ill definitely do it, just pm when and ill get back 2 u.  can we do it sometime next week though, im a little busy this week

hmm i think i could...even though i'm not going to Nats. What day of the week to you plan on having this? Tuesday and Wednesday aren't good for me.

Well yeah I'm not going to schedule this so every single person can come, but there will hopefully be plenty in the future, run by others.

I'll do the AIM Math Jam

I'll just use another s/n since you've already blocked two of them :D (i have three)

Anytime is fine for me...at least i think :ninja:

i'll do it...IF MY AIM STOPS BEING RETARDED

EDIT:haha aim address:kstan013

same here
AIM: manthgacrazed
hopefully i'll get better XD

oh, darn, i have a s/n, but i need both urs and to install aim on this laptop.  oh well, i guess im out

interesting idea.. i'll join even though i'm not going to nationals

AIM: subliminal404

hehe so far based on this thread we have leetspeak, kstan, aznbrain, and four people from indiana!

Hey leetspeak

Unblock my s/n Aznwithabrain (Please!)

[quote="SplashD"]Why don't you host it in one of the classrooms?[/quote]

i asked this w/ chapter discussion and they said no

so i'd guess, no

jorian

[quote="SplashD"][quote="diophantient"][quote="SplashD"]Why don't you host it in one of the classrooms?[/quote]
how?
[/quote]

You click the little button on the left of your screen that says "Classroom" though it might not be as private as AIM and potentially has inconvenient times where it opens.[/quote]
Not everyone has access to those though, you have to be taking an aops class and even then they can only access the one that corresponds to their level (intro, interm, woot)

yeah, and since not everyone takes classes here...i think the aim idea is the brightest yet

i won't attend. duh.

[quote="diophantient"][/quote][quote="SplashD"][quote="diophantient"][quote="SplashD"]Why don't you host it in one of the classrooms?[/quote]
how?
[/quote]

You click the little button on the left of your screen that says "Classroom" though it might not be as private as AIM and potentially has inconvenient times where it opens.[/quote][quote="diophantient"]
Not everyone has access to those though, you have to be taking an aops class and even then they can only access the one that corresponds to their level (intro, interm, woot)[/quote]

the classroom is always open isn't it?

well, no

an admin (i think patrick), told me privately

jorian

Can I join even though I'm not going to Nats?
Username: Stargirl240240   :rotfl:

Although there is no possible way that I'll do well at nationals, I'll try, if the times are good. My sn is azngirl1292

[quote="Aryth"]in. o0 Teapot 0o.
Thanks.[/quote]

Why aren't you part of the Teapot Appreciation Club?

See the following thread for the specific information about this Monday's Jam, ask specific questions there. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143375[/url]

Note that I am not writing down these AIMs, so please just send me a message asking for the invite to the jam ar 7:50 (eastern) ish, if you are blocked please send word through someone else.

[quote="jhredsox"]what if you blocked us???

jorian[/quote]
Look two posts above yours :roll:

I'll do the next one since this first one just was supposed to have started
aim: amardini

Everyone EXCEPT CARL is there  :furious:

[quote="13375P34K43V312"]OK, let's get a string of these started. Like last year, we will make it similar to the AoPS Math Jams, but unfortunately, they won't be able to sponsor any this year, evidently. I (or whoever is sponsoring) will post a countdown problem, and the first three responders will get a point, etc., etc. No extra points for explaining solutions. Different sponsorers may wish to tweak the rules a little.

I'm going to host one, if there's enough interest, sometime next week, hopefully. It will be in the evening, classtime-ish, I'll see if I can schedule it around classes (it's tough though, WOOT Monday and Thursday, NT Wedesday, C/P Tuesday). If you don't want your s/n posted here, fine, IM me (pocketaces2838). You don't even have to tell me your AoPS account.

If I am not appeared to be on at the time of the countdown, I most likely blocked you, have someone else send me an IM telling me to unblock you for the duration of the thing.

More rules shall be discussed in the actual jam if it shall occur.

Please post here if you are interested. I shall set a date if enough interest gets generated.

If possible, could a mod sticky this?[/quote]

1337 sorry to reiterate this but i cant do this because you sorta warned me and blocked me on IM....im macgyver2222222 if your interested.....even though this topic is realy old

r u seriously responding 2 this topic now? dont revive old topics, especially ones that r now totally irrelevant.


Hi,

Given that $ x$ is directly proportional to $ y$ and to $ z$ and is inversely proportional to $ w$, and that $ x\equal{}4$ when $ (w,y,z) \equal{} (6,8,5)$, what is $ x$ when $ (w,y,z) \equal{} (4,10,9)?$

I need help understanding the following solution:

[i]Solution[/i]: Because $ x$ is inversely proportional to $ w$, when all other variables are constant, $ xw$ is constant. Similarly, when the other two variables are constant, each of $ \frac{x}{y}$ and $ \frac{x}{z}$ is constant. We can combine all these by saying $ \frac{xw}{yz}$ is constant. Thus, from the problem we have:

$ \frac{xw}{yz} \equal{} \frac{(4)(6)}{(8)(5)} \equal{} \frac{3}{5}$

When $ (w,y,z) \equal{} (4,10,9)$, we find

$ x \equal{} \frac{3yz}{5w} \equal{} \frac{27}{2}$

I do not understand these phrases(the ones in [b]bold[/b]):

[i]Solution[/i]: Because $ x$ is inversely proportional to $ w$, [b]when all other variables are constant[/b], $ xw$ is constant. Similarly, [b]when the other two variables are constant[/b], each of $ \frac{x}{y}$ and $ \frac{x}{z}$ is constant. [b]We can combine all these by saying $ \frac{xw}{yz}$ is constant.[/b]

What significance does "when all other variables are constant" hold? What variables? And also why only when all other variables are constant, $ xw$ is constant?

Second, why/how is $ \frac{xw}{yz}$ a constant? So in that case will $ \frac{xz}{yw}$ or something like that also be a constant?

Your help will be appreciated, thank you :)

Regards,
limac

this was very confusing to me too when i first learned this, so i'll warn you that you may or may not get this immediately....it me several years to have a good and comfortable intuitive grasp of this.

also, pardon me for using specific numbers to explain, but i understand these things better with actual numbers than with variables

since x and y and directly proportional, doubling x will double y (or tripling, or quadrupling, or whatever you wish to multiply x by). Thus $ \frac xy \equal{} k$ for some constant k. (If you do not understanding this, try plugging in a certain random pair x,y and then change x, i.e. double x or whatever, then see what you must do to y to keep that fraction equal to the constant k.)

Now, if we add in the z is also proportion, we know that $ \frac xz \equal{} k_1$ for some constant $ k_1$, by the same reasoning as above. In fact, it may be hard to believe, but we can write $ \frac{x}{yz}\equal{}k_2$ for some constant $ k_2$. For example, let's assume x=4, y=6,z=8. To show you the significance of holding some variables constant: It's clear here that $ k_2\equal{}4/(6x8)\equal{}1/12$. Now, we can double x, so we should have to double y now to keep the fraction constant wrong? Not exactly, note that, for example, we can double x (x=8), triple y (y = 18), and multiply z by 2/3 (z = 16/3). If we plug this into the expression relating x, y and z (which is $ \frac{x}{yz}\equal{}k_2$,) you can verify that this works.

in fact, if we double x, we could have done anything to y and z as long as we double the product of y and z (since we can view the product yz as a single variable, for instance, a, and say that x is directly proportional to z) however, if we hold y to be constant (let's say $ y\equal{}y_1$ for some constant $ y_1$), then when we double x, we have to double z, since we are holding y constant (we have to double the product yz, and since we can't change y, we must change z by the same proportion as we changed x)

for w inversely proportional to x, we can write xw=constant, in fact, we can write $ \frac{xw}{yz}\equal{}k_3$ for some constant $ k_3$ (again you should plug in many values of x y z w to convince yourself if you do not understand)

simply think of the fractions  that you are asking about as ratios: $ \frac xy$ is the ratio of x to y. We cannot determine any relationship between x and y unless the other variables (in this case w and z do not change). This is because we can think of changing 3 of x y z w at a time as making 2 changes at a time. if say for another example, x = 2, y = 3, z = 4, w = 5, then the fraction $ \frac{xw}{zy}\equal{}5/6$ so... we can say that since w and x are inversely proportional, we can multiple x by 1/2 and w by 2. As another step, we can muliply x by 1/2 and y by 1/2 since they are directly proportional. this gives us x = 1/2, y = 3/2, z = 4, w = 10. plugging these into the expression relating w,x,y,z should give the same fraction 5/6. Note the changes overall: x was multiplied by 1/4, y was multiplied by 1/2, and w was multiplied by 2. We cannot say anything about the ratio of x to y as being a constant ratio. the initial ratio was 2/3, now it is 1/3. This is because we changed w. However, if we could not change w, (that is, in other words, if we held it constant) we could say x/y IS some constant (if we also held z constant, since z could mess with the ratio of x/y as another step of changing the variables around in accordance to the given relationship just as w did in the previous example)

in summary, you should plug in many many numbers for the variables and change them around if this helps. I think you may use similar reasoning to answer the other parts of your question.

for the second question: we can always combine information about relationships into one expression if they if all of them are related in some way (through as many variables as need) into one nice expression, like $ \frac{xw}{zy}\equal{}k_3$. This relationship is found by combining the 3 relationships that all have to do with x (, if x in numerator, if in direct proportion (to x), put it in denominator,and if inversely proportional, put in numerator).

i know it is hard to believe, but if you plug in x y z w into the expression and change the variables 2 at a time in accordance with the given information, you should find that the expression always stays the same (so that you will change x each time.
however, if you plug them into, say $ \frac{xz}{yw}$ as you proposed, and change them 2 at a time in accordance to the given info, that expression is not constant. I don't think any other expression that does not simplify to xw/(yz) will not ever always be constant

hope that helped and sry about it being way too long, because i have  a dim understanding of algebra

Hi hurdler,

Thank you very much for your time. I do sort of understand this now, but not completely. So, I'll just skip the understanding of this specific problem right now.

Thanks, again, for your help :)

Regards,
limac


symmetry
inequalities
inequalities unsolved

[quote="Nameless"][6,2,0] is $x^6y^2+y^6z^2+z^6x^2$ or $x^6y^2++x^6z^2+y^6z^2+y^6x^2+z^6x^2+z^6y^2$
Same question with [7,1,0] and [5,2,1]???

 :huh:[/quote]

It is the latter one since it is symmetry sum  ;)  .

[quote="pellishau"]Well, Muirhead is a given.
Schur gives $[8,0,0] + [6,1,1] \geq 2[7,1,0]$ and  $[6,1,1] + [4,2,2] \geq 2[5,2,1]$ (r=6 and 3)[/quote]

well , what about [6,2,0] ? and how u use above to solve it ? I cant see it ....


Outline a synthesis of 2,3-dibromotoluene from toluene.

how abt:
!. first protect the para group(sulphoantion) and then nitrate to get mono nitro product. now desulphonate and reduce the nitro group to amine.
2.now again sulphonate to get so3h at para pos to nh2. treat with base and protect amino from oxidation by treating with AC20 and again nitrate to get nitro group ortho to NHAc.
3. now reduce the nitro and desulphonate and remove Ac group and carry out sanmeyer reaction to get dibromo product.

[quote="VARUNARASU"].now again sulphonate to get so3h at para pos to nh2[/quote]

Wrong!

is it bcoz nh3 attacks the electrophile. in that case heat it to get so3h at para position.

The amino group is protonated, being converted in a meta director.

i never thought of that. :fool: then how do you sulfonate a ring with nh2? by the use of protecting groups?

That's right: just acylate the amino group, and the resulting acetamido group is less activating (why?).


[quote="info_math"]find all primes p such that there exist integers m and n :

     m^3+n^3-4  divides m^+n^2=p[/quote]

It is not clear!
Do you mean:

find all primes p such that there exist integers m and n such that $ p \equal{} m^2 \plus{} n^2$ and
$ {(m^3 \plus{} n^3 \minus{} 4)} | m^2 \plus{} n^2$
It is trivial for positive integers m,n though  :P

I'm sorry for my english :blush:
yes! you are right!

The correct problem is $ p | m^3 \plus{} n^3 \minus{} 4$.  I'll post my solution in a few hours.  (It's being recycled as part of a problem set due at midnight...)

How about $ p \equal{} 5$?  :)  You took $ \bmod 6$ in your last step instead of $ \bmod p$ for some reason.

Sorry.My mistake. I thought $ m^2\plus{}n^2|m^3\plus{}n^3\minus{}4$ means they have to be same mod 6.That was a wrong assumption which caused all the problems.
Thank you very much.

It's midnight (PST)!  My solution, as I had it written up:

Let $ u \equal{} mn, v \equal{} m \plus{} n$.  We want to find $ p$ such that $ m^3 \plus{} n^3 \equiv 4 \bmod p$.  Note that $ p \equal{} v^2 \minus{} 2u$; therefore,

$ u \equiv \frac {v^2}{2} \bmod p \Leftrightarrow$
$ m^3 \plus{} n^3 \equal{} v(m^2 \minus{} u \plus{} n^2) \equiv \minus{} uv \equiv \minus{} \frac {v^3}{2} \equiv 4 \bmod p \Leftrightarrow$
$ (v \plus{} 2)(v^2 \minus{} 2v \plus{} 4) \equiv 0 \bmod p$.

[b]Case:[/b]  $ p | v \plus{} 2$.  AM-QM gives

$ \sqrt {\frac {m^2 \plus{} n^2}{2} } \ge \frac {v}{2} \implies \sqrt {2p} \ge v$

so that we can only have $ v \equal{} p \minus{} 2$, and in fact this can only occur when $ \boxed{p \equal{} 5}$.  When $ p > 5$, we have $ p \minus{} 2 > \sqrt {2p}$.

[b]Case:[/b]  $ p | v^2 \minus{} 2v \plus{} 4$.  But $ v^2 \minus{} 2v \plus{} 4 \equiv 2(u \minus{} v \plus{} 2)$, hence $ p | u \minus{} v \plus{} 2 \equal{} (m \minus{} 1)(n \minus{} 1) \plus{} 1$.  Because $ m, n \ge 1$ we know that $ (m \minus{} 1)(n \minus{} 1) \plus{} 1 \ge 1$, but because $ v \ge 2$, we have

$ u \minus{} v \plus{} 2 \le u \le \frac {p}{2}$

by AM-GM, hence we never have $ (m \minus{} 1)(n \minus{} 1) \plus{} 1 \equiv 0 \bmod p$.

[b]N.B.:[/b]  While looking at the values of $ m^3 \plus{} n^3 \bmod p$ for small $ p$, it appears to be the case that the remainder is greater than $ \frac{p}{2}$, which is stronger than the solution to this problem, but I'm not sure how my solution can be tweaked to prove this more general assertion.


Let $ n,m,k,t$ be 4 positive integers satisfying $ n \geq k$ and $m \geq k$ and $ n + m - k + 1 = 2^t$. Prove that $ {m \choose k} + {n \choose k}$ is even.

Assume that ${m \choose k} + {n \choose k}$ is odd. Then one of the summands is also odd, while the other is even. Without loss of generality, assume that ${m \choose k}$ is odd. [url=http://en.wikipedia.org/wiki/Lucas'_theorem]Lucas' theorem[/url] implies that for every nonnegative index $i$, if the $i$-th binary digit of $k$ is 1 then so is the $i$-th binary digit of $m$ and thus the $i$-th digit of $m-k$ is 0. Then the equality
\[n+(m-k) = 2^t - 1\]
implies that the $i$-th digit of $n$ is also 1. Therefore, by Lucas' theorem, ${n \choose k}$ must be odd, a contradiction that prove that ${m \choose k} + {n \choose k}$ is even.


See here  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=255907 :wink:

Ine. is equivolent to:
$ xyz(x\plus{}y\plus{}z\plus{}\sqrt{x^2\plus{}y^2\plus{}z^2} \leq \frac{3\plus{}\sqrt{3}}{9}(x^2\plus{}y^2\plus{}z^2)(xy\plus{}yz\plus{}xz)$

We have:
$ \frac{1}{2}\sum_{sym}{x^2yz}\plus{}\sum_{sym}{x^3y} \geq \frac{3}{2}\sum_{sym}{x^2yz}$
by Muirhed inequlity.
or
$ \frac{3}{9}(x^2\plus{}y^2\plus{}z^2)(xy\plus{}yz\plus{}xz) \geq xyz(x\plus{}y\plus{}z)$ $ (*)$

We also have that:
$ \sum_{sym}{x^6y^2}\plus{}\sum_{sym}{x^4y^4}\plus{}2\sum_{sym}{x^3y^3z^2} \geq 2\sum_{sym}{x^3y^3z^2}\plus{}2\sum_{sym}{x^5y^3} \geq 4\sum_{sym}{x^4y^3z} \geq 4\sum_{sym}{x^4y^2z^2}$
and
$ \frac{5}{2}\sum_{sym}{x^4y^2z^2}\plus{}\sum_{sym}{x^6yz}\plus{}4\sum_{sym}{x^4y^3z}\plus{}2\sum_{sym}{x^5y^2z} \geq \frac{19}{2}\sum_{sym}{x^4y^2z^2}$
So:
$ \sum_{sym}{x^6y2}\plus{}\sum_{sym}{x^4y^4}\plus{}\frac{5}{2}\sum_{sym}{x^4y^2z^2}\plus{}\sum_{sym}{x^6yz}\plus{}4\sum_{sym}{x^4y^3z}\plus{}2\sum_{sym}{x^3y^3z^2} \plus{}2\sum_{sym}{x^5y^2z} \geq \frac{27}{2}\sum_{sym}{x^4y^2z^2}$
or
$ xyz\sqrt{x^2\plus{}y^2\plus{}z^2} \leq \frac{\sqrt{3}}{9}(x^2\plus{}y^2\plus{}z^2)(xy\plus{}yz\plus{}xz)$ $ (**)$
By summing $ (*)$ and $ (**)$ we get desired result.


So yeah.
I don't really get this problem, even though someone already explained it to me.
So anyways.
Two circles with a radii of 12 have edges that intersect each other's centers, creating a shape.
What is the area of this shape?
Well it wasn't really worded like that but whatever.

I think the problem, reworded a little, was find the area of the region contained by exactly one circle.

[hide="hint"]
Consider circle A (left-hand side) to be set A and circle B (right-hand side) to be set B, then if the area of set A is $ \# (A)$ and the area of set B is $ \# (B)$, the area of the region stated above is $ \# (A)\plus{}\# (B)\minus{}2(\# (A\cap B))$
[/hide]

[quote="alanchou"]I think the problem, reworded a little, was find the area of the region contained by exactly one circle.

[hide="hint"]
Consider circle A (left-hand side) to be set A and circle B (right-hand side) to be set B, then if the area of set A is $ \# (A)$ and the area of set B is $ \# (B)$, the area of the region stated above is $ \# (A) \plus{} \# (B) \minus{} 2(\# (A\cap B))$
[/hide][/quote]
Yeah I know, but I can't find the area of the intersection....

[quote="alanchou"]Let the two intersection points be $ P$ and $ Q$, and the centers be $ A$ and $ B$.  Then, draw $ PQ$, $ PA$, and $ QA$.  You have a triangle, and a sector.  The area of half the interestion is the area of the sector minus the area of the triangle[/quote]
Ok i'm lost.

I'll help you out, but in return go back to my Counting Problem and answer the question I asked you on my last post there (under the EDIT).

Do what alanchou said.  Try actually drawing the intersections of the two circles; it must be 2 points (if it was 1 point then the area of that intersection would be 0, which is a dumb problem).  In other words, draw 2 circles overlapping, with intersection points P and Q.  Make the centers of both the circles A and B.  Draw line segments from A to P, A to Q, and P to Q.  That forms a triangle.  Now, this triangle cuts off half of the total intersection region of the two circles.  Next, think only of Circle A (ignore Circle B completely in the picture).  Not only did you form a triangle, but you also cut off part of circle A with arc PQ and sides AP and AQ (in the shape of a piece of pie).  If you take the total area of that piece of pie (using CentralAngle/360 and that proportion equal to the AreaOfThePiece/AreaOfTheCircle) and subtract triangle APQ from it, you are left with the aforementioned half-the-total-intersection-region-of-the-two-circles piece (as denoted by chord PQ and Arc PQ).  Double this area (since it is only half of the intersection region) to get the area of the total region.

The only way to solve the areas of the piece-of-pie and triangle APQ is to know the central angle (angle PAQ) as a given fact or the length of the sector (arc PQ), in which case you would use s=(r/Theta), where Theta is the Central Angle in Radians (if you don't know radian conversion, then you probably won't even be given the sector length so don't worry about it)

[quote="alanchou"]since the circles are congruent, you actually can determine the angle from just the information given[/quote]

I think the angle of an equilateral triangle should be 60 degrees.

I don't know why, but I think that a way to solve it.

My answer is $ \approx176.9$.

By the way, is there a formula for this type of problem?

There's no real point for a formula, because many different questions can be asked.

And it's incorrect

Since Peijin now should know how to solve it, we'll let other people have a go. (He goes to the same school as me)

Wait, you're looking for the intersection of the circles right?
Or the total area of the two circles?

The question said to find the area of the regions that are contained by a circle but not in the intersection.

I will assume that he wants the area of the entire shape.  Because that is actually what the question asks, at least that's how I interpret it.

[hide="Solution"]Notice that if we draw radii from each circle's center to one intersection point, and then connect the two centers, we have three congruent radii.  Thus, we have an equilateral triangle of side length $ 12$.

To find the area of the intersection, we need to find the area of one of the circular sectors formed with the triangle and one of the circular segments.  Draw it and you'll see what I mean.  Now, to find the area of one half of the intersection, we take two times that area and subtract the area of the triangle.

The area of the sector: $ A_s\equal{}\frac{1}{6}*\pi*12^2\equal{}24\pi$

The area of the equilateral triangle: $ A_\bigtriangleup\equal{}\frac{\sqrt{3}}{4}*12^2\equal{}36\sqrt{3}$

Thus the area of one half of the intersection is: $ 2A_s\minus{}A_\bigtriangleup\equal{}48\pi\minus{}36\sqrt{3}$
The entire area: $ 96\pi\minus{}72\sqrt{3}$

Now that we know the intersection, the area of the entire shape is $ 2*144\pi\minus{}(96\pi\minus{}72\sqrt{3})\equal{}192\pi\plus{}72\sqrt{3}$

You guys can correct my arithmetic if it's incorrect.[/hide]

Since that is what you are looking for, that is correct for that.  The original problem actually made it seem like the answer was $ 0$.

[quote="mathcrazed"]I will assume that he wants the area of the entire shape.  Because that is actually what the question asks, at least that's how I interpret it.

[hide="Solution"]Notice that if we draw radii from each circle's center to one intersection point, and then connect the two centers, we have three congruent radii.  Thus, we have an equilateral triangle of side length $ 12$.

To find the area of the intersection, we need to find the area of one of the circular sectors formed with the triangle and one of the circular segments.  Draw it and you'll see what I mean.  Now, to find the area of one half of the intersection, we take two times that area and subtract the area of the triangle.

The area of the sector: $ A_s \equal{} \frac {1}{6}*\pi*12^2 \equal{} 24\pi$

The area of the equilateral triangle: $ A_\bigtriangleup \equal{} \frac {\sqrt {3}}{4}*12^2 \equal{} 36\sqrt {3}$

Thus the area of one half of the intersection is: $ 2A_s \minus{} A_\bigtriangleup \equal{} 48\pi \minus{} 36\sqrt {3}$
The entire area: $ 96\pi \minus{} 72\sqrt {3}$

Now that we know the intersection, the area of the entire shape is $ 2*144\pi \minus{} (96\pi \minus{} 72\sqrt {3}) \equal{} 192\pi \plus{} 72\sqrt {3}$

You guys can correct my arithmetic if it's incorrect.[/hide][/quote]

Ah... I made a calculation error.. But that is what I did too..

There is an extremely similar problem to this in "Introduction to Geometry". I believe it was in "Funky Areas" section....


Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ a\sqrt{4a^2\plus{}5b^2}\plus{}b\sqrt{4b^2\plus{}5c^2}\plus{}c\sqrt{4c^2\plus{}5a^2}\geq(a\plus{}b\plus{}c)^2\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ a\sqrt {4a^2 \plus{} 5b^2} \plus{} b\sqrt {4b^2 \plus{} 5c^2} \plus{} c\sqrt {4c^2 \plus{} 5a^2}\geq(a \plus{} b \plus{} c)^2
\]
[/quote]
We have

$ 3LHS \equal{} \sum a\sqrt {(4 \plus{} 5)(4a^2 \plus{} 5b^2)}\ge 4(a^2 \plus{} b^2 \plus{} c^2) \plus{} 5(ab \plus{} bc \plus{} ca) \equal{} \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{2} \plus{} \frac {5}{2}(a \plus{} b \plus{} c)^2\ge 3(a \plus{} b \plus{} c)^2 \equal{} 3RHS$

So $ a\sqrt {4a^2 \plus{} 5b^2} \plus{} b\sqrt {4b^2 \plus{} 5c^2} \plus{} c\sqrt {4c^2 \plus{} 5a^2}\geq(a \plus{} b \plus{} c)^2$
 :D

Suppose that $ x;y;z>0$.Prove that:
$ x\sqrt{y^2\plus{}4z^2}\plus{}y\sqrt{z^2\plus{}4z^2}\plus{}z\sqrt{x^2\plus{}4y^2} \leq \frac{3}{4}(x\plus{}y\plus{}z)^2$

[color=green]Posted before by hungkhtn, Solved by squaring two sides.  :)  [/color]


I made some corrections to my original one

As I said before: I type in english, but I can read korean and japanese.


algebra
polynomial
geometry unsolved
geometry

[quote="epsilon07"]In a triangle ABC  given the equation :     (8|AB|-7|BC|-3|CA|)^2=6.(|AB|^2-|BC|^2-|CA|^2)
m(A)=?[/quote]

I would think somebody may able to solve this.... :(

[quote="castigioni"]If we solve in BC the result is a complex number,so something is wrong with the given equation[/quote]

result is 60 but I couldn't find it...


Let a,b,c>0.  Prove that
$ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{8abc}  \geq   (\frac{a\plus{}b\plus{}c}{\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}})^2$

[quote="tran300491"]Let a,b,c>0.  Prove that
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc} \geq (\frac {a \plus{} b \plus{} c}{\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}})^2$[/quote]
There is proof by AM-GM. :wink:

Let a,b,c>0 Prove that
$ \sum(\frac{a}{a\plus{}b})^3 \leq \frac{3}{8}(\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca})^2$

[quote="tran300491"]Let a,b,c>0 Prove that
$ \sum(\frac {a}{a \plus{} b})^3 \leq \frac {3}{8}(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca})^2$[/quote]
By  $ [(a\minus{}b)(b\minus{}c)(c\minus{}a)]^2 \geq 0$   we have done :rotfl:


For all positive integers $ n$, to prove that:

$ \left(\frac{2}{n+1}\right)^{1/2}+1\geq\left(n+1\right)^{1/n}$.

[quote="pavel kozlov"]It's easy and nice. If $n=1$ we have $2\geq 2$, if $n\geq 2$ by Newton's bynom we have:
$(1+a)^n\geq 1+\frac{n(n+1)}{2}a^2=1+n$, where ${a=\sqrt{\frac{2}{n+1}}}$ :arrow: $1+a\geq (n+1)^{\frac{1}{n}}$.[/quote]

I think your solution is incorrect.

$(1+a)^n\geq 1+\frac{n(n+1)}{2}a^2$ is not generally true (see n=2 and a=10).

By Newton's bynom is only true the following one:

$(1+a)^n\geq 1+\frac{n(n-1)}{2}a^2$ but it cannot help to our problem..

Yes, I have the same perplexity of Conceive.
Maybe is necessary to explain some steps of the demonstration.
'Bye  :)

[quote="fuzzylogic"][quote="conceive"]I think your solution is incorrect.

$(1+a)^n\geq 1+\frac{n(n+1)}{2}a^2$ is not generally true (see n=2 and a=10).
...[/quote]

It is true for $a= \sqrt{\frac{2}{n+1}}\le 1$ and $n\ge 2$.

$(1+a)^n\geq 1+na + \frac{n(n-1)}{2}a^2 \ge 1+na^2 + \frac{n(n-1)}{2}a^2 = 1+ \frac{n(n+1)}{2}a^2$[/quote]

I admit it. Sometimes it is suitable to write each step of the solution. Thanks fuzzylogic for clarifying.

Well, it's correct!
Thanks also by me  ;)


Prove that:[tex] (x^2+3)^2(y^2+3)^2(z^2+3)^2\geq\frac{4^6}{3^6}xyz(xy+yz+zx)^3[/tex] for all real numbers x,y,z.

It's here :
http://www.ddtoanhoc.net/forum/index.php?showtopic=1114
But at there , there're no solution .

PHEW!! At last I got the solution. But it's very very ugly because I expanded them all and use only Cauchy and Schur *_*

[quote="perfect_radio"][quote="harazi"]Bleah! What is this? First, it is sufficient to prove it for $x,y,z$ positive. Next, put by homogeneity $ xyz=xy+yz+zx$ and then $ x=1/a$ and so on. You need thus to prove that if $a+b+c=1$ then $ (1+3a^2)(1+3b^2)(1+3c^2)$ is at least $64/27$. This is not difficult and I am sure I've already solved it somewhere on the forum. Anyway, this seems like an inequality, not a monster![/quote]

i don't understand the steps made here

how can you put $ xyz=xy+yz+zx$ ?

and how do you get to the form :  if $a+b+c=1$ then $ (1+3a^2)(1+3b^2)(1+3c^2)$ is at least $64/27$.??[/quote]

I would like to ask the same question too..

Guys, I don't understand what you want. Everything follows from homogeneity. What should I explain?  :?

I think $x^2$ and $3$ are of different degree?

Uppps, Alzheimer! Indeed, I was idiot, sorry.  :blush:

After 2 days i find a  solution :)

It is obviously ok if we assume that x,y,z >0
Now put a=1/x ,b=1/y ,c =1/z

The first inequality reduces to :    :prod: ( 3a :^2:  + 1 ) :^2: >= 4^6 / 3^6 ( :Sigma: a):^3:   (*)
WLOG we assume that c<=b<=a

First case  : b<=1/3 ,so c<=b<= 1/3 .Then we have : ( 3b :^2:  - 1/3 )(3c :^2:  - 1/3 ) >= 0

 => ( 3b :^2:  +1 )( 3c :^2:  +1 ) >= 4( b:^2: +c :^2: ) + 8/9  >= 2(b+c):^2: +8/9
By Bunhiacopski inequaltiy we have :
 :prod:(3a  :^2:   +1) >= (3a :^2:  +2/3 +1/3 )(4/9 + 2(b+c):^2: + 4/9 ) >= ( 2/ :rt3: a + 2/ :rt3: (b+c) + 2/3 :rt3: ):^2:   = ( 2/ :rt3: (a+b+c) + 2/3 :rt3: ) :^2:

Put a+b+c = t

LHS(*) >= (2/ :rt3: t + 2/3 :rt3:  ) :^4: =A and by expanding all it's very easy to prove that A >= RHS(*)

The equality hold if and only if x=y=z=3

Second case : b>=1/3 ,so a>=b>=1/3 ,and we prove similiarly like first case

Really a nice proof, much better than expanding and using Shur as I did.


Prove that the polynomial $P(X)=X^n+4$ factors as a product of integer polynomials iff $n$ is divisible by 4.

If 4 divides $n$, all is clear. I am interested in the converse...

Suppose $x^n+4=f(x)g(x)$, where $f,g\in \mathbb{Z}[x]$, $f(x)=x^k+a_{k-1}x^{k-1}+...+a_1x+a_0$, $g(x)=x^l+b_{l-1}x^{l-1}+...+b_1x+b_0$.
Since all roots of $x^n+4$ have modulus $4^{1/n}$, from Viet theorem we conclude $|a_0|=4^{k/n}$ and $|b_0|=4^{l/n}$, but $a_0$ and $b_0$ are integer. It follows that $n$ is even and $k=l=n/2$. Moreover, $k$ must be even, otherwise $f(x)$ has real root and in the same time $x^n+4>0$ for all $x\in \mathbb{R}$! Thus, $k=n/2$ is even, so $4\mid n$.


linear algebra
matrix
linear algebra theorems

Please help!

Prove that a square matrix whose entries are all real has an LU-factorization if and only if the origin is not contained in the numerical range W(A), where W(A)={<Ax,x> such that <x,x>=1 for all x reals}.

I have tried considering the 2x2 matrix case but got stuck. :( Thanks!

Don't waste your time.

Consider $ L\equal{}\begin{bmatrix}1&0\\1&1\end{bmatrix}$, $ U\equal{}\begin{bmatrix}1&\minus{}1\\0&1\end{bmatrix}$, $ A\equal{}\begin{bmatrix}1&\minus{}1\\1&0\end{bmatrix}$, with $ x\equal{}\begin{bmatrix}0\\1\end{bmatrix}$.


Let $x_i,y_i$ are real numbers and $a,b$ are  positive numbers.

Assume that $\sum_{i=1}^n{x_i}^2=a^2-1,\sum_{i=1}^n{y_i}^2=b^2-1$.

Prove the following inequality \[\sum_{i=1}^nx_iy_i\leqq ab-1\]

Is there some mistakes?
The inequality follows immediately from cauchy...

Thank you for your reply, Darkseer.

Certainly, this problem is not difficult. Could I trouble you to send your solution?

kunny


1) For $a,b,c$ sides of a triangle such that $a+b+c =2$ prove that

$5(ab+bc+ca) \geq 4+9abc $

2) Prove the inequality

$\sum(s-a)\frac{c-b}{a} \geq 0$

where $a,b,c$ are elements of a triangle and $c\geq b \geq a >0$ and $s$ is the semiperimeter

1)
Putting $a=y+z$, $b=x+z$, $c=x+y$, and homogenizing, the ineq becomes (after some algebra)

\[
x^3+y^3+z^3 +3xyz \geq x^2y + x^2z + y^2x + y^2z + z^2x + z^2y
\]
which is just Schurs.

2) Using the same substitution as above. We have $0 < z \leq y \leq x$, and
\[
\sum_{\mathrm{cyc}}(s-a)\frac{c-b}{a} = \sum_{\mathrm{cyc}}x\frac{y-z}{y+z} = \frac{(x-y)(x-z)(y-z)(x+y+z)}{(x+y)(x+z)(y+z)}
\]
which is clearly non-negative.


Hi,
any one can help me on this problem:
with some explanation

calculate  using differenial  resistance R=(40 + -1 ) expenditure of P when U = 53 V ??

thanks :?

Well here better translation:
count the power P  with its defect limits spent by the resistor R =(40 + - 1) using the differential when voltage U=53 V ?

thanks


Let $ a,b,c\geq0$ s.t. $ abc\equal{}1$.Prove that:
$ a^2\plus{}b^2\plus{}c^2\plus{}9(ab\plus{}bc\plus{}ca)\geq10(a\plus{}b\plus{}c)$

[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ abc \equal{} 1$.Prove that:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 9(ab \plus{} bc \plus{} ca)\geq10(a \plus{} b \plus{} c)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80197


topology
geometry
probability
calculus
Support
set theory

Are there other good and well known math forums like AOPS and mathlinks but more college mathematics oriented????? I heard from someone today in class that there is a forum whose name begins with A but i can't remember. Can anyone help me???

I really doubt that you're going to find anything better.  While it would be nice if the college forums were a little more active, there's a lot of combined expertise between the regular posters.

As the subject is "Mathematics Forum", I suppose that it is not a problem if I give some examples of sites that are very useful for highschool. See these: http://www.mateforum.ro and http://www.mategl.com. They are written in Romanian  but some of the problems are really easy to understand.

Despite the name, [url]http://www.physicsforums.com/[/url] has seven forums devoted to discussion of advanced mathematics such as a "Topology & Geometry" forum, a "Set Theory, Logic, Probability, Statistics" forum, etc. Also, there are two forums devoted to math homework help, one of which, "Calculus & Beyond", is essentially just for college students. Also, there is a 2087-post thread in the "Academic Guidance" section called "who wants to be a mathematician?" where you can ask questions about being a professional mathematician (although many of the posts in that thread are just some professor rambling about abstract algebra).

[quote="Scott H"]I would love to find a forum similar to sci.math that supports TeX font.[/quote]

How about the one you are posting on just now?


The inscribed circle of the triangle $A_1A_2A_3$ touches the sides $A_2A_3, A_3A_1, A_1A_2$ at points $S_1,S_2,S_3$ respectively. Let $O_1,O_2,O_3$ be the centers of the inscribed circles of triangles $A_1S_2S_3, A_2S_3S_1, A_3S_1S_2$ respectively. Prove that the straight lines $O_1S_1, O_2S_2, O_3S_3$ intersect at one point.

it is just the same as the incenter theorem.
TALE?of course i think not.


geometry
3D geometry
tetrahedron
trigonometry
geometry solved

Let $ABCD$ be a tetrahedron in which the opposite sides are equal and form  equal angles.

Prove that it is regular.

it looks very much like the law of cosines

i tried to prove it using the law of cosines for some triangles, but i got stuck.

You can prove it by using the scalar product, I believe.

i still can't prove it.

$39$-$45$ World War II

[quote="Sailor"]Denote by $t$ the angle between $AC$ and $BD$ then prove that
${\cos{t}}={\frac{AD^2+BC^2-AB^2-CD^2}{2\cdot{AC}\cdot{BD}}}$ and from here you can get the result![/quote]
We have \[AD^2+BC^2-AB^2-CD^2=(\overrightarrow{AD}-\overrightarrow{AB})(\overrightarrow{AD}+\overrightarrow{AB})+(\overrightarrow{CB}-\overrightarrow{CD})(\overrightarrow{CB}+\overrightarrow{CD})=\] \[=\overrightarrow{BD}(\overrightarrow{AD}+\overrightarrow{AB})-\overrightarrow{BD}(2\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AB})=-2\overrightarrow{AC}\cdot \overrightarrow{BD}.\]
That's all.


hi,
calculate all distinct solutions in natural numbers of the equation
\[ 1 \plus{} x_{1} \plus{} 2x_{1}x_{2} \plus{} ... \plus{} (n\minus{}1)x_{1}x_{2} ... x_{n\minus{}1} \equal{} x_{1}x_{2} ... x_{n}.\]

There is no nice explicit description of the solution space. However, it admits a simple recursive construction:

For $ n\in\mathbb{N}\,,\,\bar{x}: \equal{}(x_1,\ldots,x_n)\in\mathbb{N}^n$ define $ F_n(\bar{x})\equal{}1\minus{}\prod_{i\equal{}1}^nx_i\plus{}\sum_{k\equal{}1}^{n\minus{}1}\left[k\,\prod_{i\equal{}1}^kx_i\right]$ and $ S_n\equal{}\{\bar{x}\in\mathbb{N}^n\,|\,F_n(\bar{x})\equal{}0\}$. Then we have:

(1) $ S_1\equal{}\{(1)\},S_2\equal{}\{(1,2)\}$
(2) $ \forall n>1: \,S_{n\plus{}1}\equal{}\left\{(x_1,\ldots,x_{n\minus{}1},d,\frac{x_n}{d}\plus{}n)\,\Big |\,(x_1,\ldots,x_n)\in S_n\,,\,d|n\right\}$.

[i]Proof[/i]: Clearly $ S_1\equal{}\{(1)\}, S_2\equal{}\{(1,2)\}$ hold. For $ \bar{x}: \equal{}(x_1,\ldots,x_{n\plus{}1})$ observe the identity $ F_{n\plus{}1}(\bar{x})\equal{}F_{n}((x_1,\ldots,x_{n\minus{}1},x_n(x_{n\plus{}1}\minus{}n))$, which immediately verifies the claim.

Example: $ S_4\equal{}\{(1,1,1,7),(1,1,2,5),(1,1,4,4),(1,2,1,6),(1,2,3,4)\}$


linear algebra
matrix
algebra
polynomial
abstract algebra
inequalities
least common multiple

is there a way to characterize all matrices s.t. A^2 = A?

If you want to look this up somewhere, the word you are looking for is [b]idempotent.[/b]

For such a matrix, the minimal polyomial is $x^2-x.$ (That's pretty obvious, when you think about it.) The characteristic polynomial is $(x-1)^jx^k.$ Such a matrix is diagonalizable, and is similar to a diagonal matrix that has $j$ ones on the diagonal followed by $k$ zeros.

[quote="Kent Merryfield"]For such a matrix, the minimal polyomial is $x^2-x.$ (That's pretty obvious, when you think about it.) The characteristic polynomial is $(x-1)^jx^k.$ Such a matrix is diagonalizable, and is similar to a diagonal matrix that has $j$ ones on the diagonal followed by $k$ zeros.[/quote]

Just in case, the minimal polynomial is a divisor of $x^2-x$, but not necessarely that polynomial. Remember, the zero and the identity matrices are both idempotent, but they are the only to exceptions to the rule given by Dr. Merryfield.

[quote="polymorphic"]Thanks for the responses.  sorry if this is a stupid question, i don't know my linear algebra very well, but why are all the idempotent matrices diagonalizable?[/quote]

There is no need for an apology, nobody was born having learned anything.

Well, there is a theorem that says that a Matrix is diagonalizable over $\mathbb{R}$ if and only if its minimal polynomial has all its roots real, and well, it is true in the case of an idempotent matrix.

Now, Moubinool gives another name, Projection Matrices, what is another good way to look at such matrices. Note that in this case, if $Range A=W$, then $Null A=W^\perp$, and therefore, give an orthonormal base for $W$, and complete it with an orthonormal base for $W^\perp$, and then, let $B$ be the matrix of change of basis from the canonical to the one you just got, then, $BAB^{-1}$ is diagonal

[quote="djimenez"]Well, there is a theorem that says that a Matrix is diagonalizable over $\mathbb{R}$ if and only if its minimal polynomial has all its roots real[/quote]
Not exactly. The appropriate theorem: A matrix is diagonalizable over $F$ if and only if its minimal polynomial splits over $F$ as a product of distinct linear factors. Since $x^2-x=x(x-1)$, this applies to idempotent matrices.
Proof: If $M$ is diagonalizable, it has the same minimal polynomial as a diagonal matrix $D$ that it is similar to. The minimal polynomial of $D$ is clearly the least common multiple of $x-r_i$, where the $r_i$ are the diagonal elements, and this is a product of distinct linear factors.
Conversely, assume $(M-r_1I)(M-r_2I)\cdots(M-r_kI)=0$ for some $n\times n$ matrix $M.$ For any $A$, define $N(A)$ to be the dimension of the kernel of $A$. Under this definition, $N(AB)\le N(A)+N(B).$ Applying this inequality, $N(M-r_1I)+N(M-r_2I)+\cdots+N(M-r_kI)\ge N(0)=n.$ Since any collection of eigenvectors for different eigenvalues must be linearly independent, we can choose a basis form among these kernels, and the matrix is diagonal with respect to that basis.


logarithms
trigonometry
limit
calculus
calculus computations

yes lim $\frac{f(x)}{g(x)}$ = lim $\frac{f'(x)}{g'(x)}$
give other solution

L'Hospital rule doesn't belong to this forum... ;)

[quote="Fermat -Euler"]Find
$lim_{h \to 0} \frac{Ln(sinah)}{Ln(sinbh)}$
a,b are reel  ;)[/quote]

Wrong forum  ;)

[hide]$\frac{d}{dh} \ln{(\sin{(ah)})} = a \times \cos{ah} \times \frac{1}{\sin{ah}} = a cot{(ah)}$. Applying the rule of the hospital, we have:
\[ lim_{h \to 0} \frac{Ln(sinah)}{Ln(sinbh)} = lim_{h \to 0} \frac{a}{b} \times \frac{\cot{ah}}{\cot{bh}} = lim_{h \to 0} \frac{a}{b} \times \frac{\tan{bh}}{\tan{ah}}  \]

Now we know: $\frac{d}{dh} \tan{ah} = \frac{a}{(ah)^2 + 1}$. So applying the rule again and simplifying we obtain: \[ lim_{h \to 0} \frac{(ah)^2 + 1}{(bh)^2 + 1}  \]. Since both the top and bottom limits are 1, this limit is 1 (heh what else could you expect...I wish there would be some limits with interesting answers).[/hide]

:P  :P
 i want other solution whitout using hopital

[quote="Virgil Nicula"][size=134]$a>0,\ b>0,\ a\ne b\Longrightarrow \lim_{x\to 0}\frac{\sin ax}{\sin bx}=\lim_{x\to 0}\frac{\sin ax}{ax}\cdot \frac{ax}{bx}\cdot \frac{bx}{\sin bx}=\frac ab\ .$

$f(x)=\frac{\ln \sin ax}{\ln \sin bx}=1+\frac{\ln \frac{\sin ax}{\sin bx}}{\ln \sin bx}$$\Longrightarrow \lim_{x\searrow 0} f(x)=1-\frac{\ln \frac ab}{\infty}=1\ .$

[u]Otherwise:[/u] $\lim_{x\searrow 0}f(x)=\lim_{x\searrow 0} \frac{\ln \frac{\sin ax}{ ax}+\ln a+\ln x}{\ln \frac{\sin bx}{bx}+\ln b+\ln x}=\lim_{x\searrow 0}\frac{1+\frac{\ln a}{\ln x}+\frac{\ln \frac{\sin ax}{ax}}{\ln x}}{1+\frac{\ln b}{\ln x}+\frac{\ln \frac{\sin bx}{bx}}{\ln x}}=1\ .$[/size][/quote][b]Another exercise:[/b]
$m,n\in Z^*\Longrightarrow \lim_{x\to \frac{\pi}{2}} \frac{\ln \{\sin [(4m+1)x]\}}{\ln \{\sin [(4n+1)x]\}}=\left(\frac{4m+1}{4n+1}\right)^2$
([u]without l'Hospital's rules[/u]).


Find all real parameters $p$ for which the equation:

$x^{3}-2p(p+1)x^{2}+(p^{4}+4p^{3}-1)x-3p^{3}=0$

has three distinct real roots which are sides of a right triangle.

[quote="gotztahbeazn"]errr i ono if this would be it.. but yea

[hide]

let the roots be $m, n, \sqrt{m^{2}+n^{2}}$

the sum of the squares of the roots is therefore $2(m^{2}+n^{2})$

now, we find the square of the roots of the polynomial in terms of $p$

hence we have:

$[(2p)(p+1)]^{2}-2(p^{4}+4p^{3}-1) = 2(p^{2}+1)^{2}$

hence the parameter for $p$ is

$2(p^{2}+1)^{2}= 2(m^{2}+n^{2}) \implies (p^{2}+1)^{2}= m^{2}+n^{2}$

and is that it? im not sure what is meant by "parameters"

[/hide][/quote]

[hide]I did the same thing as you did.  Now take the square root of both sides of that last equation to get $p^{2}+1=\sqrt{m^{2}+n^{2}}$, so $p^{2}+1$ is a root.  Plugging $p^{2}+1$ in for $x$ eventually gives $p=0,\pm\sqrt{2}$ (it simplifies nicely).  We still have to plug in these possible values for $p$ to make sure they work.  For example, if $p=0$, the roots are $-1,0,1$, which satisfy $a^{2}+b^{2}=c^{2}$ but can't be the sides of a right triangle.  Now try $p=\sqrt{2}$ and $p=-\sqrt{2}$.  I haven't done that yet, but you can use the fact that $p^{2}+1=3$ is a root.[/hide]

[quote="matt276eagles"][hide="Hint"]Find the sum of the squares of the roots in terms of $p$.[/hide][/quote]

This is just a lovely hint.  :clap2:


Prove or disprove that for positive nonzero numbers,the following inequality holds:
$\ 2[1;1;1]+[3;00]\geq 3[2;1;0]$
An equivalent enunciation:Prove or disprove that for positive nonzero numbers a,b,c,the following ineq holds:
$12abc+2(a^{3}+b^{3}+c^{3})\geq3(a^{2}b+b^{2}a+a^{2}c+c^{2}a+b^{2}c+c^{2}b)$

[quote="andyciup"]Prove or disprove that for positive nonzero numbers,the following inequality holds:
$\ 2[1;1;1]+[3;00]\geq 3[2;1;0]$[/quote]

Did u mean cyclic sum?

http://www.mathlinks.ro/Forum/viewtopic.php?t=103773

Anyway $[3,0,0]+[1,1,1]\ge(2,1,0)$? That is just Schur....

As he seemed to state it (all sums symmetric), it fails.
$a=b=2$ $c=1$

But as a symmetric he should write that

$2(1;1;1)+(3;0;0)\geq 3(2;1;0)$

[quote="Xevarion"]Since there seems to be no consensus about that yet, I went by

[quote="andyciup"]Simetric![/quote][/quote]

At least he should write [i]symmetric[/i], instead of [i]simetric[/i].

Do you want to show that it is stronger than schur?
But the answer is no.Because schur is strongest as the coefficient changing.
Try $a=b,c \to 0$.

I am really sorry for the misunderstanding.I have edited my first post and gave a version of the inequality without the missleading brackets :)


Show that if \[ \phi (x) \equal{} n\] has one solution then it has a second solution.

This problem is [url=http://mathworld.wolfram.com/CarmichaelsTotientFunctionConjecture.html]currently open[/url].


How do I graph a radical equation?

For example:

Graph:  sqrt{4-x} -8 = y

that is not graphable...there is only one variable

$\sqrt{4-x}=8+x$
$x^2+17x+60=0$
$\frac{-17\pm\sqrt{289-240}}{2}$
$x=-5,-12$.
Two lines.  :)

If you meant something like $\sqrt{4-x}-8=y$...
First take the graph of $\sqrt{-x}=y$.  Move the graph 4 to the right to make it $\sqrt{4-x}=y$.  Move the graph 8 down to make it $\sqrt{4-x}-8=y$.

you should know what a regular square root looks like i.e. $y = \sqrt{x}$
then perform the transformations, flip over y axis, translate right 4, down 8


S  is the set of all points with coordinates  (m,n)  such that  m  and  n  are positive integers with  m < 6  and  n < 6 . Two points from  S  are chosen at random. What is the probability that the midpoint of the segment of the two points is also in the  S ? Give your answer as a common fraction in lowest term.

Answer key says 6/25, but I got 1, by thinking that set S is a right isosceles triangle with leg length 5 and the vertex at the base.  If the points are in the triangle, then how can the segment connecting them's midpoint be out of the triangle!?  It's a 45-45-90!

Oh, I was thinking that their sum had to be less than 6.  But if two points are in a square, how the midpoint of the segment connecting them be out of the square?  How does one find points like these?

The problem says (m,n) are [i]integers[/i].  So, just because the midpoint's in the bounded region doesn't mean it's in set S.

it is a square, but still if two points are in the square the midpoint is too.

Yes, it's still in the square, but not in set S.  S consists of positive  integers only.  The midpoint isn't necessarily integer coordinates.

thanks, now I can stop driving myself crazy about the problem.
Wait a second, I now get 1/4. Only 1/25 off, but still wrong

[hide="solution"]
\[\begin{array}{ccccc}\bullet & \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet & \bullet \end{array}\]
So we are finding how many segments have midpoints on this grid:

And there are
Horizontal/vertical: $2 \cdot 4 \cdot 5$
+
Diagonal: $2(1+2+4+2+1)$
+
Slanted diagonal: $6 \cdot 2$
=
$72$ such segments.

There are a total of $\binom{25}{2}$.

So the probability is $\frac{72}{\binom{25}{2}}=\boxed{\frac{6}{25}}$.[/hide]

I just listed them out.
[hide="solution"] There are 25C2 = 300 ways to choose 2 points from the set.
We can start by counting the 3-dot lines which are at 45 degree angles.  There are 18 of these.   Also, we count the 3-dot horizontal and vertical lines, of which there are 30.  Also, we count the big diagonals.  There are 2.  We also count the 5-dot horizontal and vertical lines, of which there are 10.  And finally, we count 12 more lines which are the hypotenuses of right triangles with legs 2 and 4.  In total, we have 72 lines.
72/300 = 6/25.[/hide]

It's 5x5 because 0 is not a positive integer.

[quote="usaha"]Wait, why is it a 5X5 grid?[/quote]

It is because we are considering all lattice points bound by $0<m<6$ and $0<n<6$. Then $m=1,2,3,4,5$ and $n=1,2,3,4,5$ possible. So it is actually a $5 \times 5$ grid.

i drew a 4X4 grid and got 1/5...i thought i said they were less than 5...


induction calculus derivative function algebra polynomial integration

I am trying to show that
$ \frac{d^n}{dx^n} (x^2\minus{}1)^n\equal{}0$
has n distinct real roots lying between -1 and 1. I am guessing it needs induction and Rolle's theorem.
Could you help?

Well Rolle's theorem states that in the closed interval $ [a,b]$ with $ f(a) \equal{} f(b)$ there exists a c such that $ f'(c) \equal{} 0$. Notice f(-1) = f(1). So, if we take the derivative of $ (x^{2} \minus{} 1)^{n}$ we get $ 2xn(x^{2} \minus{} 1)^{n \minus{} 1} \equal{} 0$ so the first derivative has an additional root at $ x \equal{} 0$. Again, we have f(-1) = f(1) so you can just repeatedly use Rolle's theorem and maybe find a pattern (ex: roots increase by 1 for each additional derivative).

My other idea is to expand (x^{2}-1)^{n} using the binomial theorem, then take the derivative of each term. i.e.

$ ((x^{2} \minus{} 1)^{n})' \equal{} \bigg({n \choose 1}x^{2n} \plus{} {n \choose 2}x^{2n \minus{} 1} \plus{} {n \choose 3}x^{2n \minus{} 2} \plus{} \cdots \plus{} {n \choose n}\bigg)'$. After finding the derivative I guess maybe you could repeatedly factor out x to some power, giving you your n roots.

One important fact is that if $ f(x)$ has a zero of degree $ k$ at $ a,$, then $ f'(x)$ has a zero of degree $ k\minus{}1$ at $ a.$

This is because we can write $ f(x)\equal{}(x\minus{}a)^kg(x),$ where $ g(a)\ne0.$ Then $ f'(x)\equal{}k(x\minus{}a)^{k\minus{}1}g(x)\plus{}(x\minus{}a)^kg'(x)\equal{}
(x\minus{}a)^{k\minus{}1}[kg(x)\plus{}(x\minus{}a)g'(x)].$ And note that the function in the brackets equals $ kg(a)\ne0$ at $ x\equal{}a.$

What we do is to use Rolle's theorem repeatedly.

Let $ f(x)\equal{}(x^2\minus{}1)^n.$ Note that $ f(x)$ has zeros of degree $ n$ at both $ \minus{}1$ and $ 1.$

By Rolle, $ f'(x)$ has (at least) one zero in the open interval $ (\minus{}1,1).$ And $ f'(x)$ still has zeros of degree $ n\minus{}1$ at both $ \minus{}1$ and $ 1.$

Then by Rolle again, $ f''(x)$ has at least two zeros in the open interval $ (\minus{}1,1):$ one between $ \minus{}1$ and the zero of $ f'(x)$ that we found previously, and one between that zero of $ f'(x)$ and $ 1.$ And $ f''(x)$ still has zeros of degree $ n\minus{}2$ at both $ \minus{}1$ and $ 1.$

We proceed inductively. We find that for $ 0<k<n,$ $ f^{[k]}(x)$ has (at least) $ k$ distinct zeros in the open interval $ (\minus{}1,1)$ and also zeros of degree $ n\minus{}k$ at both $ \minus{}1$ and $ 1.$ That gives us (at least) $ k\plus{}1$ intervals between successive zeros. By Rolle, each of those open intervals must contain a zero of $ f^{[k\plus{}1]}(x),$ and these $ k\plus{}1$ zeros are all distinct because the open intervals are disjoint. At the same time, if $ k\plus{}1<n,$ we still have $ f^{[k\plus{}1]}(x)$ having zeros of degree $ n\minus{}k\minus{}1$ at both $ \minus{}1$ and $ 1.$

The only difference for this argument in passing from $ f^{[n\minus{}1]}(x)$ to $ f^{[n]}(x)$ is that the zeros at $ \minus{}1$ and $ 1$ finally disappear altogether. The rest of the argument still works, and gives us that $ f^{[n]}(x)$ has at least $ n$ distinct zeros in $ (\minus{}1,1).$ However, $ f^{[n]}(x)$ is the $ n$th derivative of a polynomial of degree $ 2n,$ and is thus a polynomial of degree exactly $ n.$ It cannot have more than $ n$ zeros, and hence it has exactly $ n$ distinct zeros, all of them in $ (\minus{}1,1).$

----

Multiply by some conventionally assigned normalizing constant, and these $ n$th degree polynomials are the well-known Legendre polynomials. Call the resulting polynomial $ P_n(x).$

Fact: If $ g(x)$ is a polynomial of degree $ \le n\minus{}1,$ then

$ \int_{\minus{}1}^1P_n(x)g(x)\,dx\equal{}0.$

This can be proved using the definition of $ P_n$ given in this problem and an $ n$-fold integration by parts.

Now, the $ n$ zeros that $ P_n(x)$ has in $ (\minus{}1,1)$ become quite important objects. As roots of polynomials, they can be efficiently approximated using Newton's method. Extensive tables of these roots exist - primarily because evaluation at precisely those points lies at the heard of the efficient and accurate method of numerical integration known as Gaussian Quadrature.

----

Gaussian Quadrature on $ [\minus{}1,1]$ (with a "flat" weight):

Let $ P_n(x)\equal{}c_n\left(\frac{d}{dx}\right)^n(x^2\minus{}1)^n.$ Let $ \xi_{n,k},\ 1\le k\le n$ be the $ n$ zeros of $ P_n(x)$ in $ (\minus{}1,1).$ Find coefficients $ a_{n,k}$ such that for $ 0\le j\le n\minus{}1,$

$ \int_{\minus{}1}^1x^j\,dx\equal{}\sum_{k\equal{}1}^na_{n,k}\xi_{n,k}^j.$

(This is linear algebra: solving $ n$ linear equations for $ n$ unknowns.)

For smooth functions $ g$ on $ [\minus{}1,1],$ define

$ I_n(g)\equal{}\sum_{k\equal{}1}^na_{n,k}g(\xi_{n,k}).$

Then if $ g$ is a polynomial of degree $ \le 2n\minus{}1,$ we have that

$ I_n(g)\equal{}\int_{\minus{}1}^1g(x)\,dx$

And in more general cases, the difference between $ I_n(g)$ and $ \int_{\minus{}1}^1g(x)\,dx$ depends on the size of the $ 2n$th derivative of $ g.$

I think you can use induction and the product rule.


ratio limit function algebra polynomial LaTeX calculus

Hey guys I have some questions about limits at infinite or infinite limits. (what is the difference?)

1) lim  1/ (x^2)(x+7)
   x>0

I found the answer to be positive infinite because its just logical, 1 over 0 x 0 is 1 over 0 and 1/0 is infinite. But what is the correct mathematical way to find the answer?

2) Lim         (9-x) (2+11x) / (3-11x) (9+9x)
    x> + inf

What I did here is I multiplied everything out and got the degree in the numerator = to the degree of the denominator (x^2). Then I just took the ratio of those coefficients and I found the answer to be -1/9. Is this the right way to do it?

3) Lim              5 / (e^x-7)
    x> infinity

Lim            5/ (e^x-7)
x> -infinity
First one, I know that e to infinity is infinity so 5/ infinity - 7 is 5/infinity which is 0.
Second one, I know that e raised to -infinity is 0, so 5/0 -7 which is -5/7.

Again, what are the correct ways to do them?

4) lim           sqrt (x^2 + 7x+1) - x
   x>-infinity
i know that -infinity^2= positive infinity so square root of positive infinity is positive infinity minus -infinity is + infinity. So positive inf + inf = positive inf.

5) Lim    (-18x^2 - 29x^3)
   x> - inf
This one I'm just stuck..

6)   Lim       sqrt (x^2 + 3x) / (4-4x)
    x> -inf

Here I divided everything by x and I got -1/4 but apparently my answer is incorrect?

Thanks in advance for all the help guys! I'm nearing my exam so I would really appreciate any help I can get.

What do you mean correct mathematical way?  Do you want an $ \epsilon$-$ \delta$ argument?  That would require you explicitly find a $ \delta$, but that's not going to be too fun.

If I was grading, I would accept a response that uses the fact that $ \frac{c}{x^{n}}\rightarrow\infty$ as $ x\rightarrow 0$ and then decompose the limit in terms of these known ones.  So for example the first one, if you could do a partial fraction decomposition and write it as a sum of some constant times $ x^{\minus{}1}$, $ x^{\minus{}2}$, and $ (x\plus{}7)^{\minus{}1}$, I think that would be fine.

[quote="JRav"]What do you mean correct mathematical way?  Do you want an $ \epsilon$-$ \delta$ argument?  That would require you explicitly find a $ \delta$, but that's not going to be too fun.

If I was grading, I would accept a response that uses the fact that $ \frac {c}{x^{n}}\rightarrow\infty$ as $ x\rightarrow 0$ and then decompose the limit in terms of these known ones.  So for example the first one, if you could do a partial fraction decomposition and write it as a sum of some constant times $ x^{ \minus{} 1}$, $ x^{ \minus{} 2}$, and $ (x \plus{} 7)^{ \minus{} 1}$, I think that would be fine.[/quote]

By correct mathematical way I mean like in a test in order to get all the points you need to show all the steps right? What are those steps for each one is what I was asking.

I don't know how your TA would grade, but if I were grading, I would just want to see evidence that you know what you're doing.  If you do a partial fraction decomposition and show that every term goes to 0, for example, I would accept that as perfectly valid.  I seriously doubt that you would be expected to do an $ \epsilon$-$ \delta$ proof-or have you done in these class?

I'm looking at your third problem now.  Aside from some sloppy wording, the logic is perfectly fine.  What I mean is that I would not say $ \frac{5}{\infty\minus{}7}\equal{}\frac{5}{\infty}$, I would say something along the lines of: as $ x$ gets arbitrarily large, $ e^{x}$ is large as well, so $ e^{x}\minus{}5$ is large.

Here's the answer to the fifth one:

$ \lim_{x \to \minus{} \infty} \minus{} 18x^2 \minus{} 29x^3 \equal{} \lim_{x \to \minus{} \infty}x^2( \minus{} 29x \minus{} 18)$

Both $ x^2$ and $ \minus{} 29x \minus{} 18$ grow arbitrarily large as $ x \to \minus{} \infty$, so their product grows arbitrarily large, and the answer is $ \infty$.

[quote="helpmeplz"]
1) lim  1/[(x^2)(x + 7)]
   x>0

2) Lim         (9 - x)(2 + 11x)/[(3 - 11x)(9 + 9x)]
    x> + inf

3) Lim              5/(e^x - 7)
    x> infinity

Lim            5/(e^x - 7)
x> -infinity

4) lim           sqrt(x^2 + 7x + 1) - x
   x>-infinity

5) Lim    (-18x^2 - 29x^3)
   x> - inf

6)   Lim       sqrt(x^2 + 3x)/(4 - 4x)
    x> -inf

[/quote]

helpmeplz,

when you're typing in horizontal style, you should be observing some
of the following:

1) Place outer grouping symbols** around products in the denominators
as I have shown above in the quote box as a must because of the
order of operations.

2) Have your parentheses up against each other with no spaces, have
them up against the "/" with no spaces, and have the function
(e.g. "sqrt") up against the open parenthesis without a space.

3) Also, be consistent with the spacing of your plus and minus signs
within and between your polynomial terms and other different types
of terms.  I recommend the spacing as in the above quote box for
improved readability.

** other sets of parentheses, square brackets, and/or braces
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=--=-=-=-=-=-=-=-=-=-=

These steps are what you must and/or should do, unless you use the
Latex system, and much of that is geared for the vertical style.


Is the USAMTS website down? I tried logging in with my username and password but it told me that my password was incorrect. So I reset my password, and tried logging in again - using the new password that was e-mailed to me - this time the page could not be found.

Just wondering if it's something with my account or with the site in general.

you can log in technically, but the MyScores page doesn't work.  and it directs you there when you log in, so you should get a Not Found error.

I think this is fixed now; let me know if any other pages are missing on the USAMTS site.


wanted to know if theres an elegant way to solve this

tan(x+100) = tan(x+50)*tanx*tan(x-50)

for least possible x. all measures in degrees.

thnx in advance.


geometry
circumcircle
trigonometry
parallelogram
geometry unsolved

[quote="limes123"]Let $ D$ be a point on side $ AC$ of triangle $ ABC$ with $ AC \equal{} BC$, and $ E$ be a point on the segment $ BD$. Prove that $ \angle EDC \equal{} \angle CED$ if $ BD \equal{} 2AD \equal{} 4BE$.[/quote]
I think your problem has a mistake.

[quote="limes123"]Let $ D$ be a point on side $ AC$ of triangle $ ABC$ with $ AC \equal{} BC$, and $ E$ be a point on the segment $ BD$. Prove that $ \angle EDC \equal{} 2\angle CED$ if $ BD \equal{} 2AD \equal{} 4BE$.[/quote]
Draw a line l which passes through D and parallel to AB, a line m which passes through B and parallel to $ AC,l\cap BC\equal{}\{K\}. m\cap l\equal{}\{L\}$
We have $ \angle BKL\equal{}\angle CKD\equal{}\angle CDK\equal{}\angle CAB (1)$
But $ \frac{BE}{AD}\equal{}\frac{1}{2}\equal{}\frac{AD}{DB}\equal{}\frac{BL}{BD}$ and $ \angle ADB\equal{}\angle DBL$
so $ \Delta ADB$~$ \Delta EBL$ then $ \angle DAB\equal{}\angle EBL (2)$
From (1),(2) we obtain $ \angle LEB\equal{}\angle BKL \Rightarrow$ quadrilateral BEKL is cyclic.
$ \Rightarrow \angle DEK\equal{}\angle DLB\equal{}\angle CDK.$
Draw the circumcircle (W) of triangle DKE then CD,CK are two tangents of (W).
Let $ CE\cap (W)\equal{}\{I\}$
To have $ \angle EDC\equal{}2\angle CED$ we need to prove DI is the bisector of $ \angle CDE.$
If and only if $ \frac{CI}{CE}\equal{}\frac{CD}{CD\plus{}DE}$
But $ \frac{CI}{CE}.CE^2\equal{}CI.CE\equal{}CD^2$ then we need to prove $ \frac{CD}{CD\plus{}DE}\equal{}\frac{CD^2}{CE^2}$
$ \Leftrightarrow CD^2\plus{}CD.DE\equal{}CE^2 (1)$
Moreover, applying the law of cosin for the triangle CDE we have $ CE^2\equal{}CD^2\plus{}DE^2\minus{}2CD.ED.cos \angle CDE$
So $ (1)\Leftrightarrow CD.DE\equal{}DE^2\minus{}2CD.ED.cos \angle CDE$
$ \Leftrightarrow \frac{DE}{CD}\equal{}1\plus{}2cos\angle ADB (2)$
Applying the law of cosin again for the triangle ADB we get $ AB^2\equal{}5AD^2\minus{}4AD^2.cos\angle ADB$
So $ (2)\Leftrightarrow 1\plus{}2cos\angle ADB\equal{}1\plus{}\frac{AB^2\minus{}5AD^2}{2AD^2}\equal{}\frac{AB^2\minus{}3AD^2}{2AD^2}\equal{}\frac{3AD}{2DC}$
$ \Leftrightarrow \frac{AD}{DC}\equal{}\frac{AB^2\minus{}3AD^2}{3AD^2} (3)$
On the other hand, $ DK.AB\equal{}DK.DL\equal{}DE.DB$
$ \Rightarrow \frac{CD}{AC}\equal{}\frac{DK}{AB}\equal{}\frac{3AD^2}{AB^2}$
$ \Rightarrow \frac{CD}{AD}\equal{}\frac{3AD^2}{AB^2\minus{}3AD^2}$
Therefore (3) is true.
->QED

Here is another proof for this problem without using trigonometry :D
Similar proof 1, we have KLEB is a cyclic quadrilateral.
Let I be the intersection of CE and $ (DKE)$. DI cut BL at P.
Using Miquel's theorem we get $ IKLP$ is a cyclic quadrilateral.
Then $ \frac {ID}{DL} \equal{} \frac {IK}{LP}$
$ \frac {DE}{DL} \equal{} \frac {EK}{BL}$
$ \Rightarrow \frac {ID}{DE} \equal{} \frac {IK}{LP}.\frac {BL}{EK}$
But $ (DKIE) \equal{} \minus{} 1$ then $ \frac {DI}{DE} \equal{} \frac {KI}{KE}$
So $ LB \equal{} LP \equal{} AD$
Thus $ ADPL$ is a parallelogram $ \Rightarrow DI//AG\Rightarrow \angle CDI \equal{} \angle DAG \equal{} \angle DEI \equal{} 1/2\angle CDE$
->QED


analytic geometry
graphing lines
slope
greatest common divisor
geometry
rectangle
number theory

A line passes through $ A(1,1)$ and $ B(100,1000)$. How many other points with integer coordinates are on the line and strictly between $ A$ and $ B$?

$ \textbf{(A)}\ 0\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 9$

[hide]
This line has slope $\frac{999}{99}=\frac{111}{11}$
This is the most reduced form so both will be integers for every 11 values of x.
It would take 9 repetitions of 11 to reach 100 from 1 but the last one doesnt count since it is not between the points but rather is one of the points so there are 8. $\boxed{D}$[/hide]

[hide="Here is how I would do in this type"]
If you find the equation, you get $y = \frac{111x - 100}{11}$. So, you basically want $11|111x-100$, right? This can be rewritten as:

$111x-100 \equiv 0 \mod 11$
$111x \equiv 100 \mod 11$
$111x \equiv 111 \mod 11$
$x \equiv 1 \mod 11$

So each $x$ should be in point $1+11n$ where $n$ is a integer. [/hide]

[hide]We find the slope of the line to be $\frac{111}{11}$, and every time $x$ value increases by $11$, we always get an integer $y$ value, so the values are: $12,23,34,45,56,67,78,89,100$, and so there are $\boxed{8}$[/hide]

EDIT: Forgot it was $\text{between}$ the points

[quote="JesusFreak197"][hide="Note for this type of problem"]The number of integer points other than the endpoints is $GCF(x_2-x_1,y_2-y_1)-1$.[/hide][/quote]Sorry for the revival, but this is a great fact which I've never heard of! Does anyone have a proof of this? Thanks in advance! :)

There's a variant of this problem that pops up frequently, too - the question is if you have a $100 \times 1000$ cake sliced into square units, and you put one diagonal strip of red icing on the cake from the lower left to the lower right, how many square slices of cake get red icing?

I don't have the time right now to go through a rigorous proof, but the proof of that and the proof of "how many lattice points does the icing hit" are very similar.  I think I did it one time here a LONG time ago (we're talking months, if not years) - it basically states that if you go from $(0,0)$ to $(p,q)$, with $p$ and $q$ relatively prime, you will NEVER hit a lattice point on the interior.  You only get the two exterior lattice points.

Thus, if you go from $(0,0)$ to $(kp, kq)$, with $p, q$ relatively prime, then $k$ is the GCF of $kp$ and $kq$, and you hit an interior lattice point $k-1$ times - you repeat $p\times q$ rectangles, with the lattice points only coming there.

To get the number of slices of cake that get hit, you just examine each relatively prime rectangle, and note that there are $p + q$ slices that get hit, and you subtract off the endpoints.

[hide]We find the expression of the line is $111x-11y=100$. The integers from $1-100$ exclusive that $\equiv 1\pmod {11}$ is $\boxed 8$[/hide]


Let the base b representation of x_n be 111.....12.....25 where there are n 1's and (n+1) 2's. Prove that b = 10 if and only if the following property is true:

There exists an integer M such that for any integers n > M, x_n is a perfect square.

And it's been posted before on the forum, but I can't remember if it was solved.

see here http://www.mathlinks.ro/Forum/viewtopic.php?p=19129#p19129 . Locked!


The number of the integers a, for which the equation $x^{3}+(a-4)x^{2}+(a+4)x+9=0$ has exactly 2 different integer roots is: A)1, B)2, C)3, D)4, G)5

Since it has two integer roots and integer coefficients it has three integer roots. Hence a root has multiplicity $2$. A root can be $\pm 3, \pm 9$. This give the four values for $a$: $-1, 11,-5, 15$. Hence the four equations are:
$x^{3}-5x^{2}+3x+9 = 0 = (x-\alpha)^{2}(x-\beta)$. $\alpha = 3, \beta =-1$ works.
$x^{3}+7x^{2}+15x+9 = 0 = (x-\alpha)^{2}(x-\beta)$. $\alpha =-3, \beta =-1$ works.
$x^{3}-9x^{2}-x+9 = 0 = (x-\alpha)^{2}(x-\beta)$. None of the two possible cases works.
$x^{3}+11x^{2}+19x+9 = 0 = (x-\alpha)^{2}(x-\beta)$. $\alpha =-1, \beta =-9$ works.
Hence there are three values for $a$.

I hope not to have made mistakes  :)


Which round, in your opinion, was the hardest?

I personally think Round 3, but that's because I did horrible on it :D

I picked Round 4, because there were a few questions I had really messy solutions for, and it took me a long time to figure out that one with the parallel lines.  And it was due the day before the AIME.

I was out of town the two weeks before round four was due, and I had been procrastinating a lot prior to it.  I ended up not turning round four in, quite a shame, as I might have gotten honorable mention :( .

Definitely round 4. I spent by far the most time on it, I ended up bashing #3 with vectors and cross products, the first time I've used significantly more brute force than necessary on ANY math problem, and I lost a point on #5. And yeah, the due date didn't really help much.

Round 4 was the hardest. Maybe it was because many people, including I, was busy at that time period. And there was also the new daylight saving time in effect (earlier than usual) the Sunday before. But the questions themselves were harder as well, as they are supposed to be.

I guess #4 since I had the most trouble typing up decent solutions.  Odd, as I got my only two commendations of the year on it.

My scores speak for themselves

25, 25, 25, 13

Granted I was lacking motivation, but I would have had trouble with #2 and #3 no matter how hard I was trying.

round 4, even though i got a higher score on it. i just had more time to do it.

i had to think the hardest on round 2.  #4 on that round was a time consuming one for me.

I had the lowest score on round 1, but I think round 4 was the hardest, considering how lucky I got on it. On #3 I made some random assertions that turned out to be correct and ended up with a 3, and on #5 my solution was half-rigorous but I still got a 5. Also, I was alot better when round 4 came around, if I hadn't seen the solution to 1.4 right afterward, I probably would be able to solve it as of now.

[quote="vishalarul"]Everyone's going to call me stupid, but:

Round 1 was the HARDEST for me. That's because on 1/4, I didn't even think of using pigeonhole.[/quote]

I guess the same thing’ll happen to me too, because I also thought round 1 was the hardest. Too bad I didn’t really try #4 or 5.  :rotfl:

Otherwise, I suppose 2/4, 4/2, 4/3 were the questions that took me the longest to figure out the answer to. That would make round 4 the hardest.


let $ABC$ be a triangle and $E \in AB$ ,$F \in AC$ such that $BE=CF$ if $M,N$ be the midpoints of $BC,EF$ respectively
prove that $MN$ is paralell by internal bisector of $\angle A$.

Call $P$ and $Q$ the midpoints of $BE$ and $CF$. Then $MNPQ$ is a rhombus. And the result follows.


hi everybody
find all f:R--->R that for any real x,y
(x^5)-f(f(y))=y*f(x)
 :oops:

[quote="irantst"]find all $ f: \mathbb{R} \mapsto \mathbb{R}$ that for any real $ x,y$
$ x^5\minus{}f(f(y))\equal{}y \cdot f(x)$
 :oops:[/quote]

Filling in $ y\equal{}0$ gives us $ x^5 \minus{} f(f(0)) \equal{} 0$ , so $ f(f(0)) \equal{} x^5$ for all $ x \in \mathbb{R}$, a contradiction. So no such functions exist.


geometry
3D geometry
sphere
function
ratio
probability and stats

suppose that $ B_t$ is a brownian motion in $ \mathbb{R}^n$, starting from $ 0$. I am looking for the law of
$ \tau_1 = \inf \{t \geq 0 : |B_t| = 1\}$

Hence, one tries to compute $ E[f( \tau_1)]$ where $ f$ is an appropriate test function (eg: $ f(x) = \exp(- \lambda x)$ gives the Laplace transform, $ f(x) = \exp( i \xi x)$ gives the Fourier transform, etc..). My idea was to find a function $ g: \mathbb{R}^n \to R$, radial, $ g(|x|=1) \neq 0$, such that $ M_t = g(B_t)f(t)$ is a martingal. Hence,
$ g(0)f(0) = E[g(1)f(\tau_1)]$
and thus
$ E[f( \tau_1)] = \frac{g(0) f(0)}{g(1)}$

Ito formula gives: $ dM_t = [g(B_t)f_t(t) + \frac{1}{2} \Delta g(B_t) f(t)] dt + f(t) \nabla g(B_t) . dB_t$
Hence, in order $ M_t$ to be a martingale, the following relation must hold (for all $ \| x  \| \leq 1$):
$ g(x)f_t(t) + \frac{1}{2} \Delta g(x) f(t) = 0$

This implies that $ f(t) = \exp( - \lambda t)$ : one are trying to compute the Laplace transform of $ \tau_1$.
Now, my problem is that I know almost nothing about Bessel functions to continue from here:
$ \Delta g - 2 \lambda g = 0$
and since $ g(x) = H(\|x\|)$ (i.e: $ g$ is radial), this reads:
$ H"(r) + \frac{n-1}{r}H'(r) - 2 \lambda H(r) = 0$
with the additional condition that $ g(x) = H(\|x\|)$ is of class at least $ C^2$ on the unit ball of $ \mathbb{R}^n$, and continuous on the unit ball (hence $ H'(0)=0$). How would you compute the ratio $ L(\lambda) = \frac{H(0)}{H(1)}$ ?

Well, once you have that type differential equation, it seems like the power series decomposition will do just fine for not too large $ \lambda$. The question is if $ Ee^{\minus{}\lambda r_1}$ is your ultimate goal or you want to come back to $ r_1$ eventually.

yes, it would be great if I could come back to $ \tau_1$, to compute things like $ P[\tau_1 < t]$ for example, etc ..


rotation
geometry
angle bisector
geometry unsolved

Circle $ I$ with center $ O$ is inscribed in triangle $ ABC$. The tangent points are marked $ D$, $ E$ and $ F$ where the circle is tangent to $ BC$, $ AC$, and $ AB$, and rays $ DO$, $ EO$, and $ FO$ are drawn, their points of intersection with the triangle being marked as $ K$, $ L$, and $ M$ respectively.

Prove that two of the points $ K$, $ L$, and $ M$ will always be on one side of the triangle ABC. (That is, prove that no matter what ABC is, the points $ K$, $ L$, and $ M$ will never be strictly contained one within each side.)

What I have so far is:

[hide="My partial solution"]Let the [i]clockwise[/i] measure of $ \angle AOK$ be represented with $ \theta_1$, $ \angle BOL$ be represented with $ \theta_2$, and $ \angle COM$ be represented with $ \theta_3$. (The [i]clockwise[/i] measure of an angle $ \angle XYZ$ is defined as how far a ray $ YX$ has to rotate clockwise to coincide with $ YZ$.)

Thus, some of the angles are positive, but some may be negative. (This is the part that I still have to prove - that $ \theta_1 \plus{} \theta_2 \plus{} \theta_3 \equal{} 0$.)

In order for the points $ K$, $ L$, and $ M$ to be on one side each, $ \theta_1$, $ \theta_2$, and $ \theta_3$ would have to be all positive or all negative, thus not adding up to zero and making an invalid case.[/hide]

My partial solution the way it is would get at most 1 if it were a USAMO/IMO problem I was solving, most likely not even that...but I've only been working on it for about half an hour.

Alright, I have the second part of my proof:

[hide="The proof as a whole"]
Let the [i]clockwise[/i] measure of $ \angle AOK$ be represented with $ \theta_1$, $ \angle BOL$ be represented with $ \theta_2$, and $ \angle COM$ be represented with $ \theta_3$. (The [i]clockwise[/i] measure of an angle $ \angle XYZ$ is defined as how far a ray $ YX$ has to rotate clockwise to coincide with $ YZ$.)

[url=http://www.imagebam.com/image/31181b43712670][img]http://thumbnails2.imagebam.com/4372/31181b43712670.gif[/img][/url]
[size=67]In the image, all angles marked as red are "negative" - that is, they go counterclockwise.[/size]

Thus, some of the angles are positive, but some may be negative.

We start by proving that these three angles ($ \theta_1$, $ \theta_2$, and $ \theta_3$) must add up to zero:

Let $ \angle OAB$ = $ \alpha$, $ \angle OBC$ = $ \beta$, and $ \angle OCA$ = $ \gamma$. Also, mark the place where the bisector of $ \angle A$ intersects $ BC$ as $ P$, the place where the bisector of $ \angle B$ intersects $ AC$ as $ Q$, and the place where the bisector of $ \angle C$ intersects $ AB$ as $ R$.

Then, $ \angle POD$ = $ \theta_1$, $ \angle QOE$ = $ \theta_2$, and $ \angle ROF$ = $ \theta_3$, if we are measuring the angles clockwise.

[url=http://www.imagebam.com/image/05aea943712671][img]http://thumbnails7.imagebam.com/4372/05aea943712671.gif[/img][/url]

Draw the altitudes of $ ABC$, listing the points where they intersect $ BC$, $ AC$, and $ AB$ as $ A'$, $ B'$, and $ C'$ respectively. Then, $ \angle PAA'$ = $ \angle POD$ = $ \theta_1$. A similar relationship exists for angles $ QBB'$ and $ RCC'$, making them equal to $ \theta_2$ and $ \theta_3$ respectively.

[url=http://www.imagebam.com/image/c43a5443712672][img]http://thumbnails19.imagebam.com/4372/c43a5443712672.gif[/img][/url]

However, $ PAA'$ merely equals $ 90^\circ \minus{} \alpha \minus{} 2\beta$, which is obvious. A similar relationship applies for angles $ QBB'$ and $ RCC'$, making them equal to $ 90^\circ \minus{} \beta \minus{} 2\gamma$, and $ 90^\circ \minus{} \gamma \minus{} 2\alpha$ respectively. These three angles then sum up to $ 270^\circ \minus{} 3\alpha \minus{} 3\beta \minus{} 3\gamma$.

Since $ 2\alpha \plus{} 2\beta \plus{} 2\gamma \equal{} 180^\circ$, $ \alpha \plus{} \beta \plus{} \gamma \equal{} 90^\circ$, or $ \alpha \plus{} \beta \plus{} \gamma \minus{} 90^\circ \equal{} 0$, so $ \theta_1 \plus{} \theta_2 \plus{} \theta_3 \equal{} 270^\circ \minus{} 3\alpha \minus{} 3\beta \minus{} 3\gamma \equal{} 0$, as required.

In order for the points $ K$, $ L$, and $ M$ to be on one side each, $ \theta_1$, $ \theta_2$, and $ \theta_3$ would have to be all positive or all negative, thus not adding up to zero and making an invalid case. This is because they would all have to be rotated a small degree clockwise or counterclockwise from the original angle bisector in order for one point to correspond to exactly one side.
[/hide]

Could a mod please move this over to "Proposed and own problems"? I'm too lazy to delete both posts and make new ones.


limit
real analysis
real analysis unsolved

$ \lim\nolimits_{n\to\infty}$$ (x_n)\equal{} \infty$ and let $ (y_n)$ be such that $ lim(x_ny_n)$ belongs to $ R$.  Show that $ (y_n)$ converges to 0.

So I know that there is some K($ \epsilon$) such that both for all n[u]>[/u]K$ \epsilon$ $ x_n>\alpha$ and

$ |x_ny_n \minus{} x|< \epsilon$

I know I can manipulate this with division, but anyway I try this, I can't get the result
$ |y_n \minus{} 0|< \epsilon$

my simplification is

|$ y_n \minus{} 0$|$ < \frac{\epsilon \plus{}x}{x_n}$

Could someone please help?

If $ x_ny_n$ converges, then $ |x_ny_n|$ is uniformly bounded by some number $ M > 0$.

If $ x_n\to\infty$ as $ n\to\infty$, then for every positive integer $ N$, there exists $ k \equal{} k(N)$ such that $ n\geq k$ implies $ |x_n|\geq NM$.

What does this imply about $ |y_n|$ if we let $ N$ range over the set of positive integers?

Is there a way to do this by the definitions alone of properly divergent and convergent?

why don't you try proving the assertions in the above that aren't familiar to you?  they are all straightforward.

Look up the proof of the arithmetic laws for limits, and mimic multiplication/division. You need to show that $ x_ny_n$ is eventually not too big as an intermediate step.


AMC
AIME
geometry
inradius
USA(J)MO
USAMO
AIME I

[quote="yarakvar"]thx guys i would appreciate ur reply
ok look into the solution book volume 2 and page 20 number 48.
they say r= squareroot((s-a)(s-b)(s-c)/s))

wat does r mean?? it seems like radius but it doesn't make sense. radius from nowhere.
and how did u get the thing on the right?

and another questions on AoPS volume2, its kind of hard for me to go through it.
its not hard after looking to solutions but many questions like proofs, i can't solve by my own the first time. i generally can solve about half  including the example/exercise/ problems by my own.

does it mean i suck? do i even have a chance of scoring high in AIME?
i haven't prepared for this things in years so now i'm pretty behind. but i think i have the potential of catching up.

do usually ppl who do good on AIME and MOP, do good on these kinds of problems the first time?
or is it more like even the smartest people won't be able to do everthing from this book their first try.

each chapter takes about 2~5 hours for me to do. i do most of the questions before and try hard to solve it before looking up the answers. and i try hard to understand.

at the end i understand. but is this too much time for each chapter? is it another sign that i suck???

i plan to finish this book a month before the test. so then i can go over the past tests.
for two weeks i've done 3 chapters, now 22 more chapters to go.
so it means 7 more weeks to finish this. but now i'm gonna spend 2 hours on it a day. so it should be about 5-6 weeks. is this possible? is later chapters harder than the first ones??[/quote]

r is the inradius - the radius of the inscribed circle.

AoPS v2 is pretty difficult. I never finished it, and I was a usamo qualy last year, so maybe that puts things into perspective? It certainly does not mean you suck (i'd like to think i don't, anyway) and with enough practice I'm sure you could pull off a pretty high score on the AIME.

People rarely do well on problem types they see for the first time (problems at their difficulty level...). It's a matter of developing an approach to lots of problems and refining that to accomodate for more and more problems. I'd say even the smartest people wouldn't be able to solve every problem in AoPS v2 on their first try.

Some chapters are less relevant to scoring well on the AIME, but it'd be useful to go through the whole book. With enough time, it is definitely feasible to complete it within a month of the AIME and then do practice tests for a month. Go for it; it's probably the best prep you can get.

Well, I hope that answered some of your questions. And good luck.

hey thanks..
you gave me all the answers i needed.


Prove that
$x^3+y^3+z^3+x^2y+y^2z+z^2x+xyz=0$
has no non-zero integer solutions.

[hide]Assume it is possible.  SInce this equation is symmetric (each variable appears exactly 4 terms), without a loss of generality, assume x is even.  There's 3 odd terms left equalling 0 which is obviously false.
Assume 2 of the variables are even.  Again, it makes no difference which ones so let's say they're x and y.  There's only the $z^3$ term left equalling 0, which violates the rules.
SInce there's 7 terms, not all the variables can be odd.  The only chance is that all the variables are even.  x=2m, y=2n, z=2p.  Substituting and dividing by 8, one gets $m^3+n^3+p^3+m^2n+n^2p+p^2m+mnp=0$.  In other words, (m,n,p) are also a solution.  However, by the rules, those numbers are also divisible by 2 so an even smaller pair can be found.  This goes on forever, which eventually violates the rule that the solutions are integers.[/hide]


What is the sum of the positive whole numbers that are solutions of $ \minus{}3n \plus{}3 >\minus{}11?$

this is $ \minus{}3n>\minus{}14$
or $ 3n<14$

now you get $ n$ is from $ 1$ to $ 4$. sum is $ 10$ ($ (1\plus{}4)(\frac42)$)


Prove that among any $39$ consecutive natural numbers it is always possible to find one whose sum of digits is divisible by $11$

It is URSS 61. I think there is only one reasonable solution (which I won't post ;))

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sum+digits&t=6301

Pierre.


I bet you can't do this

cot(2x)=-3/4, x is in 1st quadrant

What is cosx , exactly?

ha dude you're good ur good

how long did that take u?

um... 5 minutes?
most people on this forum do better :)
i think this should be moved to getting started

lol

ok hey, time yourself on this one

tan(11pi/8) . what's exact answer without using calculator.

[hide="half angle formulas"]
$\sqrt{2}+1$?
[/hide]
i think you should post problems that require more lateral thinking :?

Whether you do that or not, I would advise using $LaTeX$ when you post problems...

For example: $\cot{2x}=-\frac{3}{4}$, $\tan{\frac{11\pi}{8}}$


A farmer bought $72$ plants that costed some whole dollars each. When he went back home, he checked the receipt he kept in his pocket, and it said:

$72$ plants ... $\$ \_{6}79 \_$.

He was surprised to see two digits of the price obliterated by some dirt in his pocket. What is the price of a single plant, in dollars?

[hide]$-679-$ is a multiple of 72, which factors to 8 and 9. the last 3 digits of a number divisible by 8 must be divisible by 8 which gives you

$-6792$

$6+7+9+2=24$
$2+4=6$
$9-6=3$

$36792/72=511$[/hide]


AMC
USA(J)MO
USAMO
AIME
search
probability
AIME I

Apparently if your are trying to qualify for USAMO, it is more advantageous to just keep checking your answers(until the end of 3 hours) once you solve a certain number of problems and not even attempt any of the other questions at all. Basically if you are trying to qualify from the floor, just solve 8 problems and keep checking them.

Like in 9th grade I solved 9 problems and just checked them over and over again and I ended up getting a 9 on AIME and qualified(I was able to catch all my mistakes).

In 10th grade I pretty much did the same thing (except I spent even longer checking because I was able to solve first 9 in an hour). Although that only happened because I wasn't able to solve #10 and got too discouraged to attempt anything else.

This year I was more aggressive and solved 12 of the problems but I only spent last 20 minutes to check and only got to check up to #5. As a result, I ended up making 6 careless errors and got a 6. If I didn't attempt #12 (which took me 10 minutes to find a method, 15 more minutes to search for a better method, 20 to 30 minutes to go through the calculations only to get it wrong by because of a computational mistake), I would be able to check most of the problems I solved and qualify this year.

In conclusion, just check the problems over and over until time is up instead of going for more problems.

True. However, this strategy seems to contradict the test makers' intentions. The AMC scoring changes were apparently introduced to encourage students to try more problems. The AIME encourages them to do just the opposite.

[quote="probability1.01"]Well, don't spend too much time checking. You should be able to solve a problem with at least 90% confidence if you spend 15 minutes on it. That gives you enough time for 12 problems, and of course a few problems are solved purely on insight, so you might only need 5 minutes for those. Realistically, supposing that you are capable of solving 12+ problems, it usually only takes 5-6 minutes to solve a problem and 3-4 additional minutes to check it.[/quote]

After this AIME, my confidence is like 0.8% for problem that I solve (before that it's around 80%, I usually do get one out of 5 or 6 wrong during practice due to careless errors)

[quote="probability1.01"] You should be able to solve a problem with at least 90% confidence if you spend 15 minutes on it. [/quote]

Not after you get past question 10 or so...

just pointing this out.

Just let him talk. I think Alex Anderson pointed out something about a IMO silver medalist being on his own scale or something.

[quote="probability1.01"]Well, don't spend too much time checking. You should be able to solve a problem with at least 90% confidence if you spend 15 minutes on it. That gives you enough time for 12 problems, and of course a few problems are solved purely on insight, so you might only need 5 minutes for those. Realistically, supposing that you are capable of solving 12+ problems, it usually only takes 5-6 minutes to solve a problem and 3-4 additional minutes to check it.[/quote]

Why would you want to get 12+ when, in this case, you can just do the first 7 and check them for the whole time and still qualify for USAMO? And if you are a junior or senior, well you should be able to get at least a 130-ish on the AMC, so you can just check 9 of them over and over again and you'll still qualify.

I hate the AIME.

I agree. Checking my work sufficiently is the only reason why I've been able to get into USAMO all these years; I catch at least two errors every time.

Well you don't really know how hard the AIME is until after the test, so unless you can predict the index in advance, this might be a bad strategy.

What makes the most sense to me is to work until you get to the point where your time to solve a problem is increasing exponentially. Then it makes sense to check your work. Actually you should probably read the rest of the problems to see if there are any random easy problems at the end, then check your work.

However, I think that the AMC people are more worried about people not attempting more than like 1 to 2 problems; I doubt that they are particularly worried about people who can solve 12 problems but only choose to do 10 to allow for more time to check.

My probability of making an error does not approach 0 as the time I spend checking gets large (it seems to approach something just under 10%). Your results may vary.

I wouldn't feel safe unless I had a 2 question margin and then spent a lot of time checking on top of that. That said, if you look at your performance on past AIME's (ie taken for practice) you can get a pretty good idea of how often you get a problem wrong on the first pass, and then you can make a more educated decision about whether it is best to spend time checking or doing additional problems.

The aforementioned comments about not knowing the index are also relevant; for me last year (as a sophmore) I almost ended up needing a 10, substantially higher than anyone would have guessed for the floor.

Kind of sad you didn't post this earlier, otherwise I might have gotten into USAMO... now I guess I'll just have to be content with 5 careless errors.

Now I have two years left. :(

101101101, why did you need a ten? The floor was 8, wasn't it last year? Another question, do they compare index for all people 10th grade and under over the floor (if they have to), or only for the people at the floor. For example, if the floor was 7 and someone got a 150/7 and another got a 120/8, and only one could qualify, who would?

[quote="101101101"]My probability of making an error does not approach 0 as the time I spend checking gets large (it seems to approach something just under 10%). Your results may vary.

I wouldn't feel safe unless I had a 2 question margin and then spent a lot of time checking on top of that. That said, if you look at your performance on past AIME's (ie taken for practice) you can get a pretty good idea of how often you get a problem wrong on the first pass, and then you can make a more educated decision about whether it is best to spend time checking or doing additional problems.
[/quote]

Well on practices I did to prepare for this year,  I make somewhere from 0 to 3 careless mistakes. Yet on the real thing this year I made 6 careless mistakes. Two years ago on the AIME I made 4 careless mistakes out of 9 but I was able to catch all of them. Last year I didn't make any careless mistakes in the first place so I didn't catch any during checking. This year I was hoping that I'd be able to make 2 to 3 careless mistakes(average) but apparently I made way more mistakes.

  I do think that I've gotten more careless since last year but still I did not expect more than 4 careless mistakes. Also the number of careless mistakes I make is out of my control (no matter how hard I try to be careful, I still make some careless mistakes).

I did go to MOP; hence the "almost" ^^

With my low AMC score, I would have needed a 10 to make the tiebreaker. I doubt they will ever actually use a tiebreaker, but I think the comparison is made among people at the floor. It is all described in the teacher's manual, although for all I know they have described some alternative tiebreaking mechanism. It doesn't concern me much anymore, being a junior.

I spent my time doing all of the problems and made careless mistakes on like 3 or 4 of them (then I missed them to check).
This forced me to end up with a low score.. (*sigh*, I'm being to dislike counting problems)

[quote="randomdragoon"]If you can catch 2 mistakes in 30 minutes, ask yourself, can you answer two more questions in that same amount of time? Both yield the same # of points.[/quote]

If you are careful the first time through (which I think is reasonable), you definitely profit by answering more questions. Most importantly, don't be scared by this year's test. It made people particularly prone to careless mistakes. I never check my work for any test, and always do very well (though I did make 3 careless mistakes on this AIME).

I think it depends on the person, but I am inclined to suggest that you keep on going and not check your work, even if that assumes a fundamental change in work strategy.

Checking more problems vs. Attempting more problems

This argument really depends on who you are. If you are prepared enough and are able to do up to 14 questions, then go for it. Even if you miss 4, you'll still have a 10. If you're at the 8-9 range, check over your answers.

In all honesty, I don't see why people don't just take their time with the questions.

[quote="xxazurewrathxx"]
In all honesty, I don't see why people don't just take their time with the questions.[/quote]

Unless I write a formal solution for each problem, no matter how hard I try I still make careless mistakes the first time doing the problems (I am able to catch my mistakes the second time doing them)

I suppose it's different with me. I'm pretty used to making "careless mistakes" but I've gotten used to it now. I don't really complain about it anymore.


[color=blue] [size=150]$3^{rd}$  Iberoamerican Olympiad[/size]
Lima, PERU. [1988]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 1.[/b] Los ángulos de un triángulo están en progresión aritmetica y sus alturas también. Demostrar que tal triángulo es equilátero.[/quote]
Sean $h_A,h_B,h_C$ las alturas del $\triangle ABC$ referentes a $A,B,C.$ Asumimos que tanto los ángulos $A,B,C$ como las alturas $h_A,h_B,h_C$ forman progresión aritmética con términos intermedios $A,h_A,$ respectivamente. Así:

$2A=B+C \Longrightarrow \ A=60^{\circ} \ \Longrightarrow a^2=b^2+c^2-bc \ \ (1)$

$2h_A=h_B+h_C \Longrightarrow \frac{2}{a}=\frac{1}{b}+\frac{1}{c} \ \Longrightarrow a=\frac{2bc}{b+c} \ \ (2)$

$(1) \cup (2) \ \ \frac{4b^2c^2}{(b+c)^2}-b^2-c^2+bc=0 \Longrightarrow \frac{(b-c)^2(b^2+c^2+3bc)}{(b+c)^2}=0$

Como $b^2+c^2+3bc >0,$ entonces $b=c$ $\Longrightarrow$ $\triangle ABC$ es equilátero.

[quote][b]Problema 2.[/b] Sean $a, b, c, d, p, q$ números naturales no nulos positivos tales que $ad - bc = 1$ y $\frac{_a}{^b} > \frac{_p}{^q}>\frac{_c}{^d}.$ Demostrar que $q \ge b+d,$ y si $q=b+d,$ entonces $p=a+c.$[/quote]
De $\frac{_p}{^q} >\frac{_c}{^d}$ se deduce que $pd>cq,$ entonces $pd \ge cq+1$ $\Longrightarrow$ $\frac{_p}{^q}\ge \frac{_c}{^d}+\frac{_1}{^{qd}}.$ Analogamente, de $\frac{_a}{^b}>\frac{_p}{^q}$ tenemos $\frac{_a}{^b}>\frac{_p}{^q}+\frac{_1}{^{bq}}.$ También, $\frac{_a}{^b}-\frac{_c}{^d} \ge \frac{_{b+d}}{^{qbd}}.$ Pero $\frac{_a}{^b}-\frac{_c}{^d}=\frac{_1}{^{bd}}$ $\Longrightarrow$ $q \ge b+d.$

Asumiendo que $q=b+d,$ resulta $ad-bc=1 \le d.$ Así, $ad+cd-d \le bc+cd$ y de lo anterior tenemos $d(a+c-1) \le c(b+d).$ Entonces $p \ge a+c.$ Analogamente, $ad-bc \le b.$ Ahora, $bc+b+ab \ge ad+ab$

$\Longrightarrow \frac{a+c+1}{b+d} \le \frac{a}{b} \ \wedge \ p \le a+c \Longrightarrow p=a+c.$

[quote][b]Problema 3.[/b] Probar que entre todos los triángulos tal que la distancias de sus vértices a un punto dado $P$ son $3,5$ y $7,$ el que tiene mayor perímetro es aquel que tiene incentro $P.$[/quote]
Sea $\mathcal{E}$ la elipse con focos $B,C$ y se ubica un punto $A'$ tal que la bisectriz de $\angle BA'C$ pasa por $P.$ Por propiedad reflexiva de la tangente a una cónica, se sigue que $AP'$ es la normal a la elipse $\mathcal{E}$ con focos $B,C$ que pasa por $A',$ i.e. el círculo $\omega$ con centro $P$ y radio $PA'$ es tangente internamente a $\mathcal{E}$ por $A'.$ Así, todo punto de $\omega$ excepto $A'$ yace en el interior de $\mathcal{E}$ y por ende se cumplirá $A'B+A'C \ge AB+AC.$ Razonamiento analogo permite concluir que el $\triangle ABC$ con perímetro máximo es aquel que tiene a $P$ por incentro.

[quote][b]Problema 4.[/b] Sea $\triangle ABC$ un triángulo cuyos lados son $a, b, c.$  Se divide cada lado del triángulo en $n$ segmentos iguales. Sea $S$ la suma de los cuadrados de las distancias de cada vértice a cada uno de los puntos de división del lado opuesto distintos de los vértices. Demuestre que $\frac{_S}{^{a^2+b^2+c^2}}$ es un número racional.[/quote]
Supongamos el lado $BC$ dividido en $n$ segmentos iguales mediante $n-1$ puntos $P_1,P_2,P_3 , ... , P_k.$  Así por el teorema de Stewart se obtiene la expresión general:

$(AP_k)^2= \left ( \frac{k}{n} \right ) b^2+ \left (1- \frac{k}{n} \right ) c^2-\frac{k}{n}\left (1- \frac{k}{n} \right )a^2 \ , \ k=1,2,3,4,5 , ... , n$

$(AP_k)^2= c^2+ \left ( \frac{a \cdot k}{n} \right )^2 -\frac{k}{n} (a^2+c^2-b^2)$

La sumatoria $S_A$ de los cuadrado de las las $n-1$ cevianas $AP_k$ se plantea entonces como:

$S_A = n  c^2+\frac{a^2}{n^2} \cdot \sum_{k=1}^{n}k^2 - \left (\frac{a^2+c^2-b^2}{n} \right) \cdot \sum_{k=1}^{n}k - b^2$

Note que para $k=n,$ la ceviana $AP_n$ se vuelve identica al lado $AC.$ Así  en la anterior sumatoria se ha sustraido $b^2$ puesto que no debe figurar en ésta. Desarrollando las sumatorias resultantes se obtiene la expresión:

$S_A = n  c^2+\frac{a^2}{n^2} \cdot \left [ \frac{n(n+1)(2n+1)}{6} \right ] - \left (\frac{a^2+c^2-b^2}{n} \right) \cdot \left [ \frac{n(n+1)}{2} \right ] - b^2$

$S_A=\frac{(n-1)(b^2+c^2)}{2}-\frac{(n^2-1)a^2}{6n}$

Por permutación cíclica de letras, se obtienen analogamente las expresiones:

$S_B=\frac{(n-1)(a^2+c^2)}{2}-\frac{(n^2-1) b^2}{6n} \ , \ S_C=\frac{(n-1)(b^2+a^2)}{2}-\frac{(n^2-1) c^2}{6n}$

$S=S_A+S_B+S_C= \left [ (n-1)-\frac{(n-1)(n+1)}{6n} \right ] \cdot (a^2+b^2+c^2)$

$\Longrightarrow  \frac{S}{a^2+b^2+c^2}=\frac{(n-1)(5n-1)}{6n}.$

[quote][b]Problema 5.[/b] Sea $k^3 = 2$ y $x, y, z$ racionales cualesquiera, tal que $x + y k + z k^2 >0.$ Probar que  existen racionales $u,v,w,$ tales que $(x + y k + z k^2)(u + v k + w k^2) = 1.$[/quote]
Primeramente, es facil ver que $x+yk+zk^2=0$ $\Longleftrightarrow$ $x=y=z=0.$ Así,

$(x+yk+zk^2)(x+y\epsilon k+z\epsilon^2 k^2)(x+y\epsilon ^2 k+z\epsilon k^2)=t\in \mathbb Q^*$

Siendo $\epsilon$ una raiz de $x^2+x+1=0.$ Esto se debe a que $x+yk+yk^2\ne 0$ $\Longrightarrow$  $(x,y,z)\ne (0,0,0)$ $\Longrightarrow$ $ x+y\epsilon k+z\epsilon^2 k^2,$ $x+y\epsilon ^2 k+z\epsilon k^2\ne 0.$ Así, probamos que $1,\epsilon k,\epsilon^2 k^2$ y $1,\epsilon^2 k,\epsilon k^2$ son independendientes sobre $\mathbb Q,$ en el mismo modo que lo probamos para $1,k,k^2.$ Debido a que, en general,

$(a+b+c)(a+\epsilon b+\epsilon ^2 c)(a+\epsilon^2 b+\epsilon c)=a^3+b^3+c^3-3abc.$

Por otro lado, $(x+y\epsilon k+z\epsilon^2 k^2)(x+y\epsilon ^2 k+z\epsilon k^2)=a+bk+ck^2,\ a,b,c\in \mathbb Q$ y así $(u,v,w)= \left (\frac {_a}{^t},\frac{_b}{^t},\frac {_c^}{^t} \right ).$


function
integration
calculus
calculus computations

[quote="stancioiu sorin"]Let $\ f$ be a function such that $\ f(x)=\arctan\left(\frac{1}{x}\right)$.Compute $\int_{0}^{1}f(x)dx$[/quote]

Let $J: =\int_{0}^{1}\arctan\left(\frac{1}{x}\right) dx$

integrate by parts: let $u=\arctan\left(\frac{1}{x}\right) , du=\frac{-\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}}dx =-\frac{dx}{1+x^{2}}, dv=dx, v=x$

$J=\left[ x\arctan\left(\frac{1}{x}\right)\right]_{x=0}^{1}+\int_{0}^{1}\frac{x}{1+x^{2}}dx = \left[ x\arctan\left(\frac{1}{x}\right)+\frac{1}{2}\log(1+x^{2})\right]_{x=0}^{1}$
$=\frac{1}{2}\left( \frac{\pi}{2}+\log(2)\right)$

  -Ben


Ring Theory
superior algebra
superior algebra solved

Suppsose [b]$A$[/b] = $<x^3 + x + 1>$ is a principal ideal in $Z_2[x]$ generated by $x^3 + x + 1$.

How do you find the distinct cosets in $Z_2[x] / A$? How many are they? Can you please give me an example of one of the cosets.

You need not to post it again ;)
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