Suppose you are given a 5x5 grid of pts. How many squares can be drawn on the grid? (include "tilted" squares)

$1^2+2^2+3^2+4^2+5^2=\boxed{55}$

EDIT: Whoops, i thought you meant 5x5 grid with squares in them (like a tiled square)...sorry :blush: ignore this post

[hide]When i call squares $\underbrace{a}_{b}$, i mean an $a$ number of $b  x  b$ squares.

First looking at the aligned squares, we have $\underbrace{16}_{1} + \underbrace{9}_{2} + \underbrace{4}_{3} + \underbrace{1}_{4} = 30$

Then looking at squares at a $45^\circ$ angle on the grid we have $\underbrace{9}_{\sqrt{2}} + \underbrace{1}_{2\sqrt{2}} =10$

There are two more types of squares that can be drawn:
$\underbrace{8}_{\sqrt{5}}$ squares (the ones that go two across and one up -- doubled b/c of flipped orientation) and $\underbrace{2}_{\sqrt{10}}$ squares (the one that goes three across and one up-- again, doubled b/c of orientation). This gives a total of $\boxed{50}$ squares.  [/hide]

edit: lol $2^2 \neq 3$... and i added the flipped orientation (i did this problem too fast and forgot about those). So i get $\boxed{50}$ also.

[hide]For the squares with side lengths $\sqrt{5}$ and $\sqrt{10}$, their orientation can be switched, so there are $10$ such squares.
[/hide]

[hide]$1$ with side length $4$
$4$ with side length $3$
$9$ with side length $2$
$16$ with side length $1$
$9$ with side length $\sqrt{2}$
$1$ with side length $2\sqrt{2}$
$8$ with side length $\sqrt{5}$
$2$ with side length $\sqrt{10}$
Total: $\boxed{50}$[/hide]

there are lots of different solutions so far, so obviously some people are wrong. I'm working on the problem myself and I'm trying to see patterns for a general formula. I noticed that the number of "45 degree" squares in an nxn grid is given by f(n)=(n-2)^2+(n-4)^2+...  down to your last positive square. Can anyone confirm this?

I haven't had time to check that yet, but I just drew out all the possibilites for this one, and I got exactly what LordoftheMorons did...

Alright, well, I don't know if I could really prove this, but by examining several cases of this, I've come up with a formula that will always work for this: \[ \sum_{a=1}^{n-1}a^2+\sum_{a=1}^{n-2}(n-1-a)a^2  \]If you want, I could explain what I did, but if you look at the squares you can make by groups of the slopes of their sides, it should be apparent...


vector
topology
calculus
integration
advanced fields
advanced fields theorems

Let $ G$ be a vector field defined in $ A = R^n - {0}$ with div$ G=0$ in $ A$. Show that as $ M$ ranges over all compact $ n$-manifolds in $ R^n$ for which the origin is not in $ \partial M$, the integral
$ \int_{\partial M} <G,N> dV$
where $ N$ is the unit normal to $ \del M$ pointing outwards from $ M$, has only two possible values.

The manifold either contains the origin or it does not.  In either case, use the divergence theorem.


search
function
algebra unsolved
algebra

Three non zero numbers a,b,c are said to be in a harmonic progression if 1/a + 1/c = 2/b. Find all three term harmonic progression a,b,c of strictly increasing positive integers in which a = 20 and b divides c.

PLZ provide an answer or a hint..

See http://www.mathlinks.ro/viewtopic.php?t=238347

And, btw, just using the search function with "rmo 2008" gives a lot of answers to your two last posts ...


Seven  lines are given on a plane.They are meeting at 13 points.A pair of lines meeting at 11 points,three lines at one point and four lines at the last point.Prove that two of lines are parallel to each other.

Isn't it really easy?

Let $A$ (resp. $B$) be the point where three (resp. four) of the lines meet each other.

Suppose that no line through $A$ contains $B$.
Thus, the lines are $D_1,D_2,D_3$ through $A$ and $D_4,D_5,D_6,D_7$ through $B$ and we have all our seven lines.
If no two are parallel then each of the pair $i,j$ where $i = 1,2,3$ and $j=4,5,6,7$ corresponds to an intersection point where exactly two of the lines meet, which leads to 12 such points and contradicts that there are only 11.

Thus, there is a line, say $ \Delta$ wich passes through $A$ and $B$.
Let $D_1,D_2$ be the lines through $A$ and $D_3,D_4,D_5$ be the lines through $B$. Thus, there is a remaining line, say $D$.
If no two lines are parallel, then $D$ meets each of the six other lines in pairwise distinct points (and distinct from $A$ and $B$). Moreover each of the pair $i,j$ for $i=1,2$ and $j=3,4,5$ give another point where exactly two of the lines meet. It leads to 6+6 = 12 such points, and to the same contradiction than above.

Thus, two of the lines are parallel.

Pierre.


Can you calculate the dimensions of the largest (i.e. greatest volume) rectangular (or is it cuboidal?) parcel that can be securely tied up with a piece of string $360 cm$ in length?

The string must pass twice around the width and once around the length as shown in the diagram ( please find attached ). The string can be cut and the insignificant amount used up in the knots is to be disregarded.

$x$=length, $y$=breadth, $z$=height
we have, $2\times(x+y) + 2\times2\times(y+z) = 360$, which is same as...
$x + 3y + 2z = 180$

so we just need to find maximum values of $x, y, z$

PS: edited for incorrect equation. Now its ok.

I dont think you necessarily need to find the maximums of $x,y,z$ as much as you need the maximum for $x \cdot y \cdot z$. Though i'm not sure how to do this either, i havent learned how to do maximums yet..

something says calculus

something else says i shouldnt use calculus

[quote="robinhe"][quote="xxxyyyy"]$x$=length, $y$=breadth, $z$=height
we have, $2\times(x+y) + 2\times2\times(y+z) = 360$, which is same as...
$x + 3y + 2z = 360$
[/quote]

It should be $x+3y+2z=180$, right?

[hide]$\frac{x+3y+2z}{3}\ge\sqrt[3]{6xyz}$
$60\ge\sqrt[3]{6xyz}$
$\frac{60^3}{6}\ge xyz$
$xyz\le 36000$

So the max is $36000$?
That seems a little large :?  :roll: [/hide][/quote]

Well, no it doesnt seem that large.. for example, if i just pick three random dimensions, say $x=100,  y= 20,  z=10$, we would get the volume to be $20000$ which isnt really too far off of $36000$.

[quote="t0rajir0u"]The equation looks wrong.  And calculus is unnecessary.  And needlessly difficult.

[hide="Hint"]AM-GM ;) [/hide][/quote]

you were right, i made a mistake in the equation. I have updated it now.

$x + 3y + 2z = 180$

[quote="robinhe"][quote="xxxyyyy"]$x$=length, $y$=breadth, $z$=height
we have, $2\times(x+y) + 2\times2\times(y+z) = 360$, which is same as...
$x + 3y + 2z = 360$
[/quote]

It should be $x+3y+2z=180$, right?

[hide]$\frac{x+3y+2z}{3}\ge\sqrt[3]{6xyz}$
$60\ge\sqrt[3]{6xyz}$
$\frac{60^3}{6}\ge xyz$
$xyz\le 36000$

So the max is $36000$?
That seems a little large :?  :roll: [/hide][/quote]

this is more or less correct, but if you want to be rigirous, you need to show that a box with that volume actually exists! that inequality is true, but what if there was another inequalitiy that was $V\le30000$, etc,

you would show the maximum exists by finding the dimensions that give the volume of $36000$

[quote="t0rajir0u"][quote="Altheman"]
you would show the maximum exists by finding the dimensions that give the volume of $36000$[/quote]

Which is straightforward enough.  Who wants to do it?  [hide="Hint"]What is the equality case for AM-GM?  :) [/hide][/quote]

[hide]
Equality holds when $x=3y=2z$.

$V=xyz=\frac{x^3}{6}=36000$, so $x=60$, $y=20$, $z=30$.
[/hide]


[quote="Karth"]When claiming to present an outline for the program, will the source code suffice? My source code, for example, is in Java. I will still prove how the program works and stuff though. But is it okay to only present a computer oriented solution (for multiple problems, even)? That is, will we be penalized?[/quote]

You will not be penalized for a computer oriented solution, provided that you follow the rules (in that you must prove that your program does indeed solve the problem).


quadratics
analytic geometry
graphing lines
slope
calculus
derivative
conics

Find $ a$ (element of R) such that the least value of $ 4x^2 \minus{}4ax \plus{} a^2 \minus{}2a \plus{}2$ on the interval $ 0<\equal{} x <\equal{}2$ is equal to $ 11$.

[hide="I am stupid XD - incorrect solution"]meh

Wherever we want a "least value," we'll have a slope of zero. We need to find the value of $ x$ for which the derivative is zero:

$ 8x - 4a = 0$

We also know that for that value of $ x$,

$ 4x^2 - 4ax + a^2 - 2a + 2 = 11$.

Let's rewrite that:

\begin{align*} (2x - a)^2 - 2a + 2 & = 11 \\
\cfrac{1}{16}(8x - 4a)^2 - 2a + 2 & = 11 \\
0 - 2a + 2 & = 11 \\
a & = \boxed{ - \cfrac{9}{2}} \end{align*}[/hide]

Define $ f : \mathbb{R} \to \mathbb{R} : x \mapsto 4x^2 - 4ax + a^2 - 2a + 2$
$ f(x) = (2x - a)^2 - 2a + 2$ for all $ x$.
$ f$ is increasing for $ x \ge \frac {a}{2}$ and decreasing for $ x \le \frac {a}{2}$
We consider 3 cases:
case 1: Suppose $ {\frac {a}{2} < 0}$,
then $ f$ is increasing on $ [0,2]$ with minimum (on $ [0,2]$) $ f(0) = a^2 - 2a + 2 = 11$ iff $ a = 1 - \sqrt {10}$.

case 2: Suppose $ 0\le \frac {a}{2} \le 2$,
then $ f$ is minimal is $ x = \frac {a}{2}$.
For this $ x$, $ f(x) = - 2a + 2$.
There does not exist a real number $ a$ with $ 0\le \frac {a}{2} \le 2$ and $ - 2a + 2 = 11$.
So, no solution in this case.

case 3:  Suppose $ \frac {a}{2} > 2$.
This goes the same way as in case 1.
(Solution $ a=5+3\sqrt{2}$).

Interesting , the book says $ a$ belongs to $ R \minus{} (1 \minus{}\sqrt{10} , 5 \plus{}3 \sqrt2)$

A parabola $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ is symmetrical about the line $ x\equal{}\minus{}b/(2a)$. This axis of symmetry is where the max/min will occur. So simply we solve for $ a$ such that $ f(\minus{}b/(2a))\equal{}11$, and that the leading coefficient is positive.

[quote="Conjurer"]Are you folks not understanding my question?

PeterP has understood but I dont think he has the right answer yet.[/quote]
Why don't you think PeterP's solution is correct..? I've got the same answer (and your book got it too).


Find all prime numbers $p$ such that the system

$p+1=2x^2$
$p^2+1=2y^2$

have an integral solution.

We can suppose x,y >  0

Subtract second from first.

It becomes $p(p-1) = 2(x-y)(x+y)$.

Suppose p is 3 mod 4.  Try 4y-1 = x+y = p.  It works.

Suppose p = 2.  It obviously can't work.

Suppose p is 1 mod 4.  It becomes $2k + 1 = x^2$ and $8k^2 + 4k + 1 = y^2$.  So $(x^2)^2 + (x^2 - 1)^2 = y^2$.  But it's impossible.

But it dosent work for all $p \equiv 3 \mod{4}$

Example: $3$

Indeed through casework we can improve it to $p \equiv 7 \mod{16}$

In fact the only answer is $p=7$ and it has been posted before here

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=p%5E2%2B1%3D2y%5E2&t=26718[/url]

 ;)


[quote="ckck"]The equation $p(x^{2}+1)=8x-1$ has two roots such that one root is 3 times the other root for two distinct values of p. Find the absolute value of the difference between these two values of p.[/quote]
You mean that you want to know $|p_{1}-p_{2}|$, don't you?
[hide]$p(x^{2}+1)=8x-1 \Rightarrow px^{2}-8x+p+1=0$
Applying Viète theorem, we get:
$x_{1}+x_{2}= \frac{8}{p}$
$x_{1}x_{2}= \frac{p+1}{p}$
Assume that $x_{1}=3x_{2}$
We have $\frac{p+1}{p}=3x_{2}^{2}= \frac{12}{p}$
$\Rightarrow p^{3}+p^{2}-12p=0$
Notice that $p \neq 0$ so that $p_{1}=3, p_{2}=-4$
Thus $|p_{1}-p_{2}|=7$[/hide]


$ A\equal{}\frac{p^2 \sin y \sin x}{2 \sin(y\plus{}x)}$
$ \equal{}\frac{p^2 \sin y \sin x}{2 \sin z}$
$ \equal{}\frac{p(2R \sin y) \sin x}{2}$
$ \equal{}\frac{pq \sin x}{2}$
$ \equal{}[\triangle PQR]$


If the radius of k1=1 and the radius k5=4 then k4=?

Connect each the center to the top ray (which will just be the radius). Then the radii of the circles form a geometric sequence (try to figure it out yourself - hint: Similar triangles).

So let this ratio between each radius be $ r$, then

$ k_5 \equal{} r^4 k_1$

$ 4 \equal{} r^4$

$ r \equal{} \sqrt{2}$

So $ k_4 \equal{} r^3 k_1 \equal{} \boxed{2\sqrt{2}}$


If a,b,c are positive integer and  (b- 1/a)(c- 1/b)(a- 1/c) is integer , solve a,b,c

$ \frac{(bc-1)(ca-1)(ab-1)}{abc}\in N \iff bc\equiv 1[a],ac\equiv 1[b],ab\equiv 1[c]$
then $ gcd(a,b)=gcd(b,c)=gcd(c,a)=1$ and $ a,b,c|ab+bc+ca-1$
so, $ abc|ab+bc+ca-1$
then $ ab+bc+ca-abc-1=(1-a)(1-b)(1-c)-2\ge 0$
then $ (1-a)(1-b)(1-c)\ge 2>0$
but we have $ min(a,b,c)\ge 1$
so, $ (1-a)(1-b)(1-c)\le 0$ (contradiction)

ab+bc+ca-abc-1 doesn't equal to (1-a)(1-b)(1-c)-2


ABCD is a cyclic quad in which AD = AB + CD. Prove that the bisectors of angles B and C meet on AD.

[hide]
Let point $ E$ on $ \overline{AD}$ be constructed such that $ AB=AE$ and $ ΕD=DC$. Let the angle bisector of $ \angle{B}$ intersect $ AD$ at $ I$. We will prove that $ CI$ is the angle bisector of $ \angle{C}$.

Because $ \triangle{ECD}$ is isosceles, $ \angle{CED} = 90-\frac{\angle{D}}{2}$. Because $ BI$ bisects $ \angle{B}$, $ \angle{CBI} = \frac{\angle{B}}{2} = 90-\frac{\angle{D}}{2}$ because $ ABCD$ is cyclic.

Therefore, $ BCEI$ is cyclic; the conclusion follows after noting that $ \angle{ECI} = \angle{EBI}$ and subtracting angles.

[/hide]


Let a quadrilateral $ABCD$ such that $AC = BD$. Construct outside $ABCD$ 4 isoceles triangles $AHB$, $BMC$, $CPD$, $DNA$ at $H$, $P$, $M$, $N$. And $\triangle AHB \sim \triangle CPD$, $\triangle BMC \sim \triangle DNA$. Prove that $HP$ and $MN$ are perpendicular.

I bet it works out just fine with complex numbers :).

Why this is still unsolved?
Anyway, this problem is really nice.

Let the perp. bisectors of AB and CD intersect at point E. Then we easily obtain $\triangle AEC\cong\triangle BED$. Hence, $\triangle APB\sim\triangle CPD$=>$\frac{AB}{CD}=\frac{EX}{EY}$, where $X$ and $Y$ are the midpoints of $AB$ and $CD$ respectively. Since $\triangle AHB\sim\triangle CDP$, $\frac{AB}{CD}=\frac{HX}{PY}$. Therefore $XY//HP$. Similarly, MN is parallel to the lines joining the midpoints of $AD$ and $BC$, say $WZ$. But as $XWYZ$ is a rhombus, $XY\perp WZ$ which implies $HP\perp MN$.

From this proof, we can see that the result is also true for $H,P,M,N$ lying in the interior of $ABCD$.

I should have posted this in solved problem because I solved it. I denoted $X$, $Y$, $Z$, $T$ are 4 midpoints of $AB$, $BC$, $CD$, $DA$. I proved $XZ \parallel HP$ and $YT \parallel MN$. So that' all we have to do.

[quote="Huynh Anh Hao"]I should have posted this in solved problem because I solved it.[/quote]No. A problem is never to be posted in the Solved Problems forums. It is moved there after it is discussed and solved on the forum (ie a solution is posted).

If you want to propose a problem that you have already solved, but want to see other people's solutions, post it in the Proposed & Own problem section (same for the problems you created).  :)

Name T the common point of AC and BD and E the 2nd intersection of the circles (TAB) and (TCD). Since <TAE = <TBE and <TCE = <TDE, with AC = BD we get the triangles AEC and BED congruent (a.s.a.), consequently AE = BE and CE = DE.
Further as per post #3.

[b]Particular case:[/b] [i]If H and P are the midpoints of the arcs AB and CD opposite to T on the circles (ATB) and (CTD), then H, T and P are collinear (the bisector of the angle <ATB), that is, always HP and MN, as well as the diagonals of the rhombus formed by the midpoints of the sides of the quadrilateral ABCD are parallel to the bisectors of the angle between its (equal) diagonals.[/i]

Best regards,
sunken rock


$ a_{1} \equal{}1$, $ a_{n\plus{}1} \equal{}a_{n }\plus{} \frac {1} {a_n}$

Then find the $ [a_{100} ]$.

It is similar with my latest post I think.

please solve it~

[hide="solution"]notice that from the condition we have
$ a_{i}^2\equal{}\frac{1}{a_{i\minus{}1}^2}\plus{}a_{i\minus{}1}^2\plus{}2$
now we take i from 2 to 100 and add all them up we get
$ a_{100}^2\equal{}199\plus{}\sum^{99}_{i\equal{}1}\frac{1}{a_i^2}$ now let $ n\equal{}100$
from this we get $ a_{100} \geq \sqrt{199} >14$
we will prove that $ a_{100} <15$
note that $ 1\equal{}a_1 < a_2 < a_3 <..<a_100$
and $ a_2\equal{}2 ; a_3\equal{}2,5 ; a_4\equal{}2,9$
so that $ a_{100}^2 < 199\plus{}1\plus{}\frac{1}{2^2}\plus{}\frac{1}{2,5^2}\plus{}\frac{95}{2,9^2} < 225$
then $ a_{100} < 15$ which means $ [a_{100}]\equal{}14$[/hide]


inequalities
geometry
circumcircle
incenter
inequalities proposed

$\triangle ABC,\ (b-a)(b-c)\le 0\Longrightarrow |a-c|\ \le \ \frac{s}{s-b}\cdot OI$

[b](The points O, I are the circumcenter and the incenter for the triangle ABC).[/b]

[quote="levi"]$\triangle ABC,\ (b-a)(b-c)\le 0\Longrightarrow |a-c|\ \le \ \frac{s}{s-b}\cdot OI$

[b](The points O, I are the circumcentre and the incentre for the triangle ABC).[/b][/quote]

The inequalities $s\sqrt 3\le 4R+r$ and $s^2\ge 3r(4R+r)$ are wellknown.

I offer a common solution for both using the same obvious inequality:

$\left(\forall\right)x,y,z\in R\Longrightarrow3(xy+yz+zx)\le (x+y+z)^2\ \ (*)$ I recall some familiar identities:

$ab+bc+ca=s^2+r(4R+r);\ r_a +r_b +r_c =4R+r;\ r_a r_b +r_b r_c +r_c r_a =s^2.$

Therefore, from the inequality (*) result:

$1^{\circ}.\ x:=a,\ y:=b\ z:=c\Longrightarrow 3(ab+bc+ca)\le (a+b+c)^2\Longrightarrow 3r(4R+r)\le s^2;$

$2^{\circ}.\ x:=r_a,\ y:=r_b,\ z:=r_c\Longrightarrow 3s^2\le (4R+r)^2\Longrightarrow s\sqrt 3\le 4R+r.$

[b]Remark.[/b] The inequality $1^{\circ}.$ may be used in the solution of the proposed problem.

[b]Bonny BONUS.[/b]

$B1^{\circ}. |r_a +r_b -2r_c |\le 2\sqrt {(4R+r)^2-3s^2}.$

$B2^{\circ}. \max_{x\in \{ a,b,c\}} \left| s-\frac 32 x\right| \le \sqrt {s^2-3r(4R+r)}.$

_____________________________________$Farewell,\ my\ dear\ Mathlinks.ro\ !$

[quote]
$\displaystyle Farewell,\ my\ dear\ Mathlinks.ro\ ! $

[/quote]
Why?

Here, at my daugter and the  two grand-children, I had more free time. I leave for home (Romania) after two days, where I have my job and don't have a leisure for the noble and fine lady "Matematica" (in Romanian language). We will meet in the future year. I thank for your issue !

Levi,there is a better ineqaulity than the second one you have posted.I posted when I solved that Iran'96 problem. ;)

Caezsar, you leave the trifles and you deal with the proposed inequality.


advanced fields
advanced fields unsolved

I need an example of a primitive, but not simple, $ C\sp*$-algebra. If $ H$ is infinte-dimensional $ B(H)$ works, but is there an easy separable example?

Compact operators plus multiples of the identity.


First we can considerate a function $f: R^{2}\to R$ such that $f(x,y)=4x^2+4y^2-9yx^2-9xy^2+13xy+1-4x-4y$. And now we can calculate it´s partial derivates and we´ll obtain that this function has a minimun value in $x=y=\frac{1}{3}$, since by hipothesis $1>x,y>0$ and $x+y+z=1$.

Now it´s just to prove that your ineq implies this last that I´ve mentioned. And this part isn´t difficult, we have to divide both sides by $xyz$ and write $\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$.

Κώστα καλωςήρθες .Σου έστειλα τη λύση κατευθείαν σε Private message δεν την έλαβες ?

Γειά σε όλους τους χρήστες του φόρουμ.Πρέπει να πώ πως αυτή η άσκηση την προτείνει ο Αρχιμήδης,ενας πορωμένος φίλος τον μαθηματικών.Thanks mffmartinelli.Σιλουάν ευχαριστώ για την λύση.Δυστηχώς δεν τη βρήκα.Τι λες για γενίκευση?

Ok και για τη γενίκευση λύνεται με τον ίδιο τρόπο όπως η λύση που σου έστειλα .Για; να τη βρεις τη λύση που σου έστειλα πήγαινε εκεί που λέει Private messages

διαφορετικά [url=http://www.mathematica.gr/forum/viewtopic.php?p=183315#p183315]εδώ[/url] και παρακάτω


calculus
integration
real analysis
real analysis unsolved

Find:
1. $ \int \sqrt{2\plus{}\sqrt{4\plus{}\sqrt{x}}} \,dx$
2. $ \int \frac{1}{\sqrt{2\plus{}\sqrt{1\plus{}\sqrt{x}}}} \,dx$
3. $ \int \sqrt[3]{1\plus{}\sqrt{x\minus{}3}} \,dx$

How about these? You may find solutions to: http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/powersubdirectory/PowerSub.html

[hide="Number 1"]Make the following substitutions: $ 4 \plus{} \sqrt{x} \equal{} t$, and then $ \sqrt{2\plus{}\sqrt{t}} \equal{} u$:

$ P \equal{} \int 2(t\minus{}4) \sqrt{2\plus{} \sqrt{t}}\,dt \equal{} 8 \int u^2(u^2\minus{}2)[(u^2\minus{}2)^2 \minus{} 4]\,du$

and now we have only elementary primitives.[/hide]

[hide="Number 2"]Make the following substitutions: $ 1 \plus{} \sqrt {x} \equal{} t$, then $ 2 \plus{} \sqrt {t} \equal{} u$, and finally $ \sqrt {u} \equal{} z$:

$ P \equal{} \int \frac {2(t \minus{} 1)}{\sqrt {2 \plus{} \sqrt {t}}}\,dt \equal{} \int \frac {4[(u \minus{} 2)^2 \minus{} 1](u \minus{} 2)}{\sqrt {u}}\,du \equal{} 8 \int(z^2 \minus{} 2)[(z^2 \minus{} 2^2 \minus{} 1)]\,dz$

and we have elementary primitives.[/hide]

[hide="Number 3"]Make substitution $ \sqrt {x \minus{} 3} \equal{} t$, $ x' \equal{} 2t$, and next make substitution $ \sqrt [3]{1 \plus{} t} \equal{} u$, $ t' \equal{} 3u^2$:

$ P \equal{} \int 2t \sqrt [3]{1 \plus{} t}\,dt \equal{} \int 2u(u^3 \minus{} 1)3u^2\,du \equal{} 6 \int(u^6 \minus{} u^3)\,du \equal{} 6\left(\frac {u^7}{7} \minus{} \frac {u^4}{4}\right)$

$ \equal{} \frac {6}{7} (1 \plus{} t)^{4/3} \left(t \minus{} \frac {3}{4}\right) \equal{} \frac {6}{7} (1 \plus{} \sqrt {x \minus{} 3})^{4/3}\left(\sqrt {x \minus{} 3} \minus{} \frac {3}{4}\right) \plus{} C$.[/hide]


Prove that the equation $ (x\plus{}y)^n\equal{}x^m\plus{}y^m$ has a unique solution in integers with $ x>y>0$ and $ m,n>1$.

[hide="some hints"]Let$ \gcd(x,y)\equal{}d$ and $ \frac{x}{d}\equal{}x_1,\frac{y}{d}\equal{}y_1$ then we have
$ d^{m\minus{}n}(x_1^{m}\plus{}y_1^{m})\equal{}(x_1\plus{}y_1)^{n}$.obviously the set of prime divisors of $ x_1^{m}\plus{}y_1^{m}$ and $ x_1\plus{}y_1$ are the same(note that $ m>1$).(the case $ x_1\plus{}y_1\equal{}2^k$ is trivial) we'll have two cases:
1-there exist an odd prime $ p$ such that $ v_{p}(d^{m\minus{}n})\plus{}v_{p}(x_1\plus{}y_1)<v_{p}((x_1\plus{}y_1)^{n})$ then prove that
$ \frac{x_1^{p}\plus{}y_1^{p}}{x_1\plus{}y_1}\equal{}p$ (prove this using $ v_{p}(x^{n}\plus{}y^{n})\equal{}v_{p}(x\plus{}y)\plus{}v_{p}(n)$).
the rest is solved using inequality.But the other case remaining is:
2-there doesn't exist such $ p$:it means that $ m$ must be $ 1$.(prove this using again
$ v_{p}(x^{n}\plus{}y^{n})\equal{}v_{p}(x\plus{}y)\plus{}v_{p}(n)$)[/hide]

I will instead characterize every solution $(x,y,m,n)$ given that $x,y>0$ and $m,n \geq 1$—this is the version of the problem I received.

Clearly $(x,y,m,n)=(x,y,1,1)$ works, and by size we require $n\geq m$ hence no other solution exists for $n=1$, so assume $n>1$. Then as a consequence of Zsigmondy there must exist some prime dividing $x^n+y^n$ that doesn't divide $x+y$ and thus $(x+y)^m$, unless:
[list]
[*] $(x,y)=(k,2k)$ for some $k$, and $n=3$ hence $m \in \{1,2\}$. Then either $3k=9k^3$ which is impossible or $9k^2=9k^3 \implies k=1$, which yields $(x,y,m,n)=(1,2,2,3)$ and its permutation $(2,1,2,3)$.
[*] $(x,y)=(k,k)$ for some $k$, so $2^mk^m=2k^n$ from which we find that $k$ must be a power of $2$, say $2^j$. Then
$$2^{m+mj}=2^{nj+1} \implies j=\frac{m-1}{n-m},$$
from which we extract the very ugly answer $\left(2^\frac{m-1}{n-m},2^\frac{m-1}{n-m},m,n\right)$ for all $(m,n)$ such that $n-m \mid m-1$.
[/list]
Since everything is reversible, it follows that the solutions are $(x,y,m,n)=(x,y,1,1)$ $,(1,2,2,3),(2,1,2,3)$, and $\left(2^\frac{m-1}{n-m},2^\frac{m-1}{n-m},m,n\right)$ for all $(m,n)$ such that $n-m \mid m-1$. $\blacksquare$ (clearly, this implies the original problem)

[hide=probably wrong sol]
Assume $x=y.$ Then, we have $m = n = 1.$ So, we have the solutions $\boxed{(k,k,1,1)}.$ Now, assume one of $x,y$ is 1. WLOG let it be $x.$ Therefore, we have $(y+1)^m = 1+y^n.$ For some prime $p,$ suppose that $v_p(y+1) = a,$ such that $a\geq1.$ Then, $v_p(1+y^n) = v_p(n)+a.$ So, since $(y+1)^m = 1+y^n,$ we have $$v_p(n)+a = am\implies v_p(n) = a(m-1)\implies n\geq p^{a(m-1)}.$$ Note that $p\geq2$ and since $y\geq2,$ $n\leq m\log_{2}3\leq 1.6m.$ So, $2^{(m-1)}\leq 1.6m,$ meaning $m\leq 3$ If $m = 2,$ then $y^2+2y+1 = 1+y^n$ and $n\leq m\log_23\approx3.2,$ so $n = 3.$ Solving, we have $y = 2.$ If $m = 3,$ then $y^3+3y^2+3y+1 = 1+y^n$ and $n\leq m\log_23\approx4.7,$ meaning $n=4.$ Solving, we find no solutions. Now, assume that $x\neq y\neq 1$. Therefore, there exists some prime $p$ such that $v_p(x)\neq v_p(y)\neq 0.$ WLOG, let $v_p(x)>v_p(y).$ Since $(x+y)^m = x^n+y^n,$ we have $$mv_p(y) = v_p((x+y)^m) = v_p(x^n+y^n) = nv_p(y),$$ meaning $m = n.$ Since, $(x+y)^n \geq x^n+y^n,$ we must have $n = 1.$ In conclusion, the solutions are $\boxed{(x,y,1,1),(1,2,2,3),(2,1,2,3)}.$
[/hide]


geometry
3D geometry
quadratics
algebra
polynomial

Can someone explain why this is true:

Let $ z$ be a cube root of unity, so $ z^3\equal{}1$ and $ z^2\plus{}z\plus{}1\equal{}0$.  Note that $ z^5\plus{}z^4\plus{}1\equal{}z^2\plus{}z\plus{}1$, so $ n^2\plus{}n\plus{}1$ is a factor of $ n^5\plus{}n^4\plus{}1$.

Any explaination of the last phrase would be appreciated.

$ z^5\plus{}z^4\plus{}1\equal{}z^2\cdot z^3\plus{}z\cdot z^3\plus{}1\equal{}z^2\plus{}z\plus{}1$

$ P(x) | Q(x)$ if and only if each root of $ P(x)$ is also a root of $ Q(x)$ (counting multiplicity).  Since the roots of $ x^2 \plus{} x \plus{} 1$ are the roots of $ x^3 \equal{} 1$ not including $ 1$, we conclude that $ x^2 \plus{} x \plus{} 1 | Q(x)$ if and only if $ Q(\omega) \equal{} Q(\omega^2) \equal{} 0$ where $ \omega, \omega^2$ are the two roots of $ x^2 \plus{} x \plus{} 1$.

[quote="t0rajir0u"]$ P(x) | Q(x)$ if and only if each root of $ P(x)$ is also a root of $ Q(x)$ (counting multiplicity).[/quote]

If we take $ P(x) \equal{} 2(x \minus{} 1), Q(x) \equal{} x \minus{} 1$, is the converse "if each root of $ P(x)$ is also a root of $ Q(x)$, then $ P(x)|Q(x)$ still true"?
Maybe I'm misunderstanding the definition of "divisible"...  :maybe:

By the way, is there a way to determine if the one of the two roots of a quadratic is a square of another (like in the case $ x^2 \plus{} x \plus{} 1 \equal{} 0$)?  Thank you.

That's not a converse.  "Divides" here means that the quotient is a polynomial; it does not have to have integer coefficients.  (For that condition to hold we need to require that $ P(x)$ is primitive, that is, that there is no integer greater than $ 1$ dividing all of its coefficients.)

As for your second question, try to find all the monic integer polynomials such that this is true.  (There aren't many.)

We have $ (n^5 \plus{} n^4 \plus{} 1) \equal{} (n^2 \plus{} n \plus{} 1)(n^3 \minus{} n \plus{} 1) \therefore (n^2 \plus{} n \plus{} 1)|(n^5 \plus{} n^4 \plus{} 1)$


Six distinct positive integers are randomly chosen between 1 and 2006, inclusive.  What is the probability that some pair of these integers has a difference that is a multiple of 5?

$ \textbf{(A) } \frac 12 \qquad \textbf{(B) } \frac 35 \qquad \textbf{(C) } \frac 23 \qquad \textbf{(D) } \frac 45 \qquad \textbf{(E) } 1$

[hide="Solution"]You're asked to pick 6 integers basically, and find the probability that one minus the other in any of them is going to be a multiple of 5. You realize you're working in $mod 5$, but you can't pick 6 numbers and have all of them be different in $mod 5$, so the answer is 100%, or $\boxed {E}$[/hide]

[hide]

looking at mod 5 we have $0,1,2,3,4$ . Since we are picking 6 integer , by  pigeon hole , there must be two of them with the same residue .Hence the difference between that two integer must be a multiple of 5. So the answer is $\boxed{E}$[/hide]

I love this problem.

[hide]There are essentially 5 "groups" of numbers $[1,6...2006],[2,7...2002],[3,8...2003],[4,9...2004]$ and $[5,10...2005]$ However, you must chose 6, so the answer is 1.[/hide]

[quote=Syngates]You gotta choose 6 man

[quote=sub_math]I think I'm missing something here - what if you pick {2, 4, 6, 8, 10}? Then which difference will be a multiple of 5?[/quote][/quote]

ah oops  :blush:  there's always something I don't read

[ DISREGARD THIS POST ]
Please do, I really messed up.

this question be legendary, I wish all probability questions had answer 1 or 0.

[hide=Solution]
There are $5$ possible residues modulo $5$. By Pigeonhole Principle, there must be some residue that at least $2$ of the numbers have. Therefore, the answer is $\boxed{\textbf{(E)} 1}$
[/hide]


I have a couple of questions about dissolving:

(1) I dissolve ammonium nitrate in water.  What would this equation look like?

(2) I dissolve calcium choride in water.  What would this equation look like?

(3) How is dissolving a two step process?

Thanks in advance!

3) $ \ce{A_nB_m}(s) + aq \rightleftharpoons \ce{A_nB_m}(aq)$

$ \ce{A_nB_m}(aq) \rightleftharpoons \,\,or \,\, \longrightarrow n \, \ce{A^{x + }}(aq) + m\, \ce{B^{y + }}(aq)$

These two steps are often abreviated (in the sake of brevity) by

$ \ce{A_nB_m}(s) + aq \rightleftharpoons \,\,or \,\, \longrightarrow n \, \ce{A^{x + }}(aq) + m\, \ce{B^{y + }}(aq)$

1) $ \ce{NH4NO3}(s) + aq \longrightarrow \ce{NH4+}(aq) + \ce{NO3-}(aq)$

2) $ \ce{CaCl2}(s) + aq \longrightarrow \ce{Ca2+}(aq) + 2 \, \ce{Cl-}(aq)$

Just wondering, are $ \ce{NH4NO3}(s) + aq \longrightarrow \ce{NH4+}(aq) + \ce{NO3-}(aq)$ and $ \ce{CaCl2}(s) + aq \longrightarrow \ce{Ca2+}(aq) + 2 \, \ce{Cl-}(aq)$ supposed to be endothermic or exothermic?  Also, for $ \ce{A_nB_m}(s) + aq \rightleftharpoons \ce{A_nB_m}(aq)$ and $ \ce{A_nB_m}(aq) \rightleftharpoons \,\,or \,\, \longrightarrow n \, \ce{A^{x + }}(aq) + m\, \ce{B^{y + }}(aq)$, are they endothermic or exothermic processes?  Why?

Thanks!

The first one is endothermic and the second is exothermic (pretty sure that's the case).

P.S. Using the original question you can try and deduce the balancing act that determines whether a dissolution process is endothermic or exothermic. If you do a search of the forum I'm sure you can find something because I asked a similar question before and Carcul provided a great answer.

The dissolution of gases in liquids involve the same two steps. As an exercise, write the equilibria that explain the successfull dissolution of carbon dioxide (whose molecules are non-polar) in water.


Let $n$ be an odd number and $a_2,\ldots,a_n \in \mathbb{C}$. Also, $x_1,\ldots,x_n$ are the roots of $x^n+a_2 x^{n-2}+\ldots+a_n=0$. Prove that $n x_1 x_2 \ldots x_n = x_1^n+x_2^n+\ldots+x_n^n$.

[quote="perfect_radio"]$x^n+a_2 x^{n-2}+\ldots+a_n=0$[/quote]

what does that mean?
$x^n+\sum_{i=2}^n a_i x^{n-i}=0$ or $x^n + \sum_{i=1}^{\frac n2} a_{2i}x^{n-2i} = 0$ ?
the second would be contradicting to $n$ is odd ...

[quote="Singular"]I dont think its correct.  For example, roots -1, -1, -1, -1, 1, 1, 2.

They have sum zero, product 2, and 7th powersum 126.  But $14 \ne 126$.[/quote]

sorry :( the problem is wrong. but i realized that only after i posted this topic...


induction
trigonometry
function
algebra
functional equation
algebra theorems

Who can give me a [b]very detailed[/b] proof of: If $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$, then $ f(x) \equal{} kx$?

When you say $f(x)=kx$ do you mean that k is constant for any x and y?  This looks tricky.

[quote="blahblahblah"]It's not true, you need at least one more condition (on $f$)[/quote]

Explain?

[quote="shobber"]Who can give me a [b]very detailed[/b] proof of: If $f(x+y)=f(x)+f(y)$, then $f(x)=kx$?[/quote]

Is $f(x)$ continuous? :?

Oh, someone has already pointed out that. :blush:
This is Caucy's functional equation.

As for proving it in the continuous case, let $f$ be continuous and $E\subset X$.  If $E$ is dense in $X$, then $f(E)$ is dense in $f(X)$.  Since the rationals are dense in the reals, the result would follow from proving this lemma.

[quote="gauss202"]It's only true if $f(x)$ is continuous.

This problem is possibly too difficult for the intermediate forum. It would belong better in the Pre-Olympiad or Other Problem Solving Topics forum.[/quote]

Sorry i forgot it.... :blush:

without any conditions, you can prove f(x)=kx for rational x.

by induction we can get f(nx)= nf(x) for all real x and for all positive integers n.
thus f(n)=nk for all positive integers n, where k=f(1).
in the original eqn, f(0)=f(x-x)=f(x)+f(-x), thus f(n)=nk for all integers n.
let n/m be a rational number, with n and m being integers.
then nk=f(n)=f(m*n/m)=mf(n/m).
thus f(x)=kx for all rational numbers x.

to move from rationals to reals requires another condition.  it can be any one of the following:
1. f is continuous.
2. f is monotone (i can't remember if it must be STRICTLY monotone).
3. f is bounded on an interval [a,b] (i cant remember if the interval is open or closed, or can be either).

[quote="kennyworm"]
......
by induction we can get f(nx)= nf(x) for all real x and for all positive integers n.
thus f(n)=nk for all positive integers n, where k=f(1).
......
to move from rationals to reals requires another condition.  it can be any one of the following:
1. f is continuous.
2. f is monotone (i can't remember if it must be STRICTLY monotone).
3. f is bounded on an interval [a,b] (i cant remember if the interval is open or closed, or can be either).[/quote]

Well, how do u know that $f(1)$ is a constant?

Then let $f$ be $strictly$ monotone, then how do you move from rationals to reals?

[quote="kennyworm"]1. f is continuous.
2. f is monotone (i can't remember if it must be STRICTLY monotone).
3. f is bounded on an interval [a,b] (i cant remember if the interval is open or closed, or can be either).[/quote]
4. f is measurable

[quote="shobber"]
Then let $f$ be $strictly$ monotone, then how do you move from rationals to reals?[/quote]

If $f$ is strictly monotonic, then it has limits to the left and to the right in every point, and the value of $f$ at any point is between the limit when $x$ tends to the point from the right and the limit when $x$ tends to that point from the left. Now, simply choose, for each real, a sequence of rqationals converging to it from the right, and a sequence converging to it from the left . You'll get the result.

So we see $3 \Rightarrow 1,2$ and $4\Rightarrow 1,2$.
I proved 4 for more general statement somewhere on forum.
Therefore I would like to see 3.

[quote="Myth"]Therefore I would like to see 3.[/quote]
Oh, it is so trivial. :?

It follows from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=graph+dense&t=28183]this discussion[/url].

Of course, I didn't see it. I can prove it myself.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=175963#p175963]Here[/url]'s another nice problem on the topic.

just want to complete what kennyworm wrote
if x is real we consider an open interval in the neighbourhood of x : (x-1/n,x+1/n)
where n>0
so there exists rn belongs to Q/ rn belongs to (x-1/n,x+1/n)
so r1 belongs to (x-1,x+1)
r2 belongs to (x-2,x+2)............
then rn is a sequence of rational numbers such that rn-x belongs (-1/n,1/n)
so -1/n<rn-x<1/n
then 0<lim(rn-x)as n tends to infinty<0 (sandwish theo)
then lim rn=x
bec rn is rational we can say that f(rn)=k.rn
if f is continous as n tends to infinity f(rn)=f(x)
lim f(rn)=lim f(k.rn)=k.lim(rn)=kx
so f(x)=kx

Do the trig functions also have an identity like the one y=kx has??

[b][i)][/b]Try to solve the Cauchy functional equation $(*)     f(x+y)=f(x)+f(y), \forall x,y \in {\mathbb R}  $ when
  $f:{\mathbb R}\to {\mathbb R} $ is supposed to be continuous only at the point  $x_0=0 . $ It's a nice exercise !
[The continuity on the whole real axis is a simple consequence  of the continuity at the origin.]
Then.  the equation is solved on ${\mathbb N} , {\mathbb Z} , {\mathbb Q} $ and finally (using the continuity, and the notion of real numbers) on the whole real axis $ {\mathbb R} $ .
[b][ii)][/b]  By the way, denote  $ D=(-\infty,0)\cup (0,\infty) $ . Find all  real continuous functions $ f $ $: D \to
{\mathbb R} $ which satisfy
 \[ f(x+y)=f(x)+f(y) \; \; ,\; \; \forall x,y \in D.\]
[b][iii)][/b] I remember that Cauchy equation (*)   is solved also for measurable functions.

[quote="flip2004"][b][iii)][/b] I remember that Cauchy equation (*)   is solved also for measurable functions.[/quote]
I solved it on the forum in last November. I proved, that if $2f\left(\frac{x+y}{2}\right)\leq f(x)+f(y)$ and $f$ is measurable than $f$ is convex.


how do I write combinations (i.e. "a choose b") in LaTeX
I'm using amsmath package.

Use \binom{}{}

[color=indigo]Example 1[/color]

\binom12 yields $ \binom12$

[color=indigo]Example 2[/color]

\binom6{42} yields $ \binom6{42}$

... and many others.

$ \binom{11}{4,4,2,1}\equal{}\frac{11!}{4!4!2!1!}$

Multinomial coefficient - I just had to try that out. And yes, \binom for that even though it's multinomial. (After all, we use \sqrt[3] for cube roots.)


inequalities
geometry
parallelogram
geometry proposed

Let $P$ be a point inside a triangle $ABC,$ and $A_{1}B_{2},\,B_{1}C_{2},\,C_{1}A_{2}$ be segments through $P$ parallel to $AB,\,BC,\,CA$ respectively, where points $A_{1},\,A_{2}$ lie on $BC,\, B_{1},\,B_{2}$ on $CA,$ and $C_{1},\,C_{2}$ on $AB.$ Prove that:
\[[A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}]\ge\frac{2}{3}[ABC] \]
where $[XYZ]$ denote the area of polygon $XYZ.$


Here a problem that I have no idea to solve it! :(

How many are there sub-sets of {1,2,3,...,1990} which have 31 elements and whose the sum is divisible by 5 ?

Hint: Consider the following map:

Let $k$ be the sum of the elements of $S$.  Let $m$ be the least positive residue of $k$ (mod 5), and subtract $m$ from each element of $S$.

What does $S$ mean ?
The set with 31 elements or the set with 1990 elements?

The set with 31 elements.  So for example {1,2,...,31}  would go to {1990, 1, 2, ..., 30} since 1+2+...+31 is congruent to 1 mod 5


function
complex analysis
complex analysis unsolved

show that if $f(z)$ is an entire function and satisfies $\lvert f(z) \rvert < K \lvert z \rvert^{n}$ for all $\lvert z \rvert$ sufficiently large, then $f(z)$ reduces to a polynomial.
Hint: Compute $\lvert f^{(n+1)}(z_{0}) \rvert , z_{0}\in \mathbb{C}$

Have you not seen a proof of Liouville's theorem? It is exactly the same.


function
real analysis
real analysis unsolved

Let $f$ be a real continuous function which sends every open interval in a closed interval. Then $f$ is constant.

Atile told me about this one a ling time ago, and every time I remember it, I think of a different proof :). Indeed, it's very easy, but I like it too, for some reason :).

Suppose, for instance, that $f$ has the property $f(a)<f(b)$ for some $a,b$. Let $M=\max_{[a,b]}f$, and let $b_1=\min\{t\in[a,b]\ |\ f(t)=M\}$. Then the image of the interval $(a,b_1)$ has $M$ as a limit point, but does not contain it, so it can't be a closed set.


From the hypothesis, we have $x^{2}+y^{2}+z^{2}-3=0$
So one of three variables must have the square greater or equal to 1.WLOG $x^{2}\ge 1$
*)$x\ge 1$ so $x\le y$
Then it's easy to show that$y\ge1$ and$y\le z$
By the same way, $z\ge 1$ and$z\le x$
that means $x\le y\le z\le x$
So the equalities occur. One root is $x=y=z=1$
*)$x\le-1$ Do the same we have another root $x=y=z=-1$
So the system has 2 roots.


geometry
3D geometry
number theory
relatively prime

a and b are natural numbers. what is the smallest possible value of a+b/ The answer is 1260 but I want to know how to get it

$ 5a^3\equal{}6b^4$.

since 5 and 6 are relatively prime, a must be a multiple of 6 and b must be a multiple of 5.

so lets say $ a\equal{}6v, b\equal{}5u$.

then it becomes $ 5\cdot 216v^3\equal{}6\cdot 625u^4, 36v^3\equal{}125u^4$.

let $ v^3$ be $ 125x$, and let $ u^4$ be $ 36x$.

then $ x$ is cube of something since $ 5^3\cdot x$ is $ v^3$.
also, $ x$ is in the form of $ 36\cdot$another something to the fourth.
if another something=6, something is 36. ( :rotfl: )
then $ x$ is $ 36^3$.

$ v^3\equal{}5^3\cdot36^3, v\equal{}180$
$ u^4\equal{}36\cdot36^3, u\equal{}36$

$ a\equal{}6v\equal{}6\cdot180\equal{}1080$
$ b\equal{}5u\equal{}5\cdot36\equal{}180$

$ a\plus{}b\equal{}1080\plus{}180\equal{}1260$

answer : 1260


the earth is sliced into two by its equator. With how much force could a giant pull the two slices apart?

Do you have a solution to this problem? I mean, is it possible to calculate this work in closed form?  :huh:

of course it has a closed form. otherwise I wouldnt have posted it. :lol:

actually it's L=K*sqr(M)/R*(3*pi/64).M-earth's mass ,K-constant gravitation,R-earth radius.

oups...sorry...is it 3*pi/32 instead of 3*pi/64

oops I'm very sorry I meant with what force could he pull it apart, I edited the problem too. Yes if it was work there would be no closed form. Once again sorry. :blush:

i'm not sure of how you are thinking the problem.what means work in closed form?if i consider the distance between the 2 semisphere's mass centrums then the distance is 8R/3*pi=d then F in the initial stage is F=K*sqr(M/2)/sqr(d).

'Closed form' is a mathematical term and means that the expression concerned is analytical (so it has no unsolvable integrals, unsolvable equations etc.)

This problem is now easy -- in comparison to the previous case. The maximal force of interaction between the two halves will be achieved when they are pressed together (ie. they form a ball), and in that case it is possible to calculate the force of attraction, although it seems to involve some non-enlightening integrals. Is there an elegant solution?

Btw, [b]Thaakisfox[/b], where do you get these problems from?

Well I have lots of books and the basic problems can be found there. for xmple: John David Jackson: Classical electrodynamics; priciples of movement; Theoretical physics books; etc. Some of them have my question.
Here is a solution that uses calculus only once:
The work needed for pulling the two slices apart by $d$ is :
$W=F d$. ($F$ is the force we are looking for)
This work can be calculated another way: How much work does it take to scoop out the mud from a disc of width $d$.
Lets calculate this work from the growing of the gravitational potential energy of the mud which is brought to the surface.
The gravitational accleration on the surface of earth is $g$.
So the  gravitational accleration in distance $x$ from the center of earth is : $g(x)=\frac{x}{R}g$.(R is the radius of earth).
Lets look at the mud between radii $x$ and $x+\delta x$ of volume $\delta V=d2\pi x \delta x$ and mass of $\delta m= \rho \delta V$ ($\rho=\frac{3M}{4\pi R^{3}}$ is the mass density of earth).
The  gravitational force on this "piece" of mud inside the earth is :$\frac{x}{R}\delta m \cdot g$, on the surface: $\delta m \cdot g$. during the scooping out the gravitational force changes constantly, we can count we the arithmatic middle of the force at the beginning and at the end. The total shift is :$R-x$, so the work:
$\delta W=\delta m \cdot g\frac{1+\frac{x}{R}}{2}(R-x)=\delta m g \frac{R^{2}-x^{2}}{2R}=\frac34 \frac{Mgd}{R^{4}}(R^{2}-x^{2})x \delta x$.
The total work:
$W=\sum \delta W =\frac34 \frac{Mgd}{R^{4}}\sum (R^{2}-x^{2})x \delta x$
if  $\delta x \to 0$ then we can integrate:
$W=\frac34 \frac{Mgd}{R^{4}}\int_{0}^{R}(R^{2}-x^{2})xdx=\frac{3}{16}Mgd$
So the force needed for the giant to pull the slices apart is:
$F=\frac{3}{16}Mg$

Two remarks, for now:

1) How come it's $W = F\cdot d$? Shouldn't there be an integral?

2) I understood that you're pulling away slice by slice of the Earth, but that can't be right because there are several inconsistencies... Could you please explain this method for calculating work a bit more?

Well, $d$ is so small that the force between the two slices hardly changes at all so we can say that the work needed to pull apart by $d$ is $W=Fd$.
The surface of the earth is equipotential so spreading the scooped out mud over the surface doesnt need extra work.
was this the inconsistance you were talking about?

No, it wasn't  :(  I'm confused. What do you do with the mud, from where do you take it and where do you put it?

ok.
Lets assume that we pulled the slices apart by $d$. We want to find the work we needed to do it.
Now the slices are together. Lets look at a disc of width $d$ at the meeting of the slices. this disc is full of mud. Now lets scoop all of this mud out to the surface of the earth.(Like when we dig we throw the mud out of the hole we dug). then lets spread it on the earth evenly. We dug out all of the mud in the disc, that means the two slices now are separated by a distance of $d$. The spreading of the mud around the earths surface doesnt need extra work because the surface is equipotential. So the only work we need to do is to dig all the mud out  . This is the same as if we separated the two slices.
Clear now? :)

Yes :lol: Thanks -- nice problem. I'll try to post some problems soon ;)

[quote="Djole"]Yes, it is, but $d$ is very small so $U = Fd$ can hold...[/quote]

So then why go through the trouble of integrating, then dividing?

Well, as I understand, the idea is to find the work needed to remove a disk of width (thickness) $d$ and radius $R$ from the Earth. This disk contains the equatorial plane and $R \gg d$ holds. After taking all the earth (or mud) from this disk and smearing it across the Earth's surface, you effectively get two halves of the Earth at a distance $d$ apart. Now, the integration is used to find this work by considering the work needed to move infinitesimal rings from the earth; $d$ has nothing to do with the integration itself (it is merely a constant).

Finally, after you've calculated the work needed to separate the two halves, you may find the force of their interaction; since this force will only decrease as the distance between the halves grows, you may claim that this force is sufficient to pull the two halves apart of each other.

I hope it's clearer now -- if not, I'll let [b]Thaakisfox[/b] take over the explanation. It's his problem (and solution) after all :)

OOHHH, I get it now.

But, we really should say $F = \frac{dW}{d d}$, which gives you the same answer as dividing, but seems better (at least to me) conceptually.

ok. yep I should have writed $\frac{dW}{dd}$. because there were misunderstandings with the $\frac{W}{d}$.
I'm glad you all understood it. :wink:


For what pairwise different positive integers is the value of

$ \frac{a}{a\plus{}1}$+$ \frac{b}{b\plus{}1}$+$ \frac{c}{c\plus{}1}$+$ \frac{d}{d\plus{}1}$

is an integer?

I believe all possible solutions are $ (1,3,5,11), (1,3,4,19), (1,2,9,15), (1,2,8,17), (1,2,7,23), (1,2,6,41)$ and all permutations of elements of that quadruples.
It's easy to solve this using limitation:
If $ \frac {a}{a \plus{} 1} \plus{} \frac {b}{b \plus{} 1} \plus{} \frac {c}{c \plus{} 1} \plus{} \frac {d}{d \plus{} 1}$ is an integer, so is $ \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \plus{} \frac {1}{d \plus{} 1}$. Now, we can assume WLOG $ a < b < c < d$ and just mention permutations in the final solution as all variables are obviously symmetric. Now, if $ a\geq 2$, we get $ \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \plus{} \frac {1}{d \plus{} 1}\leq \frac {1}{3} \plus{} \frac {1}{4} \plus{} \frac {1}{5} \plus{} \frac {1}{6} < 1$, so it obviously isn't an integer. That means $ a \equal{} 1$. So, we need $ \frac {1}{2} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \plus{} \frac {1}{d \plus{} 1} < \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{4} \plus{} \frac {1}{5} < 2$ to be a positive integer. Therefore, $ \frac {1}{2} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \plus{} \frac {1}{d \plus{} 1} \equal{} 1$. Now, we have
$ \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \plus{} \frac {1}{d \plus{} 1} \equal{} \frac {1}{2}$. If $ b\geq 5$, obviously we have a contradiction ($ \frac {1}{6} \plus{} \frac {1}{7} \plus{} \frac {1}{8} < \frac {1}{2}$). There are $ 3$ cases:
1) $ b \equal{} 4$. Then, we have $ \frac {1}{c \plus{} 1} \plus{} \frac {1}{d \plus{} 1} \equal{} \frac {3}{10}$, which is wrong for $ c\geq 6$. We check if $ c \equal{} 5$ gives a solution, which it doesn't.
2) $ b \equal{} 3$. Then, we have $ \frac {1}{c \plus{} 1} \plus{} \frac {1}{d \plus{} 1} \equal{} \frac {1}{4}$, which is wrong for $ c\geq 7$. We check if $ c \equal{} 6$, $ c \equal{} 5$ and $ c \equal{} 4$ give us any solutions. They do, $ (1,3,5,11)$ and $ (1,3,4,19)$.
3) $ b \equal{} 2$. That implies $ \frac {1}{c \plus{} 1} \plus{} \frac {1}{d \plus{} 1} \equal{} \frac {1}{6}$, which is wrong for $ c\geq 11$. Also, obviously, there are no solutions if $ c\leq 5$. We check for solutions in $ c \equal{} 10$, $ c \equal{} 9$, $ c \equal{} 8$, $ c \equal{} 7$ and $ c \equal{} 6$. We obtain the following solutions: $ (1,2,9,15), (1,2,8,17), (1,2,7,23), (1,2,6,41)$.

That's it. We check all the solutions, they seem to be alright, we note there are many more actual solutions due to permutations and that wraps it up. I apologize for any unintentional mistakes.


Well... what do you think Ignite168's avatar is?

Argh... I don't realize about what is the avatar of Ignite168 is go board until you told it :wallbash:  :wallbash:  :wallbash:. His signature makes me somehow don't really see it clearly (or I realized it, but somehow forgot about it  :maybe: ).

A good point to start is http://en.wikipedia.org/wiki/Go_%28board_game%29 .

it kind of looks like a dog being hung upside down


Let there be a strict positive sequence $ (a_{n})_{n\geq 1}$ such that for any $ n\geq 2$ we have \[ a_{n\minus{}1}\leq (a_{n\plus{}2}\minus{}a_{n})^{n} \leq a_{n\plus{}1}\] Calculate $ \displaystyle\lim_{x\to\infty} \frac{a_{n}}{n}$.

Maybe the problem is rather simple, but still, can you post your solution?


algebra
polynomial
superior algebra
superior algebra unsolved

We consider only monic polynomials.

Of course every polynomial $ P(x) $ commutes with $ P^{(n)}(x) := P \circ \ldots
\circ P (x) $ ( $ n $ times ), for $ n \geqslant 0 $. If this set contains
all polynomials which commute with $ P(x) $,then we call such a polynomial
poor. For each degree find many pairwise
nonequivalent polynomials which are poor and prove that.

[quote="Leva1980"]People!
Please, try to solve my problems about commuting polynomials!
They are interesting!![/quote]

They are just interesting or you need an answer for them?  :P

I solved them by myself, but I want to see another proofs.


How many whole numbers between 99 and 999 contain
1) one 0
2) exactly one 0?

[hide="Number 1"]
Let the number be $ n$, which clearly has 3 digits (I'm assuming this is 99 to 999, not inclusive).
Let $ n\equal{}\overline{abc}$.
$ a$ cannot be 0. There are 9 possible values for $ a$.
Suppose $ b$ is 0 and $ c$ is not. Then there are $ 9$ possible values for $ c$, so we have $ 9^2\equal{}81$.
Suppose $ c$ is 0 and $ b$ is not. $ 81$ more values.
Suppose both are 0. $ 9$ more values.
In total, $ 2\cdot 81\plus{}9\equal{}\boxed{171}$.
[/hide]

[hide="Number 2"]
Same thing as before, without the last case.
$ 2\cdot 81\equal{}\boxed{162}$.
[/hide]


Duke
college
Stanford
geometry
number theory
real analysis

I'm sure that both Duke and Stanford have some sort of distribution requirements for graduation and that you can find the details of their graduation requirements online.

But I felt the same way as you when I was applying to colleges (about taking the minimum number of non-math courses), however, I found once I got there that a lot of the non-math courses I took were actually really interesting and very useful. They helped me to think and argue more clearly, and they even helped strenghten my overall understanding of the place of mathematics in the perspective of humanity and science. I wouldn't shortchange these courses until you actually get to college and take some. Non-math courses in college aren't necessarily the same old thing as non-math courses you've taken in high school. Well... some are, but not all of them.

[quote="probability1.01"]Also, it's not really that fun to be doing math [i]all day[/i]. Even people who are really into math can only do so much without getting inefficient, and on the days that math just doesn't seem all that appealing, it's good to have something else to do.[/quote]

Hey! When I go to university next year I will be doing a straight maths degree in the UK. I don't think I'll have a problem doing just maths

First of all, congratulations on getting into those 2 colleges. Caltech's math department is not actually overly famous, iirc, so if you're going into pure math, you shouldn't be too upset about that :D

I live almost right next to Stanford, and have come in contact with many undergrads there. In terms of required classes, I think the IHUM and PWR are the 2 big humanities course sequences that you MUST take, no matter which major you're in. I took Stanford's 50H math sequence last year, and absolutely loved it. It was my first non HS math class, and the rigor and everything was perfect. My friend who was in 50H with me, and matriculated to Stanford this year, is taking 2nd year math classes. He said that most of them actually were a bit easier than the honors intro sequence. However, the math department at Stanford is in the rotating top 5, so I'm sure if you seek out professors and TAs as an undergrad for work beyond the courses, it will be great.
I of course, can't also live right next to Duke, so that's my reason for not having much familiarity with that school =P.

Are you coming to Stanford's Admit Weekend? If so, we should talk, since I'm also thinking of studying math at Stanford. Hope this helped.

Edit: also, to the previous posters: I'm sure you won't be doing math the entire day, even if you take multiple math course all the time. There's just so much going on in college that the only way you'll be doing math all day is if you really want to :D

[quote="Abel"]I have been accepted by Duke, Stanford and rejected by Caltech. Could someone on this forum suggest which of the two would be more suitable for pursuing pure mathematics? I am looking for a place where I will have to deal with the minimum number of non-math courses.[/quote]

I am a freshman at Stanford right now, and I have a good friend who goes to Duke, so I have a bit to say about both.

Stanford: As krusyteklown mentioned, IHUM and PWR are pretty significant humanities requirements that you can't get out of; IHUM is 3 quarters (freshman year) and PWR is 2 (one freshman, one sophomore). There are also a number of distribution requirements which can be fulfilled fairly easily by taking a few classes that satisfy multiple. The last major requirement is 3 quarters of foreign language if you don't have prior credit (AP/SAT/placement test). If you need to meet all the above requirements, you can expect to take around 11 or 12 classes at the minimum which are not major-oriented (keep in mind that they are only quarter-long as well).

Duke: I don't have as much in-depth knowledge, but I do know that Duke has a pretty diverse set of distribution requirements. In particular, there are Area of Knowledge requirements (which basically means you need to take two classes or so in a variety of different subjects) and Modes of Inquiry requirements (somewhat vague to me). The latter contains 3 semesters of foreign language and 3 semesters of writing. As far as I know, these requirements are easier to satisfy in general than Stanford's requirements, i.e. many classes satisfy 2 or more. I can't say how many classes you'll need to take which are not major-oriented, however.

I would hesitate to choose a school based on the minimum number of non-math courses you need to take -- Stanford and Duke are both excellent schools with their strengths and weaknesses. Look into them more carefully, and try to get in touch with students at each of them to learn more. I love Stanford despite its requirements (which can obviously be a pain sometimes), and my friend loves Duke. Chances are you'll be happy with either one you pick, but it's still good to pick the right one :).

I also agree with the previous posters who recommend that you shouldn't make your plans around avoiding non-math classes. I am a junior in maths right now. For the first two years, I mainly tried to get rid of distribution requirements as quickly as possible and then fill all available space with math classes. But this year I actually took some non-math classes for enjoyment, and I discovered that I am very interested in and excited by some other topics!! But I would never have known about this if I had just taken only math classes...

Thanks for the replies. How do the strengths of the departments compare with respect to abstract algebra and number theory? (Although I have no idea about how that would matter for undergrads.)

The Professor at Duke who helped me get in said that grad courses are easily accesible in the freshman year. Is this true for Stanford as well?

You can sign up for whatever classes you want at Stanford (whenever you want). The only question is whether or not you're ready for them.

As for abstract algebra and number theory, I really have no idea how they compare.

Grad courses at every school are accessible to at least some freshmen.  It all depends on who the freshmen are.

Seriously, though, at Duke it is rare for freshmen to take the intro graduate student courses on algebra and real analysis, because there are equivalent courses for undergraduates that actually go more in depth (b/c they run for two semesters instead of one).  Aside from, and beyond, those two courses, it is true that the top undergrads are comfortable in grad classes (and are often some of the best students).  However, freshmen are pretty rare in graduate student classes.  The prof might have meant graduate *level* classes.  See the "Duke Math Department" thread for more of my writings on Duke.

At Stanford, it is fairly common for top undergrads to take the standard intro grad courses (real analysis, algebra, complex analysis/topology/geometry) along with the first-year grad students (who are much stronger than at Duke).  Real analysis was the most common choice for undergrads (back in my day, at least).  Again, though, freshmen are rare.

Stanford has become very strong in number theory, recently adding 3 renowned number theorists (many good schools have zero).  Stuff like that usually doesn't matter too much for your undergraduate education, but if you're sure you want to get really hard and heavy into number theory, then Stanford is the clear choice.

General piece of advice for freshmen:  Unless you know A TON of math, don't start out with grad classes.  Even if you are well-educated in one area (e.g. algebra), you might not know as much about other things (e.g. geometry), and it's a good idea to shore up your foundations before rising to the stratosphere.  And who knows, you might find other math topics that you like more.


calculus
integration
calculus computations

[quote="pankajsinha"]Evaluate:

$ \int_{0}^{1}\frac {2x \minus{} (1 \plus{} x^2)^2 cot^{ \minus{} 1}x}{(1 \plus{} x^2)(1 \minus{} (1 \minus{} x^2)cot^{ \minus{} 1}x)}dx$[/quote]

Would substituting $ x \equal{} \cot \theta$ help. Atleast we may get rid of a term from the denominator.


[quote="zaizai-hoang"]Disprove the following inequality  :D  For all $ a,b,c\in \mathbb{R^ \plus{} }$:
\[ 2(a^4 \plus{} b^4 \plus{} c^4)^2\ge (a \plus{} b \plus{} c)(\sum a^7 \plus{} \sum ab^6)
\]
[/quote]
Try $ a\equal{}1,$ $ b\equal{}2$ and $ c\equal{}0.$ :wink:


geometry
Duke
college
summer program
MathPath
MATHCOUNTS
Mathcamp

I'm in SET. I'm going to the Latin course, session 1.

It is JHU CTY... John Hopkins University Center for Talented Youth. A great program for middle schoolers, but not for high school....(at least not the BEST).

I've been there, and my brother is going to be there this year.

I'm going there this summer, and I'm taking the Individually Paced Math Sequence...going to try to take Honors Geometry before high school starts! :)

I'm a SET member thingy and I went to CTY in 7th and 8th grade.  I didn't like it that much, but it was ok.  The first year was better.  Go to the Lancaster campus.

Sounds similar to DUKE TIP.  Anyone going there, by the way?

[quote="Phelpedo"]I used to go to CTY, but last year I went to a camp called MathPath. At 70 people, it's a relatively small camp, but a lot of math elite went like 3/4 of the winning Illinois team including Greg Gauthier. In fact, probability1.01 u were the only one from that team that didn't go! Anyway, a lot of its members are also on this board, and it's not too late to apply![/quote]

Wow!  I never realized MathPath was so... elite.  I always thought when it said gifted, if meant AAP gifted, not MathCounts gifted, so it would be boring and repetitive.  But I guess I should look into it.  And soon!

It is true that you need to develop your writing skills as well as your math, but SAT math problems are generally much easier than Mathcounts, so if you won state Mathcounts and didn't make a simple mistake on the SAT math, you should get 800 with no problem.

[quote="falcon_i"]I was really amazed by the boy from PA who scored 1600 on old SAT-I last November (and he was a 8th grade student then). I believe it is a much bigger accomplishment than winning the mathcount at the state level. All the people who loves math should pay more attentions on the language side as new SAT put more emphasis on testing your reading and writing skills.[/quote]

The highest 7th grade score nationwide this year (according to Duke) was 1540 on the old SAT.  However, 800's were achieved in both categories.  I missed an 800 by one question.  Who cares about when a train arrives in NY?

[quote="seungsoo"]Last year, I scored a 750 on the math and a 770 on the verbal on the old SAT-I. I was a 7th grader then. My twin brother Seungjun scored an 800 on the math and a 760 on the verbal, totaling 1560 :first: . He was also in 7th grade.  :lol:[/quote]
wow...  gj :first:

i went to baby cty a few years ago and a camp at phillips exeter acadeemy a couple years ago along with a very good math camp in texas called SWT mathworks. it is cheaper than mathcamp and mathpath but is still an excellent program

this year im going to duke tip i think

[quote="falcon_i"]I'm new for the JHU CTY SET, and will take an essay writing class this summer. Anyone here who scored 700 or above for the SAT-I V or 800 for SAT-I M and got into the SET? I knew there are bunch of Chinese kids who got into the SET due to their SAT-I M scores. I was really amazed by the boy from PA who scored 1600 on old SAT-I last November (and he was a 8th grade student then). I believe it is a much bigger accomplishment than winning the mathcount at the state level. All the people who loves math should pay more attentions on the language side as new SAT put more emphasis on testing your reading and writing skills.[/quote]

Well an 800 in SAT-Math is really not as difficult as qualifing for Nats IMO. My friend got an 800 but was only 7th at MC States, and my state is pretty small. I qualified for JHU SET last year, highest score in the state, and was 40th at MathCounts States(although didn't take SAT this year and qualified for Nats). The reason why few people get and 800 is b/c only 20,000 people take part in the Talent Search (7th and 8th combined) while over half a million participate in MathCounts.

On the other hand, a 1600 is quite impressive, still not as impressive IMO as National Champion, but quite impressive. I have no doubt though that if all Nationals qualifiers were given a few months to prepare for the SAT, more than a fewwould get in the 1500s.

wth... 250,000 dollars?? and 100 thousnad dollars???

[quote="falcon_i"]A quick reply before going to school. During the 20 years tenure of JHU CTY program, only one girl (age 12) got 1600 SAT, and she got a $\$250K scholarship immediately. The MC national winner can only got $\$100K maximum. BTW, most people only got a TI calculator. Money really speakes itself.[/quote]

I'm with sheepwarrior. What are you talking about?

But back to the real thing. First of all, I doubt that the girl got 250k immediately. At least not the way you're stating it.

Also, money doesn't speak for itself. The spelling bee yields more money than MC or any other competition, yet I'd prefer to be National Champion than spelling bee winner. I got a scholarship for the SAT-I amounting to more than the 3rd place prize at Nats, but I think I'd rather have the latter. Thousands of people have gotten a 1600; 800 in Math is a piece of cake for any Nationals qualifier and 800 Verbal is really just a lot of memorization. Just because MathCounts (as nonprofit) has nowhere near the funding of JHU doesn't mean it is any less of an honor.

This is the response to Phelpedo and other kids here:
First of all, I didn't participae MC, but I do admire all the people who advanced to the national level which is a great accomplishment, and I was referred by staffs from JHU to this site. Hopefully, I will have time next fall to be part of this great math exercise. (After I'm done with my new SAT next month and October). Second of all, there is always a winner for MC each year, but there is no guarantee that there will be a kid under age 13 who can score perfectly on SAT, and that's why I believe it is harder to get latter. There were over half million kids took the SAT as part of JHU CTY program, and only one girl who were able to score perfectly.  Finally, momey is important. And without $\$ $\$ $\$, there would be no MC at all next year.

I participated in the CTY talent search, and as far as I know there are only 20,000 annually.

Yes, money is important, but it is not necessarily an accurate representation. You only get 1 million for winning the Nobel prize or solving a millenium problem; the chinese track athlete who won a gold medal (liu xiang?) got 10. Some guy who won the lottery got 200 million, and I'm pretty sure he isn't 200 times as smart as a nobel prize winner.

Also, the nature of the National Competition is different from getting a 1600. The definition of a competition is that there is a winner every time, just like there is a #1 SAT score every year. I encourage you to participate in MathCounts so you can gain a real respect for the difficulty and competitiveness. As I said, I got a $5000 scholarship for highest SAT score in the state (and so did a couple of people who didn't even get that) and was 40th at MC state.

I agree that a 1600 is difficult, but it is really a lot of memorization and studying as opposed to creative and innovative thinking, and no problems are as complex, difficult, or stimulating as those found in a MathCounts National set. It's just that the SAT requires a little maturity and very few middleschoolers take it; on the other hand, upper classmen are required to and that's why a lot of them get 1600s. But I bet most of them couldn't beat Greg Gauthier in a math competition.

[quote="chess64"]Any JHU CTYers here? I really wish I could take Number Theory: [url=http://cty.jhu.edu/summer/docs/syllabi/theo_1.pdf]cty.jhu.edu/summer/docs/syllabi/theo_1.pdf[/url] but there aren't any more spots  :([/quote]

I'm surprised that this thread isn't in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=136]Other US Contests & Programs Forum[/url]. The JHU-CTY number theory course, because it only is offered for one section in each session, is notoriously hard to get into. Basically, only applicants with the enrollment priority of RETURNING JHU-CTY summer program students have a chance to get in. But, if it's any consolation, I have read a statement by an AoPSer who says that the course is actually easier than it is reputed to be, and I have read another statement by a CTYer that he took the eight-week Ross Program the next year, and two weeks of Ross > three weeks of JHU-CTY when it comes to number theory content and difficulty.

With regard to the subsequent development of this thread, getting an 800 on the SAT I math section at any age (for my son, it was at age eleven) is less difficult than aceing a specialized math contest for competitors of the same age. The SAT I has all the correct answers right there on the page for you to notice (except for a trivially small number of free-response questions), and authentic review materials from College Board are abundant. By contrast, most state math leagues, MATHCOUNTS, and AIME all require you to generate answers on your own. The time limits are often tighter on the AMC contests too. I am impressed by SAT I math section aces before, say, age sixteen, but I am more impressed by aceing an AMC test or getting a moderately high rank at MATHCOUNTS nationals.

Is anybody here on AoPS going to the Midwest Academic Talent Search award ceremony in Evanston on 4 June 2005? That's when we find out how our son placed this year (on the ACT).

From what I've known talking to number theory people at JHU Lancaster, it really isn't that difficult.  All it is is learning to use induction, working with mods, and maybe some encryption.  You could get more out of an AoPS class or just reading the AoPS books.

As for SATs, re-read every one of Phelpedo's posts :P.

Hey probability, I'm a little confused by your sig. I can't tell if you're trying to make some parts jokes or what...

Heh, it's not done but it wouldn't let me add any more.

im a SET member because I got a 720 on math when i was in 5th grade and (something Im not going to say) on verbal.

I'm taking essay crafting in baltimore. Anyone else?

im in set too, i got a 1500 this year on SAT. 780 math, 720 verbal

[quote="tokenadult"]
Is anybody here on AoPS going to the Midwest Academic Talent Search award ceremony in Evanston on 4 June 2005? That's when we find out how our son placed this year (on the ACT).[/quote]
Hmm..., you mean you do not know?

We are not going because my son will take the new SAT on June 4th and we are checking out high schools on June 3rd in PA.

[quote="Naga"][quote="Phelpedo"][ The reason why few people get and 800 is b/c only 20,000 people take part in the Talent Search (7th and 8th combined) while over half a million participate in MathCounts.
[/quote]

Much more than that Phelpedo.  The Talent search is held by CTY - JHU and Northwestern, TIP - Duke , and  RMTS.  The Duke statistics alone show that over 50,000 seventh grade students took the test:

http://www.tip.duke.edu/TS/7TSResultsSummary.pdf

The half million who participate in MathCounts do so in the training program.  A fraction go to chapter and a smaller fraction to state.

I agree that with some determination and preparation it is easier to get a perfect 800 on the Math component of SAT than to get it in MC.  There's more time for the SAT, it is multiple choice and it covers traditional class room material.[/quote]

For some reason I seem to think it's less... maybe that's the plus test.

Still, 50,000 is only 1/10 of half a million, and the fact that only small fractions make it to chapter/state is just my point. And considering that several hundred people get 400s... (for those of you who don't know, if you don't fill in any answers you'll get higher than that)

Hey probability, one of my friends at Choate knows you from CTY.  :)

[quote="Naga"][quote="tokenadult"]
Is anybody here on AoPS going to the Midwest Academic Talent Search award ceremony in Evanston on 4 June 2005? That's when we find out how our son placed this year (on the ACT).[/quote]
Hmm..., you mean you do not know?[/quote]

Yeah. We really don't know here. We know he is invited to the ceremony, but this year the statistics summary combines information from two previous years with only PARTIAL information from this year. On historical grounds (that is, by checking last year's statistics sheet, showing only one year), I would expect him to be in the top three, but maybe this was an anomalous year. I think our ceremony invitation letter said in so many words that we wouldn't know the top rank places until the day of the ceremony, which is certainly odd.

[quote="tokenadult"]

I think our ceremony invitation letter said in so many words that we wouldn't know the top rank places until the day of the ceremony, which is certainly odd.[/quote]

So did ours.  I am hoping that a perfect is in one of the top three. :)

Wish we could have made it.  I love Chicago.

Naga


Find the real roots $ a$ and $ b$ that satisfy the following equations:

$ (3a^2\plus{}b^2)(a^2\plus{}3b^2)\equal{}\frac1a\plus{}\frac{1}{2b}$
$ 2(b^4\minus{}a^4)\equal{}\frac1a\minus{}\frac{1}{2b}$

[quote="jelly119"]Vinskman, you are really [b][i]fantastic[/i][/b]!![/quote]

Thanks..

 :lol:


limit
trigonometry
calculus
inequalities
derivative
calculus computations

so when the lim of sin5x/5x, and x approaches 0, the answer is 1? can someone explain this?

Make the substitution $ 5x \equal{} \theta$ and you get the well known limit $ \lim_{\theta \to 0} \frac{ \sin(\theta)}{\theta} \equal{} 1$

right, but I don't really get why sin(theta)/theta as theta approaches 0 = 1. is there something like a proof or explanation somewhere?

Most calculus textbooks have a geometric argument, involving a wedge of a circle, a triangle containing the wedge, and a triangle contained in the wedge. From this, they derive the chain of inequalities $ \cos \theta<\frac{\sin \theta}{\theta}<1,$ valid for $ \theta$ in the first quadrant (and the fourth quadrant, by symmetry). The squeeze theorem applied to this (and the assumption that the cosine is continuous at zero) then yields the desired limit.

And we need this limit before we can derive the formulas for the derivatives of sine and cosine. That's the logical pathway if you start with a geometric, unit circle-based definition of the trigonometric functions.

The logical alternative is to define the sine and cosine in a more analytic way to begin with - as solutions to a certain differential equation, or as power series (filtered through the definition of the exponential function). These methods give rise to the calculus of sine and cosine, including this limit, fairly immediately; you need to do more work to show all the properties that you take for granted with the geometric definition.

ok, I think i understand it now. But I need to prove sinx<x<tanx for the larger proof to work. i've been trying to think about it algebraically and geometrically, but to no avail. :(


Find all possible sets of four consecutive integers having the property that the sum of the cubes of the smallest three is equal to the cube of the largest.

Source:  MATC Mathematics Club

[hide]Solutions in n where n is the first consecutive integer will fit:

n^3+(n+1)^3+(n+2)^3=(n+3)^3

After expanding and simplifying, we are left with:

n^3-6n-9=0

Factors into:

(x-3)(x^2+3x+3)=0

Thus, the only integer solution for n is 3, and the only set is [b](3, 4, 5, 6).[/b][/hide]


A massless elastic cord (that obeys Hooke’s Law) will break if the tension in the cord exceeds $ T_m$. One end of the cord is attached to a fixed point, the other is attached to an object of mass $ 3m$. If a second, smaller object of mass $ m$ moving at an initial speed $ v_0$ strikes the larger mass and the two stick together, the cord will stretch and break, but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one-dimensional collision, the cord will still break, and the larger mass will move off with a final speed of $ v_f$. All motion occurs on a horizontal, frictionless surface. Find $ v_f/v_0$.

I did it several times, but I always got $ 1/\sqrt {2}$,which is wrong.

I assume that isn't the right answer?  Could you type out your work and explain your free-body diagrams?  I'm not in the mood to attempt to solve any physics problems at the moment but if you post what you did, I could certainly take a look at it and try to catch your mistake.

Let $ v_1$ be the velocity of the $ 4m$ mass after collision 1, let $ v_2$ be the velocity of the $ m$ mass after collision 2, and let $ v_3$ be the velocity of the $ 3m$ mass after collision 3. Let the maximum extension of the spring be $ s$ such that $ T_m \equal{} \minus{}ks$.

In collision 1, momentum is conserved, $ mv_0 \equal{} 4mv_1 \implies v_1 \equal{} v_0/4$.

Between the time after the collision and before the spring breaks, energy is conserved, $ (4m)v_1^2/2 \equal{} ks^2/3 \implies s^2 \equal{} 4mv_1^2/k$.

In collision 2, momentum is conserved, $ mv_0 \equal{} mv_2 \plus{} 3mv_3$, kinetic energy is conserved because it is perfectly elastic, $ mv_0^2/2 \equal{} mv_2^2/2 \plus{} 3mv_3^2/2$, and the energy after the collision 'till before the spring breaks is conserved, $ 3mv_3^2/2 \equal{} 3mv_f^2/2 \plus{} ks^2/2$ (the maximum extension of the spring does not change between the cases), which

$ \implies v_0\equal{}v_2 \plus{} 3v_3$
$ v_0^2 \equal{} v_2^2 \plus{} 3v_3^2$
$ 3v_3^2 \equal{} 3v_f^2 \plus{} 4v_1^2 \equal{} 3v_f^2 \plus{} v_0^2/4$ (where we've substituted in the result from case 1).

Solving the first two equations gives us $ v_3 \equal{} v_0/2$, and we substitute that into the third equation, gives us $ 1/6 \equal{} v_f^2/v_0^2 \implies \boxed{1/\sqrt{6}}$?

[quote="xsk13"][quote="simon_math"]Oh,sorry,I know where I was wrong now.A stupid mistake.[/quote]

could you post your solution?  :)[/quote]
In an equation of kinetic energy,I put $ m$ in some place where it should be $ 3m$,that's why my answer is $ \frac{1}{\sqrt{3}}$ times larger than the right answer. JRav wanted me to show my work, and I did. Since I'm not good at Latex,I did it slowly. But on half of the way, I suddenly discovered my mistake.


floor function
modular arithmetic
number theory unsolved
number theory

Let $ a$ and $ b$ be integers, and let $ n\geq 1$ be a natural number. Prove that the number $ b^{n \minus{} 1}a\left(a \plus{} b\right)\left(a \plus{} 2b\right)...\left(a \plus{} \left(n \minus{} 1\right)b\right)$ is divisible by $ n!$.

Suppose $p^k\,|\,n!$ and $n!$ is not divisible by $p^{k+1}$, then $k\leq n-1$. Indeed, $k=\sum_{i\geq 1}\left\lfloor\frac{n}{p^i}\right\rfloor< \sum_{i\geq 1}\frac{n}{2^i}=n$.

We need to prove $p^{k}\,|\,b^{n-1}a(a+b)...(a+(n-1)b)$.

There are two cases:

[b]1)[/b] $p\,|\,b$. In this case $b^{n-1}$ is divisible by $p^k$.

[b]2)[/b] $(b,p)=1$. Then we can find integer $c$ s.t. $bc=1\pmod{p^{k}}$. Note that $(c,p)=1$, so it is sufficient to show that $S=c^n\cdot a(a+b)(a+2b)...(a+(n-1)b)$ is divisible by $p^k$. But
\[S=ca(ca+cb)...(ca+(n-1)cb)\equiv ca(ca+1)...(ca+(n-1))=S'\pmod{p^k},\]
and it is well known $n!\,|\,m(m+1)...(m+(n-1))$, therefore $p^k\,|\,S'$.

That's all.


geometry
3D geometry
sphere
invariant
real analysis
real analysis unsolved

I thought that I should post it in superior algebra, but I think it's more analysis:
     Let $E$ be a normed real vectorial space with a norm $N$ and suppose that the group of isometries of this space acts transitively on the unit sphere (with respect to this norm). Is it true that the norm should come from a scalar product if:
   a) the space has finite dimension b) always?

The answer to (a) is positive, I think.

The norm is uniquely characterized by its unit ball, which is a convex body in $\mathbb R^n$, centrally symmetric wrt the origin. The norm is induced by a scalar product iff this unit ball is an ellipsoid (this has been proven before, I believe; I seem to recall a thread started by sam-n where this was discussed). Also, given a centrally symmetric convex body, there is a unique ellipsoid of minimal volume centered at the same point which contains it.

Assume now that the unit ball $\mathcal B$ of $E$ is different from the minimal ellipsoid $\mathcal E$ containing it. $\mathcal B$ is invariant under every isometry (and every isometry is linear; this i true in arbitrary dimension if the isometry is also known to be onto), which means, in particular, that every isometry is volume-preserving. The image of the minimal ellipsoid $\mathcal E$ is hence also an ellipsoid (because of the linearity of the transformation) of the same volume (because the transformation is volume-preserving), i.e. $\mathcal E$ is invariant under every isometry of $E$. Naturally, $\partial\mathcal E\cap\partial\mathcal B\ne\emptyset$. Now let $p\in\partial\mathcal E\cap\partial\mathcal B$, and let $q\in\partial\mathcal B\setminus\partial\mathcal E$ (if $\mathcal E\ne\mathcal B$, such a $q$ does exist). Then, on the one hand, there must be an isometry which carries $p$ onto $q$, and, on the other hand, the image of $p$ through this isometry must belong to $\partial\mathcal E$. We now have a contradiction.

($\partial X$ denotes the border of the set $X$)

Magnificient, grobber, really super nice solution. Thanks a lot.

Hmm, someone knows the name of the topic posted by sam-n ?

I guess it's probably this one : http://www.mathlinks.ro/Forum/viewtopic.php?t=27235 . ;)


What is the general formula for the product of all positive factors of positive integer n?

Also, Suppose we have n=(2^4)(3^2)(5)

Does the sum of all the divisors of n equal:

(1+2+4+8+16)(1+3+9)(1+5)?

$f(n)$ is called a [i]multiplicative function[/i] in number theory iff is satisfies this property: $f(ab)=f(a)f(b)$ whenever $a$ and $b$ are relatively prime. A multiplicative function is completely determined by what it does at the powers of the primes.

Since the sum of the divisors of a given number is a multiplicative function, your observation about the sum of the divisors of $2^43^25$ is correct.

The product of the divisors is not a multiplicative function. Think about that a little more. Two notes: The number of factors should be involved somehow; and it may matter whether or not the number is a square.


The area of a particular square is $ x$ square units and its perimeter is $ x$ units. What is the value of $ x$?

$ X \equal{} 16$

Solution:
Write an equation:
$ X \equal{} Y^2$,
$ X \equal{} 4Y$  where $ Y$ is the length of one side.
quickly solve and you find that the only legitimate value for $ Y$ is $ 4$;therefore, the only possible value for $ X$ would be $ 16$.


[i]Given $ a,b,c\geq0$ Prove that the flowing inequality holds[/i]

$ \sqrt{\frac{a^2}{a^2\plus{}7ab\plus{}b^2}}\plus{}\sqrt{\frac{b^2}{b^2\plus{}7bc\plus{}c^2}}\plus{}\sqrt{\frac{c^2}{c^2\plus{}7ca\plus{}a^2}}\geq 1$

We can rewrite this inequality:
Given $ x,y,z\ge 0$ such that $ xyz\equal{}1$.Prove that $ \frac{1}{\sqrt{x^2\plus{}7x\plus{}1}}\plus{}\frac{1}{\sqrt{y^2\plus{}7y\plus{}1}}\plus{}\frac{1}{\sqrt{z^2\plus{}7z\plus{}1}}\ge 1$
Let $ x\equal{}(\frac{vw}{u^2})^2,y\equal{}(\frac{wu}{v^2})^2,z\equal{}(\frac{uv}{w^2})^2$
Applying Holder inequality,we obtain $ \sum(5u^6v^3\plus{}2(uvw)^2\minus{}7u^5(vw)^2)\plus{}\sum(u^6v^3\minus{}u^4v^4w)\ge 0$(Q.E.D)


function algebra domain calculus geometry derivative Asymptote

Does endpoint of function consider a critical point? Why?

[quote="JRav"]If the endpoint of the function is in the domain, then it may be considered a critical point.[/quote]
The endpoints is in the domain, but according to the Calculus 8th edition, it does not consider critical points; however, according to the Calculus with Analysic Geometry, it consider critical points.

[quote="JRav"]Is there a specific problem you have in mind?  I don't know those texts and I certainly don't have them.[/quote]
I don't have specific question in mind. Just when I read this two books, they give me a question mark.

linj, this is the way I see it: I could answer your question "yes" or I could answer "no." Would the difference between my two possible answers have any mathematical substance? I say that there is no mathematical substance to the distinction. Since there is no mathematical substance, then it is a matter of convention, and I don't particularly care which way you answer it.

What is the underlying mathematical substance? Suppose that $ f$ is a function continuous on an interval in $ \mathbb{R}.$ We can, unambiguously, define what it means for $ f$ to have either a local minimum or a local maximum at a point $ a.$ We have the following theorem:

If $ f$ has a local maximum or local minimum at $ a,$ then at least one of the following three things must occur:

1. $ f'(a) \equal{} 0,$ or
2. $ f$ is not differentiable at $ a,$ or
3. $ a$ is an endpoint of the interval that is the domain of $ f.$

This is a theorem of enormous practical value, as it gives us the procedure for finding maxima and minima. What it does is provide us with a "list of suspects" - we can find all points for which any of those three conditions is true, and as long as it is a finite list of manageable size, we may then simply go through this entire list of suspects to determine which ones are local maxima, local minima, or neither, and from that list we may directly determine the global maximum and minimum. Sure, there are examples that would destroy the utility of this, such as nowhere differentiable continuous functions, but for many functions encountered in practice, this works very well.

The second derivative test only makes sense in case #1, when $ f'(a) \equal{} 0.$ In the other two cases, there's nothing we can learn from second derivatives.

Consider $ f(x) \equal{} x^{2/3}$ on $ [ \minus{} 1,1]$ Is $ 0$ a critical point? Well, it is a point that lies on the list of suspects, a point at which #2 is true. Whether you call it a critical point or not, it is a point you'll have to investigate in searching for the extreme values - and it is the location of the global minimum.

Consider $ f(x) \equal{} \frac {2x \plus{} 1}{2x \minus{} 1}$ on $ [0,1].$ Is $ x \equal{} \frac12$ a critical point? After all the derivative doesn't exist there. Well, technically it's not a critical point, because the function isn't defined there, either. It lies outside the domain of $ f$ - and I should never have said that $ f$ was defined on $ [0,1],$ because it isn't. But what harm would come from calling it a critical point? That makes it a suspect point that you must investigate. And the outcome of that investigation will be to discover that the function has a vertical asymptote there. You cannot make any sense at all of the appearance of the graph of this function without first understanding its behavior near $ x \equal{} \frac12.$ In the end, misclassifying $ x \equal{} \frac12$ as a critical point does not harm your understanding.

Consider $ f(x) \equal{} 1 \minus{} x^2$ on $ [0,1].$ Is $ x \equal{} 0$ a place where $ f'(x) \equal{} 0?$ You could argue that either way. You could say that $ f'(0)$ doesn't exist because the function is not defined in a neighborhood of $ 0.$ Or you could say, "Let's count the one-sided derivative and that is zero." But your answer to the question of whether it's a point where $ f'(x) \equal{} 0$ makes no difference. This point is on the list of suspect points anyway simply because it is an endpoint, and it is the location of a local maximum which is the global maximum.

Consider $ f(x) \equal{} \begin{cases}2x, & 0\le x\le 1 \\
x^2 \plus{} 1, & 1 < x\le2\end{cases}.$

What are the critical points for this function? Let me not answer that question, and phrase a different question. What are the points that you must investigate if you are investigating the extreme values and the appearance of the graph of this function? There are three points: $ x \equal{} 0,x \equal{} 1,$ and $ x \equal{} 2.$ $ x \equal{} 0$ and $ x \equal{} 2$ appear on the list of suspects as endpoints. $ x \equal{} 1$ draws attention because it is a point at which the piecewise-given definition of $ f$ changes. There are no points in either $ (0,1)$ or $ (1,2)$ at which $ f'(x) \equal{} 0,$ so there are no other critical and/or suspect points. It's not obvious at first glance whether or not $ f'(1)$ exists or even if $ f$ is continuous at $ 1.$

Now, once we conduct the investigation, we discover that $ f$ is continuous at $ 1$ and it's even differentiable at that point: $ f'(1) \equal{} 2.$ And since it's a point of differentiability at which the derivative is not zero, then $ 1$ is not a critical point. That doesn't matter; I had to include it on the list of suspects anyway, because I wouldn't have known any of that if I hadn't checked.

As a side note- if you're referring to calculus textbooks, they're [b]all[/b] called something like "Calculus". Use the author(s) to distinguish between them.


Hola!!! Me gustaría que los argentinos que estemos en MathLinks firmemos este topic para saber cuantos somos, hablar de la OMA, y esas cosas... quién empieza????

(Aunque me parece que el único argentino que sigue usando MathLinks es Gieczewski).

Bueno, ya he visto a 10 argentinos en MathLinks, pero ninguno firmó. todos van a "Olympiad Section".......

Bueno, como ningún argentino va a presentarse, voy a ir poniendo la cantidad de argentinos que vea que usan MathLinks, y lo voy a ir editando: Hay 11 argentinos...

[quote="Charlydif"]¿Leopoldo?[/quote]

¿Quién? ¿Yo? No. [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=10162]Acá está Leopoldo[/url]

[quote="mcphisto"][quote="Charlydif"]¿Leopoldo?[/quote]
Charly! ni te atrevas a colgar en este foro el nombre q me pusiste! :lol:[/quote]

Quién sos???

[quote="mcphisto"]soy Franco...y obviamente por la banderita soy de Perú...aunque si le preguntas a carlos dira q soy victor[/quote]

Ahh, se conocieron en la IMO?? Bueno, yo pensé que eras argentino, en la bandera ni me fijé la verdad...


Consider the four positive real numbers $a,b,c,d$ with sum $4.$

(a) Prove that the inequality
\[a^{2}+b^{2}+c^{2}+d^{2}+ab+bc+cd+da \geq 8 \]
holds.

(b) Find the greatest real number $q$ for which the inequality
\[(a-c)^{2}+(b-d)^{2}+q(a+b)(b+c)(c+d)(d+a) \geq 16q \]
is true.

[hide="Edit"] I transformed (b) into a nicer form. The original problem was finding the largest $p$ for which
\[a^{2}+b^{2}+c^{2}+d^{2}+ab+bc+cd+da \geq 8+p\left(1-\frac{(a+b)(b+c)(c+d)(d+a)}{16}\right) \]
[/hide]

Yes, (a) is trivial, as it is the special case of (b) for $q=0.$ But this is by far not the largest possible value of $q.$ A very nice, though still weak, special case of (b) is
\[(a-c)^{2}+(b-d)^{2}+\frac{(a+b)(b+c)(c+d)(d+a)}{16}\geq 1.\]


[1] Prove that $g.c.d(a + b,a - b) \geqslant g.c.d.(a,b)$

[2] Prove that,if a and b are nozero integers,then$g.c.d.(a,b)|l.c.m.(a,b)$

[hide="Answer to 2"]Because l.c.m.$(a,b)$ is going to be the product of the highest powers of each prime in the prime factorizations of $a$ and $b$, and because the g.c.d$(a,b)$ can be represented as the product of the common lowest power of the prime factorization of $a$ and $b$, then the elements of the product of the g.c.d. would be a subset of those in the l.c.m., and thus would divide into it.[/hide]

[hide="Answer 1"]
Let $(a,b)=c$ such that $cx=a$ and $cy=b$.  Then $a+b=(x+y)c$ and $a-b=(x-y)c$.  Since both are divisible by c, $(a+b,a-b)$ is at least c and $g.c.d(a + b,a - b) \ge g.c.d.(a,b)$[/hide]