if you could choose one piece to play for an audition in which you were trying to impress the conductor of the orchestra (for which you were auditioning) what piece would you choose? (that means you must be able to play the piece  :P)

i like Harry Potter music.  My violin teacher gave it to me.  Its called "Highlights from Harry Potter"

Perhaps Meditation from Thais (Jules Massenet).  Easy to play, sounds really good...well at least the conductor would probably think it sounds good.

[quote](that means you must be able to play the piece)[/quote]

Dang it. No Tchaikovsky or Sibelius for me then. Well, if it has to be a piece I can play, then definitely Kreisler's Praeludium and Allegro. It's a pretty impressive looking piece, and a lot of fun to play.

For the piano, I would play my Rachmaninoff concerto, and probably the toccata from Prokofiev. THey're both difficult pieces.

I'd probably play this excerpt from Pirates of the Carribean. (the orchestra arrangement). It makes you shift up to 3rd position, 5th, 7th, back to 5th, then 3rd, and then to 1st really fast. And you do it really quickly. It took me a LONG time to get it in tune at tempo.

Sibelius concerto :)
which along with Bach's Chaconne and some other fun stuff, is what I will actually be doing now that college auditions are fast approaching (ughhhh).

Come to think of it, at this point, perhaps the Bach concerto in A minor...got that down pretty well at this point
But then again, if i wanted to play more of a soloistic piece, perhaps the beethoven romance in f major...
Or if i was told to just randomly try to sightread some stuff whcih i couldn't play, I'd pull out a copy of Caprice 1 or Polonaise in D Major and make some noise  :D

[quote="buzzer11"]Any of Lalo, Tchicovski (didn't spell rite :blush: ), or Sarasate.[/quote]

I totally agree.  :lol:

If you playing a violin solo, you may want to attempt, from easiest to  hardest:
Praeludium and Allegro, by Fritz Kreisler (It's actually not as impressive as harder pieces....)
Any Mozart Concerto (These show off your technique, but personally, I hate the style)
WARNING: BIG DIFFICULTLY LEVEL JUMP HERE
Gypsy Airs, or Ziguenerweisen, by Sarasate (very, very impressive)
Mendehlsson Violin Concerto 1 (commonly played by advanced students, impressive)
Bach Solo Sonatas (Pure technique. Extremely impressive, but it gets boring to practice)
Paganini 24 Caprices (Same as Bach, a little less impressive but funner to play)
Carmen Fantasy (Second most impressive piece. Technique wise, pretty difficult. Music wise, pretty fun.)
Ravel: Tzigane (The most impressive piece ever. Very difficult, I must warn you.)

Aryth, I disagree with your list entirely. Have you ever played these songs? I think a more reasonable rating of difficulties is found here: [url]http://violinmasterclass.com/repertoire.php[/url]

First of all, "any Mozart  concerto" is definitely not more difficult than Praeludium and Allegro. I played the Mozart No. 3 long before I was able to tackle the Kreisler. Mozart can be difficult to play right, but technically many of his concertos are much less demanding.

Bach is definitely NOT pure technique. Musicality and phrasing is essential to a decent performance of any of his solo violin sonatas or partitas.

I would hardly call Ravel's Tzigane the most impressive piece ever. Personally, I'd place it at around the level of Zigeunerweisen, althoug perhaps a little higher. Much more suitable for this spot would be a Paganini caprice or concerto.

But those are just my thoughts.

[quote="Rhapsody in Green"]Bach is definitely NOT pure technique. Musicality and phrasing is essential to a decent performance of any of his solo violin sonatas or partitas. [/quote]I agree completely. Bach can be pretty boring if you play it as "pure technique." A bunch of really fast and loud notes really aren't very exciting - it's the [i]music [/i]that makes for an impressive and memorable performance.

id have to say almost anything written by Vivaldi that i can play.  :P  :lol:

If you want to play a violin solo to impress, go with the Ciaconna from Bach Partita No.2 in D minor. Heck, play any of the Bach Unaccompanied movements well and you will impress.

[quote]Aryth, I disagree with your list entirely. Have you ever played these songs? I think a more reasonable rating of difficulties is found here: http://violinmasterclass.com/repertoire.php

First of all, "any Mozart concerto" is definitely not more difficult than Praeludium and Allegro. I played the Mozart No. 3 long before I was able to tackle the Kreisler. Mozart can be difficult to play right, but technically many of his concertos are much less demanding.

Bach is definitely NOT pure technique. Musicality and phrasing is essential to a decent performance of any of his solo violin sonatas or partitas.

I would hardly call Ravel's Tzigane the most impressive piece ever. Personally, I'd place it at around the level of Zigeunerweisen, althoug perhaps a little higher. Much more suitable for this spot would be a Paganini caprice or concerto.

But those are just my thoughts.[/quote]

All right, now that I'm a bit more experienced, yes, I agree that Mozart Concertos are not very difficult. However, professional violinsts would agree that it takes at least some musical knowledge and skill combined to perform any of them. Not purely skill.

And yes, I also agree that Bach is not purely technique. Especially the partitas.

Tzigane IS quite impressive, although not the most impressive. To people who are not very knowledgable about musicality, Tzigane is just a random mess of notes, and perhaps several melodies. I played it for my friends, and now they know it as the "heart-attack" piece xD. The Paganini Caprices are of great difficulty, but only playing one or two of them is not a good way to impress somebody knowledgable about technique. Each one specializes in some different skills (except perhaps the last ;) ) and knowing two skills isn't going to bring you far. I would use the Caprices more like etudes than performances pieces like Tzigane (but I'm not saying that the caprices are Etudes!!!!!) Yes, they are indeed good for performances, but only when you have other pieces that can show off other skills.

I've just heard another piece that I find kind of interesting. At the moment, it seems decently difficult and not very boring. Bartok's Violin Concerto No. 2. I can pick out some melodies, but often, it becomes really weird (like Bartok can often (Not ALWAYS!) be). And, speaking of Bartok, something not too difficult but good sounding is his Rumanian Folk Dances (the violin solo version).

The last movement of Carmen Fantasy is impressive because of the the thirds in the beginning and all of the shifting.

Otherwise, I'm not sure... there are many worlds of music for violinists.

Never ever play something from a movie for audition. Movie music are used to enhance the feeling of a movie, if the auditioner has never seen the movie, it's pretty bad. And movie music aren't technically demanding.

For Auditions, it's best that you choose something short but shows off your technique, you may choose to do only a 2 minute excerpt from a piece, like the cadenza of a concerto.

Pagannini caprices and Bach Partitas are the best for this purpose. They both show off your technique, and demands you to be musical. These pieces are not just to be played, there's a reason that no dynamics are marked in these. You are to show off your musicality by phrasing, which is very hard to do with pieces that are technically challenging.

The 2nd and 3rd last pages of the Vitali chaconne are also good because they sound hard, but they are okay to play as long as you've had over 12 hours of practicing purely on those pages, which isn't long.
-The Ten Have 'Allegro Brilliante" is good because if the auditioner doesn't play violin, it can fool them to sound like an AR piece but is really only gr.7 or 8 :P
-Mendelssohn <- common
-Bruch <- last movement would be impressive, cadenza from the 1st would also suffice
-Lalo symphonie espagnole and Sarasate Zapateado excerpts
-Zigeuner<- the fast part is the only part that you should choose.
-Tchai concerto if you can manage it :P I don't know many ppl (only 1) that have ever played this under the age of 18.

Mendelssohn (spelling) concerto it goes into high positions. then again you can go with like scales like f major

This all depends on what the audition is for.


geometry
parallelogram
circumcircle
angle bisector
power of a point
radical axis
geometry solved

Nice problem Andrei.
Let $DM$ cuts $AB$ at $X$ and the bisector of the angle $ B$ cuts $AC$ at $L$. I 'll write the proof for the case $AB<BC$ so $ L$ is between $ A$ and $ M $ and $ P$ is between $B$ and $L$. The case $ AB>BC $ can be proved in the same way. If $\angle{BAP}=\angle{BCL}$ then the triangles $ABP$ and $BLC$ are similar. So we have to prove that $\frac{BL}{BP}=1+\frac{LP}{BP}=\frac{BC}{AB} (1) $. Using Menelause's theorem for the triangle $ABL$ and line $DM$ we get $\frac{LP}{BP}=\frac{AX}{BX}\cdot\frac{ML}{AM} (2)$. Menelause's theorem for the triangle $ABC $ and line $ DM$ gives  that $\frac{AX}{BX}=\frac{CD}{BD}$ (3) (because $AM=BM$). From $(2)$ and $(3)$ we get that $\frac{LP}{BP}=\frac{CD}{BD}\cdot\frac{ML}{AM} (4)$. Putting $(4)$ in $(1)$ we get that we have to prove $1+\frac{CD}{BD}\cdot\frac{ML}{AM}=\frac{BC}{AB}$
It is given that $ \frac{CD}{BD}=1+\frac{BC}{AB}$ so our target is   $1+{\frac{ML}{AM}={\frac{BC}{AB}}\cdot{(1-\frac{ML}{AM})}}$ or
$\frac{CL}{AM}=\frac{BC}{AB}\cdot\frac{AL}{AM}.$ which  immediately follows from the property of the angle bisector.

Armo, your solution to Andrei's problem is very nice. Here is another one:

Denote the angle bisector of the angle ABC by l. Then, the point P lies on this line l.

Let the parallel to BC through A meet the parallel to AD through C at S. Then, the quadrilateral ADCS is a parallelogram, and since the diagonals of a parallelogram bisect each other, the segment DS must therefore pass through the midpoint of the segment CA, i. e. through the point M. Hence, the points D, M and S lie on one line; clearly, the point P lies on the same line (since it lies on DM).

Now note that the lines AD and l are parallel. In fact, since DB = AB, the triangle ABD is isosceles, and thus < ADB = (180 - < ABD) / 2 = < ABC / 2, while < PBC = < ABC / 2 since the line l bisects the angle ABC, and hence < ADB = < PBC and it follows that AD || l.

So we have three mutually parallel lines: AD, CS and l. We will use directed segments; hereby, these three parallel lines are assumed to have one and the same direction. The lines AS and BC are also parallel to each other, and we also require them to have one and the same direction.

Now let the parallel to BC through A meet the line l at Q. Then, the quadrilateral BCSQ is a parallelogram, and thus QS = BC (with directed segments). On the other hand, since AS || BC, Thales shows us

$\frac{BP}{QP}=\frac{BD}{QS}=\frac{BD}{BC}$.

Thus,

$\frac{BP}{PQ}=-\frac{BP}{QP}=-\frac{BD}{BC}=\frac{DB}{BC}$.

Using the algebraic rule $\frac{x}{y}=\frac{x^{\prime }}{y^{\prime }}\Longrightarrow \frac{x}{x+y}=\frac{x^{\prime }}{x^{\prime }+y^{\prime }}$, we can conclude that

$\frac{BP}{BP+PQ}=\frac{DB}{DB+BC}$,

i. e.

$\frac{BP}{BQ}=\frac{DB}{DC}$.

Now, the quadrilateral ADBQ is a parallelogram, and thus BQ = DA. Hence,

$\frac{BP}{DA}=\frac{DB}{DC}$.

Now, if the line l meets the circumcircle of triangle ABC at a point R (apart from B), then < ARB = < ACB (cyclic), hence < ARB = < ACD; also, < ABR = < ADC (since we have already proven < ADB = < ABC / 2, and thus < ADC = < ADB = < ABC / 2 = < ABR). Hence, the triangles ARB and ACD are similar, and it follows that

$\frac{DA}{BA}=\frac{DC}{BR}$.

Multiplying this with $\frac{BP}{DA}=\frac{DB}{DC}$, we get

$\frac{BP}{BA}=\frac{DB}{BR}$.

Since DB = AB, this becomes

$\frac{BP}{BA}=\frac{AB}{BR}$.

Also, clearly < PBA = < ABR. Thus, the triangles PBA and ABR are similar, and this yields < BAP = < BRA. But < BRA = < BCA (cyclic); thus, < BAP = < BCA.

$\blacksquare$

  Darij

[quote="pestich"]Construct circumcircle ABC and circle centered at B with radius AB.
     The internal and external bisectors of angle B are orthogonal.
     Point P is located on the radical axis of two circles which is divided in half
    by circles center line orthogonal to it .  The angles in question are inscribed
    angles in circumcircle of ABC  and they sit on the equal halfs of the radical axis.

    Pestich.[/quote]

Your idea looks very promising to me. But could you please explain me why P lies on the radical axis of the two circles? Maybe I'm blind, in any case I don't seem to understand it...

  Darij

[quote="Andrei"]Let $ABC$ be a triangle, and let $D$ be a point on the line $BC$ such that the point $B$ is lying between the points $D$ and $C$ and such that $DB=AB$.  Let $M$ be the midpoint of the segment $AC$. Denote by $P$ the point of intersection of the line $DM$ with the angle bisector of the angle $ABC$. Prove that $\angle BAP= \angle BCA$.[/quote]
Let N is the midpoint of AD. And K is the foot of the bisector of <ABC. MN intersects PK at I. We have I is the midpoint of PK.
And <ANM = <ADC = 1/2<ABC = <ABI.
=> ANBI inscribed. => <AIB = 90�.
=> the triangle APK has AP = AK.
=> the proof is solluted.

Extend AP to meet line BC at E. Then apply Menelaus to triangle ACE wrt to line DPM. The conclusion follows.

Huynh Anh Hao wrote:

--------------------
   Let N is the midpoint of AD. And K is the foot of the bisector of <ABC. MN intersects PK at I. We have I is the midpoint of PK.
-----------------------

  Why I is the midpoint?

  Col. Pestich

[quote="pestich"]Huynh Anh Hao wrote:

--------------------
   Let N is the midpoint of AD. And K is the foot of the bisector of <ABC. MN intersects PK at I. We have I is the midpoint of PK.
-----------------------

  Why I is the midpoint?[/quote]

Oh, this time, it's really simple: As I showed in the very beginning of post #3, we have AD || l. In other words, AD || KP. Thus, by the Thales theorem, $\frac{KI}{IP}=\frac{AN}{ND}$. But since the point N is the midpoint of the segment AD, we have $\frac{AN}{ND}=1$; thus, $\frac{KI}{IP}=1$, and the point I is the midpoint of the segment PK.

  Darij


Are the mock AIMEs during the weekend session going to be posted here after the classes end?

It's already in the handouts section of the special AIME problem seminar section of the my classes section.

I hope everyone noticed that there is a typo in the first problem, it should be "die" but "dice" :)

Please note: the mock AIME is just for those students in the Special AIME Problem Seminar.

"A fair die is tossed 5 times." is correct English grammar. However, dice would be used in "Five fair dice are tossed."

Edit: fixed a grammar error

[quote="RC-7th"]"A fair die is tossed 5 times." is correct English grammar. However, dice would be used in "Five Fair Dice are tossed."[/quote]Ooops  :blush: ... well learning :D


calculus
integration
trigonometry
abstract algebra
geometric series
complex analysis
complex analysis unsolved

As a Fourier analyst, I do recognize the Poisson kernel.

Let $a=re^{i\alpha}$ and $z=e^{i\theta}.$ Then I claim that

$\sum_{k=-\infty}^{\infty}r^{|k|}e^{ik(\alpha-\theta)}= 1+2\text{Re }\sum_{k=1}^{\infty}r^{|k|}e^{ik(\alpha-\theta)}=$

$=1+2\sum_{k=1}^{\infty}r^{k}\cos(k(\alpha-\theta))= \frac{1-r^{2}}{1-2r\cos(\alpha-\theta)+r^{2}}=\frac{1-|a|^{2}}{|z-a|^{2}}.$

These identities can be proved by summing geometric series and other algebraic manipulations.

$\int_{0}^{2\pi}\sum_{k=-\infty}^{\infty}r^{|k|}e^{ik(\alpha-\theta)}\,d\theta= \sum_{k=-\infty}^{\infty}r^{|k|}\int_{0}^{2\pi}e^{ik(\alpha-\theta)}\,d\theta=2\pi$

since $\int_{0}^{2\pi}e^{ik(\alpha-\theta)}\,d\theta=0$ for $k\ne 0.$

---

There should be multiple other approaches to proving this. I'll let someone else post an alternative or two.

[quote="Kent Merryfield"]
$1+2\sum_{k=1}^{\infty}r^{k}\cos(k(\alpha-\theta))= \frac{1-r^{2}}{1-2r\cos(\alpha-\theta)+r^{2}}.$

These identities can be proved by summing geometric series and other algebraic manipulations.
[/quote]

How to prove this identity, can you give some details?

Let $\beta=\alpha-\theta$ to save me some writing. Go back to the form

$\sum_{k=-\infty}^{\infty}r^{|k|}e^{ik\beta}$

Then $\sum_{k=0}^{\infty}r^{k}e^{ik\beta}$ is a geometric series with sum $\frac{1}{1-re^{i\beta}}.$

The sum $\sum_{k=-\infty}^{0}r^{-k}e^{ik\beta}= \sum_{k=0}^{\infty}r^{k}e^{-ik\beta}$ is then just the complex congugate of the previous sum, namely $\frac{1}{1-re^{-i\beta}}.$

Add these up and correct for the fact that we've added in $1$ twice:

$\frac{1}{1-re^{i\beta}}+\frac{1}{1-re^{-i\beta}}-1$

$=\frac{1-re^{-i\beta}+1-re^{i\beta}-(1-re^{i\beta})(1-re^{-i\beta})}{(1-re^{i\beta})(1-re^{-i\beta})}$

$=\frac{2-re^{i\beta}-re^{-i\beta}-1+re^{i\beta}+re^{-i\beta}-r^{2}}{1-re^{i\beta}-re^{-i\beta}+r^{2}}$

$=\frac{1-r^{2}}{1-2r\cos\beta+r^{2}}$

See what happens if you try to look up the words "Poisson kernel" or "Dirichlet problem on the unit disk" or "Abel summability of Fourier series."


Putnam
geometric series
calculus
calculus computations

This was from a practice packet for the Putnam

$ \displaystyle\sum_{n\equal{}0}^\infty\sum_{m\equal{}0}^\infty\frac{m^2n}{3^m(n3^m\plus{}m3^n)}$

Wooow! Very interesting problem!
My first thought were "there is no way it has a closed form".

When you have a problem with a slight broken simetry, try to subtract $ f(a,b)$ from $ f(b,a)$ or to sum, to see what you get. After we simplify we get this identity.

$ \frac{m^2 n}{3^m \left(m 3^n\plus{}3^m n\right)}\equal{}\frac{m n}{3^{m\plus{}n}}\minus{}\frac{m n^2}{3^n \left(m 3^n\plus{}3^m n\right)}$

Follows that

$ \sum _{n\equal{}0}^{\infty } \sum _{m\equal{}0}^{\infty } \frac{m^2 n}{3^m \left(m 3^n\plus{}3^m n\right)}\equal{}\frac{1}{2} \sum _{n\equal{}0}^{\infty } \sum _{m\equal{}0}^{\infty } \frac{m n}{3^{m\plus{}n}}\equal{}\frac{1}{2} \left(\sum _{m\equal{}0}^{\infty } \frac{m}{3^m}\right){}^2\equal{}\frac{1}{2} \left(\frac{3}{4}\right)^2$

The last sum can be computed deriving the geometric series and multiplying it by $ \minus{}b$.

$ \sum _{m\equal{}0}^{\infty } \frac{m}{b^m}\equal{}\minus{}b \frac{d}{db}\left(\sum _{m\equal{}0}^{\infty } \frac{1}{b^m}\right)\equal{}\minus{}b \frac{d}{db}\left(\frac{b}{b\minus{}1}\right)\equal{}\frac{b}{(b\minus{}1)^2}$


Determine the sum of the coefficients of the polynomial that is obtained if we expand the expression                             $ (1+x-3x^{2})^{2005}$ and reduce like terms.


Prove that the product of the first $ n$ natural numbers is divisible by the sum of the first $ n$ natural numbers if and only if $ n\plus{}1$ is not an odd prime.

[hide="Pretty Straightforward"]

The sum of the first $ n$ natural numbers is

$ \equal{} 1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} n \equal{} \frac{n(n\plus{}1)}{2}$

Product of the first $ n$ natural numbers is simply $ n!$.

So, the quotient is $ \frac{n!}{\frac{n(n\plus{}1)}{2}} \equal{} \frac{2(n\minus{}1)!}{n\plus{}1}$. Clearly, if $ n\plus{}1$ is an odd prime (since only even prime is 2 and it'll be divisible), then the bottom does not divide the top so $ n\plus{}1$ must not be an odd prime. QED [/hide]

I'd say the other direction requires some attention still.

9 is composite.

All of 9's proper divisors are factors of 3!.

9 doesn't divide 3!.

-----
Also, n=3 might require a [short] line for itself.

[quote="metafor"]If $ n \plus{} 1$ is composite, all of its "proper" divisors are factors of $ (n \minus{} 1)!$.[/quote]

"Proper" divisors? I'm not sure what you mean.

But as with Mathnerdmo's comment, I'm guessing I need to elaborate more on "iff" part?

[hide="More"]
I already explained why $ n\plus{}1$ is not to be an odd prime. Now, let me explain why $ n\plus{}1$ can be composite number.

Note that from my fraction, $ n\plus{}1$ is in same parity as $ n\minus{}1$. If they are both even, then it is clear that $ \frac{n\plus{}1}{2}$ is one of the factors of $ (n\minus{}1)!$ and 2 will get canceled by 2 on the top. If they are both odd, then it is clear that $ n\plus{}1 \equal{} k \cdot P$ where $ k$ is a positive odd factor and $ P$ is the quotient of it. Clearly, both $ k$ and $ P$ are less than $ (n\minus{}1)!$ and they will be canceled out. Thus, $ n\plus{}1$ can be composite and an even prime.

Therefore, $ n\plus{}1$ cannot be an odd prime, and that is only restriction. [/hide]

The "proper" divisors of a positive integer $ n$ are the factors of $ n$ that are not $ n$ (and, if you want to extend the concept to $ \mathbb{Z}$, that are not $ \minus{} n$).

EDIT:

[quote="Mathnerdmo"]9 is composite.

All of 9's proper divisors are factors of 3!.

9 doesn't divide 3!.[/quote]

If $ n \plus{} 1 \equal{} 9$, then $ (n \minus{} 1)! \equal{} 7!$

>.>

Mathnerdmo was (correctly) pointing out that "all the proper factors of $ n$ are less than $ m$" does not imply that $ n$ divides $ m!$.

[quote="metafor"]Of course not; but, then again, I never said that. I said it is true for the specific case where $ m \equal{} n \minus{} 2$, $ n > 4$ composite (of course, I didn't specify $ n > 4$ before, although the proposition is not true for $ n \equal{} 4$).[/quote] No offense meant, but I personally see the restriction to the special case as begging the question.

I was thinking more along these lines, myself:

[hide]We want to show that $ \frac {2(n \minus{} 1)!}{n \plus{} 1}$ is a integer whenever $ n \plus{} 1$ is not an odd prime.

Obviously $ n \plus{} 1 \equal{} 2$ makes the fraction an integer.
If $ n \plus{} 1 > 2$ is even, then $ \frac {n \plus{} 1}{2} \leq n \minus{} 1$, so $ \frac {n \plus{} 1}{2} | (n \minus{} 1)!$.

If $ n \plus{} 1$ is odd (but not an odd prime), then let p be the smallest prime such that $ p | (n \plus{} 1)$.  Obviously $ 3 \leq p \leq \frac {n \plus{} 1}{p} < n \minus{} 1$.
Then when $ p \neq \frac {n \plus{} 1}{p}$, they both appear in the product $ (n \minus{} 1)!$ and therefore their product $ n \plus{} 1$ divides it.
If $ p \equal{} \frac {n \plus{} 1}{p}$, then $ n \plus{} 1 \equal{} p^2$.  Since $ p \geq 3$, both $ p$ and $ 2p$ are less than $ n \minus{} 1$, which shows that $ 2(n \plus{} 1) \equal{} 2p^2 | (n \minus{} 1)!$.
[/hide]


Let $a,b,c,d$ be nonegative real numbers such that
\[a^{2}+b^{2}+c^{2}+d^{2}=1,\]
prove that
\[a+b+c+d\geq a^{3}+b^{3}+c^{3}+d^{3}+3(abc+bcd+cda+dab).\]

Multiply the left member by $a^{2}+b^{2}+c^{2}+d^{2}$ and then AM-GM

That's right, Svejk. Very good beginner :lol:

The following is harder and more beautiful:

Let $a,b,c,d$ be nonegative real numbers such that
\[a^{2}+b^{2}+c^{2}+d^{2}=1,\]
prove that
\[a+b+c+d\geq a^{3}+b^{3}+c^{3}+d^{3}+ab+bc+cd+da+ac+bd.\]

[quote="pvthuan"]
Let $a,b,c,d$ be nonegative real numbers such that
\[a^{2}+b^{2}+c^{2}+d^{2}=1, \]
prove that
\[a+b+c+d\geq a^{3}+b^{3}+c^{3}+d^{3}+ab+bc+cd+da+ac+bd. \]
[/quote]
It's equivalent to $\sum_{sym}(a^{3}b^{3}-2a^{3}bcd+a^{2}b^{2}c^{2})\geq0,$ which is nice if true.


${f(x)=\sin (x) \cos (x)+\sqrt{\sin^{2}(x)+\alpha }\cos (x)}$
${f(x)=\cos (x) \left(\sin (x)+\sqrt{\sin^{2}(x)+\alpha }\right)}$

The maximum or minimum arewhen  ${f'(x)=0}$

For simplify  ${\beta =\sqrt{\sin^{2}(x)+\alpha }}$

${f'(x)=\cos (x) \left(\frac{\sin (x) \cos (x)}{\beta }+\cos (x)\right)-\sin (x) (\beta+\sin (x))}$
${f'(x)=-\frac{(\beta+\sin (x)) \left(\beta \sin (x)-\cos^{2}(x)\right)}{\beta }}$

Since If we assume that ${\alpha >-\sin^{2}(x)}$

So ${(\beta+\sin (x)) \left(\beta \sin (x)-\cos^{2}(x)\right)=0}$

Using  ${\cos^{2}(x)=1-\sin^{2}(x)}$

${(\beta+\sin (x)) \left(\sin^{2}(x)+\beta \sin (x)-1\right)=0}$

Substituing  ${\beta }$ And to simplify ${\sin (x)=y}$

${\left(y+\sqrt{y^{2}+\alpha }\right) \left(y^{2}+\sqrt{y^{2}+\alpha }y-1\right)=0}$

Which is easy to solve
the solution is

${y=\frac{\pm 1}{\sqrt{\alpha+2}}}$

So  ${\sin (x)=\frac{\pm 1}{\sqrt{\alpha+2}}}$

${c_{1}\in \mathbb{Z}\land \alpha+2\neq 0}$
${x=-\sin^{-1}\left(\frac{\pm 1}{\sqrt{\alpha+2}}\right)+2 \pi c_{1}+\pi \lor x=\sin^{-1}\left(\frac{\pm 1}{\sqrt{\alpha+2}}\right)+2 \pi c_{1}}$

But.... Need to know if it is a maximum, minumun or nothing... but....

its borrow..

if someone knows a more easy way

[quote="lotrgreengrapes7926"]Find the x-value when $\cos x\sin x+\cos x\sqrt{\sin^{2}x+a}$ is maximized, in terms of $a$.

Sorry if this is in the wrong forum.  :blush:  I really have no idea how difficult this problem is.[/quote]
$|\cos x\sin x+\cos x\sqrt{\sin^{2}x+a}| \leqslant \sqrt{\left({\sin^{2}x+c{\text{os}}^{\text{2}}x}\right)(c{\text{os}}^{\text{2}}x+\sin^{2}x+a)}= \sqrt{1+a}$

u didnt find the x value

[quote]Find the x-value when...[/quote]

okay there's no equality case for that... I think the proposed problem is wrong and overlooked it also... it's a strange function.


1.
In triangle ABC, 3 sin A + 4 cos B = 6 and 4 sin B + 3 cos A = 1.  Find the measure of angle C in degrees.

2.
Find the number of complex numbers a+bi such that
(a+bi)^19 = (a-bi)^99.

3.
A convex quadrilateral ABCD with area 2002 contains a point P in its interior such that PA = 24, PB = 32, PC = 28 and PD = 45.  Find the perimeter of ABCD.

ive seen all of these before, so im not going to answer. but they look like they are problems given on that student worksheet in the amc10/12 teachers manual from 2002. btw, if you are just checking, the answers are there.

Not just checking, I want to see how other people did these.

WarpedKlown1335 wrote:2. Find the number of complex numbers a+bi such that
(a+bi)^19 = (a-bi)^99.

[hide]Either a+bi = 0 or we see that |a+bi| = (a2+b2)1/2=1, since that quantity to power 19 equals that quantity to power 99.  Then we note that in that case (a+bi)(a-bi)=1 so (a-bi)99=(a+bi)-99.  So (a+bi)118=1.  There are 118 solutions equally spaced around the unit circle (one of them is 1).  So there are a total of 119 solutions.

[/hide]

--Dan

WarpedKlown1335 wrote:1.
In triangle ABC, 3 sin A + 4 cos B = 6 and 4 sin B + 3 cos A = 1.  Find the measure of angle C in degrees.

[hide]Square it.  9sin :^2: A + 16cos :^2: B + 24sin A cos B = 36

16sin :^2: B + 9cos :^2: A + 24sin B cos A = 1

Add 'em

9(sin :^2: A+cos :^2: A) + 16(sin :^2: B + cos :^2: B) + 24(sin A cos B + sin B cos A) = 37

Be beautiful.

sin A cos B + sin B cos A= 1/2

sin (A+B) = 1/2

A+B = arcsin 1/2 = 30 or 150

So C = 150 or 30

try A+B=30 in the eqn.

A,B<30 (duh)

so 3sinA < 1.5 and 4cos B < 4... 5.5<6, so A+B=150 and C=30[/hide]


Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299797[/url]


Let $ a,b,c$ be nonnegative real numbers, no two of wich are zero. Prove that

$ \frac {bc}{(4a\plus{}b\plus{}c)(b\plus{}c)}\plus{}\frac {ca}{(4b\plus{}c\plus{}a)(c\plus{}a)}\plus{}\frac {ab}{(4c\plus{}a\plus{}b)(a\plus{}b)} \le \frac 1{4}$.

It's obviously trues by Am-Gm because:
$ LHS \leq\ \sum\ \frac{bc}{4(ab\plus{}bc\plus{}ca)} \equal{} \frac{1}{4}$
But I'm so eager too see your detail: why is it related to Iran 96 ? Thank you  :)

[quote="nguoivn"]It's obviously trues by Am-Gm because:
$ LHS \leq\ \sum\ \frac {bc}{4(ab \plus{} bc \plus{} ca)} \equal{} \frac {1}{4}$
But I'm so eager too see your detail: why is it related to Iran 96 ? Thank you  :)[/quote]
The following inequality is an other generalization of Iran 96:

If $ 0\le k < 1$ or $ k\ge 2$, then

$ \sum \frac {(1 \plus{} k)(4 \minus{} k)a(b \plus{} c) \plus{} 4(1 \minus{} k)bc}{(ka \plus{} b \plus{} c)(b \plus{} c)} \ge 3(3 \minus{} k)$.

The proposed inequality is for $ k \equal{} 4$. :wink:

[b]Remark.[/b] For $ 1<k\le 2\sqrt 6 \minus{}3$, the reverse inequality holds.

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers, no two of wich are zero. Prove that

$ \frac {bc}{(4a \plus{} b \plus{} c)(b \plus{} c)} \plus{} \frac {ca}{(4b \plus{} c \plus{} a)(c \plus{} a)} \plus{} \frac {ab}{(4c \plus{} a \plus{} b)(a \plus{} b)} \le \frac 1{4}$.[/quote]

The inequality

\[ \frac {bc}{(ka \plus{} b \plus{} c)(b \plus{} c)} \plus{} \frac {ca}{(kb \plus{} c \plus{} a)(c \plus{} a)} \plus{} \frac {ab}{(kc \plus{} a \plus{} b)(a \plus{} b)} \le \frac 3{2(k\plus{}2)}\]

holds for $ 0\le k\le 4$.


I'm not sure about the accuracy of this but:

the average of all the numbers you pick should be pretty close to the average of all the numbers from one to N. find the average of all the numbers you picked and then double it (roughly) and that's a pretty good idea of what N is.

[quote="Kle"]The chips in the bowl are numbered consecutively from 1 to N, one number per chip. Randomly select three chips, without replacement, from the bowl. Write down the numbers on the chips and reuturn the chips to the bowl. Repeat this at least 20 times.
Then estimate N using the samples

I just dont have any hint how to approach this problem... Will anyone guide me please? Thank you[/quote]

[hide]the expected sum of the three numbers chosen on each pick is 3 times the average of all the numbers or $3\times \frac{1+N}{2}$.

Now take the average of all the numbers and set it equal to the expression above.  Then solve for $N$[/hide]

If you had drawn 60 values simultaneously, I would suggest as an alternative that you multiply the largest number you drew by $\frac{61}{60}$.  However, I'm not sure how the possibility of drawing the same number repeatedly affects the effectiveness of this operation.

We can actually, in principle, calculate the average error from these two methods: we'd have to sum over all possible combinations of draws the errors for each method, then divide by the number of possible draws.  Unfortunately, this is difficult to actually implement (especially given the size of the desired experiment).  However, it's not difficult to do in smaller cases on Mathematica.  If we only repeat the draw 4 times (rather than 20), I find that the method offered by the previous two posters gives these average errors for the following values of $n$:

$n=3 \Longrightarrow AvErr = 0$
$n=4 \Longrightarrow AvErr = \frac{19}{64}\approx 0.297$
$n=5 \Longrightarrow AvErr = \frac{3461}{7500}\approx 0.461$
$n=6 \Longrightarrow AvErr = \frac{6111}{10000}\approx 0.611$
$n=7 \Longrightarrow AvErr = \frac{3397672}{4501875}\approx 0.755$

while my proposed method (take the largest value and scale up) gives average errors of:
$n=3 \Longrightarrow AvErr = \frac{1}{4}= 0.25$
$n=4 \Longrightarrow AvErr = \frac{343}{1024}\approx 0.335$
$n=5 \Longrightarrow AvErr = \frac{16927}{40000}\approx 0.423$
$n=6 \Longrightarrow AvErr = \frac{324447}{640000}\approx 0.507$
$n=7 \Longrightarrow AvErr = \frac{873143}{1500625}\approx 0.582$

So it seems (on relatively small amounts of anecdotal evidence) that my idea performs better; unfortunately, I don't have the computing power to do any larger examples.


The numbers in the sequence $S_{N}$ satisfy the recursion $S_{n+1}=2S_{n}+1$ for any positive integers $n$. If $S_{0}=1$ and $S_{1}=16$, write an equation for $S_{n}$ explicitly in terms of $n$.

Put $S_{n}= a2^{n}+b$ and equate coefficients.  We get $b=-1$.  Plugging in $S_{0}=1$, we get $a-1=0$, whence $a=1$.  So $S_{n}=2^{n}-1$

Oops. Typo...that should change your answer.. :blush:  sorry

The problem is impossible the way it stands now, because $S_{1}\neq 2S_{0}+1$

The problem states that the recurrence only holds for positive $n$.  I get $S_{n}=17\cdot 2^{n-1}-1$, but this doesn't hold for $n=0$.

Is this from a lecture? And if so, who's?


Count all the factores of 5 in 236!  Watch out for 25s, 50s and 125s that contain more than one factor of 5.   Since there are many more factors of 2 than factores of 5, the number of fives is the numbe of zeroes in the number.
(Factors of 2 and 5 make a ten which ends in 0)

[quote="Jongao"]Countdown Problems are easy, right?
Answer this:
The number 236! ends in how many zeros when expressed as an integer?  Explain.[/quote]

Refer to this thread:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1834830816&t=122655[/url]

for a similar problem.

[quote="AstroPhys"][hide="HINT"]
$ \left\lfloor \frac{236}{2}\right\rfloor+\left\lfloor \frac{236}{2^{2}}\right\rfloor+\left\lfloor \frac{236}{2^{3}}\right\rfloor+...$
$ \left\lfloor \frac{236}{5}\right\rfloor+\left\lfloor \frac{236}{5^{2}}\right\rfloor+\left\lfloor \frac{236}{5^{3}}\right\rfloor+...$
Until a term reaches zero in each.
Then the smaller sum will be the number of zeros because $ 2 \cdot 5 =10$.
The $ \left\lfloor x \right\rfloor$ is the greatest integer $ \le x$.
[/hide][/quote]

You don't need the first formula; it will always be less than the second one.

That's simple.  It wouldn't be much of a challenge even in the Countdown round.

[hide]There are 235/5= 47 single multiples of five.  There are 225/25=9 multiples of 25.  There is one multiple of 125.  Thus, the number of zeroes should be [size=150]57 :lol: [/size].[/hide]

not a challenge? can you get it in 5 seconds?

Actually, I probably could do it in 5 seconds. This number is pretty small and you only have to divide 3 times, 2 of which are easy.

Exactly.  Not to be conceited, but I am pretty good at countdown problems.  My average time for a semi-hard problem should be around three to four seconds.  You can ask my Georgia team coach.  But I'm not exactly amazing at the written round. :(

coughcough radioactive kimchi coughcough

But yeah, with practice you can really just take one glance at the problem and solve it after you press the buzzer.

Wait, MathGeek, what were you ranked at Nats?


let a,b,c be positive numbers  such that :a^2+b^2+c^2=2(ab+bc+ca-1)
find the minimum value of the expression :
Q=12(a^2+b^2+c^2)+3ab-8bc-19ca   :lol:


[quote="Kyoshiro"]Suppose that $ x,y,z > 0 ; x \plus{} y \plus{} z\le \frac {y \plus{} z}{3x}$
Find the maximum of $ P \equal{} \frac {x}{y \plus{} z}.$[/quote]
Consider $ y \plus{} z \equal{} t$. The problem become:
[quote]Suppose that $ x,t > 0 ; x \plus{} t\le \frac {t}{3x}$
Find the maximum of $ P \equal{} \frac {x}{t}.$[/quote]
$ x,t > 0 ; x \plus{} t\le \frac {t}{3x} \Leftrightarrow 3x^2 \plus{} 3xt \minus{} t \le 0$
We solve this inequation  :wink: It has roots: $ ( \minus{} 3 \minus{} \sqrt {21})t \le x \le ( \minus{} 3 \plus{} \sqrt {21})t$
Hence, $ \frac {x}{t} \le \minus{} 3 \plus{} \sqrt {21}$
$ \Rightarrow \max P \equal{} \minus{} 3 \plus{} \sqrt {21}$
Equality holds if and only if $ x \plus{} y \plus{} z\equal{} \frac {y \plus{} z}{3x}$

Sorry everyone :oops: !
$ x\plus{}y\plus{}z\le \frac{yz}{3x}$

Is the problem to find the maxim value of $ P\equal{}\frac{x}{y\plus{}z}$ or $ P\equal{}\frac {x}{yz}$.

[quote="Kyoshiro"]Suppose that $ x,y,z > 0 ; x \plus{} y \plus{} z\le \frac {yz}{3x}$
Find the maximum of $ P \equal{} \frac {x}{y \plus{} z}.$[/quote]
But are you sure that the ineq is correct? Because from nessbitt, $ P\ge\frac{3}{2}$


Think about all of the positive common (reduced) fractions with a denominator of 11 or less. How many of these common fractions have a value less than 7/11?

[hide="ANSWER"]Answer: 26[/hide]

nobody tell me im thinking too hard.

for each denominator, you want to find the greatest possible value of the numerator. 1-0 2-1 3-2 4-3 5-3 6-4 7-4 8-5 9-5 10-6 11-6
then do 1+2+3+4+4+5+5+6+6. it doesnt take THAT long. and where did you get 26 i got 36.

You overcounted. For example, I think you took $ \frac{2}{4}$ to be different than $ \frac{1}{2}$
Edited

[quote="(xpmath intended to write)"]You overcounted. For example, I think you took $ \frac{2}{4}$ to be different than $ \frac{1}{2}$[/quote]

Just fixing the message.

[quote="xpmath"]Whoopsie. I fixed it. Thanks
As for the problem, I gues you would just make a chart and go carefully.[/quote]

I think just counting only reduced fractions should do the trick  :wink:

Well, that's what I meant. I was just wondering whether there was a way to do it without listing out every reduced fraction...
Edit: It actually doesn't take that long to list it out.


X is an angle measured in radians lying in the range 0<=X  <=pi

Prove that SinX/(2+CosX) < 2.Sin(X/2)/[2 +Cos(X/2)]  and deduce hence prove that

   X > 3SinX/(2+CosX).

 Please try to use only simple algebra and trigonometry-no calculus or convexity arguments!

Prove that:
$\frac{{\sin x}}{{2 + \cos x}} \leq 2 \cdot \frac{{\sin \frac{x}{2}}}{{2 + \cos \frac{x}{2}}},\forall x\in [0,\pi]$

[u]Solution:[/u]
$\frac{{\sin x}}{{2 + \cos x}} \le 2 \cdot \frac{{\sin \frac{x}{2}}}{{2 + \cos \frac{x}{2}}}\mathop  \Leftrightarrow \limits_{\mathop {\scriptstyle 2 + \cos x > 0 \hfill \atop
  \scriptstyle 2 + \cos \frac{x}{2} > 0,\forall x \in R \hfill}\limits^ \uparrow  } $
$ \Leftrightarrow 2 \cdot \sin x + \sin x \cdot \cos \frac{x}{2} \le 4 \cdot \sin \frac{x}{2} + 2 \cdot \sin \frac{x}{2} \cdot \cos x \Leftrightarrow $
$ \Leftrightarrow 4 \cdot \sin \frac{x}{2} \cdot \cos \frac{x}{2} + 2 \cdot \sin \frac{x}{2} \cdot \cos ^2 \frac{x}{2} \le 4 \cdot \sin \frac{x}{2} + 2 \cdot \sin \frac{x}{2} \cdot \left( {\cos ^2 \frac{x}{2} - \sin ^2 \frac{x}{2}} \right) \Leftrightarrow $
$ \Leftrightarrow 4 \cdot \sin \frac{x}{2} \cdot \cos \frac{x}{2} \le 4 \cdot \sin \frac{x}{2} - 2 \cdot \sin ^3 \frac{x}{2}\mathop  \Leftrightarrow \limits_{\mathop {\sin \frac{x}{2} \ge 0,\forall x \in \left[ {0,\pi } \right]}\limits^ \uparrow  } $
$ \Leftrightarrow 2 \cdot \cos \frac{x}{2} \le 2 - \left( {1 - \cos ^2 \frac{x}{2}} \right) \Leftrightarrow \cos ^2 \frac{x}{2} - 2 \cdot \cos \frac{x}{2} + 1 \ge 0$
that is true.

Prove that:
$\frac{{3 \cdot \sin t}}{{2 + \cos t}} \le t,\forall t \in \left[ {0,\frac{\pi }{2}} \right]$

$f(t)=t-\frac{3*sint}{2+cost} $ then $f'(t)\geq0$, so it is increasing.
We get its minimum when t=0. So $t\geq \frac{3*sint}{2+cost}$ I hope it is true


geometry
3D geometry
sphere
calculus
derivative
absolute value
calculus computations

I've been working on this problem for a while and I can't seem to find the correct answer.

The circumference of a sphere was measured to be 70 cm with a possible error of 0.2 cm. (This means that the actual length of the edge is between 70 - 0.2 and 70 + 0.2).

(a) Use differentials to estimate the maximum error in the calculated surface area. (you can also view it as estimating the absolute value of the change of the area when the length of the circumference changes from 70 to 70 + 0.2 or from 70 to 70 - 0.2. Since we will take the absolute value of the linear approximation, either changes will give us the same answer. )
Answer: maximum possible error is...?

(b) Use differentials to estimate the maximum error in the calculated volume. (Same remarks above apply here also).
 maximum possible error is...?

I know that r=c/2pi and I know that I need to take the derivative of the surface area which is given by S=4pir^2, but I don't understand how to solve for the maximum possible error. How do I solve this kind of problem?

$ C \equal{} 2\pi r\implies\Delta C \equal{} 2\pi\Delta r$
Hence, we can get $ r$ and $ \Delta r$ using $ C \equal{} 70$ and $ \Delta C \equal{} .2$
Now, use that $ S \equal{} 4\pi r^2\implies\Delta S \equal{} 8\pi r\Delta r$
and, $ V \equal{} \frac {4}{3}\pi r^3\implies\Delta V \equal{} 4\pi r^2\Delta r$


geometry
rectangle
analytic geometry
search

I need help with 2 of the 2005 Nationals questions:

1.) [b]2005 National Sprint #30[/b]-A gecko is in a room that is 12 feet long, 10 feet wide, and 8 feet tall. The gecko is currently on a side wall (10' by 8'), one foot from the ceiling and one foot from the back wall (12' by 8'). The gecko spots a fly on the opposite side wall, one foot from the floor and one foot from the front wall. What is the length of the shortest path the gecko can take to reach the fly assuming it does not jump and can only walk across the ceiling and the walls? Express your answer in simplest radical form.

2.) [b]2005 National Target #7[/b]-A point inside a square is positioned so that the distances to the four vertices are 27, 21, 6, and x units. If x is a whole number, what is the value of x?

1.) Draw a rectangle that contains the side walls and the front wall. We don't need to worry about any of the area closer than a foot to the top of the rectangle and the same applies to the bottom of the rectangle and one of the sides. (The side that the gecko is on.) As for the other side. We don't need to worry about anything farther than a foot away from the front wall. Now that we have cut off all these pieces of the original rectangle we now have a 22 ft. by 6 ft. rectangle. Using the Pythagorean Theorum, we find that the diagonal of this rectangle (or the distance from the gecko to the fly) is $ \sqrt{520}$.

Now lets draw another rectangle. This rectangle contains the sidewalls and the floor. Just like before we don't have to worry about anything that is closer than a foot away from the top or the bottom of the rectangle. We also don't have to worry about anything closer to the end of the side of the rectangle than the gecko is. Anything closer than 7 ft. away from the other wall we don't need to worry about. Now we have an eighteen ft. by ten ft. rectangle. The diagonal of this (or the distance from the gecko to the fly) is $ \sqrt{424}$. This is a closer distance then that of the other rectangle. So $ \sqrt{424} \equal{} 2\sqrt{106}$ is the answer.

2.) Let's call the square $ ABCD$, such that the point is $ 27$ units away from $ A$, $ 21$ units away from $ B$, $ 6$ units away from $ C$, and $ x$ units away from $ D$. Let's pretend that the vertex $ A$ lies at the origin of the coordinate plane, and that the point in the square is $ (m, n)$. Let the side length of the square be $ y$. The point $ (m, n)$ is $ m$ points to the right of $ A$ and $ n$ below $ A$. We use the pythagorean theorum to discover that the distance from $ A$ to the point in the square is $ \sqrt{m^2 \plus{} n^2}$. We know that this is $ 27$, so $ m^2 \plus{} n^2 \equal{} 729$. The point is $ (y \minus{} m)$ to the left of $ B$ and $ n$ below $ B$. So by using the pythagorean theorum again, we find that the distance from the point to $ B$ is;

$ \sqrt{(y \minus{} m)^2 \plus{} n^2}$

$ \sqrt{y^2 \plus{} m^2 \minus{} 2ym \plus{} n^2}$

We know that this is equivalent to $ 21$, so;

$ y^2 \plus{} m^2 \minus{} 2ym \plus{} n^2 \equal{} 441$

The point is $ (y \minus{} n)$ above $ C$, and $ m$ to the right of $ C$, so by using the pythagorean theorum again we get;

$ \sqrt{(y \minus{} n)^2 \plus{} m^2} \equal{} 6$

$ y^2 \plus{} n^2 \minus{} 2yn \plus{} m^2 \equal{} 36$

Finally, we know that the point is $ (y \minus{} m)$ to the left of $ D$, and $ (y \minus{} n)$ above $ D$. Using the pythagorean theorum again, we get;

$ \sqrt{(y \minus{} m)^2 \plus{} (y \minus{} n)^2} \equal{} x$

$ y^2 \plus{} m^2 \minus{} 2my \plus{} y^2 \plus{} n^2 \minus{} 2ny \equal{} x^2$

We know that $ m^2 \plus{} n^2 \equal{} 729$ so;

$ 2y^2 \minus{} 2y(m \plus{} n) \plus{} 729 \equal{} x^2$

So we have the following four equations;

$ m^2 \plus{} n^2 \equal{} 729$

$ y^2 \plus{} m^2 \minus{} 2ym \plus{} n^2 \equal{} 441$

$ y^2 \plus{} n^2 \minus{} 2yn \plus{} m^2 \equal{} 36$

$ 2y^2 \minus{} 2y(m \plus{} n) \plus{} 729 \equal{} x^2$

From the first equation, we conclude that;

$ m \equal{} \sqrt{729 \minus{} n^2}$

Plugging this into the third equation, we get;

$ y^2 \plus{} n^2 \minus{} 2yn \plus{} 729 \minus{} n^2 \equal{} 36$

$ y^2 \minus{} 2yn \equal{} \minus{}693$

Plugging this into the fourth equation we get;

$ y^2 \minus{} 2ym \plus{} 729 \minus{} 693 \equal{} x^2$

$ y^2 \minus{} 2ym \plus{} 36 \equal{} x^2$ (This is Equation X)

We can also conclude from the first equation that;

$ n \equal{} \sqrt{729 \minus{} m^2}$

Plugging this into the second equation we get;

$ y^2 \plus{} m^2 \minus{} 2ym \plus{} 729 \minus{} m^2 \equal{} 441$

$ y^2 \minus{} 2ym \equal{} 288$

Plugging this into Equation X, we get;

$ 288 \plus{} 36 \equal{} x^2$

$ 324 \equal{} x^2$

Therefore $ x \equal{} 18$.  :)

#7: [url=http://en.wikipedia.org/wiki/British_flag_theorem]British Flag Theorem[/url]

For British Flag Theorum;

We can form the following equation;

$ 27^2 \plus{} 6^2 \equal{} 21^2 \plus{} x^2$

$ 729 \plus{} 36 \equal{} 441 \plus{} x^2$

$ 765 \equal{} 441 \plus{} x^2$

$ x^2 \equal{} 324$

$ x \equal{} 18$

Was my answer to the first question correct?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=938934011&t=182041]This may help.[/url] :)

In other words, yes.

[quote="mathlete5"]But the answer is 2sqrt113, not 2sqrt106!!![/quote]
Yeah, I remember posting in that topic. The correct answer isn't what I said, but rather 2sqrt113


The sum of $n$ consecutive positive odd integers, beginning with the least one, $a$, is $2004$. Compute the smallest possible value of $a$...

If the third and fourth terms of an arithmetic sequence are increased by $3$ and $8$ respectivley, then the first four terms form a geometric sequence. Compute the sum of the fourth term of the arithmetic sequence and the second term of the geometric sequence.

Explanation
[hide]We know, by using one, that n cannot equal 45 or more.  We also know that it must be at least 2 (1001+1003).  The sum sequence formula, when simplified, is 2004=n(a+n-1).  Therefore, we know that n (the number of terms) must be a divisor of 2004.  Looking at 2004, we see it equals 12*167.  Therefore, since the divisors must be between 45 and 2, they can only be 12, 6,4,3,2.  Testing these shows that 6 is the largest n where a is an odd integer.[/hide]

Hmm, good questions.
1)
[hide]
To start out, I found the equation for the sum given $a$ and $n$:

Basically, we need to find the sum of all the odd numbers from 1 to the nth odd number after a, and subtract from that the sum of all the odd numbers from 1 to the greatest odd number less than a.
Using the formula for the sum of the first n odds $= n^2$, this comes to

$(\frac{a+2n-1}{2})^2 - (\frac{a-1}{2})^2$

$= n(n + a - 1)$

Thus, we must find the smallest odd integer $a$ for which
$n(n + a - 1) = 2004$ for some integer $n$.

Also, $n$ must be even, because for the sum of odds to be an even number (2004), an even number of odds must be added.
Knowing $a$ is odd, $n + a - 1$ will always be even, so we need to find two evens whose product is 2004... great.

Well, we just need to find the two closest factors of 2004 that are both even.  Once we have them, set the larger one equal to $n + a - 1$ and we already have $n$ at that point (the smaller factor is $n$) so we can solve for $a$.

Interestingly enough, the two closest even factors of 2004 are 6 and 334.
Therefore, $6 + a - 1 = 334$
$a = 329$
[/hide]

[hide]By letting the first term = $a$, the common difference = $d$ and the common ratio = $r$, you can describe the terms in two different ways and thus have a system of equations.

$a=a\\a+d=ar\\a+2d+3=ar^2\\a+3d+8=ar^3$

solving the system, you get $a=27$, $d=-9$, and $r=\frac{2}{3}$, so the fourth term of the arithmetic is $0$ and the second term of the geometric is $18$, their sum being $\framebox[.25in]{18}$

Except now that I think about it, since you know the 2nd of the geo = 2nd of arith, then the sum equals $2a+4d$ so there's probably an easier way than solving the entire system[/hide]


quadratics
geometry
rectangle
algebra
quadratic formula

Hello everyone,

   The question is:

1. The diagrams show the first four rectangles in a pattern.

1st rectangle: L = 2, W = 4           2nd rectangle:  L = 3, W = 5   3rd rectangle: L = 4, W = 6           4th rectangle:  L = 5  W = 7

• Write an equation of the form A = that relates the area, A, of each rectangle to its width, w.

[u]My work[/u]

I substituted the area and width into the quadratic formula:

y = ax^2 + bx + c

16a + 4b + c = 8    -(1)
25a + 5b + c = 15  -(2)
36a + 6b + c = 24  -(3)

(2) - (1): 9a + b = 7    -(4)
(3) - (2): 11a + b = 9  -(5)

(5) - (4): 2a = 2
               a = 1

Sub. a = 1 into (4):

9(1) + b = 7
[color=darkred][b]b = -2 [/b][/color]  <-- My problem is here. In
the answer section, my     textbook says that b = +2, not -2.

Going on, sub. a & b into (1) would give us the equation:

y = 1x^2 - 2x + 0
   = x^2 - 2x  <-- The answer in the textbook is x^2 + 2x.

Could anyone please spot the mistake? Thanks.

the solution is actually much simpler than that. You can see that the width of each rectangle is 2 more than the length, so you can write $l=w-2$. Now you know the area of the rectange is $lw$, so
$A=(w-2)w$
$A=w^{2}-2w$
So you got the right answer. Either the textbook is wrong or it was asking for the area in terms of the length

But you are still correct, because the question asked for the area in terms of length

Thank you for your attention and replies!

I'll bring this up with the teacher in class.


What is the product of the solutions of the equation $ \minus{}35\equal{}\minus{}x^2\minus{}2x$?

Rearranging give $ x^2\plus{}2x\minus{}35\equal{}0$ Using Vieta, we get $ r_1r_2\equal{}\frac{c}{a}\equal{}\minus{}35$.


geometry
USAMTS
AMC
USA(J)MO
USAMO
similar triangles
congruent triangles

Note: This problem may be hard to you if you don't have a geometry as class curriculum or have practiced geometry a lot.

Answers must be in spoiler.

Given: AE || BD; BC || AD; AE  :cong: BC; AD :cong: BD.

Prove:
A) AC :cong: BE
B) EC || AB

Here is a solutin for A)

[hide]AD = BD which means that angle DAB = angle DBA also because it shows that they're isosceles. Also, we're given that    AE = BC, right? And AB is a reflexive side. So, if we add up that three congruency, we'll get triangle AEB = triangle BCA. And by CPCTC, we have AC = BE. [/hide]

Solutions for B)

[hide]Using the A)'s knowledge, we have AC = BE, AE = BC, and EC = EC by reflexive property. This let us to say triangle ACE is congruent to triangle BEC. By CPCTC, we have angle EAC congruent to angle CBE, which also let us to say angle CAB = angle EBA because angle EAB = angle CBA from the triangle we proved in the frist exercise and subtraction of two congruent angles form congruent angle. In another word:

angle EAB = angle EAC + angle CAB
angle CBA = angle CBE + angle EBA
angle EAB = angle CBA
angle EAC = angle CBE
So,
angle EAC + angle CAB = angle CBE + angle EBA
angle CAB = angle EBA by subtraction.

Now, that last congruency let us to say OB = OA because the triangle OAB is isosceles. And here is an interesting part comes in:

AC = OB + CO
BE = OA + BE

And since OB = OA, CO = BE also. And angle EOC = angle AOB. So, proportion of CO/OA = BE/OB and that makes them similar triangles, which have same measures of angle.

At last, this makes us to say angle ECA = angle OAB, which are alternate interior angles!!

Then means EC || AB
  [/hide]

Solutions are there for you to check, not to cheat.

Try your best and have a question, post here!

My A might be a little long, but here it goes.

A) [hide]Notice by alternating interior angles, <EAD  :cong: <CBD. By the given, AE :cong: CB, and by identity, AB :cong: AB. Since DAB is isosceles, <DAB :cong: <DBA, and such by angle addition, < CBA :cong: < EAB. So, by SAS, and then CPCTC, AC :cong: BE, as proven.[/hide]

A
[hide]1. AE || BD; BC || AD; AE congruent BC; AD congruent BD.
1. Given

2. $\angle EAD \cong \angle ADB$, $\angle ADB \cong \angle DBC$
2. Alternate Interior Angle theorem

3. $\triangle EAD \cong \triangle CBD$
3. Side Angle Side

4. $ED \cong CD$, $\angle EDA \cong \angle CDB$
4. Congruent Parts of Congruent Triangles are Congruent

5. $\angle ADB \cong \angle ADB$
5. Reflexive Property

6. $m\angle EDA = m\angle CDB$, $m\angle ADB = m\angle ADB$
6. Definition of Congruency

7. $m\angle EDA + m\angle ADB = m\angle CDB + m\angle ADB$
7. Addition Property of Equality

8. $m\angle EDA + m\angle ADB = m\angle EDB$, $m\angle CDB + m\angle ADB = m\angle CDA$
8.  Angle Addition Property

9. $m\angle EDB = m\angle CDA$
9. Substituion Property

10. $\angle EDB \cong \angle CDA$
10. Definition of COngruency

11. $\triangle CAD \cong \triangle EBD$
11. Side Angle Side

12. $AC \cong BC$
12. Corresponding Parts of Congruent Triangles are Congruent[/hide]

Would my proof be okay for things like USAMTS, or USAMO, or would the judges prefer the way Silverfalcon, or white_horse_king88 did theirs?

Yours is much more formal. It is the two column standard proof in euclidean geometry. I just condensed mine so that it wouldn't take up lots of lines. If you were to do an actual proof, yours would be the "better" one in terms of format. That is in my opinion.

There are many types of proof.

Two column is one of them and also, there is a proof called a key-proof where you just put the most important things in paragraph-format. And there is one called paragraph proof that is same as two-column but just in form of paragraph.

 :P

I have removed the attachment since most people solved it.

hallo, was ist die antworted fur diesis probleme,?

      78.66A*77c/( :sqrt: 69)b= 88.7

wie machst du diesis probleme? spater

hey,im not gode at englishe, but im a foran exchange student form gemrnay, but anywy im in america, and Joemll88(joseph lourindi) is in my classe, and joe was teaching the teacher how to correctly solve equations is that not ironic, as americans would say


[quote="sharkman"]Find functions f and g so that f(g(x)) = H.

H(x) = |2x^2 + 3|[/quote]

I'm not sure if that's "all functions" or not, but here are a few

$ f(x)\equal{}x, g(x) \equal{} 2x^2\plus{}3$ (and the opposite)
$ f(x)\equal{}x\plus{}3, g(x) \equal{} 2x^2$
$ f(x)\equal{}2x\plus{}3, g(x)\equal{}x^2$
$ f(x)\equal{}G(X), g(x)\equal{}2(G^{\minus{}1}(x))^2\plus{}3$

[quote="SimonM"][quote="sharkman"]Find functions f and g so that f(g(x)) = H.

H(x) = |2x^2 + 3|[/quote]

I'm not sure if that's "all functions" or not, but here are a few

$ f(x) \equal{} x, g(x) \equal{} 2x^2 \plus{} 3$ (and the opposite)
$ f(x) \equal{} x \plus{} 3, g(x) \equal{} 2x^2$
$ f(x) \equal{} 2x \plus{} 3, g(x) \equal{} x^2$[/quote]

Yes

[quote="SimonM"] $ f(x) \equal{} G(X), g(x) \equal{} 2(G^{ \minus{} 1}(x))^2 \plus{} 3$[/quote]

No, very little mistake, just swap  :wink: :

$ f(x) \equal{} 2(G^{ \minus{} 1}(x))^2 \plus{} 3$, $ g(x) \equal{} G(X)$, for any $ G(x)$ bijective.


calculus
integration
trigonometry
calculus computations

how quickly can you solve this (i.e. in as few lines as possible) ? :D

$ \int\;\frac{\textbf dx}{99\plus{}100\cdot\cos x\plus{}101\cdot\sin x}$

Is the substitution tan(x/2) = t enough or is there a special trick here?

[quote="Extremal"]No.
I am sure that there is special trick. because we cant calculat this integral
$ \int\frac{dx}{a\plus{}b\cos x\plus{}c\sin x}$[/quote]

With substitution $ x \equal{}\tan (t/2)$

$ \int\frac{dx}{a\plus{}b\cos x\plus{}c\sin x}\equal{} 2\int\frac{dt}{(c\minus{}b)t^{2}\plus{}2at\plus{}(c\plus{}b)}$

what is the trick misan,though the problem can be solved by taking t=tan(x/2)


geometry
parallelogram
geometry unsolved

A triangle ABC has medians mA, mB, and mC, such that mA is the line segment from A to the midpoint of BC.

m = 1/2 (mA + mB + mC)

Prove the area of ABC = 4/3 the square root of ( m(m - mA)(m - mB)(m - mC) )

here is a hint:
find a triangle with sides equal to $ \frac23m_a,\frac23m_b,\frac23m_c$ and then use heron's formula...

Let $ FACH$ be a parallelogram then $ BFCH$ also be a parallelogram. It easy to check that: triangle $ BEH$ have $ 3$ sides are $ m_a,m_b,m_c$ and $ D$ be the center point of $ BEH$.
Then:
$ S_{ABC} \equal{} 4S_{BDE} \equal{} \frac {4}{3}S_{BEH}$
Apply Heron formula we have solution!

[img]http://i227.photobucket.com/albums/dd72/TPFC/trungtuyen.png[/img]

I've also a formulae for surface of the triangle by its heights but what about the formulae for the surface of the triangle by its angle bisectors? May you give such a formulae?

It is not so beautiful but is powerful. There are formulae for sides of the triangle expressed by medians. Then replace them in Heron's formula.

Here is another very interesting proof:
[hide="Alternate Solution"]
We call the triangle $ ABC$, with medians $ AD$, $ BE$, and $ CF$, with centroid G.  Now, denote the midpoints of $ AG$, $ BG$, and $ CG$ to be $ X$, $ Y$, and $ Z$ respectively.  Notice that $ 2XF \equal{} BG \equal{} 2ZD$, $ 2DY \equal{} CG \equal{} 2XE$, and $ 2YF \equal{} AG \equal{} 2ZE$.  It is easy to see that $ [XFYDZE] \equal{} \frac {[ABC]}{2}$.  We can also easily see that the hexagon is composed of 6 triangles of equal area, each equal to $ \frac {1}{9}$ of the area of the triangle with side lengths $ m_a$, $ m_b$, and $ m_c$.  So, the hexagon is $ \frac {6}{9} \equal{} \frac {2}{3}$ of the area of the triangle.  Also the hexagon is $ \frac {1}{2}$ of the area of triangle ABC.  So our triangle as an area of $ \frac {3[ABC]}{4}$.  Thus, we can follow be Heron's to prove our result.
[/hide]


prove that if T is the Toricelli point in a triangle and O is the circumcenter then

                     $ OT^2=a^2+b^2+c^2-4A\cdot\sqrt{3}.$ ;)

There is a decent explanation here http://mathworld.wolfram.com/FermatPoints.html

Just as a matter of interest, this could be used to solve a certain IMO problem from a long time ago, to show the RHS is positive

Yes, it was posted a few topics back, and I offered a proof http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45611


How many distinct lines go through [b]atleast 2 [/b]of the points (i,j,k) in a 3-D plane if i,j,k are positive integers less than or equal to 4?

[hide="An Attempt"]
There are $5$ coordinate values per plane $i \ , \ j \ , \ \text{and}\ k$.
So the total amount of possible points is $5^{3}=125$.
Then $\binom{125}{2}=15500 \ \text{lines}$, since atleast two points are needed per line.
Though this over counts the number of lines by including lines that intersect more than two points, more than once.
[/hide]

[quote="AstroPhys"]
Though this over counts the number of lines by including lines that intersect more than two points, more than once.
[/quote]
How so?

It asks for distinct lines.
The line through the two points $(0,0,0)$ and $(1,0,0)$ is the same as the line through the points $(2,0,0)$ and $(3,0,0)$, the previous method counts that same line twice.  There are many more instances were lines are counted more than once.

[quote="AstroPhys"]
There are $5$ coordinate values per plane $i \ , \ j \ , \ \text{and}\ k$.
So the total amount of possible points is $5^{3}=125$.
Then $\binom{125}{2}=15500 \ \text{lines}$, since atleast two points are needed per line.
Though this over counts the number of lines by including lines that intersect more than two points, more than once.
[/quote]

How did you get 125?  Isnt it $4^{3}=64$?..

[quote="AstroPhys"]Including point $(0,0,0)$, makes it five and could also make it wrong.[/quote]

Positive integers.. 0 is not positive..

There are 64.  I dont think any lines go throguh 3 points.  I believe there are 4*4*3=48 lines that go through 4 points that are straightly horizontal or verticle.
2*4*3=24 lines that go throguth 4 points that are diagonals on a face of the cube
4 lines that are space diagonals of the cube.

There are 76 lines total that go trhgouh 4 points.  They are counted 4C2=6 times, we want to count only 1 time so subtract them 5 times.

$\binom{64}{2}-5\cdot76=1636$lines


function
number theory
least common multiple
relatively prime

Find the number of positive integers $ \le 25200$ that are relatively prime to $ 3600$.

(Preferably a non-brute force way)

$ \phi{(3600)}\equal{}960$

There are 960 integers $ k$ relatively prime to 3600 in every range $ 3600n \le k < 3600(n\plus{}1)$ for any integer $ n$.

Since $ \frac{25200}{3600}\equal{}7$, the answer is $ 960 \times 7 \equal{} 6720$

Read up on the phi function if you do not understand this: [url]http://en.wikipedia.org/wiki/Phi_function[/url]

i dunno wut brute force is
for this
[hide="by simple l.c.m"]prime factors of 3600 are 2,3,5. so, lcm =30. in span of first 30 numbers,(15+5+2)=22 numbers are divisible by 2,3or5 so, 8 are coprime . similarly for all such sets of 30. now, no. of sets =25200/30=840.
so, $ 8\times 840$ r coprime $ \equal{}6720$[/hide]

yes, phi is also good.


factorial
number theory open
number theory

Find all couples $ (m,n)$ of integers such that $ m(m \plus{} 1)(m \plus{} 2) \equal{} n!$

Small cases: $ m\equal{}1,n\equal{}3;m\equal{}2,n\equal{}4;m\equal{}4,n\equal{}5$.

It's kind of easy to prove that $ m(m\plus{}1)(m\plus{}2)<(m\plus{}1)!$ for $ m>4$.

So, $ m(m\plus{}1)(m\plus{}2)\ge m!$. That means $ (m\plus{}1)(m\plus{}2)\ge (m\minus{}1)!$. Again easily we can prove that this is possible only for $ m\le 5$, so it must be $ m\equal{}5$. But, that case doesn't give solutions, so all solutions are the small cases above.

How does $ m(m \plus{} 1)(m \plus{} 2) < (m \plus{} 1)!$ imply $ m(m \plus{} 1)(m \plus{} 2)\ge m!$   ?

what about trying to prove that if[b] m>8[/b], [b]m[/b] must be infinitely large?
for example if [b]m>8[/b], then [b]n>6[/b], so n is at least [b]7[/b] and [b]5040[/b] divides [b]m(m+1)(m+2)[/b] ==> [b]m≥54[/b], because we would have that [b]16 [/b]divides [b]m(m+1)(m+2)[/b], [b]35[/b] divides [b]m(m+1)(m+2)[/b] and [b]9[/b]divides [b]m(m+1)(m+2)[/b] and the smallest m that fulfills this is 54.
==> [b]27[/b] divides [b]n![/b], which implies [b]n≥9[/b], and so on...


trigonometry
vector
geometry
3D geometry
sphere

Find the net electric field at any point inside a sperical shell due to a surface charge distribution that varies with the polar angle $ \theta$ as $ \sigma \equal{} \sigma_{0}\cos \theta$ where $ \sigma_{0}$ is an constant with proper units.

Is polar angle $ \frac{\pi}{2}\minus{}\lambda$; the latitude?

You mean it turns out to be constant? :o

[quote="shreyas_patankar"]Is polar angle $ \frac {\pi}{2} \minus{} \lambda$; the latitude?[/quote]
yes

[quote="shreyas_patankar"]You mean it turns out to be constant? :o[/quote]
what turns out to be constant?

This charge distribution can be represented as a result of a small relative shift $ a$ of two uniformly charged halls of radius $ R$ whose charges are equal in magnitude and opposite in sign. Then $ \vec{E} \equal{} \frac {\sigma_0}{3\epsilon_0}\vec{k}$ where $ \vec{k}$ is the unit vector along the polar axis from which $ \theta$ is measured.

[quote="phucnv87"]
This charge distribution can be represented as a result of a small relative shift a of two uniformly charged halls of radius $ R$ whose charges are equal in magnitude and opposite in sign. Then $ \vec{E} \equal{} \frac {\sigma_0}{3\epsilon_0}\vec{k}$ where $ \vec{k}$ is the unit vector along the polar axis from which $ \theta$ is measured.[/quote]

well how does one prove that??

[quote="phucnv87"]  Then $ \vec{E} \equal{} \frac {\sigma_0}{3\epsilon_0}\vec{k}$ where $ \vec{k}$ is the unit vector along the polar axis from which $ \theta$ is measured.  [/quote]
Isn't missing a minus sign?

Imagine: a sphere with a uniform volume density of positive charge, and another sphere with an equal uniform volume density of negative charge, originally superposed to make a neutral (uncharged) sphere. The positive sphere is then displaced slightly w.r.t the negative sphere, the body of the uncharged sphere would remain neutral, but a little positive charge will appear on one side, and some negative charge will appear on the opposite side. If the relative displacement of the two spheres is small, the net charge is equivalent to a surface charge on a spherical surface that depends on $ \cos{\theta}$

Now, we know that the potential from each of the spheres of charge is the same as from a point charge, for points outside the sphere, the two spheres considered above which were displaced just slightly are actually like two point charges, so the potential is just that of a dipole.

A possible motivation to think of this would be to remember that potential of a dipole too depends on the cosine of the angle between the axis of the dipole and the radial vector. (The surface considered being a sphere, the potential doesn't vary radially.)

[quote="cyprusad"] If the relative displacement of the two spheres is small, the net charge is equivalent to a surface charge on a spherical surface that depends on $ \cos{\theta}$

[/quote]
how do u get that


I have just discovered a file on my PC where I have stored most of the MathLinks messages which were posted around 21 April 2004 and subsequently got lost when the forum moved to another server.

I am posting them back into the respective threads (although many of them would be obsolete now).

  Darij

Eum darij, the messages I posted then were meant to be temporarily :?

And you've posted quite some topics that already are on here :) you should double check before you post them.

[quote="Peter VDD"]And you've posted quite some topics that already are on here :)[/quote]

Yes, I'm trying to avoid this but I really don't know what's on on most of the subforums. The search function often fails to give out most of the threads.

  Darij


number theory
greatest common divisor
algorithm

Overly clear and difficult (imo, it is easy, but confusing) for MC. Out we get to HSB.

[quote="nutz_for2.718281828"]Compute

$GCD(GCD(GCD(1056, 4368), GCD(425, 6850)), GCD(5345, 358645))$[/quote]

difficult for MC? maybe if you're trying to use Euclid's algorithm ;)


($ab+bc+ca=3,~a,b,c > 0$) \[a^{3}+b^{3}+c^{3}+6abc \geq 9\] I have searched this problem but get nothing(solution). I thought out the solution but it is not very nice. I will post it later.
If who have own solution, post here please. Share with mathlinkers. :D

See here the last post:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63308

My ugly solution
Homogenizing, the inequality is equivalent to \[\left( a^{3}+b^{3}+c^{3}+6abc \right)^{2}\geq 3\left( ab+bc+ca \right)^{3}\Longleftrightarrow\]  \[\left( \sum_{\textrm{cyc}}a^{3}\right)^{2}+12abc\sum_{\textrm{cyc}}a^{3}+36a^{2}b^{2}c^{2}\geq 3\left( \sum_{\textrm{cyc}}a^{3}b^{3}+3abc\sum_{\textrm{sum}}ab^{2}+6a^{2}b^{2}c^{2}\right)\] First, we have \[\left( a^{3}+b^{3}+c^{3}\right)^{2}\geq 3\left( a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}\right)\] Second, the inequality remains \[12abc\sum_{\textrm{cyc}}a^{3}+36a^{2}b^{2}c^{2}\geq 3\left( 3abc\sum_{\textrm{sum}}ab^{2}+6a^{2}b^{2}c^{2}\right)\] Simplify it, we obtain \[4\sum_{\textrm{cyc}}a^{3}+6abc \geq 3\sum_{\textrm{sum}}ab^{2}\] Now we need prove the inequality above. \\ \[2\sum_{\textrm{cyc}}a^{3}+6abc \geq 2\sum_{\textrm{sum}}ab^{2}\qquad (1)\] By Schur's inequality. \[2\sum_{\textrm{cyc}}a^{3}\geq \sum_{\textrm{sum}}ab^{2}\qquad (2)\] By AM--GM or Muirhead. \\
Sum up (1) and (2), we get the desired ienquality.


If $a$, $b$, $c$ are nonnegative real numbers satisfying $a^{2}+b^{2}+c^{2}=1$, prove that

$a+b+c\leq \frac94 abc+\sqrt2$.

[quote="chien than"]Prove that:$a+b+c \leq \sqrt{2}+\frac{9}{4}abc$
where $a;b;c$ are positive real numbers satisfying $a^{2}+b^{2}+c^{2}=1$[/quote]
$\frac{9}{4}abc+\sqrt{2}\geq2abc+\sqrt2\geq a+b+c$ and the stronger inequality see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32399

I sovled it this morning.But my teacher said that I got a misstake !!
 So where is it? and If my solution is false, how to correct it ??

 [b]My solution[/b]
 Rewrite the ineq in form $a+b+c-\frac{9}{4}abc\leq\sqrt{2}$
 assume that $a\leq b\leq c \rightarrow b\leq\frac{\sqrt{2}}{2}$
 Set $a=b-x, c=b+y$. From $a^{2}+b^{2}+c^{2}=1$ we have $3b^{2}-2b(x-y)=1-x^{2}-y^{2}$
 $f(b)=-\frac{9}{4}b^{3}+\frac{9}{4}b^{(}x-y)+\frac{9}{4}bxy+3b-x+y$
 $f'(b)=-\frac{9}{4}(3b^{2}-2b(x-y))+\frac{9}{4}xy+3$
  $f'(b)=\frac{9}{4}(x^{2}+xy+y^{2})+\frac{3}{4}>0$
 So, $f(b)$ increase
 $\rightarrow f(b) \leq f(\frac{\sqrt{2}}{2})=g(x,y)$
 An equality occur when $b=\frac{\sqrt{2}}{2}$ .
 And from ${a\leq b\leq c \rightarrow a=0, c=\frac{\sqrt{2}}{2}}\rightarrow x=\frac{\sqrt{2}}{2},y=0 \rightarrow g(x,y)=\sqrt{2}$
 So, the ineq is proved
 [b][size=150]DID I GET A MISSTAKE?? WHERE??[/size][/b]

I think, the mistake here:
[quote="eagerkill"]
 $f'(b)=-\frac{9}{4}(3b^{2}-2b(x-y))+\frac{9}{4}xy+3$
[/quote]
$x$ and $y$ depends from $b.$ :wink:

[b]argady[/b], Why that is misstake?? Please more clearer!!  :maybe:


function
induction
number theory unsolved
number theory

Let be given a function: $ f:N^* \longrightarrow N$ satisfy:
	1) f is bijective.
	2) $ \forall n \in N^*$ : $ f(n+1)=f(n)-1$ or $ f(n+1)=4f(n)-1$.
Calculate $ f(2004)$

After taking a first look at it, I think (I might very well be wrong, of course), that the function could be obtained as follows (and if this is correct, then it would be very easy for us to find $f(2004)$ or $f(k)$ for any $k$ we have in mind): the sequence $f(1),f(2),f(3),\ldots$ is obtained by first writing down $1,2,3,\ldots$ and then reversing all the substrings of the form $f(2^t),f(2^t+1),\ldots,f(2^{t+1}-1)$. This means that it should look like this: $1,3,2,7,6,5,4,15,14,13,12,11,10,9,8,\ldots$.

In order to prove this we could proceed by induction, but I haven't actually tried to complete a proof. For example, it's easy to show that $f(1)=1$. Try to do that and then continue along these lines.

Yes, it looks like you're right. I rushed into it and made some mistakes, as usual... :) Sorry about that.

First of all, hi to everyone. I'm new to this forum, but already I like it so much.

Well, I found in this problem something weird that I really can't figure out and I wanted to know your advice.

So, if any solution is possible, it has to be bijective from $\mathbb{N}^*$ to $\mathbb{N}$. Then there exists univocally determined some $n\in\mathbb{N}^*$ such that: $f(n) = 0$, and subsequently: $f(n+1) = -1$. But this contraddicts the condition establishing $f$ has to act onto $\mathbb{N}$. Where do I mistake? Try to excuse my awful english...

P.S.: I assumed $\mathbb{N}^*$ denotes the set of positive integers, $\mathbb{N}$ the set of all non-negative ones. Maybe did I misunderstood your notations? Thanks for any reply.

Hi 1l4gn1rt3r0t4vl4s, and welcome to AoPS-MathLinks Forums.

Your assumption is correct, thus we can conclude that there is no such function $f$.

The problem might be valid if $f:\mathbb{N}^*\to\mathbb{N}^*$, or, using international notations, $f:\mathbb{N}\to\mathbb{N}$, where $\mathbb{N}$ should be the set of intergers $n\geq 1$.

Hi 1l4gn1rt3r0t4vl4s and welcome.

Real cool pseudo  :D

Pierre.


We can skip some unnecessary calculations by noting that each dimension of the smaller triangle is $ \frac{1}{2}$ of the larger triangle. The ratio of the areas is $ \left(\frac{1}{2}\right)^2 \equal{} \frac{1}{4} \equal{} \boxed{25\%}$.

big triangle area =4
small triangle area=1

1/4=.25 --->25%

[hide="so the answer is ..."]25%[/hide]

[size=200][hide="im big and proud"][color=#FF80FF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH[/u][/i][/b][/color][/hide][/size]

:bomb: :bomb: :bomb: :bomb:  [size=150][b][i][u][color=#FF0000]RUN FOR YOUR LIVES ITS A BOMB : )[/color][/u][/i][/b][/size]

I said 25% but it said I was incorrect.............is it because of the %?

I don't think it matters, I put in the % sign all the time.


search
quadratics
algebra
polynomial
number theory unsolved
number theory

International olympiad gold medalist?
congratulations!  :first:

if it wasnt so late and if i wasnt lazy to go to look for a pen and a sheet of paper, i would try to solve this problem. for now my answer is that such numbers dont exist. ok, i will try tomorrow if i am not lazy.

[quote="Xixas"]Does there exist infinitely many natural $ n$ such that $ n$, $ n+2$ and $ n+4$ are all squarefree.[/quote]

$ 1).$ To dziudo : Such numbers exist, obviously. $ (3,5,7)$, $ (11,13,15)$, $ (13,15,17)$, $ \ldots$

$ 2).$ To Xixas : here is maybe an idea (not a rigorous demonstration) :
Obviously , these numbers are odd (else $ 4$ would divide at least one of them).
Density of squarefree numbers in $ \mathbb{N}$ is known to be $ \frac{6}{\pi^{2}}$
Since an even squarefree number is always twice an odd one, density of odd squarefree numbers in $ \mathbb{N}$ is $ \frac{4}{\pi^{2}}$
So density of squarefree numbers in $ \{2n+1,n\in\mathbb{N}\}$ is $ \frac{8}{\pi^{2}}$
So density of non-squarefree numbers in $ \{2n+1,n\in\mathbb{N}\}$ is $ 1-\frac{8}{\pi^{2}}=0.189\ldots$

And since density of non-squarefree numbers in the set of odd positive integers is less than $ \frac{1}{3}$, there exist infinitely many odd natural $ n$ such that $ n$, $ n+2$ and $ n+4$ are all squarefree.

We could even say that there exist infinitely many natural $ n$ such that $ n$, $ n+2$, $ n+4$, $ n+6$ and $ n+8$ are all squarefree

[quote="baz"][quote="pco"]
Since an even squarefree number is always twice an odd one, density of odd squarefree numbers in $ \mathbb{N}$ is $ \frac{4}{\pi^{2}}$
So density of squarefree numbers in $ \{2n+1,n\in\mathbb{N}\}$ is $ \frac{8}{\pi^{2}}$
So density of non-squarefree numbers in $ \{2n+1,n\in\mathbb{N}\}$ is $ 1-\frac{8}{\pi^{2}}=0.189\ldots$
[/quote]

pco,
Do you mind explaining this a little.  I was not sure what was happening here.  Thanks.[/quote]

Let $ a$ the density of odd squarefree integers and $ b$ the density of even squarefree integers. We have $ a+b=\frac{6}{\pi^{2}}$
Number of even squarefree integers $ \leq x$ is about $ bx$ but is also the number of odd squarefree integers $ \leq \frac{x}{2}$, so is $ a\frac{x}{2}$.
So $ a=2b=\frac{4}{\pi^{2}}$

Then, density of (odd) squarefree numbers in the set of odd numbers is clearly twice the density of odd squarefree numbers in $ \mathbb{N}$, and so is $ \frac{8}{\pi^{2}}$.

Then density of (odd) non-squarefree numbers in the set of odd numbers is $ 1-$density of (odd) squarefree numbers in the set of odd numbers, and so is $ 1-\frac{8}{\pi^{2}}$ and this number is $ <\frac{1}{5}$

I hope (but I'm not sure) this is clearer.

I suppose you have seen some of the following threads:

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2030043826&t=125867

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1303864617&t=121178

http://www.mathlinks.ro/Forum/viewtopic.php?t=27502&search_id=1005452582&start=20

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2030043826&t=125371

In the present situation, one may actually show that the precise asymptotic density of the set $ \{n \mid n, n+2, n+4 \text{ squarefree}\}$ equals $ \frac{1}{2}\prod_{p \text{ odd prime}}\Big( 1-\frac{3}{p^{2}}\Big) > \frac{1}{2}\prod_{p \text{ odd prime}}\Big( 1-p^{-2}\Big)^{4}= \frac{2^{11}}{\pi^{8}}> \frac{1}{5}$. If anyone wants to see a rigorous proof, let me know!

More generally, for arbitrary integers $ a_{1},\ldots,a_{k}$, there are infinitely many $ n$ with $ n+a_{1}, \ldots, n+a_{k}$ all squarefree, [b] provided[/b] only that there is at least one such $ n$!  What is more, the natural density of such $ n$ in this case is positive, and is effectively computable.

One may even extend the above list to add values of quadratic polynomials, such as $ n^{2}+1$ and so on. For instance, there are infinitely many $ n$ with $ n, n+1, n+2, n+4, n+5, n+6, n^{2}+1,n^{2}+2, n^{2}+4, n^{2}+5, n^{2}+6$ simultaneously squarefree.

All this is proven by a standard sieve argument, as in Ravi B's proof in the first link above.

[quote="vess"]I suppose you have seen some of the following threads:

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2030043826&t=125867

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1303864617&t=121178

http://www.mathlinks.ro/Forum/viewtopic.php?t=27502&search_id=1005452582&start=20

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2030043826&t=125371

....[/quote]

I had not.
Quite impressive  :thumbup:

Thanks


inequalities
function
algebra unsolved
algebra

If $a,b,c$ are three real numbers and \[ a + \dfrac{1}{b} = b + \dfrac{1}{c} = c + \dfrac{1}{a} = t \] for some real number $t$, prove that $abc + t = 0 .$

thank u for this probleme .this is a way to solve the problem

   a-b=1/c-1/b
    a-c=1/a-1/b
    b-c=1/a-1/c
   so,   (a-b)(a-c)(c-a)=(b-c)(b-a)(c-a)/(abc)^2
    if a=b;a=b=c
   if not,(abc)^2=1

  and thanks:)

Case  $a=b=c$ is not  working  because  : $abc+t=0\Leftrightarrow a^3+a+\frac{1}{a}=0\Leftrightarrow a^4+a^2+1=0$ which is  not true for real $a$. So maybe  we  must  assume  $a\neq b\neq c\neq a$.

Yes. The original problem had $a \neq b, b \neq c, a \neq c$.

Here's a cool non-intuitive proof:

In $abc+t$, replace $t = a+\frac{1}{b}$

$abc+t$
= $abc+a+\frac{1}{b}$
= $ac(b+\frac{1}{c})+\frac{1}{b}$
= $act+\frac{1}{b}$
= $\frac{abct+1}{b}$

Similarly, replacing $t$ by the other two expressions we get
$abc+t =$
$\frac{abct+1}{b}= \frac{abct+1}{c}= \frac{abct+1}{a}$

But $a \neq{b}\neq{c}$. So the only way that the last inequality holds is when $abct+1 = 0$. But then $abc+t = 0$

this is my solution.

1. [(a + 1/b)(b + 1/c)(c + 1/a) = t*t*t = t^3]

2. hence, [abc + a + b + c + (1/a) + (1/b) + (1/c) + (1/abc) = t^3]

3. rearranging, [abc + (1/abc) + 3t = t^3]

4. thus, [abc = t^3 - 3t - (1/abc)]

5. substituting for [abc] we get, [t^3 - 3t - (1/abc) + t = 0]

let f(t) = t^3 - 2t

the graph of this function reaches it's min at [t = 1] (also, this value is negative) after which is progressively increases. (i.e. goes into the first quadrant) and max at [t = -1] after which is progressively decreases. (goes into the third quadrant)

now, treating (1/abc) as a constant, we see that the equation in 5. is just [f(t) - (1/abc) = 0] we need to show that there exists a value of t for which the above equation is satisfied, i.e. that the graph of [f(t) - (1/abc)] intersects the x-axis for some value of x

if (1/abc) is +ve, this will shift the graph down by (1/abc) units, no matter what the magnitude of this value is, there will be an intercept on the right hand side of the graph as it constantly increases on the right hand side

if (1/abc) is -ve, this will shift the graph above by (1/abc) units, again, there will be an intercept on the left side for sure as the graph constantly decreases on that side.

thus, there exists a real value of t such that [abc + t = 0]

(this also implies, i think, that any cubic equation with integer coffs will have at least one real root)

The problem was , $a,b,c$ are three distinct numbers and $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}=t$ then prove $abc+t=0$.
Here is my proof , : from each of the relation we get, $ab+1=tb,bc+1=ct,ac+1=at$.
Squaring gives $bt^2=abt+t=(ac+1)(b)+t=abc+b+t \implies$ $ b(t^2-1)=abc+t, c(t^2-1)=$ $abc+t,a(t^2-1)=abc+t$; since $a,b,c$ are all distinct, we must have $t = \pm1$ and $abc+t=0 \Box$

Let $f:=t-{1\over x}$ then $(a\ c\ b)$ is a cycle under $f$ and $a,b,c$ are fixed points of $f^3={{(t^3-2t)x-(t^2-1)}\over
{(t^2-1)x-t}}$. Hence $x=a,b,c$ satisfies $(t^2-1)x^2-(t^3-t)x+(t^2-1)=0$. But $a,b,c$ are distinct, hence $t=\pm 1$.
Plugging this in yields $b={1\over{t-a}}$ and $c={1\over {t-b}}={{t-a}\over {t^2-1-at}}=
{{a-t}\over{ta}}$. Hence $abc={{-1}\over t}=-t$.

Mine is similar to Rieman's
$a+\frac{1}{b}        (1)
=b+\frac{1}{c}        (2)
=c+\frac{1}{a} (3)= t   $
1-2: $a-b= \frac{1}{c}- \frac{1}{b} = \frac{b-c}{bc}$
2-3: $b-c= \frac{1}{a}- \frac{1}{c} = \frac{c-a}{ca}$
3-1: $c-a= \frac{1}{b}- \frac{1}{a} = \frac{a-b}{ab}$
If we multiply all of them,
$(a-b)(b-c)(c-a) = \frac{(a-b)(b-c)(c-a)}{(abc)^2}$
or
$(abc)^2=1$
if $a, b, c$ are all different real numbers, ok
if they are the same, we can use nickolas's solution.


modular arithmetic
quadratics
summer program
PROMYS
vector
geometry
parallelogram

Prove that for $p$ prime: \[p\ |\ 4x^{2}+1 \Rightarrow p \equiv 1 \pmod 4\]

[hide]$4x^{2}\equiv-1 \pmod p$. Then $(2x)^{2}\equiv-1 \pmod p$. So $-1$ is a quadratic residue $\bmod p$. Hence $\left(\frac{-1}{p}\right) = 1$. Then $p \equiv 1 \pmod 4$. [/hide]

[hide]
$(2x)^{2}\equiv-1\mod{p}$, so -1 is a qr mod p.  Then $p \equiv 1\mod{4}$[/hide]

Ah...same soln as Andreas.

[quote="t0rajir0u"][hide="In fact..."] This holds for $p | (x^{2}+1), p > 2$.  There is a nice way to prove this using Minkowski's Theorem  :D [/hide][/quote]

You could just say that holds by the same fact though...and that ${\frac{-1}{p}}=1 \iff p \equiv 1, 2\mod{4}$ can be proven with little difficulty using Euler's Criterion.

[quote="t0rajir0u"] This holds for $p | (x^{2}+1), p > 2$.  There is a nice way to prove this using Minkowski's Theorem  :D [/quote]

how do you prove that?

[quote="ZetaX"]It can be proven with no diificulty and no theorems.

To the solutions above, I've just to say that you proved in fact nothing. You just restated it.[/quote]

Yeah, I know...the whole point seems to be to prove that -1 is a qr mod p iff p is 1, 2 mod 4, but that's not hard to prove at all.

we can use our awesome gaussian integer skillz and be like

$p|(2x+i)(2x-i)$, then $p$ must be factorable over $Z[i]$ into $(a+bi)(a-bi)=p$, we can look at some mod 4 to show that $p\equiv 1\mod 4$, but this does not prove that $p\equiv 1\mod 4 \implies \exists \{a,b\}\subset\mathbb{Z}| a^{2}+b^{2}=p$

that is a well known theorem, but maybe i am using circular reasoning

Well, using the fact that any prime modulus has a primitive root (which isn't too hard to prove either), proving ${\frac{-1}{p}}=1 \iff p \equiv 1, 2\mod{4}$ is trivial.  If needed, I can give a proof of that...(i'll just copy it from my promys midterm write-up  :D  )

[quote="Jan"][quote="t0rajir0u"][hide="In fact..."] This holds for $p | (x^{2}+1), p > 2$.  There is a nice way to prove this using Minkowski's Theorem  :D [/hide][/quote]

Please, t0rajir0u, could you post the solution you are talking about? Thanks[/quote]

[hide="Okay"] Let $a$ be the integer such that $p | (a^{2}+1)$.

Consider the lattice $L$ spanned by the vectors $\left< a, 1 \right>$ and $\left< p, 0 \right>$.  It's not hard to show that the square of the distance of any point in this lattice from the origin is divisible by $p$.

By Minkowski's Theorem for general lattices, a convex region symmetric about the origin with area greater than $4$ times the area of the fundamental parallelogram contains a non-origin lattice point.

We produce such a region:  $\{ x, y | x^{2}+y^{2}< 2p \}$.  This is a circle of radius $\sqrt{2p}$; its area is given by

$\pi r^{2}= 2 \pi p > 4p$

Whereas the area of the fundamental parallelogram is $p$.  Hence there exists a nontrivial lattice point in this region.

This point is $(u, v)$ such that

$p | (u^{2}+v^{2})$

But such that

$u^{2}+v^{2}< 2p$

Hence, $p = u^{2}+v^{2}$ and from here the conclusion is obvious. [/hide]

On a side note, the proof given above can be generalized to show Lagrange's Four Square Theorem.

[quote=rafayaashary1]Or $p\mid \Phi_4(2x)\implies p\mid 4\text{ or }4\mid p-1$[/quote]

Pleaze what is  $ \Phi_4(2x) $ ?

The fourth [url=http://www.yimin-ge.com/doc/cyclotomic_polynomials.pdf]Cyclotomic Polynomial[/url] evaluated at $2x$ :)

If p ( a prime number)is in the form of $4k+3$, and It devides $a^2 + b^2$, then it divides both $a$ and $b$.
This can be proven by order, I'll post it later.

IF $p=4k+3,$ and $4k+2=p-1,$ then
$(2x)^2\equiv -1 \pmod p.$
And $$(2x)^{2\cdot (2k+1)} \equiv (-1)^{2k+1} $$ $$\equiv -1\pmod p.$$ Then
$(2x)^{p-1}\equiv -1\pmod p.$
But we know from Fermat theorem
$(2x)^{p-1}\equiv 1\pmod p.$
Then $p\not = 4k+3,$ It be $4k+1.$


Let $ a,b,c$ be positive numbers such that: $ a\plus{}b\plus{}c\equal{}3$.Prove that;
$ \frac{ab}{\sqrt[3]{13a\plus{}3b^2}}\plus{}\frac{bc}{\sqrt[3]{13b\plus{}3c^2}}\plus{}\frac{ca}{\sqrt[3]{13c\plus{}3a^2}} \le \sqrt[3]{\frac{9(ab\plus{}bc\plus{}ca)}{16}}$

:)

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that: $ a \plus{} b \plus{} c \equal{} 3$.Prove that;
$ \frac {ab}{\sqrt [3]{13a \plus{} 3b^2}} \plus{} \frac {bc}{\sqrt [3]{13b \plus{} 3c^2}} \plus{} \frac {ca}{\sqrt [3]{13c \plus{} 3a^2}} \le \sqrt [3]{\frac {9(ab \plus{} bc \plus{} ca)}{16}}$

:)[/quote]
It's easily by Holder and AM-GM.
This is my solution for this one:
Applying Holder inequality :
$ {\left( {\sum {\sqrt [3]{{\frac {{16{a^2}b}}{{13a \plus{} 3{b^2}}}}}.\sqrt [3]{{ab}}.\sqrt [3]{b}} } \right)^3} \le \left( {\sum {\frac {{16{a^2}b}}{{13a \plus{} 3{b^2}}}} } \right).3(ab \plus{} bc \plus{} ca)$
So we have to prove that:
$ \sum {\frac {{16{a^2}b}}{{13a \plus{} 3{b^2}}}} \le 3$
We have: $ 13a \plus{} 3{b^2} \ge 16\sqrt [{16}]{{{a^{13}}{b^6}}}$ (By AM-GM)
Since we must prove that:
$ \sum {\sqrt [{16}]{{{a^{19}}{b^{10}}}}} \le 3$ ,which is true too by AM-GM
Because $ \sqrt [{16}]{{{a^{19}}{b^{10}}}} \equal{} \sqrt [{16}]{{{{\left( {ab} \right)}^{10}}.{a^2}.{a^2}.{a^2}.{a^2}.a.1}} \le \frac {{10ab \plus{} 4{a^2} \plus{} a \plus{} 1}}{{16}}$

[quote="leedt26"][quote="quykhtn-qa1"][quote="quykhtn-qa1"]What about the following,leedt26?Can you solve it?
$ \frac {ab^2}{13a \plus{} 3b^2} \plus{} \frac {bc^2}{13b \plus{} 3c^2} \plus{} \frac {ca^2}{13c \plus{} 3a^2} \le \frac {3}{16}$
With the same condition[/quote]
leedt26,with this problem,can you solve it? :)[/quote]
No , I can't solve it, quykhtn-qa1 .Can you make some hint with this inequality? .Thanks![/quote]
dear my friend,you can solve it by Cauchy-Schwarz ineqality :)

[quote="quykhtn-qa1"][quote="leedt26"][quote="quykhtn-qa1"]What about the following,leedt26?Can you solve it?
$ \frac {ab^2}{13a \plus{} 3b^2} \plus{} \frac {bc^2}{13b \plus{} 3c^2} \plus{} \frac {ca^2}{13c \plus{} 3a^2} \le \frac {3}{16}$
With the same condition[/quote]
We can rewrite the inequality
$ \frac {ab^2}{13a \plus{} 3b^2} \plus{} \frac {bc^2}{13b \plus{} 3c^2} \plus{} \frac {ca^2}{13c \plus{} 3a^2} \le \frac {3}{16}$
as
$ \frac {{{a^2}}}{{13a \plus{} 3{b^2}}} \plus{} \frac {{{b^2}}}{{13b \plus{} 3{c^2}}} \plus{} \frac {{{c^2}}}{{13c \plus{} 3{a^2}}} \ge \frac {3}{{16}}$
By Cauchy-Schwarz inequality we have
$ \frac {{{a^2}}}{{13a \plus{} 3{b^2}}} \plus{} \frac {{{b^2}}}{{13b \plus{} 3{c^2}}} \plus{} \frac {{{c^2}}}{{13c \plus{} 3{a^2}}} \equal{} \frac {{{a^3}}}{{13{a^2} \plus{} 3a{b^2}}} \plus{} \frac {{{b^3}}}{{13{b^2} \plus{} 3b{c^2}}} \plus{} \frac {{{c^3}}}{{13{c^2} \plus{} 3c{a^2}}} \ge \frac {{{{\left( {a\sqrt a \plus{} b\sqrt b \plus{} c\sqrt c } \right)}^2}}}{{13\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right) \plus{} 3\left( {a{b^2} \plus{} b{c^2} \plus{} c{a^2}} \right)}}$
I think $ p,q,r$ works here,right. :maybe:[/quote]
Dear my friend,the way of you can't solve it  :lol:[/quote]
Yes,you're right quykhtn-qa1 . My way can't solve it.

[hide="@quykhtn-qa1"]can you post your full proof for it ,Quy?[/hide]

Dear my friends,you can solve it by Cauchy-Schwarz inequality and The sum of square :) .
I hope you can sove it :) .


if there are two boxes of matches, ten in each. and a person at random takes matches from each box, until one is empty. what is the probability that the other box has three matches remaining in it?

i have a couple of soultions....so im confused as to which is right!

Here is how I would approach the problem.

[hide="my solution"]

We're going to define [a,b] to mean the probability that at any time (obviously, for any a,b, there is only one such time b/c the total number of matches is monotonically decreasing) there are a matches in box 1 and b in box 2.  We want to find [3,0]+[0,3].

We trivially set [10,10] = 1, and any duplet where either a or b is greater than 10 to be zero.  Then we define some common-sense rules.

[a,b] = 1/2*[a,b+1]+1/2*[a+1,b].

[a,b] = [b,a] (from symmetry).

[10-a,10-b]=[a,b].  This one is a little tricky, but we can think of filling the box with ten "negative matches" and then taking away "negative matches".  At any rate, it holds true.  So we really need to find 2*[7,10].

[10,10]=1
[9,10]=1/2
[8,10]=1/4
[7,10]=1/8

So the probability is 1/4.[/hide]

ahh, ive got something quite different though =S.

one way of mine is to treat the chance of picking matches from either box has 1/2, and the other i took it as after every match pickd, the probability of choosing a match from the now used box is less. Regardless of this difference in assumptions, different methods for both assumptions i found gave me different answers. so i am lost as to where i have gone wrong in any of my methods =S

Why don't you write down what you've done in full, and then we can have a look at it?

Teki-Teki's argument is wrong because (in that notation) we do not actually have $ [a, b] \equal{} [10 \minus{} a, 10\minus{}b]$ -- just consider what that means for $ a \equal{} b \equal{} 0$.  (I think the error was in assuming that we keep going until [i]both[/i] boxes are empty, rather than just one.)

Ah - you are right.  Darn...

I suppose that the general method still works, just without the simplification.  So it would be tedious calculation, which I don't particularly feel like doing.

hmmmm.... I have a suspicion that my solution is wrong..
Here's what I did:

Every possible match-taking sequence can be represented as a permutation of 10 $ a$'s and 10 $ b$'s,
E.g. $ abaabaaabbbbababaabb$ is the sequence where the first match is taken from the first box, second from the second box, third from the first box, fourth from the first box, etc.

So we want the last four letters to be $ abbb$ or $ baaa$, since the question is the same as asking "What is the probability that the last three matches drawn are from the same box (Exactly 3)?"

The two conditions are symmetric, and in the first, we want to arbitrarily place $ 9$ $ a$'s and $ 7$ $ b$'s, so the probability is

$ \frac {\binom {7} {16} \times 2}{\binom{10}{20}} \equal{} \frac {40}{223}$


USAMTS
AMC
USA(J)MO
USAMO
summer program
PROMYS

Age 15
USCF: 1930 ish- 50 ish in the country used to be top 20 but slacked off ;( and don't play much anymore
Squash Ranking: 90 ish in the country
Math: two years ahead, silver usamts, no usamo, promys, won mathleauge for my state 2 years in a row

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214779

Moved from RT.

uuummm.... I know many people that could trump lots of your accomplishments. So don't go post them unless it's in a topic where people post their accomplishments.


function
limit
integration
calculus
derivative
probability and stats

If $ \phi(t)$ is the characteristic function of a random variable $ X$ and if $ \lim_ {t\to 0} \frac {\phi(t)\minus{}1}{t^2}\equal{}c$ then prove that $ EX\equal{}0$ and $ EX^2\equal{}\minus{}2c$.

$ \phi(0)\equal{}1$, the statements are direct consequence of the basic properties of characteristic function.

[quote="siiz"]$ \phi(0) \equal{} 1$, the statements are direct consequence of the basic properties of characteristic function.[/quote]

Can you give me some hint? :)

$ \phi (t) \equal{} E( e^{itX})\equal{}\int_{\minus{}\infty}^\infty e^{itx}dF_X(x)$

$ \to \phi (0) \equal{} E(1)\equal{}1$

$ \phi^{(k)} (t) \equal{} \int_{\minus{}\infty}^\infty x^ke^{itx}dF_X(x)$

$ \to \phi^{(k)} (0) \equal{} \int_{\minus{}\infty}^\infty x^kdF_X(x)\equal{}EX^k$

hopefully you can see that the rest is just lhospital or taylor theorem

[quote="limsup"][quote="kenn4000"]$ \phi^{(k)} (t) \equal{} \int_{ \minus{} \infty}^\infty x^ke^{itx}dF_X(x)$

$ \to \phi^{(k)} (0) \equal{} \int_{ \minus{} \infty}^\infty x^kdF_X(x) \equal{} EX^k$

hopefully you can see that the rest is just lhospital or taylor theorem[/quote]

How do we know that the characteristic function is $ k$ times differentiable?[/quote]

you don't (a priori).  the number of times the characteristic function is differentiable is related to the number of (finite) moments you have.

That's what. How do we know that the 1st and 2nd moments are finite? We have to prove that they are finite and then prove $ EX\equal{}0$ and $ EX^2\equal{}\minus{}2c$. Sorry if I am mistaken.

[quote="limsup"]How do we know that the 1st and 2nd moments are finite? [/quote]

this is where you use the condition that $ (\phi(t)\minus{}1)/t^2\to c\in(\minus{}\infty,\infty)$ as $ t\to 0$...

Can we use L Hospital rule? But for that we need differentiability of  $ \phi(t)$.  Isn't it?

the existence of $ \lim_{t\to 0} \frac {\phi(t) \minus{} \phi(0)}{t^2} \equal{} c\in( \minus{} \infty,\infty)$

implies that $ \exists \delta$ s.t. $ |x| < \delta \to |\phi(x) \minus{} \phi(0)| < K|x \minus{} 0|^2$

this condition implies existence of a second derivative at 0

Another question you should be asking yourself is why you can get away with differentiating under the integral sign.

For example (assuming I haven't messed something up here):  (EDIT: In fact, I did mess something up here -- a $ t$ disappeared somewhere when computing the derivative.  However, I'm still not convinced that you can say $ \phi^{(k)} (t) \equal{} \int_{ \minus{} \infty}^\infty x^ke^{itx}dF_X(x)$ without assuming in advance that the $ k^{th}$ moment of $ X$ exists).

(incorrect example preserved below for posterity and my own embarrassment).
[hide]Let $ X$ be defined by $ P(x \equal{} 2^k) \equal{} P(x \equal{} \minus{} 2^{k}) \equal{} 2^{ \minus{} k \minus{} 1}$ for all positive $ k$.  Then
\[ \phi(t) \equal{} \sum_{k \equal{} 1}^{\infty} \frac {e^{ \minus{} i 2^kt} \plus{} e^{i2^k t}}{2^{k \plus{} 1}} \equal{} \sum_{k \equal{} 1}^{\infty} \frac {cos(2^k t)}{2^{ k }}
\]
(note that all sums so far are absolutely convergent, so I can get away with swapping around orders of summation here).

We have $ \phi(0) \equal{} 1$, and for any $ m > 0$ and any $ t$ we have
\[ |\frac {\phi(t) \minus{} 1}{t}| \leq |\sum_{k \equal{} 1}^m \frac {\cos(2^k t) \minus{} 1}{2^{ k}}| \plus{} |\sum_{k \equal{} m \plus{} 1}^{\infty} \frac {\cos(2^k t) \minus{} 1}{2^{ k }}|
\]
For any fixed $ m$ the first sum goes to 0 as $ t$ tends to $ 0$, while the second sum is at most $ 2^{ \minus{} m \plus{} 1}$ (bound each term individually, and use the triangle inequality).   This means that $ \phi$ is differentiable at $ 0$, with derivative $ 0$.  However, $ E(X)$ doesn't exist!

[/hide]


factorial
number theory proposed
number theory

Prove that there are infinetely many solutions such that n^2+1|n! (n is natural number)

This problem has been posted a lot of times.
But i will give you a hint:

Choose n=2a² .

If $ n \equal{} 2a^2$, so $ n^2 \plus{} 1 \equal{} 4a^4 \plus{} 1 \equal{} (2a^2 \plus{} 1)^2 \minus{} 4a^2 \equal{} (2a^2 \minus{} 2a \plus{} 1)(2a^2 \plus{} 2a \plus{} 1)$.

So, how $ 2a^2$, $ 2a^2 \minus{} 2a \plus{} 1$ is a factor of $ (2a^2)!$.
Now, you can make $ 2a^2\plus{}2a\plus{}1$ multiple of 5, for example, to get new factors and so, this new factors will appear on the factorial!
The idea is that!
Try a bit more.

Sorry for my English!


Προσπαθώ εδώ και μέρες μία άσκηση αλλά δεν τα έχω καταφέρει μπορώ να πω.

Έστω συνάρτηση $ f$ συνεχής στο $ [a,b]$ και παραγωγίσιμη στο $ (a,b)$. Αν ισχύει
$ \frac{f(a)\plus{}f(b)}{2}\equal{}\frac{\int_{a}^{b}f(x)dx}{b\minus{}a}$
τότε να δειχθεί ότι υπάρχει $ z\in(a,b)$ τέτοιο ώστε

$ f'(z)\equal{}2\frac{f(z)\minus{}f(a)}{z\minus{}a}$

Ελπίζω να μην βλεπω κάτι προφανές..... :blush:

Παιδιά πρόκειται για πολύ ωραία άσκηση. Μου κέντρισε του ενδιαφέρον από την πρώτη στιγμή που την είδα. Την προσπάθησα, αλλά δεν την έλυσα.

[hide]Κατά τη γνώμη μου, η άσκηση δεν απαιτεί κάποιο βαρύ θεώρημα-ίσως να εντάσσεται και στην ύλη της Γ.
Η αρχική ιδέα είναι να κάνεις Rolle στην $ G(x)\equal{}\displaystyle f(x)(x\minus{}a)\minus{}3\int_a^x{f(t)dt\plus{}2f(a)x}$.
Κάπου κολλάει το πράγμα...Επίσης υπάρχει και μια δυνατότητα προσέγγισης με ΘΜΤΟΛ και Θεώρημα Μέγιστης και ελάχιστης τιμής αφού από αυτά τα δύο προκύπτει ότι υπάρχουν $ c_1,c_2$ τέτοια ώστε $ f(c_1)\equal{}\frac{f(a)\plus{}f(b)}{2}$ και $ f(c_2)\equal{}\displaystyle \frac{\displaystyle{\int_a^b{f(x)dx}}}{b\minus{}a}$, άρα $ f(c_1)\equal{}f(c_2)$. Τέσπα αυτές είναι διάφορες ιδέες-πιστεύω θα έχει κάποιο trick αλλά πάνω σε αυτά που προανέφερα. I wish I had more to spend on the problem.[/hide]

Προσπαθείστε την - έχει ενδιαφέρον.

Νομίζω ότι για να μην την λύνουν τα παιδιά κάτι κακό πρέπει να έχει η άσκηση
Παραθέτω μια διαισθητική λύση φέρνοντας ένα αντιπαράδειγμα (που θα αρέσει στον Γιώργο)για να τεκμηριώσω ότι τα πράγματα δεν πάνε καλά εδω

[quote="Nick Rapanos"]Παιδιά πρόκειται για πολύ ωραία άσκηση. Μου κέντρισε του ενδιαφέρον από την πρώτη στιγμή που την είδα. Την προσπάθησα, αλλά δεν την έλυσα.

[hide]Κατά τη γνώμη μου, η άσκηση δεν απαιτεί κάποιο βαρύ θεώρημα-ίσως να εντάσσεται και στην ύλη της Γ.
Η αρχική ιδέα είναι να κάνεις Rolle στην $ G(x) \equal{} \displaystyle f(x)(x \minus{} a) \minus{} 3\int_a^x{f(t)dt \plus{} 2f(a)x}$.
Κάπου κολλάει το πράγμα...Επίσης υπάρχει και μια δυνατότητα προσέγγισης με ΘΜΤΟΛ και Θεώρημα Μέγιστης και ελάχιστης τιμής αφού από αυτά τα δύο προκύπτει ότι υπάρχουν $ c_1,c_2$ τέτοια ώστε $ f(c_1) \equal{} \frac {f(a) \plus{} f(b)}{2}$ και $ f(c_2) \equal{} \displaystyle \frac {\displaystyle{\int_a^b{f(x)dx}}}{b \minus{} a}$, άρα $ f(c_1) \equal{} f(c_2)$. Τέσπα αυτές είναι διάφορες ιδέες-πιστεύω θα έχει κάποιο trick αλλά πάνω σε αυτά που προανέφερα. I wish I had more to spend on the problem.[/hide]

Προσπαθείστε την - έχει ενδιαφέρον.[/quote]
Νικο δεν εξασφαλίζεται ότι το $ c_1$ είναι διαφοετικό του $ c_2$ και το να θεωρήσεις την κατάλληλη G λύνοντας μια διαφορική εξίσωση δεν με οδήγησε κάπου καλά
Αφού είχα γράψει το αντιπαράδειγμα είδα το post που έστειλες

k. Bori efxaristoume gia to antiparadeigma. den theorisa oti ta c1 kai c2 einai diaforetika, apla egrapsa duo diaforetikous tropous prosegiseis, alla to thema theoreitai liksan!Efxarisotume polu,

Για να γίνει πιο αυστηρό το αντιπαράδειγμα θεωρείστε
$ g(x)\equal{}\frac{\minus{}1}{48}x^3\plus{}\frac{13}{48}x^2\frac{3}{16}x\frac{\minus{}23}{16} , a\equal{}1 , b\equal{}3$
Τότε η $ g$ είναι κυρτή στο $ [a,b]$ , $ g(a)\plus{}g(b)\equal{}1\minus{}1\equal{}0 ,g'(b)\equal{}\frac{5}{4}<2g'(a)\equal{}\frac{4}{3}$

Αν δεν κάνω λάθος, και η $ f(x) \equal{} x$ (σε οποιοδήποτε διάστημα [a,b]) είναι αντιπαράδειγμα.

Χίλια συγνώμη σε όλους για τον χρόνο που του σπατάλησα τζάμπα αλλά δεν το ήξερα... :blush:
Κι εμένα μου είχει φάει καμιά 10αρια μέρες και δεν είχε αποτέλεσμα.Δεν σκέφτηκα ότι θα ήταν λάθος οπότε δεν πήγε το μυαλό μου να βρω αντιπαράδειγμα.
Ελπίζω την επόμενη φορά να μην μου τύχει κάτι τέτοιο....


the last 24 game died. so im making a new one.

we start easy but make future ones harder because were doing points.
if 1 day has elapsed and no one has gotten an answer, then you must post your answer and you get a point.
no exponents, square roots, and others. just the four basic funtions.
up to 2 digit numbers. no like 999999999999999999, 7, 999999999999242 or whatever that other guy posted.
REMEBER TO POST THE NUMBERS!!!

[b]#1[/b]
1,2,3,4

oh and please post the numbers.

[u][i][b]#2[/b][/i][/u]

1,5,7,10

lets do points. and post some hard ones so people dont get it right off the bat.

[b][i][u][color=red]#3[/color][/u][/i][/b]
11,11,5,1

ZeroFive1-1
D3m0n Shad0w-1

Err, sorry, but what is 24, and how do you play?

Well, you're given  4 numbers, and you try to make 24 using addition, subtraction, multiplication, and division. But there are different styles of it...

[quote="7h3.D3m0n.117"]Hmm are fractions/decimals, etc allowed? There are different versions of 24 and it might help 24 players if they know which one they are playing.[/quote] decimals are, fractions are not

$ 2, 4, 6, 8$

Let's get into the habit of posting new ones after solving the previous one.

[quote="jeez123"]I'm not sure if this solution was already posted, anyhow:

$ 9 \plus{} 3 \equal{} 12$
$ 7 \minus{} 5 \equal{} 2$
$ 12 \cdot 2 \equal{} 24$

4 7 8 8[/quote]

8-7. 1
now we have 4 8 1
Pretty obvious now- 4-1,
Now we have 3 8
3*8
24.

The totient function usually isn't allowed in 24.

$ (7\plus{}1)(4\minus{}1)$ or $ (7\minus{}1)(4)(1)$

$ 8, 13, 19, 20$

I need these functions to solve problems, also you can say:
φ(7)*4*1*1  !!!