trigonometry
inequalities
geometry
conics
ellipse
symmetry
perpendicular bisector

Let $ABC$ an isosceles triangle, $P$ a point belonging to its interior. Denote $M$, $N$ the intersection points of the circle $\mathcal{C}(A, AP)$ with the sides $AB$ and $AC$, respectively.

Find the position of $P$ if $MN+BP+CP$ is minimum.

i hope i didn't make any mistakes any mistakes, here's my solution: we have $MN = AP \sin{A}$, therefore $T=MN+BP+CP = AP \sin{A}+BP+CP = a ( \frac{AP}{2R}+\frac{\sin{m}+\sin{n}}{\sin{(m+n)}})$ (by the sin law) , where $m,n$ are respectively the angles $\angle{PBC}$ and $\angle{PCB}$.

therefore $T$ is minimum when $\frac{AP}{2R}+\frac{\sin{m}+\sin{n}}{\sin{(m+n)}}$ is minimum, but $\frac{\sin{m}+\sin{n}}{\sin{(m+n)}}= \frac{BP+CP}{BC}\geq 1$, with equality when $P \in BC$, therefore $T$ is minimum when $AP$ is minimum, therefore $P$ is minimum when $P \equiv A$.

PS: i think there is a particulary case when $P$ is on the perpendicular bisector of $BC$, but that is trivial.

[quote="pohoatza"]
1) we have $MN = AP \sin{A}$

2) therefore $T$ is minimum when $\frac{AP}{2R}+\frac{\sin{m}+\sin{n}}{\sin{(m+n)}}$ is minimum, but $\frac{\sin{m}+\sin{n}}{\sin{(m+n)}}= \frac{BP+CP}{BC}\geq 1$, with equality when $P \in BC$, therefore $T$ is minimum when $AP$ is minimum

3) with equality when $P \in BC$, [...] therefore $T$ is minimum when $P \equiv A$.
[/quote]

1) Why? We have $AM = AN = AP$, so by the sines law, $MN = AM \sin{A}$ $\Leftrightarrow$ $MN = \sin{ANM}\cdot \frac{MN}{\sin{A}}\cdot \sin{A}$ $\Leftrightarrow$ $\angle ANM = 90^{\circle}$. But $ABC$ is isosceles!
2) $AP$, $BP$ and $CP$ are not independent!
3) So $T$ is minimum when $P \in BC$ [u]and[/u] $P \equiv A$. How come?!?

at 1) you answered yourself $AM=AP$, and $MN=2AM \sin{A}$, therefore $MN=2AP \sin{A}$ as like i said.

2)-3) i don't understand what you mean, i meant $\frac{\sin{m}+\sin{n}}{\sin{(m+n)}}= \frac{BP+CP}{BC}$ is minimum when $P \in BC$, and it's minimum is $1$, therefore now for $T$ to realise his minimun, it's necesary that $P \equiv A \in BC$, because $AP$ is the only thing that varies in the expresion.

[quote="pohoatza"]at 1) you answered yourself $AM=AP$, and $MN=AM \sin{A}$, therefore $MN=AP \sin{A}$ as like i said.
(...)
[/quote]
You didn't read after that! There's an '$\Leftrightarrow$' (equivalence) sign there.
[quote="pohoatza"]
2)-3) i don't understand what you mean.
[/quote]
[quote="pohoatza"][...] $P \equiv A \in BC$[...] [/quote]
In an isosceles triangle, $A \in BC$ ?!? That's a bit weird...

I think 3) is very clear. Obviously it's a mistake to state that when $T$ is minimum, both $P \in BC$ and $P \equiv A$ hold.

oh, now i see at the first part, i typed badly the first time, we have $MN = 2AP \sin{A}$ , not simply $AP$., but nothing changes in the rationament., having at the denominator in $E_{1}$ , $R$, which is also constant :)

for your second observations, let's take it otherwise to understand, $E_{2}$ is minimum when it's $1$, and obvious $AP \geq 0$, and because there exists such a $P$ for which the minimum $1$ is realizable, the answer is good, $P \equiv A \in BC$., and if you insist with the fact that $ABC$ is isosceles, that means $P \equiv A$, where $A$ is the midpoint of $BC$, therefore this is an degenarate "isoscel triangle", otherwise the observation that $ABC$ is isoscel is more likely there to separate a case in which the whole perpendicular bisector of $BC$, may contain the point $P$ with the minimum desired.

[quote="pohoatza"]oh, now i see at the first part, i typed badly the first time, we have $MN = 2AP \sin{A}$ , not simply $AP$., but nothing changes in the rationament., having at the denominator in $E_{1}$ , $R$, which is also constant :)
[/quote]
And also nothing changes in my "rationament" (how did you come to this word? it doesn't even exist - just [url=http://www.google.com/search?hl=en&q=define%3Arationament]check here[/url]), too! I get from your post that $\sin{ANM}= \frac{1}{2}$, which is usually false!

[quote="pohoatza"] [...] the answer is good, $P \equiv A \in BC$.[/quote]
$ABC$ is a triangle; why should $A \in BC$? $A,B,C$ are fixed points, only $P$ is varying.

in a circle $w$, let's consider a chord $MN$, thus we have $MN = 2R \sin{\widehat{MN}}$, just replace it here., and $A \in BC$ just in the equality case of the triangle ineq., because the triangle inequality is most likely strict, i will stop arquing with you about nothing.

yes, when $A$ is on $BC$ is the most particulary case when the "best" minimum is realized. for a "general" minimum, we must simply study the minimum realised in $E_{1}$ , respective $E_{2}$, and we observe that $P \equiv A$ is the "best", hope now it's clear.

That position is not the one where the minimum occurs.. think more! :!:

i hope the following is correct
[hide="answer"]
$P$ is the orthocenter of $\triangle ABC$.
[/hide]
[hide="proof"]
let $K$ and $L$ be the projections of $P$ on $AB$ and $AC$ and let $S$ be the common point of $AP$ and $MN$.
since the triangle $\triangle AMP$ is isosceles, it follows that the length of $MS$ is at least the length of the altitude from $M$, but the altitude from $M$ is equal with the altitude from $P$, that is $PK$. thus, $MS\geq PK$. a similar argument shows that $SN \geq PL$.
thus, $MN=MS+SN \geq PK+PL$.
so, $BP+CP+MN \geq (BP+PL)+(CP+PK) \geq 2h$, where $h$ equals the height from $B$ in triangle $\triangle ABC$. this fact follows from the triangle inequality.
now, note that when $P\equiv H$, it follows that $P$ is on the perpendicular bisector of $[BC]$, thus $MN\bot AP$, so in the above, $MS=PK$ and $SN=PL$. also, because $BP\bot AC$ and $CP\bot AB$, it follows that $BP+PL=CP+PK=h$, so for $P\equiv H$ the minimum value of $2h$ is indeed attained. this ends the proof.
[/hide]
[hide="note"]
my first idea was to prove that $P$ lies on the perpendicular bisector of $[BC]$. fix $r=AP$, and let $P$ vary on the circle $\mathcal{C}(A,r)$. now take the ellipse with foci $B$ and $C$, that is tangent to the circle $\mathcal{C}$. due to symmetry, the tangency point lies on the perpendicular bisector of $[BC]$. call this point $Q$. it is easy to show (triangle inequality) that for any point $P$ outside the ellipse, the inequality $BP+CP>BQ+CQ$ holds. (just take $P'$ the intersection between $CP$ and the ellipse and apply triangle inequality in $\triangle BPP'$). thus, for a fixed radius, the minimum value is attained when $P$ lies on the perpendicular bisector.
after i did this, i used the idea above, which i saw it can be easily extended. however, i think that a computational proof works from this point as well.
[/hide]

edit : small writing errors.

[quote="PhilAndrew"][color=darkred]Let $ABC$ be an isosceles triangle ($AB=AC$) and let $P$ be a mobile point which belongs to its interior. Denote the intersection points $M$, $N$  of the circle $\mathcal{C}(A, AP)$ with the sides $AB$ and $AC$ respectively. Find the position of the point $P$ when the sum $MN+BP+CP$ is minimum.[/color][/quote]
[color=darkblue][b][u]Lemma.[/u][/b] For any $x\in\mathcal R$, $a\sin x+b\cos x\le \sqrt{a^{2}+b^{2}}$ and for any $x\in\left[0,\frac{\pi}{2}\right]$, $a>0$, $b>0$ we have $\min\{a,b\}\le a\sin x+b\cos x\le\sqrt{a^{2}+b^{2}}$.
Particularly, for any $x>0\ ,\ x\sin\frac{A}{2}+\cos \frac{A}{2}\le \sqrt{1+x^{2}}$, i.e. $\boxed{\sqrt{1+x^{2}}-x\sin\frac{A}{2}\ge \cos\frac{A}{2}}$ with equality iff $x=\tan\frac{A}{2}$.
[b][u]Proof of the proposed problem.[/u][/b] Denote the middlepoint $D$ of the side $[BC]$. I"ll suppose w.l.o.g. that $DB=DC=1$, $AD=a>0$. I
n the minimum position of the mobile point $P$ we have obviously $P\in (AD)$ (prove easily !).
Denote $PD=x\in (0,a)$. Thus, $AP=a-x$, $MN=2AP\sin \frac{A}{2}=2(a-x)\sin\frac{A}{2}$, $PB=PC=\sqrt{1+x^{2}}$, $\tan\widehat{PBC}=x$.
Therefore, $MN+PB+PC$ is minimum $\Longleftrightarrow$ $(a-x)\sin\frac{A}{2}+\sqrt{1+x^{2}}$ is minimum $\Longleftrightarrow$ $\sqrt{1+x^{2}}-x\sin\frac{A}{2}$ is minimum $\Longleftrightarrow$
([u]from the above lemma[/u]) $x=\tan \frac{A}{2}$ $\Longleftrightarrow$ $\tan\widehat{PBC}=\tan\frac{A}{2}$ $\Longleftrightarrow$ $m(\widehat{PBC})=\frac{A}{2}$ $\Longleftrightarrow$ $PB\perp AC$ $\Longleftrightarrow$ the point $P$ is the orthocenter of $\triangle ABC$.[/color]


[hide="approach"] Let $ y_k = \frac {x_k}{\sum_{j = 1}^n{x_j}}$, then we are looking to find the maximum (actually the [i]supremum[/i] turns out to be more appropriate) of
\[ \sum_{j = 1}^n{\frac {y_j^2}{2y_j^2 + 1}
}\]
where $ \sum_{j = 1}^n{y_j} = 1$ and $ y_j > 0$.  Try to show that the sum is maximized when all but one or two of the $ y_j$ are arbitrarily close $ 0$. [/hide]

Thanks for idea [b]JoeBlow[/b] but still I can't finish problem. Can anyone help me with this one?


The logarithm of $27\sqrt[4]{9}\sqrt[3]{9}$ to the base 3 is:
\[ \text{(A)}\ 8\frac{1}{2} \qquad \text{(B)}\ 4\frac{1}{6} \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 3 \qquad \text{(E) none of these}  \]

[hide]
soo by definition of a logarithm..

 $3^x = 27\sqrt[4]{9}\sqrt[3]9$

Let's make it easier to look at

$3^x = 27\cdot9^{\frac{1}{4}}\cdot9^{\frac{1}{3}}$

Now let's put it all into something with 3 (yeah, great wording I know) Note: $3^2 = 9$

$3^x = 3^3 \cdot 3^{2(\frac{1}{4})} \cdot 3^{2(\frac{1}{3})}$

$3^x = 3^3 \cdot 3^{\frac{1}{2}} \cdot 3^{\frac{2}{3}}$

We know that when we multiply exponential values with the same base we can just add the exponents

$3^x = 3^{3 + \frac{1}{2} + \frac{2}{3}}$

Now we can see what x equals

$x = 3 + \frac{1}{2} + \frac{2}{3}$

Simplify the fractions

$x = 4 + \frac{1}{6}$

$(B)$[/hide]

I don't really know what log is, but I simplified the expression to [hide]3^(4 1/6)[/hide]

[quote="Phelpedo"]I don't really know what log is[/quote]
Log is logarithm.  if you had $\text{log}_{2}=64$ it means what number to the second power is 64.  The answer is 8.  That's what i think log means.


Find all pairs of integers $(x,y)$ satisfying

$1+x^{2}y=x^{2}+2xy+2x+y$

[hide="Solution"]Rearrange it as $y(x^{2}-2x-1)=x^{2}+2x-1\iff y=\frac{x^{2}+2x-1}{x^{2}-2x-1}=1+\frac{4x}{x^{2}-2x-1}$

For $y$ to be integer, we must have either $\frac{4x}{x^{2}-2x-1}=0$, which immediately gives $x=0, y=1$, or $\frac{4x}{x^{2}-2x-1}\geqslant 1\lor \frac{4x}{x^{2}-2x-1}\leqslant-1$.

Let's first find the values of $x$ for which the denominator is negative: $x^{2}-2x-1=0\iff x_{1,2}=1\pm\sqrt{2}$. Therefore $x=0,1,2$. Plugging in those values we get $y=1,-1,-7$ respectively, thus two more solutions.

For $x\geqslant 3\lor x\leqslant-1$ we have two inequalities:

1. $\frac{4x}{x^{2}-2x-1}\geqslant 1\iff 4x\geqslant x^{2}-2x-1\iff x^{2}-6x-1\leqslant 0$

The roots of the polynomial are $x_{1,2}=3\pm\sqrt{10}$, hence the inequality is satisfied for $x=3,4,5,6$. Out of those candidates, only $x=3$ gives an integer value $y=7$

2. $\frac{4x}{x^{2}-2x-1}\leqslant-1\iff 4x\leqslant-x^{2}+2x+1\iff x^{2}+2x-1\geqslant 0$

The roots of the polynomial are $x_{1,2}=-1\pm\sqrt{2}$, hence the inequality is satisfied for $x=-2,-1$. Out of those candidates, only $x=-1$ gives an integer value $y=-1$.

Therefore all the solutions are $(-1,-1),(0,1),(1,-1),(2,-7),(3,7)$[/hide]

Nice solution.

Is there any way we can solve this problem with Simon's trick? It looks too tempting to use Simon's Favorite Factoring Trick.

I just came up with a solution.

Suppose that $y=1$. Then $x=0$. And if $x=0$, then $y=1$. So we've got a solution $(x,y) = (0,1)$ and we can now suppose that $x\neq 0$ and $y\neq 1$. Consider the equation modulo $x$ to see that $y\equiv 1\bmod{x}$. Therefore, $\exists k\in\mathbb{Z}: \boxed{y = kx+1}$.
Because $y\neq 1$, we know that $k\neq 0$. Fill that $y=kx+1$ in in the initial equation to get the equivalent $kx^{3}= 2kx^{2}+4x+kx$. Because $x\neq 0$ and $k\neq 0$, this is equivalent to $x^{2}= 2x+\frac{4}{k}+1$, i.e. $\boxed{(x-1)^{2}= 2+\frac{4}{k}}$. So $2+\frac{4}{k}$ must be a perfect square. This surely means that $2+\frac{4}{k}\in\mathbb{Z}$, i.e. $\frac{4}{k}\in\mathbb{Z}$, i.e. $k\in\{\pm 4, \pm 2, \pm 1\}$. Only for $k\in\{-4,-2,2\}$ you'll get $2+\frac{4}{k}$ a perfect square.
For $k =-4$, you've got $x\in\{0,2\}$, but $x\neq 0$, so $x=2$ and thus $y=kx+1 =-7$. For $k =-2$, you've got $x=1$, so $y=kx+1 =-1$. And for $k=2$, you've got $x\in\{-1,3\}$, so $y = kx+1 \in\{-1,7\}$
All the solutions therefore are $(x,y)\in\{(0,1),(2,-7),(1,-1),(-1,-1),(3,7)\}$.

[quote="Kurt Gödel"]...Because $y\neq 1$, we know that $k\neq 1$.[/quote]

Nice one! :) But I presume the above should've been "...we know that $k\neq 0$".

Thanks for pointing that out  :P Fixed it.

I wonder if there are other solutions than these two...


f and g are two polymonial so that the complex of ther roots are the same,
and the same thing happens about f+1 and g+1.prove that :g=f

The problem is easy following the way by apply "unique theorem"

please tell me the solution because Im very weak in this part of algebra.

Does this '$f$ and $g$ are two polymonial so that the complex of their roots are the same' mean $(x-1)^{2}$ and $x-1$ have the same roots?


How do i prove that the Herschel graph which is the following one:
http://mathworld.wolfram.com/HerschelGraph.html

is NOT HAMILTONIAN?
Thanks in advance!

i figured it myself;
it's not hamiltonian because it's bipartite and has an odd number of vertices.


function
combinatorics proposed
combinatorics

Each student in a class chose a student to be the class representative (he may choose himself). There were exactly two students who chose themselves. There was exactly one pair (a,b) of distinct students such that a chose b and b chose a. Now, one year later, the class do the same thing. Is it possible that for every student x, the student who is chosen by x choose the student who is chosen by x one year ago?

when u say that they do the same thing, what is [i]thing[/i]?
is it also included [i]"There were exactly two students who chose themselves. There was exactly one pair (a,b) of distinct students such that a chose b and b chose a."[/i]


pigeonhole principle
combinatorics unsolved
combinatorics

Let $S$ be a set of $n$ points in the plane such that any two points of $S$ are at least $1$ unit apart.
Prove there is a subset $T$ of $S$ with at least $\frac{n}{7}$ points such that any two points of $T$ are at least $\sqrt{3}$ units apart.

Put the $ n$ points on the $ xy$-plane. Take a point $ P_1$ with maximum ordinate and remove all other points which are les than $ \sqrt3$ unit from $ P_1$. Among the remaining points, take point $ P_2$ with maximum ordinate and remove all other points which are less than $ \sqrt3$ unit from $ P_2$. Do this process repeatedly. If we can show that in each process at most six points are removed, than we can get $ \frac{n}7$ points such that any two of them are at least $ \sqrt3$ units apart. Assume otherwise that there are 7 points removed in some process. They are less than $ \sqrt3$ units from a point $ P$ which has the maximum ordinate. So they lie inside the semicircle with center $ P$ and radius $ \sqrt3$, but outside the semicircle with center $ P$ and radius $ 1$. Divide the semicircle into 6 equal regions, each region is bounded by two $ 30^{\circ}$ arcs of the two semicircles. By pigeonhole principle one of the regions contains two points, suppose that this region is bounded by smaller arc $ AB$ and larger arc $ CD$, so that segments $ AD$ and $ BC$ have a common point. The distance between any two points in this region is less than $ AD$. But \[ AD^2\equal{}PA^2\plus{}PD^2\minus{}2PA\cdot PD\cos30^{\circ}\equal{}1\plus{}3\minus{}2\cdot1\cdot\sqrt3\cdot\frac{\sqrt3}2\equal{}1\] so the distance between the two points in this region is less than 1, contradiction.


Given a [u][b]continous[/b][/u] fonction in $ x_0$

such that $ f(x_0)\neq0$

Prove that there exists $ \alpha > 0$ such that
$ \forall x \in ]x_{0} \minus{} \alpha, x_0 \plus{} \alpha [$ we have:

$ f(x)\neq0$

[hide]The definition of continuity in $ x_0$ give:
$ \forall \epsilon > 0$ [u]There exists[/u] : $ \alpha > 0$ such that:
$ \forall x \in ]x_0 \minus{} \alpha, x_0 \plus{} \alpha$ we have :
$ |f(x) \minus{} f(x_{0})| < \epsilon$
and we have $ f(x_0) \neq 0$
But I cant move from here[/hide]

Hint: try letting $ \epsilon\equal{}\frac{|f(x_0)|}2.$

[quote="Kent Merryfield"]Hint: try letting $ \epsilon \equal{} \frac {|f(x_0)|}2.$[/quote]

It works perfectly, Thanks  :lol:


Cine imi da si mie o rezolvare? Sa aflu si eu cum se face? Ca nu mi-a dat problema si rezolvarea tot nu o stiu...!!!

  Daca ecuatia a*(x^2) + b*x + c = 0 ,a ,b ,c apartin numerelor intregi si a>0 are doua radacini distincte in (0 , 1) atunci aratati ca a >= 5 !

And in english:  a*(x^2) + b*x + c = 0 ; a>0 ;  0<x_1<1 and 0<x_2<1  ; then results that a >= 5

Atat! Cine ma ajuta?

  Si cine ma ajuta si pe mine sa ma invete cum pot scrie in limbaj din acela, si sa imi apara matematic nu cum am scris eu?

salut cristinelu,

Ma chinui la problema ta de o jumatate de ora...am gasit deja conditiile necesare urmatoare:
\[ a>0, b<0, c>0, a>c, 2a+b>0, a+b+c>0 , b^2>4ac \]
In ceea ce priveste "scrisul matematic" trebuie sa faci
Math Forum>>Site&General informations>>LaTeX ...si dai de rubrica(mail) lui Valentin Vornicu
"How to Write in LaTeX and how to use LaTeX" .Cu ceva engleza (sau poti sa schimbi limba !) si rabdare vei reusi !
Bafta! :roll:

Un hint: foloseste faptul ca $16xy(1-x)(1-y)<1$ daca $x,y$ sunt distincte in $(0,1)$. Aplica apoi Viete si e gata. A fost mai mult decat un hint...

[quote="Cristinelu"]  Si cine ma ajuta si pe mine sa ma invete cum pot scrie in limbaj din acela, si sa imi apara matematic nu cum am scris eu?[/quote][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5261]Aici[/url] inveti limbajul, iar [url=http://www.mathlinks.ro/Forum/index.php?f=224]aici[/url] il testezi.


[quote="Diogene"][quote="Cezar Lupu"]Let $f\in C^{2}[-1,1]$ function. Prove that there is $c\in (-1,1)$ such that

                       \[ f(1)+f(-1)=2f(0)+\frac{1}{4}\cdot f''(c).  \][/quote]

Manifestly false .. For ex. $f(x)=e^x$  :D
 :cool:[/quote]

if $f(x)=e^x$ then $e+\frac{1}{e}\ =$2+$\frac{1}{4}\ e^c$,  and $c=ln(4e^2+4-8e)-1$ $\epsilon (-1,1)$ :D

[quote="JBL"]Some latexy niceness: $\in$ instead of $\epsilon$ takes fewer letters.  Is $\simeq$ the symbol you use for approximately, Ramanujan?  In the US, $\approx$ is standard.

Perhaps if $\frac14$ is replaced with a larger fraction?[/quote]

Yes   $\approx$ . I am still learning .... ;)

And not  only in US    :D

What if we replace   $\frac{1}{4}$ by $e$ ??

The result is true if $\displaystyle \frac{1}{4}$ is replaced by $1$.  The result is false for every other constant.


superior algebra
superior algebra solved

show (x+y^2,x+x^2+2xy^2+y^4) is prime ideal in R[x,y]

Simple algebra reduces our expression for the ideal to $(x,y^{2})$. Do you know what to do in this form?


Do we get confirmation emails when we send in the Usamts files and are they checked to make sure that everything works out ok? I might mail mine in for the first round cuz I odn't know anything about latex or all this file creation so there might be some hidden errors.

You should receive a confirmation email within a few days of our receiving your solutions (provided that you have your correct email in your profile, which you can check by clicking the Profile button above when you are logged in).  We will let you know if there's anything wrong with what you submit.


We have a half-circle with endpoints $A$ and $B$ and center $S$. The points $C$ and $D$ lie on the half-circle such that $\angle BAC = 20^\circ$ and the lines $AC$ and $SD$ are perpendicular. What is the angle between the lines $AC$ and $BD$?

A. 45
B. 55
C. 60
D. 67.5
E. 72

(Angles are in degrees)

[hide]I got B.

Notice that angle ASD = 70, so angle DSB = 110. Now, angle ABD is half ASD because it subtends same arc but it's on the perimeter of the circle. So, angle SDB = 35. And Let AC and BD intersect at P. Then DPA = 55. [/hide]

Since AC and DS are perpendicular, $ \angle ASD = 90^\circ - 20^\circ = 70^\circ $
So $ \angle BSD = 110^\circ $
DS is a radius, forming an isosceles triangle with BS, so $ \angle SBD = \angle SDB = 35^\circ $
In the triangle formed by D and the intersections of AC and DS and of AC and BD,
$180^\circ - 90^\circ - 35^\circ = 55^\circ $, so B is the answer.


search
Euler
combinatorics proposed
combinatorics

[quote="nhat"]Prove that if $n$ is an arbitrary positive integer then
                   $\sum_{\textstyle{k_{1},\dots,k_{n} \geq 0 \atop k_1+2k_2+\dots+nk_{n}=n}} \frac{(k_1+k_2+\dots+k_n)!}{k_1!\cdot\ldots\cdot k_{n}!}=2^{n-1}$[/quote]
Search about Euler forumla. It talks about the chain problem.


algebra
polynomial
quadratics
Vieta
sum of roots
quadratic formula

If the arithmetic mean of two numbers is 6 and their geometric mean is 10, then a equation with the the given two numbers as roots is:

A. $x^2+12x+100 = 0$
B. $x^2+6x+100 = 0$
C. $x^2-12x-10 = 0$
D. $x^2-12x+100 = 0$
E. $x^2-6x+100 = 0$

[quote="Shogia"]This situation cannot exist because AM is always greater than GM.[/quote]

No, trust me. It is possible if you consider the $a,b$ to be all numbers, instead of restricting them to all "real" numbers.

Oh, what a silly assumption to make, my bad.

[hide]Let the numbers be a and b.  $(a+b)/2$=6 so $a+b=12$ and $sqrt(ab)=10$ so ab=100
We can factor the polynomial with roots a and b as $(x-a)(x-b)$ which may be expanded to $x^2-(a+b)x+ab$  We know the values of $a+b$ and $ab$ so this polynomial is $x^2-12x+100$ so the answer is $D$[/hide]

Yes, remember AM-GM only applies to non-negative numbers.

[hide="Answer"]$\frac{a+b}{2}=6\Rightarrow a+b=12$, $\sqrt{ab}=10\Rightarrow ab=100$
$(x-a)(x-b)=x^2-(a+b)x+ab\Rightarrow x^2-12x+100\Rightarrow \boxed{D}$[/hide]

[hide]Let $j=a+b=2\cdot \text {AM}=2 \cdot 6=12$.
Let $k=ab=\text {GM}^2=10^2=100$.

The product of the roots of the quadratic

\[ ax^2+bx+c=0 \]

is $ab=k=c/a=100$.

The sum of the roots is $a+b=j=-b/a=12$. Therefore $b/a=-12$.

Thus our quadratic is $x^2-12x+100 \implies D$.[/hide]

[hide]From the arithemetic mean given, a+b=12. The sum of roots is 12, so -b=12, b=-12. From the geometric mean, given, ab=100. c=100. THe equation $x^2-12x+100$ contains the roots. [/hide]

[hide]Let $x_1$ and $x_2$ be two numbers.

$\frac{x_1 + x_2}{2}=6$

$\implies x_1 + x_2 = 12$

$\sqrt{x_1 x_2}=10$

$\implies x_1 x_2 = 100$

By Vieta's formula, we have the equation with $x_1, x_2$ as roots: $x^2 - 12x + 100 =0$

$\boxed {D}$[/hide]

[quote="Jimmy"]Just out of curiosity, what number was this?[/quote]

Solving the equation $x^2 - 12x + 100 = 0$ will give us complex roots

$delta = (-12)^2 - 4*100 = -256$

$\implies \sqrt{delta} = 16i$

Then use the quadratic formula we can find these two roots:$x_{1,2}= \frac{12\pm 16i}{2}$ which is equal to $x_{1,2}=6\pm8i$

[quote="Jimmy"]Just out of curiosity, what number was this?[/quote]

This was #3, which is in Part I.


In 2004, February 29th is a sunday. Syed wishes to keep and reuse his 2004 calendar in the next year in which Feb 29th is a sunday. Wat year'll that be?

[hide]The date moves one a year, two on leap years. So it moves five every four years. We need a factor of 4, 5, and 7. So $2144$ by which time Syed will probably be gone. [/hide]

[hide]February is the extra day of the leap year.  It occurs every 4 years. (except every 4thousandth :P).

In 1461 days, February 29th will be a Friday.  Therefore it is 5 days of the week later every leap year. LCM of 5, 7=35.

35/5=7.

7 leapyears=28 years.

Therefore in the year 2032[/hide]

I may be wrong

[hide]Instead of looking when Feb 29 will be a Sunday, look at when Feb 28 will be a Saturday. Every year, there will be 52 5/7 weeks. So every 7 years, Feb 28 will be a Saturday. But we want it to be a leap year, so it will happen every 28 years. 2004+28 = [b]2032[/b][/hide]

In my post,I mean 4 hundredth instead of f thousandth.


Convert the parametric equation to rectangular form.

x = e^t, y = e^(-t)

[quote="MellowMelon"]1. Stop posting trivial homework problems.
2. Why's this in the AMC forum?[/quote]

this is the amc forum?

It was probably moved here  :wink:

Too bad it wasn't moved to HSB

No, I'm pretty sure it wasn't moved. It didn't strike me as a homework problem though after reading sharkman's prev post, so I decided to take out the half a second to post the above response, even if in wrong forum.


vector
topology
real analysis
real analysis unsolved

Let $ S \subset R^n$ be an $ n \minus{} 1$ dimensional parametrized manifold. Given an orientation form for the surface $ S$ that is in $ \Lambda^{n \minus{} 1}(R^n)$, show that there is a $ \textbf{continuous}$ vector field $ \hat{n}: R^n \to R^n$  such that at each point of $ x \in S$, $ \hat{n}(x)$ is a vector of length one perpendicular to the tangent place of $ S$ at this point.


geometry
circumcircle
ratio
trigonometry
number theory
greatest common divisor
parallelogram

[color=darkred][size=150]Given a cyclic quadrilateral ABCD，AD intersect BC at E，AB intersect CD at F，AC intersect BD at G. H1，H2，H3，H4 is the orthocenter of ECD，EAB，GCD，GAB separately。Try to proof：H1H3H2H4 is a parallelogram.[/size][/color]

[color=darkblue][size=150]the picture is uploaded
I tried hard for using analytic geometry. but failed.
If you have some effective way to solve this by using analytic geometry,then will be better!(I am seeking...) :)[/size][/color]

[hide="Solution"]First I will change the names of points such that available for my solution.
[hide="Step 1"]
Let $ H_1, H_2, H_3, H_4$ be the orthocenters of triangles $ FBC, FBD,FAC,FAD$, respectively. We will show that $ H_1H_2H_4H_3$ is a parallelogram. $ (*)$
Note that $ H_1H_2$ and $ H_3H_4$ are perpendicular to $ FC$ so we show that $ H_1H_2 \equal{} H_3H_4$
$ H_1B \equal{} 2O_1P, H_2B \equal{} 2O_2Q, H_3A \equal{} 2O_3P, H_4A \equal{} 2O_4Q$
Therefore $ H_1H_2 \equal{} H_3H_4 \Leftrightarrow O_1O_3 \equal{} O_2O_4$
Since $ O_1O_3\perp FC, O_2O_4\perp FC$ we get $ O_1O_3//O_2O_4$
But $ O_1O_2\perp FB, O_3O_4\perp FB$ then $ O_1O_2//O_3O_4$
So $ O_1O_2O_4O_3$ is a parallelogram.
$ \Rightarrow O_1O_3 \equal{} O_2O_4$
Step 1 is done.[/hide]
[hide="Step 2"]Put $ \angle BCD \equal{} \alpha$
 Let $ H_5, H_6$ be the orthocenters of triangles $ AEB$ and $ EDC$ then prove that $ H_4H_5 \equal{} H_1H_6 (**)$
We have $ \frac {AH_5}{AH_4} \equal{} \frac {2O_5X}{2O_4W} \equal{} \frac {O_5A}{O_4A}$ (Because $ \angle XO_5E \equal{} \angle FO_4W$)
Since circumcenter and orthocenter are isogonal conjugate in a triangle we get $ \angle O_5AO_4 \equal{} \angle H_5AH_4$
$ \Rightarrow \Delta O_5AO_4\sim \Delta H_5AH_4$ with similar ratio is $ 2\cos \alpha$
Similar, $ \Delta H_1CH_6\sim \Delta O_1CO_6$ with similar ratio is $ 2\cos \alpha$
We only need to show that $ O_4O_5 \equal{} O_1O_6$
Let $ (O_5)\cap (O_4) \equal{} \{Q\}$. A well-known result: $ Q, O_1, O_4,O_5,O_6,O$ are concyclic. $ (1)$
$ \angle O_4QO_5 \equal{} \angle O_4AO_5 \equal{} 180^o \minus{} \alpha (2)$
Because $ O_6O\perp DC, O_1O\perp BC$ we obtain $ \angle O_1OO_6 \equal{} 180^o \minus{} \alpha (3)$
From $ (1), (2)$ and $ (3)$ we have $ O_1O_6 \equal{} O_4O_5$[/hide]
[hide="Step 3"]Let $ H_7, H_8$ be the orthocenters of triangle $ GAB$ and $ GCD$. We can prove that $ H_7H_2 \equal{} H_8H_3$ similar step 2. $ (***)$
Acording to [b]Steiner[/b] line we get $ H_1,H_4,H_5,H_6$ are collinear and $ H_2,H_3,H_7,H_8$ are collinear. $ (****)$
From $ (*), (**),(***)$ and $ (****)$ we are done.[/hide][/hide]

[u]http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=43&Id=17110[/u]
if you know chinese ,see the proof I mentioned in floor2,
sorry, I have little time recently, I will translate my solution when I have time

[b]Lemma.[/b] Steiner line of a cyclic quadrangle $ ABCD$ with circumcircle $ (O)$ goes through its diagonal intersection $ P.$

[b]Proof.[/b] Let $ M,N,L$ denote the midpoints of $ AD,AC,BD.$ $ \triangle MNL$ is pedal triangle of $ O$ WRT $ \triangle APD,$ thus the perpendiculars  from $ D$ and $ A$ to $ ML$ and $ MN$ meet at the isogonal conjugate $ H_1$ of $ O$ WRT $ \triangle PAD,$  which becomes orthocenter of the triangle bounded by the lines $ AB,AD,DC,$ since $ MN \parallel DC$ and $ ML \parallel AB.$ Then it follows that lines $ PH_1,$ $PO$ are isogonals WRT the angle between the diagonals $ AC,BD.$ In exactly the same way, we get  that the orthocenter $ H_2$ of the triangle bounded by the lines $ AB,$ $BC,$ $CD$ is the isogonal conjugate of $ O$ WRT $ \triangle PBC$ $\Longrightarrow$ lines $ PH_1$ and $ PH_2$ are identical $ \Longrightarrow$ Steiner line of the cyclic quadrangle $ ABCD$ goes through $ P.$
_________________________________

Let $ X,Y,Z,W$ be the orthocenters of $ \triangle EAC,\triangle EBD,\triangle AGD,\triangle BGC.$ They are collinear on the Steiner line  of the complete quadrangle $ EAGB$ and clearly $ H_2YH_1X$ and $ H_3WH_4Z$ are both parallelograms. From the previous lemma, the Steiner line $ H_1H_2$ of the cyclic quadrangle $ ABCD$ goes through $ G$ and since $ GW \parallel H_1Y$ and $ GZ \parallel H_1X,$ it follows that $ \triangle GWZ$ and $ \triangle H_1YX$ are centrally similar through the center $ S$ of the parallelogram $ H_2YH_1X.$ Hence they have $ H_1S$ as their common median on $ ZW$ and $ XY$ $ \Longrightarrow XY$ bisects $ H_3H_4$ and $ H_1H_2$ $ \Longrightarrow$ $ H_1H_3H_2H_4$ is a parallelogram.


If $ a,b$ be real numbers, then

$ \max_{c\in [0,1]}{\left|ac \plus{} b \plus{} \frac {1}{c \plus{} 1}\right|} \geq \frac {1}{4\left(3 \plus{} 2\sqrt {2}\right)},$

with equality if $ a\equal{}\frac{1}{2},b\equal{}\minus{}\frac{1}{4}\minus{}\frac{\sqrt{2}}{2},c\equal{}\sqrt{2}\minus{}1.$


limit
number theory unsolved
number theory

Let $ x_n$ there $ x_n = a^2n^2 + bn$ and $ a,b \in\ N$
find ${ lim{(\sqrt x})}$ with $ (m) = m - [m]$

$ [\sqrt {x_n}] \equal{} an \plus{} c$, were $ a^2n^2 \plus{} 2acn \plus{} c^2\le a^2n^2 \plus{} bn < a^2n^2 \plus{} 2a(c \plus{} 1)n \plus{} (c \plus{} 1)^2$.
It give $ 2ac < b\le 2a(c \plus{} 1)$ or $ c \equal{} [\frac {b \minus{} 1}{2a}], b\not \equal{} 0$ and
$ \{\sqrt {x_n}\} \equal{} (an \plus{} c)(\sqrt {1 \plus{} \frac {(b \minus{} 2ac)n \minus{} c^2}{(an \plus{} c)^2}} \minus{} 1) \minus{} an \minus{} c \equal{} \frac {b \minus{} 2ac}{2a} \plus{} O(\frac 1n).$
$ \lim_{n\to \infty }\{\sqrt{x_n}\}\equal{}\frac{b\%2a}{2a}$, were $ b\%2a\equal{}2a$ if $ 2a|b$ tlse $ b\%2a\equal{}b\mod 2a$.

[quote="Allnames"][quote="Yuriy Solovyov"]Ok: $ \frac {x_n}{a^2 n^2} = 1 + \frac {b}{a^2 n}$

$ \sqrt {\frac {x_n}{a^2 n^2}} = \sqrt {1 + \frac {b}{a^2 n}} = 1 + \frac {1}{2}\cdot \frac {b}{a^2}\cdot \frac {1}{n} + O(n^{ - 2})$

$ \sqrt {x_n} = an + \frac {b}{2a} + O(n^{ - 1})$

Finally: $ \lim_{n\to \infty}{\{\sqrt {x_n}\}} = \{\frac {b}{2a}\}$[/quote]
Sorry but what is $ O(n^{ - 1})$[/quote]

[url]http://en.wikipedia.org/wiki/Big_O_notation[/url]  :)

If $ 2a|n$, then $ \{\frac{b}{2f}\}\equal{}0$, but answer is 1.
For example $ a\equal{}1,b\equal{}2$.

Yes, you are right. We should take one more term of the series to see this:

$ \sqrt {\frac {x_n}{a^2 n^2}} \equal{} \sqrt {1 \plus{} \frac {b}{a^2 n}} \equal{} 1 \plus{} \frac {b}{2a^2 n} \minus{} \frac {b^2}{8a^4 n^2} \plus{} O(n^{ \minus{} 3})$

$ \sqrt {x_{n}} \equal{} an \plus{} \frac {b}{2a} \minus{} \frac {b^2}{8a^3 n} \plus{} O(n^{ \minus{} 2})$

When $ 2a|b$ then $ \left\{\sqrt {x_n}\right\} \equal{} \left\{Integer \minus{} \frac {\textbf{const}}{n}\right\} \to 1$ when $ n\to\infty.$


Hi, I had a major-related question:

Suppose that a person wants to become a professional mathematician and has just finished high school. Also note that the job prospects for a BS in mathematics are so bad the even though the person wants to pursue a PhD later on in another country, the family and friends of that person say that in the future that person might not be able to pursue a PhD in another country because of the harsh competition or for other problems that might occur in one's life. Anyways, the general idea (in our student's society) is that when one has a chance of getting a BS in engineering, one should go for it. Now suppose that if that person wanted to apply for a PhD in a country like say, the US, which of the options are better:

I. Double major in an engineering field (civil/industrial/chemical in a very good university or electrical/computer/mechanical in good, but less-known universities) with mathematics. Our imaginary student would take 6 years for completing this program.

II. Majoring in an engineering field (those mentioned above) and taking the core courses in mathematics (e.g. real analysis, abstract algebra). Then, taking the national entrance exam for the mathematics M.Sc. and after that, applying for the PhD. Our imaginary student would take 7 years for completing this program.

So, in the first option our applicant will finish with two BS degrees and in the second our applicant will have a BS in one field and an MS in another. In both cases, the student wants to either apply to a PhD or an MS program that has the option of entering the PhD program after the first year.

Since our student has only a few days to make his decision your immediate response would be deeply appreciated.

Also please note which engineering field would be more suitable (and easier!!) to study with mathematics.

What harsh competitions/other problems are you talking about?

If first option, always pick the very good one. It will makes you happy a lot more unless you are dead set in e/c/m major.

How good is the university in the second option?

Where are you from again? I don't understand why it takes 6 years to finish a degree.

P.S. I don't know a software (mathematica) can talk about this matter. I mean mathematica is not human you know. They (The company) could sue you for misrepresenting them. So, do change the user name if you want to be on the safe side (Unless you like to debate in court? That would be a nice thought).

The two options don't sound vastly different to me.  I guess one of them takes a year longer, you get a masters in math, and learn a bit more math.  My instincts say to go with option one.  If your goal is to get to the US to do a math PhD, you don't want to spend too long waiting to do it.  And I don't think that the masters degree makes it any easier to get admitted to a PhD program, because more years of math education just leads to a higher standard.  (Someone who does admissions at an unnamed prestigious school more or less told me this once.)

So really, you should try go to college for the minimum number of years (greater than or equal to 4) that will allow you to (1) obtain a solid undergraduate math education, and (2) convince some professors that you are very talented at math.  (Ideally, some professors with at least some background or interaction with mathematics in the US.)  Getting an engineering degree is okay, though.  I think many Americans do realize that most of the top students from certain countries get degrees in engineering.  Try to choose a prestigious specialty.

For your option II, it sounds a lot like a Pure Math Option we have for engineering students at the University of Waterloo, where it doesn't take any extra time to get your degree, but you get to take the fun math courses like Real Analysis.

That being said, is there any universities for your Undergrad that you could do something like that in?  That way you don't have to go through the 7 years, but you'd also get the degree and the courses you want.


function
abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ S$ be the set of all integer valued sequences $ (u_1, u_2, \ldots, u_k, \ldots)$, and $ e_i\equal{}(0,0, \ldots,0,1,0, \ldots)$, and $ f: S \to \mathbb{Z}$ a function satisfying $ f(u \plus{} v)\equal{}f(u)\plus{}f(v)$. Show that if $ F(e_i)\equal{}0$ for all $ i \geq 1$, then $ F$ is the trivial function.

this was here many times before, but without a clear proof ;).

let $ (a_i)$ be a integer-valued sequence. write $ a_i \equal{} 2^i c_i \plus{} 3^i d_i$ with integers $ c_i,d_i$. this is possible since $ 2,3$ and hence $ 2^i,3^i$ are coprime. since $ f$ vanishes on the $ e_i$, we have for every $ k$ the equation $ f((2^i c_i)) \equal{} f((2^i c_i)[k]) \equal{} 2^k f((2^{i \minus{} k} c_i)[k])$, where $ [k]$ indicates that the first $ k$ elements of the sequence are set to zero. since $ k$ is arbitrary, this implies $ f((2^i c_i)) \equal{} 0$. analoguously $ f((3^i d_i)) \equal{} 0$ and thus $ f((a_i)) \equal{} 0$.

besides, it is also possible to show that for all homomorphisms $ f : S \to \mathbb{Z}$ we have $ f(e_i) \equal{} 0$ for almost every $ i$. this shows that the dual of $ S$ is a free abelian group of countable rank.

Is this accurate?

First of all $ f(nu)\equal{}nf(u),n\in \mathbb {N}$.
Secondly $ (u_1,u_2,...,u_k,...)\equal{}\sum_{i\equal{}1}^{\infty} u_ie_i$.
And thirdly $ f((u_1,u_2,...,u_k,...))\equal{}f\left ( \sum_{i\equal{}1}^{\infty} u_ie_i \right ) \equal{}\sum_{i\equal{}1}^{\infty}f(u_ie_i)\equal{}\sum_{i\equal{}1}^{\infty}u_if(e_i)\equal{}0$

No, you cannot justify passing to the "infinite sum linearity" from "finite sum linearity". As a matter of fact, it doesn't necessarily hold: there are plenty of "finitely linear" mappings from $ S$ to $ \mathbb Q$ such that $ f(e_i)\equal{}0$ for all $ i$ but $ f\not\equiv 0$.

Exactly. I spent a while trying to prove it false by finding a basis- but the module isn't free.


Good job everyone who participated in MCTM!

I know some of the scores

6th grade-

Concepts- Stella 1st, Emily 9th
Problem Solving- Stella 2nd, Emily 3rd

Algebra

Individual- Shuyang 1st, Haohang 2nd
Team- Ladue (Victor, Christine, Haohang) 1st

Shut up! It's not like you couldn't just give a link to the website, which I hate.

MCTM is for losers. First place at MCTM doesn't really mean anything.

cuz we all know it's the AMAZINGLY AMAZING MATH STATE OF MISSOURI!!!!!!!

No.
blablabla cuz it was too short otherwise.

Well, if first place at MCTM meant anything about your actual math skill, then why did someone that didn't even place in problem solving beat you at nats. Also, how did someone that didn't even place in either concepts or problemsolving(math154) beat you at states?

Are you kidding me??? He pwned you 3-1 at cd!

Yeah but not written. Countdown$ \equal{}$Pointless In MO

Guys, states and everything in the past is over--it doesn't mean anything anymore. Pay attention to the future.

@mewto, Countdown=Pointless in MO would be better and more grammatically correct

[quote="MathSlayer4444"]Ah ha so that target DID kill me :noo: (as well as my lack of preparation)[/quote]
The target didn't [i]kill[/i] you, it slightly messed you up.

Number sense [i]killed[/i] me :(. Hmm oh well gj to you anyways, it shows you will be on the national mathcounts team next year ;).

[quote="MathSlayer4444"]True. I should have prepared [s]more[/s][/quote]
I definitely believe you. Of course you didn't prepare at all, and then you ended up doing really good. Then again, #SlayerIsAlsoPro, so maybe you used your pro-ness to do so well.

[quote="medianpi"]Yep.
#MedianpiIsTooLazyToStudyTherebyFailuresOccur
well at least u placed @slayer[/quote]
You placed too... :)

Oh wow look at the 2-man middle school team that got 1st...wow...they almost doubled the 2nd place team's (a 4-person team) score :o . Has anyone even [i]heard[/i] of Rockwood South before...? Wow...2pro...2pro...:noo: .

[quote="medianpi"]yep thats tru. but when u look at the overall standings, and me compared to people.....especially ladue...[/quote]
Yeah, but if you compare ME to people who got higher than me it means I got 100th percentile meaning I suck (which I do)

And it's also too bad that Alan didn't want to play "checkers" with us

[quote="number"]Oh wow look at the 2-man middle school team that got 1st...wow...they almost doubled the 2nd place team's (a 4-person team) score :o . Has anyone even [i]heard[/i] of Rockwood South before...? Wow...2pro...2pro... .[/quote]
:rotfl:

Lol, they messed up in the awards ceremony and accidently forgot to say Alan's name in the team. And now they have it messed up on the website! Okay, someone should seriously go to the Feedback / Report Errors and report that because that isn't fair for Alan :P.

I would have been :rotfl: -ing when they announced that had I not been mad at myself for screwing up Number Sense SOOOO badly.

[quote="coolrg"]Lol, they messed up in the awards ceremony and accidently forgot to say Alan's name in the team.[/quote][quote="coolrg"][...] screwing up Number Sense SOOOO badly.[/quote]
heh

(get it?)

[quote="coolrg"]
Lol, they messed up in the awards ceremony and accidently forgot to say Alan's name in the team. And now they have it messed up on the website! Okay, someone should seriously go to the Feedback / Report Errors and report that because that isn't fair for Alan.[/quote]
But Alan probably didn't do anything to benefit the team...his teammates did it all by themselves (I've heard that they made IMO a few times as well).

OMG look at the 5th place team in "Sweepstakes". Some random AP guy from Rockwood South (again?!) got a [b]0[/b] on team round but still managed to get 5th going solo. Look at the Team Borda: "92 94 71 71 [b]0[/b] 60 72 59 12 59 60 60 45 45 45..."
I think Rockwood South is up to something.

Good job, everyone!
http://mathleague.org/results/middleschool/indivresults.php?event=4045

Lenny was probably just trying to do it with his eyes closed, and Moses probably just guessed all the answers

Heh, national MATHCOUNTS is coming up.
GO MISERY

Wow that 8 on target 2pro, like only 3 people at CA got 8 on target -.-

@Below: Right I forgot oops

[quote=MathStudent2002]Wow that 8 on target 2pro, like only 3 people at CA got 8 on target -.-[/quote]

Well this is expected, MO > CA


calculus
integration
trigonometry
function
limit
calculus computations

calculate: $ \int_{\minus{}\infty}^{\plus{}\infty}\frac{x\cos x}{x^2\plus{}2x\plus{}10}\, dx$.

hello, i have got
$ \frac{1}{3}\pi(\sinh(3)\minus{}\cosh(3))(\minus{}3\sin(1)\plus{}\cos(1))$
Sonnhard.

The integral is $ Re\left( \int_{ \minus{} \infty}^{ \plus{} \infty}\frac {xe^{ix} }{x^2 \plus{} 2x \plus{} 10}\, dx\right).$
Introduce the complex function $ f(z)\equal{}\frac {ze^{iz}}{z^2 \plus{} 2z \plus{} 10}$ and integrate over the upper semicircle plus
$ (\minus{}\infty,\plus{}\infty).$ You get $ I\equal{}2\pi i \lim_{z\to\minus{}1\plus{}3i} {ze^{iz}\over (z\plus{}1\plus{}3i)}...$

you mean $ I\equal{}Re\left (2\pi i\lim_{z\to\minus{}1\plus{}3i}{ze^{iz}\over (z\plus{}1\plus{}3i)}\right )$?


percent
geometry
trigonometry
AoPS Books

Hey guys,

I have been nominated to participate in the state comprehensive tests. Since I have never done this before, I have some general questions.
[b]
1. What topics are on the test? (Is precal included? Roughly what percent of each topic is on the test?)

2. What is the normal score for placement of top-20 at state? (assuming that I will qualify for the state exam)

3. Are calculators allowed?[/b]

[b]4. What AoPS books or other math books would aide me the best in the following months and years for this competitition?[/b]

I know this is a lot of questions. (sorry) Thank you in advance. :D

From what I've seen...

Qualifying for state depends upon your region.  The top 6 or so qualify from each region.
Raleigh is hard, other places less so. Qualifying from some regions is relatively easy.
Without a working knowledge of the material contained in AoPS II, I'd venture to say you have no chance of sniffing the top 20 statewide. I think the state final contest has roughly 120 participants.
The NC state math contest website has previous years' exams to give you an idea of both questions and score distributions.
Good luck!

[quote="high_flyer10"]
[b]
1. What topics are on the test? (Is precal included? Roughly what percent of each topic is on the test?)
[/b][/quote]

Algebra, geometry, trig, combinatorics/number theory. Standard math contest stuff. Precal is included.

[quote][b]
2. What is the normal score for placement of top-20 at state? (assuming that I will qualify for the state exam)[/b][/quote]

umm... varies. Check out the results on the website? Last year I think 80-85 made top 20.

[quote][b]
3. Are calculators allowed?[/b][/quote]

Yes. EDIT: NO, APPARENTLY

[quote]
[b]4. What AoPS books or other math books would aide me the best in the following months and years for this competitition?[/b]

oh dang. lots of books help. hmm... I really don't know, I don't prepare specifically for state math contest (and therefore do suckily)... AoPS should help, and do old contests. What else?... Art and Craft of Problem Solving maybe... specific subject books (Titu and Zuming's 10x series come to mind... 102 combinatorial problems, etc, though these may be slightly higher-level)

also, Raleigh and Charlotte are usually the hardest regions to qualify from [usually only very (USAMO-qualifying? at least top 6/7 last year were at Wake Tech) strong contestants qualify].

bleargh I should actually not suck at state math contest this year :(
[/quote]

DAMIEN JIANG CALCULATORS ARE NOT PERMITTED THIS YEAR

[quote="high_flyer10"]Thanks for the feedback. A couple of things...

[url]http://courses.ncssm.edu/goebel/statecon/Inform/rules.htm[/url]

[quote]Calculator Policy:

NOTE: Beginning in 2009 the NC State High School Contest Comprehensive Division will not allow the use of any calculator.[/quote]

Is this true?

Also, how does the scoring system work at state? How much does the multiple choice worth and how much does the integer solutions worth and is there a guessing penalty? Thanks.[/quote]

WHOAH... So we can't use a calculator this year!?!? Wow, I had no idea. So are calculators banned at qualifying sites too? Can someone please confirm this? Maybe the state writers changed their minds.

Check the qualifying sites' websites for specifics on calculators. At state, multiple choice is 4 points for right, -1 for wrong, 0 for blank; short answer is 10 for right, 0 otherwise (correct me if I'm wrong? - this is what I remember)

I would like to add that this year, the test is going to be drastically different from the other years since calculators will not be allowed (in past years, many of the problems have required the use of a calculator).

Also, scoring, time limit, and the number of questions varies by year, depending upon the author of the test in that particular year.


linear algebra
matrix
algebra
polynomial
algorithm
vector
Ring Theory

Let $ M$ be a square matrix $ 2 \times 2$ over ring $ \mathbb{Z}[x]$. When there is two matrices $ P, Q \in GL(2, \mathbb{Z}[x])$ s.t. matrix $ PMQ$ upper triangular?

I can't help with upper triangular but even better is if you can get a diagonal matrix.  This requires two conditions:
1) the characteristic polynomial splits
2) for each eigenvalue $ \lambda$ of $ M$, the multiplicity of $ \lambda$ is equal to $ N(M\minus{}\lambda I)$

Then if $ Q$ is a matrix of eigenvectors, then $ Q^{\minus{}1}MQ$ is diagonal.

You can find it in 2946530.pdf

Sorry, I'm just some stupid noob. Please ignore me.

A matrix $ A$ over $ F$ is similar to an upper triangular matrix if and only if its characteristic polynomial $ p(t)$ splits completely over $ F$. In this case, the Jordan form works.

For the simpler upper triangular version, here's an algorithm:
Choose an eigenvalue $ \lambda$ of $ A$ and a corresponding eigenvector $ v$. Let $ B$ be a basis of $ F^n$ with $ v$ as its first element. Now $ B^{ \minus{} 1}AB$ has the form $ \begin{bmatrix}\lambda & * \\
0 & A'\end{bmatrix}$, where we don't care about $ *$ and $ A'$ is a $ (n \minus{} 1)\times(n \minus{} 1)$ matrix. The characteristic polynomial of $ A'$ is $ \frac {p(t)}{t \minus{} \lambda}$.
Repeat this process on $ A'$, and extend the new basis we get in this case using standard basis vectors: the next similarity uses $ \begin{bmatrix}1 & 0 \\
0 & B_2\end{bmatrix}$.
After $ n$ steps, $ A$ is similar to an upper triangular matrix with the eigenvalues in whatever order we wanted on the diagonal.

This argument also proves Schur's lemma: if we have an inner product and can normalize vectors, we can use orthonormal bases in the similarity.

On the other hand, this is actually a completely different question: allowing the two sides to be independent of each other, but only working with a ring.
We don't have a PID, but [url=http://en.wikipedia.org/wiki/Smith_normal_form]this is relevant[/url].

[Moderator edit: Fixed your LaTeX (you missed an \end{bmatrix}).]


limit
real analysis
real analysis unsolved

Find  f : $[0,\infty) \to [0,\infty)$ s.t
i) f is continuous in $[0,\infty)$
ii) f(f(x))=f(x) for all x $\in [0,\infty)$
iii) f(2x)=2f(x)
iv) $\lim_{x \to \infty}\frac{f(x)}{x}=1$

You are right. I make a serious mistake when create this problem.


trigonometry
inequalities
topology
absolute value

Let $ a, b, c, d$ be four real numbers, given
             $ a^2 \plus{} b^2 \equal{} 1$
             $ c^2 \plus{} d^2 \equal{} 1$,
             $ a*c \plus{} b*d \equal{} 0$,

then
             $ a^2 \plus{} c^2 \equal{} 1$,
             $ b^2 \plus{} d^2 \equal{} 1$,
             $ a*b \plus{} c*d \equal{} 0$.

[I meant four real numbers]

[hide="Hint"]The first two equations can be rewritten as:
$ \left \{\begin{array}{c} a = \cos \theta & b = \sin \theta & c = \cos \phi & d = \sin \phi \end{array} \right \;$

[b]Questions:[/b]

1. Why can we do this?
2. Using this relation, what does the equation $ ac + bd = 0$ mean?
3. Can we use this to write $ c$ and $ d$ in terms of $ \theta$?
4. If so, how does this help us prove what we want to prove?[/hide]

[quote="sexynsmartjenny"]Let $ a, b, c, d$ be three real numbers.[/quote]
Do you mean only three of them are real numbers?

TZF: How do we know that the absolute value those variables is less than or equal to 1?

Because their squares are less than one.

[quote]How do we know that the absolute value those variables is less than or equal to 1?
[/quote]

Let's assume that at least one of $ a$ or $ b$ is greater than 1. Also assume that $ a$ and $ b$ are real numbers.

WLOG assume $ |a| > 1$ and consider the value of $ b$.

$ a^2 \plus{} b^2 \equal{} 1 \implies b^2 \equal{} 1 \minus{} a^2$.

Now, we know that $ |a| > 1$, so $ a^2 > 1$, so the RHS is negative. This implies that $ b^2 < 0$, but this is contradicted by the trivial inequality, $ n^2 > 0 \mid n \in \mathbb R$.

Thus both $ |a|$ and $ |b|$ must be $ \le 1$.

Oops, I was reading $ a^2\minus{}b^2\equal{}1$.

this is so simple.  This is like high school stuffs, how do they expect college graduates who already took in abstract algebra and topology to solve this kind of problem?


inequalities
number theory
inequalities theorems

if $ a_1\leq a_2 \leq ... \leq a_n$
  and  $ b_1\leq b_2 \leq ... \leq b_n$  ($ a_i,b_i$ reals)
and $ (x_1,x_2,...x_n)$ is a permutation of $ (b_1,b_2,...,b_n)$
then
$ a_1b_1\plus{}a_2b_2\plus{}...\plus{}a_nb_n\geq a_1x_1\plus{}a_2x_2\plus{}...\plus{}a_nx_n$

1)Can you proove it?
2)when the equality holds?

Write the permutation as a product of transpositions and show that a transposition decreases the value of $ a_1 x_1 \plus{} ... \plus{} a_n x_n$.

use adjustment
for$ j\geqslant i$  and $ m\geqslant n$ we have the following
$ a_{m}j\plus{}a_{n}i\geqslant a_{m}i\plus{}a_{n}j$

[quote="amitsaha"]:wink: Hey I need a e-book on rearrengement inequality or inequalities in general. Can anyone tell me that?[/quote]
I believe [i]Introductions to Number Theory and Inequalities[/i] has olympiad inequalities, although I haven't read it. I also think [i]An Introduction to Inequalities[/i] could help, but it's more towards intermediate level.


search
real analysis
real analysis unsolved

[b][size=150]Let f : R → R be a differentiable function. Then f′(x) is continuous: true or false[/size][/b]

Answer for your question is [hide="here"]http://www.mathlinks.ro/viewtopic.php?search_id=464182958&t=265572[/hide]


Okay, here goes:
For every non-negative real numbers $ d,e,f$, show that: $ 2(d \plus{} e \plus{} f)\geq 3\sqrt [3]{(d \plus{} e)(e \plus{} f)(f \plus{} d)}$

[b]EDIT: i made a typo, sorry. The last term in the radical should be (f+d) insted of (f+e)[/b]

[quote="Papalakis"]Okay, here goes:
For every non-negative real numbers $ d,e,f$, show that: $ 2(d \plus{} e \plus{} f)\geq 3\sqrt [3]{(d \plus{} e)(e \plus{} f)(f \plus{} e)}$[/quote]
Try $ d\equal{}f\equal{}0$ and $ e\equal{}1.$ :wink:

Actually i created this prob and i already have a one  page solution but i wanted to see if it had any flaws/mistakes and if there where any other solutions to it, and as you proved there are :) my solution is probably just too sloppy  :P

Are you sure you didn't think of

$ 2(d \plus{} e \plus{} f)\geq 3\sqrt [3]{(d \plus{} e)(e \plus{} f)(f \plus{} d)}$ ?

[EDIT: I see you've seen it  :lol:  ]

Now, it's quite easy.

$ \frac{(d\plus{}e)\plus{}(e\plus{}f)\plus{}(d\plus{}f)}{3}\geq\sqrt[3]{(d\plus{}e)(e\plus{}f)(d\plus{}f)}$ directly from AG.

Ahh yes :) Thank you for that simple solve. Besides, all i did was build it on AM-AG, substitute some stuff, bring it to a certain form and voila! but your solution way more simpler :D


linear algebra
matrix
algebra
polynomial
vector
calculus
integration

Prove the following theorem:

Let $A \in M_{n,p}(\mathbb{C})$ and $B\in M_{p,n}(\mathbb{C})$. Then $AB$ and $BA$ share the same invertible part.
In other word, $AB \cong N \oplus C$ and $BA \cong M \oplus C$ for some nilpotent square matrices $N$ and $M$ and some same invertible square matrix $C$.
($\cong$ means similarity. and $\oplus$ means matrix direct sum.)

Remark: The [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5704]characteristic polynomial equation of AB and BA[/url], $t^{p}\cdot ch_{AB}(t)=t^{n}\cdot ch_{BA}(t)$ can be deduced from the above theorem.

The proof idea is to use $(AB)A = A(BA)$.

Some convention: Think of a matrix $E\in M_{r,s}(\mathbb{C})$ as a linear transformation $\mathbb{C}^{r}\to \mathbb{C}^{s}$ where it acts by right-multiplication on row vectors in $\mathbb{C}^{r}$.

Let $V_{i}= \mathbb{C}^{n}(AB)^{i}= \{v(AB)^{i}| v \in \mathbb{C}^{n}\},\ i = 0,1,2,\cdots$. If $V_{n}\neq V_{n+1}$, then $V_{n+1}\subsetneq V_{n}\subsetneq V_{n-1}\subsetneq \cdots \subsetneq V_{0}= \mathbb{C}^{n}$, which is impossible. Therefore we have $V = V_{n}= V_{n+1}= V_{n+2}= \cdots$. Similarly, let $W = W_{p}= W_{p+1}= \cdots, \ W_{j}=\mathbb{C}^{p}(BA)^{j}$.

$V$ is the part where the invertible part of $AB$ lives on.  $V'= V'_{n}= V'_{n+1}= \cdots, \ V'_{i}= \ker( (AB)^{i})$ is the part where the nilpotent part lives on. i.e. $\mathbb{C}^{n}= V \oplus V',\ V'AB\subset V',\ V'(AB)^{n}= 0,\ VAB = V$.

Now we want to prove (*) $VA=W,\ \dim(V) = \dim(W)$.
Proof of (*):
$VA = \mathbb{C}^{n}(AB)^{n+p}A = (\mathbb{C}^{n}A)(BA)^{n+p}\subset \mathbb{C}^{p}(BA)^{n+p}= W$.
Similarly, $WB \subset V$.
Now, $V = VAB \subset WB \subset V$, hence $WB = V$ and $\dim(W)\geq\dim(V)$. Similarly, $VA=W$ and $\dim(V)\geq\dim(W)$.

Since (AB)A=A(BA) and (*), the conclusion of the theorem follows.

Remark:
* The theorem is true for any field. Same proof.
* V and V' are called eventual image and eventual kernal of AB
* This proof is similar to the proof of the fact that : The similarity class of the invertible part of an integral square matrix A is an invariant of [url=http://www.math.ku.dk/ma/kurser/symbolskdynamik/konjug/node12.html]shift equivalence[/url]. i.e. If A and B are shift equivalent, their invertible parts are similar. It's one of the fundamental results in [url=http://en.wikipedia.org/wiki/Symbolic_dynamics]symbolic dynamics[/url].


What's your religion? Why do you or why do you not believe? Also, what branch of each religion? For example for Christianity: Cathlolic, Orthodox, Jehova's Witness, etc.

Why isn't there a "none" in your poll? :P

I really don't know if there is a god, and honestly I don't care at all. Even if there were one, how could he blame us for not knowing it? We're humans, not gods! :D

Sorry about that!  :blush: I forgot to add None. After all, I don't believe either! But you can just say that you don't believe in a reply. ;)

Atheist, but not in the traditional sense.  Basically, I have no religion.  I suppose you could call it existentialism, but not quite.  Semi-existentialism, perhaps.

Roman Catholic, attending a Roman Catholic school

I also don't believe. What does existentialism mean?

Well, I'm no expert, but existentialism is essentially the belief that life has no meaning, and that humans have no purpose other than to be.  Existentialists believe that goals such as success or enlightenment.  They feel that they should [i]seize the day[/i] to enjoy life in itself without infringnig on the rights of others to enjoy their lives (the reason why existentialists don't succumb to their every fleeting passion, as some theists occasionally assume of them and other atheists).  I would be an existentialist, but for a few reasons, the two greatest ones being:

1. I believe that success does have a purpose: so that when one desires to seize the day later (delayed gratification), one will have more options to seize.

2. I hold a code of ethics beyond simply not infringing on the ability of others to enjoy themselves.  I take many values generally referred to as "Christian" (i.e., no swearing, drinking, smoking, premarital sex, or other destructive behaviors) to heart, but primarily for reasons based in science or personal well-being.

I am not entirely certain that my description of existentialism is quite correct, so if anyone notices an error or oversight, please correct me.

I am a Roman Catholic Christian but I haven't been confirmed yet.

Weak agnostic. I see no proof for or against the existence of God, but I won't say it is impossible to prove whether or not God exists (that would make me a strong agnostic).
P.S. RobinHood3000, what do you mean by the 'traditional' sense of atheism?

i think i count as atheist/agnostic/none

[quote="RobinHood3000"]When I say not in the "traditional" sense of atheism, I mean that I don't disagree with the institution of religion as a whole.  I think religion has a more or less positive influence on the general public, and I don't attempt to convert others to my brand of atheism.  Explain, yes, but not convert.  I just don't find religion the right thing for me, and sometimes get irritated when some of the more enthusiastic Christians refer to other athestic philosophies such as Buddhism and Confucianism as "fairy-tale religions," (sad, but true) and come to view me as an amoral sociopath upon learning that I am an atheist.[/quote]
Well the vast majority of athiests are perfectly fine with religion...it's just that those who aren't are usually the most loudmouthed. Yeah, it's too bad how some conservatives of any religion can't respect beliefs different from their own...which is why I post this website whenever I have a reasonable excuse...
http://www.religioustolerance.org
It's an amazing site, though I don't think it would change the feelings of any person close-minded enough to come to the conclusion that religions different from their own were inferior in the first place...

Wow, that's an incredible site.  Thanks for the link, it is greatly appreciated.

I was born a Buddhist, but I don't really practice my religion or anything.

im an antheist, I don't believe in the existence of "god(s)"....

Wow, the atheist/religion-less population seems greater than I thought.

To hear some people tell it, one would think the world was divided into Christians and blasphemous heathens.  Welcome to BH Anonymous!

Maybe the community of math-lovers is considered a biased sample when it comes to religion...even though there is no reason why that should be...

Of course there is a reason as to why math lovers should produce a biased sample in this respect. Math people tend to be more logical than an average bear and more scientifically minded. Therefore these people are more likely to question religions that are not based on scientific evidence.

[quote="gameworld7"]um question, why do people even care about asking for others religion? shouldn't we be talking about math? its a math forum (even though this is fun and games fact.)[/quote]

WOW!!!!!! The irony isn't just dripping, not even overflowing. It's being poured out! Look who's telling other people we should be talking about math!!!!!!

[quote="Jockey"]im a religous jew, not yet sure whether i want to be conservative, modern orthodox, or even charedi. you can definitely be logical and religous. its a common misconception nowadays that you can't, but leibnitz, newton, euler, were very religous christians.[/quote]
That's not what ComplexZeta meant...he said 'more likely' to question accepted religions. 'More likely' to be skeptical of things, whether or not being skeptical is reasonable. 'More likely' doesn't mean everyone, just a plurality.

[quote="gameworld7"]um question, why do people even care about asking for others religion? shouldn't we be talking about math? its a math forum (even though this is fun and games fact.)[/quote]
Uh, Fun and Games was invented specifically so we could have some time off from math. All forums have that.

I have no religion. I don't especially believe in supreme beings, but I do believe in fate. I tend to lean towards Simfish's explanation. I read Cat's Cradle this summer, and it struck a pretty good point about religion. It stated taht all religions are harmless lies. It's good to have faith in something, especially if there are things you cannot explain. I always find it interesting to talk to my friends about religion and our views.

I've misquoted it before, and I'll misquote it again:

"Religion is the opiate of society."

I agree for the most part with Simfish, especially that religion is an overall positive influence on society.  I just don't think that organized religion is for me.

[quote="Peter VDD"]We're humans, not gods! :D[/quote]
Thx Pete!!! That makes math competitions much more exciting than if we knew it all. :D

BTW I'm sorta Buddhist  :cool:  :blush:  ;)

Of course I'm cool.  I'm not starting a flame war, either.

If I had, there'd be an exclamation point after the question mark in my post.

[quote="3X.lich"][quote="RobinHood3000"][quote="3X.lich"]BTW I'm sorta Buddhist  :cool:  :blush:  ;)[/quote]

What's wrong with being Buddhist?[/quote]

What's wrong with [b]not[/b] being Buddhist? I'm not starting a flame war but my sentence 'BTW I'm sorta Buddhist' does [b]not[/b] imply if there was something wrong with being a Buddhist. Stay cool please. :cool:  :cool:  :cool:

To clear things up, I said I'm sort of Buddhist because: 1) I have only learned bits of the Dharma;  2) I actually started learning more of Buddhism recently and never paid attention to it before.[/quote]

I think the reaction was based much on the  :blush:  smilie.
It originally looked as if you were kind of embarrased of being Buddhist. Good job on clarifying though.

I find that each unexplained thing that we figure out often leads to new unexplained things.

The more I learn about advanced mathematics and physics, the stronger is my feeling that we won't be able to satisfyingly explain everything. Think of Godel's Incompleteness Theorem. The difficult philosophical questions of the form and structure of mathematics. In physics: what is time? Causality?

I'm not religious at all. But I still find the need for faith. And in some way, maybe you can consider the unknown to be "divine." There is always something beyond the limits of our knowledge. How we reconcile our individual views of the border and our guesses about what's beyond are our own business.

I am a Sikh... it is a kind of large religion. It originated in India. Similar to Buddhism.

proud atheist :lol:

maybe a community of problem solvers are more inclined to decide for themselves what they believe, as opposed to simply following what their parents or perceived authority figures told them.

[quote="jzzsxm"][quote="Simfish"]Atheist.

People invented religion to help describe what they thought undescribable. Now that we can describe what was previously undescribable with science, I now believe that religion is merely a human invention. Especially considering that many religions treat Earth as the center of the universe when in reality, we are just on a mere planet out of many.

Now, some may take that as implying that religion is preposterous. Let me just say that I do not wish anything wrong with religion, and that I wish I believed in a religion. However, in the words of Marie Curie, "I want to believe, but I just can't, can't, can't."[/quote]

This is my exact belief.  Do you have any idea how many times I've tried to explain this to people only to have them just smile and nod?????  Man invented religion to explain things.  We can explain them now!  Religion is just sticking around to provide people hope and good morals, both of which can be achieved without religion.[/quote]

*applauding* :clap2: :clap2: :clap2:

I'm a mix between Christianity and Penguinism ;) , but I am not very religeous, so I picked none.

The smarter people you take a sampling off, the less religious they usually are.
http://en.wikipedia.org/wiki/Religion_and_intelligence

I believe in God (not a god), but I don't belong to any religion.

[quote="JBL"]What?!  Pastafarianism (Flying Spaghetti Mosterism) isn't included?  Outrage!

Sikhism is more like Hinduism than Buddhism, no?

What on earth does it mean to say "I believe in God (not a god)"?  Is it purely a reference to the fact that you happen to think that the god you believe in deserves a capital "g"?[/quote]

That means I believe in the God that they believe in in Judiaism, Christianity, Islam, etc. not just the existence of a god.


geometry
circumcircle
cyclic quadrilateral
projective geometry
geometry unsolved
circumscribed quadrilateral

A quadrangle $PQRS$ has an in-center $O$ (which implies $PQ+RS=QR+SP$). $T=PR\cap QS$. $H_{A}, H_{B}, H_{C}, H_{D}$ are respectively the orthocenters of $POQ, QOR, ROS, SOP$. Prove that $T, H_{A}, H_{B}, H_{C}, H_{D}$ are collinear.

I have solved it by Meallous( :?: Idon't know how to spell) Theorem,But very long. :blush:

Can you give the main idea by using "Menelaus" theorem?

I have found another way to solve it .Please wait me and I will try to complete my solution,and post it.I am busy now....

Thank you. And I'm looking forward to it.

The proof is quite short. Let A, B, C, D be the tangency points of PQ, QR, RS, SP with the quadrilateral incircle (O). The tangential quadrilateral PQRS and its contact cyclic quadrilateral ABCD have the same diagonal intersection $T \equiv PR \cap QS \equiv AC \cap BD.$ The tangency points A, B of PQ, QR with the incircle (O) are feet of the O-altitudes of the triangles $\triangle POQ,\ \triangle QOR,$ respectively. The quadrilateral AOBQ is a cyclic kite on account of OA = OB, QA = QB, and the right angles $\angle OAQ = \angle OBQ = 90^\circ,$ let (Z) be the its circumcircle with diameter OQ. Let X, Y be feet of the Q-altitudes of the triangles $\triangle POQ,\ \triangle QOR,$ respectively. The angles $\angle OXQ = \angle OYQ = 90^\circ$ are both right, hence $X,\ Y \in (Z).$ Since AOBQ is a kite with OA = OB, OQ bisects $\angle AOB$ and

$\angle ADB = \frac{\angle AOB}{2}= \angle QOB = \angle QXB$

Since in addition $QX \parallel AD$ (both perpendicular to OP), the points B, X, D are collinear and the lines $BX \equiv BD$ are identical. Similarly, we can show that the points A, Y, C are collinear and the lines $AY \equiv AC$ are identical. Thus $T \equiv AY \cap BX.$ By Pascal's theorem for the cyclic hexagon AQBOXY, the intersections $H_{A}\equiv OA \cap QX,$ $H_{B}\equiv OB \cap QY,$ $T \equiv AY \cap BX$ are collinear. In exactly the same way, we can show that $H_{B},\ H_{C},\ T$ are collinear, $H_{C},\ H_{D},\ T$ are collinear, and $H_{D},\ H_{A},\ T$ are collinear. As a result, all 5 points $H_{A},\ H_{B},\ H_{C},\ H_{D},\ T$ are collinear.

This is now Theorem 18 [b]a)[/b] in [url=http://www.cip.ifi.lmu.de/~grinberg/CircumRev.pdf]"Circumscribed quadrilaterals revisited"[/url]. The proof is more elementary than Yetti's above (Pascal is not used), but more configuration-dependent, so I prefer Yetti's. What was your proof, Hawk Tiger?

  darij

It's known that $ \frac{PT}{TR}\equal{}\frac{PA}{RC}$, $ \frac{QT}{TS}\equal{}\frac{QB}{SD}$. We have $ \angle AOQ\equal{}\angle BOQ$, so $ \angle APH_A\equal{}\angle BRH_B$. It implies that $ \triangle{PAH_A}\sim\triangle{RBH_B}$, i.e. $ \frac{PA}{RC}\equal{}\frac{PH_A}{RH_B}$. And notice that $ PH_A\parallel RH_B$ and $ \frac{PA}{RB}\equal{}\frac{PT}{RT}$, so by the Thales' theorem, we have $ H_A$, $ T$ and $ H_B$ are collinear.

Similarly, $ H_B$, $ T$ and $ H_C$ are collinear; $ H_C$, $ T$ and $ H_D$ are collinear. Hence $ H_1$, $ H_2$, $ H_3$, $ H_4$ and $ E$ are lie on a line. And our solution is completed.

This line has a very interesting property:
Let $ d$ the line pass through O,midpoint of PR and midpoint of PS and $ d'$ be the line pass through $ T,H_1H_2H_3H_4$.Then $ d \perp d'$  :D

[b]Lemma.[/b] If $ M$ and $ N$ are the midpoints of the sides $ AB,AC$ of  $\triangle ABC,$ then $ MN$ is the polar of the orthocenters of the triangles $ \triangle IBC , \triangle I_aBC$ with respect to the incircle $(I)$ and the A-excircle $(I_a),$ respectively.

Let $\omega$ denote  the incircle of $ PQRS,$ $ X \equiv{} PQ \cap SR$ and $ Y \equiv{} QR \cap PS.$ Line $ XY$ is the polar of $ T$ WRT  $\omega$ and according to the lemma, polars of $ H_A, H_B,H_C, H_D$ are the lines $ p,q,r,s$ joining the midpoints of $ (YQ,YP),$ $(XQ,XR),$ $(YR,YS),$ $ (XP,XS).$ Since $ p,q,r,s,XY$ concur at  the midpoint $U$ of $ XY,$ then $ H_A,H_B,H_C,H_D,T$  are collinear on the polar of $U$ WRT $\omega,$ as desired.


ratio
geometry
geometric transformation
reflection
circumcircle
inradius
trigonometry

Let $ABC$ be a right triangle where $\measuredangle A = 90^\circ$ and $M\in (AB)$ such that $\frac{AM}{MB}=3\sqrt{3}-4$. It is known that the symmetric point of $M$with respect to the line $GI$ lies on $AC$. Find the measure of $\measuredangle B$.

$\frac{AM}{c}= \frac{AM}{AM+MB}= \frac{3 \sqrt{3}-4}{3 \sqrt{3}-3}= \frac{5-\sqrt{3}}{6}$

Let K be the midpoint of IG and $N \in AC$  the reflection of M in K. $\triangle AMN$ is right, the midpoint K of its hypotenuse MN is its circumcenter, AK = MK. Let P be the midpoint of AM, $\frac{AM}{2}= AP = \frac{\frac{c}{3}+r}{2},$ where $r = \frac{bc}{a+b+c}$ is inradius of the right $\triangle ABC.$ Hence $\frac{AM}{c}= \frac{1}{3}+\frac{b}{a+b+c}= \frac{5-\sqrt{3}}{6},$ $\frac{b}{a+b+c}= \frac{3-\sqrt{3}}{6}.$ Solving $x+1+\sqrt{x^{2}-1}= \frac{a+b+c}{b}= 3+\sqrt{3},$ where $x = \frac{a}{b}= \frac{1}{\sin B},$ yields x = 2, $\sin B = \frac{1}{2},\ B = 30^\circ.$


floor function
pigeonhole principle
search
function
limit
integration
algebra

Prove or disprove:

Given $\alpha$, an irrational real number, the set $\{ k\alpha - \lfloor k\alpha \rfloor : k \in \mathbb{N}\}$ is dense in $[0,1]$.

Are there many ways of seeing this?

This is certainly true. Some people refer to this result as the "Kronecker density theorem". Hm, you could try to give a proof using the pigeonhole principle for example :)

search to Weyl criterion on the forum. It has already be discussed many times, I think ..

One of the many ways to prove this it to actually prove something much stronger - that the multiples of $k\alpha$ are not merely dense in the circle ($\mathbb{T}=\mathbb{R}/\mathbb{Z});$ they are uniformly distributed in the following sense:

If $f$ is a continuous function periodic of period $1,$ then

$\lim_{n\to\infty}\frac1n\sum_{k=1}^nf(k\alpha)=\int_0^1f(x)\,dx.$

You can do this in two steps. The first is to verify that this is true whenever $f$ is a trigonometric polynomial - that is, when $f(x)=\sum_{j=-m}^ma_je^{2\pi ijx}.$ The second is to use the trigonometric version of the Weierstrass approximation theorem, namely that any continuous function can be uinformly approximated arbitrarily well by a trigonometric polynomial.

The need for $\alpha$ to be irrational will be clear in the first step, the explicit calculation for trigonometric polynomials.

You may use the (elementary) theorem from analysis that every bounded sequence has a convergent subsequence.  This means that there exist $i,j$ such that $\{\alpha(i-j)\}<\epsilon$ (where $\{\}$ denotes the fractional part).  From there, to get arbitrarily close to our real number, take multiples.

I posted a proof a couple of months ago, but don't really feel like looking for it again.


Let $a,b,c\ge 0$. Prove that $ab^{2}+bc^{2}+ca^{2}\ge \sqrt[3]{abc}(ab+bc+ca)$.

[hide="Solution"]Substitute $x=ab^{2}$, $y=bc^{2}$, $z=ca^{2}$, then the inequality becomes $x+y+z\geqslant \sum x^{\frac{2}{3}}z^{\frac{1}{3}}$ which is trivial by am-gm. [/hide]


A large cube's faces are painted, then separated into unit cubes. 27 unit cubes have no faces painted. How many unit cubes are there in total?

Please explain this simply -- I have no geometry background.

think of the 27 cubes unpainted as itself. so it has side length 3. now picture a shell completely covering it. the shell must have one extra cube on each side (front&back) of the 27 cubes in the middle. so the answer is (3+2)^3=125


ARML
ratio
number theory
relatively prime
geometry
angle bisector

Source: ARML 1989 T4.
T4. In triangle $ABC$, angle bisectors $\bar{AD}$ and $\bar{BE}$ intersect at $P$. If $a=3$, $b=5$, $c=7$, $BP=x$, and $PE=y$, compute the ratio $x: y$, where $x$ and $y$ are relatively prime integers.

Oops, never mind for my diagram.. this is the first time I attached diagram... :blush:

This is the FOURTH time I saw the exact same problem....

[hide="Hint"]
know your angle bisector theorem and use it like crazy (or strategically, depends if you see it or not)
[/hide]

[quote="me@home"][quote="TheAmazingOne"][hide]
We know that

$\frac{AE}{CE} = \frac{AB}{BC}$
$\frac{AE}{5-AE} = \frac{7}{3}$
$AE = \frac{7}{2}$

And we know that

$\frac{BP}{PE} = \frac{AB}{AE}$
$\frac{x}{y} = \frac{7}{7/2}$
$x: y = 2: 1$
[/hide][/quote]
how do we know this: $\frac{BP}{PE} = \frac{AB}{AE}$?[/quote]

I think it mostly likely is because of the Angle Bisector Theorem.


Who here does a lot in Excel, I am trying to study it write now, and even though I use a textbook (which is old) I often get into some trouble and cant perform some commands. Like I cant create an outline, it says cannot create an outline, can someone help? I will probably have many other questions later on. :)  :(  :?  :D  :lol:  ;)  :blush:  :P  :o  :mad:  :oops:  :roll:  :!:  :?:  :idea:  :|  :rotfl:  :huh:  :ninja:

Your smilies are scaring me, but I do a little bit of Excel...if you need help, look in the "help" toolbar on Excel ;) I would tell you, but I haven't done it in a while.

That doesnt help me, I know about help in excel. Why do the emoticons scare you?

i learned how to use Excel 2000 back in middle school
now i'm learning the the new one (2004)
totally new design and everything

so u r saying it is very different? How about hte operations? Are you very good in it though, if I want to ask u some questions is that fine will u know?

i will try, but i not that good at it, bc i dont use it much.
and BTW my ed. is for Mac, so don't know about Windows.  i believe they should be identical

What I am struggling on now is Outlining, for some reason I cant perform the operation: in the toolbars dialog box, highlight the contents of the toolbar name edit box, type outlining, and then choose the New button. I dont know how to get to the toolbar dialog box. Can someone please help? Thanks

I use MS Excel to calculate my school status. I use some functions (easy ones like AVERAGE, SUM, ROUND.. etc.) to calculate my means. It's also useful for some graphics. :)

[quote="hello"]Why do the emoticons scare you?[/quote]You definitely use too many of them. Please try to limit their numbers, so that you can spare me the trouble of creating a script that does that automatically  ;)

I saw more emoticons in a post :). Usually girls are using a lot of them :P


We have that $\sum_{cyc}\left(a(a-1)+\frac{1}{4}\right)^{2}\geq 0$ so we have $\sum_{cyc}a^{2}-\sum_{cyc}a+1\geq 0$. :lol:


inequalities
number theory proposed
number theory

Prove that \[\pi(n!+2n)+\pi(n)\leq\pi(n!+n)+\pi(2n)\forall n\geq 2,\] where $\pi(x)$ are the number of primes which does not exceed $x$.

You must be kidding that this is analytic number theory! It's just a small counting argument!

Harazi, this is an easy problem just for warming up. Don't worry, mr. Panaitopol give us in course some other hard problems. I will post them. :wink:

Please post them, we really need some nice problems on this forum.

It is obviosly. If p prime, suth that $n!+n<p\le n!+2n$ then $p-n!=q$ prime, were $n<q\le 2n$, therefore $\pi(n!+2n)-\pi(n!+n)\le \pi(2n)-\pi(n).$
In this problem $n!$ can be change to \[a\prod_{p\le n}p, a\in N\] and $2n$ to $r: \ n<r< s^{2}$, were $s$ is least prime $s>n$.


Construct equilateral triangle $ ABC$ which is inscribed in $ (O)$. Let $ C'$ be the projection of $ C$ wrt $ O$. Construct incenter $ I$ of triangle $ BCC'$. The line through $ I$ and perpendicular to $ OA$ intersects $ OB$ at $ Y$. The line through $ Y$ and perpendicular to $ OB$ intersects $ OA$ at $ I_1$. Construct $ (I_1,I_1Y)$. Similarly for others circles and we are done.


Im having problem with this one....Could you give me some hint,plz

[b]Problem:[/b] Let $ a,b,c$ be positive real numbers satisfying $ abc \equal{} 1$. Prove that
$ \sum_{cyc} \frac {1}{\sqrt {b \plus{} \frac {1}{a} \plus{} \frac {1}{2}}} \ge \sqrt2 \quad (1)$

[b]Proof:[/b]
[hide]Let $ a \equal{} \frac {x}{y}, b \equal{} \frac {y}{z}, c \equal{} \frac {z}{x}$. Then we can rewrite $ (1)$ as
$ \sum_{cyc} \frac {1}{\sqrt {\frac {y}{z} \plus{} \frac {y}{x} \plus{} \frac {1}{2}}} \equal{} \sum_{cyc} \sqrt {\frac {2xz}{2xy \plus{} 2yz \plus{} xz} }$
Now the inequality is homogenous, so we may assume $ xy \plus{} yz \plus{} zx \equal{} 1$ and therefore $ \sum_{cyc} \sqrt {\frac {2xz}{2xy \plus{} 2yz \plus{} xz} } \equal{} \sum_{cyc} \sqrt {\frac {2xz}{2 \minus{} xz} }$. Now, for $ n \in [0,2]$
$ \sqrt {\frac {2n}{2 \minus{} n}} \ge \sqrt2 n \Leftrightarrow (n \minus{} 1)^2 \ge 0$. So $ \sum_{cyc} \sqrt {\frac {2xz}{2 \minus{} xz} } \ge \sum_{cyc} \sqrt2 xz \equal{} \sqrt2$.[/hide]

Edit: did this problem appear in any competition or where did you find it?

Thank you very much!!!

So,when you have homogenous inequallity you can always put symetric relation like x+y+z=2, xy+yz+zx=1,...?
This problem is from Serbia and Montenegro federal competition 2005, I-st grade i think...

@Smilley
[quote]So,when you have homogenous inequallity you can always put symetric relation like x+y+z=2, xy+yz+zx=1,...? [/quote]
Doing so is called Normalization.
Sometimes, you yourself can go for substitutions like $ a\equal{}\dfrac{x}{y}$  and similarly for $ b$ and $ c$. In that way, you can normalize $ abc\equal{}1$. Similalry, there are several other ways you can normalize in inequalties.
What is Ist grade in Serbia Olympiad?
[/quote]

[quote="Mathias_DK"][quote="Smilley"][code]\sqrt {\frac {2n}{2 - n}} \ge \sqrt2 n \Leftrightarrow (n - 1)^2 \ge 0[/code] this isn't good because left side of equivalence is equivalent to  n(n-2)>=0 which isn't true ...[/quote]
Nope.. It's equivalent to $ 2n(n \minus{} 1)^2 \ge 0$......

Btw. Equality never occours, so the ineq is sharp.[/quote]

I know that it's not equivalent to $ (n \minus{} 1)^2 \ge 0$, but, like Mathias said, it's equivalent to $ 2n(n \minus{} 1)^2 \ge 0$. I only wanted to show that my inequality is verified.

Yes, equality never occurs, but $ \frac {1}{2}$ is the best constant.  :)

Note that: $ \frac{1}{2}\plus{}b\plus{}\frac{1}{a}\plus{}\frac{1}{2}>2\sqrt{\frac{1}{2}\cdot (b\plus{}\frac{1}{a}\plus{}\frac{1}{2})}.$ ([color=blue]AM-GM inequality[/color]). So $ \frac{1}{\sqrt{b\plus{}\frac{1}{a}\plus{}\frac{1}{2}}}>\sqrt{2}\cdot\frac{1}{b\plus{}\frac{1}{a}\plus{}1}.$

Using the similar inequalities for the other two denominators of the left hand side of the required inequality we get:

$ \frac{1}{\sqrt{b\plus{}\frac{1}{a}\plus{}\frac{1}{2}}}\plus{}\frac{1}{\sqrt{c\plus{}\frac{1}{b}\plus{}\frac{1}{2}}}\plus{}\frac{1}{\sqrt{a\plus{}\frac{1}{c}\plus{}\frac{1}{2}}}>$ $ \sqrt{2}\cdot (\frac{1}{b\plus{}\frac{1}{a}\plus{}1}\plus{}\frac{1}{c\plus{}\frac{1}{b}\plus{}1}\plus{}\frac{1}{a\plus{}\frac{1}{c}\plus{}1})\equal{}$ $ \sqrt{2}\cdot (\frac{a}{a\plus{}\frac{1}{c}\plus{}1}\plus{}\frac{\frac{1}{c}}{a\plus{}\frac{1}{c}\plus{}1}\plus{}\frac{1}{a\plus{}\frac{1}{c}\plus{}1})\equal{}\sqrt{2}.$

We use that $ \frac{1}{b\plus{}\frac{1}{a}\plus{}1}\equal{}\frac{a}{ab\plus{}a\plus{}1}\equal{}\frac{a}{a\plus{}\frac{1}{c}\plus{}1}$ and $ \frac{1}{c\plus{}\frac{1}{b}\plus{}1}\equal{}\frac{1}{c(ab\plus{}a\plus{}1)}\equal{}\frac{\frac{1}{c}}{a\plus{}\frac{1}{c}\plus{}1}$ (because $ abc\equal{}1$).

[quote] What is Ist grade in Serbia Olympiad?  [/quote]
Its like 9th grade in Russia for example
By years,its 15/16

@Maxim Bogdan very simple and beutiful solution !


function
logarithms
integration
calculus
derivative
irrational number
real analysis

You all probably know the famous row:
1+1/1!+1/2!+1/3!+....+1/n!+.... ->e (has the limit the transcendental number [b]e[/b] [i])
I would ask you to prove this as easy and clear as you can (I am searching for the most elegant solution).[/i]

Many people take that to be the [i]definition[/i] of $e$, so there is nothing to prove.  Reconciling various possible definitions of $e$ has appeared many times on the forum.  I suggest you do a search.

[quote="akech"]You have $f(x) = e^{x}$, then you know that:
$f(x) = \sum_{n=0}^{\infty}\frac{f^{n}(0)x^{n}}{n!}$ (Maclaurin series)
Now let $x = 1$ and you have what you want.[/quote]
Isn't this a circular definition of $e$? Because you use $e$ to prove $e$.

I wasn't saying what I wrote is a proof, but was just restating a known fact that the analytic function in question has the indicated representation, which of course is taken as its definition.

My definition.

$\ln x = \int_{1}^{x}\frac{dt}{t}$ for $x>0$.

Then we can so the properties of this function, specifically the one-to-one property from $\mathbb{R}^{+}$ to $\mathbb{R}$.

And hence define a function $e^{x}$ as its inverse function.

It is not hard to show that the derivative of this function is itself.

Then we show that $1+x+x^{2}/2!+...=$ is its Taylor series, and furthermore, it converges to $e^{x}$.

Oh, by the way, I forgot. We define $e$ to be so that $\int_{1}^{e}\frac{dt}{t}= 1$.

[quote="Alicea"]Now prove that [i]e[/i] is transcendental! :lol:[/quote]

I dont think this is an analysis type of problem. The way to show something is transcendental is by using something called [u]diophantine approximation[/u]. One result is as follows:

[b]Theorem:[/b] Let $\alpha$ be an irrational number and for each $n\in \mathbb{Z}^{+}$ there exists $p/q \in \mathbb{Q}$ with $q\geq 2$ such as:
$\left| \alpha-\frac{p}{q}\right| \leq \frac{1}{q^{n}}$
Then $\alpha$ is a transcendental number.

See if that works here.

No, $e$ is not Liouvillian. The result that $e$ is transcendental is almost purely analytic but it can be found in many textbooks and I don't see any point in reposting it here. Hilbert's proof is quite short, elegant, and easy to understand.


[quote="PenguinIntegral"]Hi, I am PenguinIntegral, your new mod. If you want anything done/changed/discussed,
post about it in this thread.

Anything involving any kind of problem solving skills should not be here.[/quote]

Congratulations! You're a mod!!!

A note on "no problem solving", the problem solving skills can't be none to be effective, but your class mates should be able to do these problems (at least most of them).

The harder problems (relatively easy still but with more problem solving) should go on the Basics forum (they will be moved if not on level). Hopefully this, as MCrawford said, will attract more students to this forum, making this a better, more active place.

You are too smart to be here. Get out :P

[quote="hwenterprise"][quote="PenguinIntegral"]Hi, I am PenguinIntegral, your new mod. If you want anything done/changed/discussed,
post about it in this thread.

Anything involving any kind of problem solving skills should not be here.[/quote]

How to Assume your new Role as a Moderator:

Penguin Integral, do you solely swear that you will:

Respect everyone on this forum
Give equal treatment to everyone,

and

Delete spam?

If you do, good. If you don't, well, I dunno :huh:[/quote]

Why delete spam when I can ban 236factorial from this fourm?
It does the same thing in the end. :rotfl:

[quote="PenguinIntegral"]Why delete spam when I can ban 236factorial from this fourm?
It does the same thing in the end. :rotfl:[/quote]

You better be careful about what you do  ;)

Nah, I don't see the reason for blocking you on my blog, you're no harm  :P

[quote="jhollenbeck"]:rotfl: good one penguin internal! I wonder how much this forum's post count would go down if we deleted all of 236factorial's posts :rotfl:[/quote]

It's penguinintegral, one.

two, it'll go down 3016. :D


Daca se demonstreaza Goldbach, atunci merge chiar cu doua. ;)

Mie mi-a iesit suma de 4010 si 4011 si 4012 numere prime in felul urmator:$2005*2006=2005*2003+ 2005*3$ egal suma de 4010 nr prime(2005 de 2003 si 2005 de 3)
Sau $2005 *2006=(2003+2)*2006=2003*2006+ 2*2006$ suma de 4012 nr prime
Sau $2005*2006=2003*2006 + 2*2006 = 2003*2003 + 2003*3 + 2*2003+2*2006$=suma de 4011 nr prime.
In general daca un nr se poate scrie sub forma $n=(k+p)(q+r)$,cu k,p,q,r nr prime,atunci nr se poate scrie ca suma de "$k+p+q+r$" nr prime.Avem $n=kq+kr+pq+pr$ .Luand multimile $\{k;q\},\{k;r\},\{p;q\},\{p;r\}$ observam ca daca luam un numar din fiecare multime,suma dintre cele 4 nr alese reprezinta numarul de numere prime in care poate fi "spart" numarul n.

As avea si eu o intrebare va rog, trebuie sa selectez vreo optiune pentru a fi activat latexul cand postez o problema? Chiar daca scriu o kestie banala, cand trimit apare postat exact in forma in care am scris.

Gata mi-am dat seama..aveam activat codul ala BB

[quote="_NightWish_"]Sa se arate ca 2005*2006 se poate scrie ca suma de 4009 numere prime[/quote]
   Apropo de problema asta cu numere prime, demonstreaza ca nu exista o infinitate de numere impare consecutive care sa fie prime. ;)

[quote="spix"]Ar insemna ca exista $M$ astfel incat oricare ar fi $n>M$ si $n$ impar atunci $n$ prim. Fals.[/quote]
    Frumoasa solutie.Bine, eu ma gandisem la alta...mai lunga.


Let a triangle ABC and I is its incenter. AI cuts the incircle (I) at D. Prove that the tangent of (I) at D and the external bisector of angle BIC meet on BC.

Denote $X,Y,Z$ the tangency points of $(I)$ with $BC,CA,AB$ and let the internal angle bisector of $\angle BIC$ and the tangent of $(I)$ at $D$ cut  $BC$ at $V,P,$ respectively. We shall prove that $IP$ is the external bisector of $\angle BIC.$ Midpoint $M$ of the arc $BC$ of the circumcircle $\odot(ABC)$ is circumcenter of $\triangle BIC.$ Thus,  $IX$ and $IM \equiv IA$ are isogonals with respect to $\angle BIC$ $\Longrightarrow$ $IV$ bisects $\angle MIX.$  Analogously, if $R$ is the projection of $X$ on $YZ,$ the rays $XI,XR$ are isogonals with respect to $\angle YXZ.$ Lines $IM,XR$ are parallel since they are both perpendicular to $YZ$ $\Longrightarrow$ angle bisectors of $\angle MIX$ and $\angle RXI$ are parallel $\Longrightarrow$ $XD \parallel IV,$  but $IP \perp XD$ $\Longrightarrow$ $IP \perp IV$ $\Longrightarrow$ $IP$ is the external bisector of $\angle BIC.$

Dear Mathlinkers,
with my notation

Dear Mathlinkers,

1. ABC a triangle
2. (I) the incircle
3. DEF the contact triangle
4. X the point of intersection of the segment AI with (I)
5. M the foot of the I-inner bissector of the triangle BIC
6. A* the point of intersection of the tangent to (I) at X and BC.

Proof

1. According to http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=612981
DX // IM
2. A*I is perpendicular to DX
3. A*I is then perpendicular to IM
and we are done…

Sincerely
Jean-Louis

[hide="Solution"]As I am going to be referencing a previous post, we rename the points for the sake of convenience. Denote the incircle by $\omega.$ Let $D, E, F$ be the points of tangency of the incircle with sides $BC, CA, AB$, respectively. Let $N = AI \cap \omega$ and let $M$ be the foot of the $I$-internal angle bisector in $\triangle IBC.$

[b]Claim.[/b] $DN \parallel IM.$
[i]Proof.[/i] Note that $AE, AF$ are tangents to $\omega.$ Since $I$ is the center of $\omega$, it follows that $AI \cap \omega = N$ is the midpoint of $\widehat{EF}.$ Therefore $DN$ is the internal angle bisector of $\angle EDF.$ Then

[quote]We proceed by angle chasing, using the fact that $\triangle BFD$ is isoceles due to equal tangents, and quadrilateral $AEIF$ is cyclic due to right angles: \[\angle NDB = \angle NDF + \angle FDB = \frac{1}{2}\angle EDF + \frac{180^{\circ} - B}{2}\]\[= \frac{1}{4}\angle EIF + 90^{\circ} - \frac{B}{2} = \frac{1}{4} \left(180^{\circ} - A\right) + 90^{\circ} - \frac{B}{2}\]\[= 135^{\circ} - \frac{B}{2} - \frac{A}{4}.\] In addition, we have \[\angle IMB = 180^{\circ} - \frac{B}{2} - \angle BIM = 180^{\circ} - \frac{B}{2} - \frac{1}{2}\angle BIC\]\[= 180^{\circ} - \frac{B}{2} - \frac{1}{2}\left(90^{\circ} + \frac{A}{2}\right)\]\[= 135^{\circ} - \frac{B}{2} - \frac{A}{4}.\]
Therefore, we conclude that $DN \parallel IM$, as desired.[/quote]

Now, let $Z$ be the intersection of line $BC$ and external angle bisector of $\angle BIC.$ By the definition of external angle bisector, we have $IZ \perp IM.$ Therefore we have $IZ \perp DN.$ Now, let $N' \ne D$ be a point of tangency from $Z$ to $\omega.$ Since $D, N'$ are points of tangency, it follows that $DN'$ is the polar of $Z$ w.r.t. $\omega.$ Therefore $IZ \perp DN'.$ However, there is a unique point $P \in \omega$ such that $IZ \perp DP.$ Therefore we conclude that $N' = N.$ Hence, $ZN$ is a tangent to $\omega$, and so the tangent to $\omega$ at $N$, the external angle bisector of $\angle BIC$, and line $BC$ all concur at one point.
The proof is complete.[/hide]


We have $ \sqrt{\frac{4x\plus{}9}{28}}\equal{}7x^2\plus{}7x$

$ \implies\frac{4x\plus{}9}{28}\equal{}49x^4\plus{}98x^3\plus{}49x^2$

$ \implies4x\plus{}9\equal{}1372x^4\plus{}2744x^3\plus{}1372x^2$

$ \implies1372x^4\plus{}2744x^3\plus{}1372x^2\minus{}4x\minus{}9\equal{}0$

$ \implies(14x^2\plus{}12x\minus{}1)(98x^2\plus{}112x\plus{}9)\equal{}0$

$ \implies x\equal{}\left\{\frac{\minus{}12\pm\sqrt{200}}{28},\frac{\minus{}112\pm\sqrt{9016}}{196}\right\}$

$ \equal{}\left\{\minus{}\frac37\pm\frac{5\sqrt2}{14},\minus{}\frac47\pm\frac{\sqrt{46}}{14}\right\}$.

However, checking, we see that the only solutions are

$ x\equal{}\boxed{\left\{\minus{}\frac37\plus{}\frac{5\sqrt2}{14},\minus{}\frac47\minus{}\frac{\sqrt{46}}{14}\right\}}$.

Heh... the only reason why I didn't solve it was to do $ 49 x 28$ :p I really feel there may be a simplier way to solve it...


greatest common divisor
symmetry
number theory
relatively prime

Can anybody give the proof the theorem
that For x,y,z natural
The equation
$ x^3\plus{}y^3\equal{}z^3$
hasn't got solution.

Assume (x,y,z)=1, if not we can divide by the gcf cubed to make it so.
$ x^{3}\equal{}(z\minus{}y)(z\minus{}y\omega)(z\minus{}y\omega^{2})$
The terms on the RHS are relatively prime, we now use that $ z\minus{}y\omega\equal{}(a\plus{}b\omega)^{3}$
$ z\equal{}a^{3}\plus{}b^{3}$
$ \minus{}y\omega\equal{}3a^{2}b\omega\plus{}3ab^{3}\omega$
We have that 3|y.
By symmetry 3|x.
Mod 27 3|z.
But now we have that (x,y,z)=3 giving us an infinite descent/contradiction.

The terms on the RHS of $ x^{3} \equal{} (z \minus{} y)(z \minus{} y\omega)(z \minus{} y\omega^{2})$ are not necessarily coprime, at least this is not trivial. You need to prove this.
In fact, this resembles the harder case of the fermat problem $ x^p \plus{} y^p \equal{} z^p$, namely that one if $ p|xyz$.

And $ z \minus{} y\omega \equal{} (a \plus{} b\omega)^{3}$ gives $ z \equal{} a^3 \plus{} b^3 \minus{} 3ab^2$ and $ \minus{}y\omega \equal{} 3a^2b \omega \minus{} 3ab^2 \omega$.


Pedro stood at the center of a circular field that had a radius of 120 feet. He walked due north halfway to the circle. He then turned and walked due east halfway to the circle. He turned again and walked due south halfway to the circle. Finally, he turned and walked due west halfway to the circle. When he stopped, how many feet was Pedro from the center of the circle? Express your answer to the nearest foot.

I know how to get the answer but i want to know if there's an easier way...

its  only  simple  geometry  and  pythogoras  theorem.  no special methods   bring  in coordinate geometry  if  u  have knowledge of it


Prove that $ (2^n)!$ can be expressed in the form $ 2^{2^n\minus{}1}\cdot a$,where $ a$ is natural odd number

[quote="Obel1x"]Prove that $ (2^n)!$ can be expressed in the form $ 2^{2^n \minus{} 1}\cdot a$,where $ a$ is natural odd number[/quote]

$ v_2((2^n)!)\equal{}\sum_{k\ge 1} \left[\frac{2^n}{2^k}\right]$ $ \equal{}\sum_{k\equal{}0}^{n\minus{}1}2^k$ $ \equal{}2^n\minus{}1$

Hence the result.

[quote="pco"][quote="Obel1x"]Prove that $ (2^n)!$ can be expressed in the form $ 2^{2^n \minus{} 1}\cdot a$,where $ a$ is natural odd number[/quote]

$ v_2((2^n)!) \equal{} \sum_{k\ge 1} \left[\frac {2^n}{2^k}\right]$ $ \equal{} \sum_{k \equal{} 0}^{n \minus{} 1}2^k$ $ \equal{} 2^n \minus{} 1$

Hence the result.[/quote]

So $ a \equal{} 2^n \minus{} 1$.
Let's take $ n \equal{} 3$ we have $ 8! \equal{} 2^{2^3 \minus{} 1} \cdot (2^3 \minus{} 1) \implies 2^7\cdot 3^2\cdot5\cdot7 \equal{} 2^7 \cdot 7$ ???
I  think I'm not getting the solution ,please can you explain me a bit more about the formula and the problem?

[quote="Obel1x"][quote="pco"][quote="Obel1x"]Prove that $ (2^n)!$ can be expressed in the form $ 2^{2^n \minus{} 1}\cdot a$,where $ a$ is natural odd number[/quote]

$ v_2((2^n)!) \equal{} \sum_{k\ge 1} \left[\frac {2^n}{2^k}\right]$ $ \equal{} \sum_{k \equal{} 0}^{n \minus{} 1}2^k$ $ \equal{} 2^n \minus{} 1$

Hence the result.[/quote]

So $ a \equal{} 2^n \minus{} 1$.[/quote]

No. So $ (2^n)!\equal{}2^{2^n\minus{}1}a$ with odd $ a$.

Which is the required result