trigonometry
projective geometry
geometry
trig identities
Law of Sines
angle bisector
geometry proposed

Given a triangle $ ABC$, let $ r$ be the external bisector of $ \angle ABC$. $ P$ and $ Q$ are the feet of the perpendiculars from $ A$ and $ C$ to $ r$. If $ CP \cap BA \equal{} M$ and $ AQ \cap BC\equal{}N$, show that $ MN$, $ r$ and $ AC$ concur.

[hide="Solution"]Denote I be the intersection of CM and AN. We have (PIMC)=-1, so MN,AC,PQ are concurrent.[/hide]

[quote="mr.danh"]... PM meets QN at I... [/quote] :huh:  :?:

[quote="Jutaro"]Given a triangle $ ABC$, let $ r$ be the external bisector of $ \angle ABC$. $ P$ and $ Q$ are the feet of the perpendiculars from $ A$ and $ C$ to $ r$. If $ CP \cap BA \equal{} M$ and $ AQ \cap BC \equal{} N$, show that $ MN$, $ r$ and $ AC$ concur.[/quote]
- $ AC\cap r \equal{} D$, sufficient to show that: $ M,N,D$ are colinear.
- $ AN\cap CM \equal{} U; BU\cap AC \equal{} E$.
Because:
$ \noindent \angle{PBA} \equal{} \angle{QBC}\Rightarrow \triangle{PBA}\sim\triangle{QBC}\Rightarrow \dfrac{QB}{PB} \equal{} \dfrac{QC}{PA} \equal{} \dfrac{QU}{AU}\Rightarrow EB\perp PQ\Rightarrow \angle{ABE} \equal{} \angle{CBE}.$
Then, because $ AN,CM,BE$ are concurrent, $ N,M,D$ are colinear if and only if $ (AECD)$ forms an harmonic division (lemma 1 [url]http://reflections.awesomemath.org/2007_4/harmonic_division.pdf[/url]) and it's trivial (lemma 2 [url]http://reflections.awesomemath.org/2007_4/harmonic_division.pdf[/url]).

suppose without loss of generality that $ C > A$.. and denote by $ T$ the intersection of $ AC$ and $ r$... we have that $ C$ lies between $ A$ and $ T$...

we can prove that $ \dfrac{\sin BTN}{\sin CTN} \equal{} \dfrac{\sin BTM}{\sin CTM} \equal{} \dfrac{1}{2\cos\frac {A \minus{} C}{2}}$, and we're done...

*this is not immediate, i mean, it requires a trig ceva and some law of sines, but it's easy..

I cant speak and write english
The Desargues solution is the most popular
We have to prove that APM and CQN are in perspective, and we know that AP//CQ, therefore we have to prove that if X is the intersection of PC and AQ  BX is the angle bisector of <ABC o BX//CQ//AP.
Let Y the intersection of CQ and AB.
APB y CQB are similar then: AB/BC=AP/CQ
APX y CQX too: AB/BC=AX/QX
But CBY is isosceles and then BC=BY, AB/BC=AB/BY=AX/QX y BX//CQ.

[quote="Concyclicboy"]The solution based on the [b][size=100]Desargues theorem[/size][/b], is the most popular.
We have to prove that $ \bigtriangleup APM$ and $ \bigtriangleup CQN$ are perspective, and we know that $ AP\parallel CQ,$ therefore we have to prove that $ BX\parallel CQ\parallel AP,$ where $ X\equiv PC\cap AQ.$
Let $ Y$ be, the intersection of $ CQ$ and $ AB.$
$ \bigtriangleup APB,\ \bigtriangleup CQB$ are similar and then: $ \frac {AB}{BC} \equal{} \frac {AP}{CQ}.$ Also, $ \bigtriangleup APX,\ \bigtriangleup CQX$ are similar and then: $ \frac {AB}{BC} \equal{} \frac {AX}{QX}$
But $ \bigtriangleup CBY$ is isosceles and then $ BC \equal{} BY,$ and $ \frac {AB}{BC} \equal{} \frac {AB}{BY} \equal{} \frac {AX}{QX}$ $ \Longrightarrow$ $ BX\parallel CQ\parallel AP.$[/quote]
I like this solution dear [b][size=100]conyclicboy[/size][/b] and let me to try a bit simplification.

From the similar right triangles $\bigtriangleup PAB,\ \bigtriangleup QCB,$ we have also $\frac {AP}{CQ} \equal{} \frac {BP}{BQ},\ (1)$

From $AP\parallel CQ,$ by [b][size=100]Thales theorem[/size][/b], we have that $\frac {AP}{CQ} \equal{} \frac {XP}{XC},\ (2)$

From $ (1),\ (2),\ \Longrightarrow$ $\frac {BP}{BQ} \equal{} \frac {XP}{XC}\ \ ,(3)\ \Longrightarrow$ $ BX\parallel CQ\parallel AP.$

Best regards, Kostas vittas.

Why does the intersection of PC with QA being in the angle bisector imply the triangles are in perspective?

oh wait oops... for a second there I dont know why i read concyclicboy was trying to prove that PXA and QXC were in perspective...

[quote="Concyclicboy"]
We have to prove that $ \bigtriangleup APM$ and $ \bigtriangleup CQN$ are perspective, and we know that $ AP\parallel CQ,$ therefore we have to prove that $ BX\parallel CQ\parallel AP,$ where $ X\equiv PC\cap AQ.$
[/quote]

I was very confused with this.
I think instead we have to show
$ \bigtriangleup ABC$ and $ \bigtriangleup PXQ$ in perspective.
Thus when $BX \parallel AP \parallel CQ$ is proved,
descargues theorem shows that due to
$AB \cap PX = M$, $AC \cap PQ = D$ and $BC \cap XQ =N$,
 $MDN$ is collinear.

Dear Mathlinkers,
an alternative synthetic proof:
1. X the point of intersection of AP and BC,  Y of CQ and BC
2. by Pappus theorem applied to the hexagon PAMNCQP inscribed in the lines AQ and CR, XY, PQ and MN are concurrent .
3. P is the midpoint of AX, Q the midpoint of CY
4.  by considering the complete trapez  XACY, we are done.
Sincerely
Jean-Louis

[hide="My Solution using Menelaus"] Let $K$ be the intersection of $r$ and $AC$
Since $\bigtriangleup APB$ and $\bigtriangleup CBQ$ are similar we have: 1) $\frac{AP}{BP}=\frac{CQ}{BQ}$.
Then because $AP \parallel CQ$ by Thales in $\bigtriangleup KQC$ : 2) $\frac{AK}{CK}=\frac{AP}{CQ}$
Then by Law of Sines in $\bigtriangleup AMP$ and $\bigtriangleup PMB$
$\frac{AP}{\sin\angle PMA}=\frac{AM}{\sin\angle APM}$  and  $\frac{BP}{\sin\angle PMB}=\frac{BM}{\sin\angle BPM}$
Dividing these two relations and because $\sin\angle PMB=\sin\angle PMA$ (they sum 180) we get: $\frac{AM}{BM}= \frac{AP\sin\angle APM}{BP\sin\angle MPB}$  but $\angle APM+\angle MPB=90$ so $\frac{\sin\angle APM}{\sin\angle MPB}= \cot\angle MPB=\cot\angle CPQ=\frac{PQ}{CQ}$ so 3) $\frac{AM}{BM}=\frac{AP * PQ}{BP * CQ}$
By same arguments in $\bigtriangleup BQN$ and $\bigtriangleup CQN$ we have $\frac{BN}{CN}=\frac{AP*BQ}{CQ*PQ}$
$\Rightarrow\frac{AM}{BM} * \frac{BN}{CN} * \frac{CK}{AK} = \frac{AP}{BP} * \frac{PQ}{CQ} * \frac{AP}{PQ} * \frac{BQ}{CQ} * \frac{CK}{AK}=1$
Therefore by Menelaus Theorem: $M$,$N$ and $K$ are collinear. So lines $r$, $MN$ and $AC$ concur.[/hide]

Let $S = r \cap AC $, then, let $D$ be the intersection of the angle bisector of $ \angle ABC $ and $AC$, then is well known that $BD$ is parallel to $AP$ and $CQ$ (since the three are perpendicular to $r$)

Then, the problem can be read as "Prove that $AB$, $CP$ and $SN$ concur at $M$"

Now let's suppose that $N'=SM \cap BC$, we want to prove $N'=N$

By Ceva Theorem, we have that $\frac{SA}{AC} \cdot \frac{CN'}{N'B} \cdot \frac{BP}{PS} = 1$

Then, by Menelaus Theorem with $\triangle SBC$ and points $A$, $N$ and $Q$ we have that

 $\frac{SA}{AC} \cdot \frac{CN}{NB} \cdot \frac{BQ}{QS} = 1$, then,  we have  $ \frac{CN'}{N'B} \cdot \frac{BP}{PS} =  \frac{CN}{NB} \cdot \frac{BQ}{QS} $ so, it's clear that if we show that $\frac{BP}{PS} =\frac{BQ}{QS} $ then we are done, but, that would mean that $S$, $P$, $B$ and
$Q$ are on harmonic division, but that is true; let $O$ be the "ideal point" (the infinite one) Then it is clear that this happens: $SADC\frac{O}{\barwedge} SPBQ$, but,  $S$, $A$, $D$ and $C$ is a well known harmonic division (Internal and external angle bisector), so we are done


Mary jogs 16 miles in 2 hours, 24 minutes. On average, how many minutes does it take her to jog one mile?

$ 2$ hours, $ 24$ minutes is $ 2\times60 \plus{} 24 \equal{} 144$ minutes.
Thus $ 144 \div 16 \equal{} \boxed{9} \ \text{minutes}$.


algebra
polynomial
function
linear algebra
matrix

[hide]

moving everything to one side, you get

$P(2) -5=0, P(3)-10 =0, P(5)-26=0$ and etc

now you can see the general rule of thumb for this polynomial for any $x$

$P(x) - (x^2+1) = 0 \Rightarrow P(x) = x^2+1$

so $P(1) = 2$

[/hide]

I don't get how you came up with the rule $\displaystyle P(x) - (x^2+1) = 0 \Rightarrow P(x) = x^2+1$.  Is it just trial and error, or was there a method.

It looks like intuition/instict, you can't really trial and error that method. He saw the pattern first so that's what he went for.

How can $P(x)=x^2+1$ if it is a FIFTH DEGREE POLYNOMIAL?

I didn't see the x^2+1 pattern until others pointed it out, but I know how to do it now that I do see it...I think.
[hide]$P(x)-x^2-1$ has roots 2, 3, 5, -1, and -4, so it can be written as (x-2)(x-3)(x-5)(x+1)(x+4), so $P(x)=(x-2)(x-3)(x-5)(x+1)(x+4)+x^2+1$. Then P(1) is (-1)(-2)(-4)(2)(5)+1+1=-78.[/hide]

oh shoot i forgot about the 5th degree polynomial thing ugh

[quote="xxreddevilzxx"][quote="LynnelleYe"]$P(x)-x^2-1$ has roots 2, 3, 5, -1, and -4, so it can be written as (x-2)(x-3)(x-5)(x+1)(x+4), so $P(x)=(x-2)(x-3)(x-5)(x+1)(x+4)+x^2+1$. [/quote]

i dont get that last part.[/quote]

Since $P(x)$ follows the general rule for $x^2 + 1$ for the values given,

$P(x) - (x^2 + 1)$ must be equal to zero for the given values.

Then 2, 3, 5, -1, and -4 are the roots of that polynomial, the values of $x$ that satisfy the equation $P(x) - (x^2 + 1) = 0$.

$\therefore P(x) - (x^2 + 1) = (x - 2)(x - 3)(x - 5)(x + 1)(x + 4)$

[quote="LynnelleYe"]$P(x)-x^2-1$ has roots 2, 3, 5, -1, and -4, so it can be written as (x-2)(x-3)(x-5)(x+1)(x+4), so $P(x)=(x-2)(x-3)(x-5)(x+1)(x+4)+x^2+1$.[/quote]
remember you guys disregarded one important condition BUT it so happened you guys got the correct solution. Bob said $P(x)$ is not some fifth degree polynomial, but it is a [b]monic[/b] fifth degree polynomial
in her solution, it should be $P(x)=a(x-2)(x-3)(x-5)(x+1)(x+4)+x^2+1$, where $a$ is some constant. but since it is monic, $a=1$

[quote="Bob"]$P(x)$ is a monic fifth degree polynomial such that $P(2)=5$, $P(3)=10$, $P(5)=26$, $P(-1)=2$, $P(-4)=17$. Find $P(1)$. NO CALCULATORS ALLOWED[/quote]
Since $P(x)=x^2+1$ for $x=2, 3, 5, -1, -4$, $Q(x)=P(x)-x^2-1$ has roots at those same values. Since $P(x)$ has degree $5$, $Q(x)$ must have a degree of $5$. Thus, we can write it as $Q(x)=C(x-2)(x-3)(x-5)(x+1)(x+4)$ for some constant $C$, since we already know its factorization.
Then $P(x)=C(x-2)(x-3)(x-5)(x+1)(x+4)+x^2+1$, but it is a monic polynomial, so $C=1$.
And so finally, we have $P(1)=(1-2)(1-3)(1-5)(1+1)(1+4)+1+1=-78$.

Well, I took it for granted everyone could tell why the coefficient of the big thing was 1, but I guess some of my statements weren't very rigorous because of that.

I think we can do like that:
Assume $P(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ then plug the numbers in. Is that possible?

so there are 5 equations and 6 variables, is there a way to solve it?

However, since Bob said it was a monic fifth degree polynomial, $a=1$, so you have five variables with five equations, and so it is solvable.

[quote="nolachrymose"]However, since Bob said it was a monic fifth degree polynomial, $a=0$, so you have five variables with five equations, and so it is solvable.[/quote]

Okay, I was confused about that word monic.

Multiply the P(-1) function by -16 and add it to the P(2) function to cancel the b.  Continue doing this with the others until you have 4 equations with only 4 variables.  This will work to bring it down till you can find d.
The other variables are then easily solvable and you can find P(1).

Okay... after spending some time working on it in that vein, I have to say that the intuitive way is how it is meant to be solved. Otherwise, you need to work out a huge number of different equations to be able to simplify down to find just one variable...

[quote="chess64"]What does monic mean? :?[/quote]
monic means the leading coefficient of the polynomial is equal to 1

[quote="236factorial"]I would've done it like how JesusFreak197 done it, but yes, 6 variables won't work out with just 5 equations,[/quote]
Well, there is something called diophantine equations where you can solve everything in terms of something else.

[quote="JesusFreak197"]Oh, okay, so it simplifies to:

$P(2)=16b+8c+4d+2e+f=-27$
$P(3)=81b+27c+9d+3e+f=-233$
$P(5)=625b+125c+25d+5e+f=-3099$
$P(-1)=b-c+d-e+f=3$
$P(-4)=256b-64c+16d-4e+f=1041$

I'm still not sure what to do from here... I can't find an easy way to cancel out variables.[/quote]

There is something very useful called Gaussian Elimination that a mathematician should never leave home without. So go [url=http://mathworld.wolfram.com/GaussianElimination.html]learn[/url] about it.

Umm... as far as I can tell, that's what I was using already.

Or the better way, write it in matrix form and premultiply by the inverse of the matrix. You can find some websites that will invert large matrcies for you

If this type of question appears on a test where calculators are allowed, such as the AMC-12, you can just use "rref" under matrix-math.


1. Let ABC be a right-angled triangle with right angle at C. Pick a point D on the segment
BC. Let E be a point on the circumcircle x of ABD, such that DE is perpendicular to AB.
Prove that \BAE = \BEA if and only if AC is tangent to x.

5. Determine the minimum possible value of the expression |n^2−5^(4m+3)| for non-negative integers
m and n.

6. Michael’s mother likes to keep him busy with an odd form of solitaire. To set up the game she
places coins on some of the squares of a normal 8×8 chessboard. Michael plays by adding one
coin at a time, always placing coins only on squares which already have at least two adjacent
squares containing coins. (Two squares are adjacent if they share an edge, but not if they only
share a vertex.)
Michael wins when he’s placed a coin on every square of the board. What is the minimum
number of coins that Michael’s mother can place on the board to start with, so that it is still
possible for Michael to win?

These questions are NZ mathematics olympiad selection problems
I dont think that you are supposed to put these questions up like this

Yeah please don't do that ... it's not cool.


geometry
rectangle
rotation
combinatorics proposed
combinatorics

A table $ 7\times 7$ is overlapped by $ 16$ rectangles $ 1\times 3$ such that still $ 1$ cell is not yet overlapped. Find this cell!

Label the unit-squares with numbers $ 1,2,3,1,2,3,1$ for the upper row, then $ 2,3,1,2,3,1,2$ for adjacent row and so on, so that the four corner squares of the table are with label $ 1$ and any rectangle $ 1 \time 3$ covers exactly one square for each number $ 1,2,3$.
Since there are one more $ 1$-square than for the two thers numbers, the non-covered cell must have number $ 1$.
Now, any tiling of the table gives another tiling when rotate of $ 90$ degrees, thus when rotating the non-covered cell must be transformed into another cell with number $ 1$. It is easy to see that the only such cells are the corners of the table, the cells at the middle of an edge of the table and the center of the table.
Conversely, for each of he cells, it is easy to find an appropriate covering.

Pierre.


Find a set of infinite positive integers $ S$ such that for every $ n\ge 1$ and whichever $ n$ distinct elements $ x_1,x_2,\cdots, x_n$ of S, the number $ x_1\plus{}x_2\plus{}\cdots \plus{}x_n$ is not a perfect square.

[hide]3, 12, 48, 192, ... 3(4^n). If the first term is used, then a sum is equivalent to 3 mod 4. All perfect squares are equivalent to 1 mod 4 or 0 mod 4. If the second term is used, but not the first, then it is equivalent to 4(n) where n is equivalent to 3 mod 4. This can be factored into (2^2)n and n is not a perfect square. This process can be repeated indefinitely.[/hide]

We can use induction. Take the set that worked for $ k$ numbers and add another number $ m$ large enough such that both $ m$ and $ m$ plus all the $ k$ other numbers are in between two (large) consecutive squares. Now we have a set for $ k \plus{} 1$ numbers that works, because if the sum doesn't use $ m$, it can't be a square, but if it does, it must be between two consecutive squares and thus can't be a square.

For example, suppose we have the set $ {3,5,7}$ so far. Note that $ 100^2 \equal{} 10000$ and $ 101^2 \equal{} 10201$. So we can add $ 10100$ to the set, because both $ 10100$ and $ 10100 \plus{} 3 \plus{} 5 \plus{} 7 \equal{} 10113$ are between two consecutive squares. So if the sum uses only the numbers $ 3,5,7$, it can't be a square, but if it uses $ 10100$, it must be between two squares and thus can't be a square.

By the way, this is only a sketch, you can make it rigorous.

1+2+3+4+5...+n=(n)(n+1)/2
2(1+2+3+4+5...+n)=(n)(n+1)
n(n+1) will never be a perfect square
so we have (2, 4, 6, 8, 10..)
as our set
How do you dem them apples!  :D

Rob: But what if we take the subset 2,4,10? Then we get 16!

:maybe:
Ah, I see. I thought the numbers are taken in order and you cannot take a subset

the sequence $a_{n}=(a_{n-1}+1)(a_{n-1})$ works as well.

I think I have a particularly simple solution though is not a handy solution (meaning  its essence  is not applicable in many other problems, as the argument from @dgreenb801 is, cool idea btw makes a lot of sense approaching it this way to be honest). Consider the set $A=\{10, 1000, 100000, \cdots, 10^{2n+1} , \cdots \}$. Notice that no matter the choice of a subset of $A$, the sum of all the elements in the subset must end in an odd number of zeroes, hence it cannot be a perfect square.


modular arithmetic
number theory proposed
number theory

Show that there is no positive integer $ n$ for which $ 2^n \plus{} 1$ is divisible by $ 7$.

 [hide="My Proof"]

Assume $ 2^n \plus{} 1$ is divisible by $ 7$. This means
$ 2^n \plus{} 1 \equiv 0 \pmod 7 \Rightarrow 2^n \equiv \minus{} 1 \pmod 7 \Rightarrow 2^n \equiv 6 \pmod 7$.

Now consider the first few positive powers of $ 2 \pmod 7$.

$ 2^0 \equiv 1 \pmod 7$
$ 2^1 \equiv 2 \pmod 7$
$ 2^2 \equiv 4 \pmod 7$
$ 2^3 \equiv 1 \pmod 7$
$ 2^4 \equiv 2 \pmod 7$
$ 2^5 \equiv 4 \pmod 7$

and so on and so forth. We prove this is a periodicity by seeing that $ 2^{n\plus{}3} \minus{} 2^{n} \equal{} 2^n(2^3 \minus{} 1) \equal{} 7 * 2^n \equiv 0 \pmod 7 \Rightarrow 2^{n\plus{}3} \equiv 2^n \pmod 7$.
Notice the degree of this periodicity is $ 3$, and that none of the elements include $ 6$. Hence, $ 2^n$ is only congruent to $ 1, 2,$ or $ 4 \pmod 7$ for a positive integer $ n$. This contradicts the original assumption that $ 2^n \equiv 6 \pmod 7$, which concludes there is no positive integer $ n$ for which $ 2^n \plus{} 1$ is not divisible by $ 7$.[/hide]

Is there a more rigorous way to prove this aside from noticing that pattern? Specifically, how would we formally prove there is a periodicity?

EDIT: changed the proof a bit thanks to Mathias_DK :)

[quote="ark_xm"]Show that there is no positive integer $ n$ for which $ 2^n \plus{} 1$ is divisible by $ 7$.

 [hide="My Proof"]

Assume $ 2^n \plus{} 1$ is divisible by $ 7$. This means
$ 2^n \plus{} 1 \equiv 0 \pmod 7 \Rightarrow 2^n \equiv \minus{} 1 \pmod 7 \Rightarrow 2^n \equiv 6 \pmod 7$.

Now consider the first few positive powers of $ 2 \pmod 7$.

$ 2^0 \equiv 1 \pmod 7$
$ 2^1 \equiv 2 \pmod 7$
$ 2^2 \equiv 4 \pmod 7$
$ 2^3 \equiv 1 \pmod 7$
$ 2^4 \equiv 2 \pmod 7$
$ 2^5 \equiv 4 \pmod 7$

and so on and so forth. We notice the degree of this periodicity is $ 3$, and that none of the elements include $ 6$. Hence, $ 2^n$ is only congruent to $ 1, 2,$ or $ 4 \pmod 7$ for a positive integer $ n$. This contradicts the original assumption that $ 2^n \equiv 6 \pmod 7$, which concludes there is no positive integer $ n$ for which $ 2^n \plus{} 1$ is not divisible by $ 7$.[/hide]

Is there a more rigorous way to prove this aside from noticing that pattern? Specifically, how would we formally prove there is a periodicity?[/quote]
Periodicity is easily proved by $ 2^{n \plus{} 3} \minus{} 2^n \equal{} (2^3 \minus{} 1)2^n \equal{} 7 \cdot 2^n$ so $ 2^{n \plus{} 3} \equiv 2^n \bmod 7$

Edit: I just found a fun way to prove it too :). Assume $ n$ is the smallest non-negative integer such that $ 7 \mid 2^n \plus{} 1$.(It's easy to see that $ n \ge 3$) Then we have:
$ 7 \mid 2^n\plus{}1$ so $ 7 \mid 2^n\minus{}6 \equal{} 2(2^{n\minus{}1}\minus{}3)$ so $ 7 \mid 2^{n\minus{}1} \minus{} 3$ So $ 7 \mid 2^{n\minus{}1} \plus{} 4 \equal{} 4(2^{n\minus{}3}\plus{}1)$ So $ 7 \mid 2^{n\minus{}3}\plus{}1$.
But then $ 2^{n\minus{}3}\plus{}1$ is also divsible by $ 7$, and so $ n$ is not there exists no smallest integer $ n$ such that  $ 7 \mid 2^n \plus{} 1$, and therefore there exists none at all :)

[quote="t0rajir0u"][b]Extension:[/b]  For what values of $ m, n$ is it true that $ 2^m \minus{} 1$ does not divide $ 2^n \plus{} 1$?  What if we require that they have no common factors?[/quote]
Because $ (2^n \minus{} 1,2^n \plus{} 1) \equal{} 1$ we know:
$ 2^{(m,2n)} \minus{} 1 \equal{} (2^m \minus{} 1, (2^n \minus{} 1)(2^n \plus{} 1)) \equal{} (2^m \minus{} 1,2^n \minus{} 1) \cdot (2^m \minus{} 1,2^n \plus{} 1)$ And isolation gives:
$ (2^m \minus{} 1,2^n \plus{} 1) \equal{} \frac {2^{(m,2n)} \minus{} 1}{2^{(m,n)} \minus{} 1}$.

If they have no common factor: $ (2^m \minus{} 1,2^n \plus{} 1) \equal{} 1$ $ \iff$ $ 2^{(m,2n)} \minus{} 1 \equal{} 2^{(m,n)} \minus{} 1$ $ \iff$ $ (m,2n) \equal{} (m,n)$.
Thus they have no common factor if and only if: For any power of $ 2$ which divides $ m$ it also divides $ n$.


Bob is using a rope tied to his tractor to remove an old tree stump from a field.  Which method given below will result in the greatest force applied to the stump?  Assume that the tractor will exert the same amount of force using either method.  Explain your answer.

	Method A: Tie the rope to the stump and pull

	Method B: Tie one end to the stump and the other end to a nearby pole.  Then pull on the rope perpendicular to the pole at a point about halfway between the two.

[hide]After we turned this in, the teacher claimed that the tractor exerted the same amount of force on the stump in method B and the poll also exherted force therefore making the force on the stump greater.  But I think that this argument seems like of fishy.  For example, wouldn't the tractor also pull on the poll, reducing the amount of force on the stump?[/hide]

If you have a solution (especially if you are in AP Physics or something), could you write it up?  As a freshman, I know no physics.

The force exerted in method B is a function of the angle that the rope is bent.

[hide="A sketch"] Suppose the tractor exerts a force $F$, and suppose that the angle between the direction of $F$ and the rope is $\theta$.  Let the tension in the rope be $T$.  Then, assuming the tractor pulls with constant velocity,

$F = 2T \cos \theta \implies$
$T = \frac{F}{2 \cos \theta}$

In the first method we simply have $T = F$.  We have equality of the two methods iff $\theta = 60^{\circ}$.

Edit: if $\theta > 60^{\circ}$ then $\text{Method B}> \text{Method A}$, and if $\theta < 60^{\circ}$ then $\text{Method B}< \text{Method A}$, so the original problem was ill-posed.  In fact, if $\theta$ is very close to $90^{\circ}$ then $T$ becomes arbitrarily big.  (Which means that probably the rope will break.) [/hide]


\begin{eqnarray)
Ca & = & 40.08\\
C & = & 12.01\\
0_{3}& = & 16.00 \cdot 3\\
nonumber & = & 100.09 \mbox g in 1 mole of CaCO_{3}.
\end{eqnarray}

the text won't appear on my paper, please help!

And also, How do I make 2x3 table? Can you give me the code to copy and paste? thanks!

the table:

Chalk       Mass

Original     2.44g

Afterward   2.39g

1. It is vital you look at the warnings that LaTeX tells you about. If you are using TexnicCenter pressing on the ! on the top right of the toolbar gives you the line the error is on. There are 3 main errors:
a) \begin{eqnarray) must end in } not )
b) nonumber is a command so should be \nonumber
c) you can't use _ (which is for mathmode) inside \mbox which is for text. Instead of CaCO_{3} you would need to use CaCO$\$$_{3}$\$$

But let's do the whole thing properly, eqnarray is no longer used and has been replaced by align which gives better results - you must have \usepackage{amsmath} but I would hope you do anyway. Use this code:
[code]\begin{align}
Ca & = 40.08\\
C & = 12.01\\
0_{3} & = 16.00 \cdot 3\\
& = 100.09 \text{ g in 1 mole of CaCO$_{3}$}\notag.
\end{align}[/code]

2. Tables are simple to do with [url=http://www.g32.org/latable/]LaTable[/url]. You can choose any size table, type in the text you want and it generates the LaTeX code ready to copy and paste. You can use it to add borders which makes it all trivial.

thank you so much! I love creating tables!!!


I'm considering a few private high schools, and one of the things I am looking for is the number of AP courses they offer.  One of the schools offers AP french lit and AP french lang, while the other offers AP french 4 and AP french 5.  I don't get the difference between these.  Can someone clarify?  Thanks.

Also, from what math level could you start taking APs?

I believe French 4 is Language and French 5 is Literature.  Don't quote me on that, though.  Look it up on the College Board website.

AP Statistics and AP Calculus AB are the two lowest AP math courses.

AP French Language and AP French Literature are the official names of the AP exams. "French 4" and "French 5" sound like local names - the high school's own names for its courses. One might reasonably guess that those correspond to the exams, but that would be a guess.

And the highest AP course available in math is Calculus BC, which corresponds to a "Calculus II" or second-semester calculus course at many semester-system colleges.


limit
trigonometry
calculus
calculus computations

Let $ x \equal{} r \cos \theta$, $ y \equal{} r \sin \theta$.
$ \lim_{(x, y) \to (0, 0)} \frac {\sin (x^2 \plus{} y^2)}{x^2 \plus{} y^2} \equal{} \lim_{r \to 0} \frac {\sin r^2}{r^2} \equal{} \boxed{1}$.


Hi,

I am not sure why these two functions different: $ f(x) \equal{} \sqrt{\frac{2x\minus{}5}{x\minus{}8}}$ and $ g(x) \equal{} \frac{\sqrt{2x\minus{}5}}{\sqrt{x\minus{}8}}.$ Can you just rewrite $ g(x)$ like the way $ f(x)$ is written? So, why are they different?

Thank you very much for the help.

Hmmm..

Rules of surds say that $ \frac{\sqrt{a}}{\sqrt{b}} \equal{} \sqrt{\frac{a}{b}}$

So I think that they are the same function.

Question: When we deal with square root functions, do we only consider the positive root?

KrazyFK, that's what I thought at first too, but that problem was from Intermediate Algebra, that the Solution says that they are different. To be more specific, for $ f(x),$ the whole block of fraction, namely $ \frac {2x \minus{} 5}{x \minus{} 8} \ge 0.$ But for $ g(x),$ the number and the denominator had to be satisfied separately. (And this by the way is a real-valued function.)

EDIT: mathemagician, I think you posted while I was typing, so I didn't see your post. But yes, the function is real-valued, so only the positive roots are considered.

Well, it does make sense the way they put it but still intuitively it isn't all that clear how they are could be different. Like they are after all the same thing right?

They are different because they are defined over different domains.

When you are working with functions of real variables, say $ f: \mathbb{R}\rightarrow\mathbb{R}$, the square root of any negative number is not defined. $ f$ and $ g$ in your case are clearly different once you analyze carefully the domains. Note that $ \sqrt{P(x)}$ exists if $ P(x)\ge0$.

In g(x), both 2x-5 and x-8 have to be positive or zero so you can take the square roots.

However, in f(x), if both 2x-5 and x-8 are negative, then ((2x-5)/(x-8)) is positive, and so you can take the square root of ((2x-5)/(x-8)). But you cannot take the square root of 2x-5 and the square root of x-8 and then divide them because 2x-5 and x-8 are negative numbers.

A function is like a recipe. It's a rule that is applied to a certain set of ingredients. Even if the "rule" is the same, it's a different recipe if the "ingredients" are not the same.

So since the functions above have different domains (or ingredients) they are not the same function.

Another example: The functions $ f(x) \equal{} x$ and $ g(x) \equal{} (\sqrt{x})^2$ are different functions - even though they agree for all non-negative numbers. (Note also that $ f(x) \equal{} x$ and $ h(x) \equal{} \sqrt{x^2}$ are not the same function. Do you see why?)

[quote="tornado.adv4"]In g(x), both 2x-5 and x-8 have to be positive or zero so you can take the square roots.[/quote]$ x \minus{} 8$ can be zero?

@Anyone: Find the domains of $ f$ and $ g$.

$ f(x) \equal{} \sqrt {\frac {2x \minus{} 5}{x \minus{} 8}}$

Isn't the domain $ x > 8$  or $ x < \frac {5}{2}$ (giving us negative on both top and bottom $ \Rightarrow$ a positive)?

$ g(x) \equal{}\frac{\sqrt{2x\minus{}5}}{\sqrt{x\minus{}8}}.$

Then isn't the domain $ x >8$?

[quote="limac"]Hi,

I am not sure why these two functions different: $ f(x) \equal{} \sqrt {\frac {2x \minus{} 5}{x \minus{} 8}}$ and $ g(x) \equal{} \frac {\sqrt {2x \minus{} 5}}{\sqrt {x \minus{} 8}}.$ Can you just rewrite $ g(x)$ like the way $ f(x)$ is written? So, why are they different?

Thank you very much for the help.[/quote]

Look at x=0 for example. f(x) is undefined, but g(x)=sqrt(5/8)

[quote="Mewto55555"][quote="limac"]$ f(x) \equal{} \sqrt {\frac {2x \minus{} 5}{x \minus{} 8}}$ and $ g(x) \equal{} \frac {\sqrt {2x \minus{} 5}}{\sqrt {x \minus{} 8}}.$[/quote]

Look at x=0 for example. f(x) is undefined, but g(x)=sqrt(5/8)[/quote]I think you meant $ f(0)\equal{}\sqrt{\frac58}$, $ g(0)$ is undefined.

@Evolved: $ g$ is correct. Your answer for the domain of $ f$ is almost correct. See if you can catch that.

Since $ \sqrt{\frac{2x\minus{}5}{x\minus{}8}} \equal{} \frac{\sqrt{2x\minus{}5}}{\sqrt{x\minus{}8}},$ what if we re-wrote $ g(x)$ as from $ g(x) \equal{} \frac{\sqrt{2x\minus{}5}}{\sqrt{x\minus{}8}}$ to $ g(x) \equal{} \sqrt{\frac{2x\minus{}5}{x\minus{}8}}?$ Would that be "illegal" in any case?

You would have to change the domain to re-write it like that.

[quote="limac"]Since $ \sqrt {\frac {2x \minus{} 5}{x \minus{} 8}} \equal{} \frac {\sqrt {2x \minus{} 5}}{\sqrt {x \minus{} 8}},$ what if we re-wrote $ g(x)$ as from $ g(x) \equal{} \frac {\sqrt {2x \minus{} 5}}{\sqrt {x \minus{} 8}}$ to $ g(x) \equal{} \sqrt {\frac {2x \minus{} 5}{x \minus{} 8}}?$ Would that be "illegal" in any case?[/quote]"Legal" if you are practicing algebraic simplifications, "illegal" if you want to properly follow the 'rigorous' path. So, in general, it is "illegal"  :D

Oh   :P

The numerator can be equal to zero so the domain of

$ f(x) \equal{}\sqrt{\frac{2x\minus{}5}{x\minus{}8}}$

Is $ x \le \frac{5}{2}$ or $ x > 8$

Ok, that clarifies it. Interesting concept. At a first glance, it looks like they are the same, but deeper analysis proves otherwise. Love these problems :P

Thank you very much for the replies.


Suppose $A\in M_n(\mathbb C)$, and $A=[a{}_i{}_j]$. Suppose $|a{}_i{}_i|>\sum_{j\neq i}|a{}_i{}_j|$. Prove that $A$ is invertible.

Assume there is a non-zero $(x_n) \in \mathbb{C}^n$ with $Ax = 0$. Let $i$ be an index such that $|x_i|$ is maximal among the $|x_j|$. Then $|x_{ii}| = |\sum_{j\neq i}a_{ij}\frac{x_j}{x_i}| \leq \sum_{j\neq i}|a_{ij}|$, contradiction.

This lemma can be used in some proof of the Dirichlet unit thm.


Given $ a,b,c\ge 0$

Show

$ \frac{2ab}{a^2\plus{}b^2\plus{}c^2\plus{}4ab} \plus{}\frac{2bc}{a^2\plus{}b^2\plus{}c^2\plus{}4bc}\plus{}\frac{2ca}{a^2\plus{}b^2\plus{}c^2\plus{}4ca} \le 1$

Yeah; this doesn't seem very sharp.  :wink: Note that chrischris's solution also implies that equality never holds.


If the graph of $f(x)$ and the graph of $g(x)=e^x$ are symmetric about(which prep. is better?) the point $(0,1)$, then $f(x)=$______
(A) $\ln x-1$
(B) $2-e^{-x}$
(C) $e^{-2+x}$
(D) $2-\ln x$

[hide]B

I just made the transformation so that $(0,1)$ is the origin.  Then I found the inverse and made the transformation back.[/hide]


algebra
polynomial
floor function
combinatorics proposed
combinatorics

prove that
$ \sum_{i\equal{}n\minus{}k}^{\lfloor \frac{n}{2} \rfloor} {n \choose 2i} {2i \choose i} {i \choose n\minus{}k} 2^{n\minus{}2i} \equal{} {2k \choose n}{n \choose k}\equal{}{2k \choose k}{k \choose n\minus{}k}$

The identity becomes much less fearsome once you realize that

$ {n \choose 2i} {2i \choose i} {i \choose n \minus{} k}\equal{}{n \choose k} {k \choose 2i\minus{}n\plus{}k} {2i\minus{}n\plus{}k \choose i}$,

so it remains to show that $ \sum_i {k \choose 2i\minus{}n\plus{}k} {2i\minus{}n\plus{}k \choose i} 2^{n\minus{}2i} \equal{} {2k\choose n}$.

This is Theorem 2 in http://www.mathlinks.ro/Forum/viewtopic.php?t=87265 . Do you see why?

  darij


modular arithmetic
invariant
combinatorics unsolved
combinatorics

[quote="Mashimaru"]A circle is devided into 8 equal parts. In each part there is a stone. For 1 move, you can randomly choose 2 stones and change their position to the next part. Notice that a legal move is the move that 1 stone is moved in the same direction of a clock's hand and the other is moved to the opposite direction. With this rule of moving, after a limited number of moves, can you move all stones into 1 part?

Sorry for my bad English T____T[/quote]

Give a number from $ 1$ to $ 8$ to each part.
Let $ p(x)$ the part number of the stone $ x$.
Let $ S \equal{}\sum_{i \equal{} 1}^{8}p(i)$

Before any move, $ S \equal{}\sum_{i \equal{} 1}^{8}i \equal{} 36$
A move does not change $ S\pmod{8}$ (since $ p(x)$ is increased by $ 1$ for one stone and decreased by $ 1$ for another one) and  so $ S$ is always $ 36\pmod{8}$

And since $ 8$ does not divide $ 36$, it is impossible to have the eight stones in the same part.

This problem has a generality enunciation way. We can devide the circle to n parts, and do like Mashimaru told above.
The solution has 2 cases:
1) If $ n$ is an odd number, so we always have one easy way to move all the stones to one part. U can check it.
2) If not, consider the invariant
\[ S\equal{}\sum_{i\equal{}1}^{n}ia_{1}(mod~ n)\]
( $ a_{i}$ is the numbers of stones in the part $ i$ )
we can show that we can not do as required

Ps: Mashimaru is a Vietnamese? Are u?

We can easily recognize that $ \large S \equal{}\sum p(i) \equal{}\frac{n(n\plus{}1)}{2}(mod n)$ is invarient (at $ p(i)$ shows the part number of stone $ i$). So, if after some moves, all stones can be moved to part $ k (k\in N)$), then $ nk \equal{}\frac{n(n\plus{}1)}{2}$ or $ k \equal{}\frac{n\plus{}1}{2}$, it shows that $ n$ must be an odd number. And when $ n$ is an odd number, we can easily prove that there is a way to move all stones into 1 part!

@mathlink: How can you know I'm Vietnamese???? (tất nhiên là anh nói đúng ^.^)


algebra
polynomial
superior algebra
superior algebra solved

Let $K$ be a field, and $a\in K$, and p a prime number.Prove that if $a$ cannot be written as $b^p,b\in K$, then $X^p-a$ is ireducible.


[hide]After a lot of trial and error... I got
$ \sqrt{.2}^{\minus{}2}$[/hide]

Is there an easier way than trial and error?

[quote="FlyAgaric"]
Is there an easier way than trial and error?[/quote]
No

[quote="small_unicorn"]Using exactly two 2s and any of standard mathematical symbols, write down an expression whose value is five.[/quote]

Is there only one value? I know $ \sqrt{.2}^{\minus{}2}$ works...

Does this count?

[hide]$ \frac{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{...2}}}}}}{.2}$[/hide]


we have $xyz=1$ so $x^{\frac{1}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}}=1$
Multiply both sides by $1$ of the inequality we need to prove and for left side we will subst $x^{\frac{1}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}}=1$
Get the ineq equivalent to:
$\sum_{cyc}x^{\frac{4}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}}\le x^{2}+y^{2}+z^{2}$
Multiplying both sides by $2$ get:
$\sum_{sym}x^{\frac{4}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}}\le \sum_{sym}x^{2}$
Which is true by Muirhead.

[quote="vishalarul"]Let $x,y,z\leq{0}$ and $xyz=1$. Prove that:
$x+y+z\leq{x^{2}+y^{2}+z^{2}}$[/quote]I think you mean $x,y,z\ge0$.[hide]From AM-GM, \[\frac{x+y+z}{3}\ge\sqrt[3]{xyz}=1.\] From Chebyshev, \[\frac{x^{2}+y^{2}+z^{2}}{x+y+z}\ge\frac{x+y+z}{3}.\] From transitivity, \[\frac{x^{2}+y^{2}+z^{2}}{x+y+z}\ge1.\] Hence, \[x^{2}+y^{2}+z^{2}\ge x+y+z.\] [/hide]


[quote="sinankaral53"]$ a \minus{} )$ Find all prime $ p$ such that $ \dfrac{7^{p \minus{} 1} \minus{} 1}{p}$  is a perfect square

$ b \minus{} )$ Find all prime $ p$ such that $ \dfrac{11^{p \minus{} 1} \minus{} 1}{p}$  is a perfect square[/quote]
here it is the Turkish solution

It's harder to read in Turkish than to solve this problem!  :D

Since $gcd(q^{\frac{p-1}{2}}-1,q^{\frac{p-1}{2}}+1)=2$ for any odd integer $q$, if $px^2=q^{p-1}-1$ for an integer $x$ and an odd prime $p$, than either
Case1:$q^{\frac{p-1}{2}}-1=2py^2$ and $q^{\frac{p-1}{2}}+1=2z^2$ or
Case2: $q^{\frac{p-1}{2}}-2=2y^2$ and $q^{\frac{p-1}{2}}+1=2pz^2$ for some integer $y$ and $z$.
A)In Case1: $6|7^{\frac{p-1}{2}-1=2py^2}$ ->$3|py^2$. When $p=3$ we have $\frac{7^2-1}{3}=4^2$. If $p\geq 2$
then $3|y^2$ ->$9|7^{\frac{p-1}{2}-1}$ -> $3|\frac{p-1}{2}$. Let $k=\frac{p-1}{6}$. then $2z^2=7^{3k}+1=(7^k+1)(7^{2k}-7^k+1)$ implies that $7^{2k}-7^k+1$ is a perfect square. But is not possible as $(7^k-1)^2\geq7^{2k}-7^k+1\geq(7^k)^2$.
In Case2:Since 2 is a quadratic residue $mod7$, but $-1$ is not
We conclude that $p=3$ is only prime for which $\frac{7^{p-1}-1}{p}$ is a perfect square.
B)This time 2 is not a quadratic residue $mod11$ and we have only Case 2 to consider. $p|11^{\frac{p-1}{2}}+1$ thus
$2y^2=11^{\frac{p-1}{2}}-1=-2(modp)$ -> $p|y^2+1$ -> $4|p-1$.Therefore $11^{\frac{p-1}{2}}-1=2y^2$ implies $11^{\frac{p-1}{4}}+1=u^2$ which is impossible as $u-1$ and $ u+1$ cannot be both powers of 11; or $11^{\frac{p-1}{4}}+1=2u^2$ which is impossible 2 is not  a sqyare residue $mod11$.So there in no prime $p$ for which $\frac{11^{p-1}-1}{p}$ is a perfect square.

[quote="Domination1998"]Why do we have 3|py^2 ?????[/quote]
Is that too obviously ? $6|7^{ \frac{p-1}{2}}-1=2py^2$ so $3|py^2$.

[quote="Domination1998"]i wanna know why $4|(p-1)$[/quote]
Lemma: If $p=4k+3$ is a prime and $p|a^2+b^2$ then $p|a,p|b$.

ok thank you

$ 11^{\frac{p-1}{4}}+1=u^2 $

but why we can have this?????

can i know w sir????
why we must have that true???

It has been reached that $11^{\frac{p-1}{2}}-1=2y^2$, with the conclusion $11^{\frac{p-1}{4}}+1=u^2$, [b]or else [/b]$11^{\frac{p-1}{4}}+1=2u^2$.

Denote $11^{\frac{p-1}{4}} = x$; then it has been reached that $(x-1)(x+1) = x^2-1=2y^2$. Since $\gcd(x-1,x+1) = 2$, it means that either $x+1 = u^2$ and $x-1=2v^2$ for some integers $u,v$, or else $x+1 = 2u^2$ and $x-1=v^2$ for some integers $u,v$.

Rather than repeatedly asking questions, maybe spending some - even minimal - amount of time in consideration, would help you.


[quote="moogra"]There are 101 people. They shake hands with each other once. How many total handshakes are there.[/quote]
[hide="the answer"]
[hide="this is the 2nd hide tag"]
[hide="Almost there"]

Each consecutive person shakes one less time than the person before him/her. Thus since you start with 100 shakes, Thus there should be 5050 shakes....if I did my math correctly :) [/hide][/hide][/hide]

[hide]yes you are right
[hide](100-1)*100/2 should be the correct answer... or 5050[/hide][/hide][/hide]

[hide]You use the formula (n-1)(n)/2.

(100-1)(100)/2=
99*50=[size=150]5050[/size][/hide]

The formula is located on my thread "Formulas" in the MathCounts section. :P

[size=84][color=green]Hide your answers![/color][/size]

[quote="robinhe"]Two things to clarify to everyone:

[hide="Number 1"]$99 \times 50$ doesn't equal $5050$[/hide]

[hide="Number 2"]There aren't 100 people in the room.[/hide][/quote]

Yeah, your right. The answer should be [size=150]4950[/size]. Not 5050.

No it shouldn't.

[hide]There are $101$ people in the room, so the answer is $\frac{101(100)}{2}=5050$.[/hide]

[hide]Choose any pair of people from 101 people.How many choice???
The correct answer is 5050= $101*(101-1)/2$[/hide]

Why is the answer 4950? I think it should be 5050. Am I misunderstanding something?

I'm confused. First Robinhe said that it wasn't $5050$ and it was $4950$. But then he says it is $5050$. And besides it is $5050$. And this was because Goldendomer made a mistake. And why am I reviving an old thread? Anyway's, here's the logic.

There are 101 people. First shakes with 100, second with 99, third with 98, fourth with 97. Note the pattern. For the first, it is $101-1$, for the second it is $101-2$...

So for the $101^{st}$ person, it would have been zero handshakes. Note that if you drew out a chart, you would see a triangle. That triangle's area is the number of handshakes. (Sorry, I don't know how to use asymptote. :blush: )

So, the area of the triangle is half multiplied by the base, $101$, height, $100$, giving $5050$. Which is correct.

I hope this settled all confusion about the answer.

[hide="Solution 1"]
The first person shakes hands with 100 people.
Then the second person shakes hands with 99 people.
And so on....If I did my math correctly, the answer is 5050.
[/hide]
[hide="Solution 2"]
$101*50=5050$
[/hide]

The key is to notice that [b]everyone[/b] shakes hands with [b]everyone else[/b], [i]exactly once[/i].  So we must find the number of ways the people can shake hands with each other, in other words,we must find the number of pairs of people possible.  This is merely $\binom{101}{2}=\frac{(101)(100)}{(2)(1)}=\boxed{5050}$.

[quote="moogra"]There are 101 people. They shake hands with each other once. How many total handshakes are there.[/quote]
first man shaking hands a 100 people second man shaking hand 99 people because he shaked hand a first man

third man=98 because he shaked hands a first man and second man  and  fourd man=97  fifht man=96 ....... last

 man 1 shaked hands and  100+99+98+.......+2+1   100 multiplication 101/2 =5050

Hello, My previous post on this was no solution so here's my solution!
(101  2 )  =  (101   100)
                   ( 2        1   )= 5050 Shakes
Answer 5050 Shakes.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14211[/img]

 :(

D,E are the foot of the altitudes through A,B resp. $ \frac {DE}{AB} = \frac {CD}{AC} = cot\angle C$.
DE=AB.cotC is minimal when \angle C is maximal.
* If $ AB%Error. "pararallel" is a bad command.
xy$, where $ xy$ is the fixed line through C.
The prob has only a root because exist only one circle $ (\omega)$ through A,B and touches $ xy$ at M.
C moves on $ xy$, so AC or BC meets $ (\omega)$, hence $ \angle C\leq\angle AMB$.
* If AB meets $ xy$ at I. Construct two circles $ (\omega 1)$ and $ (\omega 2)$ through A,B and touches $ xy$ at M,M' resp ( from $ IM^2 = IA.IB$)
With C on ray $ Ix$, AC meets the arc AMB of $ (\omega 1)$. With C on the ray $ Iy$, AC meets the arc AM'B of $ (\omega 2)$. Those case show that $ \angle C\leq\angle AMB or \angle AM'B\leq\angle AM'B$
Now, our task is comparing angle AMB and AM'B.
$ \angle AMB = \angle AOO',\angle AM'B = \angle AO'O$, hence the bigger angle lie in the smaller circle.
Done!


There was similar mathcounts problem like this on past year. I can't remember which though.

Six bags of marbles contain 18,19,21,23,25, and 34 marbles, respectively. One bag contains chipped marbles only. The other 5 bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, how many chipped marbles are there?

A) 18
B) 19
C) 21
D) 23
E) 25

Can't belive this was #18... ;)

[hide]let the number of marbles in the bags Jane picked be a+b+c and the number of marbles in the bags  George picked be d+e. Since a+b+c = 2(d+e), so a or b or c should be an even number and other two are odd.  If one number is 18, others are 23 and 25, then (18+23+25)/2 =33. No addition results can be achieved  for George. Then 34 should be the even number.  Try 34+23+25, not work, then 34+21+25, also not work. However, 34+19+25 works.
So the answer is D. [/hide]

[hide]rather than trying to find combinations that work, here's another idea:
since Jane has twice as many marbles as George ... then the number of marbles Jane and George have combined is divisible by three...
that is, the sum of the number of marbles in five of the bags is divisible by three...
the total number of marbles is 140.... this is 2 (mod 3)... so the number of chipped marbles is also 2 (mod 3)...
we see that 23 is the only number in our list that is 2(mod 3)... so the number of chipped arbles is 23... so our answer is [b]D[/b][/hide]
[quote]Can't belive this was #18... Wink[/quote]
18 on the AHSME... it probably was because they expected people to waste time trying combinations rather than doing the elegant way that AHSME thought of

[quote="shake9991"][hide]rather than trying to find combinations that work, here's another idea:
since Jane has twice as many marbles as George ... then the number of marbles Jane and George have combined is divisible by three...
that is, the sum of the number of marbles in five of the bags is divisible by three...
the total number of marbles is 140.... this is 2 (mod 3)... so the number of chipped marbles is also 2 (mod 3)...
we see that 23 is the only number in our list that is 2(mod 3)... so the number of chipped arbles is 23... so our answer is [b]D[/b][/hide][/quote]
Wow, that's neat...

[hide]If Jane has twice the amount George had, she must of had an even number.  Therefore, she had to have had either 3 evens or 2 odds and an even.  Since there aren't three evens, she had 2 odds and an even.  Now look at what George might have had.  If George had both 19 and 21, Jane could not make a number that ends in 0 and with the remaining #s.  Then, we notice that 19+25=21+23.  The mod residues with 34 looked promising, but when I pursued it, it was a dead end.  I started mentally thinking "Too bad 34 wasn't 44 or George's marbles [b]could be a different number that actually worked.[/b]"  Then, I noticed the working 2-3 pair 18-21--19-25-34.  So, there are 23 shattered marbles in the unchosen bag.[/hide]
Is it just me or is the tip by the problem creator more confusing?


limit
algorithm
inequalities unsolved
inequalities

Let positive integers $ a_1<a_2<\cdots<a_n$ satisfying $ 1>\sum_{i\equal{}1}^n\frac{1}{a_i}$. Find the min value of $ T\equal{}1\minus{}\sum_{i\equal{}1}^n\frac{1}{a_i}$

[quote="thaithuan_GC"]Let positive integers $ a_1 < a_2 < \cdots < a_n$ satisfying $ 1 > \sum_{i \equal{} 1}^n\frac {1}{a_i}$. Find the min value of $ T \equal{} 1 \minus{} \sum_{i \equal{} 1}^n\frac {1}{a_i}$[/quote]

This seems to easy to be correct...
Let $ a_i\equal{} \frac{1}{2^i} \longrightarrow \lim_{n \to \infty} \left(\sum_{i \equal{} 1}^n\frac {1}{2^i}\right) \equal{}\frac{1}{2\minus{}1} \equal{} 1$

$ \Rightarrow \lim_{i \to \infty} T \equal{} 0$

$ n$ is fixed I think. The answer is obtained by a greedy algorithm, picking the lowest $ a_k$ possible each time, and I recall the proof involving Abel summations, but I don't remember the details.

[quote="ocha"][quote="thaithuan_GC"]Let positive integers $ a_1 < a_2 < \cdots < a_n$ satisfying $ 1 > \sum_{i \equal{} 1}^n\frac {1}{a_i}$. Find the min value of $ T \equal{} 1 \minus{} \sum_{i \equal{} 1}^n\frac {1}{a_i}$[/quote]

This seems to easy to be correct...
Let $ a_i \equal{} \frac {1}{2^i} \longrightarrow \lim_{n \to \infty} \left(\sum_{i \equal{} 1}^n\frac {1}{2^i}\right) \equal{} \frac {1}{2 \minus{} 1} \equal{} 1$

$ \Rightarrow \lim_{i \to \infty} T \equal{} 0$[/quote]
n is fixed  :) , I think this problem is correct.


[quote="TRAN THAI HUNG"]$ \sqrt {\frac {a}{{b \plus{} c}}} \plus{} \sqrt {\frac {b}{{a \plus{} c}}} \plus{} \sqrt {\frac {c}{{b \plus{} a}}} > 2$
a,b,c>0
:lol:[/quote]
It's known from 60 years least.  :D

[hide="Very classic"]
$ \sum\sqrt {\frac {a}{b \plus{} c}} \equal{} \sum\frac {a}{\sqrt {a(b \plus{} c)}} >\equal{} \sum\frac {2a}{a \plus{} b \plus{} c} \equal{} 2$
Clearly that the equality doesnt hold.[/hide]


I've plotted the primes onto a so-called "dovetailing" grid.  It is very interesting because of the appearance of linear lines throughout the grid.

Looks very nice, indeed. But could someone, please, explain to me what exactly the dovetailing grid is?
 :?  :blush:


analytic geometry
graphing lines
slope
combinatorics proposed
combinatorics

Let $\epsilon >0$ be a real number. Around  every lattice we construct a circle of radius $\epsilon$. Prove that every line passing through the origin intersects an infinite ammount of circles.

We consider only the vertical diameters of the circles. We may as well assume that the slope of the line we're looking at is irrational, since all lines of rational slope pass through infinitely many lattice points.

The problem now ammounts to showing that given an irrational $\alpha$ and some $\varepsilon>0$, there are infinitely many pairs of integers $(m,n)$ s.t. $|n\alpha-m|<\varepsilon$. This is true, since there are infinitely many rational numbers $\frac mn$ s.t. $\left|\alpha-\frac mn\right|<\frac 1{n^2}$ (a well-known theorem, due to Dirichlet, I believe).


function
real analysis
real analysis solved

I need your help with the this question:
let  f: R^2 -> R be a smooth mapping.
prove that f is non-injective (not one-to-one).

All you need is continuity. Try to prove that a continuous function from $\mathbb R^2$ to $\mathbb R$ maps some pair of antipodes of the unit circle, for example, to the same point. It should be easy to do; all you need is the intermediate value property applied properly.

That being said, you must have realized that a problem involving the idea of continuity couldn't possibly belong to a section called "Superior Algebra" of the forum, right? :) You should have posted it in the "Real Analysis" subforum.

O.K i`ll try it.
i`ll be glad if you can right me full solution.
thanks
eden

I like grobber's idea very much. I suppose that he wanted you to think of the following: the image of the circle is the closed interval and u'll need to return to the same point after argument has "run" across the circle, that means that you'll need to make a turn.

First i want to thank you both for your help.
I still dont know how to solve my question.  :blush:
Please i need more information / explanations.

Eden

Well, may be I'm wrong then, but here is what I meant: let's see how yo function behave on a circle lying in $R^2$. Let's look at the point that moves along the circle, we want to see how it's image behave. It's image belongs to the closed interval of the R (the image of a compact is a compact and the image of the associated set, or connected sets, not sure about english term is a connected set). So image point stays in this closed interval, but the circle is closed, so when the first point returns to the beginning the image point should also return to where we started, so it had to make a turn (remember it moves on a line). This means that some way the image will pass twice, i.e. it's not an injection.
In two words: the continuous image of a closed path is the closed path, in our case as the image path is 1-dimentional it will have self-intersection.

A simple way of putting that: as $z$ moves on the unit circle, look at $g(z)=f(z)-f(-z)$. When $z$ goes from $1$ to $-1$, that quantity changes sign, and since $g$ is a continuous function of $z$, it must be zero at some point.


geometry
circumcircle
trigonometry
inequalities
inequalities proposed

Prove that:
$\frac{h_a}{l_a}+\frac{h_b}{l_b}+\frac{h_c}{l_c}\le \sqrt{2(\frac{r}{R})^2+5\frac{r}{R}+6}$

Denote the second intersections $A'$, $B'$, $C'$ of the lines $IA$, $IB$, $IC$ respectively with the circumcircle of the triangle $ABC$. From the relations $\frac{h_a}{l_a}=\cos \frac{B-C}{2}=\frac{AA'}{2R}$ a.s.o. results the proposed inequality is equivalently with the following inequality: $\boxed {\ \sum AA'\le 2\sqrt{6R^2+5Rr+2r^2}\ }$. But $AA'=IA+IA'$ and $IA=\frac{r}{\sin \frac A2},\ IA'=A'B=2R\sin \frac A2$. I know (generally, [u]Erdos- Mordell[/u])  $\boxed {\ \sum IA\ \boxed {\ge}\ 6r\ }$ and $\sum \sin \frac A2\le \frac 32$, i.e. $\boxed {\ \sum IA'\le 3R\ }$. Prove easily that $3R+6r\le 2\sqrt{6R^2+5Rr+2r^2}.....????$ [i](I swallowed the bait ![/i] ).

[b]I didn't prove this inequality, but I wanted to shaw that this inequality is very strong ![/b] I must try another way for finding solution.

$l_a$ is the length of the bisector of the $\angle BAC$.

If you prove sucess that$INEQ$:$r^2+9Rr+2R^2\ge p^2$ ,then this problem is solved.But I think it's very hard because it 's stronger:
$p^2+r^2\le 2R^2+10Rr+2(R-2r)\sqrt{R(R-2r)}$.
It made me became to silly.

In the particular acute triangle $ABC$, where $a=7$, $b=6$, $c=5$
the inequality $p^2\le r^2+9Rr+2R^2$ isn't truly because
$p=9,\ r=\frac 23\sqrt 6,\ R=\frac{35}{4\sqrt 6}\Longrightarrow$ $p^2=81>\frac 83+\frac{105}{2}+\frac{1225}{48}=r^2+9Rr+2R^2\ !!!$.

[u]Remark.[/u] In the particular right triangle $ABC$, where
$a=5,\ b=4,\ c=3$ we have $p^2=r^2+9Rr+2R^2\ \ !$

Who can prove that it?
Where are people who are good of ineq($hungkhtn,vasc,...$).

It's no very strong.Ineq$4R^2+4Rr+3r^2\ge p^2$ stronger.
And I found solution.It's nice and hard. :)

The inequality is from $HI^2=4R(R+r)+3r^2-p^2$.

Yes, I known.Inequality$4R^2+4Rr+3r^2\ge p^2$have many solution.
But "give our inequality to it" is hard and I found it.hahahahahahah :lol:

It is well-known that $HI^2=4R^2+4Rr+3r^3-p^2\ge 0\ !$

You didn't understand me. :( .$Virgil Nicula$,can you prove that first ineq of mine:
$\sum \frac{h_a}{l_a}\le \sqrt{2(\frac{r}{R})^2+5\frac{r}{R}+6}$.
 ;)

[quote="tkhtn"]Yes, you can't but I can. ;)[/quote]
Hahaha........
If you realy solve it poste your solution .
And I think Virgil can solve more problems,that you can't :P  :rotfl:

The inequalities arn't too interesting for me. If don't go out quickly,then I renounce. I have not many times for it. I congratulate you because you know solve it. But I don't interest your solution. It is O.K., my dear ?

Yes, is simple and very nice. Can I wish a geometrical interpretation (or a solution) of this inequality ? This is interestingly.


calculus
integration
Euler
algebra
polynomial
logarithms
real analysis

could anyone prove that

$ \int_0^{\infty}\frac{x}{e^x\minus{}1}dx\equal{}\frac{\pi^2}{6}$

without using $ \sum\frac{1}{n^2}\equal{}\frac{\pi^2}{6}$ ?

[quote="knoppix"]could anyone [/quote]
Not me. If I were you, I wouldn't hold my breath waiting for an answer.

Do you want to say that noone knows such proof ?

The result involves the Basel Problem, it's impossible to compute the integral without that, unless you want to explicitly evaluate that series.

The two results are pretty much equivalent; a proof of this result immediately implies the solution to the Basel problem, so...

Mhmm how did Euler derive the expression? :S

[url=http://en.wikipedia.org/wiki/Basel_problem#Euler_attacks_the_problem]Euler's proof[/url], while clever, is not rigorous.  The standard modern proof is by Fourier series.

Thanks everybody for replies.

1.
In "Proofs from THE BOOK" there is a proof with integral which is counted in two ways.
Could you please explain why are you so sure that my integral can't be used in similar manner ?
Is it proved ?

2.
(t0rajir0u's last post)
yes, it uses a theorem about polynomials to infinite series!

is it possible to "complete" his proof ? (I mean, has anyone done it?)

1.  I'm not saying there isn't another proof.  But any other proof is equivalent to another proof of the solution to the Basel problem.

2.  It is possible to be rigorous about infinite products, but as far as I know it requires a lot of [url=http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem]analysis[/url].

$ \int_0^\infty \frac {x}{e^x \minus{} 1} dx \; \stackrel{t \equal{} e^{ \minus{} x}}{ \equal{} } \; \minus{} \int_0^1 \frac {\ln t}{1 \minus{} t} \; dt \equal{} \int_0^1 \left( \int_t^1 \frac {1}{1 \minus{} t} \, \cdot \, \frac {1}{x} \; dx \right) dt \equal{} \\
\equal{} \int_0^x \int_0^1 \frac {1}{(1 \minus{} t)x} \; dx \; dt \equal{} \int_0^1 \int_0^1 \frac {dx \; dy}{1 \minus{} xy}$
now check equation (29) and further at :arrow: http://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html


geometry
3D geometry
vector
induction
real analysis
real analysis unsolved

Here's a very nice result:
    We call positive cone a subset $C$ of $R^n$ for which there are some vectors $c_1,...,c_r$ such that $C$ is the set of all linear combinations with nonnegative coefficients of these vectors. Define $C^{+}$ as the set of those vectors $v$ such that for all $c$ in $C$, $<v,c>$ is nonnegative. Prove that for any positive cone $C$, $C^{+}$ is also a positive cone.

Since we can use induction on $n$, we may as well assume that the cone $\cal C$ is actually “non-degenerate”, in the sense that it’s not contained in a hyperplane, or it containes a basis $e_1,\ldots,e_n$ of $\mathbb R^n$. Then, if we take $v\in\cal C$ to be $v=\sum_{i=1}^n\frac 1ne_i$, for example, then the affine hyperplane $\cal H$ passing through $v$ (we also consider the vectors to be points in $\mathbb R^n$) orthogonal to $v$ intersects $tw$ for some $t>0$ for every $w\in\mathcal C^+$.

The intersection between $\mathcal C^+$ and $\cal H$ is a closed convex subset $\mathcal C^+_{\mathcal H}$  of $\cal H$. It must also be bounded, otherwise we could find a sequence of vectors $(w_m)\subset C^+$ of norm $1$ with $\langle w_m,v\rangle\to 0$, which means that for some large $m$ and some $i$ we’ll have $\langle w_m,e_i\rangle<0$. Another thing we notice is that $\mathcal C^+$ is the cone on $\mathcal C^+_{\mathcal H}$, i.e. the set of vectors collinear to the points in $\mathcal C^+_{\mathcal H}$ by non-negative reals. Since $\mathcal C^+_{\mathcal H}$ is the convex combination of its set of extremal points, if we show that this set of extremal points is finite, the observations here would imply the desired conclusion.

Let’s prove that every extremal point $w$ of $\mathcal C^+_{\mathcal H}$ must be orthogonal to at least $n-1$ vectors among $c_1,\ldots,c_r$. Then, the finitude of the set of extremal points will follow, since for any $n-1$ vectors, there is at most one point in $\cal H$ representing a vector orthogonal to all the $n-1$ vectors.

Suppose $w\in\mathcal C^+_{\mathcal H}$ orthogonal to at most $n-2$ vectors among the $c_i$’s. Call these $c_1,\ldots,c_{n-2}$. The orthogonal complement of the subspace generated by $c_1,\ldots,c_{n-2}$ has dimension $2$, so it cuts $\cal H$ along a line containing $w$. The set of vectors $z$ which satisfy $\langle z,c_i\rangle>0,\ \forall i\in\overline{n-1,r}$ cuts $\cal H$ along an open ball of $\cal H$ which also contains $w$. The intersections of these two sets belongs to $\mathcal C^+_{\mathcal H}$ and contains a segment in the interior of which $w$ lies, so $w$ is not an extremal point of $\mathcal C^+_{\mathcal H}$.


Find a formula for the exponential function [i]f[/i] for which f(2)=9/4 and f(-2)=4/9.

If anyone could tell me how to do this that would be great!

Thanks

Your solution is $ f(x)\equal{}\left(\frac{3}{2}\right)^{x}$.  Then $ f(2)\equal{}\frac{9}{4}$ and $ f(\minus{}2)\equal{}\frac{4}{9}$.  As to how I figured that out, I noticed that $ \frac{9}{4}$ and $ \frac{4}{9}$ are both squared versions of $ \frac{3}{2}$, with one to the negative power, and one a positive.

So so, i didn't just notice that?

Lol

Is there some systematic way to do it everytime?

However, I am pretty sure the question will always have the reciprocals there to make it easy.

thanks

An exponential function is given by $ f(x) \equal{} a \cdot b^x$, and as such has two degrees of freedom.  Two points is enough to construct a system of two equations in two variables, which you can solve for $ a, b$.

In your case, we do the following:
$ \frac{9}{4} \equal{} a * b^2$
$ \frac{4}{9} \equal{} a * b^{\minus{}2}$

This is a two variable, two equation system which can be solved standardly.

Actually, it would be more appropriate to note that taking the logarithm of both equations gives us a system of two [b]linear[/b] equations in two variables, which does have a "standard" solution.  Two generic equations in two variables does not.


farze konid$ a,b,c,d,e>0$ va darim:
$ a^2\plus{}b^2\plus{}c^2\equal{}d^2\plus{}e^2$
$ a^4\plus{}b^4\plus{}c^4\equal{}d^4\plus{}e^4$
sabet konin:
$ a^3\plus{}b^3\plus{}c^3<d^3\plus{}e^3$

che forume activi darin!!!

albate inam bayad dar nazar begiri ke alan moghe kheili badie too irano baraye bachehaye irani,chon ye seria ke tala shodano vakhte sar zadan be site ro nadaran,ye seriam alan depresan chon ya marhale 2 ghabool nashodan ya inke tala nagereftano varede marhale jadidi az zendegi shodan (konkoor)...
ye seriam ke taze varede vadie olampiad shodano taze varedano rooshoon nemishe post bezanan,vase hamin alan moghe kheili badie,bayad hagh bedi ke forum khalvat bashe...

[quote="sumita"]
che forume activi darin!!![/quote]

in forum male hame iranias,az jomle khode shoma... :wink:

mesle inke hagh ba shomast :blush:
ba in hal man say mikonam age soale ghashangi didam bezaram tu forum ta hame fek konan.
(albate ba rahe halash(age kasi hal nakard))

behtarin moghe tu in forum mogheye emtehanaye khordade ke hame az tarse emtehana be riazi panah mibaran!!!


function
real analysis
real analysis unsolved

$E,F$ two Banach spaces. Suppose that $f: E\to F$ is a $C^1$ function such that $Df(x)$ is invertible at every point.
Prove that if there is a constant $C$ such that $||(Df^{-1}(x))||<C$ for all $x \in E$, then $f$ is bijective.

for which notion of differentiability are you talking about ?( because Df(x) is a linear map in some kind of differentiability )

one says for example that $f$ is differentiable at the point $x_0$ if there is a continuous linear map $L: E \to F$ that satisfies: $|f(x_0+h)-f(x_0)-L(h)|=o(|h|)$

but then L=Df(x) , so Df(x)^-1  is also a linearr map , if it is bounded it will be null ! :? I'm i missing something ?

you're right, $L^{-1}$ is a linear map. And $||L^{-1}||$ is the norm of this linear map: $||L^{-1}||=sup_{|x|=1}|L^{-1}(x)|$. (my notation in the original post was surely a tiny bit wrong. I've edited to make it proper)


Which of the following ratios has the largest value? Answer with the letter only.
a) $ \frac{8}{17}$ b) $ \frac{9}{19}$ c) $ \frac{10}{21}$ d) $ \frac{11}{23}$

It's $ \boxed{d}$, as since these are all the greatest numerator less than $ \frac{1}{2}$, the greatest denominator will give the greatest value because they're all odd.

[hide="Some algebra"] Let $ x$ be a positive integer. We wish to determine whether $ \frac {x}{2x \plus{} 1}$ or $ \frac {x \plus{} 1}{2x \plus{} 3}$ is greater. We think that the latter is greater, so we assume that the former is greater for the sake of contradiction. Then, according to our statement

$ \frac {x}{2x \plus{} 1} > \frac {x \plus{} 1}{2x \plus{} 3}$

Cross multiplying, we have $ (2x \plus{} 3)(x) > (2x \plus{} 1)(x \plus{} 1) \implies 2x^2 \plus{} 3x > 2x^2 \plus{} 3x \plus{} 1$. Uh-oh. Since $ x$ is postive, this can't be true, which means that our original assumption can't be true. Therefore,

$ \frac {x}{2x \plus{} 1} < \frac {x \plus{} 1}{2x \plus{} 3}$, for positive integer $ x$.

Applying this to our given fraction tells us that d. is the largest.[/hide]
This can be generalized by replacing $ 2$ with $ k$ and $ 1$ and $ 3$ with $ n$ and $ n \plus{} 2$, respectively, for some positive integers $ n$ and $ k$.


Consider the inequality

$a, b, c \ge 0$

$(a^{n}+b^{n}+c^{n})^{3}\ge (a+b+c)^{2}(a^{3n-3}b+b^{3n-3}c+c^{3n-3}a)$

For $n=1$ this is trivial.

For $n=3$, if we take $a=b=1, c=2$ we get $LHS = (1+1+8)^{3}= 1000$ and $RHS = (1+1+2)^{2}(1+2+64) = 4^{2}(67) = 1072$, a counterexample.

[b]Can someone prove for $n=2$? [/b]

------------------

Also, here is more general problem:

$(a^{n}+b^{n}+c^{n})^{k}\ge (a+b+c)^{k-1}(a^{k(n-1)}b+b^{k(n-1)}c+c^{k(n-1)}a)$

Still clearly always true for $n=1$. For which $n$, in terms of $k$, will this hold? In this general case consider $n, k$ to be real numbers ($k$ is positive).

For $n=2$ it is indeed true.

By [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=stefan&t=6026]Vasc's most famous[/url] we have that $\left( a^{2}+b^{2}+c^{2}\right)^{2}\geq 3 \left( a^{3}b+b^{3}c+c^{3}a \right)$.
Now, just multiply that inequality with this one: $3 \left( a^{2}+b^{2}+c^{2}\right) \geq (a+b+c)^{2}$.


Give $k \geq 2$ and $a,b,c$ be three positive real.Prove that:
                   $\frac{b+c}{ka^2+bc}+\frac{a+c}{kb^2+ac}+\frac{a+b}{kc^2+ab} \geq \frac{18}{(k+1)(a+b+c)}$  :)  :)
My Question:Is this problem true when we have $k<2$? :?  :?

putting $a\rightarrow 0 , b=c$ we find that $k\le \frac{1}{2}$ or $k\geq 2$ I have an ugly proof for $k\geq 2$ by expanding but I will check also the case for $k\le \frac{1}{2}$

I use S.O.S to prove my problem  :lol:  :lol:


Find

\[ \sum _{r = 0} ^{n} (-2)^r \frac {{n \choose r}}{{r+2 \choose r}} \]

This question is also from IIT-JEE but is simpler than most.

[hide]There will be cases for n= odd and n= even [/hide]

[hide]This just involves a simple division. Rest should follow itself[/hide]


Let $ n \ge 2$ be a positive integer, $ x_1,x_2,\dots,x_n$ be positive real numbers and let $ A \equal{} \frac {1}{n}(x_1 \plus{} x_2 \plus{} \dots \plus{} x_n)$, $ G \equal{} \sqrt [n]{x_1x_2 \dots x_n}$. If $ M \equal{} \max\{x_1,x_2,\dots,x_n\}$ and $ m \equal{} \min\{x_1,x_2,\dots,x_n\}$, prove that
\[ \sqrt {A \minus{} G} \ge \frac {1}{\sqrt {n}}(\sqrt {M} \minus{} \sqrt {m}).
\]

WLOG $ M\equal{}x_1$ $ m\equal{}x_2$. Since $ A\geq G$ it's sufficient to show that $ nA\minus{}nG\geq x_1\plus{}x_2\minus{}2\sqrt{x_1x_2}$ $ \Longleftrightarrow$ $ \sqrt{x_1x_2}\plus{}\sqrt{x_1x_2}\plus{}x_3\plus{}x_4\plus{}...\plus{}x_n\geq n\sqrt[n]{x_1x_2...x_n}$ which is true by AM-GM.


geometry
circumcircle
geometric transformation
reflection
homothety
cyclic quadrilateral
perpendicular bisector

In convex quadrilateral ABCD, AB=BC and AD=DC. Point E lies on segment AB and point F lies on segment AD such that B, E, F, D lie on a circle. Point P is such that triangles DPE and ADC are similar and the corresponding vertices are in the same orientation (clockwise or counterclockwise). Point Q is such that triangles BQF and ABC are similar and the corresponding vertices are in the same orientation. Prove that points A, P, Q are collinear.

The problem is rather easy, I see :)
Here is my solution.
Let O be the circumcenter of the circle BEFD. It is obvious that triangles DPE, ADC and FOB are similar and have the same orientation.
So we obviously have that P is the image of O after the composition of angle BAD bisector reflection and homothety with coefficient AD/AB = AE/AF.
So the line AP is the image of line AO after the angle BAD bisector reflection.
We can prove in the same way that the line AQ is the image of line AO after the angle BAD bisector reflection.
So the points A, P and Q are collinear. Q.E.D. :)

[quote="Wizzy"]
So we obviously have that P is the image of O after the composition of angle BAD bisector reflection and homothety with coefficient AD/AB = AE/AF.
So the line AP is the image of line AO after the angle BAD bisector reflection.[/quote]

Could you please explain why that is true?

Thanks

I must confess I do not understand the first solution. An explanation would be welcome. A slightly more voluminous approach to the problem is the following:

Throughout this solution, we will suppose that $ AB < AD$. As in the previous solution, let $ O$ be the centre of the circle circumscribed to $ BEFD$ and note that triangles $ ABC$, $ BQF$ and $ DOE$ on the one hand and triangles $ ADC$, $ DPE$ and $ BOF$ on the other hand are similar. Let lines $ BQ$ and $ PE$ meet at $ R$. Triangles $ BER$ and $ DOF$ are similar, too. These results are obtained by more or less straightforward angle-chasing.
To prove that $ A$, $ P$ and $ Q$ are collinear, it will, by Menelaos' theorem applied to $ BRE$, suffice to establish that $ \frac {EA\cdot BQ\cdot RP}{AB\cdot QR\cdot PE} \equal{} 1$ or, in view of $ BQ \equal{} \frac {AB\cdot BF}{AC}$, $ PE \equal{} \frac {AD\cdot DE}{AC}$ (given the similarity of triangles emphasised above), $ \frac {RP}{QR} \equal{} \frac {AD\cdot DE}{AE\cdot BF} \equal{} \frac {AB\cdot DE}{AF\cdot BF} \equal{} \frac {AD\cdot DE}{DE\cdot AF} \equal{} \frac {AD}{AF}$ by properties of the cyclic quadrilateral $ BEFD$.
Let $ r$ be the radius of the circle circumscribed to $ BEFD$. Note that $ \frac {RP}{RQ} \equal{} \frac {BQ \minus{} BR}{PE \minus{} ER}$. But $ BQ \equal{} \frac {r\cdot BF}{DE} \equal{} \frac {r\cdot AB}{AD}$, $ PE \equal{} \frac {r\cdot DE}{BF} \equal{} \frac {r\cdot AD}{AB}$ and $ BR \equal{} ER \equal{} \frac {r\cdot BE}{DF}$, which implies that it remains to show that $ \frac {AD}{AF} \equal{} \frac {\frac {AD}{AB} \minus{} \frac {AB \minus{} AE}{AD \minus{} AF}}{\frac {AB}{AD} \minus{} \frac {AB \minus{} AE}{AD \minus{} AF}}$, which, after some tedious algebraic work and in view of $ AB\cdot AE \equal{} AD\cdot AF$ proves to be true. This establishes the required collinearity of $ A$, $ P$ and $ Q$.

My solution is the same as [b]Wizzy[/b]'s.  I hope to clarify it in the following post:
[hide="Solution"]
As before, we let $ O$ be the center of $ BEFD$ and let $ M$ and $ N$ be the midpoints of $ FB$ and $ DE$ respectively.  Now, notice that (as everyone found before), $ \triangle DOE\sim \triangle BQF$.  We see that $ BQ \equal{} QF$, so $ Q$ lies on the perpendicular bisector of $ FB$, as does $ O$.  This means that $ QM$ is an altitude of $ \triangle BQF$ and we also have that $ ON$ is an altitude of $ \triangle DOE$.  Then, $ \frac {QM}{ON} \equal{} \frac {FB}{DE}$.  Yet, $ \angle ABF \equal{} \angle ADE$ and $ \angle FAB \equal{} \angle EAD$, so $ \triangle BAF\sim \triangle DAE$.  This means that since $ M$ and $ N$ are the respective midpoints of $ FB$ and $ DE$, we have that $ \triangle AMF\sim \triangle ANE$.  This yields that $ \angle AMF \equal{} \angle ANE$.  Yet, $ \angle OMF \equal{} 90 \equal{} \angle ONE$, so\[ \angle AMO \equal{} 90 \plus{} \angle AMF \equal{} 90 \plus{} \angle ANE \equal{} \angle ANO\]Furthermore, from the similar triangles, we get that $ \frac {AM}{AN} \equal{} \frac {AF}{AE} \equal{} \frac {FB}{DE} \equal{} \frac {MQ}{NO}$.  Then, $ \triangle AMQ\sim \triangle ANO$.  This means that $ \angle MAQ \equal{} \angle NAO$.  We can also realize that $ \triangle AMB\sim \triangle AND$, so $ \angle MAB \equal{} \angle NAD$, which means that $ \angle QAB \equal{} \angle OAD$. Similarly, $ \angle OAD \equal{} \angle PAB$, so $ P$, $ Q$, and $ A$ lie on a line, that is, in fact, $ AO$ reflected over the bisector of $ \angle BAD$. [/hide]

Thanks, QuattoMaster :)
I couldn't write a solution in such detailed way because of some troubles with my computer..


Theres this famous saying, "If a tree falls in the forest, and no one is around to hear it, did it really fall?"

I have no clue if it fell or not. What do you think?

Yes:  [quote]If a tree falls in the forest ...[/quote]

I think the actual question is, did it make a sound?

[quote="D3m0n Shad0w"]"If a tree falls... ...did it really fall?"[/quote]

nice job man


\documentclass{book}
........
\begin{document}

\part{Math}

\chapter{number}

\end{document}

Part  I  will    be   on  an  odd  page (actually, page number  1)  and  chapter 1 will start  the next   odd   page .

If    I   write  the  code like this :

\part{Math}
this   is  a   paragraph
\chapter{number}

then  ' part   I  math '  will   be    on  the page  with  page number 1, ' this   is  a   paragraph'    will  be   on  the  next  odd  page,  and  'chapter 1   number'    will  start   on  the page  with  page number  5.

How to  get   the  output   that   ' part   I  math'  and 'this is  a  paragraph'  will be  on the   same   page and  chapter 1 will on the page with page number 3?

In  other words,     the  page  with page number 1  will be  like this:

Part   I

math

this is  a paragraph

[code]\documentclass[openany]{book}
\begin{document}
\part{Math \\[1cm] \normalsize\normalfont this is a paragraph}
\chapter{number}[/code]

Thank  stevem!

Each  part  will   start   on    an   odd   page,    and  the  first   chapter   of   an   part   starts  on    an   odd   page
but   other  chapter  of  an   part   will (or  will  not )   start  on    an   odd   page.

how  to  get   that?

But the [i]openany[/i] option is designed so that chapters can start on any page so that is what you get. Without it (which is the default [i]openright[/i] option) you get chapters and parts on odd pages and, where necessary, blank pages to achieve this.

With  the   openany   option   parts   and   chapters  can   start   on   any   page ;

Without it   chapters   and  parts   on  odd   pages

[code]\documentclass[openany]{book}

\let\oldpart\part
\let\oldchapter\chapter

\def\part
  {\cleardoublepage
   \def\chapter
     {\cleardoublepage
      \let\chapter\oldchapter
      \chapter}%
   \oldpart}

\begin{document}

\part{Test}

blah

\chapter{Test}
\chapter{Test}
\chapter{Test}
\chapter{Test}
\chapter{Test}

\end{document}[/code]


Denote $ a_n \equal{} f(n)$; the recurrence relation becomes $ 2a_n \equal{} a_{n \minus{} 1} \minus{} n5^n \plus{} 2^n$.
Look for $ a_n \equal{} b_n \plus{} \alpha2^n$, so $ 2b_n \plus{} \alpha2^{n \plus{} 1} \equal{} b_{n \minus{} 1} \plus{} \alpha2^{n \minus{} 1} \minus{} n5^n \plus{} 2^n$ and ask for $ \alpha2^{n \plus{} 1} \equal{} \alpha2^{n \minus{} 1} \plus{} 2^n$, i.e. $ \alpha \equal{} \frac {2} {3}$.
Now look for $ b_{n} \equal{} c_n \plus{} (n\beta \plus{} \gamma)5^n$, so $ 2c_n \plus{} 2(n\beta \plus{} \gamma)5^n \equal{} c_{n \minus{} 1} \plus{} ((n \minus{} 1)\beta \plus{} \gamma)5^{n \minus{} 1} \minus{} n5^n$ and ask for $ 10(n\beta \plus{} \gamma) \equal{} (n \minus{} 1)\beta \plus{} \gamma \minus{} 5n$ for all $ n$, i.e. $ 9\beta \equal{} \minus{} 5$ and $ 9\gamma \equal{} \minus{} \beta$, i.e. $ \beta \equal{} \minus{} \frac {5} {9}$ and $ \gamma \equal{} \frac {5} {81}$.
It follows $ 2c_n \equal{} c_{n \minus{} 1}$, so $ c_n \equal{} \frac {1} {2^{n \minus{} 1}}c_1$. Therefore $ a_n \equal{} \frac {1} {2^{n \minus{} 1}}c_1 \minus{} (\frac {5} {9} n \minus{} \frac {5} {81})5^n \plus{} \frac {2} {3} 2^n$. From $ a_1 \equal{} \frac {4|} {3}$ follows $ c_1 \equal{} \frac {200} {81}$, hence $ a_n \equal{} \frac {25} {81} 2^{ \minus{} (n \minus{} 4)} \minus{} (\frac {1} {9} n \minus{} \frac {1} {81})5^{n \plus{} 1} \plus{} \frac {1} {3} 2^{n \plus{} 1}$.


Let $gcd(a,b) = 1$, prove that $gcd(a+b,a^2+b^2) = 1$ or $2$.

Let $d=gcd(a+b,a^2+b^2)$
It's known that gcd(x,y)|gcd(Ax+By,Cx+Dy)
then $d|gcd(2ab,a+b)=gcd(2a^2,2b^2)=2$ make the result.

[quote="Skyward_Sea"]
It's known that gcd(x,y)|gcd(Ax+By,Cx+Dy)
[/quote]

so prove it!  :D

I forgot that A,B,C,D,x,y are all integers, but you should have understood...
Proof.  By Bezout's Theorem, gcd(x,y)|Ax+By, gcd(x,y)|Cx+Dy
So gcd(x,y) is their common divisor, that makes it.


geometry
area of a triangle
Heron\u0027s formula
angle bisector

In triangle $ ABC$, $ AB \equal{} 13$, $ BC \equal{} 14$, and $ AC \equal{} 15$. Let $ D$ denote the midpoint of $ \overline{BC}$ and let $ E$ denote the intersection of $ \overline{BC}$ with the bisector of angle $ BAC$. Which of the following is closest to the area of the triangle $ ADE$?

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.5 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 3.5 \qquad \textbf{(E)}\ 4$

[quote="worthawholebean"]In triangle $ ABC$, $ AB \equal{} 13$, $ BC \equal{} 14$, and $ AC \equal{} 15$. Let $ D$ denote the midpoint of $ \overline{BC}$ and let $ E$ denote the intersection of $ \overline{BC}$ with the bisector of angle $ BAC$. Which of the following is closest to the area of the triangle $ ADE$?

$ \text{(A)}\ 2 \qquad \text{(B)}\ 2.5 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 3.5 \qquad \text{(E)}\ 4$[/quote]
[hide="Solution"]
Heron's Formula gives us that the area of $ \triangle ABC$ is $ 84$.  Now, the Angle Bisector Theorem gives us that $ DC \equal{} \frac {AC\cdot BC}{AB \plus{} AC} \equal{} \frac {15}{2}$.  Also, $ EC \equal{} \frac {BC}{2} \equal{} 7$.  Hence, $ DE \equal{} DC \minus{} EC \equal{} \frac {1}{2}\implies \frac {DE}{EC} \equal{} \frac {\frac {1}{2}}{7} \equal{} \frac {1}{14}$, so (noting that $ [AEC]\equal{}\frac{EC}{BC}\cdot [ABC]\equal{}\frac{1}{2}\cdot 84\equal{}42$), we have that $ \frac {1}{14} \equal{} \frac{DE}{EC}\equal{}\frac {[ADE]}{[AEC]} \equal{} \frac {[ADE]}{42}\implies [ADE] \equal{} 3 \equal{} \boxed{C}$.  [/hide]

[hide]First, recognize BE = 6.5 and CE = 7.5 from the Angle Bisector Theorem. Next, drop an altitude to side BC as well and call the point of intersection Q. See that it forms two Pythagorean triple triangles, a 5-12-13 and a 9-12-15. You should now have right triangle AQD with base 2 and height 12. The area of this is 12. You also have right triangle AQE with base 1.5 and height 12; the area of this is 9. AQD-AQE=AED=12-9=$ 3$ or C.[/hide]

The Quatto Master's solution can be simplified. Since DE=1/2 by angle bisector, and DE is part of BC, the answer is $ 84\cdot(.5/14)$.

Call the distance between D and E x. Then by angle bisector : 15/ 7+x =13/ 7 - x
so x = 1/2 so now all we need is the height of triangle ADE. Because DE is on BC the height of ADE is the same as the height of the triangle. Use Herons on the triangle to get Area = 84
then 14 x H = 2 x 84 so the height is 12. A = B x H /2 so Area of ADE = 12 x 1/2 x 1/2 = 3

So the answer is [b]C[/b]

[hide]My approach:
We have: S(ABE)/S(AEC)=13/15(applying the formula: S(ABC)=1/2.a.b.sinA).
So, we have: S(ABE)/13=S(AEC)/15=S(ABC)/28=84/28=3(the are of triangle ABC is easily calculated by Heron's formula)
Therefore, S(AEC)=39
In addition, we have: S(ADC)=S(ADB)=42
W.L.O.G, assume that E lies between points D and C. So, S(ADE)=42-39=3(Q.E.D) :)  [/hide]

How do you know whether E will be to the left or right of point D(midpoint)?

$E$ will be to the left of $D$ since $1>\frac{AB}{AC}=\frac{BE}{EC}$

Let $BE=x$. Then, $EC=14-x$, so by the angle-bisector theorem, $\frac{13}{x}=\frac{15}{14-x} \implies$ $15x=182-13x \implies$ $x=6.5$. Hence, $ED=0.5$.
Next, note that by Heron's formula, $[ABC] = $ $\sqrt{21(21-13)(21-14)(21-15)}=$ $84$, so the height from $A$ to $BC$ is $12$. As a result, since $\Delta AED$ and $\Delta ABC$ share heights protruding from $A$, $[ADE]=\frac{0.5 \cdot 12}{2} = \boxed{3}$.

[quote=TheMaskedMagician]Alternate Solution without Herons:

This is the famous $13-14-15$ triangle. We draw the diagram below,

[asy]
size(200);
pair A=(0,0),B=(-10,-24),C=(18,-24),D=(4,-24),EE=(3,-24),F=foot(A,B,C);
D(MP("A",A,N)--MP("B",B,W)--MP("C",C,E)--cycle);
D(A--MP("F",F,S));
D(A--MP("E",EE,SW));
D(A--MP("D",D,S));
D(rightanglemark(A,F,B,30));
label("$5$",(B+F)/2,S);
label("$a$",(F+EE)/2,N);
label("$7$",(D+C)/2,S);
label("$13$",(A+B)/2,W);
label("$15$",(A+C)/2,E);
label("$12$",(A+F)/2,W);
[/asy]

Angle Bisector Theorem yields \[\frac{13}{5+a}=\frac{15}{9-a}.\] Cross-multiplying yields \[117-13a=75+15a\implies 28a=42\implies a=\frac{3}{2}.\] Note that$[ADE]=[AFD]-[AFE]$. $[AFD]=12$ and $[AFE]=9$. So $[ADE]=12-9=3$ as desired.[/quote]

You use Heron to find $AF $.


[b]Problem.[/b] [i] On $\triangle ABC$, the points $P_{1},P_{2},P_{3}$ are situated on the line segments $BC,AC,AB$, respectively. $BC=10,AC=14,AB=18$. $P_{1},P_{2},P_{3}$ divide $BC, AC,AB$ in the ratios $2: 5,3: 7,4: 11$, respectively.  Find $[P_{1}P_{2}P_{3}]$.[/i]

[hide="brute force idea"]Find the area of $\triangle ABC$ using Heron. Express the areas of triangles $CP_{1}P_{2}$, $BP_{1}P_{3}$ and $AP_{2}P_{3}$ in terms of the area of $\triangle ABC$, and subtract.[/hide]

denote the ratios p,q,r, for BP_1/BC, CP_2/CA, AP_3/AB, then the area of each corner is: $[ABC]* \frac{p}{p+1}*\frac{1}{q+1}$, it follows that the answer is

$[ABC]-[ABC]*\frac{1}{(p+1)(q+1)(r+1)}*(pq+qr+rp+p+q+r)$
$=[ABC]*\frac{1+pqr}{(p+1)(q+1)(r+1)}$

That's correct.  For the generalization, however, I used complex numbers.


Every strongly connected graph has a vertex whose removal does not disconnect the graph??

You don't need to consider a strongly connected digraph, since then it is also weakly connected (the subjacent undirected graph is connected). The result follows from the simple

[color=blue][b]Theorem.[/b] Any connected graph contains a vertex which removed does not disconnect the graph.

[i]Solution.[/i] Any connected graph contains a spanning tree. The removal of a leaf (terminal node) does not disconnect the tree.[/color]


inequalities
trigonometry
inequalities unsolved

[quote="Mather"]Prove inequality by dearative: sin2A+sin2B+1/sinC >= 4
I'm from Viet Nam[/quote]
Try $ A\equal{}B\equal{}C\equal{}\frac{\pi}{3}.$ :wink:

[quote="arqady"][quote="Mather"]Prove inequality by dearative: sin2A+sin2B+1/sinC >= 4
I'm from Viet Nam[/quote]
Try $ A \equal{} B \equal{} C \equal{} \frac {\pi}{3}.$ :wink:[/quote]

I need a solution!! Help me

[quote="hsiljak"][b]arqady[/b] has given you counterexample: your inequality isn't true! So you can't get a solution...[/quote]

Sorry very people: Prove: sin2A+sin2B+2/sinC >=4

$ \sin 2A \plus{} \sin 2B \plus{} \frac {2}{\sin 2C} \equal{} 2\sin (A \plus{} B)\cos (A \minus{} B) \plus{} \frac {2}{\sin C}$

Are $ A,\ B,\ C$ angles of a $ \triangle {ABC}$? If so, then we have

$ \sin 2A \plus{} \sin 2B \plus{} \frac {2}{\sin 2C} \equal{} 2\sin C\cos (A \minus{} B) \plus{} \frac {2}{\sin C}$

$ \leq 2\left(\sin C \plus{} \frac {1}{\sin C}\right)$.

[quote="Mather"]

Sorry very people: Prove: sin2A+sin2B+2/sinC >=4[/quote]
Try $ A\equal{}95^{\circ},$ $ B\equal{}5^{\circ}$ and $ C\equal{}80^{\circ}.$ :wink:

[quote="arqady"][quote="Mather"]

Sorry very people: Prove: sin2A+sin2B+2/sinC >=4[/quote]
Try $ A \equal{} 95^{\circ},$ $ B \equal{} 5^{\circ}$ and $ C \equal{} 80^{\circ}.$ :wink:[/quote]

Prove: Prove cos2A+cos2B+1/2cosC >= 1 with C=Max(A,B,C}< pi/2. This problem take from Tran Phuong's book. Do you know Tran Phuong ????


I have a question: When is the AIME B course going to start? I was just wondering because I enrolled by way of phone to the office, and I haven't gotten any sign that I'm actually signed up for the course.

You're all set up white_horse.  You should see everything now.  (white_horse has signed up late, so signed up by phone - he can still access everything he missed).


From 12:01 am on one day to 11:59 pm the same day, how many times do the hour and minute hands cross?

Please do not post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1700282#1700282]the same problem[/url] twice, especially after you've received a solution.


inequalities
trigonometry
function
real analysis
real analysis unsolved

For every 0<x<(pi/2) show that

(pi/2)<(x/(sinx))+sqrt(x/tanx) < 2

Graphing $\frac{x}{\sin x}+\sqrt{\frac{x}{\tan x}}$ we can see that is a strictly decreasing function in that interval. Then find values in its extrems and ...
may be a better way than operate with inequality

but i am not able to prove f '(x)<0.Iwant help on that .

For 0<x<pi/2:
$\frac{x}{\sin x}+\sqrt{\frac{x}{\tan x}}< 2 \Leftrightarrow$
$\sqrt{\frac{x}{\tan x}}< 2-\frac{x}{\sin x}\Leftrightarrow$
$\frac{x}{\tan x}< 4-4\frac{x}{\sin x}+\frac{x^{2}}{\sin^{2}x}\Leftrightarrow$
$x\sin(x) \cos(x) < 4 \sin^{2}(x)+x^{2}-4\sin(x) \Leftrightarrow$
$x \sin(2x)-4+8\cos(2x)-2x^{2}+8x\sin(x) < 0$
Develop the lefthandside into a powerseries.
This yields
$x \sin(2x)-4+8\cos(2x)-2x^{2}+8x\sin(x) = \frac{-x^{6}}{90}-\sum_{n=0}^{\infty}(-1)^{n}a_{n}x^{2n+8}$
where for all k $a_{2k}>\left(\frac{\pi}{2}\right)^{2}a_{2k-1}$
Thus the sum is pairwise a sum of positive terms.

[quote="petruspan"]
$x\sin(x) \cos(x) < 4 \sin^{2}(x)+x^{2}-4\sin(x) \Leftrightarrow$
$x \sin(2x)-4+8\cos(2x)-2x^{2}+8x\sin(x) < 0$
[/quote]
Firstly, it is $4\cos(2x)$, not $8\cos(2x)$. Secondly, what we need is $a_{2k}>\left(\frac{\pi}{2}\right)^{2}a_{2k+1}$
(not $a_{2k-1}$) and, finally, the coefficient at $x^{6}$ is actually $-\frac 1{45}$. Otherwise, the solution seems correct.

Interestingly enough, the other inequality is much simpler: omitting the square root (which makes the inequality only stronger), we see that we need to prove that
\[x\ge \frac\pi{2}\frac{\sin x}{1+\cos x}=\frac\pi{2}\tan(x/2)\]
But the left hand side is a linear function, the right hand side is a convex function, and we have an identity at the endpoints.


In $\triangle ABC$, $AB= 425$, $BC=450$, and $AC=510$. An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle. If these three segments are of an equal length $d$, find $d$.

[hide]Let the radio between the distances from each side to the point $P$ and the length of each altittude drawn to each side be $r_{1}$, $r_{2}$, and $r_{3}$.  You can see that:
$425(1-r_{1})=450(1-r_{2})=510(1-r_{3})$.

Now consider the three smaller triangles created when you draw the lines.  They have a ratio of $(1-r_{1}-r_{2})$, $(1-r_{2}-r_{3})$ and $(1-r_{1}-r_{3})$ with the orginal triangle.  Thus, since all three line segments are equal:

$425(1-2r_{1}-r_{2}-r_{3})=450(1--r_{1}-2r_{2}-r_{3})=510(1-r_{1}-r_{2}-2r_{3})$.

And if we use the equation at the beginning of this solution, you can see that $(1-r_{1}-r_{2}-r_{3})=0$.  And we can substutite to see that $r_{3}=1-r_{1}-r_{2}$.  And:

$425(1-r_{1})=450(1-r_{2})=510(r_{1}-r_{2})$

Solving the system we can express $r_{2}$ in terms of $r_{1}$ and substitute again.  Finally we see that $r_{1}=.28$, $r_{2}=.32$, and $r_{3}=.4$.  And the answer is $425(1-r_{1})=306$ if I didn't mess up.[/hide]


I am trying to prove a (not hard) combinatorical question using binomial coefficients. After some simplifying, I got a nasty expression, which I don't know how to simplify. Can anybody help and provide a general way?

$ \displaystyle\sum_{i\equal{}1}^{998}\frac{\binom{1997}{2i\minus{}1}}{4^{i}}$

general expression-

$ \displaystyle\sum_{i\equal{}1}^{t}\frac{\binom{nt}{ni\minus{}m}}{a^{i}}$

$ 0\le m\le n$

[quote="kunny"]Let $ 2i - 1 = j$, we have $ \sum_{i = 1}^{998} _{1997}C_{2i - 1}\left(\frac {1}{4}\right)^i = \sum_{j = 1}^{1995} _{1997}C_j \left(\frac {1}{2}\right)^{j + 1}$.

We are to use the equality ${ \boxed{\sum_{r = 0}^n {}_nC{}_rx^r = (1 + x)^n}\Longrightarrow \sum_{r = 0}^n {}_nC{}_rx^{r + 1} = x(1 + x)^n}$.

I have obtained the answer : $ \frac {3^{1997} - 2^{1997} - 1997*2 - 1}{2^{1998}}$.[/quote]

thanks for the quick answer kunny. I liked the use of the equality. I compared your answer to the official solution of the problem (I may post the problem in the combinatorics section tommorow) and it's not the same (actually, my answer should be the expression I posted * 2^1997). I think that your first equality is wrong, because in RHS there are twice as much terms compared to LHS.

[quote="bambaman"]$ \displaystyle\sum_{i \equal{} 1}^{t}\frac {\binom{nt}{ni \minus{} m}}{a^{i}}$

$ 0\le m\le n$[/quote]
Let $ g(x) \equal{} x^m(1 \plus{} x)^{nt}$, then
\[ f(a) \equal{} \sum_{i \equal{} 1}^tC_{nt}^{ni \minus{} m}a^{ \minus{} i} \equal{} \frac 1n \sum_{j \equal{} 1}^n g(xexp(\frac {2\pi ij}{n}),x \equal{} a^{ \minus{} 1/n}
\]
.
In first case $ n \equal{} 2,m \equal{} 1,x \equal{} \frac 12, f(4) \equal{} \frac {3^{2t} \minus{} 1}{2^{2t \plus{} 2}}.$

The following trick is probably the easiest way to solve your initial problem:
$ \frac{\left(1\plus{}\frac{1}{2}\right)^{1997}\minus{}\left(1\minus{}\frac{1}{2}\right)^{1997}}{2}\equal{}\sum_{i\equal{}1}^{999}\binom{1997}{2i\minus{}1}\left(\frac{1}{2}\right)^{2i\minus{}1}$
(only a few steps away from your sum).

Of course, this trick is not enough to solve the general problem. Rust's solution is the only way I know for that one.

I forget $ \frac 1n$ in exxpression:
\[ f(a) \equal{} \sum_{i \equal{} 1}^tC_{nt}^{ni \minus{} m}a^{ \minus{} i} \equal{} \frac 1n \sum_{j \equal{} 1}^n g(xexp(\frac {2\pi ij}{n}),x \equal{} a^{ \minus{} 1/n}
\]
.
I tdit my post.


Prove that the sum of all squares of the divisors of a natural number $ n$ is less than $ n^{\frac{5}{2}}$

Let $ \sigma_k(n)\equal{}\sum_{d|n}d^k$ is sum of k-th degree of divisors.
Obviosly $ \sigma_k(n)$ is multiplicative function. It mean $ sigma_k(nm)\equal{}\sigma_k(n)\sigma_k(m)$, when $ (n,m)\equal{}1$.
Therefore sufficiently chek your condition for $ n\equal{}p^m$, when $ \sigma_k(p^m)\equal{}1\plus{}p^k\plus{}...\plus{}p^{mk}\equal{}\frac{p^{(m\plus{}1)k}\minus{}1}{p^k\minus{}1}<p^{cm}$. Where $ c>\frac{ln(p^{(m\plus{}1)k}\minus{}1)\minus{}ln(p^k\minus{}1)}{mln(p)}$ for any $ p^m$. It is true if it is true for $ p^m\equal{}2,3$
For k=2 we get $ c>max(\frac{ln 5}{ln 2},\frac{ln 10}{ln 3})<2,322$.
Therefore $ \sigma_2(n)<n^{2,322},n>1$.


geometry
circumcircle
geometric transformation
rotation
geometry proposed

Given triangle $ ABC$. Points $ D,E,F$ outside triangle $ ABC$ are chosen such that triangles $ ABD$, $ BCE$, and $ CAF$ are equilateral triangles. Prove that cicumcircles of these three triangles are concurrent.

denote the circumcircles of ABD and ACF intersect at A and G
then angleBGC=360-240=120
So GBEC are concyclic

Let $ \{ T \} \equal{} AE \cap BF$. A $ 60^\circ$ rotation around C will map A to F and E to B, hence AE = BF and $ \angle ATF \equal{} 60^\circ ( 1 )$. From (1) we get that the quadrilaterals ADBT, BECT and CFAT are cyclic, q.e.d.

Best regards,
sunken rock

we know that in a quadrilateral  ABCD if   ∠A+∠C=180° then we can draw a circle around ABCD.      let ω_1  be the  circumcircle of ABD and ω_2  be the circumcircle of   ACF .  ω_1  intersects ω_2  at  P    ∠APB=180°-60°=120° and  ∠APC=180°-60°=120°   so  ∠BPC=120° →∠BCP+∠BEC=180°      so the point P would be on the circumcircle of of BEC

Try the generalization  :wink:

Given triangle $ ABC$ points $ D,E,F$ outside triangle $ ABC$ are chosen such that $ <BDC\plus{}<CEA\plus{}<AFB\equal{}180^0$. Prove that cicumcircles of the  triangles $ BDC,CEA,AFB$ are concurrent

Call P the second common point of the circles (BCD) and (ABF), see that $ \angle AEC \plus{}\angle CPA \equal{} 180^\circ$.

Best regards,
sunken rock


topology
real analysis
real analysis unsolved

let (An) be a decreasing, nested sequence of non empty closed subsets of a complete metric space M. further, if diameter of An tends to 0, show that intersection of all An's contains exactly one point. this must be simple for most of you here, however i'm just starting .

i have an idea of how to do it. heres wat i think. clearly we have diameter of A=0. where A is intersection of all the An's. so A contains at most one point.

 if somehow i construct a cauchy sequence (an) such that an lies in An, we have that because of M being complete , the sequence must converge to some p. so by this we have the intersection contains atleast one point. from this we get the solution. could someone give a hint as to how to construct such a cauchy sequence?

Any sequence will do, as long as $a_n\in A_n$. If $A_N$ has diameter $c$, $a_n\in A_N$ when $n\ge N$ and $d(a_n,a_m)\le c$ when $m,n\ge N$.

ofcourse. how stupid of me? u've been very patient with me, jmerry- thanks


factorial
limit
logarithms
calculus
calculus computations

Let the notation $ log_n (x)$ denote the application of the natural log n times to x, eg log^2 p = log ( log p) ).
Also let $ x^n$ denote the application of the factorial n times to x, NOT x raised to the exponent of n. eg x^3 = x!!!.

Find       $ \lim_{n \rightarrow \infty}{ a_{n} }$ where $ a_n \equal{} log_n (3^n)$

The sequence is increasing because $ \log(N!)\approx N\log N > N$ for large $ N$. On the other hand, $ \log(N!) < N^2$ for large $ N$, and the effect of squaring the argument of $ \log_n$ is miniscule when $ n$ is large. So the limit exists (the series $ \sum (a_{n\plus{}1}\minus{}a_n)$ is convergent), but I have no idea about its exact value. The first four terms are approximately
1.791759469
1.883920941
2.116177553
2.116425936
(I used Stirling's approximation to $ (720!)!$). The sequence practically stops there. I'm not sure if there is a closed form for the limit value. Daniel Velleman's article "Exponential vs Factorial" in AMM 113:8 establishes the existence of the limit, but not its value.

The above approximations were computed with Maple. MATLAB takes a typical engineering approach:
[quote]>> factorial(3)
ans =  6
>> factorial(ans)
ans = 720
>> factorial(ans)
ans = Inf[/quote]


On side BC of $ \triangle ABC$ point A1 is taken so that $ BA_{1} : A_{1}C \equal{} 2 : 1.$ What is the
ratio in which median $ CC_{1}$ divides segment $ AA_{1}$?

This needs to be shown for a general triangle, which I
do not do here.

I will come up with an answer for a triangle with conditions
that fits, but it is not exhaustive for this problem.  By
showing this, it is promising for the problem as a whole.

Note:  I CHOSE the following values to make my
computations relatively easier for me based on a
diagram that came to mind.

In the xy-plane, let the vertices of A, B, and C be
(0,0), (2,0), and (2,3), respectively.  Point A1 is at (2,2)
because it's 2/3 of the way along from B to C.  Point C1
is at (1,0) because it's the endpoint of median CC1.

The equation of the line from point A (0,0) to point A1 (2,2)
is y = x.  The equation of the line from point C (2,3) to
point C1 (1,0) is y = 3x - 3.

Set these equal to each other and solve for x, and then y.

x = 3/2
Because y = x, then
y = 3/2

The intersection point is (3/2, 3/2).

According to the problem, we need to know the ratio of the parts
of segment AA1.  So in my model, we need to know the ratio of
the length of the line segment from point A (0,0) to the point
(3/2, 3/2), compared to the length of the line segment from the
point (3/2, 3/2) to point A1 (2,2).

Using the distance formula, the distance from (0,0) to (3/2, 3/2)
is (3/2)sqrt(2).  The distance from (3/2, 3/2) to (2,2)
is (1/2)sqrt(2).

The ratio of the longer section of segment AA1 to its
shorter section, therefore, is 3:1.

In general case, using elementary geometry use:
[hide]Ceva's thereom[/hide]

Maybe you mean Menelaus' Theorem?

Let $ AA_1$ and $ CC_1$ intersect at point $ O$.
Consider triangle $ ABA_1$ and line $ AC_1$
By Menelaus' Theorem, we get
$ \frac{AC_1}{C_1B} . \frac{BA}{AA_1} . \frac{A_1O}{OA} \equal{} 1$
$ 1 . \frac{3}{1} . \frac{A_1O}{OA} \equal{} 1$
Therefore
$ \frac{A_1O}{OA} \equal{} 3 : 1$

By Ceva's Theorem:

Let $ AA_{1}$ and $ CC_{1}$ intersect at point O.
Let line $ BO$ intersect $ AC$ at $ B_{1}$
From Ceva's Thereom we have:
$ \frac{BA_{1}}{A_{1}C}\frac{CB_{1}}{B_{1}A}\frac{AC_{1}}{C_{1}B}=1$
$ \frac{2}{1}\frac{CB_{1}}{B_{1}A}\frac{1}{1}=1$
$ \frac{CB_{1}}{B_{1}A}=\frac{1}{2}$
$ A_{1}B_{1}=\frac{AB}{3}$ and $ \triangle ABO$ is similar to $ \triangle A_{1}B_{1}O$
$ \frac{AO}{OA_{1}}=\frac{AB}{A_{1}B_{1}}=\frac{1}{3}$


rotation
geometry
geometric transformation
Asymptote

No, because the center of the rotation is the origin by default. If you want to rotate around some other point, say, the same $ (6,3)$, by the same $ 30^\circ$ use [b]rotate(30,(6,3))[/b] instead. Also, the angle marked as $ 30^\circ$ is actually much smaller. Seems like either your solution, or, at least, your diagram is incorrect :(.

Ha Ha bad label. (The 30 degree is the circle)
Next question.
How do you make the angle mark bounded by the the red lines?
Also how would you color in the inside of the red section?

There are three commands that you should learn: Arc(), dir(), and fill() (try to look them up in the manual yourself). Once you've done it, look at this code
[asy]size(100);
pair A=(6,3);
real r=2.5;
pair B=A+r*dir(90), C=A+r*dir(120);
fill(A--B--Arc(A,r,90,120)--cycle, yellow);
draw(Circle(A,r));
draw(C--A--B,red);[/asy]
Since now you have the three ponts needed for the anglemark command, you can figure out how to mark the angle yourself ;),


Can anyone write out the reaction (called Walden inversion) that happens when malic acid reacts with $ PCl_5$ and then undergo inversion to form a chloride? I am really confused with this. Does a Cl leave PCl5 to attack the hydroxyl group on malic acid, which is the leaving group?

Hydroxyl groups react with $ \ce{PCl5}$ to form a good leaving group and a "free" chloride ion, which then acts as a nucleophile in an SN2 reaction.


calculus
derivative
logarithms
calculus computations

Calculate the 100th derivative of $\frac{1}{x^2+3x+2}$ at x=0 with
10% relative error

Hi, Kent,

I saw this topic a bit too late, but in case you'll ever return here. How did you do this approximation of 100! ? At Arnold's exams you are not alowed to use computer or calculator. I thought that approximation of this expression was the hardest thing in this task, but may be you know some formula. I thought about Stirling's one but it will still require some proof and estimation.

thx,

Stan

I'm sure I used a computer - probably a spreadsheet - when I posted that. But if I had to do it by hand:

By Stirling, $100!\approx \sqrt{2\pi\cdot 100}\cdot100^{100}e^{-100}.$

$\log_{10}100^{100}e^{-100}=200-100\log_{10}e\approx 200-43.43=156.57.$

Take the antilog, and we get $\approx \sqrt{2\pi}\cdot 10^{.57}\cdot 10^{157}.$

$10^{.5}\approx 3.16$ and $10^.6\approx 4$ so crude interpolation gets us that $10^{.57}\approx 3.7.$

As for $\sqrt{2\pi},$ $2.5^2=6.25$ so it's about $2.5.$ We multiply $3.7\cdot 2.5$ to get about $9.2$ So we report the answer as $9.2\times 10^{157},$ with the second significant figure dubious.

This depends critically on the fact that I have certain decimal approximations memorized. 1.414; 2.303; .866; .693. Do you recognize these numbers? $\log_{10}e\approx.4343$ is on that list.

The fact that I have all of these numbers memorized may have something to do with going to high school just before the invention of hand calculators - and learning to rely on a base-10 logarithm table as a computational aid.

Impressive. I wish I could operate logarithms and roots approximations like this.