Hello,

Prove that : \[ \big(\sqrt{2}\plus{}1\big)^{\frac{1}{100}}\plus{} \big(\sqrt{2}\minus{}1\big)^{\frac{1}{100}}\; \text{ is irrational number } .\]

[hide="solution"]
Suppose we have two real numbers $ a,b$ such that $ a\plus{}b$ is rational and $ ab\equal{}1$

$ (a\plus{}b)^2 \equal{} a^2\plus{}2ab\plus{}b^2 \Longrightarrow (a\plus{}b)^2\equal{}a^2\plus{}2\plus{}b^2$
Since $ a\plus{}b$ and $ 2$ are rational, this implies $ a^2\plus{}b^2$ is rational $ (1)$

$ (a\plus{}b)(a^2\plus{}b^2) \equal{} a^3\plus{} ab^2\plus{}ba^2 \plus{} b^3 \equal{} a^3 \plus{} a \plus{} b \plus{} b^3$
From (1), the LHS is rational and this implies $ a^3\plus{}b^3$ is rational $ (2)$

$ (a^2\plus{}b^2)(a^3\plus{}b^3) \equal{} a^5 \plus{} a\plus{} b \plus{} b^5$
Again this implies $ a^5 \plus{} b^5$ is rational $ (3)$

Therefore if $ a\plus{}b$ is rational, then from $ (1)$ and $ (3)$;
$ a\plus{}b \Rightarrow a^2\plus{}b^2 \Rightarrow a^4\plus{}b^4 \Rightarrow a^{20} \plus{} b^{20} \Rightarrow a^{100} \plus{} b^{100}$ is rational

Hence is $ a\equal{}(\sqrt{2}\plus{}1)^{\frac{1}{100}}$ and $ b\equal{}(\sqrt{2}\minus{}1)^{\frac{1}{100}}$

If $ (\sqrt{2}\plus{}1)^{\frac{1}{100}} \plus{} (\sqrt{2}\minus{}1)^{\frac{1}{100}} \in \mathbb{Q}$

Then $ (\sqrt{2}\plus{}1)\plus{}(\sqrt{2}\plus{}1) \equal{} 2\sqrt{2} \in \mathbb{Q}$ Contradiction!

[/hide]


This ones been puzzling me:

A special type of door lock has a panel with five buttons labeled with the digits 1 through 5. This lock is opened by a sequence of three actions. Each action consists of either pressing one of the buttons or pressing a pair of them simultaneously. For example, 12-4-3 is a possible combination. The combination 12-4-3 is the same as 21-4-3 because both the 12 and the 21 simply mean to press buttons 1 and 2 simultaneously.

a. How many combinations are possible?
b. Hom many combinations are possible if no digit is repeated in the combination?

----------

I started out by isolating the single digit and multiple digit combinations. There are 5^3 possibilities if you only entertain single digit actions, and 5!^3 possibilities if you only entertain multiple digit actions. So, there are 5^3*5!^3 total possibilities for a. Is this correct?

I'm not sure how to start on b. Could someone give me a hint?

There's something wrong with part a, I don't know where that $ 5!$ came from. I think you meant $ \binom{5}{2}$, which is the number of ways of choosing $ 2$ buttons. Anyways, this equals $ 10$, so the answer is $ 15^3$ (for each action you have $ 15$ possibilities: $ 5$ if you press a single button and $ 10$ if you press two buttons).

For part b, think about the following: if $ S$ means you press a single button, and $ P$ means you press two buttons simultaneously, then the actions can be $ 3S$, $ 2S$ and $ 1P$ or $ 1S$ and $ 2P$. Now count the three cases separately (for example, in the first one, you first choose which are the 3 digits you will press, and then in which order you will press them).

Thanks for your help. So, the actions for 3S would be 5 choose 3? I'm not sure how to combine that the other options. Could you show me how?

[quote="absvalue"]So, the actions for 3S would be 5 choose 3?[/quote]  More thinking, less using formulas  :wink:  Try 3S again, and don't just give a formula, give an explanation of why this formula should be right (or, really, an explanation of how to come to this formula).  Then do the same in the other cases.


calculus
inequalities
inequalities unsolved

Let $a$, $b$, $c$ be non-negative numbers such that $a+b+c = 3$. Prove that $\frac{5-3bc}{1+a}+\frac{5-3ca}{1+b}+\frac{5-3ab}{1+c}\ge ab+bc+ca$

$A=a+b+c=3, B=ab+bc+ca, C=abc$

Finnaly, we have to prove

$(27-B)C\geq 4B^{2}+11B-45$

because $B\leq 3, 3C\leq B^{2}$

we just have to prove

$(27-B)B^{2}\geq 3(4B^{2}+11B-45)$ <=> $(B-3)(B^{2}-12B+135)\leq 0$

which holds for all $B\leq 3$

i have an ugllllly solution...but if you want my idea here you have it : after expanding it remains us to prove that :

$45\geq 3\sum{a^{2}b}+3\sum{ab^{2}}+4\sum{a^{2}b^{2}}+2\sum{ab}+\sum{a^{2}b^{2}c}+6abc$

and now we multiply by 9 to obtain the same "5"-th degree:

$405\geq 3(\sum{a^{2}b}+\sum{ab^{2}})(\sum{a^{2}}+2\sum{ab})+12\sum{a^{2}b^{2}}\cdot\sum{a}+\frac{2}{3}\cdot\sum{ab}\cdot\ (\sum{a})^{3}+6abc\cdot\(\sum{a^{2}}+2\sum{ab})+9\sum{a^{2}b^{2}c}$

 :oops:

now we try to transforme 405 in something with 5-th degree as you can imagine , maybe $\frac{5}{3}\cdot(\sum{a})^{5}$, and then after some ugly calculs we have some muirhead and stuff like that :)

Let $f(x) = 1/x$.
Then, by Jensen, we have $LHS \geq  \frac{(15-3(ab+bc+ca))^{2}}{15-3(ab+bc+ca)+5(a+b+c)-9abc}= \frac{(15-3(ab+bc+ca))^{2}}{18-3(ab+bc+ca)-9abc}$.

Hmm... not sure where to go from there...

[quote="libra_gold"]because $B\leq 3, 3C\leq B^{2}$

we just have to prove

$(27-B)B^{2}\geq 3(4B^{2}+11B-45)$ [/quote]
The sign reversed,It is not true. :P

Ooooops...

I have corrected it.

$A=a+b+c=3, B=ab+bc+ca, C=abc$

Finnaly, we have to prove $(27-B)C\geq 4B^{2}+11B-45$

If $B \leq 2.25$ then, $(27-B)C\geq 0 \geq 4B^{2}+11B-45$

If $3 \geq B > 2.25$ then we can use Shur $\sum a(a-b)(a-c)\geq 0$ to get $(a+b+c)^{3}-4(a+b+c)(ab+bc+ca)+9abc\geq 0$ that is $9C \geq 12B-27(\geq 0)$

so we just have to prove $(27-B)(12B-27)\geq 9(4B^{2}+11B-45)$

<=> $(B-3)(B-2.25)\leq 0$ which holds for all $2.25<B\leq 3$

is there any mistake now?


\documentclass{book}                  ( no  the option [openany])

Each chapter will start on an odd page.  If   there  are  some   chapters  which   they   close   on  an odd page  each , then there are some blank pages  between
chapters .  But    there are   page    headers  and   page footers  on   these ' blank pages'.

How    to   remove    these  page    headers  and   page    footers   ?

Many   Thanks!

See section 15 of the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/fancyhdr/fancyhdr.pdf]fancyhdr documentation[/url].


AMC USA(J)MO USAMO geometry reflection httpbit.ly3vjUcdg

Each point in the plane is assigned a real number such that, for any triangle, the number at the center of its inscribed circle is equal to the arithmetic mean of the three numbers at its vertices. Prove that all points in the plane are assigned the same number.

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

I remember seeing some solutions of this one on the forum some time ago. I even remember posting one, but I'm sure it was quite different from this one:

Let $B,C$ be ny two points in the plane, take $A$ s.t. $ABC$ is equilateral, and take $D,E$ on the segments $AB,AC$ respectively s.t. $AB=3AD=3AE$. Now let $U,V$ be the points where the perpendiculars through $B,C$ to $BC$ cut $DE$, and, finally, let $P=BU\cap AV$.

Since the circle inscribed in $PAB$ is the same as that inscribed in $PUV$, we get $A+B=U+V$ (for a point $X$, the letter $X$ also denotes the value we assign to it). In the same way, we get $A+C=U+V$, from which we find $B=C$. Since $B,C$ were chosen arbitrarily, we're done.

Solution by mgao, Zhero, and myself:

For a point $P$, let $f(P)$ be the number assigned to point $P$.

Let $E, F$ be any two points in the plane; we want $f(E) = f(F)$. Construct regular hexagon $ABCDEF$ with center $O$. Then, $f(A) + f(C) + f(E) = 3 f(O) = f(B) + f(D) + f(F)$.

Now, let $P$ be the intersection of lines $AB$ and $CD$. By reflection through the angle bisector of $APD$, it is easy to see that $APC$ and $DPB$ have the same incenter $I$, and we have  $f(A) + f(P) + f(C) = 3 f(I) = f(D) + f(P) + f(B)$. Subtracting the equation $f(B) + f(D) + f(F) = f(A) + f(C) + f(E)$, we get the desired result $f(E) = f(F)$.

Take an equilateral triangle $ABC$ and construct triangles $XYZ$ and $DEF$ so that $XYZ$ is the medial triangle of $DEF$ which in turn is the medial triangle of $ABC$ ($AXD,BYE,CZF$ are lines and $\triangle{ABC},\triangle{DEF},\triangle{XYZ}$ share incenter $I$). Assume for the sake of contradiction that some two points in the plane don't have the same number; WLOG let $A$ and $X$ be these two points.

Note that
\begin{align}
f(A)+f(B)+f(C)=f(D)+f(E)+f(F)=3f(I).
\end{align}Also, because $\triangle{ABE}$ and $\triangle{ACF}$ share incircles and thus have a common incenter (this can be seen for example by reflecting about $AXID$ or noting that quadrilateral $AEIF$ is tangential), we have
\[\frac{f(A)+f(B)+f(E)}{3}=\frac{f(A)+f(C)+f(F)}{3}\implies f(B)+f(E)=f(C)+f(F).\]Symmetry and (1) tell us that
\[f(A)+f(D)=f(B)+f(E)=f(C)+f(F)=2f(I).\]But we clearly must also have
\[f(D)+f(X)=f(E)+f(Y)=f(F)+f(Z)=2f(I)\]by applying the same argument to figure $DZEXFY$ as we did to $AFBDCE$, so $f(A)+f(D)=2f(I)=f(D)+f(X)$ and $f(A)=f(X)$, contradiction.

Consider two points $A$ and $B$, with values $a$ and $b$ respectively. Draw the perpendicular bisector, and pick an arbitrary point $X$ on it, with value $c$. Then, let $X'$ be the incenter of $ABX$, $X''$ of $ABX'$, and so on.

The value of $X$ is $\frac{a+b+c}{3}$, $X'$ is $\frac{4a+4b+c}{9}$, $X''$ is $\frac{13a+13b+c}{27}$, and so on, with $X''...''$ where there are $n$ $'$s, having a value of $\frac{\frac{3^{n+1}-1}{2}(a+b)+c}{3^{n+1}}$.

As $n \to \infty$, we see that the value of $X''...''$ goes to $\frac{a+b}{2}$. As our function is necessarily continuous (slight perturbations of vertices slightly perturb incenter location), the midpoint $M$ of $AB$ has a value equal to $\frac{a+b}{2}$ (the sequence $X, X', X'',...$ converges to $M$).

Knowing this, we can divide a line segment $AB$ into successively smaller intervals, to find the value of a point dividing it in a certain ratio. The midpoint of $AM$ has value $\frac{3a+b}{4}$, the midpoint of that point and $A$ again is $\frac{7a+b}{8}$, and so on. Remembering that the function is continuous, and as all fractions can be written as an infinite sum/difference of fractions with denominators powers of $2$, it is clear that a point's value is the weighted average of the values of the endpoints of the segment (with the ratio of weights being equal to the ratio of the lengths, and the higher weight given to the endpoint it is nearest to).

Now, consider a triangle with vertices at $A=(0,0), B=(2,0), C=(0,2)$. Let $D,E$ be the midpoints of $AB,AC$ respectively, and let $A,B,C$ have values $a,b,c$ respectively. $D$ then has a value of $\frac{a+b}{2}$ and $E$ has $\frac{a+c}{2}$. The midpoint $O$ of $DE$ is at $\left(\frac{1}{2},\frac{1}{2}\right)$, and thus has value $o=\frac{2a+b+c}{4}$

Also, the incenter $I$ of $ABC$ is located at $(2-\sqrt{2},2-\sqrt{2})$ and has a value of $i=\frac{a+b+c}{3}$.

However, $O$ is also on the line connecting $A$ and $I$. The distance $AO$ is $\frac{\sqrt2}{2}$, and the distance $IO$ is $\sqrt2\left(\frac{3}{2}-\sqrt2\right)=\frac{3\sqrt2-4}{2}$, and $AI$ is $\sqrt2(2-\sqrt2)=2\sqrt2-2$

Then, $O$'s value of $\frac{2a+b+c}{4}$ is also equal to $\frac{a\cdot IO+i\cdot AO}{AI}=\frac{\frac{3\sqrt2-4}{2}a+\frac{\sqrt2}{2}\frac{a+b+c}{3}}{2\sqrt2-2}$ After some effort, this (should!) simplify to $a=\frac{b+c}{2}$.

As the midpoint $M$ of $BC=\frac{b+c}{2}$ as well, all points between $M$ and $A$ have value $\frac{b+c}{2}$. Reflecting $A$ over $M$ to $A'$, we get the same result for segment $AA'$. We can keep doing this infinitely, so all point on the line $AM$ have the same value. As there is nothing special about $AM$, we have that all points on a given line have the same value. As any two points are collinear, we have that all points in the plane have an equal value.

My solution :
 Let $f(M)$ be the value at point $M$.
 Let $I$ be the center of $ABC$. Let $D$ be the center of $ABI$, $E$ be the center of $ACI$, and $F$
be the center of $BIC$, and let $O$ be the center of $DEF$. Then it is easy to see that $f(I) = f(O).$
Assume we can construct $ABC$ such that $I \neq O$. Now using rotation and scaling, any two points
$P$ and $Q$ will be $I$ and $O$ hence have the same value.
Thus our proof is complete provide that we can construct $ABC$ such that $I \neq O$. This is not too
hard a geometry problem.

Let F(A) be the value assigned to point A. Let I be the incenter of triangle ABC. Let D, E, and F be the incenters of triangles AIB, BIC, and CIA respectively. One can easily verify that F(D)+F(E)+F(F) = F(A)+F(B)+F(C) = 3F(I) but also notice that if ABC is not equilateral then the incenter I' of DEF is distinct from I. Now if we consider any 2 points P and Q in the plane then we can show that F(P)=F(Q) because the configuration described above can be translated, rotated, and dilated to have P and Q correspond to I and I' respectively. Thus the only solution is F(X) = constant.

Take triangles $A_1B_1C_1$ and $A_2B_2C_2$. Let the two incenters be $I_1$ and $I_2$ respectively. By appropriate dilation and rotation of each triangle about its respective incenter, we can match up any two corresponding sides. It follows that for point $K_1\in\{A_1,B_1,C_1\}$, and any point $K_2\in\{A_2,B_2,C_2\}$, we have that $f(K_1)-f(K_2)=f(I_1)-f(I_2)$. It follows that $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$, $I_1$ and $I_2$ all have the same weight, hence we can see that all other points must also have this weight since we can perform the same scaling and rotating on any other pair of triangles with at least one vertice from the original set that we worked with. $\square$

[quote=Mewto55555]As our function is necessarily continuous (slight perturbations of vertices slightly perturb incenter location)[/quote]

It's been more than seven years but I'd like to point out this is clearly false: by this logic if "incenter" was replaced by "centroid" the function $f$ would still be continuous, but there are noncontinuous solutions in this case: any "affine-linear" function satisfies the conditions.

The shortest solution material (a.k.a. this can be reworded to three/four lines in a contest due to the sheer emptiness and simplicity of the argument) a hard P6 I've ever seen: probably the only problem which fits that category is in ($\bigstar$) of the Motivation. \\

$\color{green} \rule{25cm}{2pt}$ \\
$\color{green} \textbf{Combinatorial/Algebraic First Step.}$ Given two points $P$ and $P'$ of distance $1$. Then, if we have proven that
\[ f(P) = f(P'); \]
we are finished with the problem.

$\color{green} \rule{25cm}{0.4pt}$ \\
$\color{green} \textbf{Should be obvious, but this part is interesting nonetheless.}$ There are two ways to finish here.
[list = disc][*]1. We will prove that $f(P_1) = f(P_2)$ for any two points of the plane. Let the distance between those two points be $d$. Then, apply the same argument on a plane magnified with homothety/dilation of scale $d$. [*] 2. We'll first prove that $f(P_3) = f(P_4)$ for any two $P_3,P_4$ with distance at most $2$. To do so, form an isosceles with sides $(1,1,\text{distance of} \ P_3,P_4)$. To prove the general statement, let we wish to prove $f(P_5) = f(P_6).$ Then find a $P_7$ in the segment $\overline{P_5P_6}$ so that $d(P_5,P_7) = (\text{fractional part of their distance}) < 1$ and induct upwards to translate $P_7$ to $P_6$.  [/list]
Note that the first way puts into perspective that
\[\begin{tabular}{p{12cm}}
If we have enforced a way to force some two general points to be uniformly in order, then regardless of the size of the distance of any $P_1$ and $P_2$, the operation then can be magnified by $d$.
\end{tabular} \\ \]
Well, and the second way also puts it into perspective that
\[\begin{tabular}{p{12cm}}
No magnification is required: by translating a valid operation that forces uniform values on two points, we can actually match any two points to be equal!
\end{tabular} \\ \]
This ends this Section. $\blacksquare$ \\

$\color{red} \rule{25cm}{2pt}$ \\
$\color{red} \textbf{The versatility and wild nature of incenters.}$ This might be the most absurd-ly worded Claim I've made in my Solutions, but here goes:
\[\begin{tabular}{p{12cm}}
Given $P,P'$ with distance $1$. Draw the segment $\overline{PP'}$ and draw two perpendicular lines to that segment that passes through $P$ and $P'$. Call the strip bounded by those two lines to be the $\textit{parallel range}$ of $P$ and $P'$. Moreover, divide the $\textit{parallel range}$ equally, and call the region closer to $P$ (and $P'$, respectively) to be the $\textit{P-half-range}$ (and $\textit{P'-half-range}$, respectively.)  \\
\\
For any $Q$ which lies inside $\textit{P-half-range}$, we can prove that \\
\\
$\quad \quad \quad \quad \quad \quad \quad \quad \quad f(Q) - f(Q') = \dfrac{f(P)-f(P')}{3}.$  \\
\\
where $Q'$ is the reflection of $Q$ towards the perpendicular bisector of $PP'$.
\end{tabular} \\ \]
$\color{red} \rule{25cm}{0.4pt}$ \\
$\color{red} \textbf{Proof 2.}$ We will proof that we can find two points $A$ and $B$ so that
$Q$ is the incenter of $\triangle PAB$ and $Q'$ is the incenter of $\triangle P'AB$. \\

Let the point $\overline{PQ} \cap \overline{P'Q'} \cap \text{perp.bisector of} \ PP'$ be $X$. Then, draw a circle $\omega$ with center $X$ and radius $|XQ| = |XQ'|$. Pick two points $A$ and $B$ so that
\[ (PP'X) \cap \omega = \{A,B\} \]
Since $X \in (PAB), (P'AB)$ and $PX,P'X$ is the internal bisector of angles $\angle APB, AP'B$, this validates the reverse construction --- $Q,Q'$ is forced to be the respective incenters of the two triangle. $\blacksquare$ $\blacksquare$ $\blacksquare$ \\

$\color{blue} \rule{25cm}{2pt}$ \\
$\color{blue} \textbf{A Final Contradiction.}$ The reason I put three blacksquares above was that the conclusion should be in sight after the $\color{red} \textbf{versatility of incenters}$: pick $Q,Q'$ in the $\textit{parallel range}$ of $PP'$ --- and pick $RR'$ to be in the $\textit{parallel range}$ of $QQ'$. It should be immediately clear that $RR'$ must be in the parallel range of $PP'$. \\

Since $R,R'$ lies in the parallel range of $PP'$, then
\[ f(R)-f(R') = \dfrac{f(P)-f(P')}{3} \]
but since $R,R'$ lies in the parallel range of $QQ'$ and $QQ'$ lies in the parallel range of $PP'$,
\[ f(R)-f(R') = \dfrac{f(Q)-f(Q')}{3} = \dfrac{\frac{f(P)-f(P')}{3}}{3} \]
rendering $f(P) - f(P') = 0$ \\
We are truly done by here. $\color{blue} \blacksquare \ \blacksquare \ \blacksquare$

[hide = Motivation: Poncelet's porism and the incenter construction?]
$(\bigstar)$ The problem I was talking about is
[hide = Spoiler for some IZhO problem. If you're currently working on some IZhO problem, then bail!]
IZhO 2021 Problem 6: the smoke screen behind the intimidating and unfamiliar statement.
[/hide]
$\color{cyan} \rule{25cm}{2pt}$ \\
$\color{green} \text{Angular centers.}$ A first thought I had was to consider regular shapes, among which were equilateral triangles and squares. If $f(I) = \dfrac{f(A)+f(B)+f(C)}{3}$, I conjured that the equilateral triangle $ABC$ with midpoints $XYZ$ would do wonders against the problem. However, I couldn't find a consistent way to fixate equal values on even two points: $I(ABC) = I(XYZ)$ will imply
\[ f(A)+f(B)+f(C) = f(X)+f(Y)+f(Z) \]
and I can only extract similar information from considering the incenters of $AYZ,BXZ,CXY$ (those three incenters have the same incenter as $I(ABC)=I(XYZ)$.) \\

I tried transforming the problem statement to complex numbers and considered the unit circle: but the only statement I can get is that for a fixed incircle and circumcenter, each point $A$ in the circumcenter will be a part of exactly one triangle with incenter $I$. \\

This means I could only get so far as to get $\textbf{one equation}$ involving both $A$ and $I$ --- unless I involved some other point outside the circumcircle. Doing so would not only involve one, but $\textit{two}$; as if $\triangle ABC$ has incircle $I$ $A,B \in$ (some fixed circumcircle), then $C$ is also on this circumcircle by Poncelet's. \\

Squares also yielded disappointing results: I could categorize $I(ABC)$ up to $I(BCD)$, but the $22.5^{\circ}$ was bad already. However, while this may not explicitly cater to cartesian distance due to $I$ being an angular center of $ABC$, bisectors and perpendicular bisectors $\textbf{will}$ play a huge role in the Solution. \\

$\color{cyan} \rule{25cm}{2pt}$ \\
$\color{green} \text{Poncelet's vs Centroid Construction.}$ Since the statement present no changes if addition were replaced by multiplication, I thought that the values of the functions must be linear in some way. \\

With that being said, I remembered that if $I$ were to be replaced with $G$ as @above mentions, $f$ can be linear in some way. Namely, $f(A)$ can be of value $x(A)$, which is the abscissa of $A$. Rotated 90 degrees, $f(A)$ can be either $y(A)$ or doubly-linear: $x(A)+y(A)$. This only got worse if you count linearity of other rotations: instead of rotating by $0^{\circ}$ or $90^{\circ}$, any $\theta(A)$ = length of projection from $A$ to a line rotated by $\theta$ through the origin will do. \\

So, this indirectly invalidates the possible argument to force linearity with regularity: if $I =G$ then the statement is redundant in some way. So, I figured that it's more fruitful to consider individual points rather than whole structures; with the hopes that individual points will not lose the scarce degree of freedom available (that is, $3$ degrees of plotting $A,B,C$ before $I$ will be given to us.) \\

$\color{cyan} \rule{25cm}{2pt}$ \\
$\color{green} \text{Height and fraction-ing differences.}$ This final Section is mainly how I figured out to think more combinatorially rather than geometrically at first: suppose that I had a function (algebra comes out!) that isn't equal everywhere. \\

What could we do, then, with this information? Since $\mathbb{R}^2$ is a broad uncountable universal set, applying the same technique as if I were to prove $``\text{injectivity, or} \ f \ \text{to be constant}"$: I tried to see what would happen if $f(x) \ne f(y)$ for some two points (in contrast to injectivity, where we assumed that $f(x) = f(y)$ for two different nodes.) \\

Since those two points works best when the whole structure is symmetric in some way, I tried to apply the Poncelet construction in some way here:
[hide = My old proof for Poncelet's.]
\[\begin{tabular}{p{12cm}}
$\textbf{Poncelet's Porism, rephrased.}$ Let $\triangle ABC$ has incenter $I$, incircle $\omega$ and circumcenter $\Omega$. If $D,E$ is on $\Omega$ and $DE$ is tangent to $\omega$, then there exists exactly one $F$ so that $\triangle DEF$ has the same circumcircle and incircle with $\triangle ABC$. \\
\end{tabular} \\ \]
Proof: Construct $F$ on $\Omega$ so that $DF$ is tangent to $\omega$. Note that this location is unique, since the tangent from $D$ (other than the line $\overline{DE}$) is unique too. Now we have to prove that $\overline{EF}$ is tangent to $\omega$. \\

This seems provable, right? \\

Unfortunately, I was out of ideas on how to verify that $I$ is the right length from $D' = DI \cap (DEF)$. In other words, if I could prove that $I(ABC)$ is at distance $|D'E| = |D'F|$ from $D'$, I win. This may be doable by considering $I'$ incenter of $DEF$ and proving $I' = I$, but this kind of failed.

$\color{cyan} \rule{25cm}{0.4pt}$ \\
So, I discarded the $E$ given in the problem statement (in my mind, as well) and start from scratch. Let $DI \cap \Omega = D'$, and construct $E'$ and $F'$ so that the circle with center $D'$ radius $|D'I|$ intersect $\Omega$ at $E',F'$. First we will prove that $I$ is indeed the incenter of $\triangle DE'F'$ and $\omega$ is at the right radius to be the incircle of $\triangle DE'F'$. \\

The first one is doable by phantom points (this is now a double phantom argument.) Let $DE'F'$ be a triangle with incenter $I'$. Then, $I'$ lie on $\overline{DI}$ as the bisector of $\angle E'DF'$, and $I'$ satisfy the length relation
\[ |D'I'| = |D'E'| \ \text{or} \ |D'F'| = |D'I| \]
so $I' = I$. \\

Moreover, let $\omega'$ (this is now a triple phantom argument) has radius $r'$. Then, applying Euler's formula twice on $\triangle ABC$ and $\triangle DE'F'$, we get
\[ R^2 = OI^2-2Rr; \quad R^2 = OI'^2-2Rr' = OI^2 - 2Rr' \]
since the points $I$ and $I'$ are equal. This results in $r$ being equal to $r'$, so $\omega'$ is in fact at the right radius to be exactly equal to $\omega$. \\

We are done with this exposition as $DE'$, $DF'$ and $E'F'$ is automatically tangent to $\omega$. (This is in fact one of the strongest phantom point argument I have encountered)
[/hide]

Recalling that, I thought it'd be nice if both triangles share some points in common; what candidate should it be rather than the $D'$ of the Poncelet argument? So, if I wanted to construct $Q,Q'$ incenters of some $PAB$ and $P'AB$ --- $Q$ and $Q'$ should act nicely first: that could happen if they are reflection of themselves along the perpendicular bisector of $PP'$. \\

After doing some construction work (notice that $A,B$ is very particular: you can't just pick $AB$ parallel $PP'$ or $QQ'$. If you tried doing so, yes, $XP$ and $XP'$ will be bisectors of $\angle APB$ and $AP'B$, but $AQ$ and $BQ$ can make it so that $\angle PAQ \ne \angle QAB$. \\

Funnily enough, I noticed the Claim's conclusion right after I did the construction --- I didn't plan for the difference of $f(Q)$ and $f(Q')$ to be less than the difference of the initial points $P,P'$ beforehand. \\

Before I close this Motivation: ponder for yourself; $``\text{why does this fail if} \ I \ \text{were to be replaced with} \ G?"$
[hide = Hint]
It has to do with the title of this Section.
[hide = Give up?]
That is due to $Q,Q' = G \ (\text{centroid of} \ ABP \ \text{or} \ ABP')$ having a fixed height with respect to the perpendicular bisector of $PP'$; in other words, the locus of $G$ is not equal to the $\textit{whole parallel strip of} \ PP'.$
[/hide]
[/hide]

Tips: If you haven't drawn a diagram to comprehend all this, either you're very good in geo or you're reading this for fun (which is okay too.) Sarcasm aside, it's always good to try your hand and follow along with ample amounts of scratch paper to test out what I'm (or anyone when they share their Solutions) writing.
\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]
Final Comments: Take note that the green part $\color{green} \textbf{Combinatorial/Algebraic First Step}$ has no technical purpose towards the Solution: if I were to write this in a contest, it'd be something like
\[\begin{tabular}{p{15cm}}
If there exists two points $P,P'$ so that $f(P) = f(P')$, construct $Q,Q'$ with distance to perp.bisector of $PP'$ less than of $P$'s and construct $R,R'$ in the same way with respect to $QQ'$. \\

Note that by (some construction algorithm) we can find $A,B$ for $P,P',Q,Q'$ so that $PAB$ and $P'AB$ have incenters $Q,Q'$. So, for any pair of pair of points $(PP') (QQ')$ that satisfies the distance construction, \\
\\
$ \quad \quad \quad f(Q)-f(Q') = \dfrac{f(A)+f(B)+f(P)}{3} - \dfrac{f(A)+f(B)+f(P')}{3} = \dfrac{f(P)-f(P')}{3}$ \\
\\
which upon telescoping with $RR'$ will yield a contradiction. \\

Then, all points have the same number. We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$
\end{tabular} \\ \]
But when explaining why the possibility of letting $f(P) \ne f(P')$ in the first place lies in the fact that $\textbf{when proving a function is constant}$, an easy assumption to make especially when the identity given is Cauchy-like is to assume deviation, then contradict that deviation. \\

Once there is no deviation among a generally particular pair, this can be expanded to *magically* finish the problem.
[/hide]

Let $(O)$ be an arbitrary circle and let $\ell_1,\ell_2,\ell_3,\ell_4,\ell_5$ be arbitrary tangent lines to $(O)$ no two of which are parallel. Let the function mapping points to real numbers be $f$. Then we have
\[f(\ell_i\cap\ell_j)+f(\ell_i\cap\ell_k)+f(\ell_j\cap\ell_k)=c\]
for all distinct $i,j,k$ and some fixed $c$. Thus we have $10$ equations in $10$ variables. To solve these equations, note
\[f(\ell_i\cap\ell_j)=c-f(\ell_1\cap\ell_i)-f(\ell_1\cap\ell_j)\forall i,j\ne 1,i\ne j.\]
Let $f(\ell_1\cap\ell_i)=a_i$ so $c=c-a_i-a_j+c-a_j-a_k+c-a_k-a_i=3c-2[a_i+a_j+a_k]$. This implies $a_i+a_j+a_k$ is fixed at $c$, and so the desired follows: each of the $f(\ell_i\cap\ell_j)$ is equal to $c/3$. But this implies that any non-degenerate triangle has all its real numbers of vertices. To finish, note that for fixed $A$ and $B$, all points $C$ not on line $AB$ must have $f(A)=f(B)=f(C)$, so for particular $D\not\in\{A,B\}$ on line $AB$ and $C$ not on line $AB$, looking at $\triangle CDB$ finishes.

Edit: This is wrong due to config issues, I'll try to find a correct approach.

Edit: Here is a correct solution which turns out to basically be Evan Chen's.

Let $f$ denote the map from the plane $\mathbb R^2$ to $\mathbb R$.

Observation: For isosceles trapezoid $ABCD$ with $AB\parallel CD$ which is not a rectangle, $f(A)+f(C)=f(B)+f(D)$.

Proof: Let $BC\cap DA=X,AC\cap BD=Y$. By Pitot's Theorem, whichever one of $XCYD$ or $XAYB$ is concyclic has an incircle. Thus $f(X)+f(A)+f(C)=f(X)+f(B)+f(D)$, finishing.

Then considering regular pentagon $ABCDE$ finishes, as $f(A)+f(C)=f(B)+f(D)=f(C)+f(E)=f(D)+f(A)=f(E)+f(B)$ so $f(A)=f(E)=f(D)=f(C)=f(B)$.

[hide = Remarks]
I don't think this problem is especially hard: the idea of looking at triangles with a common incircle is fairly natural, although I did not realize at first that it was enough to look at isosceles trapezoids. However, it is tricky to approach for sure.
[/hide]

Solved with [b]nukelauncher[/b].

The problem succumbs to the following:

[color=red][b]Claim:[/b][/color]      Let \(ABC\) be a scalene triangle with incenter \(I\).     Let \(I_A\), \(I_B\), \(I_C\) be the incenters of      \(\triangle IBC\), \(\triangle ICA\), \(\triangle IAB\),     and let \(J\) be the incenter of \(\triangle I_AI_BI_C\).     Then \(I\ne J\).

[i]Proof.[/i]      Assume for contradiction \(I\) is the incenter of \(\triangle I_AI_BI_C\).     Let \(X=\overline{AI}\cap\overline{I_BI_C}\), \(Y=\overline{BI}\cap\overline{I_CI_A}\), \(Z=\overline{CI}\cap\overline{I_AI_B}\),     and let \((I_A)\), \((I_B)\), \((I_C)\) touch \(\overline{BC}\), \(\overline{CA}\), \(\overline{AB}\) at \(P\), \(Q\), \(R\).

[asy]             size(8cm); defaultpen(fontsize(10pt));             pair A,B,C,I,IA,IB,IC,X,Y,Z,P,Q,R;             A=dir(130);             B=dir(200);             C=dir(340);             I=incenter(A,B,C);             IA=incenter(I,B,C);             IB=incenter(I,C,A);             IC=incenter(I,A,B);             X=extension(A,I,IB,IC);             Y=extension(B,I,IC,IA);             Z=extension(C,I,IA,IB);             P=foot(IA,B,C);             Q=foot(IB,C,A);             R=foot(IC,A,B);

draw(incircle(A,B,C),linewidth(.4));             draw(A--I,gray);             draw(B--I,gray);             draw(C--I,gray);             draw(IA--IB--IC--cycle,linewidth(.5));             draw(A--B--C--cycle,linewidth(.5));             draw(incircle(I,B,C),linewidth(.3));             draw(incircle(I,C,A),linewidth(.3));             draw(incircle(I,A,B),linewidth(.3));

dot("\(A\)",A,N);             dot("\(B\)",B,SW);             dot("\(C\)",C,SE);             dot("\(I\)",I,NE);             dot("\(I_A\)",IA,S);             dot("\(I_B\)",IB,NE);             dot("\(I_C\)",IC,W);             dot("\(X\)",X,dir(60));             dot("\(Y\)",Y,S);             dot("\(Z\)",Z,dir(290));             dot("\(P\)",P,S);             dot("\(Q\)",Q,NE);             dot("\(R\)",R,NW);         [/asy]

Then \(\angle YII_A=\angle ZII_A\) and \(\angle YI_AI=\angle ZI_AI\),     so \(\angle IYI_A=\angle IZI_A=:\theta\),     Then \(\angle IXI_B=\angle IZI_B=180^\circ-\theta\)     and \(\angle IXI_C=180^\circ-\theta\),     so \(\theta=90^\circ\).

This implies \((I_B)\) and \((I_C)\) are tangent at \(X\),     so \(A\) lies on their radical axis, i.e.\ \(AQ=AR\).     Analgoously \(BR=BP\) and \(CP=CQ\),     so \(P\), \(Q\), \(R\) are the points where the incircle     touches the sides of the triangle.

Hence \(\overline{II_A}\perp\overline{BC}\), implying \(IB=IC\),     and so \(\angle B=\angle C\), contradiction. \(\blacksquare\)

Now in the above configuration, \begin{align*}     f(J)&=\tfrac13\big(f(I_A)+f(I_B)+f(I_C)\big)\\     &=\tfrac19f(I)+\tfrac29\big(f(A)+f(B)+f(C)\big)     =f(I). \end{align*} For any two points \(U\) and \(V\), it is possible to rotate and scale the above diagram so that \(I\mapsto U\) and \(J\mapsto V\). Hence \(f(U)=f(V)\), so \(f\) is constant.

Clean solution but hard to motivate.

For any $P \in \mathbb{R}^2$ define $f(P)$ to be the number it was assigned. We would like to show $f(X)=f(Y)$ for all $X,Y \in \mathbb{R}^2$. Indeed, consider points $A, B, C, D$ where $ABCDXY$ is a regular hexagon. Triangles $ACX$ and $BDY$ share an incenter so \[ f(X)+f(A)+f(C) = f(Y)+f(B)+f(D) \] So, it is sufficient to show that $f(A)+f(C) = f(B)+f(D)$. Indeed, if we let $AB$ and $CD$ intersect at $T$ then we notice that triangles $ACT$ and $DBT$ share an incenter. So, \[ f(A) + f(C) + f(T) = f(B) + f(D) + f(T) \Leftrightarrow f(A) + f(C) = f(B) + f(D) \] as desired.

This might just be the most cursed solution to grace the entire AoPS forum to date.

Let $f(P)$ be the number assigned to a point $P$.

Consider the following configuration. Let $A,B,C$ be (fixed) points with $C$ on segment $\overline{AB}$. and $\ell$ be a line perpendicular to and passing through segment $\overline{AC}$. Let $I$ be a variable point on $\ell$. Let $D$ be the point such that $I$ is the incenter of $\triangle ABD$, and let $E$ be the point such that $I$ is the incenter of $\triangle CDE$. The latter two points don't always exist, but they will when $I$ lies within a certain distance of $\overline{AB}$. Observe that we have
$$f(A)+f(B)+f(D)=3f(I)=f(C)+f(D)+f(E) \implies f(E)=f(A)+f(B)-f(C),$$
which is fixed.

I claim that $E$ varies along some projective algebraic plane curve as $I$ varies. That is, there exist real rational functions $f,g$ such that $E$ (on some subset of where it exists) can be parameterized as $(f(t),g(t))$ where $t$ varies in some interval. This is due to the theory of moving points: if we move $I$ linearly, we can obtain $\overline{AD}$ and $\overline{BD}$ by reflecting $\overline{AB}$ over $\overline{AI}$ and $\overline{BI}$, so $D$ has a well-defined degree. Then we can obtain $\overline{DE}$ and $\overline{CE}$ by reflecting $\overline{CD}$ over $\overline{DI}$ and $\overline{CI}$, so $E$ has a well-defined degree too. $E$ thus varies along some projective algebraic plane curve as desired.

By implicit differentiation, the value of the second derivative $\tfrac{d^2y}{dx^2}$ at $(f(t),g(t))$ is now a real rational function in $t$, so there exist finitely many points where its sign changes. Take some finite contiguous section $\mathcal{S}$ of the locus of $E$ where the curve is always concave or always convex (WLOG concave) and where no two points on this curve have the same $y$-coordinate. Fix a tiny open disc $\mathcal{D}$ lying close to but entirely above $\mathcal{S}$ (so there exists a point on $\mathcal{S}$ lying directly below any point in $\mathcal{D}$, and $\mathcal{D}$ does not intersect $\mathcal{S}$) and let $Q$ be a point directly above the center of $\mathcal{D}$ which lies very far above $Q$.

Let $P$ be any point in $\mathcal{D}$: I claim that there exist points $X,Y \in \mathcal{S}$ such that $P$ is the incenter of $\triangle QXY$. To do this, we vary a circle $\Omega$ of radius $r$ centered at $B$ and let $X',Y'$ be the points where the tangents from $Q$ to $\Omega$ first intersect $\mathcal{S}$, which exist by our definition of $\mathcal{D}$. Consider the expression $d(P,\overline{X'Y'})-r$, where $d$ denotes signed distance and is positive if $P$ lies above line $\overline{X'Y'}$ and negative otherwise ($\overline{X'Y'}$ being vertical never happens). If this expression is positive, then $\Omega$ lies entirely in the interior of $\triangle QX'Y'$, and if it's negative, then $\Omega$ lies partially outside $\triangle QX'Y'$. The expression is clearly continuous in $r$ and positive as $r \to 0^+$. Moreover, when $r$ is chosen minimally so that $\Omega$ intersects $\mathcal{S}$ (so they're tangent), because $\mathcal{S}$ is convex, the intersection point lies nonstrictly $\overline{X'Y'}$ and thus outside $\triangle QX'Y'$, so the expression is nonpositive. Thus this expression is $0$ for some value of $r$ by the intermediate value theorem, at which point $\Omega$ will be the incenter of $\triangle QX'Y'$ as desired (by choosing $(X,Y)=(X',Y')$ for this value of $r$).

Then, since $f(\mathcal{S})$ is fixed and $f(Q)$ is fixed, $f(P)$ should also be fixed, so we find that $f$ is constant on $\mathcal{B}$. Now, consider two points $A,B \in \mathcal{B}$ such that the circle with diameter $\overline{AB}$ is contained entirely in $\mathcal{B}$. Varying a point $I$ along the perpendicular bisector of $\overline{AB}$ lying inside this circle with diameter $\overline{AB}$ and letting $C$ be the point such that $\triangle ABC$ has incircle $I$, we find that the locus of $C$ is the entire perpendicular bisector minus its intersection with $\overline{AB}$, so $f$ is also constant (and equal to its constant value on $\mathcal{B}$) along the entire perpendicular bisector. Now rotating $\overline{AB}$ about its midpoint and continually applying this fact implies $f$ is constant everywhere, as desired. $\blacksquare$

Solved with [b]davidkong, TheBeast5520, RubiksCube3.1415[/b].

Let $p$ be the real number assigned to $P$. The key idea is the following picture:

[center][img]https://cdn.discordapp.com/attachments/1211416851301797979/1219047430339825778/image.png?ex=6609e164&is=65f76c64&hm=99e50d2d44f0ed03d26c64761dfc51ca53d7bcd698b55d13811b58e3a360e907&[/img][/center]

We will be using the triangles $\triangle ABC, \triangle DEF, \triangle ABG, \triangle DFG$. Clearly, $\triangle ABC$ and $\triangle DEF$ share an incenter, so $a+b+c=d+e+f$. Notice that $AXDG$ is a kite, so it has an incircle. This incircle is the incircle of $\triangle ABG$ and $\triangle DFG$, so they also share an incenter: $a+b+g=d+f+g\implies a+b=d+f$. Therefore, $c=e$.

Rotating and translating the picture to replace $C, E$ with any points $P, Q$, we get the desired result.


vector
analytic geometry
linear algebra
matrix

If $ S$ is an orthonormal basis for an n-dimensional inner product space, and if
$ (u)_S \equal{} (u_1,u_2,...,u_n)$ and $ (v)_S \equal{} (v_1,v_2,...,v_n)$ ($ (u)_S,(v)_S$ are coordinates vectors), prove that

$ \langle u,v \rangle \equal{} u_1v_1 \plus{} u_2v_2 \plus{} ... \plus{} u_nv_n$

Let $ \{e_1,e_2,...,e_n\} \in S$  is an orthonormal basis for an n-dimensional inner product space .Here $ e_1 \equal{} \begin{bmatrix}1 & 0 & 0... & 0\end{bmatrix},e_2 \equal{} \begin{bmatrix}0 & 1 & 0... & 0\end{bmatrix},...,e_n \equal{} \begin{bmatrix}0 & 0 & 0... & 1\end{bmatrix}$
 Then $ (u)_{S} \equal{} (u_{1},u_{2},...,u_{n}) \equal{} \sum_{k \equal{} 1}^{n}{u_ke_k}$,$ (v)_{S} \equal{} (v_{1},v_{2},...,v_{n}) \equal{} \sum_{i \equal{} 1}^{n}{v_ie_i}$
 then $ < u,v > \equal{} < \sum_{k \equal{} 1}^{n}{u_ke_k},\sum_{i \equal{} 1}^{n}{v_ie_i} > \equal{} \sum_{k \equal{} 1}^{n}\sum_{i \equal{} 1}^{n}u_kv_i < e_k,e_i > \equal{} \sum_{i \equal{} 1}^{n}{u_iv_i}$ by $ < e_i,e_k > \equal{} 0$ if $ i \neq k$ and $ < e_i,e_k > \equal{} 1$ if $ i \equal{} k$ ...


this is a wrong statement= =' :( [/code]

The following inequality similar to your inequality, hungkhtn:
let $ a,$ $ b,$ $ c$ and $ d$ are non-negative numbers such that $ a^{3}+b^{3}+c^{3}+d^{3}=4.$
Prove that $ a^{2}+b^{2}+c^{2}+d^{2}\geq a^{2}b^{2}c^{2}+b^{2}c^{2}d^{2}+c^{2}d^{2}a^{2}+d^{2}a^{2}b^{2},$
but proof for this last one it's very nice.

[quote="arqady"]The following inequality similar to your inequality, hungkhtn:
let $ a,$ $ b,$ $ c$ and $ d$ are non-negative numbers such that $ a^{3}+b^{3}+c^{3}+d^{3}=4.$
Prove that $ a^{2}+b^{2}+c^{2}+d^{2}+c^{2}\geq a^{2}b^{2}c^{2}+b^{2}c^{2}d^{2}+c^{2}d^{2}a^{2}+d^{2}a^{2}b^{2},$
but proof for this last one it's very nice.[/quote]

I agree with you. This kind of inequality is very nice. I will try to solve it  :) But I think it should be

$ a^{2}+b^{2}+c^{2}+d^{2}\geq a^{2}b^{2}c^{2}+b^{2}c^{2}d^{2}+c^{2}d^{2}a^{2}+d^{2}a^{2}b^{2},$

Do you solve the following similar problem, Arqady

http://www.mathlinks.ro/Forum/viewtopic.php?t=159776

As I said, it is dedicated to you, but maybe you forget it.

hungkhtn and argady, Can you post yours?

[quote="hungkhtn"]We have the following very nice and hard result

If $ a^{2}+b^{2}+c^{2}+d^{2}=1$ then
\[ ab+bc+cd+da+ac+bd\ge 24(a^{2}b^{2}c^{2}+b^{2}c^{2}d^{2}+c^{2}d^{2}a^{2}+d^{2}a^{2}b^{2}). \]
[/quote]
Actually, the following stronger inequality holds:
\[ (\sum_{sym}{ab})(\sum_{sym}{a^{2}b^{2}}) \ge 9\sum_{sym}{a^{2}b^{2}c^{2}}. \]
Equivalently, if $ a,b,c,d$ are positive numbers such that $ a^{2}+b^{2}+c^{2}+d^{2}=4$, then
\[ (\sum_{sym}\frac 1{ab})(\sum_{sym}{a^{2}b^{2}}) \ge 36. \]


Is earth science (or some similar class) usually required for applying to college?

I noticed that plenty of colleges expect you to get science credits to apply- i might be imagining it, but do some want you to take the intro HS classes in bio, chem, and physics?

This is completely depend on the local resources available to you at your school.  Colleges take into account the courses available and only want you to have taken a reasonably rigorous courseload given your choices.  For example, it is unreasonable to expect a student to have taken multiple AP exams at a school that only offers one, but it is more reasonable to expect the same of a student whose school offers nearly every test.

Edit:  You seem to be implying you have a choice between earth science and some other class.  It might be useful for you to elaborate on what choice you're facing.

Earth science (not related to AP) is a requirement at my school. A significant number of people oppose this requirement for various reasons such as they want to take more post AP classes.

Is earth science a requirement for college? See, if it is required then there is no point in opposing the requirement.

My assumption is that some colleges require a year of high school chemistry(not AP), a year of high school biology (not AP), and a year of high school physics (not AP). I was wondering if earth science is often required.

It depends on the college. A college that draws from around the country will likely be flexible, while state universities may be more specific.
An [url=http://www.universityofcalifornia.edu/admissions/undergrad_adm/paths_to_adm/freshman/subject_reqs.html]example[/url]. The UC system in California requires two laboratory science courses, including at least two of biology/chemistry/physics.

I don't know of any colleges that require earth science. People from my high school (in California) go to college all over the country, and we don't even have earth science as an option. I think earth science generally isn't a lab science, at least it isn't at my cousin's school (he's in a situation similar to yours), so it doesn't even count towards the science requirement for the University of California, or other colleges that require lab science. But like jmerry said, your state college might be different.

The general idea at most selective colleges is just to take a science course each year you are in high school (ninth through twelfth grade). What specific courses you take is much less important.


integration
geometry
3D geometry
sphere
calculus

Consider a thick metal sphere. At it centre, we have placed a point charge of +Q.
(1) Find potential energy of system

Now the charge is drained through a small aperture in the sphere.
(2) Find the work done by external forces in draining the charge.

[hide="Doubts"]
(1) Since the field is radially uniform inside the sphere.
$ U_1 \equal{} \int \frac 12 \epsilon _0 E^2dV \rightarrow \infty$
And filed outside is some finite quantity $ U_2$

But how about this approach?
PE of a 2 particle system
$ dU \equal{} \frac {1}{4\pi \epsilon _0} ( \plus{} Q)( \minus{} dQ)\frac 1R$, where -dQ is differential induced charge and R is inradius. This gives a finite anser.

(2) Solution given which i did not understand:
 Say the point charge was a sphere of radius r. After draining, the point charge will make a sphere of radius r at infinity.
So after draining the PE of the system will be $ \frac {Q^2}{4\pi\epsilon_0 r}$ The change in PE gives the work done by external forces.[/hide]

The energy inside the ball is certainly finite (assuming the ball is finite).  There is a formula for the energy of a system, as you have.  Once you plug it in, integrate, and use your limits of integration, you will get a finite answer.  Unfortunately, I don't have any physics references in front of me (such as old homeworks, tests, and texts) so I can't give you an exact answer but as long as you get your formulas straight, you'll get a sensible answer.

Your first question is very interesting.  In fact, I'm not sure about it, either.

The way I see it, you end up with a point charge of +Q in the center, an infinitesimally thin spherical shell of charge -Q enclosing it, and another shell of charge +Q enclosing them both.  Then, you can add up the potentials from five interactions:

1. The potential energy associated with the large shell configuration
2. The potential energy associated with the small shell configuration
3. The potential energy of point charge/large shell interaction
4. point charge/small shell interaction
5. small shell/large shell interaction

For number 5, one simply multiplies the charge of the small shell by the constant electric potential in the interior of the large shell.

But I don't know if this is right.

Also, your first doubt raises an interesting question that I have: the potential energy of an isolated point charge is obviously zero.  However, when one integrates the energy density of the electric field due to that charge over the infinite volume that it occupies, one gets an infinite result.  What does that mean?

[quote="genericme"]the potential energy of an isolated point charge is obviously zero.[/quote]

why?

I would say it's because the isolated point charge isn't in any potential field, and since it isn't affected by its own electric field (which is associated with its own potential field), and the potential energy of a charge is, by definition, the product of the value of the charge and the electric potential where it is situated, the potential energy is zero.

Is my reasoning incorrect?

but the potential is $ \frac{1}{r}$ ! surely this exists and is nonzero at the origin ?

okay so your answer is essentially correct to the extent that you've learned it. this is how classical electrodynamics treats point charges and it works pretty well since here the key point is "it isn't affected by its own electric field". Basically what's going on is that the potential is technically $ \infty$, but it doesn't actually affect anything, so we're just like eh screw it let's just call it $ 0$; things actually do work out this way. Now, it doesn't work perfectly, since there are embarrassing little discrepancies when calculating the energy of a field -- if we assume a continuous charge distribution, it's $ \int \frac{1}{2} \varepsilon E^2$ which is clearly a positive quantity whereas if we approximate by a point charge distribution, there the energy can be either positive or negative; the reason the former is positive is precisely this "infinite" potential energy summed over each of the infinitesimal continuum charges. If we ignore these little discrepancies, though, the theory works pretty okay. Where we do run into trouble is in the quantum theory, because there we have loop renormalization where particles' electromagnetic fields can in fact loop around and affect themselves, and in this case we do get random infinities popping up and destroying all our answers. I don't think I'll go further into this because I'm hardly an expert on quantum field theories at this point, but hopefully what I've said is roughly correct up to this point at least.

I'm afraid I can't give you a specific reference on this one -- I know I've read some sources that have talked about all this in at least some detail. I feel like Griffiths' Electrodynamics book might address the point in the context of the classical theory and I know I saw some remarks about the quantum theory somewhere, perhaps either in Griffiths' Particle Physics book or Feynman's popular science book "QED", but I can't remember anything more specific about where it was.

This is a problem from irodov.
Answer for first part is indeterminate. (infinity)

@genericme
Potential energy of a point charge(its self energy) is not defined. The point charge in question is not occupying an infinite volume. It is constrained in the field which extends from itself to the inner surface of the sphere.

I guess what went wrong in the second approach (2-particle system and all) was that you cant consider such a system(2-particle). We have to consider self energy of the sphere to calculate its potential energy. One way of saying why i went wrong maybe would be - why didnt i consider the 2 particle system -dQ and -dQ separated by dr?
We calculate the self energy of a charged sphere by considering work done by the external agent in transferring charges to it as a whole. The formula for PE for 2 particles, is well, valid only for particles.

IS there yet a solution to the problem of infinite electrostatic energy in the field of a point charge which exist in both the classical & the quantum form?


geometry
area of a triangle
Heron\u0027s formula

Suppose $L_{1}$, $L_{2}$ and $L_{3}$ are three lines on
 the co-ordinate plane whose equations are known , and
 which are known to form a triangle of finite area. the area
 of the triangle is not known ( although it can be calculated).
 we can find the acute angle between any two lines
 using the formula $tan \theta$ =modulus of $\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}$

  Now, suppose we pick two lines say $L_{1}$ and $L_{2}$.
 We can determine the obtuse angle between them and the
 acute angle between them . How do we determine which of the
 two angles belongs to the triangle? without using the other
 roundabout procedure of determining the vertices??

Actually, it depends on the third line. Without determining the vertices, you are practically hopeless.
I would do this:
(1) Determine the 3 vertices
(2) Find the lengths of the sides of the triangle (Distance Formula)
(3) Find the area by using Heron's Formula

Example.
The 3 lines are y=x, y=-2x+2, y= 2x.
Lines y=x and y=2x intersect at (0,0).
Lines y=x and y=-2x+2 intersect at (2/3, 2/3).
Lines y=-2x+2 and y=2x intersect at (1/2, 1).

The distance between (0,0) and (2/3, 2/3) is sqrt(8)/3
The distance between (0,0) and (1/2,1) is sqrt(5)/2
The distance between (1/2,1) and (2/3,2/3) is sqrt(5)/36

Applying heron's formula gets you the answer.

vishalarul i don't think you understood the problem.
 there is a one step method for the area of a triangle
 given the equation of the sides: $\frac{(\Delta)^{2}}{C_{1}C_{2}C_{3}}$. where $\Delta$ is the coefficient determinant,
 and $C_{i}$ is the co-factor of $c_{i}$ in the line $a_{i}x+b_{i}y+c_{i}= 0$. However the question is quite something else. please read it again.


Greetings,
actually I'm pretty new to this forum, but I'm glad that I've finally found some place to ask LaTeX questions, as the documentation on the internet isn't very profound. I can imagine my question isn't new, but I've searched the forum and didn't find a thread that answers my question, so:

I have read the "Latex Graphics Companion", the "Latex companion", and Lamport's "Latex: A document preparation system"; it didn't help me solve my problem:

[b]How can I simply and directly include a .TIFF, a .BMP, or a .GIF file (by means of \includegraphics) into my latex document[/b] WITHOUT having to convert it into a .EPS-file first (which eats disk-space like candy)?

Specifically, I want to do so using the \includegraphics command of the graphicx package on a windows system (TeXLive! from the new "Latex companion" installed), using pdflatex to produce the output.
It works with JPG and PNG files (which is already great, but not enough for my needs...)

I didn't really understand a thing about that graphics driver history, which is very blurry described in the above mentioned books (at least for my understanding).
- What driver does my system use? How can I find out? (I load the graphicx package withouth options)
- or rather: what driver SHOULD I use to be able to include basic common graphics format (such as TIFF)
- Why does pdflatex process JPG and PNG files, whereas the standard latex-command (producing a DVI) merely handles EPS files?

If anybody can give me some definitive information, I'd be very grateful.

.PNG format is much better than .BMP and .GIF formats. I can't detail all the reasons, but it is a much better supported format in general (browser-wise speaking). Also almost all editors (including MSPaint) have the option of saving the edited file as a .PNG file format, so you can get a mass converter (there are numerous available on the net) or use a mass command with imagemagick to convert these and your problem is solved.

I remember I tried for a long time to put .GIF files inside, untill I accidentally discovered the .PNG trick and that was it. Also I use .PDF output only, so I'm fine with .DVI not compiling anything else (really?) than .EPS.

Of course this doesn't really answer any of your questions, but I just wanted to state my opinion in this matter :)

My understanding (which is far from definitive) is that you can't import .gif's, period.  Believe me, I've tried.  I've now come around to Valentin's way of thinking and switched to .png's.

To import .tif and .bmp, I think you have to stick in a \DeclareGraphicsExtension and/or a \DeclareGraphicsRule command somewhere.

Also make sure that you have the pdftex option set for graphicx:
\usepackage[pdftex]{graphicx}
or
\PassOptionsToPackage{pdftex}{graphicx}
\RequirePackage{graphicx}
if you're using your own .sty file.

Check out Section 10.2.5 of the new (2nd) edition of The LaTeX Companion, or look for the graphicx documentation on http://www.ctan.org.

Thanks, I slowly begin to understand the way this works.
The question I should ask is, [b]which is the graphics driver (or is there a graphics driver) that supports TIFF and BMP graphics inclusion?[/b]

I agree, GIF isn't important, but I get a lot of graphs in TIF and BMP format, and I don't want to convert them each time for a latex-document.

I've looked into the pdftex.def file, and it says there that .tif is supported, however it doesn't work. And I'm not enough of a TeX-crack to edit the driver file myself.
\DeclareGraphicsRule didn't help by the way.

How do you people handle it?
Which graphics format do you use?

I still think it's best you convert all your graphics to .PNG. Saves you a ton of time & trouble :) I use only .EPS and .PNG ...

Actually, I just looked at the pdfTeX documentation, and it says:
[quote]pdfTeX supports inclusion of pictures in png, jpeg, and pdf format.[/quote]
So by implication, I guess it doesn't support tiff and bmp.

Okay, thanks.
Guess I'll stick with PNG and JPG, actually that's quite fine.

Two more things:
- Concerning the annoying BoundingBox: If I use [color=blue]\includegraphics[scale=1]{image.jpg/png}[/color] and compile it to a DVI, it asks for a bounding box. If I create a PDF (using the pdftex driver), it does not.
Is there a driver for DVI output that can determine the size of JPG and PNG on its own? Would make it easy to switch between output formats (PDF/DVI) if I merely had to change the driver for the graphicx package.

- How do you correctly use the [color=blue]\graphicspath{{/somedirectory}}[/color] command in Windows? My images are, say, in a subfolder ./images/ relative to my .tex file.
If I specify something like {{./images/}} as the argument, it's ignored (well I guess because of the forward slashes?), and if I specify {{.\images\}} or an absolute path with backslashes, it interpretes it as commands.

Thanks a lot for your help.

Are you using graphicx with the dvips driver option?  In that case, I don't know why it wouldn't be working.

About the other case: I think the problem is that Windows doesn't recognize the dot (.) in pathnames.  Becuase I always do things like
\includegraphics{Images/diagram.png}
and it works fine in Windows.

hmm.. i'm new to Latex and it seems that png is the only one that works

i have a related question: what program should i use for drawings (at geometry)? paint is just not good enough :P

It depends what driver you're using.  pdflatex should be able to import PNG, JPEG, and PDF (and maybe TIFF too).

For our books, we use Geometer's Sketchpad to construct precise geometry diagrams, and the Macromedia Studio MX family of applications (Flash, Fireworks, and FreeHand) to prepare diagrams.  We either export diagrams as .png files, or use a direct-to-PDF print utility (such as CutePDF) to create PDF files.

I've recently been experimenting with MetaGraf (which is a free Java app for LaTeX and MetaPost).  I still haven't quite figured out how to get it to do all of the things that Geometer's Sketchpad does, though.

the ones you mentioned are not free :(

when i try to install the only free one (metagraf) my comp hangs

[quote="perfect_radio"]the ones you mentioned are not free[/quote]

If you want something free, some people like [url=http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/]C.a.R.[/url], and some people like [url=http://www.mit.edu/~ibaran/kseg.html]KSEG[/url].

Alternatively you can try doing diagrams using [url=http://www.tug.org/metapost.html]MetaPost[/url].


I just wanted to mention a solution error in the Introduction to Counting and Probability book.

On problem 3.2.3i, the solution writer did the problem correctly, but made an aritematic error. [quote]... for an answer of $\frac{5!}{2!*2!}=$[b]15[/b][/quote]

I'm only mentioning it. :?

I think so You're quite right $\frac{5!}{2!.2!}=30$ ;)

Math is my life :!:

[quote="TRAN THAI HUNG"]

Math is my life :!:

[/quote]

It is almost mine. What as coincidance.

[quote="TRAN THAI HUNG"]Math is my life :!: [/quote]
why don't you use that as your signature? it'll save you time. :D


AMC USA(J)MO USAMO AIME email geometry MATHCOUNTS

:D Congratulations to all who made MOP, discuss (NOSPAM) what you thought of the cutoffs.

Maybe we should try to compile a list of BA MOP and MOP lites?
Congratulations to:
MOP:
Albert Gu
Timothy Chu
Probably O'Dorney?

MOP lite:
Patrick Yang
Ramya Rangan
Albert Wu

(still compiling)

[quote="Ubemaya"]1 off from HM cutoff, which kinda sucks since my #2 definitely deserved more than 2 points.[/quote]
dude no you didnt if i got nothing at all for my 2

<- the proud owner of the ONLY solution in USAMo that proves the even case for number 2 without the odd case. I guess i got dinged off the cutoff for that.

Now i have to eat a hat >.>
Bowei can I borrow yours?

Hmm blue cutoff was a lot lower than last year's cutoff?

[quote="CircleSquared"]What is MOP lite?

I think I am the only MOPper who solved #6 and not #4[/quote]
Nope, John Boyle did 2, 6, and nothing else. Well whoops I missed that "MOPper" part, but if you say MOPper then I guess John was a previous MOPper, and if you don't count red then that part gets invalidated.

MOP lite is one among many synonyms for red MOP.

Also, Scrambled, what the heck, bowei's hat is too awesome for eating. Nice HM though...

...I don't believe it, I still made the red/CGMO cutoff. I. Don't. Believe. It. Not that it makes a difference to anything other than my pride...

[quote="The QuattoMaster 6000"][quote="serialk11r"]
Congratulations to:
MOP:
[b]Quatto?[/b]
Ubemaya (not fat, you are definitely less fat than me)
Timothy Chu
Probably O'Dorney?
[/quote]
not quite, I seemed to have been majorly docked on one (or more) of my solutions; I got a 14  :([/quote]

Uh oh.  That's not good.  :(

You deserve a higher score than that (from previous experiences with your beastliness).

[quote="Ubemaya"]1 off from HM cutoff, which kinda sucks since my #2 definitely deserved more than 2 points.[/quote]
If you proved it without the "distinct" part, the problem is much much easier.

[quote="Ubemaya"]1 off from HM cutoff, which kinda sucks since my #2 definitely deserved more than 2 points.[/quote]

Honestly, Albert, shut up again (not that you did last time, the 'again' implies me saying it again). Who the hell cares if you're HM or not? It's the MOP experience.

And dang, Amol, as I told you before, of all people you deserve to be at MOP. I guess it's not a fair world.

[quote="Ubemaya"]No, they are equivalent.

In fact, I just asked a grader and he said a full solution for the alternate version of the problem would have gotten 7 points. So I guess I either had a hole or wrote it up badly (most likely the latter, since my writeup was not that great).

Also, dang looks like I tied O'dorney on both AIME and USAMO... (I'm not positive of what he got but he's not an HM so <= 22, so most likely 22).

Also @ Carl, yeah I don't care that much about HM anymore. I still feel like HM is better than Blue MOP because some people don't go, etc. etc., and it's not really defined. And your name doesn't go anywhere :(.

From what I heard from other people, it seems a lot of people got docked on #2.[/quote]
Isn't the quattomaster odorney? Or am I highly confused. Also I was fairly sure that the case without distincts is much easier since i interpreted it that way on the test initially, but that might be because I believed in the bogus I produced to take care of the odd cases.

[quote="not_trig"][quote="bookaholic"][quote="CircleSquared"]I remember Zuming saying that the Girls Math Olympiad was the Special Olympics of math.

I don't remember when this was, though.[/quote]
...Wait, what?!

Okay, I can't really take offense. Still, I suggest thinking through such remarks more carefully before repeating them in the future. Regardless of truth value, you don't want to get into an argument over this kind of thing.[/quote]

yeah. touchy subject much? but seriously I think it's kind of patronizing :/[/quote]
The Special Olympics is also a perfectly legitimate competition in its own right.... (This isn't an intent to justify the OPer though)

I want to get a sex change so I can get a shiny gold medal.

[quote="Scrambled"][/quote][quote="not_trig"][quote="bookaholic"][quote="CircleSquared"]I remember Zuming saying that the Girls Math Olympiad was the Special Olympics of math.

I don't remember when this was, though.[/quote]
...Wait, what?!

Okay, I can't really take offense. Still, I suggest thinking through such remarks more carefully before repeating them in the future. Regardless of truth value, you don't want to get into an argument over this kind of thing.[/quote]

yeah. touchy subject much? but seriously I think it's kind of patronizing :/[/quote][quote="Scrambled"]
The Special Olympics is also a perfectly legitimate competition in its own right.... (This isn't an intent to justify the OPer though)[/quote]

I know... actually I personally disagree with this too.  Just pointing out something that might seem out of the ordinary.

I'm not trying to be condescending, sorry if it came off that way.

[quote="not_trig"][quote="Scrambled"][quote="CircleSquared"]Is it just me or did CA dominate Red?[/quote]
they seriously have fantasy speculation for.... math?

Curling I could imagine. But MATH?

how is math even an activity

no flaming my heresy please[/quote]

we should totally have a fantasy MOP league

we take the USAMO qualifiers, assign them their point values from last year (0 if not qualified before), have a cap of 50 points for 5 people, and see how well they do this year

yep[/quote]

yeah we should do this

also i would just like to point out that parody is homage

Man that is such a good idea. Too bad it would be totally unpredictable :O

[quote="not_trig"][quote="Scrambled"][quote="CircleSquared"]Is it just me or did CA dominate Red?[/quote]
they seriously have fantasy speculation for.... math?

Curling I could imagine. But MATH?

how is math even an activity

no flaming my heresy please[/quote]

we should totally have a fantasy MOP league

we take the USAMO qualifiers, assign them their point values from last year (0 if not qualified before), have a cap of 50 points for 5 people, and see how well they do this year

yep[/quote]

This.
Is.
Pure.
Gold.

dangit that was facetious

now I have created a new and improved way for californians to waste time

[quote="Brut3Forc3"]Totally want to join the fantasy league xD

Can anyone list the CGMO team this year, or at least people from northern california? I have a stalker sheet of the MOP candidates and I'm trying to find out who made what color... in particular, I have 2 girls unaccounted for because I can't guess genders from some of these asian names...[/quote]

what asian names

i might be able to identify them

Hm, I have also figured out the list of blue moppers (and, of course, black moppers) due to boredom.

Yay never mind, David helped me figure out the rest... Now my stalker chart is complete. :)

[quote="pythag011"][quote="CircleSquared"]My school has yer class of 2012 right here with a full 1/7 of all the Red qualifiers.

Red qualifier high score of 15 was Dai Yang, I think (not from my school).  Maybe all people with the last name Yang are total beasts.

Darn, I wish my last name was Yang.

On another note, did any freshmen make blue or black?[/quote]

Mieh, Dai Yang is from SoCal.

Do 7th graders count as freshman?[/quote]
dai yang is good at chess


geometry
rectangle
calculus
integration
national olympiad

1.Find the maximum value for the area of a heptagon with all vertices on a circle and two diagonals  perpndicular.

2.How many ways are there to divide a 2*n rectangle into rectangles having integral sides,where n is a positive integer.

3.x,y,z are positive real numbers and xy+yz+zx=1.Prove that

27/4 . (x+y)(y+z)(z+x)>=((x+y)^(1/2) +(y+z)^(1/2) +(z+x)^(1/2))^2>=6.3^(1/2)

What is the radius of the circle in the first problem? We need to fix something (e.g. radius) to find the required maximum.

[quote="cosinerburc"]3.x,y,z are positive real numbers and xy+yz+zx=1.Prove that

27/4 . (x+y)(y+z)(z+x)>=((x+y)^(1/2) +(y+z)^(1/2) +(z+x)^(1/2))^2>=6.3^(1/2)[/quote]

[quote]
$x,y,z$ are positive real numbers and $xy+yz+zx=1$. Prove that

\[ \frac{27}{4}(x+y)(y+z)(z+x) \geq (\sqrt{x+y} +\sqrt{y+z} + \sqrt{z+x})^2\geq 6\sqrt{3}.  \]
[/quote]

$(\sqrt{x+y} +\sqrt{y+z} + \sqrt{z+x})^2 \leq 3 ((x+y)+(y+z)+(z+x))$

${= 6(x+y+z)(xy+yz+zx) \leq 6\cdot \frac{9}{8}}(x+y)(y+z)(z+x) = \frac{27}{4}(x+y)(y+z)(z+x)$

$(x+y+z)^2\geq 3(xy+yz+zx)=3$, so $x+y+z\geq \sqrt{3}$.

Hence (*) $(x+y)(y+z)(z+x) \geq \frac{8}{9}(x+y+z)(xy+yz+zx) \geq \frac{8\sqrt{3}}{9}$

$(\sqrt{x+y} +\sqrt{y+z} + \sqrt{z+x})^2\geq 6\sqrt{3}$  comes from AM-GM and (*).


limit
trigonometry
logarithms
algorithm
calculus
search
calculus computations

Find

 :huh:

$ i)$ $ \lim_{x\to\plus{}\infty}\frac{e^{x}}{x^{2008}}$

$ ii)$ $ \lim_{x\to 0}sin(x)ln(x)$

$ iii)$ $ \lim_{x\to 0}$ $ xe^{\minus{}x}\plus{}ln(x)\minus{}\frac{1}{x}$

i) $ \lim_{x\to\infty}\frac{e^{x}}{x^{2008}}\to\infty$, you could L'Hopital it $ 2008$ times and find that the numerator stayed at $ e^{x}$ and the denominator became a constant, or just note that $ e^{x}$ grows much much faster than $ x^{n}$ despite $ n$ looking large.

ii) and iii) are $ \minus{}\infty$, for ii) observe $ \exp{\{\sin x\ln x\}}$, and iii) is straight forward computation, notice that both $ ln x$ and $ \minus{}\frac{1}{x}$ go to $ \minus{}\infty$ , while the first term in finite. Of course both limits should observed in $ \overline{\mathbb{R}}$.

Don't $ ii)$ and $ iii)$ not exist because $ \lim_{x\to 0^{\minus{}}}\ln(x)$ does not exist?

The question makes sense if we replace $ \lim_{x\to 0}$ with $ \lim_{x\to 0^{\plus{}}}$.
If we do, note that Mr. Doe is wrong about (ii).

$ ii)$:

$ \lim_{x\to 0^{\plus{}}}\sin(x)\ln(x)$
$ \lim_{x\to 0^{\plus{}}}\frac{\ln(x)}{\csc(x)}$

Apply L'Hopital's Rule:

$ \lim_{x\to 0^{\plus{}}}\frac{\frac{1}{x}}{\minus{}\csc(x)\cot(x)}$
$ \lim_{x\to 0^{\plus{}}}\minus{}\frac{\sin(x)\tan(x)}{x}$
$ \minus{}\lim_{x\to 0^{\plus{}}}\frac{\sin(x)}{x}*\lim_{x\to 0^{\plus{}}}\tan(x)$
$ \minus{}1 * 0$
$ 0$

Is this a correct evaluation of the limit?

Correct answer, but the methods are not to my taste.

All that algebraic manipulation isn't needed, and it's asking for trouble (in the way of possible careless mistakes).

It's really a matter of simple size. Two things you should remember and are easy to remember: a logarithm goes to infinity as its argument goes to either zero or infinity, but it does so slower than any power of that argument could go to infinity. And an exponential goes to infinity as its argument goes to $ \plus{}\infty$ faster than any power of that argument goes to infinity and and exponential goes to zero as its argument goes to $ \minus{}\infty$ faster than any power of that argument goes to zero.

Read the last phrase of HilbertThm90's post above and note that he is directly quoting the second of these two principles. For problem ii), let's use the first principle.

As $ x\to0^{\plus{}},$ $ \sin x$ tends to zero - how fast? It looks just like $ x,$ that's how fast. For the purpose of this limit, we might as well have $ x.$

But if we're looking at $ \lim_{x\to0^{\plus{}}}x\ln x,$ then we note that the power of $ x$ goes to zero faster than the logarithm can go to $ \minus{}\infty,$ and hence the limit is zero.

Think this way, don't rely on tedious calculations. Be fast and accurate at the same time. (I can assure you that when jmerry wrote "Mr. Doe is wrong about (ii)" above, he hadn't spent more than 5 seconds - and no paper - calculating the limit for himself.)

[color=darkblue]My solution for the first one ;[/color]

[color=darkblue]We have  $ L =\lim_{x\to+\infty}\frac{e^{x}}{x^{2008}}= (\lim_{x\to+\infty}\frac{e^{\frac{x}{2008}}}{x})^{2008}$

Applying [b]de l' Hospital Rule[/b] we take that  $ \lim_{x\to+\infty}\frac{e^{\frac{x}{2008}}}{x}=\frac{1}{2008}\lim_{x\to+\infty}e^{\frac{x}{2008}}$=$ +\infty$

Therefore $ L =+\infty$[/color]

[quote="giannis"]
$ i)$ $ \lim_{x\to\plus{}\infty}\frac{e^{x}}{x^{2008}}$
[/quote]

Here's a non L'Hopital solution.

We know that,
$ e^{x}\geq\frac{x^{2009}}{2009!}$ for $ x>0$.

So,
$ \frac{e^{x}}{x^{2008}}\geq\frac{x}{2009!}\to\infty$.

Wow, thanks for the tips, Kent Merryfield!  I've never been told anything like that in class!  We've usually had to "show" work for calculating limits.  However, I suppose one could reason it out, but that might be lengthy when writing it down... and it might be a lot to memorize, these tricks.

It's not a lot of things to memorize: it's two basic principles, one having to do with logarithms and one having to do with exponentials. (They're related to each other, of course.) And I think a clear-eyed understanding of these principles is [i]more[/i] important to being a user of mathematics than being able to compute limits.

You're talking to a computer scientist, and he says that while algorithm A is $ O(n),$ algorithm B is $ O(\log n).$ That should be immediately meaningful to you. You should realize that $ O(\log n)$ is a whole lot smaller than $ O(n).$

But in your computation, the most wasteful thing you did was not replacing $ \sin x$ by $ x.$ If you want formal justification, you could have written:

$ \lim_{x\to0}\sin x\ln x \equal{}\lim_{x\to0}\frac{\sin x}{x}\cdot x\ln x$

and then noted that $ \lim_{x\to0}\frac{\sin x}{x}\equal{} 1.$ That's a limit you really should know.

It in turn could be considered a small subset of the knowledge of my favorite Calculus II handout, "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=801323375&t=145324]Seven Power Series to Memorize.[/url]"


number theory
greatest common divisor
relatively prime
number theory unsolved

Prove that there are infinitely many ordered triples of positive integers $ (a, b, c)$ such that $ gcd\ (a, b, c) \equal{} 1$ and $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2$ is a perfect square.

[quote="SimonM"]Hint:

Pell Equations[/quote]

Thanks!  :)

May I enquire how you know how to proceed with a problem the moment you glance at it?

I didn't. The problem looked quite hard to start with so I thought that the best way would be to fix two variables. 1 being the first one I thought of

far simpler than that:
$ a \equal{} 1,b \equal{} x,c \equal{} x \plus{} 1$ is a solution for every natural $ x$
(also note that if u pick them in such a way $ a,b,c$ are all relatively prime to each other)

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \equal{} x^2 \plus{} (x \plus{} 1)^2 \plus{} (x(x \plus{} 1))^2 \equal{} x^4 \plus{} 2x^3 \plus{} 3x^2 \plus{} 2x \plus{} 1 \equal{} (x^2 \plus{} x \plus{} 1)^2$

infact $ a \equal{} x,b \equal{} y,c \equal{} x \plus{} y$ where $ x,y$ are relatively prime will do
$ (xy)^2\plus{}(x(x\plus{}y))^2\plus{}(y(x\plus{}y))^2\equal{}(x^2\plus{}xy\plus{}y^2)^2$


We are in the process of building a calendar of math-related events, such as contests and summer programs.  The calendar is not yet public, but it is available for beta-testing [url=http://www.artofproblemsolving.com/Calendar/AoPS_Cal.php]here[/url].

As you'll probably find out, it has some kinks to be worked out.  Please help us by breaking it and posting how you broke it here.

Also, you'll see that there's a form for submitting event information.  If you know enough about any given event that you can complete the form for the event, please do so, so we can add it to the calendar.

You'll see on the calendar that we are trying to distinguish between contests that you go compete at a designated central site, and those which you mail in.  We call the former 'Onsite contests' and the latter 'Mail-in contests'.  Unfortunately, some mail-in contest folks have mistaken 'Onsite contests' for their contest, thinking the students take it on their own site.  Can anyone suggest a better short phrase for what we were calling 'Onsite contests'?

[quote="rrusczyk"]
As you'll probably find out, it has some kinks to be worked out.  Please help us by breaking it and posting how you broke it here.
[/quote]
Actually the calendar is so simple that there are very few possibilities for breaking it. It seems to work perfectly OK if you do not try to abuse its features :).

The only "abusable" feature is the date input (you cannot really abuse the "check one or more boxes" option (or am I getting too uninventive?)). It survives most of the standard "attacks" (ultra-long input lines, weird editing procedures, etc.) but accepts a couple of "undocumented" formats. You can type something like 1/2/3/43657264926439/68478/498 and it will be accepted and (in most cases) equivalent to 1/2/3 though I managed to get year 2NaN with this format once. Another format is 1/2/00000000003456 or something like that (this particular sequence is accepted and gives you year 1838 but the probability that a random sequence of this type gets accepted is about 1/5).

This weird behavior seems pretty harmless (just gives you some fancy date) but you may still want to look into it to check that it is not just a side effect of some really bad bug. ;)

Also I don't know whether it is a clever idea to demand that the "from" date be [b]strictly[/b] earlier than the "to" date. Normally one would expect it to be earlier than or equal to the "to" date (just in case somebody wants to find the events happening on some particular day).

Another questionable idea is to list a long event if its [b]starting[/b] date is not earlier than the "from" date and not later than the "to" date. I understand that it is a matter of taste but I would rather prefer listing all events whose period of duration has a non-empty intersection with the requested range of dates. On the other hand, other people may have other ideas about this.

These are all remarks I have so far. Thanks for the new feature.  :)

Wow! This is a really interesting idea. When will the calendar be fully built?


[quote="uberl33tmage"]Ok, I thought this would be a fun challenge, but im not sure if i got the right answer. Find the distance between (:$x_0$,;$y_0$) and ax+by+c=0, where ax+by+c=0 is an arbitrary line and (:$x_0$,;$y_0$) is an arbitrary point on the xy plane. Please show your work. Have fun![/quote]

This question is answered in detail on 'Intermediate topics'

check it at
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49046

Wrong forum. Not even close :P . I think this would belong in Intermediate, maybe Getting Started.

Yes, this is an extremely difficult problem. Let's wait for a moderator to move it to high school. :)

this is exactly point line distance formula...


A census man on duty visited a house in which the lady inmates declined to reveal their individual ages, but said "We do not mind giving you the sum of the ages of any two ladies you may choose". Thereupon, the census man said, "In that case, please give me the sum of the ages of every possible pair of you". They gave the sums as: 30, 33, 41, 58, 66, 69. The census man took these figures and happily went away.

How did he calculate the individual ages?

Cute  :lol:

The sum of the ages is $(30 + 33 + 41 + 58 + 66 + 69)/3 = 99$

Looking which sum gives $99$ gives $30 + 69$, $33+66$, $41 + 58$.
If, wlog, the ages are $a \leq b \leq c \leq d$ this gives $a+b = 30$, $c+d=69$, ...

This solves to $11, 19, 22, 47$

To figure out how many ladies were there, you use nC2=6, so that n=4. Let the 4 ages be a, b, c and d. Adding all the combinations give 3 ( a + b + c + d) = 297, giving a + b + c +d= 99. Simple algebra with simultaneous equations and substitution solves the problem

(a+b)+(b+c)-(c+d)-(d+a)+2(b+d)=4b.

Sorry to post that and keep the thread up, but Kab^2, try to avoid reviving such old threads which have solutions...

Use permutations to find the number of women as  1 and then assign summation such that

Nowhere. The question might have multiple solutions.

But I am in doubt that $a=2.5$. This is obviously not possible. But I could not find a mistake in solving these equations.

Can anyone help to resolve this wrong answer?

[quote=AOPSMasters]But I got the ages as $a=2.5,b=27.5,c=30.5,d=38.5$. Where am I wrong? Assuming that $a \leq b \leq c \leq d$, my equations formed are
$$a+b=30,$$
$$a+c=33,$$ $$a+d=41,$$ $$b+c=58,$$ $$b+d=66,$$ $$c+d=69$$[/quote]

$b + c$ can also be $41$. It is not necessary that $a + d < b + c$

[quote=AOPSMasters]But I got the ages as $a=2.5,b=27.5,c=30.5,d=38.5$. Where am I wrong? Assuming that $a \leq b \leq c \leq d$, my equations formed are
$$a+b=30,$$
$$a+c=33,$$ $$a+d=41,$$ $$b+c=58,$$ $$b+d=66,$$ $$c+d=69$$[/quote]

This is because you assumed $30$ to be $a+b$, and then wrongly assumed $33$ to be $a+c$ and so on. But, in the question, it has not been specified that $30, 33, 41, 58, 66$ and $69$ are the sums of ages of which two ladies respectively, like in the real case, $$a+b=30,$$ $$a+c=33, $$ $$a+d=58,$$ $$b+c=41,$$$$b+d=63,$$$$ c+d=69$$.

Hope this clarifies your doubt.

There are two possible sets of values for the individual ages of each women.

Again a cute problem
—let a,b,c,d are there indivisiul ageS
WLOG,  a≥b≥c≥d,Then a+b will max sum and then a+c
So,we have a+b=69,a+c=66, and (a+b+c+d)=(30+33+41+58+66+69)/3=99
Now let's assume as case 1 that third greatest sum is b+c। By the way a little bit of thinking we can tell third greatest sum can b+c or a+d

if,b+c=58 ,a+b=69 ,a+c=66।               Then  b-c = odd। b+c= even not possible

So a+d =58।  Now ez to solve as we have 4 equations


quadratics
inequalities proposed
inequalities

[quote="Misha123"]The right side is just:

$\sqrt{\frac{a^2+b^2}{2}}\geq\sqrt{ab}$

which is AM-QM.

Misha[/quote]

did you mean QM-GM, or AM-GM?

Misha123 meant Arithmetic Mean-Quadratic Mean supposedly


algebra
polynomial
number theory
greatest common divisor
function
domain
geometry

one of the questions in the book asks for the three 3rd and four 4th roots of 1. i know how to do that demoivre's theorem: $ e^{2\pi k}$
however, i think im misunderstanding the use of the word "roots". isn't there only one 3rd root and fourth root of 1: $ 1^{\frac{1}{3}}\equal{}1^{\frac{1}{4}}\equal{}1$?

There are $ n$ [i]complex[/i] $ n^{th}$ roots of unity that are the $ n$ complex roots of the polynomial $ x^n \minus{} 1 \equal{} 0$.  For example, the four fourth roots of unity are $ 1, i, \minus{}1, \minus{}i$.

Finding roots with DeMoirve's equations gives [i]all[/i] roots of a number, meaning with either or both real and imaginary parts.
The $ n^{\text{th}}$ roots of $ x$ raised to $ n$ all yield $ x$.
DeMoirve's also gives another explaination why the even roots of a positive number has both a positive and negative real root.
Graphing all the $ n^{\text{th}}$ roots of $ 1$ on the complex plane split a circle with radius $ 1$ centered at the origin on the plane into $ n$ congruent sections.

[quote="mihail911"]one of the questions in the book asks for the three 3rd and four 4th roots of 1. i know how to do that demoivre's theorem: $ e^{2\pi k}$
however, i think im misunderstanding the use of the word "roots". isn't there only one 3rd root and fourth root of 1: $ 1^{\frac {1}{3}} \equal{} 1^{\frac {1}{4}} \equal{} 1$?[/quote]
Even if you're just looking at real roots, -1 is another fourth root of unity, as t0rajir0u says.In fact, -1 is an nth root of unity whenever n is even.

For every even root of $ 1$ greater than or equal to $ 2$ has $ \minus{} 1$ and $ 1$ as roots, for roots $ 4k \ , \ k \in \mathbb{Z}^{\plus{}}$ there are the roots $ 1,i, \minus{} 1, \minus{} i$, et cetra.
In fact $ \text{GCD} \left(x,y \right)$ tells the number of common roots of the $ x^{\text{th}}$ and $ y^{\text{th}}$ roots of unity.  :wink:

Recall that a [b]root[/b] of a function $ f : A \rightarrow B$ is an element $ x \in A$ such that $ f(x) \equal{} 0$, if such an element exists.  We are not necessarily restricted in our choice of function, or its domain and codomain; however, it is necessary to specify the function if we are to be precise about what we mean by "root."  Roots are also commonly called [b]zeros[/b], for obvious reasons.

If this definition seems unfamiliar or unrelated to the more familiar concept of "square roots," then it should be pointed out that a square root of a number $ n$ corresponds to the particular choice of function, namely $ f(z) \equal{} z^2 \minus{} n$.  Initially, when we are introduced to square roots, we are told things like $ \sqrt{4} \equal{} 2$, and whatnot.  But it is only in the context of functions do we really start to learn what this means.

For suppose $ n \equal{} 4$.  Then the square roots of 4 are those values for which $ f(z) \equal{} z^2 \minus{} 4 \equal{} 0$.  Obviously, $ z \equal{} 2$ is a solution.  But we cannot simply discuss solutions without considering the domain in which we are investigating the function.  For instance, if we let $ f$ be a function over the positive integers, then $ z \equal{} 2$ is the [b]only[/b] solution.  If $ f$ is a function over the real numbers, then $ z \equal{} 2$ and $ z \equal{} \minus{}2$ are both solutions.

The customary definition of the surd symbol $ \sqrt{n}$, then, is that we are in fact talking about the function $ f(z) \equal{} z^2 \minus{} n$, where the domain consists of the set of nonnegative real numbers; i.e., $ z \ge 0$.  As we begin to realize by studying complex analysis, this restriction on the domain of $ f$ is somewhat unnatural, as it is made to preserve the injectivity of this particular mapping.  But there is no inherent reason to restrict the domain in this manner.

The situation becomes more interesting when we talk about cube roots.  For instance, suppose we wanted to evaluate $ 8^{1/3}$.  Obviously 2 is an answer, since $ 2^3 \equal{} 8$.  However, if we apply the aforementioned definition of root, then we would like to find [b]all[/b] solutions to the equation $ f(z) \equal{} z^3 \minus{} 8 \equal{} 0$.  Since we already know $ z \equal{} 2$ is a solution, then we might write (either by long division, synthetic division, or factoring via difference of cubes)

$ f(z) \equal{} (z \minus{} 2)(z^2 \plus{} 2z \plus{} 4) \equal{} 0$.

We can now use the quadratic formula to solve for the remaining roots, and we find

$ z \equal{} \frac{\minus{}2 \pm \sqrt{\minus{}12}}{2} \equal{} \minus{}1 \pm \sqrt{\minus{}3}$.

Hundreds of years ago, mathematicians looked at the above result and simply dismissed it as nonsense.  Nowadays, we understand complex numbers and write

$ z \equal{} \minus{}1 \pm i \sqrt{3} \equal{} \{\minus{}1 \plus{} i \sqrt{3}, \minus{}1 \minus{} i \sqrt{3}\}$.

This demonstrates to us that there exist two additional cube roots of 8.  Together with $ z \equal{} 2$, we have three distinct numbers whose cube is equal to 8.  With only a rudimentary understanding of square, cube, or n(th) roots, we would not make such a conclusion.  Still, it is tempting to believe that the root $ z \equal{} 2$ is somehow more "valid" an answer to our question than a result as seemingly nebulous or untrustworthy as "$ \minus{}1 \plus{} i \sqrt{3}$."  But we will see shortly that this belief is ill-founded.

When we talk about n(th) roots of unity, we are looking at the function $ f(z) \equal{} z^n \minus{} 1$ over the complex numbers.  From the Fundamental Theorem of Algebra, we know that a polynomial of degree $ n$ has exactly $ n$ complex roots (counting multiplicities).  Therefore, we know that there are $ n$ $ n(th)$ roots of unity.  From Euler's and De Moivre's formulas, we see that numbers of the form $ z \equal{} e^{2\pi i k/n}$, where $ k$ is any integer, satisfy the relation $ f(z) \equal{} 0$, since

$ f(e^{2\pi i k/n}) \equal{} e^{2\pi i k} \minus{} 1 \equal{} (\cos 2\pi k \plus{} i \sin 2\pi k) \minus{} 1 \equal{} 0$.

Because of the periodicity of the sine and cosine functions, we observe that the set $ \{e^{2\pi i k/n}\}_{k\equal{}\minus{}\infty}^\infty$ is actually only comprised of $ n$ distinct elements, for which we may select the representatives $ k \in \{0, 1, 2, \ldots, n\minus{}1\}$.

The roots of unity show us an important fact, one which, in some sense, requires us to undo some of the learning we were told in elementary arithmetic:  that [b]no root is any more special than the others[/b].  We may wish to impose a simple "rule" that always gives us a unique answer when we ask for, say, "the" 5th root of $ 23028\plus{}11228i$.  In fact, there are 5 such roots, and none of them are inherently more "special" than the other four.  If you have a computer algebra program like Maple or Mathematica, you will find all roots are complex (one of them just happens to be a Gaussian integer), but there is no way to say "pick the nonnegative one" as we did before with square roots, nor is there a natural way to generalize such a choice (e.g., could you justify a rule saying "pick the one whose argument as measured counterclockwise from the positive real axis is least?").


plase move on to the attachment below.
outline needed. :)

Who can do this?
Please show me.
Thanks :P

:)
can $ a_{n}$ be expressed with $ a_{1}$ and $ a_{2}$?


calculus
integration
geometry
geometric transformation
rotation
calculus computations

Let $ t$ be a negative real number and $ D$ be the part bounded by the curve $ y\equal{}2^{2x\plus{}2t}$, the curve $ y\equal{}2^{x\plus{}3t}$ and the $ y$ axis.

(1) Find the volume $ V(t)$ generated by rotation of $ D$ about the $ x$ axis.

(2) When $ t$ moves in the range of negative real number, find the maximum value of $ V(t)$.

MMkay then that makes the maximum occur when $ t \equal{} \minus{} \frac {1}{2}$. Finding this zero involves some rather beautiful factoring if I say so myself. Nice problem kunny encore encore.

edit: typos  :D


[hide="solution"]
the given expression can be written as
$ tan^2(\frac{\pi}{16} \plus{} tan^2(\pi \minus{}\frac{\pi}{16}) \plus{}tan^2(2\pi/16)\plus{}.................$ and so on

or,
$ tan^2(\pi/16)\plus{}cot^2(\pi/16)\plus{}tan^2(2\pi/16)\plus{}cot^2(2\pi/16)\plus{}tan^2(3\pi/16)\plus{}cot^2(3\pi/16)\plus{} 1$   .........$ (1)$
here $ tan^2(4\pi/16) \equal{}1$

now, $ tan^2{\theta} \plus{} cot^2{\theta} \equal{} \frac{sin^2{\theta}}{cos^2{\theta}} \plus{} \frac{cos^2{\theta}}{sin^2{\theta}}$

or,$ \frac{1\minus{}2cos^2{\theta}sin^2{\theta}}{cos^2{\theta}sin^2{\theta}}$

multiplying by 4 and simplifying,
it becomes,$ \frac{4\minus{}2sin^2{2\theta}}{sin^2{2\theta}}$

again simplifying,it becomes, $ \frac{6\plus{}2cos{4\theta}}{1\minus{}cos{4\theta}}$..............$ (2)$

now put values of  $ tan^2{\theta} \plus{} cot^2{\theta}$ as obtained in (2) in expression for (1)

solving,I got the given expression to be $ 35$
the procedure is rught,not sure about calculations :oops: [/hide]

[hide="Alternative"]
We can construct a polynomial with roots, $ \tan 0, \tan (\pi/16), \tan (2\pi/16), \ldots, \tan (7\pi/16), \tan (9\pi/16), \ldots , \tan (15\pi/16)$.

For these values of $ \theta$

$ \tan 16\theta \equal{} 0 \equal{} \frac {\Im (1 \plus{} ix)^{16} }{\Re (1 \plus{} ix)^{16}}$

where $ x \equal{} \tan \theta$

$ \Rightarrow \Im (1 \plus{} ix)^{16} \equal{} 0$

$ \minus{} 16x^{15} \plus{} 560x^{13} \minus{} \ldots \equal{} 0$

Dividing by x, removes the solution $ x \equal{} \tan 0 \equal{} 0$, then letting $ u \equal{} x^2$ makes the roots of the polynomial in u, the terms in the sum we're trying to find.

Then by viete's the sum of the roots is $ \minus{} \frac {560}{ \minus{} 16} \equal{} 35$
[/hide]


geometry
circumcircle
Euler
rotation
geometric transformation
geometry proposed

Hi everyone ,
I have a question about the simson line , can anyone show me a way to solve it ?

-Consider a triangle ABC with circumcenter O , circumcircle (O).
(A) Given X on (O) prove that  OX is parallel to the Simson line of X with respect to ABC iff

                         3  XA  = B - C       ,  XA : arc between points X , A

(B)Prove that there exactly 3 points X on (O) such that part (A) holds.Show that these points constitue a equilateral triangle.
(C) Prove that the Simson lines in question concur in the Euler point of ABC.

I think for the part (A) we can use that the Simson line of X is parallel to AY where [XY] is the unique chord of perpendicular to BC.For part (C) if we can show that these lines are perpendicular to each other , then we can show that these lines concur on the Euler line.For part (B) I have no idea :D  :D  :D
Thanks.

Let $ D \in BC$ be foot of the A-altitude, $ AD \perp BC.$ Let perpendicular to $ BC$ through $ X \in (O)$ cut $ (O)$ again at $ U.$ $ AU$ is direction of Simson line $ s_X$ with pole $ X.$ Assume $ 3 \measuredangle AUX \equal{} \measuredangle ABC \minus{} \measuredangle BCA.$

$ \measuredangle OXU \equal{} \measuredangle OXA \minus{} \measuredangle UXA \equal{}$

$ \equal{} \left( 90^\circ \minus{} \frac {_1}{^2} \measuredangle AOX \right) \minus{} \measuredangle UCA \equal{}$

$ \equal{} (90^\circ \minus{} \measuredangle AUX) \minus{} (\measuredangle UCB \plus{} \measuredangle BCA) \equal{}$

$ \equal{} (90^\circ \minus{} \measuredangle AUX) \minus{} (\measuredangle UAB \plus{} \measuredangle BCA) \equal{}$

$ \equal{} (90^\circ \minus{} \measuredangle AUX) \minus{} (\measuredangle UAD \plus{} \measuredangle DAB \plus{} \measuredangle BCA) \equal{}$

$ \equal{} (90^\circ \minus{} \measuredangle AUX) \minus{} \left[\measuredangle AUX \plus{} (90^\circ \minus{} \measuredangle ABC) \plus{} \measuredangle BCA\right] \equal{}$

$ \equal{} (\measuredangle ABC \minus{} \measuredangle BCA) \minus{} 2 \measuredangle AUX \equal{} \measuredangle AUX$

It follows that $ XO \parallel AU \parallel s_X.$ Similarly, let $ Y, Z \in (O)$ and let perpendiculars to $ CA, AB$ through $ Y, Z$ cut $ (O)$ again at $ V, W,$ respectively. Then $ BV, CW$ are directions of Simson lines $ s_Y, s_Z$ with poles $ Y, Z.$ Assuming $ 3 \measuredangle BVY \equal{} \measuredangle BCA \minus{} \measuredangle CAB$ and $ 3 \measuredangle CWZ \equal{} \measuredangle CAB \minus{} \measuredangle ABC,$ we get $ YO \parallel BV \parallel s_Y$ and $ ZO \parallel CW \parallel s_Z.$

When $ X$ moves around the entire circumcircle $ (O)$ counter-clockwise, line $ XO$ is rotated around $ O$ by $ 360^\circ$ counter-clockwise. Since $ \measuredangle AOX \equal{} 2 \measuredangle AUX \equal{} \minus{} 2 \measuredangle DAU,$ line $ AU$ is rotated by $ 180^\circ$ clockwise, so that mutual rotation of the lines $ XO, AU$ is $ 3 \times 180^\circ,$ in which they become parallel exactly 3 times, at angular intervals of $ XO$ equal to $ \angle X_2OX_1 \equal{} \angle X_3OX_2 \equal{} \angle X_1OX_3 \equal{} 120^\circ.$ Therefore, exactly 3 poles $ X \equiv X_1, Y \equiv X_2, Z \equiv X_3 \in (O)$ exist, forming an equilateral $ \triangle XYZ,$ such that the Simson lines $ s_X, s_Y, s_Z$ of $ X, Y, Z$ are parallel to $ XO, YO, ZO.$ The equilateral triangle can be verified by direct calculation:

$ \measuredangle YZX \equal{} \measuredangle YCX \equal{} \measuredangle YCB \plus{} \measuredangle BCA \plus{} \measuredangle ACX \equal{} \minus{} \measuredangle BVY \plus{} \measuredangle BCA \plus{} \measuredangle AUX \equal{}$

$ \equal{} \minus{} \frac {_1}{^3} (\measuredangle BCA \minus{} \measuredangle CAB) \plus{} \measuredangle BCA \plus{} \frac {_1}{^3} (\measuredangle ABC \minus{} \measuredangle BCA) \equal{}$

$ \equal{} \frac {_1}{^3} (\measuredangle CAB \plus{} \measuredangle ABC \plus{} \measuredangle BCA) \equal{} 60^\circ.$

Let $ H$ be orthocenter of $ \triangle ABC.$ The Simson line $ s_X \parallel XO$ cuts the segment $ HX$ at its midpoint $ K,$ it is midline of the $ \triangle OHX,$ it cuts the segment $ OH$ at its midpoint $ N,$ which is the 9-point circle center. Similarly, $ s_Y, s_Z$ cut $ OH$ at its midpoint $ N.$

Thank you very much  :lol:   It really helped me to understand  the issue.

let $X'$ the intersection of the perpendicular to $BC$ through $X$ with$(O)$ ,$ A'$ the the reflection of $A$ in $BC$-bisector   ; first if an other point $X_1$ verify $3 AX'X_1=\angle AX'A'  (^*)$ then  $3\alpha \overset{notation}{=  } 3 (\angle AX'X-\angle AX'X_1)=0 \ [\pi] \implies \alpha=\pm \frac{\pi}3$ then it exist exactly  3 points verifying $(^*)$ so $(A)\implies (B)$
for convenience ,WLG assume that $X$ is on the arc$AA'$ that not containing $B$
$\underline{(A) [\implies]:}$
it s known that $AX'$ is the direction of the simpson line of $X$ so $AX' \parallel  XO$ then $\angle AX'X=\angle X'XO=\angle XX'O$ thus $\angle AA'X=\angle XX'O$ but $X'X\perp AA'$ then $X'O\perp A'X $ implies $OX'$ is the bisector of $\angle XX'A'$ hence  the required  result .
$\underline {(A) [\Longleftarrow ]:}$
let $X''$ be the the intersection of the bisector of $A'X$ and $(O)$ we have $\angle AA'X=\angle XX''O$ but $X''O\perp A'X$ then $X'X \perp AA'$ so $X'=X''$ and $\angle AX'X=\angle XX'O=\angle  OXX' \implies OX\parallel AX'$
for the question $(C)$ it suffices to remark that $\delta$, the simson line bisects $HX$ then it bisects also $OH$ since in the triangle $OHX$ we have $HX\parallel \delta$ therefore  it pass through the ninepoint-center


superior algebra
superior algebra unsolved

Hello, I need help to prove that if $ I_1, \ldots, I_k$ are pairwise comaximal ideals, then $ I_1 \cap \cdots \cap I_{k\minus{}1}$ and $ I_k$ are comaximal ideals.

Thanks

I'm assuming $ R$ is a commutative ring with identity.

For $ 1\leq j < k$, write $ 1\equal{}a_j \plus{} b_j$ with $ a_j\in I_j$ and $ b_j \in I_k$.  Multiply all of these equations together.


Hello! Can someone help me whith this?

Let the function f that  x^2*f(y)=yf(x) for every x,y.

The question is to prove that the limit of function f while x->0 is f(0). I find that this limit is 0. Is this true or false?
If the result of the limit is true. How we can find the f(0) ?

thanks!

The limit is f(0) = 0.  We can prove this as follows:

Let x be a non-zero number.  Then taking x = y in  x^2*f(y) = y*f(x) yields

x^2*f(x) = x*f(x),

which, since x is non-zero, implies that

x*f(x) = f(x).

Now, if x is not equal to 1, we have that f(x) = 0; since we're considering the behaviour of f(x) as x gets arbitrarily close to zero, we can then conclude that the limit of f(x) as x tends to 0 is 0.

Now, if we set x = 0 and y = 2 in x^2*f(y) = y*f(x), we have that

0^2*f(2) = 2*f(0),

ie. f(0) = 0,

which is the limit of f as x approaches 0.  This concludes the proof.

We can also look at the pair $ (2,1)$; this gives us $ f(1)\equal{}0$ as well, and the only function that works is the zero function. A homogeneous condition would have been more interesting.

Thanks for your help !

My approach to this was: Let y be a non zero number. Then from the given yields that
f(x) = [f(y)/y]*x^2. Take the expression f(y)/y as a constant we have that the limit of  f  as x tends to zero is 0.
About  the f(0) maybe i jumped in to conclusion.
It is easy to find f(0) if we set  x= 0 and y=f(0) !


Hi, I had sent an email to ask about this, but it's been a while, so I think I'll ask here:
Are we allowed to scan documents and email them like the signed indemnity form, the medical form (+ list of medication, etc) and so on instead of sending them by mail so that I have more time?
Thanks

I think we just replied to your email, but yes, that's fine.