geometry
parallelogram
circumcircle
incenter
geometric transformation
reflection
Euler

If A', B' and C' are the contact points of the incircle of triangle ABC with sides BC,CA and AB respectively, let A" and B" be the antipodes of A' and B' wrt the incircle of ABC and let A3 = AA"^BC and B3=BB"^CA. Prove that IOJP is a parallelogramm, where I , O, J and P are the circumcenters of A"B"C', ABC, A3B3C and A'B'C respectively.

PS
Actually the configuration has a LOT of nice properties.

Not unexpected at all, at least to me  :?

Here is a quick sketch of a solution:

Define not only $A_{3}=AA^{\prime \prime }\cap BC$, $B_{3}=BB^{\prime \prime }\cap CA$, but also $C_{3}=CC^{\prime \prime }\cap AB$. Then, it is easy to see that the points $A_3$, $B_3$ and $C_3$ are the points where the A-excircle, the B-excircle and the C-excircle of triangle ABC touch the sides BC, CA, AB, respectively.

The triangle A'B'C' is the pedal triangle of the incenter I of triangle ABC. Hence, the points A' and B' lie on the circle with diameter CI, and the circumcenter P of triangle A'B'C is the midpoint of the segment CI.

On the other hand, since the points $A_3$, $B_3$ and $C_3$ are the points where the A-excircle, the B-excircle and the C-excircle of triangle ABC touch the sides BC, CA, AB, respectively, the triangle $A_3B_3C_3$ is the pedal triangle of the Bevan point V of triangle ABC. By the same argument as before, we see that the circumcenter J of triangle $A_3B_3C$ is the midpoint of the segment CV.

[The [i]Bevan point[/i] V of triangle ABC is defined as the circumcenter of triangle $I_aI_bI_c$, where $I_a$, $I_b$, $I_c$ are the excenters of triangle ABC. It can be easily seen that this point V lies on the perpendiculars from $I_a$, $I_b$, $I_c$ to the lines BC, CA, AB, respectively, or, in other words, on the lines $I_aA_3$, $I_bB_3$, $I_cC_3$. Hence, the triangle $A_3B_3C_3$ is the pedal triangle of the Bevan point V. Another property of the point V is that it is the reflection of the incenter I in the circumcenter O of triangle ABC. This all follows from chasing angles and noting that the circumcircle of triangle ABC is the nine-point circle of triangle $I_aI_bI_c$.]

Altogether, the points P and J are the midpoints of the segments CI and CV. Now, remembering that the point V is the reflection of the point I in the point O, it's not a big deal anymore to show that the quadrilateral IOJP is a parallelogram.

  Darij

It's equivalent to this: if $X_b,X_c$ are the points where the excircles of $ABC$ corresponding to $B,C$ touch $CA,AB$ respectively, and $P$ is the point where the $a$-external bisector cuts the circumcircle the second time, then $A,X_b,X_c,P$ are concyclic. This follows easily from the congurence of the triangles $PX_cB,\ PX_bC$ (this congruence follows from $PB=PC,BX_c=CX_b,\angle X_cBP=\angle X_bCP$).

Terribly sorry if Darij did the same thing :).

[quote="grobber"]It's equivalent to this: if $X_b,X_c$ are the points where the excircles of $ABC$ corresponding to $B,C$ touch $CA,AB$ respectively, and $P$ is the point where the $a$-external bisector cuts the circumcircle the second time, then $A,X_b,X_c,P$ are concyclic.[/quote]

Hmm... I don't get the equivalence! As for the concyclicity of the points A, $X_b$, $X_c$ and P, it is a special case (with d = s - a) of Lemma 3 in my "The Mitten point as radical center" note (at [url=http://de.geocities.com/darij_grinberg]my geometry website[/url], as usual).

Could you explain to me why this concyclicity is equivalent to the problem? Sorry if I'm blind...

  Darij

[quote="grobber"]That's because we have to show that the line of centers of the circles $(AX_bX_c),(ABC)$ must be shown to be parallel to the $a$-bisector, meaning that we must show that $AT\perp$ to the $a$-bisector, where $A=(AX_bX_c)\cap(ABC)$, which is equivalent to showing that $P=(AX_bX_c)\cap(ABC)$.[/quote]

I think you have too much points denoted by P  :)

Anyway, this is a nice proof of OJ || IP (where P is Sprmnt21's P). What about IO || JP ?

  Darij

I guess this is where I should say "sorry, I forgot about that" :). Looks like they weren't equivalent after all..

Don't worry, actually you found a nice proof of a part of the assertion and an interesting relationship between Sprmnt21's problem and the concyclicity of the points A, $X_b$, $X_c$ and P.

  Darij

[quote="darij grinberg"]Not unexpected at all, at least to me  :?  Darij[/quote]

Indeed the inspiration derive mostly from aome of your messages  :) .

Now I will go to carefully read yours answers.

[quote="darij grinberg"]Not unexpected at all, at least to me  :?

Here is a quick sketch of a solution:

Define not only $A_{3}=AA^{\prime \prime }\cap BC$, $B_{3}=BB^{\prime \prime }\cap CA$, but also....

Another property of the point V is that it is the reflection of the incenter I in the circumcenter O of triangle ABC. This all follows from chasing angles and noting that the circumcircle of triangle ABC is the nine-point circle of triangle $I_aI_bI_c$.]

Darij[/quote]

This seems a sort of "reverse engineering" of my thought.

Just in the latter paragraph, as I don't know Bevan, I used Euler theorem, applied to the ex-centers triangle, to establish that I,O and V ( hortocenter, centroid and circum center respectively of IaIbIc) are collinear and IO=VO.

[quote="sprmnt21"]This seems a sort of "reverse engineering" of my thought.

Just in the latter paragraph, as I don't know Bevan, I used Euler theorem, applied to the ex-centers triangle, to establish that I,O and V ( hortocenter, centroid [i][I bet you mean: nine-point center - DG][/i] and circum center respectively of IaIbIc) are collinear and IO=VO.[/quote]

As you have probably noted, what you are doing here is just the most usual way to establish the properties of the Bevan point V.

[quote="sprmnt21"]Un [i]addendum [/i] to the problem.

First fact.
If one rotate of 180 ABC around its circumcenter O s.t. the transformed traingle is A'B'C', we have that IC'//VC, as V is the half rotation of I around O too.

Second fact.

Let XYZ be a triangle and R and S the orthogonal projections of a point T(in the same plain of XYZ) on XY and YZ respectively. Let R'S' be the simmetric of RS wrt the external bisector of angle Y, then TY is orthogonal to R'S'.

Proof.
Given that RTSY is cyclic, is easy to derive the claim.

These two facts togeter constitute a proof of the beautiful problem of Homid.[/quote]

Ah, you mean the one at http://www.mathlinks.ro/Forum/viewtopic.php?t=6472 . Thanks, that's a very nice application of the Bevan point! Another problem about the Bevan point is the following: http://www.mathlinks.ro/Forum/viewtopic.php?t=5680

  Darij


limit
induction
algebra proposed
algebra

Let $\{x_n\}$ be a real sequence defined by:
\[x_1=a,x_{n+1}=3x_n^3-7x_n^2+5x_n\]
For all $n=1,2,3...$ and a is a real number.
Find all $a$ such that $\{x_n\}$ has finite limit when $n\to +\infty$ and find the finite limit in that cases.

Let $f(x)=3x^3-7x^2+5x$.The sequence is defined by $x_{n+1}=f(x_n),x_1=a$

[color=blue][b]For a given a point $x_n$ on $x$ axis, we can obtain a point $x_{n+1}$ by using graph $y=f(x)$ and $y=x$.[/b][/color]

I want to sketch the these graphs,but I can't do by using Tex,that's why I want  someone to sketch.
Repeating this procedure,yeilding

$a<0\ \cdots \lim_{n\to\infty} x_n=-\infty$

$a=0\ \cdots \lim_{n\to\infty} x_n=0$

$0<a<\frac{4}{3}\ \cdots \lim_{n\to\infty} x_n=1$

$a=\frac{4}{3}\ \cdots \lim_{n\to\infty} x_n=\frac{4}{3}$

$a>\frac{4}{3}\ \cdots \lim_{n\to\infty} x_n=\infty$

Therefore the range of $a$ is $0\leqq a\leqq \frac{4}{3}$.

[quote="kunny"]

$0<a\leqq 1\ \cdots \lim_{n\to\infty} x_n=1$

$1<a<\leqq \frac{4}{3}\ \cdots \lim_{n\to\infty} x_n=\frac{4}{3}$
[/quote]
hmm....may be i'm wrong :? .but my result is if $0<a<\frac{4}{3}$ then $\lim\limits_{n \to \infty}x_n=1$
[b]kunny[/b] can you post your solution! :)

[quote="Kapi"]
hmm....may be i'm wrong :? .but my result is if $0<a<\frac{4}{3}$ then $\lim\limits_{n \to \infty}x_n=1$
[b]kunny[/b] can you post your solution! :)[/quote]
I think you are right.The answer is:
[tex] a=0: \lim x_n=0; [/tex]
[tex] 0<a<\frac{4}{3}:\lim x_n=1; [/tex]
[tex] a=\frac{4}{3}:\lim x_n=\frac{4}{3} [/tex] :)

what happens with $a=1$, we have $x_2=3-7+5=1$, so all terms are $1$, right?

Who knows the solution for the case : $ 0 \leq a \leq 4/3 $ ?

To [b]Kapi[/b],you are right. :)  :)  I have just edited.

kunny

Here's a solution when $1 \le a < \frac{4}{3}$.  Let $c = 3a^2 - 4a + 1$.  Note that $0 \le c < 1$.  By induction on $n$, we can show that
\[1 \le x_n \le 1 + (a - 1) c^{n - 1}.\]
Taking limits shows that $\lim x_n = 1$.

[quote="Myth"]Is it your intuition, kunny? ;) :lol:[/quote]

No.I have  my justification. :D

For the case [tex] 0<a<1 :[/tex]
There exits $ m $ satisfying: [tex] x_{m-1}<\frac{1}{3} [/tex] and [tex] \frac{1}{3} \leq x_{m} \leq 1 [/tex]
(Or $ x_{n} <\frac{1}{3} $, $\forall n \Rightarrow 0< \lim x_n< \frac{1}{3}$ (contracdition) since $\lbrace x_n \rbrace$ increases in $(0,1)$)
We show that [tex] x_{m+1} \in [1,\frac{4}{3}) [/tex],and this case was solved by [b]Ravi B[/b].
[tex] x_{m+1}-1 =(x_m-1)^2(3x_m-1) \geq 0 \Rightarrow x_{m+1} \geq 1 [/tex].
[tex] x_{m+1}-\frac{4}{3}=3{x_m}^3-7{x_m}^2+5{x_m}-\frac{4}{3} \leq [/tex]
[tex] 3 \cdot ({\frac{5}{9}})^3-7 \cdot ({\frac{5}{9}})^2 + 5 \cdot ({\frac{5}{9}}) -\frac{4}{3} <0 \Rightarrow x_{m+1} < \frac{4}{3} [/tex]
So it was done. :)

Oh, my friend's just has a solution for a in (0,1):
if a in (0,1) we try to prove that exist x{n} is in {1,4/3}, so a in (0,1) satisfies.

If $a<0$ then $x_n<0$ , so take $x_n=-a_n$ then the eq. becomes $a_{n+1}=3a^3_n+7a^2_n+5a_n$, then $\limits_{n\to\infty} x_n=-\infty$
 Let $f(x)=3x^3-7x^2+5x$, then $f'(x)=9x^2-14x+5$, $f'(x)<0$ for $\frac{5}{9}<x<1$, then $f(x)$ is decreasing in $\frac{5}{9}<x<1$.
$x_{n+1}>(2\sqrt{15}-7)x^2_n$, again$x_{n+2}>(2\sqrt{15}-7)x^2_{n+1}\Rightarrow{\frac{x_{n+2}}{x_{n+1}}>(2\sqrt{15}-7)x_{n+1}\Rightarrow{x_n}}$ has a limit for $x>0$.
Let the limit is $L$, then $L=1,\frac{4}{3}$.
Now, for $0<a\le{1}$, the limit is $1$.
           for $1<a\le{\frac{4}{3}}$, the limit is $\frac{4}{3}$.
          for $a>\frac{4}{3}$ ,the limit is $+\infty$, since $f$ is increasing.
           for $a=0$, the limit is $?$


The following is said to be unparseable/dangerous:

[tex]\textrm{Prove that for } $0 \leq a,b,c \leq 1, \frac{a}{b+c+1} + \frac{b}{c+a+1} + \frac{c}{a+b+1} + (1-a)(1-b)(1-c) \leq 1$[/tex]

However, splitting it into two parts seems to work fine.  The only thing idifferent is that TeX mode ends after the + sign, and begins again at (1-a)...

[tex]\textrm{Prove that for } 0 \leq a,b,c \leq 1, \frac{a}{b+c+1} + \frac{b}{c+a+1} + \frac{c}{a+b+1} +[/tex] [tex] (1-a)(1-b)(1-c) \leq 1[/tex]

Hm. I don't know why that doesn't work, but it looks better this way:

[tex]\displaystyle{\text{Prove that for }0 \leq a,b,c \leq 1,}[/tex]

[tex]\displaystyle{\begin{equation*}\frac{a}{b+c+1} + \frac{b}{c+a+1} + \frac{c}{a+b+1} + (1-a)(1-b)(1-c) \leq 1\tag*{}\end{equation*}.}[/tex]

[quote="rrusczyk"]The string may be too long.  Currently, it's set at 500 characters.  If nothing goes too wrong in the first week or so, I'll increase it.[/quote]

Aha.  But Simon's way looks better anyway.  I'll just do that.


geometry
inequalities
circumcircle
inradius
trigonometry
3D geometry
pyramid

Let $ R$, $ r$ be respectively the circumradius and inradius of a regular $ 1986$-gonal pyramid. Prove that \[ \frac{R}{r}\ge 1\plus{}\frac{1}{\cos\frac{\pi}{1986}}\] and find the total area of the surface of the pyramid when the equality occurs.

I have another version of the problem from John Scholes.

P is a pyramid whose base is a regular 1986-gon, and whose sloping sides are all equal.
Its inradius is r and its circumradius is R. Show that R/r ≥ 1 + 1/cos(π/1986). Find the total
area of the pyramid's faces when equality occurs.

Solution posted here.

http://voducdien.wordpress.com/2011/10/21/math-problem-not-solved-in-25-years-bonus-solution/


Are we to form reaction which is always -ve?

Is the reaction which has [i]least -ve[/i] cell potential the main reaction? Or what is?

The reaction which occurs in a given electrode is the one that obbeys the following two criteria:

1) It has the greatest reduction potential (for the cathode) or the lowest reduction potential (for the anode).

2) It is kinetically feasible.

Here's a good practice problem for you:

Guven the standard potentials $ E^o(Na^\plus{}/Na) \equal{} \minus{}2.71\,\,V$, $ E^o(H_2O/H_2) \equal{} \minus{}0.83\,\,V$, $ E^o(O_2/HO^\minus{}) \equal{} \plus{}0.40\,\,V$, predict the products that can be obtained by electrolysis of NaHO aqueous solutions.

Sir if we follow rule 1, then we might get a cell with positive emf. But then whats the point of electrolyzing , if we arent running a non spontaneous rxn.

I think you are making some confusion about an electrolytic cell. Look closely the exercise I gave and try to solve it.


If you are asking for the limit (as $n$ tends to infinity) of a sequence $x_{n}$, then I think $x$ is defined as the limit if:

For every $\epsilon > 0$, there exists a positive integer $N$ such that $|x_{n}-x|<\epsilon$ for every $n>N$. [Not all sequences will have limits]

Or is this an olympiad question?  :maybe:

[quote="Karth"]What exactly does:

$\lim x_{n}= \lim y_{n}= x$

mean?[/quote]

$x_{n}$ is the n-th number in a sequence of numbers all named $x$

Likewise, $y_{n}$ is the n-th number in a sequence of numbers all named $y$.

$\lim x_{n}$ is sloppy shorthand notation for $\lim_{n\to\infty}x_{n}$, which is a very formal way to say: There is no last number in this sequence, but if there were, it would be equal to the number on the right hand side of the "=" sign.

Read in English, $\lim_{n\to\infty}x_{n}$ means "The limit of $x_{n}$ as $n$ becomes arbitrarily large (i.e. as big as you can think to make it)."

What your equation says is: The limit of $x_{n}$ equals the limit of $y_{n}$, which equals (the number) $x$.

Personally, I wouldn't have chosen $x$ as the name of this limit because the sequence you are trying to evaluate is already called $x$.

Yeah, this is an olympiad level question (from Engel). Thanks for the help  :) The notation gets pretty confusing sometimes...


I just did this problem a minute ago and I liked it a lot. So, I'm posting here. :D

Simplify $\sin (x-y) \cos y + \cos (x-y) \sin y$

A) 1
B) $\sin x$
C) $\cos x$
D) $\sin {x} \cos {x}$
E) $\cos {x} \cos {2y}$

My answer:

[hide]I think this problem was actually identity that $\sin ({u+v}) = \sin {u} \cos {v} + \cos {u} \sin {v}$. Using this:

We can determine that u = x-y and v = y for this case.

So, it's basically $\sin ({x-y+y})$ or $\sin {x}$[/hide]

I'm pretty sure most people have seen this before. :)

[hide]
Let $x-y=z$. From here this becomes trivial

\[\sin (x-y) \cos y + \cos (x-y) \sin y=\sin z \cos y + \cos z \sin y\]

We now recognize this as the expansion of $\sin(\alpha+\beta)$ and it becomes

\[\sin(z+y)=\sin(x-y+y)=\sin x\]

[/hide]

I was thinking about simplifying sin(x-y) and cos(x-y) until I saw your equation :P . Silly me :P .

Put y=0.
Simplifies to $\sin x$ and none of the other answers give an equivalent expression when y=0.

 :lol:


Which of these triples could NOT be the lengths of the three altitudes of a triangle?

$\text{(A)} \ 1, \sqrt 3, 2 \qquad \text{(B)} \ 3,4,5 \qquad \text{(C)} \ 5,12,13 \qquad \text{(D)} \ 7,8, \sqrt{113} \qquad \text{(E)} \ 8,15,17$

[hide]Let $\ a,b,c$ be the altitudes of the triangle.  $\ \frac 1a + \frac 1b > \frac 1c$[/hide]

How can I get my post to say hint, instead of Click hear to reveal hidden content?

[code]
[hide=hint]Put your hint here[/hide]
[/code]


[quote="anirudh"]Can someone post some amc8 problems? Thanks

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46626]$\text{maththon}$[/url][/quote]

http://www.unl.edu/amc/e-exams/e4-amc08/archive8.html

The Middle School forums, "The Basics" and "Classroom Math" probably has a lot of AMC 8 level questions.


[hide="hint"]
show that $ADCEF$ is cyclic. what is the center of its circumcircle?
[/hide]

[hide="answer"]
$30^\circ$
[/hide]


Hi all,

At a point in the proof of a theorem, my professor stated the following without proof (it should be apparent, apparently):

If $ A \subset E \subset B$, where $ A$ and $ B$ are measurable sets of finite measure, then $ m(A)\equal{}m(B)$ implies that $ E$ is measurable.

It is entirely possible that I'm missing something here, but it's not obvious to me.  Could someone explain?

It is true for a complete measure. Since B\A is a null set, E\A is a negligible set and the set E\A becomes measurable. Hence E is measurable.

Do you know that $ m(B \setminus A)\equal{}0$ because $ m(B \setminus A)\equal{}m(B)\minus{}m(A)\equal{}0$?  I didn't know that that was necessarily true, though it makes perfect sense.  How do you conclude that $ E \setminus A$ is measurable, though?  Just because it's contained in $ B \setminus A$?

Yes. Recall the definition $ (X, \Sigma, \mu)$ is complete if and only if $ S \subseteq N \in \Sigma \mbox{ and } \mu(N) \equal{} 0 \implies S \in \Sigma$.

like mathav said, if $ m(C)\equal{}0$ then $ C$ is measurable
 and since $ E\equal{}A\bigcup E\minus{}A$ then E is measurable

However you can use definition of measurable set to prove $ E$ is measurable but it is bit complicated.


take a look at these and if you are good, you can intuitively know which one of the following is largest.
2^(1/2),3^(1/3),4^(1/4),5^(1/5),6^(1/6),7^(1/7)
prove that if the sequence goes on and on, 3^1/3 is the greatest and that each term is smaller than the one preceding it after 3.

this has a nice elementary solution.

Is it because [hide] [tex]\lim_{x\rightarrow{\infty}} {x}^{\frac{1}{x}}=1[/tex] and thus the sequences approaches 1 getting smaller and smaller as x gets larger and larger, approaching  :inf: . [/hide]

AHHH I can't get it.

JS1527 wrote:Is it because [hide] [tex]\liminf {x}^{frac\{1}{x}}=1[/tex] and thus the sequences approaches 1 getting smaller and smaller as x gets larger and larger, approaching  :inf: . [/hide]

what?

i think u messed up your texing of the equation

How do I tex it correctly?  I am just starting how to do this.

EDIT: Okay I got it, but I still would like to put the x->inf underneath "lim" Oh well.

The problem is equivalent to showing that [tex]n^{n+1} > (n+1)^n[/tex]

BTW, hover over this to see the source code:

[tex]\displaystyle \lim_{x\to\infty}x^{\frac{1}{x}} = 1~~or~\lim_{x\to\infty}\sqrt[x]{x} = 1[/tex]

Edit: Also, it could potentially be proven by binomial expansion since all terms are less than n^n, and the last term is only 1.  However, that is somewhat messy.

JS -- while your statement about limits is correct, it does not contain any information about which terms would be larger than others, etc.

Remember that a limit tells you almost nothing about [i]how[/i] a function or sequence approaches a given value.  Limits tell you about how things act "near infinity" -- but it turns out that that means nothing for any particular range.  That is, if I have a limit of something going to 1, I don't know anything about what that something looks like when x = 1,000,000, for example.  Just because [tex]x^{\frac{1}{x}}[/tex] goes to 1 does not mean that [tex]1000000^{\frac{1}{1000000}}[/tex] isn't some large number.  It tells you that "eventually" we will get close to 1, but we don't get to know when eventually might actually be.

this is my very elementary proof

[hide]:rt2: is about 1.41. if 3^(1/3)is less than  :rt2: then the fractiion you must multiply it by (3/2) should be less than  :rt2:  but 1.5 isnt less than  :rt2:.  4^(1/4) = 2^(1/2).  if 5^(1/5) is greater than 4^(1/4), then 5/4 is greater than  :rt2: . so that means that 5^(1/5) is between 5/4 and :rt2:.  so to get to 6, you have to multiply it by6/5 which is greater than 5^(1/5) so 6^(1/6) is somewhere between 6/5 and 5^(1/5).  So on.[/hide]

Scrambled wrote:this is my very elementary proof

[hide]:rt2: is about 1.41. if 3^(1/3)is less than  :rt2: then the fractiion you must multiply it by (3/2) should be less than  :rt2:  but 1.5 isnt less than  :rt2:.[/hide]

I don't follow this at all.  How do you show that 3^(1/3) > 3/2 (this isn't true)?

Quote:[hide]  4^(1/4) = 2^(1/2).  If 5^(1/5) is greater than 4^(1/4), then 5/4 is greater than  :rt2: .[/hide]

Again, it is not clear why this is true.

Quote:[hide]you have to multiply it by6/5 which is greater than 5^(1/5) so 6^(1/6) is somewhere between 6/5 and 5^(1/5).  So on.[/hide]

I don't see why you know 6/5 > 5^(1/5) (this isn't true)

It would help significantly if you explained more, so we don't misunderstand.

for the first one, i said that 3/2 is greater than  :rt2: so that 3^1/3 is greater than 2^1/2 (3^1/3 will lie between 3/2 and  :rt2:   and will be slightly closer to :rt2:.  I have to turn in now so ill answer the rest later

I think Scrambled's idea, is rather clever, but I don't think he explained it very well.  I'll try to explain, *much* more verbosely.  [If I don't get it, Scrambled, just let me know.]

[hide]Let [tex]x=\sqrt{2}\approx 1.414[/tex].  Let [tex]y=\sqrt[3]{3}[/tex].  Then

[tex]\\

x\times x=2 \\

y\times y\times y=3[/tex]

Let us assume [tex]x \ge y[/tex]:

[tex]\\

y\times y\times y = 3 \\

x\times x\times y \ge 3 \\

(x\times x)\times y \ge 3 \\

2\times y \ge 3 \\

y \ge \frac{3}{2} \\

y \ge 1.5 > 1.414 \\

y \ge 1.5 > x \\

y > x

[/tex]

But this is inconsistent with our assumption, so our assumption must have been false.  Therefore, we know [tex]x < y[/tex].  Let's verify that case:

[tex]\\

y\times y\times y = 3 \\

x\times x\times y < 3 \\

2\times y < 3 \\

y < \frac{3}{2} \\

x<y < \frac{3}{2} \\

\sqrt{2} < y < 1.5

[/tex]

Check!  Now let's compare the fourth and fifth roots:

Let [tex]v=\sqrt[4]{4}\approx 1.414[/tex].  Let [tex]w=\sqrt[5]{5}[/tex].  Then

[tex]\\

v\times v\times v\times v=4 \\

w\times w\times w\times w\times w=5[/tex]

Let's assume [tex]v\le w[/tex]:

[tex]\\

w\times w\times w\times w\times w=5 \\

v\times v\times v\times v\times w\le 5 \\

4\times w\le 5 \\

w \le \frac{5}{4} \\

w \le 1.25 < 1.414 \\

w \le 1.25 < v \\

w < v

[/tex]

This is inconsistent with our assumption, so our assumption is false.  Therefore, [tex]v>w[/tex].

[tex]\\

w\times w\times w\times w\times w=5 \\

v\times v\times v\times v\times w > 5 \\

4\times w > 5 \\

w > \frac{5}{4} \\

w > 1.25 \\

1.25 < w < v \\

1.25 < w < \sqrt{2}

[/tex]

Everything is consistent, and we have put a bound on w.

Let [tex]u=\sqrt[6]{6}[/tex].

[tex]\\

w\times w\times w\times w\times w=5 \\

u\times u\times u\times u\times u\times u=6

[/tex]

Let's assume [tex]w\le u[/tex]:

[tex]\\

u\times u\times u\times u\times u\times u=6 \\

w\times w\times w\times w\times w\times u\le 6 \\

5\times u\le 6 \\

u \le \frac{6}{5} \\

u \le 1.2 < 1.25 \\

u \le 1.2 < 1.25 < w \\

u < w

[/tex]

This is inconsistent with our assumption, so our assumption is false.  Therefore, [tex]w>u[/tex]

It looks like we can continue in this manner to show that each successive term must be smaller than the previous.[/hide]

OK, now I can see where he's going.  Nice!  But I still don't understand how to prove that every term is smaller than the previous.

nr1337 wrote:Here is a different solution:

[hide]
We have 6 numbers:
2^(1/2)
3^(1/3)
4^(1/4)
5^(1/5)
6^(1/6)
7^(1/7).

Raising all of these to the 7! = 5040 power, we get

2^2520
3^1680
4^1260
5^1008
6^840
7^720.

The number of digits in a number n = 1+[log_10 n] where [n] is the greatest integer less than or equal to n.  The number of digits in the numbers are

759,
802,
759,
705,
654,
608.

802 is the greatest here, so the cube root of 3 is the largest.[/hide]blah. this was intended to be solved without a calculator.

[quote="nr1337"][quote="probability1.01"]Uhhh... doesn't that require a calculator?[/quote]

Obviously.  (actually, you can use slide rules  :lol:)[/quote]

Calculators often have a nice ^ function, which can possibly be used to calculate numbers like 5^(1/5).  =)

[quote="probability1.01"][quote="Scrambled"]x/(x-1)>(x+1)/x (this is easily shown true).  the nth root of the nth term will be somewhere between x/(x-1) and  :rt2:. x/(x+1) is smaller than both,[/quote]

Again, I'm missing why that shows what we're looking for.  It seems that x/(x+1) is less than both x^(1/x) and (x+1)^(1/(x+1)), so I don't understand why that would show anything.  Perhaps you could elaborate again?[/quote]just read over rcv's post replacing the numbers with variables and making powers of the variable instead of making it look like multiplication of those terms


prove that there are infinite values of n where n(n+1) can be represented as a^2 + b^2 (a,b,n are non negative integers)

Well this is not the correct problem of $RMO 06$.
The correct one is:-
Q. Prove that there exist infinitely many natural numbers $n$ such that
$n(n+1)$ can be expressed as a sum two squares in two different ways.

Proof is easy:-
Let $n$ $=$ $4p^{4}$  where $p \geq 2$then
   $n(n+1)$ $=$ $(4p^{4})^{2}$ + $(2p^{2})^{2}$
                  $=$ $(2p^{2}(2p^{2}-1))^{2}$ + $(4p^{3})^{2}$


All --

I don't know what your thoughts on this are, but perhaps we should form a thread on USAMO practice problems, where we could help one another and bounce ideas off one another.  Obviously, USAMO problems are challenging enough so that no one can know all the theorems and techniques involved, but a thread like this would help us all prepare.

Thoughts?

JB

Sounds like a pretty good idea. Anything to help me do better on the USAMO :lol:

[quote="xxreddevilzxx"]I second that... but i think it will be better if we had a Math-Jam-like discussion...[/quote]

I thought about that.  The problem is that
A) MathJams are restricted to a single time, and that time is limited.  This way, we all have time to think about problems posted, and give better responses and ideas.

B) MathJams are usually scheduled after the competition date.

If we can convince the administration to set up a USAMO practice MathJam, I'd be fine with that, but I think a forum string would be an excellent complement.

JB

I believe this idea was tossed around a bit last year. The objections to USAMO-level MathJams were that

1) In one hour, it is difficult to cover a significant number of problems.
2) The content is spread out over such a wide range of materials that it is impossible for any select few problems to provide significant insight into the world of olympiad problems.

A thread might work. Go for it.

While this is a good idea, I think it might be more effective to study in small groups.  But what do I know anyway.

If a theorem is used in MO treasures, but is not a "main point" theorem (i.e. a lemma used to prove a main point theorem), then do we need to prove it on USAMO to use it as a lemma, or can we just use the theorem?

JB

Unless the lemma is obvious from the theorem I think you would need to prove it.

For example, the lemma $\frac{a^2}{x}+\frac{b^2}{y} \ge \frac{(a+b)^2}{x+y}$ used to prove Cauchy-Schwarz would probably need a quick proof on the USAMO.

whether it is obvious or not, whatever is on a lemma should be somewhat proven in a quick manner... i think it will score a more definite point for your score.

[quote="MithsApprentice"]Actually, paladin, I think that's well-known enough to escape a mandatory proof. If you know how to prove it, go for it though...[/quote]

Well maybe... I wouldn't know so I would've proved it on a competition. It reduces to the trivial inequality so its easy anyway.

Paladin -- That's actually the exact theorem I was looking at :-)

JB

Haha. Well I used it as an example because I had the exact same question ;)


[quote="weatherbottle"]Source: Massachusetts Mathematics League December 2005

The point $P(a, b)$ is the first quadrant lattice point on $17x-23y-2=0$ closest to the origin. Find the area of the triangle with vertices $P$, $(4,0)$ and $(26,1)$.[/quote]

[hide]The smallest point with positive integer values for both x and y is
when rearranging the equation, $y=\frac{17x-2}{23}$. So $17x\equiv 2 \bmod{23}$ and $17x-2\equiv 0 \bmod{23}$. The smallest integer values are $(15,11)$ and the area of the triangle is $115.5$   [/hide]


[b]Problem 4.[/b] Let $p$ be prime, such that $p^2$ divides $2^{p-1}-1$. Proove that for any natyral $n$ the number $(p-1)(p!+2^n)$ has at least three different prime divisors.

[i]Aleksandar Ivanov[/i]


Prove that $ \lvert{a \plus{} b \minus{} c}\rvert \plus{} \lvert{b \plus{} c \minus{} a}\rvert \plus{} \lvert {c \plus{} a \minus{} b}\rvert \ge \lvert {a \minus{} b}\rvert \plus{} \lvert{b \minus{} c}\rvert \plus{} \lvert{c \minus{} a}\rvert$   :P

[hide]
First, we may assume at least one of $ a,b,c$ is nonnegative; otherwise, negate all three and the inequality is unchanged.

WLOG $ a$ is maximal among $ a,b,c$. Then $ a\ge 0$ because at least one of $ a,b,c \ge 0$.

Say $ a \plus{} b \minus{} c < 0$. Then, $ c \le a < c \minus{} b \implies b < 0$. Then, $ b \plus{} c \minus{} a < c \minus{} a \le 0$ and $ c \minus{} b \plus{} a > 2a \ge 0$. Also, $ a \minus{} b > 0, a \minus{} c > 0$ because $ a$ is maximal, and $ c \minus{} b > a \ge 0$. The inequality then becomes
\[ ( \minus{} a \plus{} ( \minus{} b) \plus{} c ) \plus{} (( \minus{} b) \minus{} c \plus{} a) \plus{} (c \plus{} a \plus{} ( \minus{} b)) \ge (a \plus{} ( \minus{} b)) \plus{} (( \minus{} b) \plus{} c) \plus{} (a \minus{} c)\]

\[ \Leftrightarrow c \plus{} a \plus{} 3( \minus{} b) \ge 2a \plus{} 2( \minus{} b) \Leftrightarrow c \minus{} b \ge a\]
but this follows from our assumption.

Similarly, if $ a \plus{} c \minus{} b < 0$, we merely switch $ b$ and $ c$ in the above argument and the inequality holds.

Finally, say $ a \plus{} b \minus{} c \ge 0$ and $ a \plus{} c \minus{} b \ge 0$. This means $ a \ge |b \minus{} c|$.

Our inequality becomes
\[ (a \plus{} b \minus{} c) \plus{} (a \plus{} c \minus{} b) \plus{} |b \plus{} c \minus{} a| \ge (a \minus{} b) \plus{} (a \minus{} c) \plus{} |b \minus{} c|\]

\[ \Leftrightarrow |b \plus{} c \minus{} a| \ge |b \minus{} c| \minus{} b \minus{} c\]
But since $ a \ge |b \minus{} c|$, it suffices to show
\[ |b \plus{} c \minus{} a| \equal{} |a \minus{} b \minus{} c| \ge a \minus{} b \minus{} c \ge |b\minus{}c| \minus{} b\minus{}c\]
but this is obvious.
[/hide]

$\sum \lVert{b + c - a\rVert=\frac{1}{2}\sum(\lVert {c + a - b}\rVert +
 \lVert{a + b - c}\rVert)}$
$ \ge\frac{1}{2}\sum \lVert{( {c + a - b})-(a + b - c}) \lVert$
$=\lVert {a - b}\rVert + \lVert{b - c}\rVert + \lVert{c - a}\rVert$

[b]A similar problem[/b]：
Prove that $ \lVert{a + b - c}\rVert + \lVert{b + c - a}\rVert + \lVert {c + a - b}\rVert \ge 2(\lVert a \lVert+\lVert b \over+\lVert c \lVert).$


Let a_1,...,a_n (n >= 3) be positive real numbers such that
[b]a 2 _1 + ... + a 2 _n < 3(a_1 + ... + a_n) 2 /(3n-1)[/b]
Prove that a_i, a_j, a_k are three sides of a triangle for all 1 <= i<j<k <= n
It's not difficult than IMO2004/P4
 :)


Let $ a$ and $ b$ be natural numbers such that $ b>a>1$ and $ b$ divides $ a^2\plus{}1$. Prove that $ b\minus{}a>\sqrt{a}$.

We have $ b|(b \minus{} a)^2 \plus{} 1$ so $ (b \minus{} a)^2 \plus{} 1\geq b\geq a \plus{} 1$.
$ \rightarrow b\minus{} a \geq\sqrt{a}$.
Clear equality impossible.


function
Putnam
real analysis
real analysis unsolved

(O.Skaskiv) Prove that every solution of differential equation
\[ f''(x)\plus{}xg(x)f'(x)\plus{}f(x)\equal{}0\]
where $ g(x)\ge 0$ for all $ x\in\mathbb{R}$ a) is bounded in $ \mathbb{R}$; b) has bounded in $ \mathbb{R}$ derivative.

$ \frac {d}{dx}(f(x)^{2}\plus{}f'(x)^{2})\equal{}\minus{}2xg(x)f'(x)^{2}$. Hence, the nonnegative function $ f(x)^{2}\plus{}f'(x)^{2}$ is non-decreasing on $ (\minus{}\infty,0]$ and nonincreasing on $ [0,\plus{}\infty)$ and, thereby, bounded. I suspect that the problem came right from the exercise section in some ODE textbook :P.

Part (a) was also a Putnam problem: 1997 B2.


How many solutions (x, y) where x and y are integers are in the equation (1/x) + (1/y) = 1/2007?
This is from the Philippine Mathematical Olympiad Area Level.

It equavalent to $ y\equal{}2007\plus{}\frac{2007^2}{x\minus{}2007}$, let d is divisor $ 2007^2\equal{}3^4*223^2$, then $ x\equal{}2007\plus{}d,y\equal{}2007\plus{}\frac{2007^2}{d^2}$. Therefore number of solutions is $ 2*5*3\equal{}30$.

Rewrite the equation as $ (x\minus{}2007)(y\minus{}2007)\equal{}2007^2$. The number of divisors of $ 2007^2\equal{}3^4\cdot223^2$ is $ 2\times(5\times3)\equal{}30$ (positive and negative). Hence there are 30 solutions.

expounding gunardi's solution, the number of integer solutions of this problem is

A(x) = (2a+1)(2b+1)(2c+1)..., where A(x) is the number of solutions and a, b, c, and many more are the exponents of the prime factors of x. (actually, x must be expressed in the form (p1^a)(p2^b)...)


Determine all rhombuses with side length $ 2a$ such that there is a circle cutting the segment of length $ a$ from each side of the rhombus.

I don't quite understand your problem, can you clarify please?

Wich is the segment of lenght $ a$ from each side??

 :)

I think the problem is to find all rhombs which have the property that if they are intersected with a circle, the 4 sinews cut from the sides are exactly half as long as the sides.


I was watching the Today show today when I learned that Simon and Randy where making fun of two autistic adults. Making fun not only of there singing, but of there facial and phisical feartures. One of them was overwaght (Simon made fun of this right away), And the other had big eyes (Simon discribed him has "a baboon") Has Simon was making fun of them, you could see in there eyes exactly how much this hurt them. I was wondering if other aopsers agreed that American Idol has gotten out of hand. I think so.

Just don't watch it. I haven't for years and I don't plan on watching it anytime soon.

In Oregon we were snowed in and missed a week of school during the first two episodes of american idol, so I watched both of them.... who were the autistic adults? the mother/daughter team?

[hide="One Word"]Ratings[/hide]

EDIT: I wanna go on AI and sing BADLY, just to get on national TV.  That's what they want, anyway.

The names slip my mind, but there where two males, One of the age of 20, the other 23. There where on hte other night.(I will research who there are)

1) It's in no way against the law

2) When you sign over the release to be on American Idol, you sign over your soul. If you do so without proper means to make good of this deal (A large penguin suit and the lyrics to "Ice Ice Baby"), then you are an idiot.

He didn't really make fun of the fat guy...I found him to be rather supportive

[quote="mysmartmouth"][hide="One Word"]Ratings[/hide][/quote]

most definetely

it depends.

some are obviously people going on to be funny, but then again some people really think they can sing.  i think they need to find a way to still be very funny in their critism without it being this overboard.

it doesnt bother me personally [although my moms being telling me i had no feelings since i was like 3 :| ] but i definatley see a problem and think they need to tone it down a little.

As Carlos Mencia eloquently puts it: DER DER DERRRRR

If you even know anything about AI, you would know it's stupid to go on it as they will ridicule you.

[quote="h_s_potter2002"]I dunno how obvious this is, but

they actually have like 3 rounds of preliminary auditions, before you can actually sing in front of the judges.  So only the really good people and the really bad people are chosen to sing for them.[/quote]

very true. IT would take to long to listen to that much.

Also, they are being more mean this year just to get the show's ratings up higher.  :o

well they never sugarcoated anything

but you're right dont have to be so mean, just go "next" imo

How bad would the show be if they just went 'next'? You would get to say someone sing badly and that'd be it... people watch partly for the insults afterwards.

Not that I think that it's actually a good show or anything...

most of those people just sing to get on national television

so it doesnt matter if you make fun of them

In summary, no case at all can be made for better treatment of contestants (VOLUNTEERS) on such a TV show... thus judging never gets out of hand.


What is the number of square centimeters in the surface area
of a rectangular prism whose length is 3 cm, whose width
is 4 cm, and whose height is 5 cm?

[asy]import three; currentprojection=orthographic(1/2,-1/2,1/2);
size(101);
real x = 5,y=3,z=4;
label("5",(x/2,0,0),S); label("3",(x,y/2,0),SE); label("4",(0,0,z/2),W);
draw((0,0,0)--(0,0,z)--(x,0,z)--(x,0,0)--cycle3);
draw((0,0,z)--(0,y,z)--(x,y,z)--(x,0,z));
draw((x,0,0)--(x,y,0)--(x,y,z));[/asy]

ok
most people know how to get there, but for those who don't, i'll make it a bit more detailed

$ SA \equal{} 2(l \times w) \plus{} 2(w \times h) \plus{} 2(l \times h)$
where $ SA$=surface area, $ l$=length, $ w$=width, and $ h$=height

so, we plug the numbers in

$ SA \equal{} 2(l \times w) \plus{} 2(w \times h) \plus{} 2(l \times h)$
$ SA \equal{} 2(3 \times 4) \plus{} 2(4 \times 5) \plus{} 2(3 \times 5)$
$ SA \equal{} 2(12) \plus{} 2(20) \plus{} 2(15)$
$ SA \equal{} 24 \plus{} 40 \plus{} 30$
$ SA \equal{} 94$


logarithms
integration
calculus
calculus computations

this is fairly simple to show: :D

$\sum_{k=1}^\infty\;\left[ \;\left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{k}\right) \cdot\frac{1}{k^{2}}\;\right] \;=\;\boxed{2\;\zeta{(3)}}$

Here is my method of attack.

We have:

$\frac{1}{2}\ln^{2}(1-z)=\sum_{n=1}^{\infty}\frac{H_{n}}{n+1}z^{n+1}$ , valid for $|z|<1$.

It follows that

$\frac{1}{2}\int_{0}^{1}\frac{\ln^{2}(1-z)}{z}dx=\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^{2}}=\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}-\zeta(3)$.

Thus we only need to calculate the integral

$\int_{0}^{1}\frac{\ln^{2}(1-z)}{z}dz=\int_{0}^{1}\frac{\ln^{2}x}{(1-x)}dx=\sum_{m=0}^{\infty}\int_{0}^{1}x^{m}\ln^{2}xdx=2\zeta(3)$.
The desired result follows.


inequalities
function
inequalities proposed

I use your problem:^^
For any positive real numbers $ a,b,c$ we have
$ (\frac{a}{a\plus{}b})^2\plus{}(\frac{b}{b\plus{}c})^2\plus{}\frac{c}{c\plus{}a})^2\le\frac{3(a^2\plus{}b^2\plus{}c^2)}{4(ab\plus{}bc\plus{}ca)}$
We need prove that
$ \frac{3(a^2\plus{}b^2\plus{}c^2)}{4(ab\plus{}bc\plus{}ca)}\plus{}2\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\le \frac{11}{4}(*)$
Let $ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\equal{}x$,with $ x\ge 1$
$ \Rightarrow (*)\Leftrightarrow \frac{3x}{4}\plus{}\frac{2}{x}\le\frac{11}{4}$
$ \Leftrightarrow (3x\minus{}8)(x\minus{}1)\le 0$ Q.E.D

[quote="PHONGVAN"]Let $ a,b,c \geq\ 0$ .Prove that
 ${{{{ (\frac {a}{a + b}})^2 + (\frac {b}{b + c}})^2 + (\frac {c}{c + a}})^2 + 2(\frac {ab + bc + ca}{a^2 + b^2 + c^2}})\leq \frac {11}{4}$[/quote]
Nice, thach. :)
In my new updated of nothing1.pdf, the following inequality is proved with the Cauchy Schwarz:
\[ \sum \frac{a^2}{a^2+kab+b^2} +\frac{2k+1}{k+2}\cdot \frac{ab+bc+ca}{a^2+b^2+c^2} \le 2\]
for all $ a,b,c \ge 0$ and $ k \ge 1.$
The case $ k=1$, it becomes
\[ \sum \frac{a^2}{a^2+ab+b^2} +\frac{ab+bc+ca}{a^2+b^2+c^2} \le 2\]
This inequality I have posted in this forum. :)

[quote="vo thanh van"]
$ \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)} \plus{} 2\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\le \frac {11}{4}(*)$
[/quote]
Try $ a\equal{}13^{12^{1988}},b\equal{}c\equal{}1$ ;)

[quote="vo thanh van"]I use your problem:^^
For any positive real numbers $ a,b,c$ we have
$ (\frac {a}{a \plus{} b})^2 \plus{} (\frac {b}{b \plus{} c})^2 \plus{} \frac {c}{c \plus{} a})^2\le\frac {3(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)}$
[/quote]

How to prove this one ?

Thank you very much.

Solution: :)
We need prove
$ \frac{3}{4}(a^2\plus{}b^2\plus{}c^2)\le \sum\frac{(ab\plus{}c(a\plus{}))a^2}{(a\plus{}b)^2}\equal{}\sum\frac{a^3b}{(a\plus{}b)^2}\plus{}\sum\frac{a^2c}{a\plus{}b}$
We have $ \sum\frac{a^3b}{(a\plus{}b)^2}\le \sum\frac{a^3b}{4ab}\equal{}\frac{1}{4}(a^2\plus{}b^2\plus{}c^2)$
It remains to show that
$ \frac{1}{2}(a^2\plus{}b^2\plus{}c^2)\ge \sum\frac{a^2c}{a\plus{}b}\equal{}\sum(ac\minus{}\frac{abc}{a\plus{}b})\equal{}ab\plus{}bc\plus{}ca\minus{}abc\sum \frac{1}{a\plus{}b}$
which is true by Schur inequality.Q.E.D

[quote="Sunkern_sunflora"]How did u check that?????!!!!!! :what?:
 :omighty:[/quote]
It's just a joke, Sunkern :)
I meant that we can let $ a \rightarrow \plus{}\infty, b\equal{}c\equal{}1$.

Oh! I was thinking day and night what kind of calculators you guys use! :lol:

[quote="vo thanh van"]I use your problem:^^
For any positive real numbers $ a,b,c$ we have
$ (\frac {a}{a \plus{} b})^2 \plus{} (\frac {b}{b \plus{} c})^2 \plus{} \frac {c}{c \plus{} a})^2\le\frac {3(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)}$
We need prove that
$ \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)} \plus{} 2\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\le \frac {11}{4}(*)$
Let $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \equal{} x$,with $ x\ge 1$
$ \Rightarrow (*)\Leftrightarrow \frac {3x}{4} \plus{} \frac {2}{x}\le\frac {11}{4}$
$ \Leftrightarrow (3x \minus{} 8)(x \minus{} 1)\le 0$ Q.E.D[/quote]

I don't think this solution is correct.

Why $ 3x \minus{}8 \leq 0$

[quote="manlio"][quote="vo thanh van"]I use your problem:^^
For any positive real numbers $ a,b,c$ we have
$ (\frac {a}{a \plus{} b})^2 \plus{} (\frac {b}{b \plus{} c})^2 \plus{} \frac {c}{c \plus{} a})^2\le\frac {3(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)}$
We need prove that
$ \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)} \plus{} 2\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\le \frac {11}{4}(*)$
Let $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \equal{} x$,with $ x\ge 1$
$ \Rightarrow (*)\Leftrightarrow \frac {3x}{4} \plus{} \frac {2}{x}\le\frac {11}{4}$
$ \Leftrightarrow (3x \minus{} 8)(x \minus{} 1)\le 0$ Q.E.D[/quote]

I don't think this solution is correct.

Why $ 3x \minus{} 8 \leq 0$[/quote]
Yeah, that's why I gave my counterexample above, manlio. :)

[quote="can_hang2007"][quote="PHONGVAN"]Let $ a,b,c \geq\ 0$ .Prove that
 ${{{{ (\frac {a}{a + b}})^2 + (\frac {b}{b + c}})^2 + (\frac {c}{c + a}})^2 + 2(\frac {ab + bc + ca}{a^2 + b^2 + c^2}})\leq \frac {11}{4}$[/quote]
Nice, thach. :)
In my new updated of nothing1.pdf, the following inequality is proved with the Cauchy Schwarz:
\[ \sum \frac {a^2}{a^2 + kab + b^2} + \frac {2k + 1}{k + 2}\cdot \frac {ab + bc + ca}{a^2 + b^2 + c^2} \le 2
\]
for all $ a,b,c \ge 0$ and $ k \ge 1.$
The case $ k = 1$, it becomes
\[ \sum \frac {a^2}{a^2 + ab + b^2} + \frac {ab + bc + ca}{a^2 + b^2 + c^2} \le 2
\]
This inequality I have posted in this forum. :)[/quote]
Are there someone interested in it? :(

[quote="manlio"]

I don't think this solution is correct.

Why $ 3x - 8 \leq 0$[/quote]
Yes,but if $ a,b,c$ are side of a triangle,my solution will true.
Now,I'm giving a different solution:
${{{{ (\frac {a}{a + b}})^2 + (\frac {b}{b + c}})^2 + (\frac {c}{c + a}})^2 + 2(\frac {ab + bc + ca}{a^2 + b^2 + c^2}})\leq \frac {11}{4}$
${ \sum(1-(\frac{a}{a+b})^2)\ge 2\frac {ab + bc + ca}{a^2 + b^2 + c^2}}+\frac{1}{4}$
${ \Leftrightarrow \sum \frac{b^2}{(a+b)^2}+2\sum\frac{ab}{(a+b)^2}\ge 2\frac {ab + bc + ca}{a^2 + b^2 + c^2}}+\frac{1}{4}$
We use this problem:
$ \left( \frac {a}{a + b}\right)^2 + \left( \frac {b}{b + c}\right)^2 + \left( \frac {c}{c + a}\right)^2 + \frac {2abc}{(a + b)(b + c)(c + a)} \ge 1$
in here : http://www.mathlinks.ro/viewtopic.php?t=218478
We need prove that
${ 2\sum\frac{ab}{(a+b)^2}\ge 2\frac {ab + bc + ca}{a^2 + b^2 + c^2}}+\frac{5}{4}-\frac {2abc}{(a + b)(b + c)(c + a)}$
WLOG,we assume $ a+b+c=1$,let $ ab+bc+ca=\frac{1-q^2}{3},abc=r(1\ge q\ge 0)$,we have $ r\ge max{0,\frac{(1+q)^2(1-2q)}{27}}$
And then we consider the function $ f(r)=...$ this solution is very long,I can't post it but I think it's true :)


Given any triangular region, prove that there is always a line that goes through the triangle dividing it into two polygons of equal area and perimiter.

EXTENSION: Prove that this works for any region.

[hide="Silly"] Clearly the line passes through the centroid.  Consider the function $f(P)$ where $P$ is a point on the perimeter of the region that calculates the difference in the perimeter of the area "clockwise" of $P$ and "counterclockwise" of $P$.  Clearly

- $f(P)$ is continuous, and
- $f(P)$ is the negative of $f(Q)$ where $Q$ is the point opposite $P$ with respect to the centroid

Hence somewhere between $P, Q$ we must have $f(R) = 0$. [/hide]

[quote="t0rajir0u"][hide="Silly"] Clearly the line passes through the centroid. [i](snip)[/i][/hide][/quote]
A line parallel to a side of an equilateral triangle that passes through its centroid creates polygons with area ratio 4:5. This counterexample isn't mine, though; JBL brought this up in an older thread discussing a similar topic.

Perhaps moving the line so that the perimeters are always equal would work better. You just need to move the intersection points along the triangle at the same rate.


function
geometry
3D geometry
sphere
real analysis
real analysis unsolved

This is a very cool problem:
    What are the subsets $A$ of the sphere $S^1$ such that one can find a continuous function $f$ defined on $R^2$ with values in $R^2$ such that $A$ is the image of the unit closed ball by $f$ and $f$ is the identity on $A$?

It can easily be seen that these sets are precisely the compact arcs of the circle.

Now what's the answer to the same question for $\mathbb S^2$ and $\mathbb R^3$ instead of $\mathbb S^1$ and $\mathbb R^2$ respectively?

P.S.

Why not call it "circle" instead of "sphere $\mathbb S^1$"? :)


Prove that for all positive reals $ a,b,c,d$,

$ \frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{d^2}\plus{}\frac{d^2}{a^2}\geq \frac{a\plus{}b\plus{}c\plus{}d}{\sqrt[4]{abcd}}$

[quote="dgreenb801"]It's just Muirhead. Multiplying both sides by $ a^2b^2c^2d^2$, we have
$ \sum a^4c^2d^2 \ge \sum a^{11/4}c^{7/4}d^{7/4}b^{7/4}$[/quote]

I don't think it's symmetric, though...  :maybe:

[quote="dgreenb801"]It's just Muirhead. Multiplying both sides by $ a^2b^2c^2d^2$, we have
$ \sum a^4c^2d^2 \ge \sum a^{11/4}c^{7/4}d^{7/4}b^{7/4}$[/quote]

This is incorrect. The lefthand side is a [i]cyclic[/i] sum, not a [i]symmetric[/i] sum; therefore, you cannot apply Muirhead's Theorem.

However, there is a solution using AM-GM inequality;

$ 7 \frac {a^2}{b^2} \plus{} 5 \frac {b^2}{c^2} \plus{} 3 \frac {c^2}{d^2} \plus{} \frac {d^2}{a^2} \geq 16\frac {a}{\sqrt [4]{abcd}}$,

Adding cyclically and dividing by 16, we gain the desired result  :)

Yep, qwerty414 is right. Nice solution as well.  :)

[quote="dgreenb801"]Can we use AM-GM to prove any cyclic sum to be greater than the symmetric (or possibly even cyclic) sum is the exponents of the first sequence majorize the second?[/quote]see here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=103761 :)


Prove that there exist x,y (x,y are irrationals) that the number x^y is a rational number.

Let x = e
Let y = Ln 2

Therefore, x^y = e^(Ln 2) = 2 which is rational

Right Grey Gauss but there is an onother solution. x=y=square root of 2

It's well known that $\sqrt2 ^{\sqrt 2}$ is not rational, even not algebraical!
You mean $x= \sqrt2 ^{\sqrt 2} , y=\sqrt 2$ I think.

Sorry you have right. I make the mistake because I was in hurry

Actually, there are infinitely many solutions of the form:

Let x = e
Let y = Ln a, where a is a rational number

Then, x^y = e^(Ln a) = a, which as stated above is a rational number.

Basicly you proved nothing special, since you just used some well-known theorems. I don't see anything smart in that.
This problem was posted in a hellenic competition some years ago. And it was posted to the juniors.
So they have to solve it, in an elementary way. e.g. solutions where they accept that $ln(2)$ is irrational, where not accepted unless they give the proof of that. The problem is [b]easy[/b] and [b]smart[/b]. So try to find such a solution.  ;)

Moreover, it's not the right place for this problem, it should be in the intemidiate forum.

Yes, the "clever" solution

If $a=\sqrt{2}^{\sqrt{2}}$ is rational, then we are done. If not, then consider $\sqrt{2}^a = \sqrt{2}^2 = 2$ and we are now done.

This solution doesn't rely on any special knowledge.

That was the solution that the contestants should have thought.
Very easy, but smart indeed.
Anyway, it's better now.


Three numbers are randomly selected, with replacement, from the set of integers 1 through 100 inclusive.  What is the probability that the product of the three numbers selected will be even?

[hide]

the only possible way to get odd number is pick all three odd numbers.
since the chances of picking odd number is 50/100 (=1/2) for all three turns.

 (1/2) * (1/2) * (1/2) = 1/8

so probability for product to be even is 1 - (1/8) = 7/8

[/hide]


Due to the problems on the database on the server of the Art of Problem Solving, I did not get chance to do any of qualifying finish of COP the second.

Therefore, the questions are now due next Tuesday, 14th.

Remember, you must get all right to be qualified.

http://dhost.info/jwec05/Silverfalcon/2nd%20COP%20r1.PDF

Regards,

Is problem 5 still $\sqrt{7+\sqrt{2+\sqrt{4+\sqrt{x}}}}=3$?


AMC USA(J)MO USAMO geometry AIME inequalities blogs

So I have this friend - er, this hypothetical friend - who almost certainly qualified for USAMO (hypothetically, we'll say he made a 10 on the AIME as a sophomore) but isn't very good at geometry (or basic arithmetic, but that's an aside  ;) ).

We'll say he has a reasonable grasp of combinatorics, isn't completely hopeless at number theory, took the Olympiad Inequalities course, and so on. How would he - hypothetically, of course - go about preparing for the USAMO geometry-wise?

Thanks!

For geometry problems, construct large diagrams and construct them precisely with a ruler and compass (or with a geometry rendering program). Accurate diagrams will help prevent your wasting time trying to prove things that aren't true, or will suggest things that might be true that might help your solution.

As well, make sure you use all the information given. Sometimes when you're stuck that can help. But make sure you don't use the same piece of information in multiple ways, such as using a circle to find two radii congruent and in power of a point (at least especially on the same lengths), that can result

Don't get too messy; use elegant things like similar and congruent triangles, congruent angles, parallel lines, circles, and relatively simple (no degree higher than 2, preferably linear though) equations based on length. Don't overruse Pythagorean Theorem (this is what can get you into 4th degree equations) or trig.

Finally, just make sure you know all the basic theorems about triangles, circles, etc. In particular, know what a cyclic quadrilateral is and know theorems about them (ie Ptolmey's Theorem).

Hmm.. did this friend get a 10 on the USAMO in 2004, and did he, perhaps, at some point, that is, comment on shares of sanity?

Anyway, geometry was a subject that magically worked for me last summer, but I think there are several steps that can really help:

1) Learn all the theorems (and their proofs, of course). There are only a few that matter... Menelaus, Ceva, Power of a Point. Also be proficient at angle chasing.

2) Do a lot of problems, and look at a lot of solutions. It's great to struggle with a problem on your own, but for geometry it's more helpful study-wise to give up once you're stuck (usually means within 30 mins if you don't solve it). Of course, as you get better, maybe you can spend more time; this is for the easy problems.

To His Highness the Master Problemsolver of GA :omighty: I recommend Kiran Kedlaya's Geometry Unbound. I forget where I got it from, but it may be found at http://billydorminy.homelinux.com/Math/geometry.pdf .

I'm not sure if it is the proper difficulty, as I havenae tried to do any USAMO geometry myself or even study GU indepth, but it's a really fun book. Kiran says that it is similar to Geometry Revisited, which is a wonderful book save that it's in small print. lol.

Billy

Yeah, practice definitely pays off -- you can never do enough problems, so here's a few hundred if you're interested:

http://students.imsa.edu/~tliu/Math/planegeo.pdf  :)

Oh man, I've been looking for several thousand geometry problems to do, and I just couldn't find it, but with this I am complete!

But seriously, that's pretty cool, thanks

OMG, Thanks So Much Tony!!! yeah I felt like jumping on the bandwagon :)

hurray for thazn.

...now time to print the 485 pages using school printer..  :rotfl:

Haha you mean 495 pages.. glad you guys like it

Btw, I found it here: http://www.math.su.se/~mleites/Prasolov/pr-Plane/ and just compiled everything (there was an error in one of the files though..) There's lots of other stuff like solid geo on there too...  :)

If you don't want to read Tony's 495 pages, just check out my blog! Our math club is finishing up geometry this week, after four weeks of fun! Go back a once (twice soon?) and find the Power of a Point lecture, that was the first.

Yeah, I write a whole lot less seriously than these other guys do. Haha, see what time it is? That's why...


Let N be a positive integer. Define the sequence a0, a1, a2, ... by a0 = 0, an+1 = N(an + 1) + (N + 1)an + 2( N(N+1)an(an+1) )^1/2. Show that all terms are positive integers.


If $a,b,c,d$ are positive real numbers such that $a^2+b^2+c^2+d^2=1$ prove that

$\frac{a+b-1}{a+b}+\frac{a+c-1}{a+c}+\frac{a+d-1}{a+d}+\frac{b+c-1}{b+c}+\frac{b+d-1}{b+d}+\frac{c+d-1}{c+d}\leq 0$

It is equilibium to:
\[ 4\leq \sum \frac{1}{a+b} \]
By AM-HM:
\[ \sum \frac{1}{a+b} \cdot \sum(a+b) \geq 16 \].
\[ \sum \frac{1}{a+b} \geq \frac{8}{a+b+c+d} \]
Since: $a+b+c+d \leq \sqrt{4(a^2+b^2+c^2+d^2)}=2$, so:
\[ \sum \frac{1}{a+b} \geq 4 \]

[quote="silouan"]I think it is equivalent to $\sum\frac{1}{a+b}\geq 6$ not   $4$[/quote]
but shobber's solution also works.do you think so?
 :)


a fulcrum is placed under the center of a board. inorder for two objects to balance on the board, the distance (in feet ) of each
object  from the center of the board must vary inversely with its weight (in pounds) . what is the distance of each object from the center of the board ?

     a (7.5lb)___________5ft_______________b (12.5 lb)

the distance between two objects is 5ft

[hide="Solution"]
Place the fulcrum at a point $ x$ ft away from $ A$ and $ 5\minus{}x$ ft away from $ B$.

The net torque applied on the system must be 0 if there is rotational equilibrium:

$ \tau_{\text{net}}\equal{}7.5x\minus{}12.5(5\minus{}x)\equal{}0$.

Solving, $ x\equal{}3.125$.

Therefore, $ A$ is $ \boxed{3.125}$ ft away from the fulcrum, and $ B$ is $ \boxed{1.875}$ ft away.
[/hide]


function
geometry
rectangle
limit
real analysis
real analysis unsolved

I didn't know where to post this. I'm not really sure it fits here.

Prove that for every $n\ge 1$, the graph of a monotonic function $f:[0,1]\to[0,1]$ with $f(0)=0,f(1)=1$ can be covered by $n$ rectangles with sides parallel to the axes and area $\frac 1{n^2}$.

Yeah; there's no real analysis here.

Extend $f$ by defining $f(x)=x$ for $x>1$. Let $x_0=y_0=0$.
Let $x_1=\inf\{x: xf(x)\ge\frac1n\}$ and let $y_1=\frac1{nx_1}$. Let ${x_2=\inf\{x: (x-x_1)(f(x)-y_1)\ge\frac1n\,x>x_1}$ and let $y_2=y_1+\frac1n\cdot\frac1{x_2-x_1}$ Continue similarly until we have defined $x_1,x_2,\dots,x_n$. For each $n$, $\lim_{x\to x_n^-}f(x)\le y_n\le\lim_{x\to x_n^+}f(x)$. The rectangles with opposite corners $(x_i,y_i)$ and $(x_{i+1},y_{i+1})$ clearly have area $\frac1n$, and they clearly cover the graph from $0$ to $x_n$.

By AM-GM, $x_n+y_n=\sum_k (x_{k+1}-x_k)+(y_{k+1}-y_k)\ge2n\cdot\frac1n=2$. If $x_n\ge1$, we are done. If $y_n>1$, we must have $x_n=y_n$ by the relation to $f$, and we are done.


We have triangle $ABC$ and heights $AD$, $BE$ and $CF$ with $D \in [BC]$ $E \in [AC]$ and $F \in [AN]$. Let $M$ be the midpoint of $AE$ and $N$ a point on $[AC]$ such that $ME=NE$. Let $P$ be the midpoint of $[CH]$ Show that $M$, $N$, and $P$ are concyclic in a circle with the center at $H$ and also points $A$, $E$, $B$ and $D$ are conclyclic in a circle with the center at the midpoint of $[AB]$

*$H$ = orthocenter

[quote="icx"]Show that $M$, $N$, and $P$ are concyclic in a circle with the center at $H$ [/quote]
Could you check it please? I think it's not true.
[quote="icx"]$A$, $E$, $B$ and $D$ are conclyclic in a circle with the center at the midpoint of $[AB]$
[/quote]
It's certaintly, because $\hat{BEA}= \hat{BDA}= 90^{o}$ so $E$ and $D$ are lie in a circle which has $AB$ to be the diameter

I'm not sure.... may be I was wrong ... But this is my figure....


[b]Prove that the solution in positive (distinct) integers of equation  $\begin{array}{|c|} \hline x^3+y^5=z^7\\ \hline \end{array}$ is
\[ \left\{\begin{array}{lll} x&: =&x(u,v,p) = u(u^3+v^5)^{35p+30}\\ y&: =&y(u,v,p)= v(u^3+v^5)^{21p+18} \\ z&: =&z(u,v,p)= (u^3+v^5)^{15p+13}\\ \end{array}\right. \;  \]
where $(u,v,p)$ is an arbitrary point in  ${\mathbb N}^* \times {\mathbb N}^* \times {\mathbb N}\; .$ [/b]

How to prove it?

Is it contain hard method?


[quote="April"]Some polygon can be divided into two equal parts by three different ways. Is it certainly valid that this polygon has an axis or a center of symmetry?[/quote]

Oh, very nice! I am the author of this problem! I do know only one example. If you have found another, please, write me.

Thank you!
Sergey Markelov

What do you mean that the polygon can be divided into two equal parts? Do you mean congruent parts or equal areas? What do you mean by different ways? Divided by a line in different ways, or some other way?

It is known that given any direction, there exists a line parallel to this direction which cuts the polygon into two eqal parts. Therefore, please be more explicit in the statement of the problem.


Given that the ratio of $3x-4$ to $y+15$ is constant, and $y=3$ when $x=2$, then, when $y=12$, $x$ equals:

$\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)} \ \frac72 \qquad \text{(E)} \ 8$

[hide]Is it... really... 8?  since 3 to 12 is 4 and then 2*4 = 8??  I tried doing the ratio thing but I got 7/3... :blush:[/hide]

No, I think it has to be 7/3.  Maybe it was that rather than 3/7?

Because the ratio is constant between 3x-4 and y +15, not x and y.

I was getting 7/3 as well, is there something wrong with the answers or am I just doing something wrong? =)

[hide]Substitute the given values of x and y; the common ratio is $\frac{1}{9}$. When y=12, $3x-4=3$, $x=\frac{7}{3}$, which is strange. [/hide]

That's pretty strange... Could it have anything to do with the fact that the problem says when x is_, then y is_ and asks for when y is_, then x is_? hmmmmm........ strange :|

How is that strange? The answer is $\frac{7} {3}$. Case solved.

If you look at the smallprint at the bottom of the initial post, it says it's edited on March 18 (today). All the time before that, the choice $B$ has been ${3\over 7}$ instead of ${7\over 3}.$


induction
number theory unsolved
number theory

$F_n$ is the Fibonacci sequence $F_0 = F_1 = 1$, $F_{n+2} = F_{n+1} + F_n$. Find all pairs $m > k \geq 0$ such that the sequence $x_0, x_1, x_2, ...$ defined by $x_0 = \frac{F_k}{F_m}$ and $x_{n+1} = \frac{2x_n - 1}{1 - x_n}$ for $x_n \not = 1$, or $1$ if $x_n = 1$, contains the number $1$

Assume that at some point, we have that $ x_n = 1 $. Set it as $ y_0=1 $.
We shall construct the sequence $ {y_i} $ as follow: $ y_{n+1} = \frac{y_n+1}{y_n+2} $ so if $x_i=y_j$ then $x_{i+1}=y_{j-1}$.
So, all $x_0$ satisfying are the numbers of the sequence ${y_i}$.
We can easily prove by induction that $y_i= \frac{F_{2i}}{F_{2i+1}}$.
We shall prove that except for {k;m}={2i;2i+1}, there are no other pairs of k,m satisfying.
Assume that there are k' and m' such that $F_{k'}.F_{2i+1}=F_{m'}.F_{2i}$.
This is false due to the fact that: $F_x=F_2.F_x<F_4.F_{x-2}<...<F_1.F_{x+1}=F_{x+1}$ which gives that there does not exist different pairs {i,j} and {k,l} such that $F_i.F_j=F_k.F_l$ (when $2 \leq i,j,k,l$).


There are $n$ citizens in the city Burg. Each man knows 1 unique news.

1. Every evening some people call to other people and tell all the news they know to each other (each person perform at most 1 conversation each evening). How many [b]evenings[/b] do they need in order each man will know all news from other men.

2. How many [b]calls[/b] do they need in order to exchange all the news?

At least your answer is incorrect for $n=2$.

ok, here is what happened, there are always 2 persons who simultaneously know everything for the first time, so after that We only need $n-2$ more calls, so we need at least $2n-3$ calls. Pick two persons, $A$ and $B$. if everyone but $B$ calls $A$ ($n-2$ calls), then $A$ calls $B$, ($1$ call) and then $B$ calls everyone but $A$, ($n-2$ calls ) we get 2n-3 is indeed enough! ;)

For the first one i remember it was something with the highestpower of $2$ less than $n$, but i dont really remember, i just know a friend of mine solve it one time...

[quote="Pascual2005"]about the secound one: if $A$ is the first guy to know everything, then in that moment everyone, but $A$, has called at least one time, and from here, everyone but $A$ must receive at least a call, so they need at least 2n-2 calls. If everyone calls $A$, and then $A$ calls everyone, we see $2n-2$ calls are enough.[/quote]
But each call involves two men! For example, if $B$ and $C$ have not all news, then after a call $B-C$ they could know them all. And we need only 1 call here, not 2. Am I missing something?
I can correctly prove that to the moment then $A$ have known all the news there were at least $n-1$ calls.

The first problem is really refers to powers of 2. It is not that hard.
It was used in ARO, round 4 of 5, grade 9.

ok, i was solving another problem then
sorry

Actually, my remark about the first problem is not true.
It is an easy problem if $n=2^k$. But I have just discovered that this problem is from ToT 1980, grade 9-10. They asks to solve the problem for $n=64$, $n=55$ and $n=100$. And $n=64$ costs 5 points and $n=100$ costs 12 points.

yes i also knew it was from ToT.

Anyway if we call our number $f(n)$ i Think I can prove $f(2n)=f(n)+1$ (divide the guys in 2 groups of n people, every group knows everything from its partners in $f(n)$ days, and finally, they learn everything from the other group in the $f(n)+1$ day) also $f(n+1)<f(n)+2$. i dont know iff $f(2n+1)=f(n)+1$ or something else...

hi
i think i saw this problem
i think if $n>3$ then  they need $2n-4$ calls.
for $n=4$ we do this
$a_1 - a_2$ , $a_3 - a_4$ , $a_1 - a _3$ , $a_2 - a_4$ so for $n=4$ is true.
suppose for $n=k$ we need $2k-4$ calls
for $n=k+1$ suppose that $a_{k+1}$ is new person.
first $ a_1 - a_{k+1} $ so $a_1$ has the new news . as we assumed $a_1 , .... , a_k$ can exchange all the news in $2k-4$ calls.so if one of them calls $a_{k+1}$ everyone has all the news.
so for $n=k+1$ we need :
$1+2k-4+1=2(k+1)-4$
so if $n>4$ then $2n-4$ calls are enough!

If you decide to give up, there is a [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=137&forum=DCForumID4#14]reference on CTK Exchange[/url] (see message #14 in the thread) to a paper that claims to show that you cannot get away with fewer than $2n-4$ calls. Another proof may be found on [url]http://www.cs.cornell.edu/vogels/Epidemics/[/url] (look for "Gossips and telephones" article). Of course, it is always more fun to try to solve the problem by yourself... :)


I am planning to take the October SAT II in Chemistry.

I want to know how much detail and rigor is need for the exam. I've seen the questions - they are easy. I'm asking about the topics - how many different, which ones and in what detail should they be done. Also, I want to know what needs to be done regarding laboratory experience, experiments etc. Do we need to know all radical tests? Where can I get information about practicals done in US schools(I live in India)

Please help! I can clarify my question if you want.

Come on guys!!!
I am also planing to take the sat test yhis fall!!!!!!!
Cold someone provide me with some useful information??


geometry
3D geometry
sphere
analytic geometry
Stanford
college
similar triangles

Two spherical balls lie on the ground touching. If one of the balls has a radius of 8 inches, and the point of contact is 10 inches above the ground, what is the radius of the other ball?

wouldn't the answer be the same [hide]if they were two circles lying on a line? if so, I extend the line connecting the two centers to intersect the line the circles are resting on. This forms similar triangles, which I can use to solve for the radius of the big ball.[/hide]

I don't think so. I can disprove it by[hide]the point of contact of two same spheres touching the ground would be at the length of the radius from the ground, so it would be at 8 inches, which is less than 10 inches. So the second sphere must be slightly larger, so the first ball's upper hemisphere would touch the second sphere.[/hide]

I say this because I'm sure ComplexZeta won't post problems that easy.

yes...I realized that... I don't see how that makes a difference though...some more help please?

well imagine you having a bowling ball and a baseball. (obviously the bowling ball is bigger than the baseball) When you place them on a flat surface, the baseball rolls "under" the bowling ball. This could be imagined even more easily if you imagine a [b]huge[/b] sphere and a small one. The small one would litterally almost be "under" the bigger one, so the point of contact would almost be at the top, which has a greater distance from the ground than the radius (equator).

Did this help?

ok....so one of them's bigger than the other...so?

If it's the same, then what's the answer? If it isn't the same, can you see why not?

I think we oughta plot everything on a plane and go from there...although I tried it and it got a bit weird...though that's probably cause I don't know what Im doing, not because that is the wrong way of going about solving the problem.

[quote="Tare"]I suggest actually finding two balls of different sides and tring this problem. (It doesn't have to have the radius of 8 inches, as long as one's noticablly larger than the other)
It's much easier to actually see it than to imagine it. I hope this works, because I'm out of ideas...[/quote]

ok, so let's say you do all that. But then, why don't you just take the cross section, with the plane you're using be the one that is perpendicular to the ground and contains both the centers of the spheres. Wouldn't that just reduce it to two circles?

Yes. Same as Fiertycoon's approach, which is the right way to solve. Since you know the radius of one circle (8) you get the equation (x doesn't really matter since you can move it anywhere) of the first sphere and you know the point of contact is at y=10, so you can get the answer from there.

Well, will anyone solve it, or will they just say that it can be done?

Can we use proportions here?

Is the answer [hide]12.5[/hide]?

using my method which seems to be wrong although no-one will clearly explain why, we get R=[hide]40/3[/hide]

MysticTerminator, your answer and method are both correct.

thank you. *takes deep bow amid wild cheers from the female members of the audience*

MysticTerminator, would you mind posting your solution, because I, for some reason, got a different answer. Here is my general method:

[hide]I expressed the smaller circle in terms of x and y, then plugged in y=10 to get the x-coordinate of the intersection point of the two circles. I then expressed the larger circle in terms of x, y, and r, then plugged in y=10 and the x-coordinate which I found and solved for r.[/hide]

Why doesn't this work?

I'm guessing you might have centered the smaller circle at the origin, in which case you should plug in y=2 instead.

My method was to say HEY! let's use similar triangles. If you read my earlier posts you would know which similar triangles I am talking about, but I will explain it again. Let the center of the smaller circle be A, let the center of the bigger circle be B, let the perpendiculars from A and B to the floor be C and D respectively, and let AB meet CD at O. Also let the point of intersection of the two circles be P and its perpendicular to the floor be Q. Then OAC :sim: OPQ :sim: OBD. This means (let OA=x) that x/8=(x+8)/10=(x+8+R)/R. x turns out to be 32 and R=40/3

Ok, I understand your solution. I also understand why mine is wrong now as well (I fumbled up the equation of the larger circle--I shifted its x-coordinate by the length of the intersection + the length of its radius, but that radius length is not what I should have put in). Thanks.

hmm,

the Cartesian plane solution appeals to me a lot, cuz there's less thinking involved

[quote]the Cartesian plane solution appeals to me a lot, cuz there's less thinking involved[/quote]

Actually, I think its considerably more difficult. I never fixed my aforementioned problem after I had found it.

o
ic
but i tink its rather easy....
or am i wrong?

[quote]but i tink its rather easy....
or am i wrong?[/quote]

Dunno. Im not the sorta guy that can solve every single olympiad problem, but try it out and you will see the difficulties...

[color=cyan]No.  Do not use coordinates for this.   They make your life very hard, since you need three variables (the x coordinate, y coordinate, and radius of the other sphere).  If you just do it synthetically, you save yourself all the work of elliminating variables.

Now, take this problem and generalize.  (We can leave out the sphere part).  Given a circle sitting on a table and another circle tangent to it.  If the first circle has radius r, and the point of tangency of the is a distance h from the line, what is the radius of the other circle?  (0<h<2r.  It might be interesting to solve for h outside that range, too.  But maybe not.)

This is the "converse problem" of the more traditional: if two circles of radius r1 and r2 are tangent to each other and both sitting on a table, what distance from the table is the point of tangency of the circles?[/color]

[quote="JBL"]
Now, take this problem and generalize.  (We can leave out the sphere part).  Given a circle sitting on a table and another circle tangent to it.  If the first circle has radius r, and the point of tangency of the is a distance h from the line, what is the radius of the other circle?  (0<h<2r.  It might be interesting to solve for h outside that range, too.  But maybe not.)[/quote]

Ok...if we let the other radius be R, then if h>r, using similar triangles again (the similar triangles are caused by extending the line of centers to meet the line they're sitting on at point O, and dropping perpendiculars to the line from the centers. x is the distance from O to the center of the first circle), x/r=(x+r)/h=(x+r+R)/R. Using the fact that a/b=c/d means a/b=c/d=(a-c)/(b-d), we get (h-r)/r=(R-h)/R, which can be solved for R=rh/(h-2r), I think.

If h<r, that means the first circle is bigger, so we do similar triangles the other way, and let x be the distance from O to the center of the second circle. Now similar triangles gives us:

x/R=(x+R)/h=(x+R+r)/r. Using the fact that a/b=c/d means a/b=c/d=(a-c)/(b-d), we get (r-h)/r=(h-R)/R, which is in fact the same as before.

"Alternatively" (in fact it's the same as similar triangles) we can also express the sinus of the angle between the two centers and horizontal :

sin (this_angle) = (R-r)/(R+r) = (h-r)/r

With r, R the two radius and h the height.

Okay, I have the answer here

[hide]All you have to do is draw a diagram.  If you use the similar right triangles, you will end up with an equation for the radius of the circle.  If you solve it you will end up with 40/3 [/hide]

What kind of problem is this? WHere did you get it from? This is interesting, I haven't seen one of these before.

Hm. I don't remember where I got it from now. It might have been from Stanford Math Tournament. Then again, it might not have been from Stanford Math Tournament. Sorry.


[quote="vmsvl"]$72(cosx+2sin2x)=25cos4x+7$[/quote]

[hide="Hint"]

$e^{i\theta}= \cos\theta+i\sin\theta$[/hide]


complex analysis
complex analysis unsolved

Assume that  roots of $f(x)=0$ tends  to zero.Where$f(x)=\sum_{n=1}^{\infty}\frac{a_{n}}{z^{n}}$,and assume that
 $g(z)=\sum_{n=1}^{\infty}b_{n}z^{n}$.
What we can say about roots of $f(x)+g(x)=0$. ($a_{n},b_{n}$ are real numbers)
(Can we use Rushe's theorem.)

Without additional assumptions you can hardly say anything. Indeed, if you take $f(z)=z(e^{1/z^{2}}-1)$ and $g(z)=z$, then $f(z)+g(z)=ze^{1/z^{2}}$ will have no roots at all.


It seems that Mr. Yanco has taken the liberty of posting several more old MAML Olympiad tests, some with solutions, both level one and level two.

They can be found here: [url]http://www.matharchive.org/olympiad/[/url]

A question: Does "contest director" mean that the listed person was the chief problem writer that year as well?

[quote="perfect628"]Huh, he's also reversed the order on the page...odd.[/quote]

Let's hope it's not to put his own name at the top!  :wink:  Probably not.

Anyway, thanks alot, Mr. Yanco. It looks like there are two typos in the 2002-03 Level 2 test, in problem 1a those relations are wrong, they should probably read $ 2^{2}\plus{}3^{2}\plus{}2^{2}3^{2}\equal{}7^{2}$, etc. In problem 4c, there is a mysterious $ v$ in the exponent of the numerator, on the LHS. Doesn't that $ v$ have to be 1, making it kind of trivial?

I decided to keep ten L1 contests online, which seems more than enough, plus as many corresponding L2 contests as I could find.  (I was digging through the "Worcester Academy archive" and found those L2's, along with whatever solutions you see.)

Because the ten contests online are the ten most recent that I have, it makes sense (to me) to put them at the top of the page, not the bottom -- otherwise you'd have to scroll past 33 lines to get to the contests OR I'd have to delete 33 contest directors' names, but I like the list of names.

To answer your other question, typically the Contest Director hires someone to write the contest questions (plus quite a few extras), tries them, then picks through them, trying to come up with a good mixture of topics.  So the name you see is the name of the person who selected the questions, not necessarily who wrote them.  This year I wrote [looking...] fifteen of the 25 questions, which was unusual, but I thought it sounded fun.  Two questions came from the NYCIML, eight from a friend.

As for errors in the contests, it happens.  Check out the very first word of the 2003-04 contest.  If memory serves, there were problems on that contest with Q3, Q5, and Q25.  As you know, I blew the mass points problem this year, but I uploaded the corrected contest.  ([i]What[/i] error?)

I hope you like the L2 contests as you prepare for L2.  There are some neat questions in there.


[quote="Arne"]Let $\mathcal{A}: V \rightarrow V$ be a nilpotent linear transformation of index $k$. Show that \[ \{0\} \subsetneq \ker{\mathcal{A}} \subsetneq \ker{\mathcal{A}^2} \subsetneq \ker{\mathcal{A}^3} \cdots \subsetneq \ker{\mathcal{A}^{k - 1}} \subsetneq \ker{\mathcal{A}^k} = V.  \][/quote]

if $x \in \ker A^m$, $A^{m+1}x = A0 = 0$, so $x \in \ker A^{m+1}$.
$A^k = A^{m+1} A^{k-m-1}$
Pick a v not in $\ker A^{k-1}$, then $A^{k-m-1}v \in ker A^{m+1}$, but not in $ker A^m$

Nice, didn't know that one. I'll think about it.

Isn't it a general result that as soon as $N(T^k)=N(T^{k+1})$ the nullity of all higher powers is also equal to this?

Yeah, I somehow proved that a while ago, but I didn't know it is a "general result" :)

I'm sorry, I did not stress in my last post that I did not include T being nilpotent.
I am still correct then, or am I?

Of course you are right, since we have $T^{n+2}=T(T^{n+1}) = T(T^n) = T^{n+1}$. ;)

[quote="alekk"]another classical result:
$u_n=rank(A^{k})-rank(A^{k+1})$
Study $u_n$ (is it monotonous ..)[/quote]

Could you give a hint for a proof of the fact that the sequence is monotonous?

[quote="alekk"]another classical result:
$u_n=rank(A^{k})-rank(A^{k+1})$
Study $u_n$ (is it monotonous ..)[/quote]

Suppose $B_k: R(A^k) \to R(A^k)$ is defined by the map $x \to Ax$. Now, by dimension theorem, $rank(A^k) = nullity(B_k) + rank(B_k) = nullity(B_k) + rank(A^{k+1})$, so $nullity(B_k) = rank(A^k) - rank(A^{k+1})$. I will try to prove $nullity(B_k)$ is decreasing.
If $x \in N(B_{k+1})$, then $x \in R(A^{k+1}) \subset R(A^k)$ so that $Ax = 0$ So $nullity(B_{k+1}) < nullity(B_k)$. When we have the first m so that $nullity(B_m) = {0 }$, it becomes apparent that all the remaining $u_n = 0$ for $n \ge m$
Conclusion: $u_n$ is decreasing. Is the proof okay?


Which is the larger of the 2 numbers

$100^{117}$ or $117^{100}$?

No log tables allowed

My answer is 100^117
Divide them we get,[(1.17)^100/(100^17)] whch we can write as
1.17^2[(1.17)^7]^14 / (100)^17   Now 1.17^7 is approx equal to 2
so the expression becomes approx equal to 1.3*2^14/(100^17) whch is obviously less than 1..Hence we get  100^117 > 117^100

my answer is 100^ 117.

1.  100^117 equals 10^334.
2.  117^100 = (100+17)^100. Expanding by binomial theorem. Greatest term would be 100^100 = 10^200. Rest of the terms will only be added to it.

Hence, first one having greater number of digits is larger one.

a^b should be larger than b^a whenever
$(a-e)^2<(b-e)^2$ and $\frac{b-e}{a-e}>0$
for all positive numbers a and b


ratio
number theory solved
number theory

Prove or disprove the following statement:
Each positive ration number can be written in the form:
$\displaystyle \frac {a^2+b^3}{c^5+d^7}$ where $a,b,c,d$ are positive integer.

True. Let $\frac{p}{q} $ a rational positive number. Take $ a=p^{18}\cdot q^{87}, b=p^{12}\cdot q^{58}, c=p^7\cdot q^{35}, d=p^5\cdot q^{25} $

Good!!
Have you ever done it???? Harazi :D


Let $ a,b,c,d$ be non-negative reals. Prove that

$ 3(a^2\plus{}b^2\plus{}c^2\plus{}d^2)\plus{}2(abc\plus{}abd\plus{}adc\plus{}dbc)\minus{}4(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\minus{}1)\geq0$

[quote="Ligouras"]Let $ a,b,c,d$ be non-negative reals. Prove that

$ 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \plus{} 2(abc \plus{} abd \plus{} adc \plus{} dbc) \minus{} 4(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \minus{} 1)\geq0$[/quote]

sorry

[quote="Ligouras"]Let $ a,b,c,d$ be non-negative reals. Prove that

$ 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \plus{} 2(abc \plus{} abd \plus{} adc \plus{} dbc) \minus{} 4(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \minus{} 1)\geq0$[/quote]

Nice!Generally,we can prove that:

 \[ \frac{3(k\plus{}1)}{2}(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \plus{} 2(abc \plus{} abd \plus{} adc \plus{} dbc)\]
\[ \minus{} (k\plus{}3)(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd )\plus{}4\ge{0}\]
holds $ \forall {a,b,c,d}\ge{0}$ if and only if $ k\ge{0}.$


can anyone here tell me what synthetic geometry is?

hello, i think, this will help you
http://wiki.answers.com/Q/What_is_synthetic_geometry
Sonnhard.


2006 yılı 2. aşama sorularını foruma eklesek iyi olur... ayrıca sınav bayağı zormuş herkes 2-3 civarında yapmış.

Merhaba Eren :)
Bizim forum'a da takılmanı bekleriz.
Biz 2007'yi bile koyduk, yapıldığı hafta sonu :) Diğer haftasonuna da çözümleri yetiştirmeye çalışıyoruz. Siteye profildeki www'dan ulaşabilirsiniz...

Arkadaşlar 2007'yi burada yayınlasak iyi mi olur? Ben İngilizceye çevirdim kendi çapımda :), sonra yazıya geçirmeye üşendim de :) Eğer iyi olacaksa koyalım National Olimpiads bölümüne...

Bence iyi olur. Ben de üşendim bu sene çevirmeye soruları özelllikle 6 nın çevirisini yapmak çok zor geldi. Kaysi abi eğer sana zor geliorsa bana ingilizce çeviriyi yolla latexleme işini de ben yapayım.

Takım seçme sınavı için üşenmemişsin.
Seni matematikolimpiyati.org'a transfer etsek nasıl olur :)
Yazın biraz daha boş olacağımı düşünüyorum, çeviri işlerine o zaman dalarım. Şimdi derslerle beraber zor oluyor. Ama çevirilerde de önce lik İngilizce'den Türkçe'ye olur. Önce kendi insanım :)


Show that if $n \in N$, $n>1$ and $r$ is the number of the dissimilar prime divisors of $n$, then:
$S(n) = \sum\limits_{{d \mathord{\left/ {\vphantom {d n}} \right. \kern-\nulldelimiterspace} n}} {\left| {\mu (d)} \right|} = 2^r$
[code]P.S: S is a multiplicative function[/code]

[hide]Since $\mu$ is multiplicative and the absolute value is multiplicative, so is $|\mu(d)|$. It is a fact that $g$ multiplicative implies $\sum_{d|n} g(d)$ is multiplicative. This makes $S$ multiplicative.
Now only need to test $S$ for prime powers. Let $p$ be prime $a$ be a positive integer, $S(p^a)=\sum_{k=0}^a |\mu(p^k)|=1+1+0+0+\ldots+0=2$. From multiplicativity, $S(n)=2^r$.[/hide]


Let $ ABCD$ be a convex quadrilateral.$ M,N,P,Q$ be midpoint segment $ AB,BC,CD,AD$.Construct $ K$ inside $ ABCD$ such that $ S(AMKQ) \equal{} S(BMKN) \equal{} S(NKBC) \equal{} S(QKPD)$. Is $ K$ existed unique??????? :)

$ AC, BD$ meet at $ E$, $ G$ is the centroid of $ ABCD$ . Then $ G$ is the midpoint of $ EK$ .

Dear Lym. Can you post your solution???? :)

Sorry, $ K$ is not existed unique, when $ AB//CD$, you can found $ K$ can be on ouside of $ ABCD$ .

Well.What a interesting problem!!!! :).Can you post solution for 2 case above?? :)

Now I am lazy to type, next I will or Others may give you a solution.

Thank you :).This is mine.Let $ I,J$ be inside $ ABCD$ such that $ QIMA,PINC,NJMB,PJQD$ be parallelogram.Construct trapezoid $ QIKP,PKJN$.We'll get $ K$ to be point satisfy the problem.I'll post solution later  :)


[quote="realnumber"]$ x_1 \equal{} x_2 \equal{} ... \equal{} x_{11} \equal{} 0.001$[/quote]

 :!: please prove it!!!!! :!:


group theory
abstract algebra
superior algebra
superior algebra unsolved

The following holds as a corollary of Artin's theorem: if $E$ is a field and $G$ is a finite subgroup of $Aut(E)$, then $G=Aut(E/E^G)$.

Does the result still hold if $G$ is not supposed to be finite ?

Hi.

In general this is wrong. Take $E=\mathbb{Q}(\sqrt2,\sqrt3,...,\sqrt p,...)$ ober $\mathbb{Q}$. This is a galoisextension with galoisgroup $Gal(E|\mathbb{Q})=\mathbb{Z}_2^{\mathbb{N}}$ (the direct product of countable many copies of $%Error. "mathh" is a bad command.
{Z}_2$)
If you take the group $G=\mathbb{Z}_2^{(\mathbb{N})}$ (the direct [i]sum[/i] of countable many copies of $%Error. "mathh" is a bad command.
{Z}_2$), you get $E^G=\mathbb{Q}$ and so $Gal(E|E^G)=Gal(E|\mathbb{Q})\neq G$

Gockel.


calculus
integration
number theory
prime factorization
prime numbers

[quote="DaReaper"]$1$. what is the differrnce between a factor and an integral factor

$2$.what is an integral factor

$3$.how do you find how many integral factors a number has

          $*$for example: how many integral factors does $144$ have?[/quote]
Not really good with definitions but I can answer 3. In order to find how many integral factors you prime factorization a number. So say your example. 144 prime factorized is $2^{4}\cdot3^{2}$. So then add the exponents by one and multiply them together. So it is 4+1 and 2+1 and multiply. So 144 has 15 integral factors. Try finding how much 360 has.

[quote="DaReaper"]Actually $144$ has $30$ factors....this question came from a pizitz test review from 2005 (it was ciphering $4.4$)

i had the answer i was just wondering how to do it...[/quote]
That makes no sense. You sure you got the right answer key. 144 is a perfect square so it should have an odd number of factors.

[quote="DaReaper"]well, it was the key that last years toournament was based off off, and there were about 15 schools there...and there were no protests agiainst the question...so...im not sure

*but for the prime factorization thing.. how do you know what two numbers and exponents you need to prime factorize a number? like how did you get $2^{4}\cdot3^{2}$ ?[/quote]
I did a quick factor tree in my head. But I'll do it out. $144=12\cdot12$.       $12=4\cdot3$.     $4=2^{2}$. So 12 is $2\cdot2\cdot3$. There's another 12 so the prime facorization os 144 is $2^{4}\cdot3^{2}$.

*ohh yea forgot about that.. we learned that is like 5th grade oops  :blush: *

**but how is it that the answer is 30.. but you say it is 15...

so which one is it!!?!?!?     :mad:

okay, first of all, i would like to say whoever didnt know to protest this question is *stupido*.

but heres how you prime factor.

just keep dividing until all you have is prime numbers.
so for $144$, divide by 3.  so thats 3*48.  48=3*16.  so then we have 3*3*12.  16=4*4.  so we have 3*3*4*4.  4=2*2.  so 144=3*3*2*2*2*2, or $2^{4}3^{2}$.

so $(4+1)(2+1)=15$

the reason you add one is simply choose.  since any different combination of prime numbers will be a different number, you can choose to use $4^{2}$, $4^{1}$ or $4^{0}$ and so on for the other numbers.

[quote="DaReaper"]$1$. what is the differrnce between a factor and an integral factor

$2$.what is an integral factor[/quote]

In the normal context, "factor" means "integral factor."

[b]To those who are wondering why the answer is $30$ and not $15$:[/b]

Negative integers are also integers ;)


Euler
geometry
3D geometry
sphere
topology
vector
analytic geometry

I have a question regarging Hirsch's "differential topology":proof of thm 5.2.3.

It is well known that $ S^n$ has two charts: stereographic projection from north pole P and south pole Q.

$ \sigma: S^n\minus{}P\minus{}>R^n$ and $ \tau: S^n\minus{}Q\minus{}>R^n$.

Let $ f$ be the vector filed on $ S^n\minus{}P$ whose representation via $ \sigma$ is identity vector filed on $ R^n$. Then $ f(x)\minus{}>0$ as $ x\minus{}>P$, so define $ f(P)\equal{}0$. Thus $ f: S^n\minus{}>TS^n$ has zeros only at $ P$ and $ Q$.

In $ \tau$ coordinates, $ f$ is vector field $ x\minus{}>\minus{}x$ on $ S^n\minus{}Q$. So $ f$ is $ C^{\infty}$.

My question is: using chart $ \sigma$, $ f(x)\equal{}x$, it is easy to see $ f(Q)\equal{}0$. But as $ x\minus{}>P$, $ f(x)$ should tend to $ \infty$ in $ R^n$ since this is stereographic projection from P. Why $ f(x)\minus{}>0$ ?
Obviously let $ f(P)\equal{}0$ will not make $ f$ continuous. How come $ f$ has two zeroes?

[quote="mlok"]$ f(x)\to 0$ at $ P$ because of the chain rule.[/quote]

I see. Although the chart coordinate $ x$ is getting large as we move toward pole.
But the transport of structure by the chart will take, approximately, $ x\minus{}>\frac{1}{x^2}$
as can be seen from a simple plane geometry sketch. So the length of the
vector is tending to $ 0$ as $ x\minus{}>\infty$.

near the other pole, the transport of structure is approximately $ x\minus{}>x/2$, so it is also a zero
for this vector field.


geometry
3D geometry
pyramid
incenter
circumcircle
tetrahedron
geometry proposed

The insphere of triangular pyramid $ SABC$ is tangent to the faces $ SAB,$ $ SBC$ and $ SAC$ at points $ G,$ $ I$ and $ O$ respectively. Let $ G$ be the intersection point of medians in the triangle $ SAB,$ $ I$ be the incenter of triangle $ SBC$ and $ O$ be the circumcenter of triangle $ SAC.$ Prove that the straight lines $ AI,$ $ BO$ and $ CG$ are concurrent.

In this tetrahedron, we: $ |SA|=|SB|=|SC|=|AB|=|BC|=|AC|\Rightarrow$ the points $ I, \ G, \ O$ are centroids of the faces. And medians $ AI, \ BO, \CG$ are concurent in the centroid of the tetraedron. [b]For main solution, view romanian topics:[/b] http://www.mathlinks.ro/viewtopic.php?p=1334056#1334056


Find all real numbers $ x,y,z$ such that $ x\plus{}y\plus{}z\equal{}1$ and $ x^{4}\plus{}y^{4}\plus{}z^{4}\equal{}xyz$

[hide]We have that $ x, y, z$ are the roots of $ a^3\minus{}(x\plus{}y\plus{}z)a^2\plus{}(xy\plus{}yz\plus{}zx)a\minus{}xyz\equal{}0$.  We multiply by $ a$, plug in $ x$, $ y$, and $ z$ for $ a$ and add them to get

$ (x^4\plus{}y^4\plus{}z^4)\minus{}(x\plus{}y\plus{}z)(x^3\plus{}y^3\plus{}z^3)\plus{}(xy\plus{}yz\plus{}zx)(x^2\plus{}y^2\plus{}z^2)\minus{}(xyz)(x\plus{}y\plus{}z)\equal{}0$.

We plug in what we know to get

$ xyz\minus{}(x^3\plus{}y^3\plus{}z^3)\plus{}(xy\plus{}yz\plus{}zx)(x^2\plus{}y^2\plus{}z^2)\minus{}xyz\equal{}0$

$ (xy\plus{}yz\plus{}zx)((x\plus{}y\plus{}z)^2\minus{}2(xy\plus{}yz\plus{}zx))\equal{}(x^3\plus{}y^3\plus{}z^3)$

$ \minus{}2(xy\plus{}yz\plus{}zx)^2\plus{}(xy\plus{}yz\plus{}zx)\equal{}3xyz\plus{}(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}zx)$

$ \minus{}2(xy\plus{}yz\plus{}zx)^2\plus{}(xy\plus{}yz\plus{}zx)\equal{}3xyz\plus{}((x\plus{}y\plus{}z)^2\minus{}3(xy\plus{}yz\plus{}zx))$

$ \minus{}2(xy\plus{}yz\plus{}zx)^2\plus{}(xy\plus{}yz\plus{}zx)\equal{}3xyz\plus{}1\minus{}3(xy\plus{}yz\plus{}zx)$

$ \minus{}2(xy\plus{}yz\plus{}zx)^2\plus{}4(xy\plus{}yz\plus{}zx)\equal{}3xyz\plus{}1$

$ \minus{}2(xy\plus{}yz\plus{}zx\minus{}1)^2\equal{}3xyz\minus{}1$

Let $ (xy\plus{}yz\plus{}zx)\equal{}k$.  We have $ \minus{}2(k\minus{}1)^2\equal{}3xyz\minus{}1$

$ xyz\equal{}\frac{1}{3}\minus{}\frac{2}{3}(k\minus{}1)^2$.

So $ x, y, z$ are the roots of $ a^3\minus{}a^2\plus{}ka\minus{}(\frac{1}{3}\minus{}\frac{2}{3}(k\minus{}1)^2)\equal{}0$.

I don't have time to find the bounds on $ k$ right now, but one solution is $ (1/3, 1/3, 1/3)$.[/hide]

[hide="Solution"]Many equations can be more easily dealt with when one equalizes the degrees of all the terms (this is known as homogenization and it is useful mainly because $ f(ta_1,ta_2,\ldots,ta_n) \equal{} t^nf(a_1,a_2,\ldots,a_n)$ which allows us to assume things like $ \sum a_1 \equal{} 1$ by "scaling" because only the [i]ratio[/i] of the terms matters). Here, for example, $ x^4 \plus{} y^4 \plus{} z^4 \equal{} xyz(x \plus{} y \plus{} z) \equal{} x^2yz \plus{} xy^2z \plus{} xyz^2$ (so now the degrees of both sides are equal). But $ \frac {x^4 \plus{} x^4 \plus{} y^4 \plus{} z^4}4\ge\sqrt [4]{x^8y^4z^4} \equal{} |x^2yz|\ge x^2yz$ from AM-GM on $ x^4,x^4,y^4,z^4$, and similarly, $ \frac {x^4 \plus{} y^4 \plus{} y^4 \plus{} z^4}4\ge xy^2z$ and $ \frac {x^4 \plus{} y^4 \plus{} z^4 \plus{} z^4}4\ge xyz^2$. Adding up these inequalities gives $ x^4 \plus{} y^4 \plus{} z^4\ge x^2yz \plus{} xy^2z \plus{} xyz^2$, but we already know that $ x^4 \plus{} y^4 \plus{} z^4 \equal{} x^2yz \plus{} xy^2z \plus{} xyz^2$, and equality in AM-GM only occurs when $ x \equal{} y \equal{} z$, which means that $ x \equal{} y \equal{} z \equal{} \frac13$.[/hide]

I'm not sure, but there could be solutions with (2) negative numbers our of x, y, and z. In this case, AM-GM fails.

I'm pretty sure AM-GM works here, because $ x^4,x^4,y^4,z^4$, for example, are all nonnegative, and the absolute value of something is always greater than or equal to that something...

Or we can directly prove it...let $ a,b,c,d$ be reals, so $ a^4\plus{}b^4\plus{}c^4\plus{}d^4\minus{}4abcd\equal{}(a^2\minus{}b^2)^2\plus{}(c^2\minus{}d^2)^2\plus{}2(ab\minus{}cd)^2\ge0$.

Wow, that was so much simpler than my solution.  But here it is anyway, if you can struggle through the whole thing.

[hide][hide="lemma"]Lemma: If a cubic $ x^3 \plus{} kx^2 \plus{} lx \plus{} m$ has three real roots, then $ 27m^2 \minus{} 18klm \minus{} k^2l^2 \plus{} 4k^3m \plus{} 4l^3\leq 0$.  If the cubic has two equal roots, then equality occurs. (This is not the only time equality occurs, though.)  If the cubic has only one real root, then the inequality sign is reversed.

[hide="proof"]Proof: Let $ x^3 \plus{} kx^2 \plus{} lx \plus{} m \equal{} (x^2 \plus{} 2ax \plus{} a^2 \plus{} c)(x \minus{} b)$ and equate coefficients when $ c > 0, c \equal{} 0$ or $ c < 0$.[/hide][/hide]

We know that $ x, y, z$ are roots of $ a^3 \minus{} a^2 \plus{} ka \minus{} (\frac {1}{3} \minus{} \frac {2}{3}(k \minus{} 1)^2)$.  We apply the lemma where $ k\equal{}\minus{}1$, $ l\equal{}k$, and $ m\equal{}(\frac{1}{3}\minus{}\frac{2}{3}(k\minus{}1)^2)$.

$ 27(\frac {1}{3} \minus{} \frac {2}{3}(k \minus{} 1)^2)^2 \minus{} 18k(\frac {1}{3} \minus{} \frac {2}{3}(k \minus{} 1)^2) \minus{} k^2 \plus{} 4(\frac {1}{3} \minus{} \frac {2}{3}(k \minus{} 1)^2) \plus{} 4k^3\leq 0$

$ 3(1 \minus{} 2(k \minus{} 1)^2)^2 \minus{} 6k(1 \minus{} 2(k \minus{} 1)^2) \minus{} k^2 \plus{} 4(\frac {1}{3} \minus{} \frac {2}{3}(k \minus{} 1)^2) \plus{} 4k^3\leq 0$

$ 3(1 \minus{} 4(k \minus{} 1)^2 \plus{} 4(k \minus{} 1)^4) \minus{} 6k(1 \minus{} 2k^2 \plus{} 4k \minus{} 2) \minus{} k^2 \plus{} \frac {4}{3} \minus{} \frac {8}{3}k^2 \plus{} \frac {16}{3}k \minus{} \frac {8}{3} \plus{} 4k^3\leq 0$

$ 3(1 \minus{} 4k^2 \plus{} 8k \minus{} 4 \plus{} 4k^4 \minus{} 16k^3 \plus{} 24k^2 \minus{} 16k \plus{} 4) \minus{} 6k \plus{} 12k^3 \minus{} 24k^2 \plus{} 12k \minus{} k^2 \plus{} \frac {4}{3} \minus{} \frac {8}{3}k^2 \plus{} \frac {16}{3}k \minus{} \frac {8}{3} \plus{} 4k^3\leq 0$

$ 3 \minus{} 12k^2 \plus{} 24k \minus{} 12 \plus{} 12k^4 \minus{} 48k^3 \plus{} 72k^2 \minus{} 48k \plus{} 12 \minus{} 6k \plus{} 12k^3 \minus{} 24k^2 \plus{} 12k \minus{} k^2 \plus{} \frac {4}{3} \minus{} \frac {8}{3}k^2 \plus{} \frac {16}{3}k \minus{} \frac {8}{3} \plus{} 4k^3\leq 0$

$ 12k^4 \minus{} 32k^3 \plus{} \frac {97}{3}k^2 \minus{} \frac {38}{3}k \plus{} \frac {5}{3}\leq 0$

$ 27(12k^4 \minus{} 32k^3 \plus{} \frac {97}{3}k^2 \minus{} \frac {38}{3}k \plus{} \frac {5}{3})\leq 0$

$ 4(3k)^4 \minus{} 32(3k)^3 \plus{}97(3k)^2 \minus{} 114(3k) \plus{} 45\leq 0$

$ (4(3k)^2 \minus{} 24(3k) \plus{} 45)(3k \minus{} 1)^2\leq 0$

$ (4(3k \minus{} 3)^2 \plus{} 9)(3k \minus{} 1)^2\leq 0$.  The only solution to this inequality is $ k \equal{} \frac {1}{3}$ so the only triple that works is $ (x, y, z) \equal{} (\frac {1}{3}, \frac {1}{3}, \frac {1}{3})$[/hide]

Lots of boring algebraic manipulation, especially the proof of the lemma.  I saved you the trouble of having to read it, because I know it's true.