parallel to one of the coordinate axes, for the origin to (3,4,5) without passing through the point (2,3,2)

Also...
There are 2n players in a chess tournament.  The first round consists of pairing the players to participate in n matches with every player playing one match.  In terms of n, how many ways can this pairing take place?
I'm probably missing something here, but how did these people come up with 2n!/n^nn!   I haven't learned LaTex yet, so I'll write that out in English...2n factorial divided by n to the nth power multiplied by n and that product is also a factorial...Explanation please?  Thanks...

1st problem: First calulate all the ways from (0,0,0) to (3,4,5), these are $\binom{12}{3 \, 4 \, 5} = \frac{12!}{3!4!5!} = 27720$. then get the number of ways passing thru (2,3,2): there are $\binom{7}{2\,3\,2} = 210$ ways from (0,0,0) to (2,3,2) and $\binom{5}{1 \, 1 \, 3} = 20$ ways from (2,3,2) to (3,4,5). Combining these in any way gives $210\cdot 20 = 4200$ ways going thru (2,3,2), so we're left with 23520 ways.

2nd: you mean $(\frac{(2n)!}{n^nn})!$? I donno how they came there, but i'd say, first you have to form the pairings, which is possible in $\frac{(2n)!}{n!2^n}$ ways

(2n!): distribute the players into positions 1,2,3,...,2n, where 1 plays against 2, 3 ./. 4 etc.
/n!: since ordering of the n pairs among each other does not matter.
/2^n: since ordering in each if the n matches does not matter.

I hope i got it right.

greez

luqash


geometry
geometric transformation
reflection
symmetry
analytic geometry

Farmer Ben has a herd of sheep at $ (8,3)$ and he wants to take them to the barn at $ (2,7)$. However, he must first have the take a drink at the river which runs along the x-axis, and then to the shearing stations which run along the y-axis. What is the length of the shortest route that Farmer Ben can take (in simplest radical form)?

does the farmer need to go to the river and then the sheering station in that order???

[hide="HUGE hint"]
Use reflections.
The shortest distance between two points is a straight line... however, you need to hit any point on the river before you proceed.
[/hide]

[hide="Not sure if this is the shortest"] The farmer could just travel to the origin, completing both tasks at one point, then traveling to his final destination. The distance he travels is $ \sqrt73 \plus{} \sqrt53 \equal{} 15.8241 units$[/hide]

yes i believe that is correct.  Although i have been wrong before, happened last in the late 80's lolol  :rotfl:  :rotfl:  :rotfl:

hmm, I found it possible to make it travelling only $ 10\sqrt{2}$

[hide="reflections"]
To put it shortly, reflect $ (2,7)$ across the y-axis, then the x-axis, and find the distance between this point and $ (8,3)$
[/hide]

i thought about doing that but dont understand why you would reflect it over the axises

@ the problem with this is that the distance from $ (\minus{}7,\minus{}2)$ - (I believe this is the reflected point you were talking about?!?) to $ (8,3)$ is already $ 10\sqrt2$, but the distance from that point to the starter was not included in $ 10\sqrt2$. That distance is $ 10\sqrt2$, again. So the total distance traveled is $ 20\sqrt2$

Look at the big happy underlined HUGE hint... Shortest distance between two points in a plane is a straight line... Go prove it now =) lol... Yo. wait a second. So he goes to the river, shearing stations, and THEN the barn? Ohhh... I see how it is =).

Two reflections to do this xD.

EDIT: Answer is $ 10\sqrt{2}$.

[hide="Solution"]Note that... he first travels to the x-axis, and to minimize the distance traveled to the shearing stations and then to the barn, he must obviously travel to a point above the x-axis.  We first reflect the barn's point over the x-axis so that we can minimize the distance traveled from the river to the stations to the barn (lol).  This new point is $ (\minus{}2,7)$.  Now, we reflect it over the y-axis to find the shortest possible length.  $ (\minus{}2,\minus{}7)$ is the new point.  The distance is $ 10\sqrt{2}$.
[/hide]
I'm sorry... I was too lazy to write up a good solution xD

Mathcrazed, you didn't travel to (8,3)! From your newest point, (-2,-7) to (8,3), that's a lot of extra ground to cover. So its like a lot more.

[quote="happyme"]So lets say the barn was at (5,5), then the reflections would end up at (-5,-5) and the distance would be:

$ \sqrt {(8 \minus{} ( \minus{} 5))^2 \plus{} (3 \minus{} ( \minus{} 5))^2}$

$ \sqrt {(13)^2 \plus{} (8)^2}$

$ \sqrt {169 \plus{} 64}$

$ \sqrt {233}$

So would that be correct (since 233 is prime)?[/quote]

Yep, that's correct

I wonder how this would turn out if we extended the problem to 3-dimensional space...

Well, I figure you could always have symmetry over the $ z$ axis, so aside from the fact that the distance formula would be different (drawing cooler right triangles would be required), it don't think it would be too different.

darkmatter, we are just using symmetry to turn the path into a straight line (versus finding the optimal distance to each axis and making more than one line).

[quote="darkmatter47"][quote="infinity4ever"]No, its not.  It just looks like more, but in reality remember that they have to stop by the river.[/quote]  Wait, so does that mean that we don't count the distance he gets to travel on the river?? From (-2,-7) to (8,3), thats $ 10\sqrt2$. But if you don't the distance on the river...... Otherwise, what do you mean? Excuse my slow mind if it's obvious.   :maybe:[/quote]

LOL, think of it this way:  :D

the sheep stop by the river to drink, not stampede across it!  remember, this reflection is just theoretical.  they are not actually "hopping" across the river.

Sorry   :blush:  , let me get the solution straight. So you start at (8,3), then go to (-2,-7), then to (2,7) ?
By straight line, you mean a line going through (-2,-7), the x axis, the y axis and (2,7)?

Ermm... No.  That's the theoretical shortest way. Remember that the shortest distance between any two points is a straight line... Let's say I started at 1,1 and wanted to go to the x-axis and back in the shortest possible distance.  Now obviously this is 2.  But let's do it another way.  Reflect the point 1,1 over the x-axis.  Then we see that the distance between the new coordinate and old coordinate is 2.  This further proves the point.  Similarly for this problem.

Oh, ok. I get it. You're just theoretically proving that this kind of a distance; y'all didn't actually find points at which to calculate distances..... right?  :)

Bah, forgive the newbishness here, but I couldn't help reading through the thread and being a bit confused by the logic here...I do understand the part that the shortest length between any two points is a straight line. That length has been found. However, about [i] why [/i] that makes sense...

Assuming that the line from (8,3) to (-2,-7) slices through the x-axis and y-axis, it makes logical sense that the farmer can, without crossing the river, take that exact same path the same length to wherever he intercepts the x-axis.

Then, from there, on the line from the original spot to the reflected point, he goes a set distance downwards, still on the same path, until he intersects the y-axis. By now, he's crossed the river. However, that same segment in which he continues downwards to hit the y-axis can be, once again, [i] reflected [/i] back over the x-axis, so he can head up from the place where he touched the river to the corresponding point on the y-axis, still taking the same distance as he would have had he followed the direct path. Am I good so far?

And lastly, from there, the distances from where he touches the y-axis is the same to the final resulting point: either on the line to the reflected point, or from the axis to (2,7). Ehh...'tis all?


geometry
ratio
trigonometry
AMC
AIME
number theory
relatively prime

In triangle $ABC, AB=\sqrt{30}, AC=\sqrt{6},$ and $BC=\sqrt{15}.$  There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$ and $\angle ADB$ is a right angle.  The ratio \[ \frac{\text{Area}(\triangle ADB)}{\text{Area}(\triangle ABC)} \] can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

Hint
[hide]
$[\triangle{ACM}]=[\triangle{AMB}]$ because their bases and altitudes are equal. All we need to find now is $\frac{AM}{AD}$ to find out the area ratio between $\triangle{AMB}$ and $\triangle{ADB}$[/hide]

This is actually a very easy problem, but calculation here is just a pain... if anyone has a better solution, please post!

[img]http://www.kalva.demon.co.uk/aime/soln/diag9613.jpg[/img]

[hide]By Stewart's Theorem, we can find $AM$, which is $\frac{\sqrt{57}}{2}$. Obviously, as stated in the problem, $CM = MB = \frac{\sqrt{15}}{2}$. Let $MD = x$. We can state that:

$(\frac{\sqrt{57}}{{2}}+x)^2 + BD^2 = 30$
$(x^2 + BD^2) = \sqrt{15}{4}$

Solving this dreadful equation, we end up having $x = \frac{4\sqrt{57}}{19}$. Since $[ACM] = [AMB]$ because these triangles have the bases of same length and they have the same height, all we need to do is to find the ratio:

$\frac{1}{2} + \frac{[BED]}{2[ABE]}$

Simplifying, and after a very happy and exciting result at the end when everything cancels out, we end up with $27/38$. So, our final answer is $27 + 38 = \boxed{65}$.[/hide]

By the way, the image is from Kalva. And again, someone please post a nicer solution... this problem took me almost 30 minutes to do because of the calculations!!

$BM^2-AB^2-AM^2=-2(AM)(AB)\cos DAB$ by law of cosines

$\frac{AB^2+AM^2-BM^2}{4AM^2}=\frac{AB*\cos DAB}{2*AM}=\frac{AD}{2*AM}=\frac{[ADB]}{[ABC]}$,

by stewart: $4AM^2=57$, answer follows

[ADB]/[ABC]=([BFD]+[AFB])/2[AFB]= 1/2([BFD]/[AFB])+1/2. So we want to find [BFD]/[AFB].
By Stewart's, AM=$ \sqrt{57}/2$. Law of cosines finds angle DAB, so since DAB is right, it's easy to find AD and DF. But [BFD]/[AFB]=DF/AF since they have verticle angles.

I thought this was easy for an AIME 13.

Prove easily that $ \frac {[ABD]}{[ABC]}\equal{}\frac {b^2\plus{}3c^2\minus{}a^2}{8m_a^2}$ , where $ 4m_a^2\equal{}2(b^2\plus{}c^2)\minus{}a^2$ .

In the our particular case obtain : $ \left\|\begin{array}{c}
a\equal{}\sqrt {15}\\\
b\equal{}\sqrt 6\\\
c\equal{}\sqrt {30)}\end{array}\right\|$ $ \implies$ $ \frac {[ABD]}{[ABC]}\equal{}\frac {27}{38}$ , where $ (27,38)\equal{}1$ and $ 27\plus{}38\equal{}65$ .

Could anyone give the figure?

I can't understand the prolem.

Thanks in advance.

kunny

[size=117][color=darkblue][b][u]The enunciation is only for Kunny ![/u][/b][/color][/size]

[quote] [color=darkred]Let $ ABC$ be a triangle with $ AB = \sqrt {30}$ , $ AC = \sqrt {6}$  and $ BC = \sqrt {15}$ .  Denote $ \|\begin{array}{c} M\in BC\ ,\ MB = MC \\
\ D\in AM\ ,\ BD\perp AM\end{array}$ .

The ratio $ \frac {[ABD]}{[ABC]}$ can be written in the form $ \frac mn$ , where $ m$ and $ n$ are relatively prime positive integers.  Find the sum $ m + n$ .[/color] [/quote]
[hide="Proof."]
Denote $ AM = m_a$ and $ x = m(\widehat {BAD})$ . Therefore, $ MB = MC\ \Longleftrightarrow\ c\sin x =$ $ b\sin (A - x)\ \Longleftrightarrow \boxed {\ \tan x = \frac {b\sin A}{c + b\cos A}\ }$ .

$ AD = c\cdot\cos x$ and $ [ABD] = \frac 12\cdot AB\cdot AD\cdot\sin x$ $ \implies$ $ [ABD] = \frac {c^2\sin 2x}{4}$ . Using the relation $ \sin 2x = \frac {2\tan x}{1 + \tan^2x}$ obtain $ \sin 2x =$

$ \frac {2b\sin A(c + b\cos A)}{(c + b\cos A)^2 + (b\sin A)^2} =$ $ \frac {2b\sin A(c + b\cos A)}{c^2 + 2bc\cdot\cos A + b^2}$ $ \implies$ $ \sin 2x = \frac {2b\sin A(c + b\cos A)}{4m_a^2}$ . Thus, $ [ABD] = \frac {c^2}{4}\cdot\frac {2b\sin A(c + b\cos A)}{4m_a^2} =$

$ \frac {c\cdot bc\sin A\cdot (c + b\cos A)}{8m_a^2}$ $ \implies$ $ \frac {[ABD]}{[ABC]} = \frac {2c(c + b\cos A)}{8m_a^2} =$ $ \frac {2c^2 + 2bc\cdot\cos A}{8m_a^2}$ $ \implies$ $ \boxed {\ \frac {[ABD]}{[ABC]} = \frac {b^2 + 3c^2 - a^2}{8m_a^2}\ }$ .

[b]Remark.[/b] I used the well-known relations $ 4m_a^2=2(b^2+c^2)-a^2$ , $ [ABC]=\frac 12\cdot bc\sin A$ and $ a^2=b^2+c^2-2bc\cdot\cos A$ .[/hide]

Thank you very much, Virgil Nicula.

Here is my solution.

Construct parallerogram $ AEBC$ such that $ AM \equal{} ME$, we have $ \frac {[ADB]}{[ABC]} \equal{} \frac {AD}{2AM}$ as Altheman said.

By Pappus's therem, $ AB^2 \plus{} AC^2 \equal{} 2(AM^2 \plus{} BM^2)\Longleftrightarrow 30 \plus{} 6 \equal{} 2\left(AM^2 \plus{} \frac {15}{4}\right)$ $ \therefore AM \equal{} \frac {\sqrt {57}}{2}$.

Thus Apply to cosine law for $ \triangle{ABM}$, we have $ \cos \angle{BAM} \equal{} \frac {(\sqrt {30})^2 \plus{} \left(\frac {\sqrt {57}}{2}\right)^2 \minus{} \left(\frac {\sqrt {15}}{2}\right)^2}{2\cdot \sqrt {30}\cdot \frac {\sqrt {57}}{2}} \equal{} \frac {81}{2\sqrt {30}\sqrt {57}}$. $ AD \equal{} \sqrt {30}\cos \angle{BAM} \equal{} \frac {81}{2\sqrt {57}}$.

$ \therefore \frac {[ADB]}{[ABC]} \equal{} \frac {AD}{2AM} \equal{} \frac {27}{38}$, yielding $ m \plus{} n \equal{} 27 \plus{} 38 \equal{} \boxed{65}$.

Let $AD$ intersect $BC$ at $E$. Stewart's Theorem gives $AE = \frac{\sqrt{57}}{2}$. Next, LoC gives $\cos \angle AEC = \frac{8}{\sqrt{95}}$, so $\sin \angle AEC = \sin \angle DEB = \frac{\sqrt{31}}{\sqrt{95}}$. Note that $\cos \angle DEB = \cos \angle AEC = \frac{8}{\sqrt{95}}$, so $\frac{ED}{\frac{\sqrt{15}}{2}}=\frac{8}{\sqrt{95}}$, and $[DEB] = \frac{ED \cdot \frac{\sqrt{15}}{2} \cdot \frac{\sqrt{31}}{\sqrt{95}}}{2} = \frac{3\sqrt{31}}{19}$. Also, $[AEC]=\frac{\frac{\sqrt{15}}{2} \cdot \frac{\sqrt{57}}{2} \cdot \frac{\sqrt{31}}{\sqrt{95}}}{2} = \frac{3\sqrt{31}}{8}$. To finish, we have $[BAE] = [AEC]$, so $[ADB]=[BAE]+[DEB]=\frac{81\sqrt{31}}{152}$. Next, $[ABC]=2[AEC]=\frac{3\sqrt{31}}{4}$. Dividing gives $\frac{27}{38} \implies \boxed{65}$.

Coordbash

Let $B(0,0)$
$C(\sqrt{15},0)$
$A(\frac{13\sqrt{15}}{10},\frac{\sqrt{465}}{10})$
Let $E$ be the intersection of $AD$ and $BC$
$E(\frac{\sqrt{15}}{2},0)$
The equation of line $AE$ is $y=\frac{\sqrt{31}}{8}x-\frac{\sqrt{465}}{16}$
So the equation of line $BD$ is $y=\frac{-8\sqrt{31}}{31}x$ (perpendicular)
The two lines intersect at $D(\frac{31\sqrt{15}}{190},-4\frac{\sqrt{465}}{95})$
$[ADB]=\frac{BD*DA}{2}=\frac{81\sqrt{31}}{152}$
$[ABC]=\frac{3\sqrt{31}}{4}$ by shoelace
Divide them to get $\frac{27}{38}$ --> $\boxed{65}$


superior algebra
superior algebra unsolved

hello,

consider the cycle (123...n) and the transposition (1 k)

When do these two generate the full group of permutations on n symbols?

I have already proven that if k-1 and n are coprime, it generates the full group, but is the converse also true?

I think so. It looks like if $d=(k-1,n)$, then all the permutations generated by the given cycle and transposition share the proeprty that the distance between any two symbols with difference $k-1$ is a multiple of $d$, and this property clearly can't hold for all permutations when $d>1$.


[quote="Cyber-Shreyas"]Excellent Sum!!!

Answer is ZERO.

Why? Let us number the pulleys from top to bototm.

Consider the Mass attached to pulley 1. Let its accelaration be a upwards. Then consider the mass attached to Pulley 2. WRT Pulley 2, it should also have an accelaration a upwards but the pulley has accelaration a downwards. So the net motion is that the body does not move at all. So the other block does not move at all as well.

Note that such a system is not physically possible and infinite recursions are much more common in Maths and CSC.[/quote]

No, the ans is wrong
Can u xplain how u got mass attached 2 pulley 2 also has acc. a ? :D

shreyas's soln is perfect.
sairam, i think you didn't read what he said correctly:
[quote][b][u]WRT Pulley 2[/u][/b], it should also have an accelaration a upwards[/quote]

wrt pulley 2, the acceleration of the second mass will be 'a' only.
this is because it essentially reduces to the SAME system as the original case, except that the whole system is now moving down with acceleration 'a'
key word(s) is wrt here
:)

PS: are u saying soln is wrong or ans is wrong?

Guyz
I don think it is exactly the same system again wen viewed frm the movable pulley. :?:
I think wat u r takin is d accn of d 2nd bloc( d bloc attd 2 d second movable pulley) wrt to the movable pulley.
Bt I want evrythin frm the ground frame.
Now wen u go  2 the frame of the 3rd pulley u r goin relativ again. :arrow:
When I came acros dis jus thought it was a brute prob n solvd it. :o
I shud confess, nice approach. bt not exactly able 2 point out the bug..
Maybe som1else will do better justice.

Btw d ans is g/2.

Hey
The tension in the string attached to the second bloc n that in the 1 attached to the 1st bloc r not the same.
So replacing by the same system is wrong!! :!:

No, No. My soln is wrong. :( The mistake is not in the system but in the surroundings. :D. Actually When I say wrt to pulley, I forget that Pulley two itself is in an accelarating frame. So its a non-inertial frame. so ans is wrong.

Okay, I just saw that this system is called an infinite atwood machine and was searching the net for it.

http://www.fks.sk/~bzduso/Fyzika/BAUPC/sol43.pdf

Heres how I solved it :P

Let T be the tension in the first string, T/2 is that in the second n so on.
Now we can use work done by tension is zero coz tension is an internal force 2 the system.
so we get T.x1+T/2*x2+T/4*x3+...=0 were x1, x2.. denote displ. of ith pulley
now
differntiate twice
a1+a2/2+a3/4+..=0
so a1=-a2/2-a3/4-.. ---1
now eqn of mot for d blocs are
T-mg=ma1
T/2-mg=ma2
T/4-mg=ma3...
plug in in eqn 1
n uget
a1=g(1/2+1/2^2+1/2^3....) - T/m(1/2^2+1/2^2*1/2^2+1/2^3*1/2^3.....)
givs
a1=g-T/3m
and
a1=T/m-g
solvin
T= 3mg/2
a1= g/2
 :D


hey everyone, can anyone tell me how to find the length of a arc on a circle, knowing the radius and center angle?? thanks for any help

search on google  Arc Length

you can find easily

http://en.wikipedia.org/wiki/Arc_%28geometry%29

Length of arc = L
Radius=r
centre angle=c

$L=rc$


A free particle of mass $ m$ moves in one-dimensional space in the interval $ 0\leq x$, with energy $ E$. There is a rigid wall at $ x\equal{}0$. Write down a time-independent wavefunction, $ \varphi(x)$, which satisfies these conditions, in terms of $ x$ and $ k$, where $ k$ is the wave vector of the motion.

This is a multiple part question, and I think this is just pre-part (a) as a simple thing. I could do all the other parts if I could figure this out. I think I must be overcomplicating it, though. When I try to impose the boundary conditions on $ \varphi(x)\equal{}Ae^{ikx}\plus{}Be^{\minus{}ikx}$ I get that $ \varphi(x)\equal{}0$. This is clearly not the answer.

I think by 'rigid wall' we have to assume $ \phi (x)\equal{}0$ $ \forall x\leq 0$
Then for $ x\geq 0$ we solve the free particle equation with initial condition $ \phi (0)\equal{}0$.


calculus
integration
geometry
calculus computations

Compute
\[\iint_{D}\frac{x^{2}}{y^{2}}dxdy \]
where $D$ is area bounded by $x=2$,$y=x$,$xy=1$.

[hide="Is this correct?"]
[i]Method 1:[/i]
$yx=1$ and $y=x$ meet at $x=1$ thus we have to integrate between $x=1$and $x=2$. In this
interval $y$ varies between lower $y=1/x$ to upper $y=x$ so we have
\[\iint_{D}\frac{x^{2}}{y^{2}}dxdy=\int_{1}^{2}x^{2}\int_{y=1/x}^{y=x}\frac{1}{y^{2}}dydx=\int_{1}^{2}\frac{1}{x}-x^{5}dx \]
which is $\ln2+\frac{1-64}{6}$ but this doesnt make sense since the area is poitive an much smaller. Of the order of 1 aprrox.
[color=red]please help[/color]

[/hide]

Thanks
Is it possible to do this with variable substitution?

A substitution is possible but since it is easy to solve it in its original form I wouldn't bother looking for a substitution or transformation.


i don't now how to send symbole mathimatics  i use mathtype logiciel

If I understand you correctly, you want to use a math formatting language?  Check out the LaTeX tutorial on the left sidebar.

[quote="Fermat -Euler"]where can i downoald LATEX :!:  :!:[/quote]

I think you have it automatically.

When writing a post on the site, you just go in and out of mathmode (to write in TeX) by using the dollar signs $\$.

If you are looking for an editor for your own computer, there are download instructions in our LaTeX area.


I am trying to create documents with Times font, including the parts in math mode.  Right now, I am using the mathptmx package, which for the most part creates the effect I want, but some of the punctuation still uses the Computer Modern font while in math mode.  In particular, commas, parentheses, and square brackets look different depending on whether or not they are in math mode.  How can I get the Times versions of the punctuation to be used in math mode as well?

I had a similar issue when I was using the mathpazo package.  The commas were fine (Palatino style both in math mode and not), but parentheses and brackets were in CM if in math mode.

Yes mathptmx does use Computer Modern for some symbols. The easiest way round this is to use \text{,} instead of a comma and of course you can use
\newcommand{\comma}{\text{,}}
allowing you to use \comma in maths mode.

Thanks you, that works fine.  Is there some other package which uses the correct font automatically?  I ask because using \text{,} messes with the default LaTeX spacing.

You can adjust the spacing yourself so for example
\newcommand{\comma}{\text{,}\,}
looks better, or use \hspace.

As for other fonts, have a look at [url=http://www.tug.dk/FontCatalogue/mathfonts.html]Fonts with math support[/url], [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=psfchoice]Choice of scalable outline fonts[/url] and [url=ftp://tug.ctan.org/pub/tex-archive/info/Free_Math_Font_Survey/survey.html]A Survey of Free Math Fonts for TeX and LaTeX[/url].
Do be aware that fonts are complicated so commercial versions like [url=http://www.micropress-inc.com/fonts/tmmath/tmmain.htm]TM Math Fonts[/url] which can be superior to the free ones, are far from cheap.


$ a,b,c,d$are positive real numbers,prove that
$ (\frac{a\plus{}b\plus{}c\plus{}d}{4})^4\geq (\frac{ab\plus{}bc\plus{}cd\plus{}da}{a^\frac{2}{3}\plus{}b^\frac{2}{3}\plus{}c^\frac{2}{3}\plus{}d^\frac{2}{3}})^3$$ \geq[\frac{abc\plus{}bcd\plus{}cda\plus{}dab}{(ab)^\frac{1}{2}\plus{}(bc)^\frac{1}{2}\plus{}(cd)^\frac{1}{2}\plus{}(da)^\frac{1}{2}}]^2$ $ \geq abcd$

[quote="Jianxing113725"]Anyone to help me?And how about the generalized:
$ a_1,a_2,...,a_n$are all positive numbers,denote that
$ S_1 = a_1 + a_2 + ... + a_n,S_2 = a{}_1a{}_2 + a{}_2a{}_3 + ... + a{}_na{}_1$
$ S_k = a{}_1a{}_2...a_k + a{}_2a{}_3...a_k_ + _1 + ... + a{}_na{}_1...a_k_ - _1$
Prove that
$ [\frac {S_k}{(S_k_ - _1)^\frac {n - k}{n - k + 1}}}]^^n - k + 1$$ \geq[\frac {S_k_ + _1}{(S_k)^\frac {n - k - 1}{n - k}}}]^^n - k$

 :?:  :?:[/quote]
Well, all I can help is that:
1) the inequality is homogenous; so we can assume $ a+b+c=1$ or for the general case we can assume $ S_1=1$
2) It seems familiar to Holder, I think holder will be able to prove it; even though currently I do not have any solution...... :(
 :huh:
Ah, $ 1-\frac{n-k}{n-k+1}=\frac{1}{n-k+1}$ made me think of holfer... it might be a good choice.
 :oops:
I am still trying hard to solve this problem. :rotfl:

[quote="Jianxing113725"]:blush: In fact,I don't know whether is right or not...Just my hypothesis .[/quote]
I think it has a high chance to be correct a small java program gave me same result for thousands of tests indicating that is correct at least for the first one

[quote="Jianxing113725"]Anyone to help me?And how about the generalized:
$ a_1,a_2,...,a_n$are all positive numbers,denote that
$ S_1 = a_1 + a_2 + ... + a_n,S_2 = a{}_1a{}_2 + a{}_2a{}_3 + ... + a{}_na{}_1$
$ S_k = a{}_1a{}_2...a_k + a{}_2a{}_3...a_k_ + _1 + ... + a{}_na{}_1...a_k_ - _1$
Prove that
$ [\frac {S_k}{(S_k_ - _1)^\frac {n - k}{n - k + 1}}}]^^n - k + 1$$ \geq[\frac {S_k_ + _1}{(S_k)^\frac {n - k - 1}{n - k}}}]^^n - k$

 :?:  :?:[/quote]

 :blush: I think it isn't the same inequlity to the 1#...I have made a mistake when using $ S_k$...
very sorry,Potla,waste a lot of your time... :(

Anyone hele me to solve the first inequlity....I'm stucked in it many days....


search
articles
geometry
email
number theory
linear algebra

[quote="ADMann94"]I'm interested in finding problems to research in mathematics that someone at my level can handle (i.e. contest problem experience, college classes in number theory, linear algebra, diff EQ's and multivariable calculus).  I was thinking if there were some relatively elementary journals I would be able to plow through - because most professional ones are pretty tough reading! - I might have a shot at finding some good projects.  Anyone have any ideas?  Thanks a bunch.

Alex[/quote]

In this specific case, I would recomend you to try to find a person who can guide you. Most of the time, the open problems are not located just like that (most of the time) to be found, and when they are, usually just to understand what the problem says is complacated enough, and requires lots and lots of concepts that probably you don't have. It can be frustrating for a Grad student, then, (I assume you are either in High-School or are a Collegue Freshman or so) for someone in a less prepared position, the frustration is a lot better.

If you find some body on a collegue or university near by (I would strongly recomend not to try that on community college, just experience, may be yours can be different, but I doubt it) that can give you advises, and even give you something to work in, even when it may be small. I know at least two guys that have published pretty descent articles recently, one as a HS senior and the other as a College Freshman, but that required a lot of work, and a lot of guideness.

Also, look around the forum. I know there have been postings about some open problems in elementary number theory and elementary geometry (of course, here elementary is a very subjective word. Consider everything that a Math Contest participant on HS should know as elementary, what does not mean it is not challenging, because actually it is).

Finally, if you don't find anyone, email me, probably I could ask my advisor and he may have some ideas.

Best,


I'm glad that there are now new moderators in the Moems section. Maybe this forum can be rejuvenated now.   :D


calculus
integration
logarithms
calculus computations

Evaluate $ \int_1^e \frac{(x\plus{}1)(x\plus{}2\minus{}\ln x)}{(x\plus{}\ln x)^2}\ dx$.

hello kunny, do you have a solution for this integral? I can only solve it numerically.
Sonnhard.


abstract algebra
group theory
Ring Theory

Source: my confused thoughts.

Clearly, there are countable abelian groups isomorphic to one of their proper subgroups, e.g. Z ~ 2Z. There are infinite rings isomorphic to proper subrings as well, e.g. F2IN ~ {x \in  F2IN : x0 = 0}. But this example ring is not countable. So my question is: Are there countable rings which are isomorphic to one of their proper subrings (let us call these rings fractal rings, I like this name, and algebra could someday become popular if it involves fractals =)?

EDIT: 3X.lich probably is right... so if some mod ever comes over here, could you move it to Advanced/Algebra? Thanks!

EDIT2: Forget it... those rings exist. an example: {x \in F2IN : {i : x_i = 1} is finite}.

This topic should be moved to at least the Advanced Section/Algebra.

is it that hard? it looks to me like some basic algebra (since I found the other examples very quickly), and i'm a beginner myself, but if you say so...

The content of this problem is what makes it nonbeginner. It's hardly fair if you compare your level with the audience of this forum, the beginner's forum.

I got no clue even what this is about? what are these "rings"? how to "fractal" them? what is this for a question????

I think this should be in the college playground somewhere, elseway i don't think many people will even understand the question :?


limit
logarithms
real analysis
real analysis theorems

I want to say if (X(n)) and (Y(n)) are convergent sequences in R(n), where (X(n)) -> X and (Y(n)) -> Y then the sequence M= ((X(n))^(Y(n))) converges to X^Y.

However I am at a loss as to how to prove this.  Any help?

Suppose that $x_{n}\to X> 0,y_{n}\to Y$ and that $x_{n}>0$ for all $n$. Put $z_{n}=x_{n}^{y_{n}}$. We have that
\[\lim_{n\to\infty}\log z_{n}=\lim_{n\to\infty}\left( y_{n}\log x_{n}\right) = \left( \lim_{n\to\infty}y_{n}\right)\left( \lim_{n\to\infty}\log x_{n}\right) \]

\[= \left( \lim_{n\to\infty}y_{n}\right)\log \left( \lim_{n\to\infty}x_{n}\right) = Y\log X \]
since $\log$ is a continuous function. It follows that $\lim_{n\to\infty}z_{n}=\lim_{n\to\infty}x_{n}^{y_{n}}=X^{Y}$.


Is there a name for the line that connects the midpoints of two sides of a triangle?

Thanks. I thought it was that, but there's no entry for Midline on Mathworld.... :? so I wasn't sure. :)

Is there something wrong with asking what the name of the line is?? :?

I dont see anything wrong with the initial question.  He just asked about terminology.  There are also many other reasons why someone would ask this question...not just one problem on the USAMTS.

It is often refered to as the midsegment of a triangle.


The chickens and pigs in Farmer McCoy's barn have a total of 50 heads and 170 legs. How many pigs are in the barn?

say all the animals are pigs
how many legs would there be?

obviously, there are 50 animals, so if all the animals are pigs, then there would be 200 legs
this is too many, since we only have 170 legs
so there are 30 extra legs if they were all pigs

now consider this: if we switched one pig with a chicken, by how much would the number of legs decrease?
the answer is 2, since pigs have 4 legs and chickens have 2

so, we go back to the part where we have 30 extra legs
we know that for every pig we swap out and replace with a chicken, we subtract two legs
therefore, if we replace 15 pigs with 15 chickens, we resolve the problem of the excess legs

thus, we have a total of 50-15 pigs, or $ \boxed{35}$ pigs

instead of using reason like that, you could use equations.
Let $ p\equal{}\text{number of pigs}, c\equal{}\text{number of chickens}$
We have \[ p\plus{}c\equal{}50\] and \[ 4p\plus{}2c\equal{}170\implies2p\plus{}c\equal{}85\] Taking the difference between the two equations gives \[ 2p\plus{}c\minus{}p\minus{}c\equal{}85\minus{}50\implies p\equal{}\boxed{35}\]

vallon22's way is actually very effective. But instead of over-counting with pigs, why don't we under-count with chicken so we could skip a step (no subtraction at end). Chickens have 2 legs. If all the animals are chickens, there would be $50\cdot2 = 100$ legs. Thus, there would be 70 legs needed. Pigs have 4 legs. If we replace a chicken with a pig, we get 2 more legs. Thus $\frac{70}{2} = \boxed{35}.$

This should be your logic in ftw...
chicken = 2 legs
pigs = 4 legs
okay, 2 legs more
all chickens = 100 legs (ez multiplication)
70 legs
70/2 = 35
The confusion of mental math with algebra is skipped (keeping track of coefficients and constants).


[b][size=100][color=DarkBlue]A quadrilateral $ ABCD$ is given and let be the points $ E\equiv AB\cap CD$ and $ M\equiv AC\cap BD.$ Let $ Z,\ H$ be, two points on the side-segment $ AB$ $ ($ $ Z,$ between $ H,$ $ B$ $ ),$ such that $ ZH / ZE \equal{} AB / AE.$ Prove that the line segment $ PM,$ where $ P\equiv CZ\cap DH,$ bisects the segment $ BH.$[/color][/size][/b]

Kostas Vittas.

[hide]For the easy construction of the point $ H,$ when the point $ Z$ has been defined, we take $ AE' \equal{} ZE$ and we draw $ BB'\parallel EE'.$

Then, the point $ H,$ where $ ZH \equal{} AB',$ give us the relation as the problem states.

So, we have that $ \frac {ZH}{ZE} \equal{} \frac {AB'}{AE'} \equal{} \frac {AB}{AE}.$[/hide]

Thank you for your file. It is difficult to draw the image of this problem :(

PROOF：
PH/PD*DC/CE*ZE/ZH=1
PH/PD*DM/BM*BN/HN=1
If BN=NH , then  DC/CE* ZE/ZH =DM/BM (1)
We have ZE/ZH=AE/AB
If (1) is true DC/CE* AE/AB*BM/DM=1 (2)
We notice that △BDE and transversal AMC ,(2) is true
DONE！

Very good dear [b][size=100]plane geometry[/size][/b] and thank you a lot for your interest and the sweet proof, based on [b][size=100]Menelaus theorem[/size][/b].

Your solution is simpler than the one I have in mind based on a Lemma, as direct application of the [b][size=100]Desarques theorem[/size][/b] and I will post it here later, as an alternative approach.

Best regards, Kostas Vittas.

Thank you for your nice problem ,dear vittasko :)
I am waiting for your solution to learn more

Though the points $ B,\ E,$ we draw two lines parallel to the line segment $ PD$ and we denote as $ X,\ Y,$ their intersection points from the line segments $ PM,\ PC,$ respectively and we will prove that the points $ A,\ X,\ Y,$ are collinear.

$ \bullet$ Let $ C',\ D'$ be, two points on the line segments $ PC,\ PD$ respectively, such that the line connecting them, passes through the point $ E.$

It is easy to show, based on the [b][size=100]Desarques theorem[/size][/b], that the triangles $ \bigtriangleup MCD,\ \bigtriangleup M'C'D',$ where $ M'\equiv AC'\cap BD',$ are perspective, because of the colinearity of the points $ A\equiv MC\cap M'C',\ B\equiv MD\cap M'D'$ and $ E\equiv CD\cap C'D'.$

So, we conclude that the line segments $ CC',\ DD',\ MM',$ are concurrent at one point, here the point $ P\equiv CC'\cap DD'.$

Hence, the point $ M'\equiv AC'\cap BD',$ when the line segment $ C'D'$ passes through the point $ E,$ always lies on the line segment $ PM.$

$ \bullet$ We can see now, the line segment $ EY\parallel PD,$ is the limited case of the line segment $ EC'D',$ turned around the point $ E.$

So, the points $ Y,\ X,$ are the limited case of the points $ C',\ M'$ respectively, when the point $ D',$ is coincided with the point at infinity.

In this limited case of the line of $ EY\parallel PD,$ the point $ X,$ as the intersection point of $ AY,$ from the line through $ B$ and parallel to $ EY$ $ ($ as the intersection point of the diagonals of the limited quadrilateral $ ABYD_{\infty}$ $ ),$ lies also on the line segment $ PM.$

That is, the points $ A,\ X,\ Y,$ are collinear.

$ \bullet$ From $ BX\parallel EY$ $ \Longrightarrow$ $ \frac {BX}{EY} \equal{} \frac {AB}{AE}$ $ ,(1)$

From $ PH\parallel BX$ $ \Longrightarrow$ $ \frac {NB}{NH} \equal{} \frac {BX}{PH}$ $ ,(2)$

From $ PH\parallel EY$ $ \Longrightarrow$ $ \frac {PH}{EY} \equal{} \frac {ZH}{ZE} \equal{} \frac {AB}{AE}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ NB \equal{} NH$ and the proof is completed.

Kostas Vittas.


geometry
geometric transformation
rotation
3D geometry
tetrahedron

What is Burnside's Lemma? Does any one know any sample problems that uses it?

[url=http://en.wikipedia.org/wiki/Burnside%27s_lemma]The Wikipedia article[/url] contains both a statement and an example problem.

Does anyone know any more problems involving the lemma?

The coloring question can be asked with any number of colorings and with any polyhedron whose rotation group is known (such as the Platonic solids).  For example, a recent AMC question asked about colorings of the tetrahedron with $ 3$ colors, which is an easy enough question that it can be handled with brute force, but Burnside's Lemma is a nice overarching principle.

Burnside's Lemma can also be used to prove Fermat's Little Theorem.  One asks how many distinct ways there are to arrange $ p$ beads of $ a$ colors (where $ p$ is prime) onto a bracelet.  The rotation group of the bracelet has order $ p$; on the other hand, a given coloring is fixed by a given non-identity rotation if and only if every color is the same, and there are $ a$ such colorings.
The identity rotation fixes every coloring.  It follows that the number of arrangements is

$ \frac{1}{p} \left( a(p\minus{}1) \plus{} a^p \right)$

from which it follows that $ p | a^p \minus{} a$.


Let $ a,b,c \geq 0$ satisfy $ a\plus{}b\plus{}c\equal{}1$. Prove that :

$ (2ab\minus{}1)^2\plus{}(2bc\minus{}1)^2\plus{}(2ca\minus{}1)^2\plus{}5abc \geq 2\plus{}3(a\minus{}b)(b\minus{}c)(c\minus{}a)$

 :roll:

[quote="forevercontinue"]Let $ a,b,c \geq 0$ satisfy $ a \plus{} b \plus{} c \equal{} 1$. Prove that :

$ (2ab \minus{} 1)^2 \plus{} (2bc \minus{} 1)^2 \plus{} (2ca \minus{} 1)^2 \plus{} 5abc \geq 2 \plus{} 3(a \minus{} b)(b \minus{} c)(c \minus{} a)$

 :roll:[/quote]
It's equivalent to $ (a^2\plus{}b^2\plus{}c^2)^2\minus{}3(a^3c\plus{}b^3a\plus{}c^3b)\plus{}3(a^3b\plus{}b^3c\plus{}c^3a\minus{}abc(a\plus{}b\plus{}c))\geq0,$ which is obvious.

[quote="arqady"][quote="forevercontinue"]Let $ a,b,c \geq 0$ satisfy $ a \plus{} b \plus{} c \equal{} 1$. Prove that :

$ (2ab \minus{} 1)^2 \plus{} (2bc \minus{} 1)^2 \plus{} (2ca \minus{} 1)^2 \plus{} 5abc \geq 2 \plus{} 3(a \minus{} b)(b \minus{} c)(c \minus{} a)$

 :roll:[/quote]
It's equivalent to $ (a^2 \plus{} b^2 \plus{} c^2)^2 \minus{} 3(a^3c \plus{} b^3a \plus{} c^3b) \plus{} 3(a^3b \plus{} b^3c \plus{} c^3a \minus{} abc(a \plus{} b \plus{} c))\geq0,$ which is obvious.[/quote]

Dear arqady, your proof is a way I created this inequality , too.

But, It 's so clear and easy to find that way, sorry.  :( :|

I will try it again and be careful with creating inequality later .  :|


I use TeXnicCenter, MikTex compiler.
I use to open the pdf files with Foxit Center instead of Acrobat
Reader. The reason is that Acrobat Reader slows down my computer.
So, I create a file in TeXnic, but there's a difficulty.
I went to Profiles-->Viewer and I changed the path from Acrobat to
Foxit.exe
When I try to view the file, Foxit shows a message, "open the file". I
open it and can view the file.

But I get a message
 DocOpen("%bm.pdf")][FileOpen("%bm.pdf")] cannot execute the command.

I add that this DDE command is under view object's output , server:
acroview, Topic: control
The same command for Forward search
and for close document, [DocClose("%bm.pdf")] .

So how can I change this command or else?
I read on net that Foxit uses tm.pdf, I changed that to no effect.
Still the error
I guess I have to change also acroview into what?

Thank you

Try this taken from http://consult.wikidot.com/latex#foxit (whch also says ([i]version 2.2 works better with TeXniCCenter than the version 3.0[/i])
[quote]Create a batch file that closes Foxit and open it again. The batch file below saved as foxit_restart.bat works with WinXP Prof.

TASKKILL /FI "WINDOWTITLE eq %2.pdf*"
START "C:\Programme\Foxit Software\Foxit Reader\Foxit Reader.exe" %1
Profiles
Viewer
Pfad der Anwendung, the "" must be used because of space character in the Path:

"C:\Programme\Foxit Software\Foxit Reader\foxit_restart.bat"
Keep a space character at the begin of the command "%bm.pdf" %tm.

Projektausgabe betrachten
Command line
Command: "%bm.pdf" %tm

Suche in Ausgabe
Command line
Command:

vor Compilierung Ausgabe schliessen
Not close[/quote]Use Google to translate the German.

thank you stevem, it works
I didn't need to translate because it was clear from TeXniCCenter's window
I won't need the annoying Acrobat Reader anymore

In fact, an easy solution is to give the path to Foxit Reader in the Wizard when you install first time Miktex as seen here:
http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Downloads
The Wizard is also accessible via Build-->Define Output Profiles


One writes 268 numbers around a circle, such that the sum of 20 consectutive numbers is always equal to 75. The number 3, 4 and 9 are written in positions 17, 83 and 144 respectively. Find the number in position 210.

[quote=tµtµ]

now it's a matter of some boring work, for $k = 7$, $x_{210} = x_{83} = 4$[/quote]

How ? what is the value of i

[hide=My Solution]
Let $n_1, n_2, n_3, \cdots n_{267}, n_0$ be the 268 numbers.  First, we will prove that a number twenty positions after said number would be the same number.

[b]Lemma: $n_{i \pmod{268}} = n_{i+20 \pmod{268}}$ for integers $1 \le i \le 268$[/b]

We know that $\sum_{k=1}^{20} n_{a_k} = 75$, where $a_k = i+k-1 \pmod{268}$.  By subtracting $n_{a_1}$ and adding $n_{a_{21}}$ from both sides, we must have $75 - n_{a_1} + n_{a_{21}} = 75$.  Therefore, $n_{a_1} = n_{a_{21}}$, so $n_{i \pmod{268}} = n_{i+20 \pmod{268}}$. $\blacktriangleright$

Thus, we know that $n_{i \pmod{268}} = n_{i+20m \pmod{268}}$, where $m$ is an integer.  If $m = 14m_1$, then $i + 20m \equiv i + 280m_1 \equiv i + 12m_i \pmod{268}$.  If $m_1 = 22m_2$, then $i + 12m_1 \equiv i + 264m_2 \equiv i - 4m_2 \pmod{268}$.  Note that $4$ is a factor of $268$, so we can know that if $a \equiv b \pmod{4}$, then $n_a = n_b$.

With this in mind,
$$\sum_{k=1}^{20} n_k = 75$$
$$5(n_1 + n_2 + n_3 + n_4) = 75$$
$$5(3 + n_2 + 4 + 9) = 75$$
$$5(16 + n_2) = 75$$
$$16 + n_2 = 15$$
$$n_2 = -1.$$
Thus, since $n_2 = n_{210}$, we must have $n_{210} = \boxed{-1}$.
[/hide]


Could someone please help me how to change my avatar?  I have saved a picture as .gif, .jpeg, and .png formats, and none of them seem to load correctly onto this website.

The picture is 80 pixels wide by 100 tall...  The filesize is less than 21KB... I can't seem to find what is wrong with it.

Alternately, I tried using ImageShack to host the picture, but it couldn't find the avatar.

Yay! First post ever!

But could someone please point me in the right direction on what to do?

Thanks for noticing. There was a minor fault of our own, not setting the right permissions for avatar uploading.

You can upload your avatar now, using the profile menu: http://www.mathlinks.ro/Forum/usercp.php?mode=editprofile&sub=avatar

Thanks a bunch!  It worked on the first try!

The chicken WILL eat you.  :D

Yes, my avatar has problems as well.  I think it has something to do with the fact that when I was an 12-year old AoPS newbie, I thought it would be cool to have the admin symbol as my avatar...and then now it says "You are not allowed to display an avatar."  I am much more responsible now.  Can I have an avatar back?  Please?

I cannot click the link you gave me... When i do my computer freezes... :(

I also tried to change it from the profile tab and it says that the avatar has to be liss than 80x100 pixels.


how are level 1 MMOs scored? :maybe:
also where to find past MML problems for practice? :huh:

The level 1 MAML was 25 problems, 6 points for correct answers, 2 for blanks, 0 for wrong answers. In previous years, there were 30 problems, and the scoring distribution was either 5-2-0 or 5-1-0 (look at scores archives, I forget which).

The MML's aren't on the internet but you can always look at past state and New England meets, which are harder but have the same flavor. I think the WoCoMaL's might be similar, but I'm not sure. But your school probably has some kind of an archive of problems, just ask for them.

[quote="CatalystOfNostalgia"]

The MML's aren't on the internet [/quote]

To the contrary =)

http://matharchive.org/mml/index.html

Hmmm, I guess I never looked hard enough.

[quote="calc rulz"]To the contrary =)[/quote]
Worcester Academy is joining GBML this year, so maybe there'll be a matharchive.org/gbml directory soon.  It primarily depends what the league gives me, and whether they're OK with putting old materials online.


These are some out-of-the-box type questions that just happen to be part of an assignment of mine.  I have no idea what approach to take. Here are the two questions:

1. Sam borrowed some money on Sept. 1, 1998.  On April 1. 2002 he decided he wanted to pay back the loan.  The accumulated interest on the loan was $4234.66. What was the original amount of the loan if the interest rate was 7%/a compounded quarterly. (The only thing I can think of is to use the compound interest formula A=P(1+i)^{n}, but I don't have the amount or the principal.) 2. Michelle's parents and grandparents started an Education fund the day she was born. they alternated making deposits of
$400 and $600 respectivly, in the investment every six months in which the interest rate was 7.8%/a compounded annually. Michelle's parents made the first deposit of
$400 on the day she was born, her grand parents made the next deposit of $600 six months later, six months after that her parents made a deposit of$400, and so on.  Michelle's parent's later inherited some money and on her 12th birthday they decided to deposit $2500. AFter that they continued depositing$400 each time.  How much money will be in the investment on her 18th birthday?  Including the deposit made that day?  (Other than calculating each year, all I can think of is to somehow combine two equations . . . . .)

I would really appreciate any help on this :roll:

Edit and re-latex please. I'll see what I can do to help.
As of now, you have an unparseable forumula.


http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem

1. In the section that says "proof", where it says "given residue class $ S (mod m)$, call $ r$ the member of $ R$ in this class"

[b]What are they saying exactly?
[/b]

Among other things I'm confused about what does it mean by "the member of $ R$?" What is special about "the member" that they are referring to and which member of R are they talking about and what's the point of introducing all these R's...

2. Then it says "Each member of the residue class can be written as $ am \plus{} r$ for an integer $ a$. Since $ r$ is in the form $ bn$."

[b]What are they saying exactly? [/b]

What residue class are they talking about and now is r some arbitrarily chosen element from the set R? Why can each member of the residue class can be written as $ am \plus{} r$ for an integer $ a$?Why do we introduce these variables?

3. Does anyone have another simple explanation?

I really can't understand the "proof" where you circle numbers or whatever mentioned here in a previous post: http://mathforum.org/library/drmath/view/62084.html

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=791088239&t=205817

Here is a good proof:
[url]http://mathforum.org/library/drmath/view/62084.html[/url].

That's the one where you circle numbers.

Sorry  :oops: . I didn't realize that was the one you just posted. Anyways, any number under a circled number can be made because the circled number is a multiple of one of the numbers, and each number under the circled number is a multiple more of the other number. In fact, since all the multiples of one number are circled, if it is possible to score a number, than it would already be under one of the circled numbers because it would have to be a multiple of the second number more than the first number. So any number above a circled number can't be made.

Yeah... It's just is this last step rigorous?

"Now, if you put everything together and think about it a little, you
should be able to see that the last impossible number is the one just
above the last circled number."

I am interested in the aops proof, which I don't understand, as well because it appears to be rigorous for some reason.

Why is that not rigorous? Every step can be proven without saying it looks true. You want the highest number above a circled number, because we have [b]proven[/b] that numbers above circled numbers can't be made, and numbers below circled numbers can be made. Even though it looks like a fun, easy proof, it is still correct.

Gee now I agree with you after thinking about it.

Now... how does this one work?
http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem

http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem

1. [quote]Each member of the residue class can be written as $ am \plus{} r$ for an integer $ a$.[/quote]

Why can each member of the residue class can be written as $ am \plus{} r$ for an integer $ a$?

2. [quote]call $ r$ the member of $ R$ in this class.[/quote]

What does it mean by "the member?"

3.What exactly are we trying to do in the proof? Why do we need this set $ R \equal{} \{0, n, 2n, 3n, 4n ... (m \minus{} 1)n\}$ and what does it do for us exactly?

4. Why do we think of using $ R \equal{} \{0, n, 2n, 3n, 4n ... (m \minus{} 1)n\}$ for the proof? What is something special about R that we know will help us in the proof?

[quote]Each member of the residue class can be written as $ am \plus{} r$ for an integer $ a$.[/quote]

By definition of a [[residue class]].

[quote]call $ r$ the member of $ R$ in this class.[/quote]

The element of $ r$ in $ R$ such that $ r \equiv s \pmod{m}$, since we know each of $ R$ is distinct $ \pmod{m}$. (analogous to the columns above; picking a column, there is only one circled number)

[quote]3.What exactly are we trying to do in the proof? Why do we need this set $ R \equal{} \{0, n, 2n, 3n, 4n ... (m \minus{} 1)n\}$ and what does it do for us exactly?[/quote]

The same reason why you circled those numbers above. You show that: (a) in each residue class, all of the numbers greater than $ r \in R$ can be expressed in the form $ am \plus{} bn$, (b) in each residue class, all of the numbers less than $ r \in R$ cannot be expressed in the form $ am \plus{} bn$. (again, analogous above to showing that all numbers below a circled number do not work, while above do work)

[quote]4. Why do we think of using $ R \equal{} \{0, n, 2n, 3n, 4n ... (m \minus{} 1)n\}$ for the proof? What is something special about R that we know will help us in the proof?[/quote]

The same reason why you circled those numbers above. What's special about $ R$ is shown in the proof (namely distinctness $ \mod{n}$ and maximal in each residue class).


induction
counting
distinguishability
function
binomial coefficients
combinatorics proposed
combinatorics

Prove [i]by induction[/i] Vandermonde's identity.

This is easy. We want to prove that \[\sum_{j}\binom{m}{j}\binom{n}{r-j}=\binom{m+n}{r}\] We proceed by induction on $m$. The case $m=0$ is clear. Now suppose we have the result for some $m$. We wish to show that \[\sum_{j}\binom{m+1}{j}\binom{n}{r-j}=\binom{m+n+1}{r}\] Now we use Pascal's formula: \[\sum_{j}\binom{m+1}{j}\binom{n}{r-j}=\sum_{j}\left[\binom{m}{j}+\binom{m}{j-1}\right]\binom{n}{r-j}=\sum_{j}\binom{m}{j}\binom{n}{r-j}+\sum_{j}\binom{m}{j-1}\binom{n}{r-j}\] Using the induction hypothesis we get \[\sum_{j}\binom{m}{j}\binom{n}{r-j}+\sum_{j}\binom{m}{j-1}\binom{n}{r-j}=\binom{m+n}{r}+\binom{m+n}{r-1}\] And again by the Pascal's formula we have \[\sum_{j}\binom{m+1}{j}\binom{n}{r-j}=\binom{m+n}{r}+\binom{m+n}{r-1}=\binom{m+n+1}{r}\] Hence the induction step is complete, and we are done. :)

There is a straightforward inductive proof (there is nothing as stupid as "prove with...").

Induction start $m+n=0$: left to the reader.

Induction step: $\binom{m+n}r$ counts the number of ways to select $r$ things from $m+n$ distinguishable things.
$\binom{m}{j}\binom{n}{r-j}$ is the number of ways to select $j$ of that things from the first $m$, and $r-j$ from the other $n$. Thus $\sum_{j}\binom{m}{j}\binom{n}{r-j}$ is the number of ways to select $r$ thing from $m+n$, giving the result.

Anyone claiming that this is not inductive has to tell me one thing: what is a proof by induction and what not¿
And note: to define natural numbers, summation etc., you need induction, thus you can't define it by something like "the proof is wrong without that axiom".

[quote="ZetaX"]And note: to define natural numbers, summation etc., you need induction, thus you can't define it by something like "the proof is wrong without that axiom".[/quote]

Also, if you look at the formal derivation, you could simply toss in the axiom of induction as the first step and never use it again, magically transforming every proof into a proof by induction.

I think the use of Pascal's formula in an induction proof of some expression involving binomial coefficients is rather "natural", because Pascal's formula \[\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}\] is a recurrence relation for the binomial coefficients, which relates a quantity in function of $n$ to others in function of $n-1$, just what we need to use the induction hypothesis.


Let x, y, z be positive reals with xyz = 1. Prove that

:Sigma:(x + y)/(x + xy + y) :ge: 2.

Easy 8-)

OK !
For all x, y we have : (x - xy + y)/(x + xy + y) :ge: 1/3.
This gives an alternative solution :)


If Arccos(x)+Arccos(y)=pi/2, show that x^2+y^2=1.

[hide="Since..."]

$\arccos(y)=\frac{\pi}{2}-\arccos(x)$ and $\cos \theta= \frac{\pi}{2}-\sin \theta$, $x=\sin \theta$ and $y= \cos \theta$ for some $\theta$... and then we use the identity $\sin^{2}\theta+\cos^{2}\theta=1$[/hide]


suppose a and b are real numbers such that the roots of the cubic equation ax^3-x^2+bx-1=0 are all positive real numbers . Prove that( 1) 0<3ab<=1 (2) b>=3^(1/2)

[quote="tamal banerjee"]suppose a and b are real numbers such that the roots of the cubic equation $ ax^3 \minus{} x^2 \plus{} bx \minus{} 1 \equal{} 0$ are all positive real numbers . Prove that( 1) $ 0 < 3ab\leq1$ and (2) $ b\geq\sqrt 3$[/quote]

Let $ f(x) \equal{} ax^3 \minus{} x^2 \plus{} bx \minus{} 1$ and so $ f'(x) \equal{} 3ax^2 \minus{} 2x \plus{} b$

1) first part ($ 0 < 3ab\leq 1$)
Since $ f(0) \equal{} \minus{} 1$, we need $ a > 0$ (else we would have a negative root).

We need also that $ f'(x)$ has real root(s) and so a positive discriminant and so $ 3ab\leq 1$. The two roots $ r_1\leq r_2$ of the equation $ f'(x) \equal{} 0$ need to be positive, and so is their product $ r_1r_2 \equal{} \frac b{3a}$ and so $ b > 0$ and so $ 3ab > 0$

Hence the result

2) second part ($ b\geq \sqrt 3$)
I failed to find a simple demo for this part :

If $ f'(r) \equal{} 0$, we have $ 3ar^2 \minus{} 2r \plus{} b \equal{} 0$, and it's easy to establish $ f(r) \equal{} \frac {1}{9a}(2(3ab \minus{} 1)r \plus{} b \minus{} 9a)$ and so $ 27a^2f(r) \equal{} 2(3ab \minus{} 1)(3ar) \plus{} (3ab \minus{} 1) \plus{} 1 \minus{} 27a^2$

Let $ u \equal{} \sqrt {1 \minus{} 3ab}$ (and so $ 27a^2f(r) \equal{} \minus{}2u^2(3ar) \minus{}u^2 \plus{} 1 \minus{} 27a^2$) . We have $ 3ar_1 \equal{} 1 \minus{} u$ and $ 3ar_2 \equal{} 1 \plus{} u$

We need that $ 27a^2f(r_1)\geq 0$ and so : $ \minus{} 2u^2(1 \minus{} u) \minus{} u^2 \plus{} 1 \minus{} 27a^2\geq 0$ and so $ (u \minus{} 1)^2(2u \plus{} 1)\geq 27a^2$ and so $ 3a\leq \frac {(1 \minus{} u)\sqrt {1 \plus{} 2u}}{\sqrt 3}$

So $ b \equal{} \frac {1 \minus{} u^2}{3a}\geq \sqrt 3\frac {1 \plus{} u}{\sqrt {1 \plus{} 2u}}$ $ \geq \sqrt 3$

Hence the result

(2) Let the three positive real roots be $ p,q,r$
Then $ \begin{cases} pqr\equal{}p\plus{}q\plus{}r\equal{}\frac{1}{a} \\ pq\plus{}qr\plus{}rp\equal{}\frac{b}{a} \end{cases}$

So we are going to prove $ (\frac{pq\plus{}qr\plus{}rp}{pqr})^2 \ge 3 \frac{p\plus{}q\plus{}r}{pqr}$
$ \Leftrightarrow (\frac{1}{p}\plus{}\frac{1}{q}\plus{}\frac{1}{r})^2 \ge 3 (\frac{1}{pq}\plus{}\frac{1}{qr}\plus{}\frac{1}{rp})$
$ \Leftrightarrow \frac{1}{p^2}\plus{}\frac{1}{q^2}\plus{}\frac{1}{r^2} \ge \frac{1}{pq}\plus{}\frac{1}{qr}\plus{}\frac{1}{rp}$
$ \Leftrightarrow \frac{1}{2}[(\frac{1}{p}\minus{}\frac{1}{q})^2\plus{}(\frac{1}{q}\minus{}\frac{1}{r})^2\plus{}(\frac{1}{r}\minus{}\frac{1}{p})^2] \ge 0$, which is obviously true.

[quote="stephencheng"](2) Let the three positive real roots be $ p,q,r$
Then $ \begin{cases} pqr \equal{} p \plus{} q \plus{} r \equal{} \frac {1}{a} \\
pq \plus{} qr \plus{} rp \equal{} \frac {b}{a} \end{cases}$

So we are going to prove $ (\frac {pq \plus{} qr \plus{} rp}{pqr})^2 \ge 3 \frac {p \plus{} q \plus{} r}{pqr}$
$ \Leftrightarrow (\frac {1}{p} \plus{} \frac {1}{q} \plus{} \frac {1}{r})^2 \ge 3 (\frac {1}{pq} \plus{} \frac {1}{qr} \plus{} \frac {1}{rp})$
$ \Leftrightarrow \frac {1}{p^2} \plus{} \frac {1}{q^2} \plus{} \frac {1}{r^2} \ge \frac {1}{pq} \plus{} \frac {1}{qr} \plus{} \frac {1}{rp}$
$ \Leftrightarrow \frac {1}{2}[(\frac {1}{p} \minus{} \frac {1}{q})^2 \plus{} (\frac {1}{q} \minus{} \frac {1}{r})^2 \plus{} (\frac {1}{r} \minus{} \frac {1}{p})^2] \ge 0$, which is obviously true.[/quote]

Quite nice :)

thanks to stephencheng and pco . stephencheng's solution for the second part is very nice .


I don't know if anyone else has thought of this before, but predominately facebook, as well as other sites, have tools where you can sort of "tag" people in your posts. Before I go on further, for all of you that are unfamiliar with this idea...this means that when a user that is tagged in a post or topic gets a notification of that new topic when they enter the site, through some sort of message system. I don't know what others think of this (and that's why I'm posting this) but I think it might add an awfully cool twist to the AoPS forum. In addition, it might expedite the time for a certain response if you're asking a specific crowd for instance.

I don't think I would like to get the popup message like "your response is requested for topics 1,2,4,8,16,64" every time that I enter AoPS (much less a notification like that by e-mail) but if the list is sitting quietly somewhere and can be retrieved by clicking on some button when I want to look at it, I'm in favor of the idea. At least, it is a rather convenient alternative to sending PMs like "Can you, please, look at ...", which is the current way to attract somebody's attention to specific topics :).

I agree with fedja - as long as the list isn't being "shoved in your face", I like it.

[quote="fedja"]1,2,4,8,16,64[/quote]  Not 32, though?  Or maybe it is [url=http://www.research.att.com/~njas/sequences/A013133]this sequence?[/url]  :lol:

if you're familiar with the way facebook, or other websites have it...they just have a bar, such as the private message thing on this site, that tells you you've been "tagged"...i.e. not a popup message

It could be something like with private messages - a blue box that turns yellow.

any other comments from anyone else? incl/ admins...

[quote="yunhua98"]I think its a great idea, but you should probalbly PM a admin.[/quote]Don't ever do that with "ideas" that you might have. It will most likely result in us terminating your PM privileges :) We do read this forum quite often :P

As for the tagged idea, it's a nice idea, and we will try to implement it in the future. The basic implementation would invoke the user choosing how the notification is sent: either through email, PM, screen notification or simply no notification, but a list of such tags.

Sorry about that, didn't know we couldn't, sorry.

But still, it was a great idea.


inequalities
geometry
function
inequalities theorems

Can anyone proof the Youngs inequality with elementary mathematics :?:
(i mean without the use of analysis)

It's obviously true by Weighted AM-GM; $ \frac{1}{p}\plus{}\frac{1}{q}\equal{}1$
\[ (u^{p})^{\frac{1}{p}}(v^q)^{\frac{1}{q}}
\leq  \frac{u^p}{p}\plus{}\frac{v^q}{q}\]

Geometrical explanation gives the area of two functions $ y \equal{} f(x)$ and the inverse function $ y \equal{} f^{ \minus{} 1}(x)$.

[url=http://mathworld.wolfram.com/YoungsInequality.html]See here[/url]


Hi,

I would like to ask you if you know any weblink to download the student's books of Mathematics for Romania. Maybe in the ministry of education or something like that.

Also, if you have any weblink with Romanian books for Mathematics is aso very good.

Thanks a lot,
Alexandros

Hello,

The current syllabuses can be found on the ministry of education site at [url]http://curriculum.edu.ro[/url] (in Romanian).

For Romanian math books, as well as undergraduate textbooks, you can check GIL publishing house at [url]http://gil.ro[/url].


factorial
combinatorics solved
combinatorics

Any combinatorial proof that $\frac{{\left( {5n} \right)!}} {{40^n n!}}$ is always an integer?

first notice that $40|5!$, so if we prove that $\frac{(5n)!}{(5!)^n*n!}$ is an integer, we are done.  imagine we have $5n$ different objects and we want to arrange them in a line in any way possible.  what we are going to do is to divide them into $n$ "blocks" of $5$ objects.  now we are going to choose an order for the $n$ blocks (we have $n!$ ways).  Then we have to give an order to the objects in each block ($(5!)^n$ ways).  if we do this for every way of dividing the $5n$ objects in $n$ blocks of $5$, we will achieve every permutation possible ($(5n)!$).  therefore, $(5!)^n{n}!|(5n)!$.


combinatorics proposed
combinatorics
graph theory

Let $ G$ be a $ 2$-connected nonbipartite graph on $ 2n$ vertices. Show that the vertex set of $ G$ can be split into two classes of $ n$ elements such that the edges joining the two classes form a connected, spanning subgraph.

[i]L. Lovasz[/i]

sorry,but I don't understand the problem. what does it mean a connected spanning subgraph on the edges between the two sets of the partition?

In any way we split the vertex set $V$ of $G = (V,E)$ into two classes $A$ and $B$, the set of edges $E(A)\cup E(B)$ contained in $A$ or $B$ is non-empty (since the graph is non-bipartite), thus the edges crossing between $A$ and $B$ make up a proper subgraph $G'$ of $G$.

The problem asks to find such a split, with $|A|=|B|=n$, such that $G'$ is connected, and spanning $G$ - meaning the vertex set of $G'$ (the ends of the edges crossing between $A$ and $B$) is equal to $V$. Notice that neither connectivity, nor the spannning property are guaranteed for a random such split.

if we have $ v_{0} $ connected with $ v_{1},...,v_{2n-1} $ and $ v_{1} $ connected with $ v_{2} $ what is the partition?

That graph isn't 2-connected (though in fact for $n=2$ the partition $02 \mid 13$ manages to work anyhow).


search
geometry
circumcircle
parallelogram
ratio
cyclic quadrilateral
geometry unsolved

let ABCD is a Cyclic quadrilateral that: (AB,CD)=E and (AD,BC)=F.
M and N are midpoints of AC and BD .if AC>BD prove that:
                     \[ \frac{MN}{EF}\equal{}\frac{1}{2}(\frac{AC}{BD}\minus{}\frac{BD}{AC})\]

Let $ L$ be the midpoint of $ EF,$ which lies on the Newton line $ MN$ of the complete quadrangle $ ABCD.$ Since $ E,F$ are conjugate points WRT the circumcircle $ (O)$ of $ ABCD,$ then the circle $ (L)$ with diameter $\overline{ EF}$ belows to the orthogonal pencil defined by $ (O)$ and the axis $ EF.$ Thus $ (L)$ is orthogonal to $ (O)$ $\Longrightarrow$ $ (L)$ is also orthogonal to the circle $ (O')$ with diameter $ OK,$ where $ K \equiv AC \cap BD$ is the pole of $ EF$ WRT $ (O)$ $\Longrightarrow$ power  of $ L$ to $ (O')$ equals $ LE^2 \equal{} LF^2.$ Therefore

$ LE^2 \equal{} LN \cdot LM \equal{} (LM \plus{} MN)LM \Longrightarrow \frac {MN}{LE} \equal{} \frac {LE}{LM} \minus{} \frac {LM}{LE} \ (\star)$

Let $ P,Q,R$ be the midpoints of $ BC,CE,EB.$ It is clear that $ ERPQ$ is a parallelogram and $ L \in RQ.$ By Menelaus theorem for $ \triangle PQR$ cut by the Newton line of $ ABCD,$ we have

$ \frac {LM}{LN} \cdot \frac {RN}{RP} \cdot \frac {PQ}{MQ} \equal{} 1 \Longrightarrow \frac {LM}{LN} \equal{} \frac {EC}{ED} \cdot \frac {EA}{EB} \equal{} \left ( \frac {EA}{ED} \right)^2 \equal{} \left ( \frac {AC}{BD} \right)^2$

Combining with the power of $ L$ to $ (O'),$ we get  $ \frac {LM}{LE} \equal{} \frac {AC}{BD}$

Substituting the previous ratio into the expression $ (\star)$ yields

$ \frac {MN}{LE} \equal{} \frac {BD}{AC} \minus{} \frac {AC}{BD} \Longrightarrow \frac {MN}{EF} \equal{} \frac {1}{2} \left | \frac {AC}{BD} \minus{} \frac {BD}{AC} \right|.$


superior algebra
superior algebra solved

G is a group of nonzero real numbers under multiplication
Define the map F:G-->G so that F(x)=x^4, is F a homomorphism.

If G is the group of real numbers under addition, is F still a homomorphism?

-Representations...& Characters

Multiplication: yes, apply simply the definitions.

Addition: consider $ F(2)\equal{}F(1\plus{}1)$.


trigonometry
induction
algebra unsolved
algebra

Prove (no induction) that:

$ cos ^ {2n}x \equal{} \frac{1}{2^{2n\minus{}1}}(cos2nx \plus{} {2n \choose 1} cos2(n\minus{}1)x \plus{} ... \plus{} {2n \choose {n\minus{}1}} cos2x \plus{} \frac{1}{2} {2n \choose n})$

thanks by advance

It follows directly from the identity $ \cos x \equal{} \frac{e^{ix}\plus{}e^{\minus{}ix}}{2}$:
We multiply both terms by $ 2^{2n}$ and get -
$ LHS\equal{}2^{2n}(\frac{e^{ix}\plus{}e^{\minus{}ix}}{2})^{2n}\equal{}(e^{ix}\plus{}e^{\minus{}ix})^{2n}\equal{}\sum_{j\equal{}0}^{2n} e^{ixj}e^{\minus{}ix(2n\minus{}j)} \binom{2n}{j}\equal{}
\sum_{j\equal{}0}^{2n} e^{2ix(j\minus{}n)} \binom{2n}{j}\equal{}\sum_{j\equal{}0}^{n\minus{}1}\plus{}\binom{2n}{n}\plus{}\sum_{j\equal{}n\plus{}1}^{2n}\equal{}\sum_{j\equal{}0}^{n\minus{}1} (e^{2ix(n\minus{}j)}\plus{}e^{2ix(j\minus{}n)})\binom{2n}{j}\plus{}\binom{2n}{n}\equal{}\sum_{j\equal{}0}^{n\minus{}1} 2\cos 2(n\minus{}j)x\binom {2n}{j}\plus{}\binom{2n}{n}\equal{}RHS$


aNyone doing it? 6th contest just passed.

What was your total score? And team?

[quote="undefined117"]uhh the indiana math league we took a few months ago with 40 questions? i got a 40/40, and so did mwpl11.. several others got 39s.

--8th grade team

don't know the team scores though. you might ask mwpl11[/quote]

Our team score was 195. Honey Creek's was 194. You might want to look at this link. it has all the information/scores:
[url]http://www.mathleague.com/reports/2006_07/grade678/IN_8.HTM[/url]

team score for us was 176. we got pwned by honey creek. i got 37.

[quote="nat mc"]i wa talking about high school[/quote]
not middle

uhh..i dont think carmel high school does this..

well what does it matter what division we're talking about? the subject said INML, and that could mean either high school or middle  school.

While we're on the subject, :D me got first in 7th grade!!
Btw, why doesn't sycamore do the INML?

our school got second :(

tied with Honey-Creek   :D

There weren't any ties with us...you can check the site:
[url]http://www.mathleague.com/reports/2006_07/grade678/IN_6.HTM[/url]
[url]http://www.mathleague.com/reports/2006_07/grade678/IN_7.HTM[/url]
[url]http://www.mathleague.com/reports/2006_07/grade678/IN_8.HTM[/url]

It says that our school tied for second on the last link. (i think)

oh sorry I wasn't looking at the third row (2nd place 2nd row...lol i'm dumb...)


What is the probability that when 5 cards are drawn from a 52-card pack that at least 3 of them are spades?

[hide]
We can choose 5 cards in $ \binom{52}{5}$ ways.

There are 13 spades in the pack. Favorable hands are those with 3,4 or 5 spades. In total there are $ \binom{13}{3}\binom{39}{2} \plus{} \binom{13}{4}\binom{39}{1} \plus{} \binom{13}{5}$ favorable hands.

Thus the probability is $ \frac{\binom{13}{3}\binom{39}{2} \plus{} \binom{13}{4}\binom{39}{1} \plus{} \binom{13}{5}}{\binom{52}{5}}$
[/hide]


vector
function
induction
linear algebra

Never went on convexity before, so i'm not sure about this proof. I would appreciate any help or different proofs.

Definition: A subset $ X$ of an Vector Space $ V$ is called convex if $ u,v \in X \Rightarrow [u,v]\subset X$ where $ [u,v]=\{(1-t)u+tv \ | \ 0 \leq t \leq 1\}$ is the line joining $ u$ and $ v$.

If $ a,b,c \in \mathbf{R}$ show that $ X=\{(x,y) \in \mathbf{R}^2 \ | \ ax+by\leq c\}$ is convex
[hide]Let $ u=(u_1,u_2)$ and $ v=(v_1,v_2)$ in $ X$ and pick $ w \in [u,v]$ then we have $ w=(1-t)u+tv$ for some $ t$ where $ 0 \leq t \leq 1$.

Hence $ w=\left((1-t)u_1+tv_1, (1-t)u_2+tv_2  \right)$. Then we see that

\begin{eqnarray*}
a[(1-t)u_1+tv_1]+b[(1-t)u_2+tv_2] &=& a(1-t)u_1+atv_1+ b(1-t)u_2+btv_2\\
&=& (1-t)(au_1+bu_2)+t(av_1+bv_2)\\
&=& au_1+bu_2-t(au_1+bu_2)++t(av_1+bv_2)\\
&\leq & c-c+c=c
\end{eqnarray*}

since we know that $ au_1+bu_2 \leq c$, $ av_1+bv_2\leq c$ and $ t\leq 1$.
[/hide]

Your proof is correct. There is a generalization: if $ f\colon V\to\mathbb R$ is a convex function, then the set $ \{x\in V\colon f(x)\le c\}$ is convex for any $ c\in \mathbb R$. The converse is false.

Very Thanks mlok! When i first tried i got $ \leq 2c$ because i used $ (1 \minus{} t)\leq 1$ and $ t \leq 1$. I don't created a new post because this is very related, sorry if i'm wrong. I don't see how to do this one:

Let $ X \subset V$ be convex and $ \lambda_1,\ldots,\lambda_n$ be nonnegative reals with $ \lambda_1 \plus{} \cdots \plus{} \lambda_n \equal{} 1$. Show that if $ v_1,\ldots,v_n$ are in $ X$ then  $ \lambda_1v_1 \plus{} \cdots \plus{} \lambda_nv_n$ stills in $ X$.

Use induction on $ n$ and the identity $ \sum_{i\equal{}1}^n \lambda_i x_i\equal{}
(1\minus{}\lambda_n)\sum_{i\equal{}1}^{n\minus{}1} \frac{\lambda_i}{1\minus{}\lambda_n} x_i \plus{} \lambda_n x_n$.


Four snails walk on a plane [b]randomly[/b] with velocities [b]different[/b] in [b]magnitude and direction[/b] such that no two directions be parallel.
Now they may or may not encounter each other (encounter is one of a friendly type and it is assumed that there is no time lapse or change in velocity of any snail) and atleast theoretically, a maximum of $ ^4C_2 = 6$ encounters are possible.

Now, given that 5 encounters have already taken place, which of the following is [b]true[/b]?
$ A)$ $ 6{}^t{}^h$ encounter not possible.
$ B)$ $ 6{}^t{}^h$ encounter inevitable.
$ C)$ $ 5{}^t{}^h$ encounter only not possible!
$ D)$ Can't say anything about $ 6{}^t{}^h$ encounter.

Please explain what you mean by encounters are not friendly, and what happens to after each encounter. Because the way you have stated it I say D.

$ Encounter$ means only a meeting and there is no collision or any change in direction of that sort.

No. You could solve it from the Physics point of view. You consider a snail as your reference point.
Now, since $ 5$ encounters have occured, let us consider there are $ 2$ snails having had $ 3$ encounters each.
Now, the $ path$ of $ Snail2$ passes [b]through[/b] $ Snail1$ and of other snails, say, $ Snail3$ and of $ Snail4$ too do so (since they have encountered each other).
Thus we find [b]all the paths being co-incident[/b] from the [b]point of reference of $ Snail1$ only[/b]. Hence, since velocities are different, all six encounters are inevitable.

You may also solve this using General Relativity.


[b][u]Subiectul 1[/u][/b]
[hide]http://www.mathlinks.ro/viewtopic.php?t=240176[/hide]

[b][u]Subiectul 2[/u][/b]
[hide]http://www.mathlinks.ro/viewtopic.php?p=1321867#1321867[/hide]

[b][u]Subiectul 3[/u][/b]
[hide]http://www.mathlinks.ro/viewtopic.php?p=1321871#1321871[/hide]

[b][u]Subiectul 4[/u][/b]
[hide]http://www.mathlinks.ro/viewtopic.php?t=240178[/hide]

subiectele la cls 12 nu ai cumva  :lol: ?

subiectele de la matematica de la proba individuala pentru clasele 9-12 sunt pe http://www.mategl.com


I think your inequality is wrong... try $ (1,1,1)$  :wink:  I think we should replace $ 9$ by $ 3$. It is easy to prove by "p,q,r technique"  :)

I mean that:
Prove the following inequality for $ a,b,c\in\mathbb{R^{ \plus{} }}$:
$ \sqrt {3(ab \plus{} bc \plus{} ca)}(9(a \plus{} b \plus{} c)^2 \plus{} ab \plus{} cb \plus{} ca)\leq 9((a \plus{} b \plus{} c)^3 \plus{} abc)$

Put $ a\plus{}b\plus{}c\equal{}p,ab\plus{}bc\plus{}ca\equal{}q\equal{}3,r\equal{}abc$. Using Schur's inequality we have the inequality is equivalent to:
\[ \frac{(p\minus{}3)(8p^2\minus{}3p\plus{}3)}{9}\ge 0\]
It is alway true  :wink:

[quote="zaizai-hoang"]  I think we should replace $ 9$ by $ 3$. It is easy to prove by "p,q,r technique"  :)
 [/quote]
I'm sorry,thank you for solution.


[quote="TRAN THAI HUNG"]$ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$ $ x,y,z \in R$
Prove that
$ x \plus{} y \plus{} z \le 2 \plus{} xyz$[/quote]

[hide]
You can do casework on the number of nonegative reals.

1. 3 or 1 real numbers are nonegative. Then we can use QM-AM to get that $ 2 \ge x \plus{} y \plus{} z.$ Since two or 0 numbers are negative, xyz is nonegative and we are done.

2. 0 real numbers are nonegative. Then, the left side is negative, but using QM-AM-GM we have that the right side is positive.

3. 2 real numbers are nonegative. WLOG, let z be negative. We know that $ x^2 \plus{} y^2 < 2,$ which by QM-AM-GM implies that $ x\plus{}y < 2$ and $ xy < 1.$ Therefore, we know that since z is negative, $ z < xyz.$ Adding this and $ x\plus{}y < 2,$ we get the desired inequality[/hide]

[hide]Here's my solution(without bashing it, of course):
We have: (2-2xy)(2-2xz)(2-2yz)>=0 with x,y,z belongs to R
Therefore, we have: 1>=x^2y^2z^2-xyz(x+y+z)+(xy+yz+xz).
Multiplying 2 by both sides we have: 2>=2(x^2y^2z^2-xyz(x+y+z)+(xy+yz+xz)) [b](1)[/b]
On the other hand, our inequality is equivalent to: (x+y+z-xyz)^2=<4(if  x+y+z-xyz=<0, then it's obvious. So we just care about the case when it's>0)
This inequality can be rewritten as: 2+2(xy+yz+xz)-2xyz(x+y+z)+x^2y^2z^2=<4, which means we need to prove:
2>=2(xy+yz+xz)-2xyz(x+z+y)+x^2y^2z^2, which is obviously true because of (1)
The equality occurs when x=0,y=z=1 or x=0,y=z=-1 and their permutations(Q.E.D)
@Pythag011: You seem really like bashing complicated inequality.Probably I'll propose some inequalities which can't be bashed by hand for you :wink: [/hide]

[quote="pythag011"]
1. 3 or 1 real numbers are nonegative. Then we can use QM-AM to get that $ 2 \ge x \plus{} y \plus{} z.$[/quote]
Nope, try $ x\equal{}y\equal{}z\equal{}\sqrt{\frac{2}{3}}$


number theory
abstract algebra
superior algebra
superior algebra unsolved

let a>1 be an integer. Prove for any positive integers n,d that divides n iff a^(d)-1 divides a^(n)-1. conclude in particular that the field of degree (p^d) is contained in the field of degree (p^n) iff d divides n.

This sounds like some number theory problem...but I just got stuck.....can anyone help?

actually....I can do the first part....
but I got stuck on the conclude part..

If you have the first part, look at the multiplicative groups in those fields. (They have order $p^{d}-1$ and $p^{n}-1$ since you throw out $0$). Now use Lagrange's Theorem to check if one can be in the other.

For the first part,

Say $d|n$ then you can write $n=dk$

Thus, it is easily seen that,
$a^{d}-1|a^{dk}-1$

Because you can write,
$a^{dk}-1=(a^{d})^{k}-1$
And apply the expansion identity,

$x^{n}-y^{n}=(x-y)(x^{n-1}y+...\mbox{ whatever })$


function
inequalities unsolved
inequalities

Let $ f : N \rightarrow N$ be a function satisfying $ kf(n) \leq f(kn) \leq kf(n) \plus{} k \minus{} 1$ for all $ k, n \in N$.

(a) Prove that $ f(a) \plus{} f(b) \leq f(a \plus{} b) \leq f(a) \plus{} f(b) \plus{} 1$ for all $ a, b \in N$.

(b) If f satisfies $ f(2007n) \leq 2007f(n) \plus{} 2005$ for every $ n \in N$, show that

there exists $ c \in N$ such that $ f(2007c) \equal{} 2007f(c)$.

We have for all $ a,b \in N$

$ abf(a) + abf(b) \leq af(ab) + bf(ab) = (a + b)f(ab)$

$ \leq f((a + b)ab) \leq abf(a + b) + ab - 1$

Therefore $ f(a) + f(b) \leq f(a + b)$ for all $ a,b\in N$.

There is a similar way to prove second inequality.


Find all couples of polynomials $(P,Q)$ with real coefficients, such that for infinitely many $x\in\mathbb R$ the condition \[ \frac{P(x)}{Q(x)}-\frac{P(x+1)}{Q(x+1)}=\frac{1}{x(x+2)}\]
Holds.
[i] Nikolai Nikolov, Oleg Mushkarov[/i]

$\frac{P(x)}{Q(x)}-\frac{P(x)+1}{Q(x+1)}=\frac 1{x(x+2)}$$\Longrightarrow x(x+2)(P(x)Q(x+1)-P(x+1)Q(x))=Q(x)Q(x+1)$
for infinite many $x\in R$ , so the equality holds for all $x\in R$, because of $P,Q$ are polynomials.
I denote $F(x)=\frac{P(x)}{Q(x)}$, so $F(x+1)=F(x) - \frac12.(\frac 1x-\frac 1{x+2})$,..., $F(x+n+1)=F(x+n)- \frac12.(\frac 1{x+n}-\frac 1{x+n+2})$,
so $F(x+n+1) =F(x)-\frac12.(\frac 1x+\frac 1{x+1}-\frac 1{x+n+1}-\frac 1{x+n+2})$
Make $n\rightarrow \infty$ and you obtain $F(x)-\frac12.(\frac 1x+\frac 1{x+1})=C (constant)\Longrightarrow F(x)=\frac{2Cx^2+2(C+1)x+1)}{2x(x+1)}$ and so on ...
 :cool:

Two questions

[quote="Diogene"] , so the equality holds for all $x\in R$, because of $P,Q$ are polynomials.  [/quote]

Really? I can't quite see it.
[quote="Diogene"]
and you obtain $F(x)-\frac12.(\frac{1}{x}+\frac 1{x+1})=C (constant)$[/quote]

Why constant?


geometry
3D geometry
sphere
abstract algebra
topology
real analysis
real analysis unsolved

assume that $\mathbb{S}^2 $ is 2-dimensional sphere and $A$ is its finite subset .
a)if $|A|=1 or 2 $ then $\mathbb{S}^2-A$ could be consider as a topological group.
b)if $|A|\geq 3 or |A|=0 $ then $\mathbb{S}^2-A$ may not be consider as a topological group.
(i know some part of it but i'm not able to solve the whole.)

[moderator edit for (b)]

For the easy part (I mean (a)):

If we take out one point, what remains is isomorphic to the plane $\mathbb R^2$, which is a topological group with the addition.

If we take out two points, the space is isomorphic to the punctured plane, and this can be regarded as the topological group $(\mathbb C^*,\cdot)$ in an obvious manner.

P.S.: Maybe there's something wrong with the statement? :? You wrote "could be considered a topological group" in both cases.

In (b), I believe you meant to say that the respective space cannot be a topological group, right? If $|A|\ge 3$, then the fundamental group of the space is a free group with at least two generators, so, in particular, it's not abelian. However, the fundamental group of a topological group is abelian (I'm looking at an exercise in Munkres' "Topology, a first course" which says precisely this :)), so in this case, $\mathbb S^2\setminus A$ can't be a topological group.

I've had an exam in algebraic topology today and the problem of proving that $S^2$ is not a topological group came up.
It was revealed to me that the problem asks to prove that there is no group structure on $S^2$ that makes it into a topological group with its natural topology. Also the problem of proving that no matter what topological structure we place on $S^2$ it nas no topological group structure is false. This is because between $S^2$ and $S^1$ there is a bijection and $S^1$ is a topological group.

Now to the problem:
[b]Lemma[/b]: Let $f:S^2\to S^2$ a continuous map. Then $f\circ f$ has a fixed point. Moreover if f has no fixed points then it sends a point to its antipodal.

Proof: If $f$ has no fixed points then it is homotopic to the antipodal map.
Let $F:S^2\times I\to S^2$, $I=[0,1]$, $F(x,t)=\frac{(1-t)f(x)-tx}{||(1-t)f(x)-tx||}$ It is easy to see that if f is continuous and has no fixed points then $F$ is continuous and in fact a homotopy between $f$ and $-id_{S^2}$ (the antipodal map).
In similar fashion if $f$ sends no point to its antipodal then $f$ is homotopic to the identity map.
Now if $f$ has no fixed points and sends no point to its antipodal, then $f$ is homotopic to $-id$ and to $id$. But this means that the identity map and the antipodal map are homotopic. However $deg(id_{S^2})=1$, $deg(-id_{S^2})=-1$ and $deg$ is homotpy invariant. This completes the proof of the second part of the lemma.
If $f\circ f$ has no fixed points then it is homotopic to the antipodal so it has deg $-1$. But $deg(f\circ f)=deg(f)^2>0$ which is a contradiction.

Proof of the problem. Assume $(S^2,\cdot )$ is a topological group. Define $f_g:S^2\to S^2$, $f_g(x)=gx$. Then $f_g$ is a homeomorphism of $S^2$, with it's inverse $f_{g^{-1}}$.
$f_g(x)=x\Rightarrow gx=x\Rightarrow g=1$. So $f_g$ has fixed points iff $g=1$. $1$ denotes the neutral element of $(S^2,\cdot )$.
By the lemma, $f_{g^2}=f_g\circ f_g$ has a fixed point so $g^2=1\ \forall g\in S^2$.
Let $g,h\in S^2\setminus\{1\}$. Then $f_{gh}=f_g\circ f_h$. Because of the choice of $g$ and $h$, $f_g,f_h$ have no fixed points so they are homotopic to the antipodal. So $deg(f_{gh})=deg(f)deg(h)=(-1)(-1)=1$. If $f_{gh}$ had no fixed points then it had $deg=-1$ which would have been impossible. So $f_{gh}$ has fixed points meaning $gh=1\Rightarrow g=h$. So $S^2\setminus\{1\}$ has at most one element, so $S^2$ is finite, which is a contradiction.


Three protons and three electrons are to be placed at the vertices of a regular octahedron of edge length $ a$.  We want to find the energy of the system, that is, the work required to assemble it starting with the particles very far apart.  There are essentially two different arrangements.  What is the energy of each?

[hide]$ \minus{}3.879e^{2}/a$ and $ \minus{}2.121e^{2}/a$[/hide]

what rae ur attempts...and answer this...what is the essential contituent of this energy

I drew an octahedron and placed charges at the vertices and used the potential energy formula for each pair and summed together, but I never arrived at the answers given.

did u take the distances between them and the charges(positive or negative) right

Yes, I think I did.  I believe I took every pair and every distance.

Am I forgetting pairs?  Can someone provide a setup?

[quote="JRav"]Three protons and three electrons are to be placed at the vertices of a regular octahedron of edge length $ a$.  We want to find the energy of the system, that is, the work required to assemble it starting with the particles very far apart.  There are essentially two different arrangements.  What is the energy of each?

[hide]$ \minus{} 3.879e^{2}/a$ and $ \minus{} 2.121e^{2}/a$[/hide][/quote]

I'm not going to attempt to solve this, but by my count there are 20 different arrangements for how to place the charges at the vertices. Of these, I suspect ten give one energy, and the remaining ten give the other energy. (Because you can't distinguish between one electron and another, nor can you distinguish between one proton and another.) Nonetheless, I will give you the formula and suggest an approach by which you can solve all twenty cases if needs be.

First, the potential energy of a charge $ q$ at a location $ \mathbf{x}$ due to a lattice of other charges is given by:

\[ U\left(\mathbf{x}\right) \equal{} q \Phi\left(\mathbf{x}\right) \]

$ \Phi\left(\mathbf{x}\right)$ is the potential of the lattice of charges at location $ \mathbf{x}$, in units of volts.

\[ q \equal{}\begin{cases} \minus{}e & \text{if an electron @ }\mathbf{x} \\ \plus{}e & \text{if a proton @ }\mathbf{x}\end{cases}\]

Finally,

\[ \Phi\left(\mathbf{x}\right) \equal{}\frac{1}{4\pi\varepsilon_0}\sum_{i\equal{}1}^{N}\frac{q_i}{\left|\mathbf{x}\minus{}\mathbf{x}_i\right|}\]

$ \mathbf{x}$ is the vector location of charge $ q$, and $ \mathbf{x}_i$ is the vector location of charge $ q_i$.

$ \varepsilon_0$ is the permitivity of free space = $ 8.854\times 10^{\minus{}12}$ Coulomb/(volt$ \cdot$ meter)