Euler ratio limit logarithms function trigonometry

What's your favorite constant?  I'm probably going to get mostly pi's and e's, but...

wasnt there another topic lyk dis a while ago?
pi's cool, it has DIFF numbers

the golden ratio has always intrigued me. pi was always fun because in middle school i was one of the two people in my sixth grade class that knew what it was, but the golden ratio is just wonderful. i just always thought it was snazzy how it appeared EVERYWHERE, art, nature, random math, the human body.... my friend did a math project on the golden ratio in the human face, and he graphed a person's facial adherance to the golden ratio against a rating of their "beauty" (which he polled people on), and he actually found, quite interestingly, that the closer you were to the magic number, the more "beautiful" you were thought. of course, there were some glitches. he did some measurements of jlo i think, and she was WAY off. at least, i believe it was jlo. it was some famous girl that is supposed to be pretty.

Apry's Constant (a.k.a. [tex]\zeta(3)[/tex]) is a neat number. I have been trying to work out its value in closed form for years literally. I have given it some pretty decent attempts, but apparently other people have already considered all of them before.

[quote="sunnyray"]wasnt there another topic lyk dis a while ago?
[/quote] Yes...but i guess it's a popular discussion topic

I like e because:

1) Most people (at least math people) know about it.
2) It's transcendental.
3) It's useful and ubiquitous.
4) It's easier to memorize than pi: 2.71828182845904523536028747135266.
5) There's less competition at memorizing e.

[quote="JS1527"]Hmm.  Being the novice here I will ask, so that I may be educated about all of the constants, what is the Euler-Mascheroni Constant?  Catalan constant? Zeta(3)? Yikes.[/quote]

Euler Mascheroni Constant:
[tex]
\displaystyle\gamma = \lim_{n \to \infty} \left(\sum_{k = 1}^{n} \frac{1}{k} - \ln n\right)
[/tex]

I can't remember exactly what Catalan's constant is though.  I know it's used for estimating combinitorial functions and is approximately equal to 0.915969954.

[quote="nr1337"]Euler Mascheroni Constant:
[tex]
\displaystyle\gamma = \lim_{n \to \infty} \left(\sum_{k = 1}^{\infty} \frac{1}{k} - \ln n\right)
[/tex][/quote]

Actually you should write

[tex]\displaystyle\gamma=\lim_{n\to\infty}\Bigg(\sum_{k=1}^n\frac{1}{k}-\log n\Bigg).[/tex]

Whoa!!! The Euler-Mascheroni constant's actually beating  :pi: ! Sweet!
Hooray for biased sampling! :P

[quote="monkeyman"]by golden ratio don't you mean phi?  Anyway that is what I put, cause you sound so educated when you say it.  Anyways we should do other non-constants too like theta and stuff.  Btw that is another cool word.  And is infinity considered constant?  I mean it will always be infinity, but the are different values/levels of invinity.  Anyway I think it is kindof like zero is neither positive or negetive (ya, It should be neither constant nor non-constant.[/quote]

I believe that a constant must be a real number.  Infinity isn't a specific number so I don't think that you can call it a constant.

Why does a constant need to be real? Is i not a constant?

I guess that is just how it is defined.  I am not really sure.  Here is a link to mathworld's definition [url]http://mathworld.wolfram.com/Constant.html[/url]

Hehe, pi because I don't know what the Euler-Mascheroni and Catalan constants are. :)

Golden ratio, but $\sqrt{141}$ is way cool!

hey, $\frac{\sqrt2}{2}$ is cooler than $\sqrt2$, just because of all the cool trig stuff.  i chose $\pi$, because everyone loves $\pi$, and i know a bunch of digits...but anyway, funny how all the choices are irrational.

[quote=DuoDuoling0]14yEaroLdBumP[/quote]

Dude thats like a 14 year bump....

Also, I'd personality go with golden ratio cause if you have on ur face it's supposed to mean ur pretty or something like that.

Square root of -1. Seems cool and creepy to me


If $ \frac{b}{a} \equal{} 2$ and $ \frac{c}{b} \equal{} 3$, what is the ratio of $ a\plus{}b$ to $ b\plus{}c$?

$ \textbf{(A)}\ \frac13 \qquad \textbf{(B)}\ \frac38 \qquad \textbf{(C)}\ \frac35 \qquad \textbf{(D)}\ \frac23 \qquad \textbf{(E)}\ \frac34$

hello, from $ \frac{a}{b}\equal{}$ we get $ b\equal{}2a$ and from $ \frac{c}{b}\equal{}3$ we have $ c\equal{}3b$ so the searched ratio is $ \frac{a\plus{}b}{b\plus{}c}\equal{}\frac{3a}{4b}\equal{}\frac{3a}{8a}\equal{}\frac{3}{8}$. The right answer is $ \text{B}$.
Sonnhard.


[hide]
Rewrite the equations as
\[ 0\equal{}(a\minus{}2)^3\plus{}(a\minus{}2)\minus{}1996\equal{}(b\plus{}1)^3\plus{}(b\plus{}1)\plus{}1996\]
In particular, adding gives
\[ 0\equal{}(a\plus{}b\minus{}1)((a\minus{}2)^2\minus{}(a\minus{}2)(b\plus{}1)\plus{}(b\plus{}1)^2\plus{}1)\]
Because $ x^2\minus{}xy\plus{}y^2\equal{}\left( x\minus{}\frac{y}{2}\right)^2\plus{}\frac{3y^2}{4}\geq 0$ for all $ x,y$, we have $ a\plus{}b\equal{}\boxed{1}$.
[/hide]


which polygons can we draw with ruler and compas? :?:

i know that we can draw $2^{2^n}+1$ sides polygons.( ;) )
but i want to know about the others.

This should answer your question : http://copernicus.subdomain.de/constructible_polygon :)

thanks i will look there.thanks alot. :D


geometry
trapezoid
quadratics
AMC
AIME
circumcircle
Pythagorean Theorem

Two parallel chords in a circle have lengths $10$ and $14$, and the distance between them is $6$. The chord parallel to these chords and midway between them is of length $\sqrt{a}$ where $a$ is

[asy]
// note: diagram deliberately not to scale -- azjps
void htick(pair A, pair B, real r){ D(A--B); D(A-(r,0)--A+(r,0)); D(B-(r,0)--B+(r,0)); }
size(120); pathpen = linewidth(0.7); pointpen = black+linewidth(3);
real min = -0.6, step = 0.5;
pair[] A, B; D(unitcircle);
for(int i = 0; i < 3; ++i) {
A.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)[0]); B.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)[1]);
D(D(A[i])--D(B[i]));
}
MP("10",(A[0]+B[0])/2,N);
MP("\sqrt{a}",(A[1]+B[1])/2,N);
MP("14",(A[2]+B[2])/2,N);
htick((B[1].x+0.1,B[0].y),(B[1].x+0.1,B[2].y),0.06); MP("6",(B[1].x+0.1,B[0].y/2+B[2].y/2),E);[/asy]

$\textbf{(A)}\ 144 \qquad
\textbf{(B)}\ 156 \qquad
\textbf{(C)}\ 168 \qquad
\textbf{(D)}\ 176 \qquad
\textbf{(E)}\ 184$

[hide]Let $AB$ be the chord of length 10, $CD$ be the chord of length 14, $E$ be the point on $CD$ such that $AE=6$, $F$ be the point on $CD$ such that $BF=6$. Then $CE=FD=2$. Extend $BF$ intersecting the circle at $G$ and let $FG=x$. By Power of a Point,
$6*x=12*2\Rightarrow x=4$.
Since $\angle ABG$ is a right angle and $A$ and $G$ are on the circle, the diameter $AG=10\sqrt{2}$ by Pythagorean Theorem, so the radius is $5\sqrt{2}$.
Let $O$ be the center of the circle, $H$ be the midpoint of $CD$. Then $CH=7$ and $\angle OHC$ is a right angle. Then $OH=1$, again by Pythagorean Theorem.
The chord midway between $AB$ and $CD$ is a distance of 3 away from each, so 2 away from $O$. Using the Pythagorean Theorem one more time, half the length of the chord is equal to $\sqrt{46}$. Then $\sqrt{a}=2\sqrt{46}\Rightarrow a=184$. [b](E)[/b][/hide]

Excellent solution :first:

I too got $184$ but my solution was a lot more complicated and messy than yours since I tried to get everything in terms of $x$ and then solved equations, I think your solution was very good though.

[quote="adrenalinefix36"]Nice problem, I used a different means. [hide] Firstly I attempted to find the median of the trapezoid i contructed with parallel sides 14 and 10; obtaining 12. Putting it aside, I then subtracted (14-10)/2 to find the length difference on each side of the 14 and 10, which is 2 of coarse. Then I contructed a right triangle using the 6 and the 2 to get (40)^1/2. Next i cut this value in half with a machete and used power of the point theorem, which was all i knew. Even though it is obvious without saying, I will clarify the i used a value c along with the given length of median 12. my equation looked like (c)(12+c)=(40)^1/2)(40)^1/2)/4 so
(c)(12+c)=10. Then i used quadratic to find value of c. After this i converted to decimal since i'm lazy. I then multiplied be two since the medians length of 12 doesnt include the c(there are two), and added twelve. Then i squared the value to get 184.[/hide] This goes to show that i wouldn't do well on AIME lol. Atleast I was so late liking math that I'm too late anyways.[/quote]

good solution but remember you cant use calculators on amcs anymore (so no converting to decimal on your calculator).

[hide]
Let the radius of the circle be $ r$, then we have either
$ \sqrt{r^2\minus{}25}\equal{}6\minus{}\sqrt{r^2\minus{}49}$
or
$ \sqrt{r^2\minus{}25}\equal{}6\plus{}\sqrt{r^2\minus{}49}$
Solving them, we get $ r\equal{}\sqrt{50}$ from the first equation. Plug it back in, $ \sqrt{r^2\minus{}25}\equal{}5$. So the distance from the midway chord to the center is $ |\frac{6}{2}\minus{}5|\equal{}2$, the length of it equals to $ 2\cdot \sqrt{50\minus{}2^2}\equal{}\sqrt{184}$[/hide]

[quote]good solution but remember you cant use calculators on amcs anymore (so no converting to decimal on your calculator).[/quote]
Actually I do appreciate the help in you telling me that we can't use calculators on the AMC. To utter the word remember, may not be the right word as I have never competed in the AMC. Secondly I have not recently looked at the rules as to know to remember(which means that you couldn't have known). Sorry to sound like an butt, but language is very confusing in this way. I must also say that i am a culprit.

[hide="By Pythagorean it boils down to"]$ \text{QM[}10,14 \text{]}^2\plus{}6^2\equal{}\boxed{184}$[/hide]

[quote="TZF"][hide="By Pythagorean it boils down to"]$ \text{QM[}10,14 \text{]}^2 \plus{} 6^2 \equal{} \boxed{184}$[/hide][/quote]

What does $ \text{QM[}10,14 \text{]}^2$ mean?  Thanks!

Sorry, by QM I meant the "Quadratic Mean of"

You forgot to divide by 2 :P $ \frac {10^2 \plus{} 14^2}{2} \equal{} \frac {296}{2} \equal{} 148$

But your method is right.

[hide="My approach"]
[asy]unitsize(4mm);
pair O=(0,0), J=(-5*sqrt(2),0), K=(-5,0), L=(1,0), M=-J;
pair A=K+(0,5), B=K-(0,5), C=L+(0,7), D=L-(0,7);
pair P=(-2,sqrt(46)), Q=(-2,-sqrt(46)), X=(-2,0);
draw(Circle(O,5*sqrt(2)));
draw(A--B); draw(C--D); draw(J--M); draw(P--Q);
label("$A$",A,NW);
label("$B$",B,SW);
label("$C$",C,N);
label("$D$",D,S);
label("$J$",J,W);
label("$K$",K,NE);
label("$L$",L,NE);
label("$M$",M,E);
label("$P$",P,N);
label("$Q$",Q,S);
label("$X$",X,NE);
label("5",(K+A)/2,E);
label("5",(K+B)/2,E);
label("7",(L+C)/2,E);
label("7",(L+D)/2,E);
label("3",(K+X)/2,S);
label("3",(L+X)/2,S);
label("$x$",(J+K)/2,S);
label("$y$",(L+M)/2,S);
label("$n$",(P+X)/2,E);
label("$n$",(Q+X)/2,E);[/asy]
Let $ \overline{JM}$ be the diameter of the circle that is perpendicular to the chords, which we will call $ \overline{AB}$ and $ \overline{CD}$, and let the intersection points be $ K$ and $ L$ respectively.  Let $ \overline{PQ}$ be the chord of length $ \sqrt{a}$, which hits $ \overline{JM}$ at $ X$.  Let $ x\equal{}JK$, $ y\equal{}LM$, and $ n\equal{}PX$.

Then by Power of a Point on $ K$ and $ L$, $ x(y\plus{}6)\equal{}25$ and $ (x\plus{}6)y\equal{}49$.  Subtracting one from the other yields $ y\minus{}x\equal{}4$.  Then $ x(x\plus{}10)\equal{}25$.  Also, Power of a Point on $ X$ yields $ n^2\equal{}(x\plus{}3)(x\plus{}7)\equal{}x(x\plus{}10)\plus{}21\equal{}25\plus{}21\equal{}46\Rightarrow PQ\equal{}2\sqrt{46}\equal{}\sqrt{184}$.
[/hide]

[quote="dgreenb801"]Nice problem!
Let the trapezoid be ABCD, where AB=10, CD=14, the given chord be FG, and the circumcenter O. By Pythag, $ AD \equal{} BC \equal{} 2\sqrt {10}$. By Ptolemy on ABCD (or Pythag on AEC, where E is the altitude from A to CD), we find the diagonals are $ 6\sqrt {5}$. So $ [DAC] \equal{} (6)(14)/2 \equal{} 42 \equal{} (14)(2\sqrt {10})(6\sqrt {5})/4R$, from which we find the circumradius $ R \equal{} 5\sqrt {2}$.  So by Pythag we find the height of ODC is 1. So the height of OFG from O is 2. So by Pythag the answer is $ \sqrt {184}$.

314.math, QM is the quadratic mean, which is the square root of the average of the squares of a group of numbers.[/quote]
I follow this all the way up to
[quote]
Pythag we find the height of ODC is 1. So the height of OFG from O is 2. So by Pythag the answer is $ \sqrt {184}$.[/quote]

Can somebody explain this?

[hide="@Aequilipse"][img]http://i40.tinypic.com/10wo9dg.jpg[/img][/hide] (I think that r got cut out of the picture but it should say $x = 1 \implies r = 2 \sqrt{5}$)


Here is an old and easy problem that looks 'tough' and this should be fun for you starters :)

Find prime numbers $a,b,c$ such that $a^b+1=c$. Please post your explanation and let's start!

Do $a,b,c$ have to be unique? If not:
[hide]$a = 2$
$b = 2$
$c = 5$[/hide]

[hide]
$a^b+1=c$ Since $c$ must be odd, $a^b$ must be even, which means $a$ must be even. The only even prime is 2. And we can easily see that 2 works for $b$ also. :D [/hide]

[hide="Solution"]

  Evidently, $c$ must be some prime greater than two, therefore, it must be an odd prime.  If $c$ is odd then $a^{b}$ must be even, meaning $a$ is even.  Since $a$ must be prime, its only possible value is two.  Now consider the expression $2^{b} + 1 = c$.  Powers of 2 in modulo 3 leave residues as follows: 2, 1, 2, 1 ....  Thus $2^{b}$ must leave a remainder of 1, meaning b is also even ( if it left a remainder of 2, $c$ would need to be divisible by 3, making it in fact 3, which is impossible to satisfy.  Again this means $b$ must also be 2.  Hence our only possible set of values is:

$a=2$

$b=2$

$c=1$
[/hide]

[quote="G-UNIT"][hide="Solution"]

  Evidently, $c$ must be some prime greater than two, therefore, it must be an odd prime.  If $c$ is odd then $a^{b}$ must be even, meaning $a$ is even.  Since $a$ must be prime, its only possible value is two.  Now consider the expression $2^{b} + 1 = c$.  Powers of 2 in modulo 3 leave residues as follows: 2, 1, 2, 1 ....  Thus $2^{b}$ must leave a remainder of 1, meaning b is also even.  Again this means $b$ must also be 2.  Hence our only possible set of values is:

$a=2$

$b=2$

$c=1$
[/hide][/quote]

Err... ${2^2} + 1 = 5$, not $1$. Furthermore, $1$ is not prime  :P

Well this is sort of an easy problem that you can just use guess and check for and like lotrgreengrapes7926 said $a^b$ must be even for $c$ to be odd. So you can just work from there.


function
geometry
3D geometry
sphere
geometry unsolved

If orthogonal projection of convex body $F$ on any plane is a circle, is it true that $F$ must be a ball?
I suppose the answer is yes, but I don't know how to prove it.

I think it's true.

First try to show that all the disks (the projections of the body on all planes) have equal radii (call the common value of the radius $r$). It should also be rather easy to show that the perpendiculars to the planes raised in the centers of the disks are concurrent, so let the concurrence point be called $O$.

It follows that all points are in the closed ball $\mathcal B(O,r)$, because if we have $OX>r$ for some $X$, then the projection of the body on a plane parallel to $OX$ would not be a disk of radius $r$. Now fix a great circle of the ball and take a plane parallel to the plane of the great circle. Since the projection on this plane is a disk with radius $r$, it means that all the points of the great circle belong to the boundary of our body. Since this holds for all great circles, we get the desired result: the boundary of the body is the boundary of the ball.

It's easy to see that it must contain all of its boundary points. For any two points on the surface, let A and B be the farthest apart (A and B must exist by the theorem of Wierstrauss i think, which says that any bounded, continuous function on a closed interval or region attains a maximum and a minimum). Project onto any plane parallel to line AB. Then A and B must be at the diameter of this circle, since they are farthest apart. Sorry, i have to run to class... someone want to finish for me?

Ok, posting from class... Anyway, consider any circle with AB as diameter, and extend this circle in a perpendicular direction to make a cylinder. From projection in the direction of this cylinder, we see that the figure must be contained in the cylinder. So the figure must be contained in the union of all such cylinders, which is a sphere with diameter AB. If P was a point on the sphere but not on the circle, then the projection in a plane through OP (O=mdpt of AB) would not be a circle. So it must be the sphere with diameter AB.

Your solutions are nice. Your observation, grobber, that all circles must be congruent is
actually Russia'71. One can also notice that for any $k$ planes we can find a convex body which is
not a ball and all the projections of which (on given planes) are circles. Just cut a small enough piece
of ball.


Is there any easy way to find the average number of numbers you need to pick before repeating when picking out of n numbers?
I got
$n^{-1}\sum_{j=2}^{\infty}\left((j^{2}-j)\prod_{k=0}^{j-2}\frac{n-j}{n}\right)$
but I can't figure out how to simplify.

the product is independent of $k$ ?

so you can simplify this product

i think put $k-n$ as a misspelling, but have a n outside the limit


A circular sheet of paper with radius of $ 6$ cm is cut into three congruent sectors. What is the height of the cone in centimeters that can be created by rolling one of the sections until the edges meet? Express your answer in simplest radical form.

ok if its cut into 3 congruent sectors they each have a 120 degree angle. now the area of one of those sectors is $ 6^2\pi\cdot\frac {1}{3} \equal{} 12\pi$. now we know tht this is equivalent to the lateral surface area, which is $ \pi\cdot r\cdot6$ where r is the radius of the cone. now from this we can tell that $ r \equal{} 2$ since the 2 other latex-ified equations hafta be equivalent. now we realize, HEY! $ 6^2 \equal{} h^2 \plus{} r^2$! where r is the radius of the base and h is the height. now we knwo $ r \equal{} 2$ so just plug it in to get $ h \equal{} \sqrt {36 \minus{} r^2} \equal{} \sqrt {32} \equal{} \boxed{4\sqrt {2}}$


geometry
trapezoid
analytic geometry
ratio
perpendicular bisector
angle bisector
projective geometry

Let $ ABCD$ be an isosceles trapezoid with $ AD \parallel BC$. Its diagonals $ AC$ and $ BD$ intersect at point $ M$. Points $ X$ and $ Y$ on the segment $ AB$ are such that $ AX \equal{} AM$, $ BY \equal{} BM$. Let $ Z$ be the midpoint of $ XY$ and $ N$ is the point of intersection of the segments $ XD$ and $ YC$. Prove that the line $ ZN$ is parallel to the bases of the trapezoid.

[i]Author: A. Akopyan, A. Myakishev[/i]

x-axis is the trapezoid midline parallel to the bases and y-axis is common perpendicular bisector of the bases. $ U, V$ are midpoints of $ DA, BC.$ Let $ DA \equal{} 2u, BC \equal{} 2v, UV \equal{} 2h.$ At time $ t \equal{} 0,$ points $ P, Q \in AB$ pass through $ A, B$ in the opposite directions with constant velocities proportional to $ u, v$; let their x-coordinates be $ x_P \equal{} u(1 \minus{} t)$ and $ x_Q \equal{} v(1 \plus{} t).$ Equation of the line $ AB$ yields their y-coordinates, and the y-coordinate of the midpoint $ R$ of $ PQ$:

$ y \plus{} h \equal{} \minus{} \frac {2h}{u \minus{} v}(x \minus{} u)$ or $ y \minus{} h \equal{} \minus{} \frac {2h}{u \minus{} v} (x \minus{} v),$

$ y_P \equal{} \frac {2hut}{u \minus{} v} \minus{} h$ and $ y_Q \equal{} h \minus{} \frac {2hvt}{u \minus{} v}$ $ \Longrightarrow$ $ y_R \equal{} ht.$

Equations of the lines $ DP, CQ$ are then

$ y \plus{} h \equal{} \frac {2ht}{(u \minus{} v)(2 \minus{} t)} (x \plus{} u)$ and $ y \minus{} h \equal{} \minus{} \frac {2ht}{(u \minus{} v)(2 \plus{} t)}(x \plus{} v).$

Multiplying by $ (2 \minus{} t), (2 \plus{} t),$ respectively, and adding the 2 equations eliminates $ x,$ yielding the y-coordinate of their intersection $ S$:

$ (y \plus{} h)(2 \minus{} t) \plus{} (y \minus{} h)(2 \plus{} t) \equal{} 2ht\ \Longrightarrow\ y_S \equal{} ht$

Since $ y_S \equal{} y_R,$ it follows that $ RS \parallel DA \parallel BC.$ Since $ \frac {AM}{BM} \equal{} \frac {DA}{BC} \equal{} \frac {u}{v},$ the points $ P, Q, R, S$ coincide with $ X, Y, Z, N,$ respectively, at the same time and $ ZN \parallel DA \parallel BC.$

[b][size=100][color=DarkBlue]EQUIVALENT PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ X,\ Y$ be, the intersection points of its sidelines $ AB,\ AC,$ from the lines through $ C,\ B$ respectively and parallel to the line segment $ AN,$ as the internal angle bisector of $ \angle A.$ We denote the points $ D,\ E$ on the side-segment $ BC$ such that $ CD \equal{} AC,\ BE \equal{} AB$ and let be the point $ Q\equiv DX\cap EY.$ Prove that the line through $ Q$ and parallel to $ AN,$ bisects the segment $ DE.$[/color][/size][/b]

[b][size=100][color=DarkBlue]GENERAL PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ P$ be, an arbitrary point on the line segment $ AN,$ as the internal angle bisector of $ \angle A$ $ ($ $ A,$ between $ P$ and $ N\in BC$ $ ).$ We denote the points $ X\equiv AB\cap PC,\ Y\equiv AC\cap PB$ and let $ D,\ E$ be, two points on the side-segment $ BC$ such that $ CD \equal{} AC,\ BE \equal{} AB.$ Prove that the line segment $ PQ,$ where $ Q\equiv DX\cap EY,$ bisects the segment $ DE.$[/color][/size][/b]

I will post here later, the solution of the general problem I have in mind.

Kostas Vittas.

[quote="vittasko"][b][size=100][color=DarkBlue]EQUIVALENT PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ X,\ Y$ be, the intersection points of its sidelines $ AB,\ AC,$ from the lines through $ C,\ B$ respectively and parallel to the line segment $ AN,$ as the internal angle bisector of $ \angle A.$ We denote the points $ D,\ E$ on the side-segment $ BC$ such that $ CD \equal{} AC,\ BE \equal{} AB$ and let be the point $ Q\equiv DX\cap EY.$ Prove that the line through $ Q$ and parallel to $ AN,$ bisects the segment $ DE.$[/color][/size][/b][/quote]
[b][size=100]PROOF.[/size][/b] – (see the schema [b][size=100]t=215967[/size][/b])

From $ QM\parallel CX$ $ \Longrightarrow$ $ \frac{MD}{CD} \equal{} \frac{QM}{CX}$ $ ,(1)$

From $ QM\parallel BY$ $ \Longrightarrow$ $ \frac{ME}{BE} \equal{} \frac{QM}{BY}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac{MD}{ME}\cdot \frac{BE}{CD} \equal{} \frac{BY}{CX}$ $ \Longrightarrow$ $ \frac{MD}{ME} \equal{} \frac{AC}{AB}\cdot \frac{BY}{CX}$ $ ,(3)$

But, from $ BY\parallel CX$ $ \Longrightarrow$ $ \frac{AC}{AB} \equal{} \frac{CX}{BY}$ $ ,(4)$

Hence, from $ (3),$ $ (4)$ $ \Longrightarrow$ $ \frac{MD}{ME} \equal{} 1$ $ \Longrightarrow$ $ MD \equal{} ME$ and the proof is completed.

[quote="vittasko"][b][size=100][color=DarkBlue]GENERAL PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ P$ be, an arbitrary point on the line segment $ AN,$ as the internal angle bisector of $ \angle A$ $ ($ $ A,$ between $ P$ and $ N\in BC$ $ ).$ We denote the points $ X\equiv AB\cap PC,\ Y\equiv AC\cap PB$ and let $ D,\ E$ be, two points on the side-segment $ BC$ such that $ CD \equal{} AC,\ BE \equal{} AB.$ Prove that the line segment $ PQ,$ where $ Q\equiv DX\cap EY,$ bisects the segment $ DE.$[/color][/size][/b][/quote]
[b][size=100]PROOF.[/size][/b] - Let be the point $ Z\equiv BC\cap XY$ and it is easy to show that this point is the harmonic conjugate of $ N,$ with respect to $ B,\ C$ $ ($ from the complete quadrilateral $ PXAYBC$ $ )$ and so, we conclude that the line segment $ AZ,$ is the external angle bisector of $ \angle A$ of the given triangle $ \bigtriangleup ABC.$

So, we have $ \frac {NB}{NC} \equal{} \frac {ZB}{ZC} \equal{} \frac {AB}{AC}$ $ ,(1)$

$ \bullet$ We see that the pencils $ X.PYQM,\ Y.PXQM,$ have the line segment $ XY,$ as their common ray and the points

$ P\equiv XP\cap YP,\ Q\equiv XQ\cap YQ,\ M\equiv XM\cap YM$ are collinear and so, they have congruent [b][size=100]double ratios[/size][/b] ( = [b][size=100]cross ratios[/size][/b] ).

Hence, we conclude that $ (X.PYQM) \equal{} (Y.PXQM)$ $ ,(2)$

The pencils $ X.PYQM,\ Y.PXQM,$ are intersected from the sideline $ BC$ of $ \bigtriangleup ABC$ and so, we have:

$ (X.PYQM) \equal{} (C,Z,D,M)$ $ ,(3)$ and $ (Y.PXQM) \equal{} (B,Z,E,M)$ $ ,(4)$

From $ (2),$ $ (3),$ $ (4)$ $ \Longrightarrow$ $ (C,Z,D,M) \equal{} (B,Z,E,M)$ $ ,(5)$

From $ (5)$ $ \Longrightarrow$ $ \frac {DC}{DZ} \div \frac {MC}{MZ} \equal{} \frac {EB}{EZ} \div \frac {MB}{MZ}$ $ \Longrightarrow$ $ \frac {AC}{DZ} \cdot \frac {1}{MC} \equal{} \frac {AB}{EZ} \cdot \frac {1}{MB}$ $ \Longrightarrow$ $ \frac {AB}{AC} \equal{} \frac {MB}{MC}\cdot \frac {ZE}{ZD}$ $ ,(6)$

From $ (1),$ $ (6)$ $ \Longrightarrow$ $ \frac {ZB}{ZC} \equal{} \frac {MB}{MC}\cdot \frac {ZE}{ZD}$ $ \Longrightarrow$ $ \frac {ZB}{ZC} \equal{} \frac {MB}{MC}\cdot \frac {ZB \plus{} AB}{ZC \minus{} AC}$ $ ,(7)$

From $ \frac {ZB}{ZC} \equal{} \frac {AB}{AC},$ it is easy to show that $ \frac {ZB \minus{} AB}{ZC \minus{} AC} \equal{} \frac {AB}{AC} \equal{} \frac {ZB}{ZC}$ $ ,(8)$

From $ (7),$ $ (8)$ $ \Longrightarrow$ $ \frac {ZB \minus{} AB}{ZC \minus{} AC} \equal{} \frac {MB}{MC}\cdot \frac {ZB \plus{} AB}{ZC \minus{} AC}$ $ \Longrightarrow$ $ \boxed{\frac {MB}{MC} \equal{} \frac {ZB \minus{} AB}{ZB \plus{} AB}}$ $ ,(9)$

$ \bullet$ Let $ M'$ be, the midpoint of the segment $ DE$ and we will prove that $ M'\equiv M.$

We put $ BC \equal{} a,\ AC \equal{} b,\ AB \equal{} c$ and then we have:

$ BD \equal{} a \minus{} b$

$ CE \equal{} a \minus{} c$

$ DE \equal{} a \minus{} BD \minus{} CE \equal{} b \plus{} c \minus{} a$

It is easy to show now, that $ M'B \equal{} BD \plus{} \frac {DE}{2} \equal{} \frac {a \minus{} b \plus{} c}{2}$ $ ,(10)$ and $ M'C \equal{} CE \plus{} \frac {DE}{2} \equal{} \frac {a \plus{} b \minus{} c}{2}$ $ ,(11)$

From $ (10),$ $ (11)$ $ \Longrightarrow$ $ \boxed{\frac {M'B}{M'C} \equal{} \frac {a \minus{} b \plus{} c}{a \plus{} b \minus{} c}}$ $ ,(12)$

From $ (9),$ $ (12),$ it is enough to prove that:

$ \frac {ZB \minus{} AB}{ZB \plus{} AB} \equal{} \frac {a \minus{} b \plus{} c}{a \plus{} b \minus{} c}$ $ \Longrightarrow$ $ \frac {ZB \minus{} c}{ZB \plus{} c} \equal{} \frac {a \minus{} b \plus{} c}{a \plus{} b \minus{} c}$ $ \Longrightarrow$ $ \boxed{ZB \equal{} \frac {ac}{b \minus{} c}}$ $ ,(13)$

But the relation $ (13)$ is true from $ \frac {ZB}{ZC} \equal{} \frac {AB}{AC} \equal{} \frac {c}{b},$ where $ ZC \equal{} ZB \plus{} a$

So, we conclude that $ M'\equiv M$ and hence, the point $ M\equiv BC\cap PQ$ is the midpoint of the segment $ DE$ and the proof is completed.

Kostas Vittas.

Proof of initial problem:
Let {P}=AD∩CY and Q = BC∩DX. Obviously, AD/BC=AM/BM  ( 1 ), since CM=BM. We have also AP/BC=AY/BY=(AB-BM)/BM  ( 2 ). From (1) and (2) we get  PD/BC=(AB-BM+AM)/BM  ( 3 ); similarly CQ/AD=(AB-AM+BM)/AM  ( 4 ). From (3) and (4) we get
PN/CN=PD/CQ=(AB-BM+AM)/(AB+BM-AM)  ( 5 ). By straight calculations we get AZ/BZ= (AB-BM+AM)/(AB+BM-AM) ( 6 ), so PN/CN=AZ/BZ , q.e.d.

Proof of general problem:
Let {M}=DX∩PE and {Q}=EY∩PD. Applying Menelaus in triangle PNC with the transversal BAX we get  PX/CX=c.AP/(b+c).AN ( 1 ), after having taken into account that AN was the angle bisector of ÐA; similarly PY/CY=b.AP/(b+c).AN  ( 2 ) (a, b and c being the sides of triangle ABC).
Next, apply Menelaus to triangle PCE with the transversal DMX and get   bc.AP/(b+c)(b+c_a).AN ( 3 ); similarly, apply Menelaus to triangle PBD with the transversal EQY and get for  PQ/QE the same value  ( 4 ). From (4) and (3) we get that DM and EQ intersect on the median from P of the triangle PDE.

Best regards,
sunken rock


calculus
integration
trigonometry
complex numbers
real analysis
real analysis unsolved

Prove that if the real part of $ x$ is positive then

$ \int\limits_{0}^{1}\frac{t^{x\minus{}1}}{1\plus{}t}dt\equal{}\frac{1}{2}\sum\limits_{n\equal{}0}^{\infty}\frac{n!}{x(x\plus{}1)(x\plus{}2)\cdots(x\plus{}n)}\frac{1}{2^{n}}$.


induction
number theory solved
number theory

Problem. Let $a_1,\ldots,a_n \in \mathbb{N}$, none of which is a perfect square. Prove or disprove that the number $\sqrt{a_1} + \cdots + \sqrt{a_n}$ is always irrational.

Alternative Formulation. A non-zero real number is called "polyquadratic" if it is the rational linear combination of numbers of the form $\sqrt{m}$, where $m>1$ is a square-free integer. Prove or disprove that each polyquadratic number is irrational.

I'm not sure about this, but I think that induction on the cardinal number of the set $S$ formed by the prime divisors of $\sqrt{a_i},\ i\in \overline{1,n}$ might work. Assume we have shown it for all polyquadratic numbers s.t. $|S|\le k$ and we're trying to show it for $|S|=k+1$ (I considered the second formulation, in which $a_i$ are square-free and we're looking at linear combinations of them over $\mathbb Q$). In this case we have $r_1\sqrt{a_1}+r_2\sqrt {a_2}+\ldots+r_t\sqrt{a_t}+r_{t+1}\sqrt{pb_{t+1}}+\ldots+r_n\sqrt{pb_n}+r=0$, where everything starting with $r$ is rational, $a_i,\ b_i$ are square-free and $pb_i=a_i$ are the only $a$'s which are divisible by $p$, which is the $k+1$'th prime factor in $|S|$.

We put everything which involves $p$ on the right side, square it, and $p$ comes out as a factor of rational coefficients, and we're left with an expression involving only those $a_i$'s which don't have $p$ as their factor, thus having $|S|=k$, and we have already shown that this can't hold if we have square roots, and so all products of two $a_i$'s which didn't have $p$ as a factor must be perfect squares, i.e. $a_ia_j$ is a perfect square for all $i,j\in \overline{1,t}$. Remember that $a_i$ were square-free, and this means that all of them must be the same (for $i,j\in \overline{1,t}$.

I'm sure we're close, but I don't know how to continue (I haven't thouight about this; I thought I had solved it, and only now am I realizing that I haven't). Who knows, maybe this doesn't even lead to the conclusion.

Sorry, but I think what I did is wrong: when we put part of the expression on the right and we square, the square roots might be reduced and we might end up with an identity without obtaining a contradiction.

I already solved this problem on the forum.
See :
http://www.mathlinks.ro/viewtopic.php?t=4496&highlight=rational

Pierre.

That's a very cool solution Pierre, but the last argument makes me wonder if it follows from there that if $\displaystyle \sum_{i=1}^nr_i\sqrt{a_i}$ is rational then $a_i$ are perfect squares, no matter what the rationals $r_i$ are (take $a_i$ to be square-free positive integers).

I'm asking because you used the fact that the numbers are positive, and I think this stronger thing holds: $\sqrt{a_i}$ are independent over $\mathbb{Q}$. It's probably obvious from the solution you gave, but I'm not sure.

Yes, Pierre, I have to say that this is a very nice proof.   :)  And, yes, a (very slight) modification of Pierre's proof implies linear independency over $\mathbb{Q}$ of the numbers $\sqrt{m}$ for which $m$ is a square-free integer. Moreover, (this is not so obvious) the proof can be modified to show that, for each $k \geq 2$, the set of reals of the form $m^{\frac{1}{k}}$, where $m \in \mathbb{N}$ is not an $m$th power, is
linearly independent over $\mathbb{Q}$ (which was the original problem proposed by P. Mihailescu).

--Vesselin


let  E and F in R.
 let theses functions

f: E------>F  and  g: G------>E
such that for all  (x in G)   f(g(x)) = x^2+1

1) we have f and g are bijective . find lg^-1(f^-1(x))l for all x in F.

f^-1 is the bijection reciproque of f

[quote="greatestmaths"]it seems to me you can't answer for this problem :rotfl:[/quote]

Maybe you could be more precise in your problem :

What is G ?
What is the meaning of the first statement : we must consider / prove / anything else / .. that f and g are bijective ?

Thanks for any precision.


[hide]Numerele sunt de forma 4k+3 deci niciun patrat perfect.  :) [/hide]

In legatura cu aceasta problema:
Demonstrati ca in multimea ${1,11,111,....,11111..11}$ exista un numar divizibil cu $p$-prim, unde $p$ este numarul de elemente ale multimii, diferit de $2$ sau $5$.

[quote="adiba"]Aratati ca in multimea {11,111,1111,....,111...1,.....} nu exista niciun patrat perfect.[/quote]
O solutie pentru clasa a V-a!
Notam u(k)=ultima cifra a lui n= k mod 10.
Daca u($n^{2}$ )=1, atunci $u(n)\in{1,9}$. Luand toate cazurile posibile :11, 21, ...91, 19, 29, ...99 (!) constatam ca, de fiecare data, penultima cifra a patratului este para , deci nu poate fi 1.
Ca  'generalizare':
1) nu esista nici un patrat perfect de forma 1111....1n1, unde n este cifra impara
2) nu exista nici un patrat perfect, mai mare ca 10, care sa aiba toate cifrele identice

SAU
Daca n ar fi patrat perfect, radacina patrata a numarului ar fi de forma 10k+1 sau 10k-1. Patratul acesteia este (aplicand patratul binomului) de forma 20p+1. Asadar n=20p+1, de unde n-1=20p, adica n-1 este divizibil cu 20, deci si cu 4. Ori n-1 se termina in 10, care nu este divizibil cu 4. Contradictia demonstreaza ca presupunerea facuta este falsa, deci...


function
complex analysis
complex analysis unsolved

I do not know if $ f$ is an analytic function on the annulus $ \{ z \in \mathbb C : 1<|z|<2\}$, then $ f$ extends to a meromorphic function on the disk $ \{ z \in \mathbb C : |z|<2\}$.

Thank you  :).

[quote="leshik"]No, this fact is not true.[/quote]

Could you provide some tips?

Just consider the function $ sin(\frac{1}{1\minus{}z})$ :) Is the rest clear?


function
number theory
totient function
superior algebra
superior algebra unsolved

How do I proof that phi(m) = phi(2m) only if m is odd?

If $m$ is even, then any integer that is coprime to $m$ is also coprime to $2m$. Since $m+1$ is one of such integers, and it is not counted by $\phi(m)$, it follows that $\phi(2m)>\phi(m)$.

Now suppose that $m$ is odd. If $k$ is coprime to $m$ and $1\le k\le m$, then exactly one of the integers $k$ and $k+m$ is coprime to $2m$. Hence $\phi(2m)\ge \phi(m)$. Conversely, if $k$ is coprime to $2m$ and $m<k\le 2m$, then $k-m$ is coprime to $m$. This implies $\phi(m)\ge \phi(2m)$.

This is merely a special case of a [url=http://en.wikipedia.org/wiki/Euler%27s_totient_function]more general formula[/url].

This webpage http://libraryofmath.com/euler-phi-function.html has what you are looking for.

There are some basic properties of the Euler's Phi Function with some examples.


How many integers $ n$ satisfy $ n^2 \leq 5$?

The proper way. :P

$ n^2 \leq 5$

$ n \leq \sqrt {5}$ and $ n \geq \minus{}\sqrt5$, thus, the answer is $ \boxed{5}$.

Actually ernie if what you say is true then n=sqrt5.

You meant -sqrt5<n<sqrt5 so n ranges from -2 to 2 (inclusive)


I am having trouble understanding 10.3.6 in Intro to Algebra. I looked at the solution, but I didn't understand it at 9=(srt3)(3 x srt3) and after. Could someone please explain it in a different way? Thank you for your kind explanation(s). :surrender:

I have no clue what the problem is but if you need to understand $ 9\equal{}(\sqrt{3})(3\sqrt{3})$, here it is:  Let's say you pull out the 3, then you have $ 3(\sqrt{3})(\sqrt{3})$.  If you multiply the two terms in parentheses, you get $ 3(3)$, which is obviously 9.


Out of 12 colored balls, 3 are colored red, 4 are colored green, and the rest are colored blue. How many ways are there to arrange these 12 balls in a line given the condition that no blue ball may be adjacent to another blue ball?

[hide]Denote the places of non-blue balls by a _ .

_ _ _ _ _ _ _

There are 8 spots inbetween these non-blue balls and we can assign 5 blue balls so that's $\binom85$ ways.

The 7 _'s are possible places to insert 3 red balls and the final 4 balls will fill in the missing spots so that's $\binom73$ ways.

Multiply the two together to get $1960$.

[/hide]


All we need to show is that in $\mathbb Z_p[X]$ we have the identity $(1+x)^{p-2}=1-2x+3x^2-\ldots-(p-1)x^{p-2}$. If we multiply both sides with $(1+x)^2$ we immediately get the conclusion.


linear algebra
matrix
combinatorics proposed
combinatorics

This is a pretty simple exercise (and it probably belongs to the linear algebra section) but I think it's a cute mathematical fact :).

Prove that if the adjacency matrix of a graph has distinct eigenvalues, then the automorphism group of the graph is abelian.

let $A$ be the adjacency matrix. For each $g\in \text{Aut}(G)$ we can associate a permutation matrix $P_g$ that satisfies
$P_gAP_g^{-1}=A$. Now since $A$ has all distinct eigenvalues, each eigenspace has dimension $1$. Now if $\{v_1,v_2,\ldots,v_n\}$ is a basis of eigenvectors then by the above commutativity it follows that $P_gv_i$ is also an eigenvector of $A$ with the same eigenvalue as for $v_i$, so it follows that $P_gv_i=\lambda(g)v_i$ for all $i$. The upshot is that w.r.t the basis $\{v_1,\ldots,v_n\}$ all the matrices $P_g$ are diagonal and hence commute with each other(due to simultaneous diagonalization).
actually with a little more work one can actually show that each element $g$ as above in fact satisfies $g^2=1$.

actually no more 'work' is actually necessary-afterall the matrix $A$ being real symmetric, has only real eigenvalues.and since $P_g$ is also being diagonalized over $\mathbb{R}$ the diagonal entries have to be $1$'s or $-1$'s.


real analysis
geometry
3D geometry
topology
real analysis unsolved

$ E \in \mathcal{R}$
The normal definition of lebesgue outer measure is
$ m^*(E) \equal{} \inf \{\sum_1^\infty m(E_i), E \subset \cup_{i \equal{} 1}^\infty E_i, E_i \mbox{is finite open set}\}$

if we define m* like this.

$ m^*(E) \equal{} \inf \{\sum_1^\infty m(E_i), E \subset \cup_{i \equal{} 1}^\infty E_i, E_i \mbox{is finite open set},m(E_i) < \varepsilon\},$

$ \varepsilon$ is any fixed positive real number.

The two definition are equivalent?

Yes, because any open set can be written as the union of (almost) disjoint open sets with small measure. For example, divide the space into equal cubes of sidelength $ \epsilon$, and intersect $ E_i$ by such cubes (slightly enlarged, so that the boundary is also covered.)

[quote="HilbertThm90"]Is that first one a standard way of developing measure? We used the $ E_i$'s being intervals, so that we could talk about length. Then we developed the concept of measurable from outer measure. It seems like that one uses measure in the definition outer measure.[/quote]

Both are right.
Open set  consists of disjoint open intervals.


There's this thing on facebook that let's you create pools and stuff for march madness.  And I have a pool called "I actually don't know anything about basketball."

So if you have a facebook, and don't know anything about basketball, you should join.


quadratics
function
algebra
polynomial
special factorizations

Two quick questions here...

What is the formula for Trinomials? I'm pretty sure ax^2 + bx + c is the one but my teacher also gave us ax^2 + mx + n + c. What is that for?

Thanks in advance guys!

What do you mean "the formula for trinomials"?  A trinomial is an expression with three terms; usually it denotes a quadratic equation $ ax^2 \plus{} bx \plus{} c$.  This is the general form of a quadratic equation, if that's what you meant.

As for the second expression, I'm not sure what you mean.  It's just the first expression with $ m$ substituted for $ b$ and $ n \plus{} c$ substituted for $ c$.  Did you mean $ ax^3 \plus{} mx^2 \plus{} nx \plus{} c$?  That's a cubic, not a quadratic, equation.

ive never seen any functions denoted like the cubic equation.  When a function is a degree greater than 2 we use the polynomail notation or something like that where n is the degree of the funtion which is $ a_nx^n \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} a_{n\minus{}2}x^{n\minus{}2} \plus{} ...  a_0x^0$  Atleast this is how ive always seen cubic equations and quartic equaitons and so on.  As for the $ ax^2 \plus{} mx \plus{} n \plus{} c$ i dont really know!

$ ax^2 \plus{} bx \plus{} c$ [i]is[/i] polynomial notation.  It doesn't matter what we name the variables; we can just as easily call it $ a_2 x^2 \plus{} a_1 x \plus{} a_0$ and it's still a quadratic polynomial.

yes i was just saying that i have never seen the other notations for polynomial functions becuase my school teaches us the other notation for functions with degree greater than 2 so that is why i have never seen that ax^2 + mx + n + c or whatever.

Is there any context you can provide us for the form $ ax^2 \plus{} mx \plus{} n \plus{} c$? I wouldn't be surprised if it was being used to explain [[completing the square]] (if $ n \equal{} \frac{m^2}{4a}$) or a similar concept.

hmm i suppose that might be it, now that i think about it it does kinda make some sence how one might use it to explain completing the square!


inequalities
combinatorics unsolved
combinatorics

In an international meeting of $n\geq k$ participants, $m$ languages are spoken. We know that:
- Any $k$ participants speak a common language.
- No language is spoken more that by the half of the participants.
What is the least value of $n$?
(Actually it's from http://www.mathlinks.ro/Forum/viewtopic.php?p=242552, but with the given constants it's quite easy, so I'd like to see some more ideas   :) )

Exactly the same reasoning as was shown by Pascual works here as well - the only difficulty is to find some contraints to have the following inequality valid:

$m \binom{l}{k} \geq \binom{2l}{k}$

where $n=2l$

I couldnt' afford it now though  :oops:

Yeah, but how do you know it will work, or is it obvious to you?
(In the case $m=3$ and $k=14$, it's easy to show, as pengshi mentioned)


floor function
number theory
relatively prime
Divisor Functions

If $n$ is composite, prove that $\phi(n) \le n- \sqrt{n}$.

If $n$ is composite it has a prime divisor less than or equal to $\sqrt{n}$, and there are at least $\frac{n}{\sqrt{n}}$ numbers less than $n$ divisible by this divisor and thus not relatively prime to n. So $\phi(n) \le n - \sqrt{n}$ with equality iff n is the square of a prime.

Just a minor correction. There are at least $\lfloor n/\sqrt{n}\rfloor$ numbers divisible by that divisor, so we are not quite done. However if there is another prime divisor, then $\phi(n)\le n-\lfloor\sqrt{n}\rfloor-1<n-\sqrt{n}$. In case there is no other prime divisor, then $n=p^k$ ($k\ge2$), $\phi(n)=p^k-p^{k-1}\le p^k-p^{k/2}=n-\sqrt{n}$.

To prove $$\varphi{(n)} \leq n - \sqrt{n} $$

$$n = \displaystyle \prod_{1 \leq i \leq r}{P_i^{\alpha_i}}, \hspace{0.2cm} \ni P_i > P_j, i < j$$
$$\varphi(n) \leq n - \sqrt{n} $$
$$\varphi(\displaystyle \prod_{1 \leq i \leq r}{P_i^{\alpha_i}}) \leq (\displaystyle \prod_{1 \leq i \leq r}{P_i^{\alpha_i}}) - \sqrt{\displaystyle \prod_{1 \leq i \leq r}{P_i^{\alpha_i}}} $$
$$\displaystyle \prod_{1 \leq i \leq r}\left[{P_i^{\frac{\alpha_i}{2}}{\cdot \left(\frac{P_i - 1}{P_i}\right)}}\right] \leq \displaystyle\prod_{1 \leq i \leq r}{P_i^{\frac{\alpha_i}{2}}} - 1$$
$$\displaystyle\prod_{1 \leq i \leq r}{P_i^{\frac{\alpha_i}{2}}}\left[1 - \displaystyle \prod_{1 \leq i \leq r}{\left(\frac{P_i - 1}{P_i}\right)}\right] \geq 1$$
As, $$\left(\frac{P_i - 1}{P_1}\right) \cdots \left(\frac{P_r - 1}{P_r}\right) \leq \left(\frac{P_r - 1}{P_r}\right)$$
$$\displaystyle\prod_{1 \leq i \leq r}{P_i^{\frac{\alpha_i}{2}}}\left[1 - \displaystyle \prod_{1 \leq i \leq r}{\left(\frac{P_i - 1}{P_i}\right)}\right] \geq P_r^{\frac{\alpha_1 + \cdots + \alpha_r}{2}}\left[1 - \left(\frac{P_r - 1}{P_r}\right)\right] = P_r^{\frac{\alpha_1 + \cdots + \alpha_r}{2} - 1} \geq 1$$

$\textbf{EQUALITY}$ Holds when $n = P^{2}$


[quote="Ngông"]Let $x,y,k,t \in Z_{+}$ such that: $x^{k}=y^{t}$
Prove that  there exists $a,m,n \in Z_{+}, gcd(m,n)=1$ such that $x=a^{m}, y=a^{n}$[/quote]
Let $m=\frac{t}{(k,t)},n=\frac{k}{(k,t)},a=x^{1/m}$, then $a\in Z$.

$x^{k}=y^{t}$
$p$ prim , p|x <=> p|y
and $kV_{p}(x)=tV_{p}(y)$
<=> $\frac{t}{GCD(t,k)}|V_{p}(x)$ and $\frac{k}{GCD(t,k)}|V_{p}(y)$
and $\frac{V_{p}(x)}{\frac{t}{GCD(t,k)}}=\frac{V_{p}(y)}{\frac{k}{GCD(t,k)}}$
<=> there exist s ; $Vp(x)=\frac{ts}{GCD(t,k)}$ and $V_{p}(y)=\frac{ks}{GCD(t,k)}$
<=> thre exist a; $x=a^{\frac{t}{GCD(t,k)}}$ and $y=a^{\frac{k}{GCD(t,k)}}$
=>  there exists $(a,m,n)\in N^{3}$, $x=a^{m},y=a^{n}$ with $m=\frac{t}{GCD(t,k)},n=\frac{k}{GCD(t,k)}$ and we have $GCD(n,m)=1$


Choose a random point $p$ inside a triangle such that $p$ trisects the area of the triangle into three equal pieces.

1. Does $p$ always have to be the Gravity $G$ of the triangle to possess this property?

2. Find other points (inside or outside of the triangle) that can trisect a triangle.

1: P is the centroid if it is within the triangle. You can use area ratios to show that AP is median of BC.

2: Let P be within $\measuredangle{BAC}$ and AP bisects BC. Let the midpoint of BC be D. Let P be the symmetrical point of G with respect to D, then triangle PAB, PBC and PAC are with the same area. Similarly you can get another 3 points. So there are a total of 4 points which satisfy your condition.


Do you know where I could read about set theory to learn what the following symbols mean? For example, what do the following mean:

$ \mathbb{Z}$, $ \mathbb{N}$, ...

What are other abbreviations for things such as

if and only if
for all reals
for all integers
...

Try [[Set]] or search "Set Theory" on Wikipedia.

[url]http://en.wikipedia.org/wiki/Naive_set_theory[/url]
Look through that

As for your last questions, [url=http://en.wikipedia.org/wiki/Table_of_logic_symbols]this page[/url] may be useful.

the abbreviation for if and only if is iff.


geometry
geometric transformation
reflection
vector
trigonometry
inequalities
area of a triangle

11.6 Construct for each vertex of the quadrilateral of area $S$ a symmetric point wrt to the diagonal, which doesn't contain this vertex. Let $S'$ be an area of the obtained quadrilateral. Prove that $\frac{S'}{S}<3$.
([i]L. Emel'yanov[/i])

First we introduce notations:

[color=blue][b]Problem.[/b] Given a quadrilateral ABCD with area S. Denote by A', B', C', D' the reflections of the points A, B, C, D in the lines BD, AC, BD, AC, respectively. Let S' be the area of the quadrilateral A'B'C'D'. Prove that $\frac{S^{\prime}}{S}<3$.[/color]

In order to solve this problem, we switch to using [i]directed areas[/i] (else, we could get in troubles with non-convex quadrilaterals). We will denote by $\left[P_1P_2...P_n\right]$ the [i]directed[/i] area of a polygon $P_1P_2...P_n$, and, in particular, by $T=\left[ABCD\right]$ and $T^{\prime}=\left[A^{\prime}B^{\prime}C^{\prime}D^{\prime}\right]$ the directed areas of the quadrilaterals ABCD and A'B'C'D'. Then, the non-directed areas S and S' of these two quadrilaterals are the absolute values of these directed areas: S = |T| and S' = |T'|.

Let the diagonals AC and BD of the quadrilateral ABCD intersect each other at the point F. It is almost obvious that the diagonals A'C' and B'D' of the quadrilateral A'B'C'D' intersect at F, too, but we won't need this in the rest of the proof :D . Instead, let's denote by t = < AFB the directed angle < AFB; hereby, "directed angle" means a directed angle modulo 360° (namely, this directed angle < AFB is the angle between the vectors $\overrightarrow{FA}$ and $\overrightarrow{FB}$) and not a directed angle modulo 180°. Then, < BFC = 180° - t, < CFD = t and < DFA = 180° - t. Since sin (180° - t) = sin t, we then have sin < AFB = sin < BFC = sin < CFD = sin < DFA = sin t.

Since A' is the reflection of A in the line BD, we have FA' = FA and < BFA' = < AFB. Since B' is the reflection of B in the line AC, we have FB' = FB and < B'FA = < AFB. Thus, < B'FA' = < B'FA + < AFB + < BFA' = < AFB + < AFB + < AFB = t + t + t = 3t, and < A'FB' = - < B'FA' = - 3t.

Now, since the area of a triangle equals $\frac12$ $\cdot$ the product of two sides $\cdot$ the sine of the angle between these sides, we have

$\left[FAB\right]=\frac12\cdot FA\cdot FB\cdot\sin\measuredangle AFB$;
$\left[FA^{\prime}B^{\prime}\right]=\frac12\cdot FA^{\prime}\cdot FB^{\prime}\cdot\sin\measuredangle A^{\prime}FB^{\prime}$.

Since FA' = FA and FB' = FB, we thus have

$\frac{\left[FA^{\prime}B^{\prime}\right]}{\left[FAB\right]}=\frac{\sin\measuredangle A^{\prime}FB^{\prime}}{\sin\measuredangle AFB}=\frac{\sin\left(-3t\right)}{\sin t}=\frac{-\sin\left(3t\right)}{\sin t}$,

so that $\left[FA^{\prime}B^{\prime}\right]=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FAB\right]$. Similarly, $\left[FB^{\prime}C^{\prime}\right]=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FBC\right]$, $\left[FC^{\prime}D^{\prime}\right]=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FCD\right]$, $\left[FD^{\prime}A^{\prime}\right]=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FDA\right]$. Thus,

$T^{\prime}=\left[A^{\prime}B^{\prime}C^{\prime}D^{\prime}\right]=\left[FA^{\prime}B^{\prime}\right]+\left[FB^{\prime}C^{\prime}\right]+\left[FC^{\prime}D^{\prime}\right]+\left[FD^{\prime}A^{\prime}\right]$
$=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FAB\right]+\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FBC\right]+\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FCD\right]+\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FDA\right]$
$=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left(\left[FAB\right]+\left[FBC\right]+\left[FCD\right]+\left[FDA\right]\right)=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[ABCD\right]$
$=\frac{-\sin\left(3t\right)}{\sin t}\cdot T$,

and $\frac{T^{\prime}}{T}=\frac{-\sin\left(3t\right)}{\sin t}$, so that

$\frac{S^{\prime}}{S}=\frac{\left|T^{\prime}\right|}{\left|T\right|}=\left|\frac{T^{\prime}}{T}\right|=\left|\frac{-\sin\left(3t\right)}{\sin t}\right|=\left|\frac{\sin\left(3t\right)}{\sin t}\right|$.

We have to prove that $\frac{S^{\prime}}{S}<3$; by the above, this is equivalent to $\left|\frac{\sin\left(3t\right)}{\sin t}\right|<3$. But this inequality is easy to prove:

$\left\vert \frac{\sin \left( 3t\right) }{\sin t}\right\vert =\left\vert\frac{\sin \left( 2t+t\right) }{\sin t}\right\vert =\left\vert \frac{\sin\left( 2t\right) \cos t+\cos \left( 2t\right) \sin t}{\sin t}\right\vert=\left\vert \frac{\sin \left( 2t\right) \cos t}{\sin t}+\cos \left(2t\right) \right\vert $
$=\left\vert \frac{2\sin t\cos t\cos t}{\sin t}+\cos \left( 2t\right)\right\vert =\left\vert 2\cos ^{2}t+\cos \left( 2t\right) \right\vert \leq\left\vert 2\cos ^{2}t\right\vert +\left\vert \cos \left( 2t\right)\right\vert<3$,

since $\left|2\cos^2 t\right|\leq 2$ and $\left|\cos\left(2t\right)\right|\leq 1$ with equality only for t being a multiple of 180°, what is not possible in our case since t is the angle < AFB between the diagonals AC and BD of the (non-degenerate) quadrilateral ABCD.

Proof complete.

  Darij

Я сечас умру от тоски. :(

Let $\alpha$ be the acute angle betwen the diagonals, then after the reflection the angle between diagonals becomes $3\alpha$ or $3\alpha-\pi$, moreover the length of diagonals are preserved. It follows that
\[\frac{S'}{S}=\left|\frac{\sin 3\alpha}{\sin\alpha}\right|=|3-4\sin^2\alpha|<3.\]


calculus
integration
trigonometry
calculus computations

$ \displaystyle\int^\pi_0 \sqrt {\sin{x} \minus{} \sin{x^3}} dx$

I was using trig formula $ \sin{3x} \equal{} 3\sin{x} \minus{} 4\sin^3{x}$ but I got no where. please help!

EDIT:
I am so sorry. I looked up my hw and it is actually $ \sin^3{(x)}$.

$ \displaystyle\int^\pi_0 \sqrt {\sin{(x)} \minus{} \sin^3{(x)}} dx$

hello, use that
$ \sin(x)\minus{}\sin(x)^3\equal{}\sin(x)\cos(x)^2$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, use that
$ \sin(x) \minus{} \sin(x)^3 \equal{} \sin(x)\cos(x)^2$.
Sonnhard.[/quote]

Exactly, then it becomes
$ \int^\pi_{0}\cos{x}\sqrt{\sin{x}}dx$

which equals 2/3*(sinx)^(3/2) from 0 to pi

Beware that $ \sqrt{1\minus{}\sin^2x}\equal{}\vert \cos x\vert$ and $ \cos x$ changes sign crossing $ \pi/2$

To the OP, I assume the sine on the right side of the integral should be $ \sin^3{x}$ and not $ \sin{x^3}$?

hello, i also think he meant
$ \int_{0}^{\pi}\sqrt{\sin(x)\minus{}\sin(x)^3}\,dx$.
Sonnhard.

I am so sorry. I looked up my hw and it is actually $ \sin^3{(x)}$. :blush:

$ \displaystyle\int^\pi_0 \sqrt {\sin{(x)} \minus{} \sin^3{(x)}} dx$


[quote="JRav"]$ 2\log(4) \plus{} \log(7) \minus{} 1 \equal{} \log(x)\Rightarrow\log(112) \minus{} 1 \equal{} \log(x)\Rightarrow\log(112) \minus{} \log(1) \equal{} \log(x)\Rightarrow x \equal{} 112$[/quote]

$ 1 \neq \log(1)$ :)

The replacement should have been rather $ 1 \equal{} \log(10)$, so the final answer would be $ x \equal{} \frac {112}{10} \equal{} \frac {56}5 \equal{} 11.2$.

[quote="neutralvisitor"]find x if log4+log4+log7-logx =1
              2      2     2     2
    2logx-xlog(1+y)=3
             2
[/b] the log are all base 2 for the first question
the second question xlog7 is base 2[/quote]

Ah that makes much more sense. Now I know why the random 2's were there.

(If you want to know, this is what he tried to post.)
[code]
find x if log4+log4+log7-logx = 1
            2    2    2    2
    2logx-xlog(1+y)=3
             2
[/code]

$ \text{\LaTeX}$ified:
\[ \text{Find}\ x\ \text{if}\ \log_2 4 + \log_2 4 + \log_2 7 - \log_2 x = 1.
\]
Solution:
\begin{align*} \log_2 4 + \log_2 4 + \log_2 7 - \log_2 x & = 1 \\
\log_2 4 + \log_2 4 + \log_2 7 - 1 & = \log_2 x \\
2\log_2 4 + \log_2 7 - \log_2 2 & = \log_2 x \\
\log_2 \left(\frac {4^2 \cdot 7}2\right) & = \log_2 x \\
\frac {4^2 \cdot 7}2 & = x \\
\frac {112}2 & = x \\
56 & = x \end{align*}
I still don't understand the second problem though. :huh:


For LHS use AM-GM Inequality  then prove all the term of LHS Are Less than Terms in RHS
This inequality Purposed in Competitions 1995.


conics
ellipse
ratio
analytic geometry
graphing lines
slope
trigonometry

Let $w_{1}$ and $w_{2}$ denote the circles $x^{2}+y^{2}+10x-24y-87=0$ and $x^{2}+y^{2}-10x-24y+153=0$, respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_{2}$ and internally tangent to $w_{1}$. Given that $m^{2}=p/q$, where $p$ and $q$ are relatively prime integers, find $p+q$.

The problem is written wrong... because 0 works.  In fact, any point outside $w_2$ can be made the center of a circle and make the radius so it's tangent to $w_2$

I'm also trying to figure out the significance of the first circle. Possibly the line must be externally tangent to both circles, or it contains the center of the the first circle and is externally tangent to the second?

I haven't seen the real problem, but I would guess it wants the new circle to be externally tangent to $w_2$ and internally tangent to $w_1$, since $w_1$ surrounds $w_2$.

[hide="solution"]the problem is externally tangent to $w_2$ and internally tangent to $w_1$

lets find the locus of centers for circles that are tangent as above

if we have point $A=(x_0,y_0)$, then $A$, the center of the circle and the point of tangency are collinear (why?), for the circle to be tangent, it has to be on those 2 lines (1 line per circle), and the distance from $A$ to both of the points of tangency are equal

to find those 2 lengths, (use the radii and use the distance from the point to the centers to find the lengths)
then setting them equal, we get

$16-\sqrt{(x+5)^2+(y-12)^2}=\sqrt{(x-5)^2+(y-12)^2}-4$

squaring, rearranging, and repeating, we eventually get

$-\frac{3}{4}(x^2-100)=(y-12)^2$

plugging in $y=ax$, and we notice that the minimum value will occur when that line is tangent below the graph, simplifying, we get

$(a^2+\frac{3}{4})x^2-24ax+69=0$, we want only one solution because it is tangent, so setting the discriminant equal to 0, we get

$a^2=\frac{69}{100}$, so $m+n=\boxed{169}$

$\mathbb{QED}$
[/hide]

Just to add a little note: the locus of the centers is in fact an ellipse, (distance to w1 is 16-r, distance to w2 is 5+r, total remains at 21) so you know the focii (centers of w1 and w2), just find a point on it, (-10,12) works well since its on the major axis and you can generate the equation slightly cleanlier(?) :D

Can somebody please edit the question.  There's a typo.

In the 2nd sentence, it should say "Let m be the smallest POSITIVE value..." not "possible value."

The typo is rather confusing because if m can be negative, then it can be made arbitrarily small.

Thanks.

Hello, how would you finish this solution (if possible)?
[hide="my (incomplete) solution"]
Let the centers of the new circle, circle $w_1$, circle $w_2$ be $A$, $O_1$, $O_2$, respectively.
Let the radius of the new circle be $r$ and let $\angle O_1O_2A = \theta$.
We have that $\triangle O_1O_2A $ has:
\begin{align*}
O_1O_2 &= 10 \\
O_2A &= r+4 \\
AO_1 &= 16-r \\
\angle O_1O_2A &= \theta
\end{align*}

Then, we wish to find $A$ such that the ratio of its $y$-coordinate to $x$-coordinate is minimized, given that $\theta$ ranges from $0$ to $180$ (we will tighten the minimum later, the max is because any larger angle would result in a point that has a higher slope than that at $180^{\circ}$).

First, we will find $r$ in terms of $\theta$:
by LoC:
\begin{align*}
(16-r)^2 &= (r+4)^2+10^2-2(10)(r+4)\cos{\theta} \\
\Rightarrow 140 &= 40r-20(r+4)\cos{\theta} \\
\Rightarrow r &= \frac{140+80\cos{\theta}}{40-20\cos{\theta}} = \frac{7+4\cos{\theta}}{2-\cos{\theta}}
\end{align*}

Now, we find $A$ by rotating $(r+9, 12)$ by $(180+\theta)^{\circ}$ degrees about $O_2$. We get $A=(-(r+4)\cos{\theta}+5, -(r+4)\sin{\theta}+12)$
Substituting the above expression or $r$, we have: $A=\left(\frac{10-20\cos{\theta}}{2-\cos{\theta}}, \frac{-15\sin{\theta}-12\cos{\theta}+24}{2-\cos{\theta}}\right)$

Now, we want:
\begin{align*}
\min{\frac{\frac{-15\sin{\theta}-12\cos{\theta}+24}{2-\cos{\theta}}}{\frac{10-20\cos{\theta}}{2-\cos{\theta}}}} &= \frac{3}{10}\min{\frac{-5\sin{\theta}-4\cos{\theta}+8}{1-2\cos{\theta}}} \\
\text{(1)}\;\;\;\;&= \frac{3}{10}\left(\min{\frac{5\sin{\theta}-6}{-1+2\cos{\theta}}}\right)+\frac{3}{5}\\
\text{(2)}\;\;\;\;&= -\frac{3}{10}\left(\max{\frac{-5\sin{\theta}+6}{-1+2\cos{\theta}}}\right)+\frac{3}{5}
\end{align*}

Now, because $a$ is positive, we see that $60<\theta\le180$.  On $(60,90]$, we use (1): the numerator is decreasing (in magnitude) and the denominator is increasing (in magnitude). Thus, it is overall decreasing and the minimum is at 90.
How do you find the value on $(90,180]$? (This is the range where the answer lies, but I don't know how to evaluate it).
[/hide]

Please respond.  Thank you!

This could be helpful http://en.wikipedia.org/wiki/Steiner_chain#Elliptical.2Fhyperbolic_locus_of_centers

Completing the square to get
(x+5)^2 + (y-12)^2=256
(x-5)^2+(y-12)^2=16
Therefore, w1 is centered at (-5,12) with radius=16 and w2 is centered at (5,12) with radius 4. set F as the center of the third circle and set r as its radius. Then its evident that P is on the ellipse with center (0,12) and foci at the center of w1 and center of w2. the major axis is 20, and the minor axis is 10 sqrt 3. the equation of the ellipse is then
x^2/100 +(y-12)^2/75=1
by cauchy, 1=(-ax)^2/100a^2 + (y-12)^2/75 >= (y-ax-12)^2/(100a^2+75)=144/(100a^2+75)
so 144/(100a^2+75)<=1
a^2>=69/100 so m^2=69/100 and our ans is 169


I know that earlier on in the year, there had been discussions about making merchandise for AoPS. Is this still a consideration?

Before doing so, it may be wise to do a loose poll of the members here to see who would really consider buying merchandise.

True...but I think most ppl wouldn't say its not a possibility


A cylindrical resistor of radius $3.9 mm$ and length $2.3 cm$ is made of material that has a resistivity of $2.4 * 10^{-5} m$. What are (a) the magnitude of the current density and (b) the potential difference (in V) when the energy dissipation rate in the resistor is $0.55 W$?

cant you just use ohms law? I dont know i'm mechanics right now next semester we will cover electrcity and magnetism


probability
integration
conditional probability
real analysis
real analysis unsolved

A certain store stocks in boxes of condoms such that:
[list]
[*] Each box has a certain success rate associated with it, and all condoms inside that box work with that particular success rate
[*] The success rate of any box is a random variable that is uniformly distributed from 0 to 1.[/list]
You buy a box of $n$ condoms. Given that you use all but one of the condoms, and they all work, what's the probability that the very last one will work, in terms of $n$?

what do you mean with succes rate and how the success rate of any box is a random variable that is uniformly distributed from 0 to 1? it succed or not, right?

Conditional probability: if the box has probability $p$ of success associated with it, the probability that all $n-1$ other condoms succeed is $p^{n-1}$.  The value attached to this probability is the probability our last condom succeeds as well, which is also $p$.  So we just need $\frac{\int_{0}^{1}p^{n}dp}{\int_{0}^{1}p^{n-1}dp}= \frac{\frac1{n+1}}{\frac1{n}}= \frac{n}{n+1}$.  This agrees with common sense in the only case where common sense applies, $n = 1$.


geometry
circumcircle
quadratics
trigonometry
Pythagorean Theorem
algebra
geometry proposed

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

The heptagon has to be as close to a regular heptagon as possible, with the constraint that 2 diagonals have to be perpendicular. Thus it has 3 equal longer sides and 4 equal shorter sides. The simplest idea is to put together half of a regular hexagon and half of a regular octagon with the same circumcircle. Unfortunately, such heptagon does not have any perpendicular diagonals. Perhaps the order of the longer and shorter sides can be changed to make 2 diagonals perpendicular, but that's guessing.

Assume the diagonals $BE \perp GD$ of a cyclic heptagon $ABCDEFG$ are perpendicular, $x = AB = GA,$ and $BC = CD = EF = FG.$ We are looking for such $x$ that $DE = x$ as well. For any $DE,$ the heptagon is symmetrical with respect to the line $OA,$ where $O$ is the circumcenter. Thus $DE \perp AO,\ BG \perp AO$ and the perpendicular diagonals $BE \perp GD$ meet at $P \in AO.$ Let $R$ be the heptagon circumradius. Let $BG$ meet $AO$ at $Q.$ Using Pythagorean theorem, we get

$BG = 2x \sqrt{1 - \frac{x^2}{4R^2}},\ \ \ QA = \frac{x^2}{2R}$

Since the triangle $\triangle BPG$ is isosceles right, $PQ = \frac{BG}{2} = x \sqrt{1 - \frac{x^2}{4R^2}}$ and $PA = PQ + QA = \frac{x^2}{2R} + x \sqrt{1 - \frac{x^2}{4R^2}}.$ Denote $\Sigma = \sqrt{1 - \frac{x^2}{4R^2}}.$ Power of the point $P$ to the circumcircle $(O)$ is

$PA \cdot (2R - PA) = 2R \cdot PA - PA^2 =$

$= x^2 + 2Rx \cdot \Sigma - \frac{x^4}{4R^2} - \frac{x^3}{R} \cdot \Sigma - x^2 \left(1 - \frac{x^2}{4R^2}\right) = 2xR \left(1 - \frac{x^2}{2R^2}\right) \cdot \Sigma$

On the other hand, power of the point $P$ to the circumcircle $(O)$ is $PB \cdot PE,$ where $PB = PQ \sqrt 2 = x \sqrt 2 \cdot \Sigma$ is known. Assuming $DE = x,$ then from the isosceles right triangle $\triangle DPE,$ we get $PE = DE \frac{\sqrt 2}{2} = \frac{x \sqrt 2}{2}.$ Hence $PB \cdot PE = x^2 \cdot \Sigma.$ Thus from $PB \cdot PE = PA \cdot (2R - PA),$ we get an equation for $x$:

$2xR \left(1 - \frac{x^2}{2R^2}\right) \cdot \Sigma = x^2 \cdot \Sigma$

Reducing by $x \neq 0,\ \Sigma \neq 0$ and solving the resulting quadratic equation for $\frac x R,$

$2R \left(1 - \frac{x^2}{2R^2}\right) = x,\ \ \ \frac{x^2}{R^2} + \frac{x}{R} - 2 = 0,\ \ \frac x R = -2$ or $+1$

The only acceptable root is $x = R > 0,$ hence the heptagon $ABCDEFG$ is indeed obtained by rearranging the sides of half a regular hexagon and half a regular octagon. Therefore, its area is $S_7 = \frac{S_6 + S_8}{2} = R^2 \left(\frac{3 \sqrt 3}{4} + \sqrt 2\right).$

sorry,  :? I can't understand the first two sentences in your post.is it just a supposition or proof?
however,  I have a simple solution: :)
the heptagon is $A{}_1A{}_2A{}_3A{}_4A{}_5A{}_6{}_7$
let $\alpha_i=A{}_iOA{}_{i+1}$
then suppose that$A{}_1A{}_4\perp A{}_2A{}_5$ then $\alpha_1+\alpha_3+\alpha_4=\pi ,\alpha_2+\alpha_5+\alpha_6+\alpha_7=\pi$
and the area og the heptagon is
$\sum_{i=1}^7 {\frac12 R^2 \sin{\alpha_i}}\leq 3\sin(\frac13(\alpha_1+\alpha_3+\alpha_4)+4\sin\frac14(\alpha_2+\alpha_5+\alpha_6+\alpha_7)=\frac12R^2(\frac{3\sqrt{3}}{2}+2\sqrt{2})$
there is another two case ,use the similar method we got the maxium of the area(there is another max number for a cae but it is smaller than this one,you can check it) is$\frac12R^2(\frac{3\sqrt{3}}{2}+2\sqrt{2})$


I was wondering if anyone read these books and if you did, what did you think of them? What are they all about, I want to read them because I think they relate to economics and are on the bestseller list.

Freakonomics is very interesting - I recommend it.  It's basically about an economist who takes a more practical view of various matters and finds ways to test for, say, sumo wrestler cheating, or teachers cheating on their students' standardized tests, etc.  He approaches many economic problems by considering the incentives of the actors, which in my experience is a much safer predictor of people's actions than some theoretical construct which maximizes total 'utility' of a group of people (and is why the recent expansion of eminent domain sanctioned by the Supreme Court is going to be a mess).

actually just a week or two ago i found freakonomics at my house and i turned to a random page and i read about the sumo wrestlers and teachers cheating, and the logic behind the author's argument was pretty clever! i just might get around to truly read the whole thing, for it seems like a very enjoyable book!  :)

I am waiting for it and reserved in the library.


trigonometry
trig identities
Law of Cosines

Can anybody provide a proof for Stewart's Theorem?

Oops hehe i meant to post it in Intermediate Topics  :oops:

Google, Wikipedia, AoPSWiki. At least one of those have the answer. Also see AoPS 2.

Since you didn't think to use those ever-present resources, just use the law of cosines, and the fact that cos(x)+cos(180-x)=0

Stewart's theorem is that in a triangle ABC, M on BC we have $ AM^2\cdot BC\equal{}AB^2\cdot MC\plus{}AC^2\cdot MB\minus{}BC\cdot BM \cdot CM$ . Aply the law of cosinus and you will have $ AM^2\equal{}AB^2\plus{}BM^2\minus{}2AB\cdot BM \cdot cosB$ and $ AC^2\equal{}AB^2\plus{}BC^2\minus{}2AB \cdot BCcosB$ ,therefore it results :D (sorry for my english)


geometry
3D geometry
sphere
calculus
integration
analytic geometry
articles

It is given a spheric cap charged uniformally with a density charge equal to s.
Calculate the force by which it repels an electric charge q which is put in its center.

Is a "spherical cap" the upper half of a sphere?

Yes it is but not full, it has surface but no volume, it's a cap.I don't know the proper word in english :blush:

How about call it a "hemispherical shell"?

If there is some clever way to use Gauss's law to solve this problem, I don't know it.  I am stuck with just using Coulomb's law and doing an integral.

I am going to be very explicit/detailed about how I set up my integral, because I wanted to think this through and get it down for the record.  Physicists would never explain their solution like this.  If this makes my explanation horribly unclear, I apologize.  And if it makes the problem seem difficult, please believe me---it is only my poor explanation that is difficult to understand, and somebody else will tell you how simple the problem is.

Let's use the spherical coordinate system shown in the Wikipedia article http://en.wikipedia.org/wiki/Spherical_coordinates.  Let $ r > 0$ be the radius of our hemisphere, and let's consider the particular hemisphere $ H \equal{} \{(r, \theta, \phi) | 0 \le \theta \le \frac{\pi}{2}, 0 \le \phi \le 2\pi \}$.

By symmetry, the $ x$ and $ y$ components of the force on our test charge are $ 0$.

Let $ \{\theta_i | 1 \le i \le N \}$ be a partition of the interval $ [0, \frac{\pi}{2}]$, and let $ \{\phi_j | 1 \le j \le M \}$ be a partition of the interval $ [0, 2\pi]$.  Let $ \Delta \theta_i \equal{} \theta_{i\plus{}1} \minus{} \theta_i$ and let $ \Delta \phi_j \equal{} \phi_{j\plus{}1} \minus{} \phi_j$.  Let $ R$ be the rectangle $ [0, \frac{\pi}{2}] \times [0, 2\pi]$.  Let $ P_{ij} \equal{} \{(r,\theta, \phi) | \theta_i \le \theta \le \theta_{i\plus{}1}, \phi_j \le \phi \le \phi_{j\plus{}1} \}$.  If our partitions have very small meshes, then the area of $ P_{ij}$ is approximately $ (r \Delta \theta_i)*(r \Delta \phi_j) \equal{} r^2 \Delta \theta_i \Delta \phi_j$, and the z-component of the force $ P_{ij}$ exerts on our charge is approximately $ \minus{}\frac{qsr^2 \Delta \theta_i \Delta \phi_j}{r^2} \cos \theta_i \equal{} \minus{}qs \cos \theta_i \Delta \theta_i \Delta \phi_j$.  Adding the contributions from all the different $ P_{ij}$, we get that the z-component of the force on our charge is approximately $ \Sigma_{i,j} \minus{}qs \cos \theta_i \Delta \theta_i \Delta \phi_j \approx \minus{}\int_R qs \cos \theta d\theta d\phi \equal{} \minus{}\int_0^{2\pi}\int_0^{\frac{\pi}{2}}  qs \cos \theta d\theta d\phi \equal{} \minus{}2\pi qs$.

By using partitions with smaller and smaller meshes, our approximations become better and better, and in the limit as the meshes of our partitions approach $ 0$ we have exact equality.

It's interesting (perhaps alarming) that my answer doesn't depend on $ r$.  Is there some mistake?  As $ r$ gets bigger, there is more surface charge pushing on our test charge, but the surface charge is further away, and apparently these two effects cancel each other perfectly so that the force on our test charge doesn't change.

Thank you for posting such a detailed answer but there are several things I didn't understand :maybe:
Why have you chosen zone H that way? Shouldn't we have one of the angles from 0 to pi if we consider an half of the hemispherical shell, that is a quarter of the whole sphere? The electric charge is in the center of the largest diameter

Was the surface given in this problem only $ \frac14$ of a complete sphere?  Perhaps I misunderstood.  I thought it was $ \frac12$ of a complete sphere.

The surface I described should be $ \frac12$ of a complete sphere.

No you didn't, it is half a sphere indeed but I thought that because of symmetry we shoul consider the half of this half,  that is one quarter :blush:
its OK

Oh, I see.  You could do it that way too.  Good idea.

[hide] $ \; F \equal{} {{qs} \over {4\epsilon_0}}$ [/hide]

Here is the solution:
Let the $ z$ axis be perpendicular to the "base" of the half sphere, and go through its center...
The charge on a small piece of the half sphere is (We use spherical coordinates of course...):
\[ dq \equal{} sR^2\cdot \sin\theta \cdot d\theta\cdot d\phi
\]
(as usual $ \theta$ is the angle between the $ z$ axis and the radius $ R$ is the radius of the sphere, and $ \phi$ is the angle sweeping on the horizontal..)

So the force by this small element on the small charge:
\[ dF \equal{} \dfrac{1}{4\pi\varepsilon_0}\dfrac{q\cdot dq}{R^2}\cos\theta \equal{} \dfrac{qs}{4\pi\varepsilon_0}\cdot \sin\theta\cos\theta\,d\theta\,d\phi
\]
So the total force:
\[ F \equal{} \dfrac{qs}{4\pi\varepsilon_0}\int_0^{\frac {\pi}{2}}\sin\theta\cos\theta\,d\theta \int_0^{2\pi}d\phi \equal{} \boxed{\dfrac{qs}{4\varepsilon_0}}
\]

I'm glad people chimed in on this, since my solution was wrong!

Here is what I now believe to be the correct solution:

The charge on a small piece of our surface is $ s(Rd\theta)(R\sin\theta d\phi) \equal{} sR^2\sin\theta d\theta d\phi$, as Thaakisfox astutely points out.  The z-component of the force on our test charge due to this small piece of our surface is $ \frac{\minus{}1}{4 \pi \epsilon_0}\frac{qsR^2\sin\theta d\theta d\phi }{R^2}\cos\theta \equal{} \frac{\minus{}1}{4 \pi \epsilon_0}qs \sin\theta \cos \theta d\theta d\phi$.  The z-component of the force on our test charge due to the entire surface is therefore $ \frac{\minus{}qs}{4 \pi \epsilon_0}\int_0^{2\pi}\int_0^{\frac{\pi}{2}}\sin\theta \cos \theta d\theta d\phi \equal{} \frac{\minus{}qs}{4\epsilon_0}$.  (By symmetry, the x and y components of the force are $ 0$.)

What about me?   (he he he , just joking)

you too or you three?  (another joke , sorry)


In my college class for math teachers, this is the problem I am stuck on:

What is the number of degrees in the measure of the obtuse angles formed by the hands of a standard clock at 2:48 p.m?

Its been awhile..... :blush:

At 2:48, the hour hand is at 2, which is equavalent to 10 minutes. The minute hand is obviously at 48 minutes, which means there's a 38 minute difference between them. Note that each minute equals 6 degrees (360/60 = 6). 38 times 6 is 228. However, you want to find the obtuse angle, not the reflex angle. If you look at the clock, you'll notice that there are two angles formed - a larger angle and a smaller angle. If you subtract the larger angle from 360 degrees (the number of degrees in a circle), you get [b]132 degrees.[/b]

[quote]What is the number of degrees in the measure of the obtuse angle
formed by the hands of a standard clock at 2:48 p.m?[/quote]
Assuming they want the [b]exact[/b] angle, it's trickier . . . [hide]
Let $12:00  = 0^o.$

The minute hand is at $\text{48 minutes} \;= \;\frac{48}{60}\times360^o \;= \;288^o$
That is, it is at $72^o$ [u]before[/u] "12".

At exactly 2:00, the hour hand is at $60^o.$
At 2:48, the hour hand is [b]between[/b] the "2" and the "3".
There are $30^o$ between the hour-marks.

The minute hand has gone $\frac{48}{60} = \frac{4}{5}$ of the way around the face.
Hence, the hour hand has gone $\frac{4}{5}$ of the way from "2" to "3".
It has moved $\frac{4}{5}\times 30^o = 24^o$

So the hour hand is at $60^o + 24^o = 84^o$ ([u]after[/u] "12")

Therefore, the angle between them is: $72^o + 84^o = 156^o$[/hide]

Whoops...

(sorry about that, I got very very confused and died)


All of them? (Does the expected value have to be an integer?)

edit: change in problem.

you're in calc (according to your name, but i don't know)
i'm not
show me how you did it because you're probably right.

[quote="lasagnaman"]Not meaning to hijack this completely, but as it seems we already have an answer....

I was just wondering, what would be the relative probability of him knowing 26 answers compared to, say, him knowing all 30 answers or him not knowing ANY answers GIVEN THAT he scored 0. If someone gets a score of 0, is it really more likely that they knew a bunch of them and purposefully answered them wrong or is it more likely that they guessed on most/all of them and got them wrong?

And how come he has a 4/5 chance to get it right when he guesses? Isn't there only 1 right answer?[/quote]

He has a $\frac{4}{5}$ chance of getting it [i]wrong[/i].

Probability of not knowing any answers = $\frac{\frac{4}{5}^{30}}{4.99504...} = .000247...$

Probability of knowing all of them = $\frac{1}{4.99504...} = .20019...$

you might want to read your own question. you said he has a 4/5 chance of getting it right if he guesses.

[quote="calc rulz"]A certain multiple choice test is graded as follows:

0 points for an incorrect answer
1 point for a blank answer
5 points for a correct answer
30 questions
5 choices per question

Bob takes this test as follows: if he knows the answer to a question, he gives a wrong answer. If he doesn't know the answer to a question, he guesses, with a $\frac{4}{5}$ probability of getting it wrong. If he gets a total score of 0, what is the EV of number of questions that he knew the answer to?[/quote]

Edit: oops :blush:


calculus
integration
trigonometry
limit
complex analysis
complex analysis unsolved

Using residue theorem:
$ \int_{+\partial{D}}{\frac{z^2-\pi z}{(1-e^{2iz})\sin{z}}dz}$
$ D=\{z=x+iy: -\frac{\pi}{2}\le{x}\le{\frac{3\pi}{2}},|y|\le{2}\}$

Hi.  There's a methodical technique to calculating residues.  First determine the order of the poles.  Let's see, $ p_0=0$ and $ p_1=\pi$.  They're not removable since the limits: $ \mathop\lim\limits_{z\to\{p_0,p_1\}} \frac{f(z)}{g(z)}$ do not exists.  But they are simple since: $ \mathop\lim\limits_{z\to\{p_0,p_1\}} z\frac{f(z)}{g(z)}\ne 0$
So, my book has a table to follow.  This is the one I used:

For $ \frac{f(z)}{g(z)}$ in which $ f(z)$ has a zero of order $ k$ and $ g(z)$ order $ k+1$ at zero $ p_0$, then:

$ \mathop\text{Res}\limits_{z=p_0}{\frac{f(z)}{g(z)}}=
(k+1)\frac{f^{(k)}(p_0)}{g^{(k+1)}(p_0)}$

Alright, that works for both of them so I get:

$ \mathop\oint\limits_{\partial D^{+}}\frac{z^2-\pi z}{(1-e^{2iz})Sin(z)}=2\pi i(-\pi i/2-\pi i/2)=2\pi^2$

Here's the numerical check:

[code]
In[93]:= NIntegrate[((z^2 - Pi*z)/((1 - Exp[2*I*z])*Sin[z]))*Pi*I*
       Exp[I*t] /. z -> Pi/2 + Pi*Exp[I*t], {t, 0, 2*Pi}]

Out[93]= 19.739208802177707 + 3.2811531269771876*^-12*I
[/code]


[hide="Hint"]
$2^{p}= (1+1)^{p}={p \choose 0}+{p \choose 1}+{p \choose 2}+\ldots+{p \choose p}$
[/hide]


$ A$ and $ B$ are non-zero digits for which $ \underline{A468B05}$ is divisible by $ 11$. What is $ A \plus{} B$?

A+B-1=11, since the alternating sum of a number must be divisible by 11 for the number to be divisible by 11. If A+B-1=0, then one of A and B must be 0. So A+B= $ \boxed{12}$


How do you compute the standard derivation on a TI-84?
There must be something simpler than entering in all those numbers...

i dont think this is the right forum for this...

I believe it should go to the Art and Sciences part of the forum, to the Computer Science and informatics.

Stat -> calc -> 1-var statistics then press a button for one of the lists (L1, L2, etc. they are above the 1-6 buttons).  This displays a bunch of information about the numbers on that list, including standard deviation.

edit- to create lists, go to stat -> edit and enter the data

The Other Problem Solving Topics forum is the conventional place for calculator-related questions, to the extent such a place exists.


How many digits are in the product of 27 and $ 10^{23}$?

$ 1 \times 23$ has $ 2$ digits.  But the $ 1$ has $ 24$ zeroes after it, so the answer is $ 26$.

Actually, since there are 2 digits in 27, and 23 zeroes, our answer is $ 23 \plus{} 2 \equal{} 25$.

Oh, I thought it was $ 10^{24}$.  :blush:


An application of this problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=586&t=290285 :)
[quote="who"]:roll: ,due to cecil,the proposer,there is sol,using nth dimensional cube and resistor arrangement configuration,which seems very stange :o
 you can find it in mathematical olyppiad challenges-titu,razvan.
 anyway its standard using induction,provided you dont loose hope :)[/quote]
Original problem from Kvant (Prof.Titu said it in his book-Problems from the book)

Here is a solution using generating functions.

Let $a_n=\frac{n+1}{2^{n+1}} \left(\frac{2}{1}+\frac{2^2}{2}+\cdots + \frac{2^{n+1}}{n+1}\right)$. Then its pretty clear that $a_n$ satisfies the recursion $2(n+1)a_{n+1}=(n+2)a_n+2(n+1)$. Thus, if we let $b_n=\sum_{i=0}^n \binom{n}{i}^{-1}$, we want to show $2(n+1)b_{n+1}=(n+2)b_n+2(n+1)$.

Note $n!b_n=\sum_{i+j=n} i!j!$. Thus, we define $F(x)=\sum_{n\ge 0} n!x^n$. Then $n!b_n$ is the coefficient of $x^n$ of $F(x)^2$. Let $f(x)[j]$ denote the coefficient of $x^j$ of a general function $f$. We want to show $$2F(x)^2[n+1]=2(n+1)!+(n+2)F(x)^2[n]$$ Note $2(n+1)!=2F(x)[n+1]$. Thus we want to show $2F(x)^2-2F(x)[n+1]=(n+2)xF(x)^2[n+1]$. Note that $(n+2)g(x)[n+1]=(xg(x))'[n+1]$, so we want to show $$2F(x)^2-2F(x)[n+1]=(F(x)^2x^2)'[n+1]$$ In general, this means we want to prove the identity $$2F(x)^2-2F(x)=2F(x)F'(x)x^2+2xF(x)^2$$ or $F(x)-1=F'(x)x^2+xF(x)$, or $$F(x)(1-x)=F'(x)x^2+1$$ We can verify that for $n\ge 2$, the coefficient of $x^n$ on both sides is $(n-1)\cdot (n-1)!$, and modulo $x^2$ both polynomials are $1$.

The $n$-dimensional resistor problem came up in a physics class I take so I might as well share the physical solution, which is quite fascinating.

[quote="E&M"]In an $n$-dimensional cube, there is a resistor of resistance 1 ohm on each edge. Compute the effective resistance between $A=(0,0,\ldots, 0)$ and $P=(1,1,\ldots,1)$.[/quote]

We compute said resistance in two different ways using physics and we equate the resulting expressions to prove the given identity (which finally gives a higher reason to the truth of this weird identity!)

Method 1: We can group resistors in layers depending on their "distance" from the origin. Let the vertices with $k$ ones and $n-k$ zeros be "layer $k$". By symmetry, all these vertices have the same potential, so we can collapse them into a single vertex (the "bridge building" trick). Thus we have $n$ layers of parallel resistors, each contributing $\frac{1}{\dbinom{n}{k}(n-k)}$ by Kirchoff's Law, for a total of $R_n=\frac1n\sum_{i=0}^{n-1}\dbinom{n-1}{k}^{-1}$.

Method 2: We want to compute the resistance by injecting a fictitious current, so we are motivated to relate this to smaller versions of the problem by pulling out the current at different points. We can do it like so: Consider the two $n-1$ dimensional cubes given by $\{(x_1,x_2,\ldots,x_{n-1},0)\}$ and $\{(x_1, x_2, \ldots, x_{n-1},1)\}$  where each $x$ is $0$ or $1$. Furthermore let $C=(0,0,\ldots, 0,1)$ and $D=(1,1,\ldots,1,0)$ be two diametrically opposite points, each on a different smaller cube.

Take $B$ as the ground and we inject a current of 1 ampere at $A$. Let $V_n$ be the potential at $A$

We first notice that, by symmetry, the current leaving $A$ is spread onto its $n$ adjacent neighbours and likewise for the currents entering $B$. Thus, the potentials at $C$ and $D$ are $V_n-\frac1n$ and $\frac1n$ respectively. Similarly, if we inject a current at $C$ and take $D$ as the ground, the potentials at $A,B,C,D$ will be $V_n-\frac1n, \frac1n, V_n, 0$ respectively.

By superimposing the two configurations (inject at $A$, $C$, exit at $B,D$), we get the effective potential of $A$ above $D$ is $2V_n-\frac2n$.

We compute the potential of $A$ over $D$ with another argument: by symmetry, each vertex on a $n-1$ dimensional cubes and its corresponding vertex on the other one have no net current. Thus, it's as if there were no connection between the two cubes, so the potential of $A$ above $D$ is the same as the $n-1$ dimensional case of the problem, which would be $V_{n-1}$.

Thus we have the recursion $2V_n-\frac2n=V_{n-1}$.

Solving, we have $V_n=\sum_{i=1}^n\frac{1}{i2^{n-i}}$.

By Ohm's Law, $V=RI$, but $I=1$, so $R_n=V_n$.
Equating the two quantities, we get the desired proof of the given identity.

Consider a different inductive proof:

[b]Lemma 1:[/b] $\sum_{k=0}^{n} \frac{k}{\binom{n}{k}}=\frac{n}{2}\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}$.

$\emph{Proof: }$ Since $\binom{n}{k}=\binom{n}{n-k}$, we have

\begin{align*}\sum_{k=0}^{n} \frac{k}{\binom{n}{k}}&=\sum_{k=0}^{\lfloor n/2 \rfloor}\frac{k}{\binom{n}{k}}+\frac{n-k}{\binom{n}{n-k}}\\&=\sum_{k=0}^{\lfloor n/2 \rfloor}\frac{n}{\binom{n}{k}}\\&=\frac{n}{2}\sum_{k=0}^{n}\frac{1}{\binom{n}{k}},\end{align*}

as desired. $\square$

[b]Lemma 2: [/b] $\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}=\frac{n+1}{2n} \left(\sum_{k=0}^{n-1}\frac{1}{\binom{n-1}{k}}\right).$

$\emph{Proof: }$ Using some elementary factoring, we have:

\begin{align*}\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}&=\sum_{k=0}^{n-1}\frac{1}{\frac{n}{n-k}\binom{n-1}{k}}\\&=\sum_{k=0}^{n-1}\frac{\frac{n-k}{n}}{\binom{n-1}{k}}\\&=\sum_{k=0}^{n-1}\frac{1-\frac{k}{n}}{\binom{n-1}{k}}\\&=\left(\sum_{k=0}^{n-1}\frac{1}{\binom{n-1}{k}}\right)+\frac{1}{n}\left(\sum_{k=0}^{n-1}\frac{k}{\binom{n-1}{k}}\right).\end{align*}

Using lemma 1, the above simplifies to

\begin{align*}&\left(\sum_{k=0}^{n-1}\frac{1}{\binom{n-1}{k}}\right)+\frac{n-1}{2n}\sum_{k=0}^{n-1}\frac{1}{\binom{n-1}{k}}\\&=\frac{n+1}{2n}\sum_{k=0}^{n-1}\frac{1}{\binom{n-1}{k}},\end{align*}

as desired. $\square$

At this point, we proceed by induction on $n$. As a base case for $n=1$, we have $\frac{1}{\binom{1}{1}}=1=\frac{2}{2^2}\cdot \frac{2}{1}$ as desired. Proceeding to the inductive step, we assume for some $n=k\geq 1$ that $\sum_{j=0}^{n}\frac{1}{\binom{k}{j}}=\frac{k+1}{2^{k+1}}\sum_{j=0}^{k}\frac{2^j}{j}$, and we show that this is also the case for $n=k+1$:

By lemma 1, we have

\begin{align*}\sum_{j=0}^{k+1}\frac{1}{\binom{k+1}{j}}&=\frac{k+2}{2(k+1)}\sum_{j=0}^{k}\frac{1}{\binom{k}{j}}\\ &=\frac{k+2}{2(k+1)}\cdot \frac{k+1}{2^{k+1}}\sum_{j=0}^{k}\frac{2^j}{j}\quad \text{by the inductive hypothesis}\\&=\frac{k+2}{2^{k+2}}\sum_{j=0}^{k}\frac{2^j}{j}, \end{align*}

which completes the inductive step. Hence, we are done. $\blacksquare$

The crux of this problem is noting that if $S_n$ is the LHS, then we can rewrite $$2S_{n+1}=2\sum_{k=0}^{n+1} \frac{k!(n+1-k)!}{(n+1)!}=2+\frac{1}{n+1}\sum_{k=0}^{n} 2\frac{k!(n-k)!}{n!} (n+1-k)=2+\frac{1}{n+1}\left(\sum_{k=0}^{n} \frac{k!(n-k)!}{n!} (n+1-k)+\sum_{k=0}^{n} \frac{k!(n-k)!}{n!} (k+1)\right)=2+\frac{n+2}{n+1}S_n$$ after which the result is easy by induction

Similar to [b]math154[/b]'s second solution, but who cares :D

Recall from the beta integral that
\[
\binom{n}{k}^{-1} = \frac{k!(n-k)!}{n!} = (n+1)\int_0^1t^k(1-t)^{n-k}\,dt,
\]
and so
\begin{align*}
\sum_{k=0}^n\binom nk^{-1} &= (n+1)\int_0^1\sum_{k=0}^nt^k(1-t)^{n-k}\,dt = (n+1)\int_0^1(1-t)^n\sum_{k=0}^n\left(\frac{t}{1-t}\right)^k\,dt\\
                                                    &= (n+1)\int_0^1(1-t)^n\left(\frac{1 - (\frac{t}{1-t})^{n+1}}{1 - \frac{t}{1-t}}\right)\,dt = (n+1)\int_0^1\frac{(1-t)^{n+1} - t^{n+1}}{1-2t}\,dt.
\end{align*}
Furthermore,
\[
\frac{n+1}{2^{n+1}}\sum_{k=0}^n\frac{2^{k+1}}{k+1} = \frac{n+1}{2^{n+1}}\int_0^1\sum_{k=0}^n2^{k+1}t^k\,dt = \frac{n+1}{2^n}\int_0^1\frac{1-(2t)^{n+1}}{1-2t}\,dt.
\]
By substituting these integral representations into the original equality, it suffices to show that
\[
\int_0^1\frac{(1-t)^{n} - t^{n}}{1-2t}\,dt = \frac{1}{2^{n-1}}\int_0^1\frac{1-(2t)^n}{1-2t}\,dt\qquad(*)
\]
for each positive integer $n$.

This is surprisingly tricky.  The motivation is to observe that the right hand integrand in $(*)$ has a singularity at $t = \tfrac12$, hence can be written as the sum of two integrals
\begin{align*}
&\;\;\;\;\int_0^{1/2}\frac{1-(2t)^n}{1-2t}\,dt + \int_{1/2}^1\frac{1-(2t)^n}{1-2t}\,dt\\
&= \int_0^{1/2}\frac{1-(2t)^n}{1-2t}\,dt + \int_{0}^{1/2}\frac{1-(2(1-t))^n}{2t-1}\,dt\\
&= 2^n\int_0^{1/2}\frac{(1-t)^n - t^n}{1-2t}\,dt.
\end{align*}
But observe that the new integrand is symmetric about $t = \tfrac12$.  Thus, this integral also equals $2^{n-1}\textstyle\int_0^1\tfrac{(1-t)^n-t^n}{1-2t}\,dt$, which, upon dividing by $2^{n-1}$, is precisely the left hand side of $(*)$.

Here's a combinatorial solution that looks a lot like the E&M solution probably because Markov Chain transitions look pretty similar to Kirchhoff's Laws.

[quote=Problem]
On the hypercube $\{0, 1\}^n$, two length $n$ paths traveling along the edges from $(0, 0, \ldots, 0)$ to $(1, 1, \ldots, 1)$ are selected uniformly at random. Find the expected number of vertices that they share.
[/quote]

[b][color=red]Approach 1:[/color][/b] For each $0 \le k \le n$, there are exactly $\tbinom nk$ points $(x_1, x_2, \ldots, x_n)$ on the hypercube with $x_1 + x_2 + \ldots + x_n = k$, and there is an equal probability that a length $n$ path from $(0, 0, \ldots, 0)$ to $(1, 1, \ldots, 1)$ passes through any one of them (clearly it must pass through exactly one of them). Thus, the two paths intersect on the plane $x_1 + x_2 + \ldots + x_n = k$ with probability $\tbinom nk^{-1}$, i.e.
\[ \mathbb E[\text{intersections}] = \sum_{k=0}^n \mathbb P [\text{intersect on } x_1 + x_2 + \ldots + x_n = k] = \sum_{k=0}^n \binom nk^{-1}. \]

[b][color=red]Approach 2:[/color][/b] Let $A_n$ be the answer. Clearly the two paths always share the point $(0, 0, \ldots, 0)$; for the rest of the points, we casework on the second vertex of the two paths:
[list]
[*] [b][color=darkorange]Case 1: they are the same[/color][/b]. There is a $\tfrac1n$ chance that this happens. The rest of the paths start and end on opposite vertices of a $(n-1)$-dimensional hypercube, so they intersect an expected $A_{n-1}$ times.
[*] [b][color=darkorange]Case 2: they are different[/color][/b]. There is a $\tfrac{n-1}n$ chance that this happens. WLOG assume that the first path's second vertex is $(1, 0, 0, \ldots, 0)$ and the second path's second vertex is $(0, 1, 0, \ldots, 0)$. Then consider the paths induced on an $(n-1)$-hypercube by ignoring the first and second coordinates of the first and second paths, respectively. They intersect an expected $A_{n-1}$ times, and our original paths intersect when and only when these induced paths intersect at a point with first coordinate is $1$ (since both the first and second coordinate must be $1$ for the original paths starting at $(1, 0, 0, \ldots, 0)$ and $(0, 1, 0, \ldots, 0)$ to intersect).

But on average half of the intersections of two randomly chosen length $n-1$ paths from $(0, 0, \ldots, 0)$ to $(1, 1, \ldots, 1)$ on $\{0, 1\}^{n-1}$ will have first coordinate $1$ (this is by symmetry under the transformation $\mathbf{x} \mapsto (1, 1, \ldots, 1) - \mathbf{x}$), so our original paths intersect an expected $\tfrac12 A_{n-1}$ times.
[/list]
It follows that
\[ A_n = 1 + \frac1n A_{n-1} + \frac{n-1}n \cdot \frac12 A_{n-1} = 1 + \frac{n+1}{2n} A_{n-1} \implies \frac{A_n}{n+1} = \frac1{n+1} + \frac12 \cdot \frac{A_{n-1}}n.\]
Clearly $A_1 = 2$, so
\[ \frac{A_n}{n+1} = \frac1{n+1} + \frac12 \left(\frac1n + \frac12 \left( \frac1{n-1} + \left(\cdots \left( \frac13 + \frac12 \cdot \frac22 \right) \right) \right) \right) = \frac1{2^{n+1}} \left(\frac{2^{n+1}}{n+1} + \frac{2^n}n + \cdots + \frac{2^3}{3} + 2^2 \right).\]
Therefore,
\[ A_n = \frac{n+1}{2^{n+1}} \left( \frac{2}{1} + \frac{2^2}{2} + \cdots + \frac{2^{n+1}}{n+1} \right).\]

Comparing results, we are done. $\blacksquare$

We induct on $n$; base cases are obvious. For the inductive step, we upon simplification we want to show that$$\sum_{k=1}^n \frac 1{{n \choose k}} - \frac 1{{n+1 \choose k}} = \frac n{2^{n+2}} \sum_{k=1}^{n+1} \frac{2^k}k.$$
[b][color=#f00]Claim.[/b][/color] [Main Step] We have$$\frac 1{{n \choose k}} - \frac 1{{n+1 \choose k}} = \frac k{n+1} \cdot \frac 1{{n \choose k}}.$$
[i]Proof.[/i] Simply use Pascal.
\begin{align*}
\frac 1{{n \choose k}} - \frac 1{{n+1 \choose k}} &= \frac{{n+1 \choose k} - {n \choose k}}{{n \choose k}{n+1 \choose k}} \\
&= \frac{{n \choose k-1}}{{n \choose k}{n+1 \choose k}} \\
&= \frac{k \cdot (n-k)! k!}{(n+1)!} \\
&= \frac k{n+1} \cdot \frac 1{{n \choose k}},
\end{align*}as needed. $\blacksquare$

Now,
\begin{align*}
\sum_{k=0}^n \frac k{n+1}  \cdot \frac 1{{n \choose k}} &= \frac 1{n+1} \sum_{k=0}^n \frac k{{n  \choose k}} \\
&= \frac 1{2(n+1)} \sum_{k=0}^n \frac{k+n-k}{{n \choose k }} \\
&= \frac{n}{2(n+1)} \sum_{k=0}^n \frac 1{{n \choose k}} \\
&= \frac n{2(n+1)} \cdot \frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1} \frac{2^k}k \\
&= \frac n{2^{n+2}} \sum_{k=1}^{n+1} \frac{2^k}k,
\end{align*}as required.


modular arithmetic
algebra proposed
algebra

The sequence $ (a_{n})_{n\geq 1}$ is defined by $ a_{1} \equal{} 1, a_{2} \equal{} 3$, and $ a_{n \plus{} 2} \equal{} (n \plus{} 3)a_{n \plus{} 1} \minus{} (n \plus{} 2)a_{n}, \forall n \in \mathbb{N}$. Find all values of $ n$ for which $ a_{n}$ is divisible by $ 11$.

By subtracting, we have
\[a_{n+2}-a_{n+1}= (n+2) (a_{n+1}-a_{n}). \]
Iterating gives
\[a_{n+2}-a_{n+1}= (n+2)!. \]
By summing, we get
\[a_{n}= n!+(n-1)!+\cdots+1!. \]
So we have an explicit formula for $a_{n}$.  From here, the problem should be straightforward.

We could also look at the sequence modulo $11$, since it's clear that it's eventually periodic. According to Ravi B and Rust, in this case it seems that we are lucky and we don't have to perform that many computations before hitting the periodical part.

[quote="perfect_radio"]We could also look at the sequence modulo $11$, since it's clear that it's eventually periodic ...[/quote]
You are right but how we know without [b]Ravi B[/b] that we would be lucky before making 55 steps in the worst case instead of 11 ones?   :wink:

To be honest, there's no way I'd follow that method by hand at home (I'd use a computer instead).
But in contest, with $7$ points (or $10$?) guaranteed, I'd probably go all the way.

The bottom line is that the problem is really easy and doesn't say that much.

There is a reason why the sequence is constant (modulo 11) after 11 steps.  If we plug $n = 9$ into the recurrence, we see that $a_{11}\equiv a_{10}\pmod{11}$.  If we plug in $n = 10$, we see that $a_{12}\equiv a_{10}\pmod{11}$.  And so on.


modular arithmetic
number theory unsolved
number theory

Given a prime number $ p\neq 5$ and positive integers $ a,b,c$ such that $ p$ divides $ a\plus{}b\plus{}c$ and $ a^5\plus{}b^5\plus{}c^5$. Prove that

$ p$ divides $ (a^2\plus{}b^2\plus{}c^2)(a^3\plus{}b^3\plus{}c^3)$.

show that $ a^5\plus{}b^5\plus{}c^5$ is $ (a^2\plus{}b^2\plus{}c^2)(a^3\plus{}b^3\plus{}c^3)$ plus a multiple of $ a\plus{}b\plus{}c$...

[quote="pleurestique"]show that $ a^5 \plus{} b^5 \plus{} c^5$ is $ (a^2 \plus{} b^2 \plus{} c^2)(a^3 \plus{} b^3 \plus{} c^3)$ plus a multiple of $ a \plus{} b \plus{} c$...[/quote]

But I'm pretty sure that's not true..

[quote="pleurestique"]show that $ a^5 \plus{} b^5 \plus{} c^5$ is $ (a^2 \plus{} b^2 \plus{} c^2)(a^3 \plus{} b^3 \plus{} c^3)$ plus a multiple of $ a \plus{} b \plus{} c$...[/quote]
Let $ a \equal{} 1,b \equal{} 1,c \equal{} 2$.

Proof of original question:

Since $ a \plus{} b \plus{} c \equiv 0 \pmod{p}$ we have $ c \equiv \minus{} (a \plus{} b) \pmod{p}$.
Then $ 0 \equiv a^5 \plus{} b^5 \plus{} c^5 \equiv a^5 \plus{} b^5 \minus{} (a \plus{} b)^5 \equiv \minus{} 5ab(a^3 \plus{} b^3 \plus{} 2a^2b \plus{} 2b^2a) \pmod{p}$.
Since $ p \neq 5$, we get that $ ab(a \plus{} b)(a^2 \plus{} ab \plus{} b^2) \equiv 0 \pmod{p}$.

We have $ a^2 \plus{} b^2 \plus{} c^2 \equiv a^2 \plus{} b^2 \plus{} ( \minus{} a \minus{} b)^2 \equiv a^2 \plus{} ab \plus{} b^2 \pmod{p}$ so we are done if $ a^2 \plus{} ab \plus{} b^2 \equiv 0 \pmod{p}$.

If not, then by Euclid's lemma $ ab(a \plus{} b) \equiv 0 \pmod{p}$.

It is well known that $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \equiv 0 \pmod{p}$, so $ a^3 \plus{} b^3 \plus{} c^3 \equiv 3ab( \minus{} a \minus{} b) \equiv \minus{} 3\cdot 0 \equiv 0 \pmod{p}$.
Alternatively we could expand $ a^3\plus{}b^3\plus{}(\minus{}a\minus{}b)^3 \equal{} \minus{}3ab(a\plus{}b) \equiv 0 \pmod{p}$.

you're right. I took a quick glance at it and thought it was something else...


Suppose a hexagon $ABCDEF$, such that $BC+FE=AD$, $AB+ED=FC$ and $CD+FA=EB$. Then prove that the relation holds for such a hexagon:
$\frac{AF}{CD}=\frac{ED}{BC}=\frac{AB}{FE}$.

davron latipov

[color=red][size=75][will be waiting for the solution hopefully, sorry if I made a mistake somewhere...][/size][/color]

[quote="Davron"][color=darkred]Suppose a hexagon $ABCDEF$, such that $BC+FE=AD$, $AB+ED=FC$ and $CD+FA=EB$.
Then prove that the relation holds for such a hexagon: [/color]$\boxed{\frac{AF}{CD}=\frac{ED}{BC}=\frac{AB}{FE}}$.[/quote]Please, verify if is correctly the packed relation.

... also me, I was thinking that the relation is not true. Actually the problem was given to me by my friend when we were talking and walking ... so I don't remember the correct relation. Sorry !

davron latipov

I think much of what you can say follows from the equality case of http://www.mathlinks.ro/Forum/viewtopic.php?t=71336 ...

  darij

Yes, very nice! For the equality in Darij's proof we need:
1) opposite sides parallel in the hexagon.
2) $\vec{AB}+\vec{CD}+\vec{EF}=\vec{BC}+\vec{DE}+\vec{FA}=0$
So obviously $\frac{AB}{DE}=\frac{CD}{FA}=\frac{EF}{BC}$
This result is amazing. Do you feel like it's a bit similar to IMO2003/3?

Of course. It has been given at an Intl. contest:
http://www.mathlinks.ro/Forum/viewtopic.php?p=415159#p415159

Davron even posted in that topic :)

Yes you are right perfect radio but I had understood there nothing, since I don't know nothing about Vectors :( ... so I don't know what to do know.

anyway, thank you very much all
eagle ~ I think that it is amazing but I can't understand nothing... sorry

davron latipov


Prove that if G is the centroid of triangle ABC and AB + GC = AC + GB, then ABC is isoscles triangle.

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=121223[/url], which is just your problem "times three".


search
combinatorics unsolved
combinatorics

In a $ 15\times 15$ grid, we will in numbers $ 1,2,...,225$, each once. In each row, we color the five largest numbers red, and in each column, we color the five largest numbers blue.

Prove that there are at least $ 25$ numbers colored both red and blue.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205318
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=335000168&t=202012
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=335000168&t=230660

I found these by searching for the words "225 red blue."