The following is part of a bigger problem in the advanced problems forum. I don't even know whether or not  it is possible, but here's the problem:

Write either of the following two equations in the form (product) = (constant), where  k is a constant.

(1 - k)xyz + k(xy + xz + x + yz - y - z) = k + 1

(1 - k)(xy)^2 + k(xy^2 + yx^2 + x - y) = k + 1

hello, do you mean the following form
$ xyz\minus{}kxyz\plus{}k(xy\plus{}xz\plus{}yz\plus{}x\minus{}y\minus{}z)\minus{}k\equal{}1$
$ xyz\plus{}k(xy\plus{}xz\plus{}yz\plus{}x\minus{}y\minus{}z\minus{}xyz\minus{}1)\equal{}1$
$ k(xy\plus{}xz\plus{}yz\plus{}x\minus{}y\minus{}z\minus{}xyz\minus{}1)\equal{}1\minus{}xyz$
$ \frac{1\minus{}xyz}{xy\plus{}xz\plus{}yz\minus{}xyz\plus{}x\minus{}y\minus{}z\minus{}1}\equal{}k$
$ (1\minus{}xyz)(xy\plus{}xz\plus{}yz\minus{}xyz\plus{}x\minus{}y\minus{}z\minus{}1)^{\minus{}1}\equal{}k$
with $ (xy\plus{}xz\plus{}yz\minus{}xyz\plus{}x\minus{}y\minus{}z\minus{}1)\ne0$
Sonnhard.


set theory
advanced fields
advanced fields theorems

What books/articles on the subject do you recommend?

I'm mainly looking for a proof of the fact $ \left| A \times A \right| \equal{} \left| A \right|$, but other information on the subject won't hurt.
[A post by v0id reminded me of this.]

[url=http://www.amazon.com/Basic-Set-Theory-Azriel-Levy/dp/0486420795]Basic Set Theory[/url] by Azriel Levy has that theorem in chapter 5 on the axiom of choice. It is an exercise, but it gives huge hints as to how to do it. It does provide the proof of a related theorem ($ a^2 \equal{} a$ for all infinite cardinals $ a$ using axiom of choice). It also has a full chapter on cardinals and a full chapter on ordinals. Despite "Basic" being in the title, I've found the book challenging to read; I wouldn't be surprised if it's too easy for others though.

An old book, with an old style of exposition, but useful, and containing the proofs of such equivalents of axiom of choice as $ u \plus{} v \equal{} uv$ for infinite u,v; $ a\times a \equal{} a$ for infinite a, etc:
Robert R. Stoll, [i]Set Theory and Logic[/i].

Have a look [url=http://books.google.com/books?id=3-nrPB7BQKMC&pg=PA124&lpg=PA124&dq=axiom+of+choice+%22axa+a%22+for+all+infinite+a&source=web&ots=0jaAKORXDa&sig=CmSr53rkVioLLw0qDv3tncEW7RY]at this page[/url].

The Book of Numbers
by John H. Conway, Richard Guy

chapter 10


[hide="Hint"] Tangent of both sides. [/hide]
For an investigation of the generalization:

http://mathworld.wolfram.com/Machin-LikeFormulas.html

[hide] \[\tan\left(x+y\right)=\frac{\sin x\cos y+\sin y\cos x}{\cos x\cos y-\sin x\sin y}=\frac{\tan x+\tan y}{1-\tan x\tan y}\] Letting $x=\arctan\frac{1}{3}$ and $y=\arctan\frac{1}{2}$ gives \[\tan\left(\arctan\frac{1}{3}+\arctan\frac{1}{2}\right)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=1.\] Therefore $\arctan\frac{1}{3}+\arctan\frac{1}{2}=\arctan 1=\frac{\pi}{4}.$[/hide]

Nice solution, but generally, $\tan x=1\implies x={\pi\over 4}+k\pi, k\in\mathbb{Z}$, so you still have to add an easy proof that $k$ is exactly zero in this case.

Hint
Take tangent of both sides

Also
Make sure you know the relative size of the angles compared to the 45 degree angle.

[b]Elementary solution.[/b] See the attached diagram.

Choose an arbitrary length $a$ and construct $AB=5a$. Then find $M\in AB$ such that $AM=2a$ and $MB=3a$. Then draw the perpendicular $MC$ such that $MC=6a$. By the construction, $\tan\angle ACM={1\over 3}$ and $\tan\angle BCM={1\over 2}$, hence $\angle ACB=\arctan{1\over 2}+\arctan{1\over 3}$.

By Pythagoras we have $AC=2a\sqrt{10}$ and $BC=3a\sqrt{5}$. In $\triangle ABC$ draw the altitude $AN$. Then

$[ABC]=\frac{AB\cdot MC}{2}=\frac{BC\cdot AN}{2}\implies AN=\frac{5a\cdot 6a}{3a\sqrt{5}}=2a\sqrt{5}$

Then by Pythagoras again we have $NC=\sqrt{AC^{2}-AN^{2}}=\sqrt{40a^{2}-20a^{2}}=2a\sqrt{5}$.

Therefore, $AN=NC$ and $\angle ANC=90^\circ$, which means that $\triangle ANC$ is an isosceles right, hence $\angle ACB=45^\circ$.

Here's mine:

[hide]Create three congruent squares $ABFE$, $BCGF$, and $CDHG$.  WLOG let them be unit squares.  We can see that $m\angle DEH=\arctan\frac{DH}{EH}=\arctan\frac13$.
Construct points $I$ and $J$ such that two new squares, $EBGJ$ and $BIDG$, each with side length $\sqrt2$, are formed.  We have rectangle $EIDJ$, and we can see that $m\angle IED=\arctan\frac{ID}{IE}=\arctan\frac{\sqrt2}{2\sqrt2}=\arctan\frac12$.
Note that $m\angle IED+m\angle DEH=m\angle IEH$, clearly a $45^\circ$ angle, or in radians, $\frac{\pi}4$.  Substituting in gets us $\arctan\frac13+\arctan\frac12=\frac\pi4$.
$\boxed{\text{Q.E.D.}}$[/hide]

To tell the truth, I got this from [url=http://puzzles.com/PuzzlePlayground/RaysThroughTheSquares/RaysThroughTheSquaresSol.htm]here[/url].

[quote="JBL"][quote="amcavoy"]You can use series as well.  Take the power series for the inverse tangent evaluated at $1/2$ and $1/3$.  It is the same as the inverse tangent series evaluated at $1$, which is \[\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\cdots=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{2n+1}.\] [/quote]
I don't see how you get this -- would you mind giving a fuller explanation?[/quote]
I actually just assumed that it would be the same.  It is the same, of course (using the first proof), but I can't show that the sum of series is the same directly.

The series probably don't sum nicely.  I doubt manipulating the series would lead to an easy solution.

Yeah, the series in question just happens to be the series which converges to $\pi$ slower than most other $\pi$ formulas.  :(

Many have seen this pondering over bathroom tiles (or any place where you have sqaure grid pattern)

Start with $O$  (0,0)
Go  2 diagnal units  $A$ (2,2)
Turn right and go 1 (diagonal unit),$B$ (3,1)
Go starignt down and reach  $C($ 3,0)
Angle  AOB is  = arctan (1/2) and Angle BOC =  arctan (1/3)  and their sum is 45 degrees.


geometry
circumcircle
incenter
Euler
angle bisector
congruent triangles
perpendicular bisector

Given triangle ABC with O,I and H are the circumcenter, incenter and orthocenter respectedly. Prove that : $IO=IH$ <--> $\angle A=60$ or $\angle B=60$ or $\angle C=60$.
Was something wrong ???? :?

WLOG, assume that the angle $\measuredangle B = \measuredangle ABC = 60^o$. Let $D, E, F$ be the feet of altitudes $AD, BE, CF$ to from the vertices $A, B, C$ to the sides $a = BC, b = CA, c = AB$ and $K, L, M$ the midpoints of the sides $BC, CA, AB$. The angles $\measuredangle BAD = \measuredangle BCF = 90^o - 60^o = 30^o$. Let $P, Q$ be the intersections of the angle bisector of the angle $\measuredangle B$ with the altitudes $AD, CF$. Since the angles $\measuredangle ABP = \measuredangle BAP = 30^o$ and $\measuredangle BQK = \measuredangle CQK = 30^o$, the triangles $\triangle ABP$ and $\triangle BCQ$ are both isosceles. Consequently, the perpendicular bisectors $OM, OK$ of the sides $AB, BC$ pass through the points $P, Q$, respectively, the triangles $\triangle BPD$, $\triangle BPM$, $\triangle APM$ are congruent and the triangles $\triangle BQF$, $\triangle BQK$, $\triangle CQK$ are also congruent. The points $M, D$ are the corresponding points of the congruent triangles $\triangle BQF$, $\triangle BQK$, because the lines $HD, OM$ are perpendicular to their corresponding sides $BF, BK$ and they concur at the point $P$ on their common side $BP$. Consequently, the intersections $H, O$ of the corresponding lines $HD, OM$ with the corresponding sides $QF, QK$ are also the corresponding points of these 2 congruent triangles. It follows that the common side $BQ$, identical with the bisector $BI$ of the angle $\measuredangle B$, is the perpendicular bisector of the segment $HO$. Hence, this angle bisector bisects the angle $\measuredangle HIO$ and is perpendicular to the side $HO$ of the triangle $\triangle HOI$. This means that this triangle is isosceles and $IH = IO$. The line $HO$ is the Euler line of the triangle $\triangle ABC$ and if the angle $\measuredangle B = 60^o$, the bisector $BI$ of this angle intersects the Euler line at the midpoint $N$ of the segment $HO$, identical with the center of the 9-point circle of the triangle $\triangle ABC$.

BTW, it is amusing that the vertex I of the triangle $\triangle HOI$ always points toward the $60^o$ angle, like an arrow. It always points to the mid-sized angle (even if this angle is not $60^o$ and $IH \neq IO$) and the $60^o$ angle must always be the mid-sized one.


Let $ABCD$ be a convex quadrilateral and let $M,N$ be points on the sides $AB$ and $BC.$ The line segments $DM$ and $DN$ intersect $AC$ at $K$ and $L$ and the lines $BK$ and $BL$ intersect the sides $AD$ and $CD$ at $R$ and $S,$ respectively. Suppose that $AK=KL=LC$ and

\[
\left[ ADM\right] =\left[ CDN\right] =\frac{1}{4}\left[
ABCD\right] .
\]
Prove that
\[
\left[ ABR\right] =\left[ BCS\right] =\frac{1}{4}\left[
ABCD\right] .
\]

The given conditions tell us that $\frac{AM}{MB}=\frac{CN}{NB}$ (from $[ADM]=[CDN]$ and $[ADM]=[ADK]+[AKM],\ [CDN]=[CDL]+[CLN]$). This last relation also shows that $\frac{AM}{MB}=1$ and $[ADM]=\frac 12\cdot[ABCD]$. We thus find $K,L$ to be the centroids of $ABD,CBD$ respectively, and the conclusion easily follows.


You are not allowed to post this problem. Please do not discuss it until the end of October.

[quote=nmtvn]Suppose: a, b, c are 3 lenghts of sides of a triangle, prove that:
 $ \frac{a}{\sqrt{a^{2}\plus{}3bc}}\plus{}\frac{b}{\sqrt{b^{2}\plus{}3ca}}\plus{}\frac{c}{\sqrt{c^{2}\plus{}3ab}}\geq\frac{3}{2}$[/quote]

See here: http://www.artofproblemsolving.com/community/c6h163238

[quote=Swatch][quote=FERZIO]Let $ a+b+c = 1.$
Hence, by Jensen we obtain: $ \sum_{cyc}\frac{a}{\sqrt{a^{2}+3bc}}\geq\frac{1}{\sqrt{a^{3}+b^{3}+c^{3}+9abc}}.$
Id est, it remains to prove that $ 4(a+b+c)^{3}\geq9(a^{3}+b^{3}+c^{3}+9abc).$
Let $ a = y+z,$ $ b = x+z$ and $ c = x+y.$ Then
$ 4(a+b+c)^{3}\geq9(a^{3}+b^{3}+c^{3}+9abc)\Leftrightarrow$
$ \Leftrightarrow32(x+y+z)^{3}\geq9\sum_{cyc}\left((x+y)^{3}+3(x+y)(x+z)(y+z)\right)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(7x^{3}-6x^{2}-6x^{2}z+5xyz)\geq0\Leftrightarrow$
$ \Leftrightarrow6\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)+\sum_{cyc}(x^{3}-xyz)\geq0,$
which true by Schur and Muirhead.[/quote]

Just wondering, why can you let $a+b+c=1$? What if $a=0.1$, $b=0.1$, $c=0.8$? Those wouldn't be the sides of a triangle.[/quote]

assume that the perimeter is $k$
$a+b+c=k$ then there is a triangle with side $\frac{a}{k}+\frac{b}{k}+\frac{c}{k}=1$

[quote=nmtvn]Suppose: a, b, c are 3 lenghts of sides of a triangle, prove that:
 $$ \frac{a}{\sqrt{a^{2}\plus{}3bc}}\plus{}\frac{b}{\sqrt{b^{2}\plus{}3ca}}\plus{}\frac{c}{\sqrt{c^{2}\plus{}3ab}}\geq\frac{3}{2}$$[/quote]
Let $x,y,z$ be positive real numbers. [url=https://artofproblemsolving.com/community/c6h2988570p26815362]Prove that[/url] $$\frac{x}{\sqrt{z^2+3xz}}+ \frac{y}{\sqrt{x^2+3xy}}+ \frac{z}{\sqrt{y^2+3yz}}\geq \frac{3}{2}$$
Let $a,b,c>0$. [url=https://artofproblemsolving.com/community/c6h1822770p12181037]Prove that:[/url]
$$\dfrac{a}{\sqrt{b^2+3c^2}}+\dfrac{b}{\sqrt{c^2+3a^2}}+\dfrac{c}{\sqrt{a^2+3b^2}} \geq \frac{3}{2}$$


integration
calculus
calculus computations

I am stuck on part a.
http://www.haverford.edu/math/rmanning/teaching_materials/candysched114.pdf

The $ \frac{R}{1000}\plus{}50$ cents/candy is a constant, because to produce 2 million candies in 200 days, you need to produce 10000 candies/day, so you just stick that into $ R$ meaning it costs 60 cents/candy.  Hence it costs
$ .6\times 2\mbox{ million}\equal{}1200000\mbox{ cents}\equal{}\$12000$
to actually produce the candy.
But you have to store the candy.  The storage cost is changing though, because as $ t$ increases, so does the amount of candy.  So $ A\equal{}A(t)$.  This means
$ C(t)\equal{}12000\plus{}\frac{1}{2}\int_{0}^{200}A(t)\,dt$
I may not have done this correctly.

The trouble that I am having is in trying to determine $ A(t)$ . The storage cost is confusing since in the later parts, there it asks to lower the production costs which is only possible from lowering the storage cost.

anybody? I was thinking about doing a double integral for the storage cost. Does that work?

bump. help me out! So for storage, i was thinking about using the integral of $ C(t)/2$ from $ 0$ to $ t$. Does this work?


calculus
integration
trigonometry
function
LaTeX
calculus computations

Can someone please help with these three integrals?

1. Indefinite integral of:

         t(t-10)^10 dt

There certainly must be a way to evaluate it by substitution or something fast rather than evaluating every term one by one.

2. Indefinite integral of:

            (x cos ( :sqrt: (x^2+1)))/( :sqrt: (x^2+1)) dx
I don't know what to substitute for u to make the expression easier.

3. Indefinite integral of:

  ((sin (2x)^2))((cos (2x))^3) dx

Integration by parts may not work here, but is there a way to simply this statement so we may use integration by substitution?

Thanks to anyone who gives suggestions.  :blush:

Boy, that's hard to read! keta, when you read this answer, take a look at the TeX code. (Just hold your mouse over what you want to see.)

1. $\int t(t-10)^{10}dt$

Let $u=t-10,\,du=dt,\,t=u+10.$ Then there are just two terms to evaluate. Alternatively, try integration by parts.

2. $\int \frac{x\cos\sqrt{x^2+1}}{\sqrt{x^2+1}}dx$

Try $u=\sqrt{x^2+1}.$ This problem is rigged for that to work.

3. $\int \sin^2(2x)\cos^3(2x)\,dx$.

$du = \cos(2x)\,dx$ (or maybe $du=2\cos(2x)\,dx.$) That's a substitution; I'll let you figure out what $u$ is and why it works. Hint: with trig functions, there are always identities.

There are three basic families of methods for finding antiderivatives. In order of importance:
1. Algebra.
2. Substitution.
3. Integration by parts.

--

Valentin - about that other possible forum we were talking about?

[quote="Kent Merryfield"]Boy, that's hard to read! keta, when you read this answer, take a look at the TeX code. (Just hold your mouse over what you want to see.)
[....]
Valentin - about that other possible forum we were talking about?[/quote]Better yet keta, just [b]click[/b] on the $\LaTeX$ code to see it exactly in a new window :)

Kent, let's talk on PM, ok? :)


Given are two even spheres. one is hanging from the sealing on an  heat insolating thread, the other one is standing on a heat insolating floor. We apply the same amount of heat to both. Will their temperatures be the same, or not after they got the heat?

I think we already had this one on this forum (or maybe I'm getting senile). This is a classic olympiad problem (from the first IPhO, I believe). In short: due to heat expansion, the potential energy of the second one will increase and so its final temperature will be a bit lower than the first one, whose potential energy decreases.

I think that's ok, since this is a low-traffic forum ;)


geometry
circumcircle
parallelogram
trapezoid
geometric transformation
reflection
complex numbers

could you post your solution?
thank you  [b]CatalystOfNostalgia[/b]

does anyone have idea?

[quote="CatalystOfNostalgia"]Let ABC be a triangle, with P and Q arbitrary points on CA, AB, respectively. Let PQ meet the circumcircle of ABC at X and Y. Prove that the midpoints of BP, CQ, PQ, and XY are concyclic.[/quote]
very nice,but hard.
if this problem was problem IMO 2009, I think the exam would be very interseting
can you post solution?

[quote="yetti"]$ K, L, M$ are midpoints of $ BP, CQ, PQ.$ $ A', B', C'$ are midpoints of $ BC, CA, AB.$ $ Z$ is midpoint of $ XY.$ $ MK \parallel (QB \equiv AB)$ and $ LM \parallel (CP \equiv CA)$ $ \Longrightarrow$ $ \angle LMK \equal{} \angle CAB.$ $ OZB'P, OZC'Q$ are cyclic on account of right angles at $ B', C', Z.$ Let $ (O_P), (O_Q)$ be their circumcircles. Midlines $ A'B', C'A'$ cut $ (O_P), (O_Q)$ again at $ E, F.$ Perpendicular bisectors $ OC', OB'$ of $ AB, CA$ cut $ (O_P), (O_Q)$ again at $ S, T.$ $ ES, FT$ cut $ AB, CA$ at $ U, V.$ $ OS \perp PS$ $ \Longrightarrow$ $ PS \parallel AB \parallel A'B'$ $ \Longrightarrow$ $ B'PSE, APSU$ are isosceles trapezoids. $ \triangle ABS$ is isoscelses with $ AS \equal{} BS$ $ \Longrightarrow$ $ BSPU$ is parallelogram, $ K \in (SU \equiv SE)$ and similarly, $ L \in (TV \equiv TF).$  Then $ MKEL$ is iscosceles trapezoid, cyclic, $ E \in \odot (MKL)$ and similarly, $ F \in \odot (MKL).$

From $ (O_P), (O_Q),$ $ \angle PZE \plus{} \angle FZQ \equal{} \angle CPS \plus{} \angle TQB \equal{} 2 \angle CAB$
From $ \odot (MFE),$ $ \angle FEM \plus{} \angle MFE \equal{} \angle FKM \plus{} \angle MKE \equal{} 2 \angle LMK \equal{} 2 \angle CAB$
$ \Longrightarrow$ $ \angle EZF \equal{} \angle EMF$ $ \Longrightarrow$ $ Z \in \odot (MFE) \equiv \odot(MKL).$[/quote]
dear [b]yetti[/b], here is a picture for your powerful solution, your solution is very very impressive to me  :)

[b]Lemma[/b]: Let $ R,M,N$ be the midpoints of $ XY,CQ,BP$. Then $ \triangle APQ$ and $ \triangle RNM$ are inversely similar.

[i]Proof of the Lemma[/i]:
[hide="Complex number bash"]Let $ a,b,c,x,y$ be the complex numbers $ A,B,C,X,Y$, and suppose WLOG that the circuncircle of $ \triangle ABC$ is the unit circle. Then we have that:

$ p\equal{}\frac{ac(x\plus{}y)\minus{}xy(a\plus{}c)}{ac\minus{}xy}$

$ q\equal{}\frac{ab(x\plus{}y)\minus{}xy(a\plus{}b)}{ab\minus{}xy}$

since $ P\equal{}AC\cap XY$, $ Q\equal{}AB\cap XY$, and $ AB,AC,XY$ are chords of the unit circle.

We also have that:

$ r\equal{}\frac{x\plus{}y}{2}$

$ m\equal{}\frac{c\plus{}q}{2}\equal{}\frac{abc\minus{}xyb\plus{}ac(x\plus{}y)\minus{}xy(a\plus{}c)}{2(ac\minus{}xy)}$

$ n\equal{}\frac{b\plus{}p}{2}\equal{}\frac{abc\minus{}xyc\plus{}ab(x\plus{}y)\minus{}xy(a\plus{}b)}{2(ab\minus{}xy)}$

Using the above we get:

$ \frac{p\minus{}a}{q\minus{}a} \equal{}\frac{ac(x\plus{}y)\minus{}xy(a\plus{}c)\minus{}a^2c\plus{}xya}{ac\minus{}xy} \times$
$ \frac{ab\minus{}xy}{ab(x\plus{}y)\minus{}xy(a\plus{}b)\minus{}a^2b\plus{}xya}$

$ \overline{\frac{p\minus{}a}{q\minus{}a}}\equal{}\frac{\frac{1}{ac}(\frac{1}{x}\plus{}\frac{1}{y})\minus{}\frac{1}{xy}(\frac{1}{a}\plus{}\frac{1}{c})\minus{}\frac{1}{a^2c}\plus{}\frac{1}{xya}}{\frac{1}{ac}\minus{}\frac{1}{xy}}\times$
$ \frac{\frac{1}{ab}\minus{}\frac{1}{xy}}{\frac{1}{ab}(\frac{1}{x}\plus{}\frac{1}{y})\minus{}\frac{1}{xy}(\frac{1}{a}\plus{}\frac{1}{b})\minus{}\frac{1}{a^2b}\plus{}\frac{1}{xya}}$

$ \equal{}\frac{a(x\plus{}y)\minus{}a(a\plus{}c)\minus{}xy\plus{}ac}{axy\minus{}a^2c}\times \frac{axy\minus{}a^2b}{a(x\plus{}y)\minus{}a(a\plus{}b)\minus{}xy\plus{}ab}\equal{}\frac{ab\minus{}xy}{ac\minus{}xy}$

But we also have that:

$ \frac{n\minus{}r}{m\minus{}r}\equal{}\frac{abc\minus{}xyb\plus{}ac(x\plus{}y)\minus{}xy(a\plus{}c)\minus{}ac(x\plus{}y)\plus{}xy(x\plus{}y)}{ac\minus{}xy}\times \frac{ab\minus{}xy}{abc\minus{}xyc\plus{}ab(x\plus{}y)\minus{}xy(a\plus{}b)\minus{}ab(x\plus{}y)\plus{}xy(x\plus{}y)}$

$ \equal{}\frac{ab\minus{}xy}{ac\minus{}xy}$

From this we get that $ \overline{\frac{p\minus{}a}{q\minus{}a}}\equal{}\frac{n\minus{}r}{m\minus{}r}$, which proves the lemma.[/hide]

Let $ S$ be the midpoint of $ PQ$. Suppose WLOG that $ S$ lies between $ R$ and $ Q$. Then we have that $ \angle RSN\equal{}\angle RQB\equal{}\pi\minus{}\angle AQP\equal{}\pi\minus{}\angle RMN$, so $ RSNM$ is cyclic, as desired ($ SN\parallel QB$ since $ S,N$ are the midpoints of $ PQ,PB$).

Man this is awesome!!

Anyway, this problem is also a generalization of this classic problem (perhaps old IMO?)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218486

Hello!
Nice problem. Here is my solution.

$ O$ is the circumcenter of $ ABC$.
$ K,L,M$ are the midpoints of $ BP,CQ,PQ$.
$ Z$ is the midpoint of $ XY$.
$ MK\parallel{}QB\Rightarrow MK\parallel{}AB$     (1)
$ ML\parallel{}PC\Rightarrow ML\parallel{}AC$     (2)
(1), (2) $ \Rightarrow\angle KML\equal{}\angle BAC$     (3)

$ P',Q'\in XY$ such that $ ZP'\equal{}ZP$ and $ ZQ'\equal{}ZQ$ $ \Rightarrow P'X\equal{}PY$, $ P'Y\equal{}PX$, $ Q'X\equal{}QY$, $ Q'Y\equal{}QX$.
$ \left\{A'\right\}\equal{}BP'\cap\left(O\right)$ and $ \left\{Q"\right\}\equal{}A'C\cap XY$.
$ \left(XYPQ\right)\equal{}A\left(XYPQ\right)\equal{}\left(XYCB\right)\equal{}A'\left(XYQ"P'\right)\equal{}\left(XYQ"P'\right)\Rightarrow$
$ \Rightarrow\left(XYPQ\right)\equal{}\left(XYQ"P'\right)\Rightarrow$
$ \Rightarrow\frac{PX}{PY}: \frac{QX}{QY}\equal{}\frac{Q"X}{Q"Y}: \frac{P'X}{P'Y}\Rightarrow\frac{PX}{PY}\cdot\frac{P'X}{P'Y}\equal{}\frac{Q"X}{Q"Y}\cdot\frac{QX}{QY}\Rightarrow$
$ \Rightarrow\frac{Q"X}{Q"Y}\cdot\frac{Q'Y}{Q'X}\equal{}1\Rightarrow\frac{Q"X}{Q"Y}\equal{}\frac{Q'X}{Q'Y}\Rightarrow Q"\equal{}Q'$

$ \left\{E\right\}\equal{}OZ\cap A'B$ and $ \left\{F\right\}\equal{}OZ\cap A'C$
$ ZK\parallel{}P'B\Rightarrow ZK\parallel{}EB\Rightarrow\angle OZK\equal{}\angle A'EF$     (4)
$ ZL\parallel{}Q'C\Rightarrow ZL\parallel{}FC\Rightarrow\angle OZL\equal{}\angle A'FE$     (5)
(4), (5) $ \Rightarrow\angle KZL\equal{}\angle BA'C\Rightarrow\angle KZL\equal{}\angle BAC$     (6)

(3), (6) $ \Rightarrow\angle KML\equal{}\angle KZL\Rightarrow$ the points $ K,L,M,Z$ are concyclic.

Best regards, Petrisor Neagoe :)

[b][u]A simple equivalent enunciation[/u].[/b] Let $ ABC$ be a triangle with the circumcentre $ O$ . For two points $ P\in AC$ , $ Q\in AB$ consider

the midpoints $ M$ , $ N$ of the segments $ [BP]$ , $ [CQ]$ respectively and the projection $ R$ of $ O$ on $ PQ$ . Prove that $ \widehat {MRN}\equiv \widehat {BAC}$ .

Let $M$, $M_b$, $M_c$ and $M_a$ be the midpoints of $XY$, $BP$, $CQ$ and $PQ$, respectively. Note that $M_a M_b \|AB$ and $M_a M_c \| AC$. Therefore $\angle{M_b M_a M_c} = \angle{BAC}$ and it suffices to show that $\angle{M_b M M_c} = \angle{BAC}$. Now let $P'$ be the reflection of $P$ in $M$ and $Q'$ be the reflection of $Q$ in $M$. Note that since $BP' \| MM_b$ and $CQ' \| MM_c$, it follows, if $BP'$ and $CQ'$ intersect on the circumcircle $\Gamma$ of $\triangle{ABC}$, then by arc-angle theorem $\angle{M_b MM_c} = \angle{BAC}$. Therefore it suffices to show that $BP'$ and $CQ'$ meet at a point on $\Gamma$.

Let $CM$ and $BM$ meet $\Gamma$ at $R$ and $S$, respectively. Now let $RQ$ intersect $\Gamma$ at $T$ and $ST$ meet $AC$ at $P''$. It follows by Pascal's theorem that $Q$, $M$ and $P''$ are collinear which implies that $P''$ lies on $XY$ and hence that $P'' = P$. Hence $S$, $P$ and $T$ are collinear. Now let $TM$ meet $\Gamma$ at $Z$ and let $CZ$ intersect $XY$ at $Q''$. By the Butterfly theorem, it follows that $MQ'' = MQ$ since $M$ is the midpoint of $XY$. Hence $Q'=Q''$ and $Z$, $Q'$ and $C$ are collinear. By the same argument, $B$, $P'$ and $Z$ are collinear. Hence $BP'$ and $CQ'$ meet at the point $Z$ on $\Gamma$, as desired.

Dear Mathlinkers,
you can also see my own synthetic proof in English on

http://perso.orange.fr/jl.ayme   vol. 10  A generalization of problem 2....

Sincerely
Jean-Louis


what is the remainder if 4^101 s divided by 101.it is tenth grade question

[hide="Hint (kind of a giveaway...)"]
Corollary of Fermat's little theorem
[/hide]
[hide="Solution"]By the corollary of Fermat's little theorem:
\[ 4^{101}\equiv \boxed{4}\pmod{101}\]
[/hide]


Joe is riding in a car down a straight road.  On one side there are electric poles that are evenly spaced with 100 feet between them.  Joe plays a weird game where he blinks every time the front of the car reaches an electric post and every time the back of the car reahes an electric post.  He realizes that if the car goes at a constant rate, he also blinks at a constant rate.  How long is the car?

This answer can be reached intuitively pretty quickly, but to give an algebraical proof...

[hide="Answer"]Say that the length of the car is $x$, so the time that the front of the car takes to go $x$ feet is the same as the time that the front of the car takes to $100-x$ feet. Therefore, $100-x=x \Rightarrow 2x=100 \Rightarrow x=50$.[/hide]

[quote="G-UNIT"][hide] I believe their are 2 lenghts.

The first is 50 feet, heres a visual aid to show us why this is correct:

0{x} - - - - - - - - - - - - - - - - - - - - - 100f

Lets say $x$ is the car that must travel to the electric pole at 100ft.  Since only the front must reach the end, the car must travel $ 100-x$ feet in order for the front of the car to touch the electric pole.  Now lets make another visual aid where the front of the car has reached the 100ft line.

0 - - - - - - - - - - - - - - - - - - - - - - - - - {x}100ft

Clearly in order for the back of the car to reach the 100ft line, it must travel $ x$ feet.  However each distance must be equal meaning:

$ 100-x = x $

$ 100 = 2x $

$ 50 = x$

  So the car must be 50 ft.

 Case 2: The car is 100ft

 Our visual aid in case 1, only worked when "100" was bigger than  $x$ , but what happens when $x=100$?  Then the front and and back will both meet at electric poles simultaneously, satisfying our condition that the boy in the car must blink at a constant rate ( 1 blink every 100 ft).  [/hide][/quote]
very good G-UNIT! You figured out that there is a little more to this problem, but there is still more

[quote="nat mc"][quote="JesusFreak197"]Ohh... nice solution. But if that's the case, then the car's length can be any multiple of 50 feet... correct?[/quote]
yep thats correct, any multiple of 50 will work[/quote]

ahhhhh, right, lengths over 100 occured to me but I immediately dismissed the idea because the rate in which the boy would blink would always remain the same (either 1 or 2 blinks a minute) but nice job on picking that up! =)


linear algebra
matrix
algorithm
algebra
linear equation
linear algebra unsolved

Consider the following iteration formula for estimating $ A^{ \minus{} 1}$ of an $ n\times n$ symmetric positive definite matrix $ A$,
\[ X_{n \plus{} 1} \equal{} X_n \plus{} q(AX_n \minus{} I),\qquad(*)
\]
where $ q$ is a constant and $ I$ is the $ n\times n$ identity matrix.
i) Under what conditions on $ q$ (in terms of the eigenvalues of $ A$), the iteration formula $ (*)$ converges to $ A^{ \minus{} 1}$?
ii) What should the value of $ q$ be so that the convergence is as fast as possible?

We choose on $ \mathcal{M}_n(\mathbb{R})$ the norm associated to $ \parallel{}.\parallel{}_2$, that is to say: $ \parallel{}A\parallel{}=\sqrt{\rho(AA^T)}$ where $ \rho(U)$ is the spectral radius of $ U$.
$ spectrum(A)=\{\lambda_i\}_i$ where $ 0<\lambda_1\leq\cdots\leq\lambda_n$ and $ \parallel{}A\parallel{}=\lambda_n$.
1) Let $ X_n=A^{-1}+Y_n$. Then $ Y_{n+1}=(I+qA)Y_n$ and $ \parallel{}Y_{n+1}\leq\parallel{}I+qA\parallel{}\parallel{}Y_n\parallel{}$. A SUFFICIENT condition for convergence is $ \parallel{}I+qA\parallel{}<1$ or $ sup\{|1+q\lambda_i|\}_i<1$.
ii) For the chosen norm the best choice for $ q$ is $ q=-2/(\lambda_1+\lambda_n)$. Indeed for this value we obtain the minimum value of $ \parallel{}I+qA\parallel{}$.

if you start with $ X_0 = \alpha A$, $ \alpha \in \mathbb{R}$, where $ A = UDU^t$ ($ U$ orthognal, $ D = diag(\lambda_1, \ldots, \lambda_n)$), then the problem is much easier: just study when does the sequence: $ u_{n + 1} = u_n + q( lambda_i u_n + 1)$ converge (fast). Hence one tries to find $ q$ such that $ \max_i |1 + q \lambda_i|$ is as small as possible. As loup blanc found, this gives:
$ q_{opt} = - \frac {2}{\lambda_{min} + \lambda_{max}}$ with a rate of convergence:
$ r = \frac{C(A)-1}{C(A)+1}$ where $ C(A)=\frac{\lambda_{max}(A)}{\lambda_{min}(A)}$
one also sees (with $ X_0 = \alpha X$) that the algorithm converges IFF $ - \frac {2}{\lambda_{max}}< q < 0$

Now, the problem is that one does not know, a priori, anything about $ C(A)=\frac{\lambda_{max}(A)}{\lambda_{min}(A)}$ so it may quite difficult to find the right coefficient $ C$. It would be interesting too to find a good way (conditioning) to decrease this number $ C$. In practice, one does not really care about $ A^{-1}$: one just needs to solve a linear equation
$ Ax=b$
But it is the same as $ UAx=Ub$ where $ U$ is an "easy invertible" matrix (ie: diagonal / triangular / etc.. ). So if one finds a good matrix $ U$ such that $ C(AU) \ll C(A)$, it would be much better ..


search
number theory proposed
number theory

Prove that for any $k > 0$ there exists $m > 0$ such that for all positive integers $n > m$ we have $d(2^n) > k$ (where $d(t)$ is the sum of the digits of $t$ in decimal representation).

A more general result here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=307511878&t=1666

Pierre.


Post your Fields Medal Prediction. Who is going to win the Fields Medal 2010? What is his area of research?

My prediction:
Artur Avila

I've seen Jacob Lurie mentioned as a candidate at various other forums/blogs. However, as someone over at Math Overflow mentioned, I'm not capable of judging his work, so I can't really say for sure how it stacks up.

Some of his work can be seen at his website here: http://www.math.harvard.edu/~lurie/

His Ph.D thesis, or "DAG" as it is sometimes called, gets a good bit of publicity over at Overflow.

You can also see a video of him lecturing on TQFT here:

http://lab54.ma.utexas.edu:8080/video/lurie.html

Artur Avila too.
He researches at dynamical systems.

August 19-27 right here in Hyderabad!!! I'm going!!

Maybe Ben Green? I can't really judge though...

Hm, I've never really followed the Fields Medal winners, to be frank, but I was wondering how one finds "candidates". Is there a formal pool of people that the Fields Medal will be awarded to, or is it just anyone (not literally anyone, of course) who has made significant contributions to the field?

Ngô Bảo Châu and Ben Green (In my humble opinion) have a very good chance of winning this year's Fields Medal.


Let $x,y$ be reals.Prove that $\left|\frac{1}{x^2+4}-\frac{1}{y^2+4}\right|\leqq\frac{1}{8}|x-y|$

let $ f(x) \equal{} \frac{1}{x^2\plus{}4}$
intermediate value theorem show $ f(x)\minus{}f(y) \equal{} f'(c)(x\minus{}y)$ where c belong to $ [x,y]$
but $ |f'(x)|\le \frac{1}{8}$ then the result.


Find the sum of the two smallest positive integers, each of which is a perfect square, a perfect cube, and a perfect fourth power.

I know 1 is one of them, what's another one?

If a number is a perfect 4th power, it is also a perfect square.

$n=x^3=y^4.$

So $2^{12}=4096?$

And the sum is 4097.

Are you sure this is from 1999-2000? I remember this problem from a mathcounts practice I had this year.  :?

well, maybe you practiced the round from 1999-2000


http://www.zippyvideos.com/

What do you think? I only saw 1 video, but they're cool!

Its kind of similar to Google Video and You Tube.  Still good though.


I've just registered for olympiad geometry. I missed the first class sessions. From the class transcript, I understand a forum for this class was created. However, I cannot see it. Is this normal?

Thanks in advance.


trigonometry
inequalities unsolved
inequalities

If α、β、θ∈(-90,90),and tanα+tanβ+tanθ=9,find the maximum of cos2α+cos2β+cos2θ.

May be easy for you ...but I can't...

Aren't anyone to help me... :( I extremely want to know the solutions....

hello, do you mean
If $ \alpha,\beta,\theta\in (\minus{}90^{\circ},90^{\circ})$,and $ \tan(\alpha)\plus{}\tan(\beta)\plus{}\tan(\theta)\equal{}9$,find the maximum of $ \cos(2\alpha)\plus{}\cos(2\beta)\plus{}\cos(2\theta)$.
Sonnhard.

Yes，exactly.I try to solve an inquality,finnally,have to find the max of that ...
Could you give the solution ...

:mad: Why there aren't anyone to help me...I have tried it several days already.I really need a solution...

hello, you can use the method of Langrange multipliers, defining
$ f(x,y,z,\lambda)\equal{}\cos(2x)\plus{}\cos(2y)\plus{}\cos(2z)\plus{}\lambda(\tan(x)\plus{}\tan(y)\plus{}\tan(z)\minus{}9)$. Now you must solve the system
$ f_x\equal{}0$
$ f_y\equal{}0$
$ f_z\equal{}0$
$ f_\lambda\equal{}0$.
Sonnhard.

:blush: If without calculus...How to do that?

$ a\equal{}\tan x$.
$ x^2\plus{}1\equal{}\tan^2x\plus{}1\equal{}\frac{1}{\cos^2x}$

We see that $ \cos 2x\equal{}2\cos^2x\minus{}1\equal{}\frac{2}{1\plus{}x^2}\minus{}1$

So the problem is $ x\plus{}y\plus{}z\equal{}A$, find the max of $ \sum\frac{2}{1\plus{}x^2}\minus{}1$

I believe that we can solve this by solving $ \alpha x\plus{}\beta \ge \frac{2}{1\plus{}x^2}$ for a double root at $ x\equal{}\frac{A}{3}$

Er....I solved it by shebyshev finally...


limit
real analysis
real analysis unsolved

Let $ S$ be a bounded set in $ \mathbb{R}^n$; let $ f: S \to \mathbb{R}$ be Riemann integrable over $ S$. Let $ D$ be the set of points of $ S$ at which $ f$ fails to be continuous and let $ E$ be the set of points $ x_0$  of Bd $ S$ at which the condition $ \lim_{x \to x_0} f(x) \equal{} 0$ fails to hold.

I'm in the process of showing that $ D$ and $ E$ have measure zero (Munkres outlines the proof) and I've shown that $ f_S$ ($ f$ on $ S$ and 0 elsewhere) is continuous for $ x_0 \notin D \cup E$ and not continuous on that set except possibly at isolated points of $ S$.

Why is $ B$, the set of isolated points of $ S$, necessarily countable?

Isolated points belong to disjoint (small!) balls. Each ball contains points with rational coordinates.


Give $n \in N$
Prove that there exit three different elements $a,b,c$ in the set {${n^2 , n^2 +1 ,...., n^2,n^2 +[3+\sqrt{n}] }$} so that $c \mid ab$

If you mean that a, b and c are different integers in $\{n^2, n^2+1, n^2+2, \ldots, n^ 2 + [\sqrt{n}\,] + 3\},$ then your statement is only true for $n \leq 2$.


Functional Analysis
advanced fields
advanced fields unsolved

A [i]von Neumann algebra[/i] is a self-adjoint algebra of operators on a complex Hilbert space which contains the identity operator and is closed in the strong operator topology.

Prove that a von Neumann algebra is commutative iff it contains no non-zero operators $A$ such that $A^{2}=0$.

I can do the trivial part. :blush: Suppose that the algebra is commutative and $A^{2}=0$. Then $\langle Ax,A^{*}y\rangle=\langle A(Ax),y\rangle=0$ for any $x,y\in\mathcal{H}$, which means that the ranges of $A$ and $A^{*}$ are orthogonal. Since also $AA^{*}=A^{*}A$, it follows that $A^{*}A=0$, hence $A=0$.

now the converse: we have $ x^{2}\equal{}0\Rightarrow x \equal{} 0$: this implies (in every ring), that the idempotent elements are central, because if $ p^{2}\equal{}p$, then easy calculations show $ (px\minus{}pxp)^{2}\equal{}0\equal{}(xp\minus{}pxp)^{2}$. the claim follows since the projections generate a dense subspace of the von neumann algebra (use borel functional caluclus with self adjoint elements, which generate the algebra).

This is valid in a more general context of a C-* algebra.
It is a theorem of Kaplansky.


inequalities
trigonometry
inequalities proposed

Prove that in any acute-angled triangle the following inequality holds:

  $ \frac {a \plus{} c}{\cos B} \plus{} \frac {b \plus{} a}{\cos C} \plus{} \frac {c \plus{} b}{\cos A}\geq4(a \plus{} b \plus{} c)$

We have, by Cauchy-Schwarz, that $ \left(\frac {b \plus{} c}{\cos A} \plus{} \frac {a \plus{} c}{\cos B} \plus{} \frac {a \plus{} b}{\cos C}\right)\geq \frac{(2a\plus{}2b\plus{}2c)^2}{(b\plus{}c)\cos{A}\plus{}(a\plus{}c)\cos{B}\plus{}(a\plus{}b)\cos{C}}$. But $ (b\plus{}c)\cos{A}\plus{}(a\plus{}c)\cos{B}\plus{}(a\plus{}b)\cos{C}\equal{}a\plus{}b\plus{}c$ because $ a\equal{}b\cos{C}\plus{}c\cos{B}$, and etc


limit
logarithms
real analysis
real analysis unsolved

Suppose that $a_{0}, a_{1}, \ldots$ is a sequence of positive numbers and

$\lim_{n\to\infty}a_{n}= k > 0$

Is it always true that

$\lim_{n\to\infty}\frac{\log\left(\prod_{j=0}^{n}a_{j}\right)}{n}= \log(k)$ ?

If so, prove it. If not, provide a counterexample.

Yes. Just convert to sums and apply continuity of the logarithm; this is the familiar theorem that $\lim \frac1n\sum_{k=1}^{n}b_{k}=L$ when $\lim b_{n}=L$.

Thanks. I shouldn't do math when I'm short on sleep -- I find myself doubting things that should be as ingrained as my parents' birthdays...


$ I$ is incenter. $ \triangle DEF$is pedal triangle of $ I$ with repect to $ \triangle ABC$
Obviously there exist two circle which tangent to quadrilateral $ BDIF$ and $ CDIE$ respectively.
Then, their common internal tangent line (not $ DI$) pass through $ A$.

Please post your beautiful solution :roll:

Thank you.


When I click "New Topic" in the Olympiad Forums, I always see the "Source" field, immediately below the "Subject" field.

When I click "New Topic" in the Middle School MATHCOUNTS Forum, I see the "Source" field.  When certain other users click "New Topic" in the Middle School MATHCOUNTS Forum, they do not see the "Source" field.

When I click "New Topic" in the Middle School Basics Forum, I do not see the "Source" field.

If it's designed to work this way, can you point me to the rule for when we do or do not see the "Source" field.  If it's an oversight, I would like to suggest that the "Source" field be available on all the forums which are expected to contain posted problems.  Thanks!

This issue has been discussed before. See http://www.mathlinks.ro/Forum/viewtopic.php?t=23778

[quote="rcv"]
If it's designed to work this way, can you point me to the rule for when we do or do not see the "Source" field.  If it's an oversight, I would like to suggest that the "Source" field be available on all the forums which are expected to contain posted problems.  Thanks![/quote]

I'll note that the main reason we left it off the forums that first-time users typically see on AoPS is that it caused a lot of confusion for them.  We hope to eventually have a thorough 'how to use AoPS' tutorial someday which will hopefully clear up issues like this.


As in the subject line suggests, maybe this post is a little more to do with computer programming than math (but as my math book said that the original purpose of the computer was to do mathematics and calculations faster - something surely mathematical).

Anyway, my question is this: [b]How do computers (AND calculators) add, subtract, multiply, and divide - I was thinking that they add (and subtract) and multiply (and divide) much like humans: adding or multiplying the numbers one by one, but I heard somewhere that they also have something to do with turning circuits on and off - please help!!????? How do they do logs and fractional exponents????????[/b]

[hide="Theory"]
Tiny magic gnomes who slave away in your calculator to your every calculation need.
[/hide]

Really, though, [url=http://answers.yahoo.com/question/index?qid=20070415031820AAqLalU]check this out.[/url].

A modern electric calculator actually doesn't have many parts in it. If you take one apart, you'll find a plastic backplate, screws, a lithium battery, keyboard membrane, display, keys, an on/off switch, a processor chip, and keyboard sensors.

This is a pretty good picture from explainthatstuff.com:

[img]http://www.explainthatstuff.com/insidecalculator.jpg[/img]

Input's the keyboard with a bunch of keys with a keyboard membrane underneath and a circuit under that; processor's the chip that does all the work; output is the display that shows all the numbers and answers; and the power source is the battery or solar panel.

As you press a key on your calculator, you're compressing the keyboard membrane. Each key as a little rubber button underneath it, and an empty space beneath that, so when you press the key, it squashes the rubber button against the membrane. The button then pushes and makes electrical contact with two keyboard sensors and the circuit detects it. The processor chip then figures out what key you pressed, and a circuit inside it activates certain segments on the display, depending on the key you pressed. The same thing happens as you press more number keys, but something different occurs if you press a $ \plus{}, \minus{}, \div,$ or $ \times$ key. Like let's say you hit the $ \times$ key. The calculator will take the number you originally pushed and store it in a thing called a register. Then, after you type in another number and hit $ \equal{}$, it adds the two numbers together and displays the result on the screen.

If you study the numbers on a calculator screen, you'll probably notice that it's made up of 7 segments. The calculator displays a certain combination of segments that resembles the number you pressed. This is why it's difficult for a calculator to show letters, and when they do, it's not as clear as a computer screen, which uses pixels.

You've probably read about how computers use binary numbers(0 and 1.) Calculators use the same thing. They have millions of little electronic switches called transistors that are either on or off. To store decimal numbers, the calculator switches off a series of transistors in a binary pattern.

Thanks a lot!  :D

But maybe I should make my question more clear...  :blush:

[b]My question was not what happens to the number you input to the calculator (although that was very interesting, thanks very much :) ) but rather does the calculator memorize the fact that 1 + 1 = 2, 3+6 = 9, 4 X 6 = 24, etc - the basic operations add and multiply and the first place values - so that when it does a more complex addition or multiplication problem - 11+1 = then the can take 1+1 = 2 and 1 + 0 = 1 and then put them together to get 12????????? If not, how do they do it???? Do computers do the same thing????? If no, what do they do???? And what's up with the binary thing - I still don't really get it? Any help or insight into this problem is greatly appreciated. I thank you in advance for your help.[/b]

For addition, they only memorize these four steps:
0 + 0 → 0
0 + 1 → 1
1 + 0 → 1
1 + 1 → 0 and carry

Internally, they do not do ANYTHING in base 10. When we input a base 10 number, a program converts that to base 2 before the computer does anything with it. Again, there are many sources on the Internet available to help you with your query.

Thanks! :)

Like explainthatstuff.com? Are there any more recommendations?

[quote="theax"]but they would also have to memorize how to convert regular base 10 numbers into base 2 numbers and back, right??????????

So: do the computers have to memorize the 3+6 = 9 and 4 X 9 = 36 and stuff like that or do they memorize the binary (and conversion back and forth) or BOTH?

Again: [b]My question was not what happens to the number you input to the calculator (although that was very interesting, thanks very much Smile ) but rather does the calculator memorize the fact that 1 + 1 = 2, 3+6 = 9, 4 X 6 = 24, etc - the basic operations add and multiply and the first place values - so that when it does a more complex addition or multiplication problem - 11+1 = then the can take 1+1 = 2 and 1 + 0 = 1 and then put them together to get 12????????? If not, how do they do it???? Do computers do the same thing????? If no, what do they do???? And what's up with the binary thing - I still don't really get it? Any help or insight into this problem is greatly appreciated. I thank you in advance for your help.[/b][/quote]
[hide]informally, you could define "add1" like this(for positive integers lets say, just to illustrate the point.)
Look at the last digit of x. if it is a 0, out put x, but with the last digit replaced with 1. if it is a 1, output x, with the last digit replaced with a 2. etc. if it is a 9, output put a 0 connected to the end of 1+ what remains when you take away the last digit.

EX. Add1(1)=2; ADD1(3) = 4; Add1(999) = Add1(99) connected to 0 = Add1(9) connected to 00 = Add1(0) connected to 000 = 1 connected to 000=1000
So it memorizes the special facts 0+1=1,1+1=2,2+1=3,....,8+1=9, and then i tell it what to do in the special case where the last digit is 9, or basically i tell it how to carry over.

Now, once we have adding 1 defined, we can define what it means to add two numbers x and y:

define "add"
imput variables(x,y)
evaluate Count(x,y,0)

Define count
imput variables(x,y,z)
does z=y? if so output x.
otherwise, evaluate count(add1(x)),y,(add1(z))

EX. Add (3,2)=count(3,2,0){z does not equal y, so}= count(4,2,1) = count(5,2,2) {z=y so output is x, or 5}=5. So we have defined what it means to add two natural numbers. Now we define what it means to multiply two natural numbers, using the fact that a*b=a+a*(b-1):

define multiply
imput(x,y)
does y=0? if so output 0.
otherwise, evaluate add(x, multiply(x,(y-1)))

EX. 2*3. does y=0? no, so evaluate add(2, 2*(3-1))=add(2,2*2=add(2,(add(2,2*1) = add(2,add(2,add(2,2*0)=add(2,add(2,add(2,0) = add(2,add(2,2)= add(2,4) = 6.

(i skipped over defining "-" but it is not hard)

So in a nutshell, first you teach it(the computer) how to add1, and then build everything off of that.[/hide]

This is how i learned to do it in a programming course i took. Notice the use of recursion (where you use a function in it's definition); it is a standard tool.

But that is not how computers do it. When you add 536870912 to 536870912, it will not perform the "add1" operation 536870912 times.

wait, they convert the numbers into binary and then use logic gates to find the answer, right???


inequalities
vector
rearrangement inequality

Let $ f: N \minus{} > N$ be an injective function.Prove that $ a_n \equal{} \sum_{i \equal{} 1}^n \frac {1}{k f(k)}$ is convergent.

[hide]Let $ \sigma_n(1)$ be the maximum of the set $ \{f(1),f(2),\ldots,f(n)\}$ and define $ \sigma_n(k)$, for $ k\le n$ recursively so that $ \sigma_n(k)$ is the maximum of the set $ \{f(1),f(2),\ldots,f(n)\} \minus{} \{\sigma_n(1),\sigma_n(2),\ldots,\sigma_n(k\minus{}1)\}$. Then by the rearrangement inequality, \[ a_n \equal{} \sum_{k \equal{} 1}^n \frac {1}{k f(k)}\le\sum_{k \equal{} 1}^n \frac {1}{k\sigma_n(k)}.\] Since $ f$ is injective, then $ \sigma_n(k)\ge k$, so then \[ a_n\le\sum_{k \equal{} 1}^n\frac{1}{k^2}.\] As $ a_n$ is increasing and bounded above by $ \frac{\pi^2}{6}$, then the sequence converges.[/hide]

Is it possible to use the Cauchy Schwartz inequality?

(I'm asking because the two sequences are infinite)

Yeah, it still works

Let $ \langle x,y \rangle$ be the inner product of vectors $ x$ and $ y$. Cauchy-Schwartz** states that $ \big| \langle x,y \rangle \big| \leq \| u \| \| v \|$ for vectors $ u$ and $ v$ of a real or complex inner product space, including infinite dimensional ones.

** [i]Squaring this identity and writing it in terms of orthonormal components gives the more familiar form[/i]


I seem to be having problems and I am just going to list examples on the problems I do need refreshing in, but be advised thier may be many.

Please if you can help....Thanks!

1. [u]5x+2y[/u]
       35x+14y (Simplify)

2. [u]8      +    x+3[/u]
      7+2x        2x=7 (Perform operation and Simplify)

3. [u]x-2  =  1[/u]
        4x       6 (Solve)

4. y=2x+5
a.(0,  ) B.(3,  )

Hello Exteme Angel Baby (interesting name..), welcome to the forums. Before anyone can help you they need to be able to understand what you're asking. When writing fractions, don't try to be fancy and make it take up two lines and underline the top part. Just say (2x + 3)/(4x + 6), or whatever. Some of your problems don't make any sense. Also I'm not sure what problem 4 is asking. You just need to fill in the y-coordinates of those two points that are on the given line? Finally, this stuff looks very simple so you might want to post it on the middle school forum.

[hide="Click here for some hints"]

1. $\frac{5x+2y}{35x+14y}$

For this one, see if you can transform the numerator by multiplying by a constant to make something similar to the denominator.

2. I don't understand the question, sorry. Try writing it in form like Antonio suggested.

3. $\frac{x-2}{4x} = \frac{1}{6}$

For this one, I would say to try cross multiplying. Multiply the numerator of the left side with the denominator of the right side, and set that product equal to the denominator of the left side multplied by the numerator of the right side. In other words:

$\frac{a}{b} = \frac{c}{d}$

is the same as
$ad = bc$

4. For the last question, I believe it is asking to subsititute the given vlaues and find the y coordinate. Yoiu are given the x coordinate for the set point (x,y)
So take that number and plug it into the given equation

I hope these help. Post again if you would like more explanation. [/hide]

8 + x+3
7+2x 2x=7

Do you mean

(8+x+3)/(7+2x) with 2x=7

If so 2x=7 so x=3.5

Now plug in and substitute


abstract algebra
group theory
superior algebra
superior algebra unsolved

Write $ G\equal{}Z/6Z$. Prove or disprove that $ Aut(G)\equal{}Inn(G)$. Thank you for the solution before.

I don't know anything about it, so I ask the problem here. Please explain to me briefly. Thanks in advance.

Something Ian Stewart wrote in the introduction for Polya's "How to Solve It": How do you expect to solve a problem if you don't understand what the question is?

$ Aut(G)$ is the group of automorphisms of $ G$, that is the group of all isomorphisms $ \phi: G \to G$. Every element $ x$ of $ G$ defines such an automorphism $ \phi_x$ via conjugation: $ \phi_x(g) \equal{} x^{\minus{}1} g x$. $ Inn(G)$ is the group of all such inner automorphisms.

When $ G$ is abelian, $ Inn(G)$ is obviously trivial. I have a suspicion that you're misquoting the question, and that it's actually asking about automorphisms of $ S_6$, the symmetric group on 6 elements. Of all the symmetric groups, $ S_6$ is the only one that does have outer automorphisms. That's a fairly hard question, but it's a lot more reasonable than asking about inner automorphisms of an abelian group.


vector
geometry
abstract algebra
superior algebra
superior algebra unsolved

If f and g are in the dual space of a vector space V over a field F, such that f(v)=0 implies g(v)=0, prove that g=cf for some c in F.

Oh.....well, that's a problem from our textbook...frankly I wouldn't know whether or not it is true or false.

I can't think of a counterexample off hand, though...

It's certainly true for finite-dimensional spaces. (assign an inner product, use some geometry)

So... what's wrong with this?

So, $ \ker f \subset \ker g$, so one would figure we can factor $ g \equal{} hf$ for some map $ h : F \to F$; therefore $ g \equal{} cf$ for some $ c \in F$?

Why would Ker(f) be contained in Ker(g), wouldn't it be the other way around?

[quote="Nukular"]So, $ \ker f \subset \ker g$, so one would figure we can factor $ g \equal{} hf$ for some map $ h : F \to F$; therefore $ g \equal{} cf$ for some $ c \in F$?[/quote]
I'm quite unclear on how we figure this.  I can see how we can factor $ g$ through the quotient map $ V\to V/\ker{f}$, but not through $ f$ itself.

But isn't $ f$ the map $ V \to V / \ker f$? It has kernel precisely $ \ker f$. (I'm going by eyeball here and might be completely wrong of course -- haven't done any work out on paper).

f is the map V-->F.
I think the question is asking if we are in the case such that f(v)=0 implies g(v)=0, then only can we say that g=cf, c in F.

OK, rather I was pointing out that the first isomorphism theorem gives an isomorphism $ V / \ker f \to Im(f) \subset F$. If $ f$ is nontrivial, then it is surjective onto $ F$ and so $ V / \ker f \cong F$ by the induced map $ \widetilde{f}$; since $ g$ factors through $ V / \ker f \equal{} F$, the result would follow, no?

Stepping thru: assume $ f$ is nontrivial, as otherwise $ g$ is also trivial.

take $ \pi : V \to V / \ker f$,
the map $ f : V \to F$.

Now there is a unique map $ \widetilde{f} : V / \ker f \to F$ such that $ f \equal{} \widetilde{f} \pi$; by the first isomorphism theorem $ \widetilde{f}$ is an isomorphism.

We have a factorization $ g \equal{} k \pi$ for some map $ k : V / \ker f \to F$. Then $ k\widetilde{f}^{\minus{}1} : F \to F$ must be multiplication by $ c$ for some constant $ c$; let $ \mu_c : F \to F$ denote this map. Then $ g \equal{} (c \widetilde{f}) \pi \equal{} cf$.

Ohhh, okay.

Can you explain this further? It's not coming that easily to me.

[quote="Nukular"]$ \ker f \subset \ker g$ because if $ f(v) \equal{} 0$ then $ g(v) \equal{} 0$; in other words every $ v \in \ker f$ is also in $ \ker g$.[/quote]

Sure:

Claim $ \ker f \subset \ker g$:

Proof: Let $ v \in \ker f$. Then $ f(v) \equal{} 0$ (By definition). Therefore $ g(v) \equal{} 0$ (given property). So $ v \in \ker g$ (By definition). Therefore $ \ker f \subset \ker g$.

[quote="Nukular"]Now there is a unique map $ \widetilde{f} : V / \ker f \to F$ such that $ f \equal{} \widetilde{f} \pi$; by the first isomorphism theorem $ \widetilde{f}$ is an isomorphism.

We have a factorization $ g \equal{} k \pi$ for some map $ k : V / \ker f \to F$. Then $ k\widetilde{f}^{ \minus{} 1} : F \to F$ must be multiplication by $ c$ for some constant $ c$; let $ \mu_c : F \to F$ denote this map. Then $ g \equal{} (c \widetilde{f}) \pi \equal{} cf$.[/quote]

I don't see where the products for $ f \equal{} \widetilde{f} \pi$ and $ g \equal{} k \pi$ come from.

In general if $ h : A \to B$ is a linear map and $ C \subset A$ is a subspace satisfying $ C \subset \ker h$, then it follows that there is a factorization $ h : A \to A / C \to B$. I am just using that factorization.


linear algebra
matrix
algebra
polynomial
complex numbers
superior algebra
superior algebra solved

I ussualy do not like to generalize problems. Anyway, there are problems that become really interesting in the general case. In Romania there is a tradition of generalizations. So, I will generalize something.
   Let p a prime. Find all numbers n such that we can find nxn matrices A,B with complex entries such that (AB-BA)^p=I.
   Source: adapted by Harazi from lots of particular cases.

Don't you like this one? I think it is much more interesting than all particular cases.

Moubinool proved that for a matrix M, there are matrices A,B (all 3 square matrices) such that
M=AB-BA iff Tr(M)=0.

SO the problem comes to:
Find all n such that there exists a matrix M, TrM=0 such as Mp=I.

The second condition tells us that all the eigenvalues of M are p-unity roots, and the first condition tells us that the sum of the eigenvalues is 0.
-----------------------------------------------------------------------------------
So the problem comes to:
For what n there are x1,...,xn complex numbers such as xip=1 for all i, and \sum1 \leq i \leq n xi=0.

n must be divisible by p. If it is, the complex roots obviously exist. Suppose that we have x_i such that (x_i)^p=1 and sum x_i=0. Let eps=cos(2pi/p)+isin (2pi/p). Let a_i the number of indices j such that x_j=eps^i. Then a_0+a_1*eps+...+a_p-1*eps^p-1=0. Thus the polynomial a_0+a_1*X+....+a_p-1*X^p-1 has a common root with the ireducible polynomial 1+X+...+X^p-1 and thus the two polynomials are associated in divisibility, from where a_0=...=a_p-1 and thus p divides n. I really like this problem.


My second and last one for the today

A triangle has sides of lengths a, b, and c (a >=b>= c) and area = 1. Then b can not be less than

Is that the question?  Find the lowest value of $ b$ that will work?

Wait quick question, whats that symbol in the question?  :huh:

Ignore that must be some "junk"..It just means $ a>\equal{} b >\equal{} c$

I'm not sure if this is right.

[hide]$ b \ge \sqrt{2}$.

We have the area of the triangle being $ \displaystyle\frac{bc\sin \theta}{2}$. We want to find the minimum $ b$ such that $ \displaystyle\frac{bc\sin \theta}{2} \ge 1$. We can maximize $ \sin \theta$ by setting it equal to 1. Then $ bc \ge 2$. Since $ b \ge c$, $ \boxed{b \ge \sqrt{2}}$[/hide]

[quote="darkdieuguerre"]I'm not sure if this is right.

[hide]$ b \ge \sqrt {2}$.

We have the area of the triangle being $ \displaystyle\frac {bc\sin \theta}{2}$. We want to find the minimum $ b$ such that $ \displaystyle\frac {bc\sin \theta}{2} \ge 1$. We can maximize $ \sin \theta$ by setting it equal to 1. Then $ bc \ge 2$. Since $ b \ge c$, $ \boxed{b \ge \sqrt {2}}$[/hide][/quote]

Its perfect. One small point, i guess it was given that area is $ \equal{}$ to 1 and not  $ > \equal{}$ 1. But still it does not change the solution.

Nice work


Solve for x

$\sqrt{x+\sqrt{x+\sqrt{x}}}=4$


[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers and let
$ A = \sum \frac {b^2 + c^2}{2a^2}, \quad \quad B = \sum \frac {2a}{b + c}.$
Prove that $ 4A + 69 \ge 27B.$
Equality holds if and only if $ (a,b,c) \sim (1,1,1)$ or $ (a,b,c) \sim (2,1,1).$
:)[/quote]
SOS kill's it! :lol:
$ 4\sum_{cyc}\frac {b^2 + c^2}{2a^2}+69\geq27\sum_{cyc} \frac {2a}{b + c}\Leftrightarrow$
$ \Leftrightarrow4\sum_{cyc}(\frac {b^2 + c^2}{2a^2}-1)\geq27\sum_{cyc} (\frac {2a}{b + c}-1)\Leftrightarrow$
$ \Leftrightarrow2\sum_{cyc}(\frac{c^2-a^2}{a^2}-\frac{a^2-b^2}{a^2})\geq27\sum_{cyc}(\frac{a-b}{b+c}-\frac{c-a}{b+c})\Leftrightarrow$
$ \Leftrightarrow2\sum_{cyc}(\frac{a^2-b^2}{b^2}-\frac{a^2-b^2}{a^2})\geq27\sum_{cyc}(\frac{a-b}{b+c}-\frac{a-b}{a+c})\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a-b)^2(\frac{2(a+b)^2}{a^2b^2}-\frac{27}{(a+c)(b+c)})\geq0.$
Let $ a\geq b\geq c$ and $ S_c=\frac{2(a+b)^2}{a^2b^2}-\frac{27}{(a+c)(b+c)}.$
Hence, $ S_a=\frac{2(b+c)^2}{b^2c^2}-\frac{27}{(a+b)(a+c)}\geq\frac{2(a+c)^2}{a^2c^2}-\frac{27}{(a+b)(b+c)}=S_b$
and $ S_b\geq0$ because $ 2(a+c)^2(a+b)(b+c)\geq4(a+c)^3c=$
$ =\frac{1}{2}(a+a+2c)^3c\geq\frac{1}{2}\cdot27a^2\cdot2c\cdot c=27a^2c^2.$
Hence, $ \sum_{cyc}S_c(a-b)^2\geq(S_b+S_c)(a-b)^2.$
Thus, it remains to prove that $ S_b+S_c\geq0.$ But $ S_b+S_c\geq0\Leftrightarrow$
$ \Leftrightarrow(2(b^2+c^2)a^2+4bc(b+c)a+4b^2c^2)(a+b)(a+c)(b+c)\geq27b^2c^2(2a+b+c)a^2,$
which is true by AM-GM too. :wink:

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers and let
$ A \equal{} \sum \frac {b^2 \plus{} c^2}{2a^2}, \quad \quad B \equal{} \sum \frac {2a}{b \plus{} c}.$
Prove that $ 4A \plus{} 69 \ge 27B.$
Equality holds if and only if $ (a,b,c) \sim (1,1,1)$ or $ (a,b,c) \sim (2,1,1).$
:)[/quote]

It's my old problem, my friend. I found it when I made powerful a result of Vasc. :)

We can prove also this problem by $uvw$ because this inequality is equivalent to $f(w^3)\geq0$, where $f$ is a concave function.
$f(w^3)=-25w^9-150u^3w^6+163uv^2w^6-108u^4v^2w^3+54u^2v^4w^3+12v^6w^3+162u^3v^6-108uv^8$.

[quote=can_hang2007]Let $ a,b,c$ be positive real numbers and let
$ A \equal{} \sum \frac {b^2 \plus{} c^2}{2a^2}, \quad \quad B \equal{} \sum \frac {2a}{b \plus{} c}.$
Prove that $ 4A \plus{} 69 \ge 27B.$
Equality holds if and only if $ (a,b,c) \sim (1,1,1)$ or $ (a,b,c) \sim (2,1,1).$
:)[/quote]

We have
\[LHS-RHS=\frac{1}{8}\sum{\frac{(7ca+25ab+c^2+7bc)(a+b-3c)^2(a-b)^2}{a^2b^2(a+c)(b+c)}}+\frac{21(a-b)^2(b-c)^2(c-a)^2}{abc(a+b)(b+c)(c+a)}\ge{0}\]


1.) Show the if $ n$ is prime, then $ (n\minus{}2)! \equiv 1 mod n$.

2.) Show that if $ n > 4$, and $ n$ is composite then, $ n|(n\minus{}1)!$.

[quote="guile"]1.) Show the if $ n$ is prime, then $ (n \minus{} 2)! \equiv 1 mod n$.

2.) Show that if $ n > 4$, and $ n$ is composite then, $ n|(n \minus{} 1)!$.[/quote]
[hide="1"]
By Wilson's Theorem, we have that $ (n \minus{} 1)!\equiv \minus{} 1\bmod n$ and also \[ n \minus{} 1\equiv \minus{} 1\bmod n\implies ( \minus{} 1)(n \minus{} 2)!\equiv (n\minus{}1)(n\minus{}2)!\equiv (n\minus{}1)!\equiv \minus{}1\bmod n\implies (n \minus{} 2)!\equiv 1\bmod n\] by multiplying both sides by $ \minus{}1\bmod n$. [/hide]

[hide="2"]
Let $ a$ and $ b$ be factors of $ n$, such that  $ ab \equal{} n$, $ a$ and $ b$ are integers, and $ 1 < a,b < n$.  Therefore, we have that $ 1 < a,b\le (n \minus{} 1)\implies ab|(n \minus{} 1)!\implies n|(n \minus{} 1)!$. [/hide]
EDIT: Sorry, [b]guile[/b], I made a typo.  Solution edited.

Additional information on The QuattoMaster 6000's #2 solution: $ a|(n\minus{}1)!$ because there are at least $ b\minus{}1$ factors of $ a$ in $ (n\minus{}1)!$ and likewise at least $ a\minus{}1$ factors of $ b$ in $ (n\minus{}1)!$. Note that for $ n\equal{}4,a\equal{}b\equal{}2$, there is only one factor of 2 in $ (4\minus{}1)!$, which shows why $ n\equal{}4$ fails to satisfy problem 2.

Can you please explain why  $ n \minus{} 1\equiv \minus{} 1\bmod n\implies ( \minus{} 1)(n \minus{} 2)!$?


algebra
function
domain
polynomial
calculus
integration
superior algebra

is an atomic ring a common term ?
i mean i couldnt find it but here
http://books.google.com.au/books?vid=ISBN155608000X&id=WzbzxytdAuUC&pg=RA4-PA276&lpg=RA4-PA276&ots=iTgkBFpUs4&dq=%22atomic+ring%22&sig=3Ih6gXpvAKaFo473T6iEfwSj14k#PRA4-PA276,M1
i had on the test a problem wich was saying to proove that
$R=\mathbb{R}+X\mathbb{C}[X]$ is an atomic ring
(in my lecture notes i have that an atomic ring should be the
one in wich every element is supposed to decompose under a product
of irreducible terms.
coleagues of mine suggested that one should use
bounded factorization domain notion to deal with this problem.
the issue that i was having is that i didnt knew what that was.
and i thought i could use the fundamental theorem of algebra
to proove this,by just saying that
if $f\in R$ then
$f=\prod_{r_{i}\ is \ root\ of f}(X-r_{i})$
otherwise written as
$f=a\prod_{i=1}^{n}(X-r_{i})\prod_{i=1}^{m}(X-c_{i})$
wher $r_{i}$ are real roots and $c_{i}$ are complex ones.
so the real roots cause divisors of the form $(X-r_{i})\in R$
but the complex ones $(X-c_{i})\notin R$,but i know there was
also a therorem wich stated that if a polynomial has a complex
root it has also its cojugate so $(X-c_{i})(X-\hat{c_{i}})=X^{2}-X(c_{i}+\hat{c_{i}})+c_{i}\hat{c_{i}}$
now this is in $R$ and i think its also irreducible...but i cant proove it
can anyone help with either the bounded factorization proof or
the one ive tried ?
(if one can proove that $(X-c_{i})(X-\hat{c_{i}})$is irreducible then
the problem ,i think is settled).

since you found no information about it: http://eom.springer.de/A/a013880.htm

[quote]An element of the ring is called an atom, or an extremal element, or an indecomposable element, if it cannot be decomposed into a product of non-invertible elements. An integral domain is an atomic ring if and only if each non-zero non-invertible element is a product of atoms.[/quote]

Firstly, note that the polynomials of the form $aX+b$, with $a \in \mathbb C, \, b \in \mathbb R$, are atoms in $\mathbb A = \mathbb R+X \cdot \mathbb C[X]$ (the non-invertible elements are the complex polynomials of degree $\geq 1$ with real "free term").

Secondly, $\mathbb A$ is clearly an integral domain.

Thirdly, each non-zero non-invertible element is a product of atoms. Indeed, let $f \in \mathbb A$, $f(X) = X^{k}\cdot \prod_{i=1}^{n}\left( X-\alpha_{i}\right)$, where $k \geq 0$ is the multiplicity of $0$ in $f$ and $\alpha_{i}$ are the non-zero roots (possibly with repetitions). Next, do some algebra:
\[f(X) = X^{k}\prod_{i=1}^{n}\alpha_{i}\left( \frac{X}{\alpha_{i}}-1 \right) = X^{k}\cdot \left( \alpha_{1}\alpha_{2}\ldots \alpha_{n}\right) \cdot \prod_{i=1}^{n}\left( \frac{X}{\alpha_{i}}-1 \right) . \]
Case 1: $k=0$.
By Viete, $\alpha_{1}\alpha_{2}\ldots \alpha_{n}= (-1)^{n}\cdot \textrm{free term}\in \mathbb R \setminus \left\{ 0 \right\}$. Thus, $\alpha_{1}\alpha_{2}\ldots \alpha_{n}$ is an atom. The linear factors are clearly atoms, so that's it.

Case 2: $k \geq 1$.
Write $f(X) = \underbrace{X \cdot \ldots \cdot X}_{k-1 \, \textrm{times}}\cdot \left( \alpha_{1}\ldots \alpha_{n}X \right) \cdot \prod_{i=1}^{n}\left( \frac{X}{\alpha_{i}}-1 \right)$. These are all atoms :)


[quote="Drunken_Math"]Compute the last two digits of this number:

$X=1+11+111+1111+11111+\cdots+1111\cdots1111$

Where the last number has 100 1's.[/quote]

[hide]
Since we're just finding the last 2 digits, only the last two digits of the sequence would matter.

So we have need to find the last 2 digits of $1 + 11\cdot99$.
That is equal to $1089+1 = 1090$

The last 2 digits of that is $\boxed{90}$
[/hide]


Find all functions $f: \mathbb R \to \mathbb R$ which satisfies
\[f(f(x+y))=f(x+y)+f(x) \cdot f(y)-xy\]
for all reals $x$ and $y$.

Switching x and y in the equation, $ f(f(x\plus{}y))\equal{}f(x\plus{}y)\plus{}f(x)f(y)\minus{}y$, which in combination with the original equation yields x=y, which is clearly not true for all real x and y.

I'm sorry. I have fixed the problem  :blush:

[quote="rogueknight"]find all f that: f R->R:
$ f(f(x \plus{} y)) \equal{} f(x \plus{} y) \plus{} f(x)f(y) \minus{} xy$[/quote]

Let $ P(x,y)$ be the assertion $ f(f(x \plus{} y)) \equal{} f(x \plus{} y) \plus{} f(x)f(y) \minus{} xy$

$ P(x \plus{} y,0)$ $ \implies$ $ f(f(x \plus{} y)) \equal{} f(x \plus{} y) \plus{} f(x \plus{} y)f(0)$ and so (subtracting $ P(x,y)$) we get the new assertion $ Q(x,y)$ : $ f(x \plus{} y)f(0) \equal{} f(x)f(y) \minus{} xy$

$ Q(f(0), \minus{} f(0))$ $ \implies$ $ f(f(0))f( \minus{} f(0)) \equal{} 0$ and so $ \exists u$ such that $ f(u) \equal{} 0$ and then $ Q(x \minus{} u,u)$ $ \implies$  $ f(x)f(0) \equal{} u(u \minus{} x)$

If $ f(0)\neq 0$ $ f(x) \equal{} ax \plus{} b$ and plugging this in original equation, we get $ f(x) \equal{} x$, which is wrong since then $ f(0) \equal{} 0$ but we supposed $ f(0)\ne 0$ in this line

So $ f(0) \equal{} 0$ and $ Q(x,y)$ becomes $ f(x)f(y) \equal{} xy$ and so either $ f(x) \equal{} x$, either $ f(x) \equal{} \minus{} x$ and plugging this in original equation, we get $ f(x) \equal{} x$

hence the unique solution $ \boxed{f(x) \equal{} x}$ $ \forall x$


If $ a_0,a_1,.......$ are real numbers such that
\[ (1 \plus{} z)^n  \equal{} a_0  \plus{} a_1 z \plus{} a_2 z^2  \plus{} ... \plus{} a_n z^n
\]
for all complex numbers $ z$, then find the value of
\[ (a_0  \minus{} a_2  \plus{} a_4  \minus{} a_6  \plus{} ....)^2  \plus{} (a_1  \minus{} a_3  \plus{} a_5  \minus{} a_7  \plus{} ....)^2
\]

[hide="Big hint"]Put $ z\equal{}i$[/hide]


Because I'm so bad at Physics, I thought I'd enlist the help of some AoPSers.

A jar sits on top of a piece of paper... if the paper is pulled quickly, the jar does not move.  Why?

Using formulas, prove there is a relationship between temperature and the kinetic energy fo molecules.  What is this relationship?

If you throw a ball up in a car moving at a constant velocity, where does it land?  Why?  If you throw a ball up in a car that is accelerating, where does it land?  Why?

Thanks!

I've done my best to answer; some may be wrong:

[quote="Ragingg"]Because I'm so bad at Physics, I thought I'd enlist the help of some AoPSers.

A jar sits on top of a piece of paper... if the paper is pulled quickly, the jar does not move.  Why?[/quote]

I don't know, but I think maybe when you pull, you jerk the jar up a little to temporarily create zero normal force.

[quote]Using formulas, prove there is a relationship between temperature and the kinetic energy fo molecules.  What is this relationship?[/quote]

The relationship is that they are proportional.  I don't know what formulas they want us to use (I think specific heat formulas are probably what they're looking for).

[quote]If you throw a ball up in a car moving at a constant velocity, where does it land?  Why?  If you throw a ball up in a car that is accelerating, where does it land?  Why?[/quote]

The ball is moving at the same velocity as the car horizontally when it is thrown.

[quote="Ragingg"]
Using formulas, prove there is a relationship between temperature and the kinetic energy fo molecules.  What is this relationship?

[/quote]

Well i cant prove it using formulae but I know that temperature is equal to the the average kinetic energy of all the particles in a given sample... i dunno if that helps  :?

[quote="mna851"]For Question 1:

The reason why the jar would move if the paper is pulled slowly is because when the paper is moved the static frictional force between the paper and the jar causes the jar to move.  If you exert a force that is greater than the static frictional force the jar will not move.[/quote]

Oh, right, this makes sense.  Only you should note that the force must actually be quite fast to give the friction little time to act (since it acts even if you overcome it).

i think we can explain problem one by the knowledge of engegy, like what my teacher did in our olympiad course.
It is like a bullet shoot into an object with different speed.

Does inertia have anything to do with question 1?

[quote="Ragingg"]:)

That would be it Stupidkid.... Inertia for Problem 1.  Except I don't know why that makes sense.[/quote]
w00t, I have never taken a physics course (other than a physics chapter in 8th grade science) so I don't understand all the force and friction things people were talking about.  I just thought since inertia is when objects tend to stay at rest, so if not enough force is exerted on the bottle (e.g. pulling paper quickly generates little force) it will still stay at rest.
Well that s just my guess because i m not good at phyics either.

It's so nice to find physics on a math forum--I'm thrilled. My major is physics contest  :D but in my school physics is not popular --so changed into math

[quote]A jar sits on top of a piece of paper... if the paper is pulled quickly, the jar does not move. Why? [/quote]
1)using the formula:momentum law
mv=Ft
the force F is friction--depends on miu and N, is unchanging. Mass m is unchanging--the only way to make v(velocity) smaller is to decrease time t. When you pull the paper quickly which creats a tiny t--the jar moves slightly or, can be considered unmoved.
2) Or you can just think about inertia. (physics is internal consistant that you can explain it in so many ways)

[quote]Using formulas, prove there is a relationship between temperature and the kinetic energy of molecules. What is this relationship? [/quote]

use the formula  E=i/2 KT (i=3,5,7, depends on what kind of molecule is there)
while T is temperature, E is the kinetic energy of molecules, K is a constant concluded by experiments.
According to the formula: The kinetic energy of molecules increases as the temperature increases.
And actually the "official" definition of temperature is defined by the kinematic energy of the molecules! :P

[quote]If you throw a ball up in a car moving at a constant velocity, where does it land? Why? If you throw a ball up in a car that is accelerating, where does it land? Why? [/quote]

If you throw a ball up in a car moving at a constant velocity
It lands back to your hand since you and the ball have the same velocity--same with the car moving at a constant velocity--according to the law of inertia.

If you throw a ball up in a car that is accelerating, where does it land? Why?
It lands back (behind the point it was launched) because when the car accelerates, the ball is still moving with a constant velocity as it did before(it doesn't accelerate when flying in the air) Then now the car is actually moving faster than the ball--so it lands back.(In a not inertial system, you can add a inertial force--which is a imagine force--with an opposite direction to acceleration. In this question, you can suppose there is a force acting on the ball backwards,so... :D )

*The law of inertia only applies to inertial system--which moves at a constant velocity or at rest.

[quote]If you throw a ball up in a car moving at a constant velocity, where does it land? Why? If you throw a ball up in a car that is accelerating, where does it land? Why?[/quote]

This concerns Newton's first law of inertia.
[quote]An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.[/quote]

meaning that, if the car is not accelerating, the ball has no other force acting upon it other than the momentum (from the throwing) and gravity. However when the car accelerates, the force acting upon the ball is unbalanced, resulting in that the ball hits back on your face (you face being the unbalancing force acting on the ball to keep it from flying out of the back of the car to stop the backward acceleration of the ball while the car accelerates forward.
Same idea with you sitting in a car hitting the brake The force of the road on the locked wheels provides the unbalanced force to change the car's state of motion, yet there is no unbalanced force to change your own state of motion. Thus, you continue in motion, sliding along the seat in forward motion. A person in motion tends to stay in motion with the same speed and in the same direction ... unless acted upon by the unbalanced force of a seat belt.

[quote="Ragingg"]A jar sits on top of a piece of paper... if the paper is pulled quickly, the jar does not move.  Why?[/quote]

this is definately a newton's first law question. use newton's first law to explain it.

A PHYSICS FORUM!!!!!!!!!!!!!!!!!!!!

For the accelerating car, you cannot truly answer that question because you are not measuring from an inertial frame. The answer involves general relativity and depends on the gravitational field vector compared to the acceleration vector of the car. I suppose you could make a lot of assumptions but in general that is a much more advanced problem than you should be cooncerned with.

THe relationship is KE = 3/2kT, but this is not actually true, and a correct proof involves quantum statistical mechanics. This simplification comes directly from molecular kinetic theory and the proof is straight forward, you can find it on Physics World. It comes from something called the Maxwell-Boltzman speed distribution, which is long and you can also look it up. A fairly hand waved method can be found here http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/molke.html#c1


A box contains 10 dice, 5 are crooked and produce '6' with probability 0.4. The other 5 dice are fair. A die is selected at random from the box and thrown twice - producing a '6' each time. find the probability that the dice is crooked.

This one and your defective bottle question are both classic Bayes formula questions, about conditional probability.

We have events $ C,$ that we choose a crooked die, (with its complement, $ \overline{C},$ that the die is fair, and we have event $ XX,$ of rolling a six twice, along with its complement, $ \overline{XX},$ of rolling anything other than a double six. Here's what we know:

$ P(C) \equal{} \frac12$ and $ P(\overline{C}) \equal{} \frac12.$ (Five of the ten dice are crooked.)

$ P(XX\mid C) \equal{} (.4)^2 \equal{} \frac {4}{25}.$

$ P(XX\mid \overline{C}) \equal{} \left(\frac16\right)^2 \equal{} \frac {1}{36}.$

From this, $ P(XX\cap C) \equal{} \frac12\cdot\frac {4}{25} \equal{} \frac {2}{25}$ and $ P(XX\cap\overline{C}) \equal{} \frac12\cdot\frac1{36} \equal{} \frac1{72}.$

Let me put what we know into a table. The cells in the interior of the table contain joint probabilities - that is, the probabilities of intersections:

$ \begin{array}{c|c|c|c} & C & \overline{C} & \text{Total} \\
\hline XX & \frac {2}{25} & \frac1{72} & \\
\hline \overline{XX} & & & \\
\hline\text{Total} & \frac12 & \frac12 & 1\end{array}$

The rows and columns must add up. Filling in the blanks, we get

$ \begin{array}{c|c|c|c} & C & \overline{C} & \text{Total} \\
\hline XX & \frac {2}{25} & \frac1{72} & \frac {169}{1800} \\
\hline\overline{XX} & \frac {21}{50} & \frac {35}{72} & \frac {1631}{1800} \\
\hline\text{Total} & \frac12 & \frac12 & 1\end{array}$

Now the question asked was the probability that the die was crooked, given that the double-six happened. That's $ P(C\mid XX)$ and can be calculated as

$ P(C\mid XX) \equal{} \frac {P(C\cap XX)}{P(XX)} \equal{} \frac {\frac2{25}}{\frac {169}{1800}} \equal{} \frac {144}{169} \approx 85\%.$

Of course, you don't have to make that whole table - I put that there to enhance the explanation. Do look up "Bayes' Formula" in your textbook.

[color=green][I apologize for not quite knowing enough LaTeX to get the tables right - can someone tell me how to put in the horizontal lines?][/color]

Actually, there's no need to put in vertical lines if you use \begin{tabular} instead of array.

One thing to note: without that [i]a priori[/i] estimate of the probability of a defective die, we simply cannot work this problem. Without the specific setup of this problem, if we threw some die (background undetermined) and it came up six twice in a row, and then asked the question, "Is that a loaded die?" then we simply don't have enough information to answer the question.

Change the original setup so that only one of the original 10 dice is loaded and the other 9 are fair. Repeat the computation exactly as above, and the probability that the die is crooked given event $ XX$ is now $ \frac{16}{41},$ or about $ 39\%.$ If we say that $ 99.9\%$ of all dice are fair and only one in a thousand is crooked in this exact way, then the probability of a crooked die given $ XX$ is about $ .006,$ or well under $ 1\%.$

The calculations in the that last paragraph are the same as the calculations in my post above; I've just tweaked some numbers.


calculus
derivative
function
trigonometry
vector
geometry
3D geometry

I like this problem:

A half-circle is placed on the ground with the diameter lying along the ground. A marble is placed at the very top, and a negligible initial force is applied to the marble. At what angle from vertical does the marble fly off the semicircle?

We are on a planet, so there is gravity, and it will fly off at some point. Assume no friction or air-resistance, or differing gravitational strength in the marbles path or...zzz....

I believe I got the answer to be arccos(2/3), but I would like someone else to verify this.

Cheers,

Rofler

I solved an identical problem from Kleppner and Kolenkow in mechanics and got that same result by finding the kinetic energy at a particular height, and then calculating the required centripetal acceleration v^2/R.

I don't remember the number, and the book's back in my dorm -- I just remember doing it and what I got. Sorry!

I've attached a time-lapse image of the situation in my previous post which illustrates the angle to be found.

Wow, centripetal acceleration... that would have made my solution simpler... anyways, do post if you can recall your solution. Because I forgot that (or don't know exactly how I would use centripetal acceleration, since the net force on the ball is never towards the center of the circle...), I calculated the trajectory of the ball at a given point if it were in freefall, and superimposed that graph on the cartesian plane.

The graph will always lie under the surface of the circle until it reaches a critical angle at which point the second derivative at the point of consideration on both the circle and the trajectory are equal. I figured that this point was the point where the ball would feel no inclination to stay on the circle, and merely follow freefall. So, equating the second derivatives at any point in terms of theta, yielded the value after heavy computations theta=arccos(2/3).

What I found interesting about this solution was that it had (almost) nothing to do with the fact that the path was a circle. The following is essentially the same process used with the circle, but for a more general graph. Let us say that the graph of the path that the ball had to follow had equation $ y \equal{} f(x)$.

Then we would have the speed $ |v(a)|$ at a given point $ (a,f(a))$, to be given by equating kinetic and potential energy, yielding $ |v(a)| \equal{} \sqrt {2|g|(f(0) \minus{} f(a))}$

$ \frac {v_y(a)}{v_x(a)} \equal{} f'(a)$, so $ \frac {v(a)^2}{v_x(a)^2} \equal{} 1 \plus{} f'(a)^2$, and $ v_x(a)^2 \equal{} \frac {2|g|(f(0) \minus{} f(a))}{1 \plus{} f'(a)^2}$.

The free-fall trajectory function would have second derivative $ \minus{} |g|$ if the horizontal component of the velocity was 1, but since that graph is effectively stretched horizontally by a factor of $ v_x(a)$ (which may include a horizontal flip, but this will not affect it), the second derivative is then $ \frac { \minus{} |g|}{v_x(a)^2} \equal{} \minus{} \frac {1 \plus{} f'(a)^2}{f(0) \minus{} f(a)}$. This is to be equated with the second derivative of the function $ y \equal{} f(x)$ at the point $ a$ to get that the points of [b]apparent weight 0[/b], which would be when these 2 are equal, that is:

$ f''(a) \equal{} \frac {1 \plus{} f'(a)}{f(a) \minus{} f(0)}$, which can be solved numerically, or exactly in the case of the circle.

If am in grade 11, and I spent 9 months on this problem because only recently did I learn about kinetic and potential energy, and even then I had to figure out a sufficient condition for the marble to fly off. Is there a way, in general, to find the apparent weight at any point on the graph (assuming the marble does not fly off before then)? I would have done it myself, but the only way I could think would be to find the formula for where the marble is at any given time, which requires solving a nasty differential equation that probably would not yield anything elegant.

Cheers,

Rofler

Edit: By the way, where did you get a program to do that picture? Or is it by hand?

Hey, Rofler, I don't think the solution needs to be that complicated.

Conservation of Energy:
\[ (1) U_{KE} \plus{} U_{PE} \equal{} mgR\]

\[ (2) U_{KE} \equal{} mgR (1 \minus{} \cos{\theta}) \equal{} \frac12 m v^2\]
After that we set the condition that $ F_n \equal{} 0$ when the ball leaves the sphere.  That means the only force providing centripetal acceleration is gravity.  Hence,
\[ (3) \frac {v^2}{R} \equal{} g \cos {\theta}\]
Using equation (2) to express $ \frac {v^2}{R}$, we solve for $ \cos {\theta}$, and obtain:

$ \cos{\theta} \equal{} \frac23$

There are several things that could be added to this problem to make it more interesting:

1.  Suppose instead of a mass with negligible size we had a spherical ball, rolling without sliding on top of the half cylinder... Where does the ball leave the surface?

2. Suppose we had a coefficient of kinetic friction $ \mu_k$... where does the mass leave the surface?

I find that (mg)/cos(theta) = m(v^2/r) rather than mgcos(theta) = m(v^2/r)

I'm not quite sure what I'm doing incorrectly.

Fw is pointing vertically downwards and is perpendicular to the x-axis and the centripetal force (m * v^2/r) is pointing towards the center of the circle.

sin (90-theta) = Fw/Fc where Fc is the centripetal force.
cos (theta) = Fw/Fc
Fc = Fw/cos (theta)

so somehow i get division rather than multiplication

Here is the reasoning:

- THe only force acting on the block is the normal force and the gravitational force.
- When the block leaves the surface, the normal force = 0, so all the centripetal force is due to gravitational force, in fact, it is due to the component of the gravitational force directed toward the center of the circle.

To get that component, you project the gravitational force onto the radius, and get a multiplication by $ \cos {\theta}$

[b]Picture:[/b] I just drew some circles in PowerPoint, pressed Print-Screen, and saved in MSPaint.

[b]Solution:[/b] While the ball is on the hemisphere, there are two components in the radial direction: $ N \; \mathbf{\hat{r}}$ and $ \minus{} mg \cos \theta \; \mathbf{\hat{r}}$. Their sum is the centripetal force $ \minus{} mv^2/R \; \mathbf{\hat{r}}$.

At $ \theta \equal{} 0^ \plus{}$, $ N \equal{} mg$. As $ \theta$ increases, $ \minus{} mv^2/R$ is strictly increasing while $ \minus{} mg \cos \theta$ is strictly decreasing, so $ N$ must be strictly decreasing. Then, we can find a unique point where $ N \equal{} 0$ to be the final point of contact, since after this, we'd need a negative normal force.

$ mg(1 \minus{} \cos \theta) R \equal{} \frac {1}{2} m v^2 \implies \minus{} mv^2/ r \equal{} 2mg(\cos \theta \minus{} 1)$.

We know that this is equal to $ \minus{} mg \cos \theta \plus{} N \equal{} \minus{} mg \cos \theta$, so we get $ 2mg \cos \theta \minus{} 2g \equal{} \minus{} mg \cos \theta \implies \cos \theta \equal{} 2/3$.

[quote="integration"]
To get that component, you project the gravitational force onto the radius, and get a multiplication by $ \cos {\theta}$[/quote]

How can your gravitational force be pointing towards the center of the circle though?  Isn't it always pointing downwards?  My question is how did you get that
$ \cos {\theta} x g$ is the component of the gravitational force that is directed towards the center.

If the gravitation force mg is directed downwards, and ($ 90\minus{}\theta$) is the angle between the horizontal and the ball, then wouldn't the force that is directed radially inward be mg DIVIDED by sin ($ 90\minus{}\theta$)?

Based on my picture, the "height" of the vector triangle is mg, and the angle is 90-theta.  As a result, an equation relating the "height", angle, and "hypotnuse" would be sin(90-theta) = mg/"hypotnuse"

The "hypotnuse" is the centripetal force, which would equal mg/cos(theta)

What I don't understand is what I'm doing wrong here.

Consider the attached image.

The dotted downwards black arrow is the weight. You can express it as the sum of two perpendicular components, one tangential to the surface of the sphere, and one pointing towards the center of the sphere, which I've shown in solid black.

The red vector is the normal force.

It should be clear from here that the radial force caused by gravity is indeed $ \minus{}mg \cos \theta \; \mathbf{\hat {r}}$ and not $ \minus{}\frac{mg}{\cos \theta} \; \mathbf{\hat{r}}$.

In this case, the axis is tilted, so that the mg is actually the hypotenuse.

Don't forget, part of gravity is pushing the ball off the half-circle, which is why there is a component of gravity tangent to the half circle.

Pretend the circle's outline is a curve.  Then, find the direction of radial and tangential acceleration.  These make up the components of your force, which, in this case, is only gravity.

Thanks, I understand that explanation.

What's somewhat confusing me is why I get a nearly opposite equation with a very similar method.  Let's extend the vectors you drew, such that the mg force touches the x-axis and the radially inward component of the gravity force touches the center of the circle.  Before, cos(theta) =centripetal force / mg.  Now, cos (theta) = mg / centripetal force because the centripetal force vector is now the "hypotnuse."

Two things:

[b]Thing 1:[/b] You need to understand the distinction (and relation) between the geometric length of a vector when represented by an arrow, and the magnitude of the vector.

The geometric scale to which I've drawn the arrows is arbitrary compared to the lengths in the problem. It has to be, since force has units Newtons, and distance has units meters, so the scale to which I've drawn the lengths does not define the scale to which I must draw the force vectors. I could have drawn the force vectors with a shorter scale, or a longer scale, if I wanted.

However, this does not the fact that vectors have a well-defined magnitude. The magnitude in this case is the number of Newtons of force represented by that vector. The length of the vector must be proportional to the magnitude of the vector according to the scale to which they're drawn.

This means that while you can certainly drawn the weight vector as going all the way down to the bottom of the hemisphere (which has no special significance), you can't simultaneously extend the radial component to the center of the hemisphere because by doing so, you are not preserving proportionality.

[b]Thing 2:[/b] You need to understand what's going on when you break a force down into components. The idea is that force behaves like a vector, and there is a particular way of adding them. The different vector components of the force must add to the total force. You can use the head-to-tail method for adding vectors geometrically.

In the case of my drawing, the radial and tangential components add to the total weight. Try adding them head-to-tail to convince yourself of this.

On the other hand, look at what you did, with the radial component going all the way to the center of the hemisphere and the overall vector going only to the base. Now, try to draw the vector that, when added to the radial component, will give you the total vector. In fact, it goes along the base of the hemisphere, directly horizontally, and is thus clearly not the tangential component you were looking for.

Your radial arrow and the horizontal arrow you drew do, in fact, add up to the total gravity vector. However, there is no use in decomposing the gravitational force this way in this case, because the components are not perpendicular, and thus have interdependence. It is important for this problem that we summarize the entire radial component of the acceleration in one vector, and the entire 'everything else' (which [i]must[/i] be tangential) in another vector.

@candyhyperalert:  gravitational force is down, as always, so mg is not pointing in the radial direction, rather in the negative y direction.

Since part of this problem can be done by decomposing gravitational force into radial and tangential components, we do so.

Wow, 16 replies to such a little problem! Thanks for explaining how to use centripetal acceleration guys, it made alot of sense. Also, I just downloaded the Introduction to Mechanics by K&K and it is one of the most well-written books I have ever read! I will have to finish it before school starts after christmas break.

Thanks a bunch all of you,

Rofler

Lol, at least 5 people have asked me to give them the download link for the book since I made that post. Keep them coming :P

Cheers,

Hunter


MATHCOUNTS
AMC
AIME
AMC 12
AMC 10
USA(J)MO
USAMO

are you doing mathcounts next year?
Or are you like me going to high school?

Thankfully, eighth grade.
One more year of MathCounts :D

Thankfully?  Nationals was fun, but I can't go again.

But you are national champion, so if you can't go you won't be expected to win again (like Neal Wu)

I just finished 7th grade so I'll be back. Seeing as I got about the same rank as Kevin Chen did last year, I hope I improve as much :D.

It's sad... I'll be in high school next year, so I can't do Mathcounts again, but I (hopefully) am going to help coach my middle school's team!!!

[quote="Hamster1800"]I'm not. I had my three years. Next year I get to cram for AIME instead of nats :D

Must...Make...USAMO....[/quote]

I don't think you should worry about that...you will make it...

Me, however, not so good

[quote="robinhe"][quote="Zmastr"]Do you recommend me doing the AIME or the AMC 10 next year in high school?[/quote]

You can do both in the same year.  AMC 10--->AIME sometime/usually/maybe not.[/quote]

Maybe he means AMC12 or AMC10, in which case the best idea is to do the AMC12.

If your high school offers both the a and b tests, take both AMC 10 and 12. That way, you can qualify on 12 with the lower requirement and (possibly) also have a very high index with the AMC 10 in case (like this year) there were many people who made the floor and they have to add in an index requirement (but they removed it this year after putting it in). If your school doesn't offer both, it doesn't matter that much, but I'd say take AMC 12 if you can get 120+ on it and definitely if you can get 130-140+

[quote="pianoforte"]no more mathcounts for me *sigh

oh well, time to study for amc's....(yay for AoPS offering amc books!)[/quote]

darn, I checked the kansas amc website, and I don't see that my highschool has anyone on the honor or merit roll $\implies$ they don't do AMC competitions :( . I guess I'll have to convince them to...

I think he means that taking the USAMO is great prestige whether or not you can do well. Its like saying "why take AMC if you know you cant win an IMO medal"

I'm saying you can at least prepare for it instead of being so happy you made it and you don't try at all.

[quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="kyyuanmathcount"]You mean AIME right (American Invitational Math(ematcis) Exam)?

Goals :

AMC 12 (a or b) : 138 or so?

AIME: pulling a Sachin.

USAMO: MO(S)P? Probably not.[/quote]

You mean Math(ematics), right? :rotfl:

I think your goals are too high.
Try focusing on stuff that is like AMC 10 level, first and then move your way up.

I mean 138 on AMC 12 is kind of hard and pulling a Sachin is like impossible, unless you are Sachin.[/quote]
I agree, just like the 42-46 at Nationals (which would've gotten you first or second written), don't set too high goals.[/quote]

Wait, who was it who wanted 47 at nationals?

I don't know, but it wasn't me.  I think it was Ignition.

[quote="kyyuanmathcount"][/quote][quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="kyyuanmathcount"]You mean AIME right (American Invitational Math(ematcis) Exam)?

Goals :

AMC 12 (a or b) : 138 or so?

AIME: pulling a Sachin.

USAMO: MO(S)P? Probably not.[/quote]

You mean Math(ematics), right? :rotfl:

I think your goals are too high.
Try focusing on stuff that is like AMC 10 level, first and then move your way up.

I mean 138 on AMC 12 is kind of hard and pulling a Sachin is like impossible, unless you are Sachin.[/quote]
I agree, just like the 42-46 at Nationals (which would've gotten you first or second written), don't set too high goals.[/quote][quote="kyyuanmathcount"]

Wait, who was it who wanted 47 at nationals?[/quote]

Hey, at least he was joking.
You seriously expected 42-46...

Guess I'm focusing on the AIME and AMC 10 from now on.

[quote="Klebian"][/quote][quote="kyyuanmathcount"][/quote][quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="kyyuanmathcount"]You mean AIME right (American Invitational Math(ematcis) Exam)?

Goals :

AMC 12 (a or b) : 138 or so?

AIME: pulling a Sachin.

USAMO: MO(S)P? Probably not.[/quote]

You mean Math(ematics), right? :rotfl:

I think your goals are too high.
Try focusing on stuff that is like AMC 10 level, first and then move your way up.

I mean 138 on AMC 12 is kind of hard and pulling a Sachin is like impossible, unless you are Sachin.[/quote]
I agree, just like the 42-46 at Nationals (which would've gotten you first or second written), don't set too high goals.[/quote][quote="kyyuanmathcount"]

Wait, who was it who wanted 47 at nationals?[/quote][quote="Klebian"]

Hey, at least he was joking.
You seriously expected 42-46...[/quote]
I guess I was slightly naive (try spelling it backwards) to think I wouldn't "deprove" from my practice scores.

About your signature kyyuanmathcount, were you thinking of 6^2+8^2=10^2 at National Mathcounts.

Yes, I am doing mathcounts next year. Thank god, or I will have to live forever with a 112th place finish. :(

Goals for next year:
MC Chapters:46
MC States:46
MC Nationals:40 (sha right)
AMC8:25 (duh)
AMC10: 150 (prolly not)
AMC12 if I can take it:130
AIME: 8
USAMO: Like...11? :rotfl:
IMO :rotfl: : 43. yeah that's right.

Was it neccessary to make two posts, one that was totally unrealistic and another that was totally semiunrealistic?

[quote="jclarkemathL314159"]MC Chapters: 45 (I assure you, I will make some dumb mistake)
MC States: 42-44
MC Nationals: 32-37
AMC8: 24-25 :D
AMC10: 130-145
AMC12, again, if I can take it: 115-130
AIME: 6-7
USAMO: Pft.[/quote]

All of these are really really close to what I did this year...

Yay, I still have another year left for mathcounts!  Of course, I will do all of the other contests, also.

sadly, i must say goodbye to mathcoutns.


inequalities
search
inequalities proposed

Hi,
here is an easy inequality. I did not bother to search the forum so please do not behead me if it has already been posted.
As said it is quite easy but as my knowledge of inequalities is pathetic, I'm just interested in some other solutions

Prove that for all nonnegative real numbers $x,y,z$ statisfying $x+y+z=1$ the following inequality holds:
\[ \sqrt{12xyz}+x^2+y^2+z^2\leq 1  \]

Seen in some poland olympiad...

Posted many times :D

Sorry  :blush:
May I ask you for a link?

THX

Anyway, here is my solution:

    We have  \[ \sqrt{12xyz}+x^{2}+y^{2}+z^{2}=2\sqrt{3xyz(x+y+z)}+x^{2}+y^{2}+z^{2}. \]

    But, now using the well-known inequality: $xy+yz+zx\geq \sqrt{3xyz(x+y+z)},$ the result follows. ;)

[quote="cezar lupu"]But, now using the well-known inequality: $xy+yz+zx\geq \sqrt{3xyz(x+y+z)},$ the result follows. ;)[/quote]

Embarrassing that I did not know that one  :blush: :oops:
Thanks anyway, my solution is much longer  :oops:

Socrates, can you tell me which year was this inequality given in Poland? :)

Hi Cezar!

My solution is almost like yours(homogenization)

A nice(but easy) ineq, well known.. :)

See [url]http://www.mimuw.edu.pl/~chel/Olimp/1etang00.html[/url]
 :)


Where can I find a complete collection of ARML problems? Book, websites?

At least two books have been published so far, with the most recent one containing the 1989-1994 problems. I think it is possible to buy this book from Amazon, but the previous one(s) are out of print as far as I know. There is a new book of the 1995-2003(?) problems that should be available for purchase at ARML.

In addition, NYSML, the parent-competition of ARML, should soon be offering a new book of contests covering roughly the same period (1994-present).  While slightly easier than ARML, it is good to have and helps make up for the fact that there is a very finite collection of old ARML problems.

There are 2 earlier ARML-NYSML books than the one Simon mentioned, but they are both out of print.

if anyone has any specific questions, ive got the green book (arml and nysml 1983-1988) and the blue one (arml and nysml 1989-1994)

I have the NYSML-ARML from 1973~1982 as well as the two later published books. You still can get the 1989~1994 from amazon.com. let me know if you need information from the earlier editions.

I just searched Amazon and found only one copy of 89-94 ARML-NYSML book - "only" $191.08 - rare indeed!!!!!!!!!!!!!!

Also, found something by John Arml !

[quote="mathnovice"]Where did you get the 1983-1988 one?[/quote]

two years ago they were selling them at penn state

[quote="Yooper"]I just searched Amazon and found only one copy of 89-94 ARML-NYSML book - "only" $191.08 - rare indeed!!!!!!!!!!!!!![/quote]

Jeez, I've got a weird feeling having two hundred bucks worth of math problems sitting just off my left wrist. =P