Trish and Carla are asking each other trivia questions. Each time one of the girls gives a correct answer, the other gives her one cent. At the end of the game, Trish had gotten three questions correct, and Carla had three more pennies than when she started. How many questions were answered correctly?

If there is money flow, either Trish got a question correct, or Carla got a question correct. Thus, if we know how many Trish got correct as well as the net money change, we could calculate how many Carla got correct. Since Trish got three questions correct, and Carla had a three-penny gain, our equation is $ \minus{}3\plus{}x\equal{}3$, and solving this, $ x\equal{}6$. Adding the two quantities together, $ 3\plus{}6\equal{}\fbox{9}$.

quicker way: if carla gained 3 pennies, she got 3 questions MORE right than trish. trish got 3 so carla got 6. 3+6=9

[hide]3+6=9 questions answered correctly.[/hide]


inequalities
geometry
geometric transformation
reflection
rectangle
perimeter
area of a triangle

Given a quadrilateral $Q$ with sides $a, b , c , d$ and area $S$.

If $a \leq b \leq c \leq d$, prove that  $S\leq \frac{3\sqrt{3}}{4}{c^2}$.

Determine when does the equality sign hold.

[b]HINT[/b]( a very big hint  :idea: )
[hide]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=856[/img][/hide]

[quote="shobber"]I don't know the meaning of your graph, but it can be solved using AM-GM.

Another remark is, this inequality is called the Popa Inequality.[/quote]

Well, i would like to see how it can be done. Can u show how?

I think it's when a=b=c, the equality sign holds.

In the luimichael's figure, let O be the mid-pt. of AD.

Thus angle AOB, angle BOC, angle COD are = pi/3 rad

i.e. 3 equilateral triangles.

The area S, is thus = $(3*sqrt(3))/4$

Whenever a < c slightly,

by heron's formula, and you got s=a+a+c/2, which must be larger than 3a/2, therefore one of the original equilateral triangle would have S'/3 > S/3.

If the four sides are given, then the area is largest if and only if the quadrilateral is cyclic. Thus if we can prove that the inequality true for cyclic quadrilaterals, then it is also true for any other quadrilaterals.

We know that $S^2=(p-a)(p-b)(p-c)(p-d)$. Also $(p-d)=a+b+c-p$. Hence:

$S^2=(p-a)(p-b)(p-c)(a+b+c-p)$

$S^2=3^3\cdot\prod\left( \frac{1}{3}p-\frac{1}{3}a \right) \cdot(a+b+c-p)$

$S^2 \leq 3^3 \left( \frac{\frac{2}{3}(a+b+c)}{4} \right)^4$

$S^2 \leq 3^3 \left( \frac{c}{2} \right)^4$

Therefore we get:
\[ S \leq \frac{3\sqrt{3}}{4}c^2  \]
With equality holds if and only if $a=b=c$.

very good! shobber.

Here is my solution. ( also using isoperimetric property.)
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=856[/img]
[hide]    Reflect quadrilateral $ABCD$ along the longest side $AD$, we get the convex hexagon $ABCDC'B'$ as shown in the figure  ( degenerate cases like convex pentagon or rectangle can be regarded as limiting cases),and the perimter of the hexagon is constantly $2(a+b+c)$.

    By the isoperimetric property, a hexagon with constant perimeter has maximum area if and only if the hexagon is a regular hexagon.

Therefore,  $2S\leq \frac{3\sqrt{3}}2(\frac{a+b+c}3)^2$.
Since $a \leq b \leq c$, we have $S \leq \frac{3\sqrt{3}}4 c^2$.

Equality sign holds if and only if $a = b = c$ and $d = 2a$.[/hide]


induction
ceiling function
search
combinatorics unsolved
combinatorics

Given 365 cards, in which distinct numbers are written. We may ask for any three cards, the order of numbers written in them. Is it always possible to find out the order of all 365 cards by 2000 such questions?

Suppose I already know the order of i cards, where 3^(k-1)=<i<3^k.

I find the number of questions needed to know the position of an (i+1)th card relative to the i cards.
If k=1, then it takes only 1 question.
Suppose that it takes m questions where k=m. Then for k=m+1, I use one question in this way:
I pick two cards so that there are at most (3^m)-1 cards less than the smaller of the chosen cards, between the two chosen cards, and more than the larger chosen card. I can do this since I have at most 3^(m+1)-1 cards. Then I pick the extra card. Now I will know which group that card belongs to, and it will take m further question to determine its exact location relative to the rest of the cards.

So, by induction, it will take k more questions to add one more card to our knowledge of relative positions.

Then it will take 1+(2*3)*2+(2*3^2)*3+(2*3^3)*4+(2*3^4)*5+123*6=1831 questions to determine the relative order of the 365 cards. Which is what we want.

Is this correct?

The same problem with $ 400$ cards and $ 2035$ questions was Argentina IMO TST 2006 problem 2

I think my proof is different from the ones above, it's kind of like a triple merge sort from computer science. hopefully I didn't make some silly mistake.

Suppose we have $3n$ cards. We divide the cards into $3$ groups and sort each of them independently, so that we obtain one group $a_1<a_2...<a_n$, another $b_1<b_2...<b_n$ and the last group $c_1<c_2...<c_n$. So I ask a question about $a_1,b_1,c_1$, and whichever is the smallest must be the smallest in the entire list, WLOG $a_1$, then I ask a question about $a_2,b_1,c_1$ so whichever is the smallest out of these will be the next smallest in the entire list, and I can keep adding $1$ element in order, and with $3n-2$ questions the entire list will be in order (since for the last $3$ I add them all at once). Now just keep applying the above strategy and you get the recursion $f(3n)=3f(n)+3n-2$

I'm too lazy to count for $365$, but counting for $405$ isn't too hard so I just did that and you need to know that for $5$ cards you can do it in $4$ questions (the strategy is not hard)
For $405$ it would take $403 + 3(133) + 9(43) + 27(13) + 81 (4) = 1864$, and so for $365$ we can do it in less than $2000$ questions for sure.

Edit: I manned up and did the calculation for $365$, it gives $1655$ questions, now its just a question of whether or not my proof is right. I should also mention the proof by Yossarian is essentially like a binary search but ternary, so I guess both strategies come from computer science.

[quote=Pedram-Safaei]yes it is always posible.suppose that $f(n)$ be the least number of questions that required for find orders of $n+1$ number such that we know order of $n$ of them,$g(n)$ be the minimum number of such questions that require for finding order of $n$ number,so it is obvious that $g(n+1)\leq g(n)+f(n)$.
now prove that $f(n)=f(\left \lceil \frac{n-2}{3} \right \rceil)+1$ and then we find that $g(365)\leq 1828$ and the desired result.[/quote]

How do you prove that?


Circle (x-6)^2+y^2=25 intersects (x-1)^2+(y-2)^2-8 in two points. Find the slope between the two points.

Alternatively, you could find the slope of a line perpendicular to the line segment between the centers of the circles.\[ \frac{0\minus{}2}{6\minus{}1}\equal{}\frac{\minus{}2}5\]\[ m\equal{}\frac52\]


how come the odd of winning for Box Play in Pick 3 is 1:167 ??

should that be 1:220 ??

you pick three numbers from 0 - 9, numbers can be repeated and order doesn't matter.

Three repeated numbers: $10$ possible way
(ie. 000,111,222, .... )

Two repeated numbers : $10 \cdot 9=90$ possible way
(ie. 001, 303, 448, etc  Note:001,010,100, are the same)

no repeated numbers: ${10 \choose 3}= 120$ possible way
(ie. 123,256,746,021, etc)

$10+90+120 = 220$

Am I doing something wrong here??

Odds of $p : q$ mean that the probability of winning is $\frac{p}{p+q}$.

I don't quite understand the problem.  How do you win?

Oops.. my previous entry was completely wrong. I owe an apology to New Mexico state lottery officials.  :blush:

If kelvin911's description of the rules is correct, then the probability of winning depends on the player's combination of numbers (it will be different for $111$, $112$, and $123$), so it's not clear what should be understood by "odds of winning" in that case.

Oh, sorry, that $p : q$ notation means different things depending on whether it's a probability or a ratio.  It would be more appropriate to use a fraction to remove all doubt.

In any case, [b]kelvin911[/b], (edit:) I still don't see how your approach works.  Possible ways to do what?

it's like 649 lotto... you pick 3 numbers from 0 - 9,
for example, you can pick 471, 836, 005, 494, 111, etc

and the computer randomly pick three.  the odd of winning it is 1:167 by their table, but i wonder how you get that?

i got 1:220 because there are 220 possible winning numbers

someone told me that because for straight play (which the order matters), the odd is 1:1000, because $_{10}P_{3}= 1000$ and ${\frac{_{10}P_{3}}{3!}= \frac{1000}{6}= 166.666... = 167}$

but i am not convinced because of the repeated number is allowed

In the New Mexico version you are supposed to pick three [b]distinct[/b] numbers to play the "6-way". Then the chance to win is $6/1000\approx 1/167$.
  Does your lottery have a similar provision? And even if it doesn't, the player should still pick three distinct numbers to maximize the chance to win.

maybe i misunderstand the Box Play definition....  can you fully explain what's Box Play and your version 6 way Box Play ?

[url=http://www.nmlottery.com/Pick3/PickThree.htm]New Mexico's Three-Number Game[/url]

by looking at that table, i still dont think the odd should be 1 in 166.67

i mean there are 220 different possible outcomes  i know you can divided by 6 if repeated number is not allow.  but then it is simply $10\choose 3$ which is 120 or $\frac{10 \cdot 9 \cdot 8}{6}$

[quote="mlok"]The crucial point is that there are $1000$ possible (and [b]equally likely[/b]) outcomes, namely outputs of the computer picking three numbers. If you choose three distinct numbers playing "6-way", then $6$ of the outcomes are favorable to you. Hence $p=6/1000$.[/quote]

i think the cruial point is also you have to play with 3 DIFFERENT numbers.  otherwise the odd is not 6:1000

[b]You[/b] have to play with three different numbers, but I see nothing on the website indicating that [b]the lottery[/b] does.  In other words, no matter which play style you pick, the lottery picks three numbers that may or may not be different.

if you play with 3 distinct numbers then your odd is $\frac{3!}{10^{3}}= \frac{6}{1000}$
if you play with 2 same number and 1 different (ie 112) then the odd is $\frac{\frac{3!}{2!1!}}{10^{3}}= \frac{3}{1000}$
if you play with 3 same numbers then the odd is $\frac{1}{1000}$


A set $C$ of positive integers is called good if for every integer $k$ there exist distinct $a, b \in C$ such that the numbers $a+k$ and $b+k$ are not relatively prime. Prove that if the sum of the elements of a good set $C$ equals $2003$, then there exists $c \in C$ such that the set $C-\{c\}$ is good.

A nice application of the Chinese Remainder Theorem :)
[i]Claim:[/i]
 A set $ C$ is [i]good [u]iff[/u][/i] there exists a prime number $ p$ such that for each remainder $ i\equal{}\overline{0,p\minus{}1}$ modulo $ p$, there exist $ 2$ different numbers in $ C$, both congruent to $ i\pmod p$.
[i]Proof:[/i]
Assume firstly that such a prime number $ p$ exists for a set $ C$. Let $ a_i,b_i\in C$, $ i\equal{}\overline{0,p\minus{}1}$ such that $ a_i\neq b_i$ and $ a_i\equiv b_i\equiv i\pmod p$. Then for any integer $ k\equiv s\pmod p$, the numbers $ A\equal{}a_{\minus{}s}$ and $ B\equal{}b_{\minus{}s}$ (indices taken modulo $ p$) satisfy $ p|\gcd(k\plus{}A,k\plus{}B)$, proving that $ C$ is [i]good[/i].
For the other direction, suppose we have a good set $ C$ and no such prime $ p$ exists. Let $ a\minus{}b\equal{}d$ be the greatest difference of two numbers from $ C$. Let $ 2\equal{}p_1<p_2<\ldots<p_t\le d<p_{t\plus{}1}<\ldots$ be the sequence of prime numbers. According to our assumpion for each $ i\equal{}\overline{1,t}$, there exists a remainder $ r_i$ upon $ p_i$ which is met at most one time among the numbers in $ C$. By the Chinese Remainder Theorem, the system of simultaneous equations $ x\equiv \minus{}r_i\pmod p_i$, for $ i\equal{}\overline{1,t}$ admits an integer solution (actually, infinitely many). Let $ k$ be a solution. Because $ C$ is a [i]good[/i] set, there exist $ a\neq b\in C$ such that $ g\equal{}\gcd(k\plus{}a,k\plus{}b)\neq1$. Let $ q$ be a prime number dividing $ g$. Then $ q\le g\le d$ so $ q\equal{}p_j$, for some $ j\equal{}\overline{1,t}$. Then $ k\equiv\minus{}r_j\pmod q$. Because $ q|k\plus{}a$ and $ q|k\plus{}b$, we must have $ a\equiv b\equiv r_j\pmod q$. This is a contradiction to the choice of $ r_j$.
[i]Back to the problem:[/i]
Since $ C$ is a good set, according to our claim, there exists a prime number $ p$ and pairwise different numbers $ a_0,\ldots,a_{p\minus{}1},b_0,\ldots,b_{p\minus{}1}$ such that $ a_i\equiv b_i\pmod p$. The sum $ S$ of these numbers is congruent to $ 2(0\plus{}1\plus{}\ldots\plus{}p\minus{}1)\equal{}(p\minus{}1)p\equiv0\pmod p$. It is clear that $ p<2003$. Because $ 2003$ is a prime number, $ 2003\neq S$, which means that $ C$ has at least one more element (which complete(s) the sum $ S$ till $ 2003$). Since the set $ C'\equal{}\{a_0,a_1,\ldots,a_{p\minus{}1},b_0,b_1,\ldots,b_{p\minus{}1}\}$ is [i]good[/i], any element of $ c\in C\setminus C'$ can be removed so that $ C\minus{}\{c\}$ is a [i]good[/i] set.


Since nobody else has tried this, I guess I will take the first shot and get proven wrong.
[hide]Let $ z=2y$.  Then:
$ \frac{z^{3}}{2}+\frac{\sin(z)}{2}+a=0$
$ z^{3}+\sin(z)+2a=0$
$ -z^{3}-\sin(z)=2a$

And

$ x^{3}+\sin(x)=2a$

So $ x^{3}+\sin(x)=-z^{3}-\sin(z)$

If we let $ f(a)=a^{3}+\sin(a)$, we notice that $ a$ is an odd function and is strictly increasing from $ \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$.  So that means $ x=-z$.

So $ \cos(2y+x)=\cos(x+z)=\cos(x-x)=\cos(0)=1$[/hide]


geometry
3D geometry
pigeonhole principle
number theory
prime factorization

1. If n points are placed at random inside a unit cube, what is the lowest n guaranteeing that two of these points are no more than rt3/2 apart?

2. There exists two 2-digit number such that the difference between their sum and their product is 2009. What is the larger 2-digit number?

3. Han stands at (-2,0) on a plane. Han's cake sits at (2,0) on the plane. Centred at (0,0) there is an impenetrable circle of radius 1. If Han is starving and infinitesimally thin, how far must he walk to reach his cake?

4. A restaurant offers 10% off the price of the bill that costs at least 10 dollars, 20% off the price of a bill that costs at least 20 dollars, and so on. Everything on the menu costs an integer number of dollars. What is the most you will pay?

5. Julie asks Lu to think of a 3-digit number (abc) where a,b and c represent digits in base 10 in the order indicated. Julie then asks Lu to form the numbers (acb), (bca), (bac), (cab) and (cba), to add these five numbers, and to reveal there sum, N. Lu tells Julie that N=2609. Play the role of Julie and determine the original 3 digit number (abc).

6. An equalateral triangle has vertices at (0,0), (a,5) and (b,13). Find the area of the triangle.

7. Find a 9-digit number, made up of only the digits 2 and 9, that leaves a remainder of 29 when divided by 2^9.

I got the majority of them but using disgusting methods. Can you please post up any solutions to these Q. Thanks

[b]Q5:[/b]

We have $ 122a\plus{}212b\plus{}221c\equal{}2609$

Adding the number $ (abc)$ to both sides we get:

$ 222a\plus{}222b\plus{}222c\equal{}2609\plus{}(abc)$

$ 222(a\plus{}b\plus{}c)\equal{}2609\plus{}(abc)$

Since $ (abc)$ is an integer, we now know that $ 2609\plus{}(abc)$ must be a multiple of 222.

Possible values of (abc) are: $ 277$, $ 499$, $ 721$, $ 943$.

Checking the sums of rearrangements of these numbers shows that of these, only 943 has the required sum.

Thus the answer is $ (abc)\equal{}943$

[quote="addikaye03"]6. An equalateral triangle has vertices at (0,0), (a,5) and (b,13). Find the area of the triangle.[/quote]
An equilateral has all sides equal so $ \sqrt{b^{2}} \equal{} \sqrt{a^{2}\plus{} 25}$    (*)
Now the area of an equilateral triangle $ \equal{} \frac{\sqrt{3}s^{2}}{4}$ where $ s \equal{}$ side length
now drop a perpendicular from (a,5) to the y axis giving us another way to represent the area as $ \frac{ab}{2}$
Setting
$ \frac{\sqrt{3}s^{2}}{4} \equal{} \frac{ab}{2}$
$ 2a \equal{} \sqrt{3}b$
$ a \equal{} \frac{\sqrt{3}b}{2}$
Sub the above into (*) we have $ b^{2} \equal{} \frac{3b^{2}}{4} \plus{} 25$ or $ b \equal{} 10$
Therefore the area is $ \frac{\sqrt{3}b^{2}}{4} \equal{} 25\sqrt{3}$

[hide="Question 1"]Use the pigeonhole principle. Suppose we divide the unit cube into eight congruent cubes of side $ \frac{1}{2}$. Then, if we had nine points, there must be two points in some half-cube. Since the diagonal of the half-cube is $ \frac{\sqrt{3}}{2}$, nine points is sufficient. Arranging eight points at the corners shows that nine is necessary as well. [/hide]

[hide="Question about #2"]For #2, do you mean that there are two 2-digit numbers $ a$ and $ b$ such that $ |a\plus{}b\minus{}(\sqrt{a}\plus{}\sqrt{b})|\equal{}2009$? That seems impossible...[/hide]

[hide="Question 3"]Imagine a rubber band stretching along Han's path. Clearly it will go along the circumference of the circle for a time. Draw perpendiculars from the center of the circle to the points where the path strays away from the circle. Using Pythagoras, we find the path not along the circle to be $ 2\sqrt{3}$. Since we have two $ 30\minus{}60\minus{}90$ triangles formed, we find that the path along the circle is $ \frac{1}{6}$ the circumference, so it is $ \frac{\pi}{3}$. We have an answer of $ \frac{\pi}{3}\plus{}2\sqrt{3}$. [/hide]

The following solution isn't stellar, but it works, I guess, and it's not too much computation.

[hide="Question 7"]Let's suppose the last two digits are 29. We find that seven digits, each 2 or 9, followed by 2 zeros, is divisible by 512. Divide by 100; the leftmost 7 digits must be divisible by 128.

Now subtract 2222222, leaving a number with 7 digits, each 0 or 7.

2222222 leaves a remainder of 14 mod 128, so we're looking for the 7 digits (of 0s and 7s) to be 114 mod 128.

Mod 128, we find that:
7 is 7
70 is 70
700 is 60
7000 is 88
70000 is 112
700000 is 96
7000000 is 64

This isn't as much computation as it looks -- you can actually derive the entire thing above by just dividing 7000000 by 128 using long division -- the other numbers come out along the way (do you see why?).

The sum of all of these remainders 497, just under 512, so we know that we need a sum of some of these to be equal to one of:
114; 242; 370; (498, but we realize that this is over 497, so it can't be this)

By trial and error (not too bad given the relative sizes of the numbers), we find 70+60+112 = 242, so our 7 digit number is 70770, so we add 2222222 to this, multiply by 100, and add 29 to get 229299229[/hide]

[b]Q4.[/b]

Obviously, anything at least 100 dollars will cost nothing, so we can discount these possiblities.

The greatest amount you will have to pay for each discount will be the maximal element which achieves this discount.

Our possiblities are therefore a 19, 29, 39, 49, 59, 69, 79, 89 or 99 dollar bill.

Working out the amount you need to pay for each of these, we see that the greatest amount to pay is for a 59 dollar bill, or $ 29.5$ dollars.

For #2, do you mean that there are two 2-digit numbers $ a$ and $ b$ such that $ |a \plus{} b \minus{} (\sqrt {a} \plus{} \sqrt {b})| \equal{} 2009$? That seems impossible...

Sorry everyone i fixed number 2[/quote]

Call the two numbers $ x$ and $ y$. We have $ xy\minus{}x\minus{}y\equal{}2009$

We can rearrange this to $ x\plus{}2009\equal{}y(x\minus{}1)$

$ y\equal{}\frac{x\plus{}2009}{x\minus{}1}\equal{}1\plus{}\frac{2010}{n\minus{}1}$

Thus for $ x$ and $ y$ to both be integers, we need $ (x\minus{}1) \mid 2010$ and $ x\minus{}1$ to be sufficiently large such that $ y$ is two digits.
The prime factorization of 2010 is $ 2 \cdot 3 \cdot 5 \cdot 67$ and from this we infer that the only two digit factors are:

$ 10,15,30,67$

Of these, $ 10$ and $ 15$ are too small to give two digit values for $ y$

$ x\minus{}1\equal{}30$ gives $ (x,y)\equal{}(31,68)$
$ x\minus{}1\equal{}67$ gives $ (x,y)\equal{}(68,31)$

Thus the two numbers are $ 31$ and $ 68$.


geometry
incenter
inradius
search
trapezoid
circumcircle
geometry unsolved

Given triangle $ABC$ incenter $I$, circle center $(I,t)$(with $t>r$-inradius) cuts triangle formed a hexagon, prove that three lines which are through the midpoints of opposite sides of that hexagon are concurrent.

We can have an easy solution based on the follows Lemmas:

[b][size=117]LEMMA  1.[/size][/b] – A triangle $ABC$ is given and let $I$ be, it’s incenter. We draw a circle taken as center the point $I,$ with radius bigger than the inradius of $\bigtriangleup ABC,$ which intersects the sidelines $BC,$ $AC,$ $AB,$ at pairs of points $D',$ $D''$ and $E',$ $E''$ and $F',$ $F'',$ respectively. Prove that $D'D'' = E'E'' = F'F''$ $($ easy to prove $).$

[b][size=117]LEMMA  2.[/size][/b] – In every inscribed hexagon of which three it’s non-adjoined sides are congruent, the lines connecting the midpoints of it’s opposite sides, are concurrent at one point.

I think that the above Lemmas are well known results, but I am not good in references. So, I will search my notes about their proofs and I will post them here next time.

Kostas Vittas.

Dear gemath and all my friends.

The Lemma 2, is a special case of another general one, which has already been published by [b][size=109]Nikos Kyriazis[/size][/b] $($ more friends of mine who are strong geometers, named [b][size=109]''Nikos''.[/size][/b]  Nikos Dergiades, Nikos Iosifides, Nikos Kyriazis, Nikos Rapanos,...$).$

Please see at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131804

So, we can solve your problem as follows:

$\bullet$ It is easy to prove that $D'D'' = E'E'' = F'F''$ and that the isosceles triangles $\bigtriangleup ID'D'',$ $\bigtriangleup IE'E'',$ $\bigtriangleup IF'F'',$ are similar $($ here is also congruent $).$

Hence, based on the Nikos Kyriazis’s result, we conclude that the lines connecting the midpoints of the opposite sides of the convex hexagon $D'D''E'E''F'F'',$ are concurrent at one point and the proof is completed.

Kostas Vittas.

PS. The proof of [b]Kyriazis[/b] result is another story...  :wink: and I don’t know if it is a well-known fact. So, I am eager to learn if there is any reference about it.

 $($ Some times we think a solution of a problem, as an our discovery, but many times somebody else has done the same. As Francois Rideau said, ''nothing is new under the geometry sun'' $).$

[b][size=117]PROOF OF THE LEMMA  2.[/size][/b]

Let $ABCDEF$ be, an inscribed hexagon of which $AB = CD = EF.$

We denote as $K,$ $L,$ $M,$ $K',$ $L',$ $M',$ the midpoints of $AB,$ $CD,$ $EF,$ $DE,$ $AF,$ $BC$ respectively and we will prove that the segment lines $KK',$ $LL',$ $MM',$ are concurrent at one point.

$\bullet$ Because of $AB = CD$ we have that the quadrilateral $ABCD,$ is a trapezium and so, $BC\parallel KL$ and similarly $AF\parallel KM$ and $DE\parallel LM.$ Hence, the triangles $\bigtriangleup K'LM,$ $\bigtriangleup L'KM$ and $\bigtriangleup M'KL,$ are isosceles and we will prove that are also similar.

From $BC\parallel KL$ $\Longrightarrow$ $\angle BM'K = \angle M'KL$ $,(1)$ and from $AF\parallel KM$ $\Longrightarrow$ $\angle AL'K = \angle L'KM$ $,(2)$

It is easy to show that the quadrilaterals $AKOL',$ $BKOM',$ are cyclic taken as diameters the congruent segments $OA,$ $OB,$ respectively and so, their circumcircles are congruent. Hence, because of $KA = KB,$ we have that $\angle BM'K = \angle AL'K$ $,(3)$

From $(1),$ $(2),$ $(3)$ $\Longrightarrow$ $\angle M'KL = \angle L'KM$  $,(4)$

From $(4),$ by cyclically consideration, we conclude that the triangles $\bigtriangleup K'LM,$ $\bigtriangleup L'KM$ and $\bigtriangleup M'KL,$ are similar.

We have now, the configuration of the triangle $\bigtriangleup KLM$ and three similar isosceles triangles are erected outwardly to it, with bases it’s sidelines.

So, as well known $($ [b]Kiepert[/b], or [b]Jacobi[/b], or ''[b]Isogonic[/b]'' theorem...$),$ we conclude that the segment lines $KK',$ $LL',$ $MM',$ are concurrent at one point and the proof is completed.

Kostas Vittas.


function
limit
real analysis
real analysis unsolved

Indicator function for a set, $ S$, is defined by: $ \chi_{S}(x) \equal{} 1$ if $ x\in S$ and $ 0$ if $ x\not\in S$.
a)  Let $ I \equal{} [a,b]\subset\mathbb{R}$ be a compact interval.  Show that $ \lim_{x\rightarrow \pm\infty}\widehat{\chi}_I(x) \equal{} 0$

b)  Let $ \{I_j\}_{j \equal{} 1}^{n}$ be a finite collection of compact intervals in $ \mathbb{R}$, and let $ \{c_j\}_{j \equal{} 1}^{n}$ be a finite collection of constants.  let $ g \equal{} \sum_{j \equal{} 1}^{n}c_j\chi_{I_j}$. Using part a, show that $ \lim_{x\rightarrow \pm\infty}\widehat{g}(x) \equal{} 0$

Please fully justify your solution, (i think its an easy problem but fully justifying everything takes some work).
also I would like to know how we can use this to come up with a different proof of Reiman-lebesgu lemma.

Let's spell it correctly: Riemann-Lebesgue.

Searching for that works, but turns up lots of false positives.

A good discussion, including my favorite proof, is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=91119]here[/url]. See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=131611]this thread[/url] for an outline of the density arguments.


group theory
abstract algebra
inequalities
geometry
geometric transformation
number theory proposed
number theory

Let's say a positive integer $ n$ is [i]atresvido[/i] if the set of its divisors (including 1 and $ n$) can be split in in 3 subsets such that the sum of the elements of each is the same. Determine the least number of divisors an atresvido number can have.

120 is [i]atresvido[/i], so then minimal number of divisors must be at most 16.

It is easy to prove that if $ n$ is a number of the form $ p^{k}$, $ pq$, $ p^{2}q$ and $ pqr$ ($ p,q,r$ distinct primes), then $ n$ can not be [i]atresvido[/i]. My conjecture is for satisfies this condition, $ n$ must be at least of the form $ p^{3}qr$ ($ p,q,r$ distinct primes).

Really I'm trying to make a relation between the [i]atresvido[/i] condition and the descomposition of a group into Sylow p-subgroups product.

i have done this by brute force.checking all the cases and making some inequalities with the sum of divisors formula.

It is obvious that the sum of divisors of $ N$ is at least $ 3N$ if $ N$ is atresvido, since one of the subsets must contain $ N$.

The sum of divisors of any positive integer $ N$ with $ 2n$ divisors has as an upper bound $ \sum_{k \equal{} 1}^{n}\left(k\plus{}\frac{N}{k}\right)$, and if it has $ 2n\plus{}1$ divisors (i.e., if $ N$ is a perfect square), then the upper bound is $ \sqrt{N}\plus{}\sum_{k \equal{} 1}^{n}\left(k\plus{}\frac{N}{k}\right)$.  This comes from the fact that, for any two divisors $ 1\leq d < D\leq\sqrt{N}$, then $ d\plus{}\frac{N}{d}\geq D\plus{}\frac{N}{D}$.  Then,

1) any $ N$ atresvido must be the product of at least 4 (not necessarily different) primes.  Otherwise, it has at most 8 divisors (includind 1 and itself), and it must hold that $ 3N\leq10\plus{}\frac{25N}{12}$, or $ N\leq\frac{120}{11}< 11$.  It is a trivial exercise to show that no $ N\leq10$ can be atresvido.

2) any $ N$ atresvido with at most 15 divisors must be less than 100, because if it has exactly 15 divisors, then it must hold that $ 3N\leq\sqrt{N}\plus{}\frac{1089}{420}\plus{}28$, or equivalently, $ \frac{171N}{420}\leq28\plus{}\sqrt{N}$.  If $ N\geq100$, then $ \frac{42N}{420}\geq\sqrt{N}$, and $ \frac{129N}{420}>\frac{126N}{420}\geq30 > 28$.  Checking the squares below 100, we find no atresvido number, so if any atresvido number has less than 16 divisors, it has at most 14.

3) any $ N$ atresvido with at most 14 divisors must be less than 70, because it must hold that $ 3N\leq\frac{1089N}{420}\plus{}28$, or equivalently, $ N\leq\frac{28\cdot420}{171}<\frac{28\cdot420}{168}\equal{} 70$.

So, since no perfect square below 100 may be atresvido, the only possible candidates for atresvido numbers with less than 16 divisors are numbers less than 70, which are the product of at least 4 (not necessarily different) primes, and which are not perfect squares.  This leaves 24, 32, 40, 48, 54, 56 and 60, which can be shown not to be atresvidos.

So no number may be atresvido with less than 16 divisors, or 16 is the minimum.

PS: I am waiting to see the official solution, because I cannot come up with anything simpler than this...

yes your solution is more elegant then mine. but checking is easy i think. if you have 30 minutes you can do it in some way.

[quote="anonymous1173"] but checking is easy i think[/quote]

The way that I used to check whether it could be atresvido or not is just finding all their divisors, adding them up, and seeing that they did not amount to thrice the number...    :oops:   If anybody cares to suggest a simpler way...

[hide]
Let $ n$ be atresvido. The divisor $ n$ appears in one of the three subsets, so we have $ \sigma (n)\geq 3n$. ($ \sigma$ is sum of positive divisors).

We have the bound
\[ \frac{\sigma (n)}{n}<\prod_{p|n}\frac{p}{p\minus{}1}\]
where the product runs of primes. As a result, $ n$ must have at least three distinct prime divisors (if it had at most two, the upper bound would be at most $ \frac{2}{1}\cdot\frac{3}{2}\equal{}3$, wihch is not possible. Now in the case $ n\equal{}pqr$ or $ n\equal{}p^{2}qr$ where $ p,q$ distinct primes, we have
\[ \frac{\sigma (n)}{n}\leq\frac{p^{2}\plus{}p\plus{}1}{p^{2}}\cdot\frac{q\plus{}1}{q}\cdot\frac{r\plus{}1}{r}\]
\[ \leq\frac{7}{4}\cdot\frac{4}{3}\cdot\frac{6}{5}\equal{}\frac{14}{5}<3\]
which is also not possible. So either $ n$ has at least four distinct prime divisors, or $ n$ is a product of three distinct prime divisors, in which case either one prime is raised to the third power or greater or at least two primes are raised to the second power or greater. Because of this, $ n$ has at least
\[ \min\{ 2^{4},4\cdot 2^{2},3^{2}\cdot 2\}\equal{}16\]
positive divisors. But we have $ n\equal{}120\equal{}2^{3}\cdot 3\cdot 5$ is atresvido, since
\[ n\equal{}120\equal{}60\plus{}40\plus{}20\equal{}\sigma (120)\minus{}60\minus{}40\minus{}20\]
and $ \sigma (120)\equal{}360$. So the minimum is $ \boxed{16}$.
[/hide]

[quote="daniel73"]
PS: I am waiting to see the official solution, because I cannot come up with anything simpler than this...[/quote]

I propose this problem  :lol: , and the solution i have is not short ( i analize the number of primer factors of an atresvido number) maybe the jury found a shorter solution.

Pd: ¿ do you know what means "atresvido"  :maybe: ? , i propose the problem with the name "tripartite".

$ Tipe$

it is just a play in words between "tres" (three) and "atrevido" (daring) -  I guess.  Translations are provided free of charge for our non-spanish-speaking friends all over the world  :lol:

I actually proposed a loose translation for the problem for this forum where instead of calling the number "atresvido", it could be called "threetening" in english.  Kind of lame, actually  :oops:  I just wanted to keep the part of setting a pseudo-fun name based on a variation of the name of 3 in the language in which the problem was stated...

[quote="Jutaro"][quote="tipe"]
I propose this problem  :lol: , and the solution i have is not short ( i analize the number of primer factors of an atresvido number) maybe the jury found a shorter solution.
[/quote]

So Tipe, you proposed this problem!! :D Very nice! Congratulations, I really liked it! :P[/quote]

I concur, very nice problem.  Congratulations on getting it selected, Tipe!!!

I'm very happy that the jury selected my problem  :D . The last year (IbMO-2006) problem 5 was also from Perú (the problem of geometry).

$ Tipe$

[quote="scorpius119"]
Let $ n$ be atresvido. The divisor $ n$appears in one of the three subsets, so we have $ \sigma (n)\geq 3n$. ($ \sigma$ is sum of positive divisors).

We have the bound
\[ \frac {\sigma (n)}{n} < \prod_{p|n}\frac {p}{p \minus{} 1}
\]
where the product runs of primes. As a result, $ n$ must have at least three distinct prime divisors (if it had at most two, the upper bound would be at most $ \frac {2}{1}\cdot\frac {3}{2} \equal{} 3$, wihch is not possible. Now in the case $ n \equal{} pqr$ or $ n \equal{} p^{2}qr$ where $ p,q$ distinct primes, we have
\[ \frac {\sigma (n)}{n}\leq\frac {p^{2} \plus{} p \plus{} 1}{p^{2}}\cdot\frac {q \plus{} 1}{q}\cdot\frac {r \plus{} 1}{r}
\]

\[ \leq\frac {7}{4}\cdot\frac {4}{3}\cdot\frac {6}{5} \equal{} \frac {14}{5} < 3
\]
which is also not possible. So either $ n$ has at least four distinct prime divisors, or $ n$ is a product of three distinct prime divisors, in which case either one prime is raised to the third power or greater or at least two primes are raised to the second power or greater. Because of this, $ n$ has at least
\[ \min\{ 2^{4},4\cdot 2^{2},3^{2}\cdot 2\} \equal{} 16
\]
positive divisors. But we have $ n \equal{} 120 \equal{} 2^{3}\cdot 3\cdot 5$ is atresvido, since
\[ n \equal{} 120 \equal{} 60 \plus{} 40 \plus{} 20 \equal{} \sigma (120) \minus{} 60 \minus{} 40 \minus{} 20
\]
and $ \sigma (120) \equal{} 360$. So the minimum is $ \boxed{16}$.
[/quote]

Sorry why  \[ \frac {\sigma (n)}{n} < \prod_{p|n}\frac {p}{p \minus{} 1}
\]

[quote="csbonny"]

Sorry why
\[ \frac {\sigma (n)}{n} < \prod_{p|n}\frac {p}{p \minus{} 1}
\]
[/quote]

It's not too hard to see.

You should spend more time with it.

It's a basic & trivial problem if you are being in training for olympiad.

เอาเล้ยยยยยยยยยยย บิ๊กเชียร์อยู่

[quote="csbonny"][quote="scorpius119"]
Let $ n$ be atresvido. The divisor $ n$appears in one of the three subsets, so we have $ \sigma (n)\geq 3n$. ($ \sigma$ is sum of positive divisors).

We have the bound
\[ \frac {\sigma (n)}{n} < \prod_{p|n}\frac {p}{p - 1}
\]
where the product runs of primes. As a result, $ n$ must have at least three distinct prime divisors (if it had at most two, the upper bound would be at most $ \frac {2}{1}\cdot\frac {3}{2} = 3$, wihch is not possible. Now in the case $ n = pqr$ or $ n = p^{2}qr$ where $ p,q$ distinct primes, we have
\[ \frac {\sigma (n)}{n}\leq\frac {p^{2} + p + 1}{p^{2}}\cdot\frac {q + 1}{q}\cdot\frac {r + 1}{r}
\]

\[ \leq\frac {7}{4}\cdot\frac {4}{3}\cdot\frac {6}{5} = \frac {14}{5} < 3
\]
which is also not possible. So either $ n$ has at least four distinct prime divisors, or $ n$ is a product of three distinct prime divisors, in which case either one prime is raised to the third power or greater or at least two primes are raised to the second power or greater. Because of this, $ n$ has at least
\[ \min\{ 2^{4},4\cdot 2^{2},3^{2}\cdot 2\} = 16
\]
positive divisors. But we have $ n = 120 = 2^{3}\cdot 3\cdot 5$ is atresvido, since
\[ n = 120 = 60 + 40 + 20 = \sigma (120) - 60 - 40 - 20
\]
and $ \sigma (120) = 360$. So the minimum is $ \boxed{16}$.
[/quote]

Sorry why  \[ \frac {\sigma (n)}{n} < \prod_{p|n}\frac {p}{p - 1}
\][/quote]

Thats a good bound.. I wonder how many people solved the problem since deriving that inequality is hard and this is a problem 5.  Anyways, csbony, it is well known that if $n= p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n}$ then ${\sigma (n)}= (\frac {p_1^{\alpha_1+1}-1}{p_1-1})(\frac {p_2^{\alpha_2+1}-1}{p_2-1})...(\frac {p_n^{\alpha_n+1}-1}{p_n-1})$

So $[ \frac{\sigma (n)}{n}<\prod_{p|n}\frac{p}{p-1} ]$ is equivalent to proving that

$({p_1^{\alpha_1+1}-1})({p_2^{\alpha_2+1}-1})....({p_n^{\alpha_n+1}-1}) \le (p_1^{\alpha_1+1})(p_2^{\alpha_2+1})....(p_n^{\alpha_n+1})$                 which is true.

Sorry for reviving this old thread, but I will like to see new solutions to the problem.


algebra
system of equations
algebra proposed

Solve the system of equations:
$ (1\plus{}4^{2x\minus{}y}).5^{1\minus{}2x\plus{}y}\equal{}1\plus{}2^{2x\minus{}y\plus{}1}$
$ y^3\plus{}4x\plus{}ln(y^2\plus{}2x)\plus{}1\equal{}0$

[quote="mr.danh"]Solve the system of equations:
$ (1 + 4^{2x - y}).5^{1 - 2x + y} = 1 + 2^{2x - y + 1}$
$ y^3 + 4x + ln(y^2 + 2x) + 1 = 0$[/quote]
[quote="Dr Sonnhard Graubner"]hello, by substituting $ 2x - y = t$ we have the first equation in the form
$ 5^{1 - t} + 4^t\cdot5^{1 - t} - 2^{t + 1} - 1 = 0$. By defining a function
$ f(t) = 5^{1 - t} + 4^t\cdot5^{1 - t} - 2^{t + 1} - 1$ we get from the first derivative
$ f'(t) = - 5^{1 - t}\ln(5) + 4^t\cdot5^{1 - t}(\ln(4) - \ln(5)) - 2^{1 + t}\ln(2)$ that
$ f(t)$ is strictly monotonously decreasing, so we have $ t = 1$ asthe uniquely solution of $ f(t) = 0$, thus we have $ 2x - y = 1$. By substituting $ 2x = y + 1$ in the
second equation and defining $ g(y) = y^3 + 2y + 3 + \ln(y^2 + y + 1)$ we get:
$ g'(y) = \frac {3y^4 + 3y^3 + 5y^2 + 4y + 3}{y^2 + y + 1}$. The numerator is always positive, because $ h(y) = 3y^4 + 3y^3 + 5y^2 + 4y + 3$ has at the point
${ y_{min} = - \frac {1}{12}(135 + 4\sqrt {3001})^{\frac {1}{3}} + \frac {31}{12(135 + 4\sqrt {3001})^{\frac {1}{3}}} - \frac {1}{4}}$ a global minimum
with $ g(y_{min}) \geq2 > 0$ thus the solution of our system is $ x = 0,y = 1$.
Sonnhard.[/quote]


[quote="maksmaks24"]$ a_1, a_2, ..., a_n > 0$

Prove that
\[ (a_1^n \plus{} n \minus{} 1)(a_2^n \plus{} n \minus{} 1)\cdot...\cdot(a_n^n \plus{} n \minus{} 1)\geq(a_1 \plus{} a_2 \plus{} ... \plus{} a_n)^n\]
[/quote]
$ (a_1^n \plus{} n \minus{} 1)(a_2^n \plus{} n \minus{} 1)\ldots(a_n^n \plus{} n \minus{} 1)$

$ \equal{}(a_1^n \plus{} 1\plus{}1\plus{}\ldots\plus{}1)(1\plus{}a_2^n \plus{} 1\plus{}1\plus{}\ldots\plus{}1)(1\plus{}1\plus{}a_3^n \plus{} 1\plus{}1\plus{}\ldots\plus{}1)\ldots(1\plus{}1\plus{}\ldots\plus{}1\plus{}1\plus{}a_n^n)$

with $ n\minus{}1$ numbers of $ 1$ in each $ (a_k^n\plus{}1\plus{}1\plus{}\ldots\plus{}1)$ , then , by holder :

$ (a_1^n \plus{} 1\plus{}1\plus{}\ldots\plus{}1)(1\plus{}a_2^n \plus{} 1\plus{}1\plus{}\ldots\plus{}1)(1\plus{}1\plus{}a_3^n \plus{} 1\plus{}1\plus{}\ldots\plus{}1)\ldots(1\plus{}1\plus{}\ldots\plus{}1\plus{}1\plus{}a_n^n)\geq(a_1 \plus{} a_2 \plus{} ... \plus{} a_n)^n$


Let $x,$ $y$ and $z$ are positive numbers such that $xy+xz+yz=1.$ Prove that
$\frac{x^{3}}{1+9y^{2}xz}+\frac{y^{3}}{1+9z^{2}yx}+\frac{z^{3}}{1+9x^{2}yz}\ge \frac{(x+y+z)^{3}}{18}.$

[hide="solution"]by holder, $LHS\ge \frac{(x+y+z)^{3}}{3(3+9xyz(x+y+z))}\ge \frac{(x+y+z)^{3}}{3(3+3(xy+yz+zx)^{2})}=RHS$

where the last inequality follows from $(xy+yz+zx)^{2}\ge 3((xy)(yz)+(yz)(zx)+(zx)(xy))=3xyz(x+y+z)$.[/hide]

Nice, nayel! :lol:
Try it:
Let $x, y$ and $z$ are positive numbers such that $xy+xz+yz=1.$ Prove that
$\frac{x^{3}}{1-4y^{2}xz}+\frac{y^{3}}{1-4z^{2}yx}+\frac{z^{3}}{1-4x^{2}yz}\ge \frac{(x+y+z)^{3}}{5}.$ :wink:


A garden plot is 60 feet by 40 feet.  A walk 2.5 feet wide is to be placed around this plot in such a way that the corner pieces are portions of a circle of radius 2.5 feet with each corner of the plot as center.  What is the total area of the walk?

[hide]Each corner is a quarter of a circle, so all of the corners added up is the same as a circle.  The sides are 60 or 40 long, and 2.5 wide.

(60*2.5)+(60*2.5)+(40*2.5)+(40*2.5)+((2.5)2*pi) =

500+6.25(pi)[/hide]


Let [tex] p [/tex] be a prime, [tex] p \geq 5 [/tex] and [tex] \lbrace{r_i} \rbrace_{i=1}^k :1 \leq r_1<r_2<...<r_k\leq p-1, k\geq 3 [/tex] that satisfies
  [tex] \forall 1 \leq i <j \leq k, \exists l \in \lbrace {1,2,...,k} \rbrace: r_i+r_j \equiv r_l ( mod p-1 ) [/tex]
Prove that [tex] \sum\limits_{i=1}^k{r_i} \vdots \frac{p-1}{(\sqrt{2})^{1+(-1)^k}} [/tex]  :)

I think you can replace that $p-1$ with any even natural $n$ (perhaps with some small exceptions), but I might be mistaken. :?


algebra polynomial function inequalities logarithms integration calculus

Hi,

I'm wondering if there's a constant C>0 such that

$ \sup\limits_{n\geq 1} \left\vert \sum_{k\equal{}1}^n  \frac{z^k}{k}  \right\vert \geq \frac{C}{|1\minus{}z|}$

for all complex such that  |z|=1 and  z $ \neq$ 1.

Of course, C should'nt depend on z.

thanks

This topic lies in the field of Complex Analysis.It seems to me that such constant does not exist.I'll give a sketch of the proof.
Indeed, consider the polynomials $ P_n (z)\equal{}(1\minus{}z)\sum\limits_{i\equal{}1}^{n} \frac{z^k}{k}$. It's easy to check that $ |P_n (z) \minus{}P_m (z)|\le C(\frac{1}{n}\plus{}\frac{1}{m} )$ for $ |z|\le 1$  and universal constant $ C$.This implies that sequence  $ P_n (z)$ uniformly converges to some continious function $ f(z)$. This means that (Arcella lemma)  the set of functions $ P_n(z)$ is  uniformly continous.
But this give us the following: for each $ \varepsilon>0$ there exists $ \delta$ s.t in $ \delta$- neighborhoud of point $ z\equal{}1$  $ |P_n (z)|<\varepsilon$ (because $ P_n (0)\equal{}0$).This implies a contradiction.

I thank you.

My question was very simple. So there's an other one :

Does there exist constants c>0 and C>0 such that

$ \minus{}c\ln(|1\minus{}z|)       )  \leq    \sup\limits_{n\geq 1} \left\vert  \sum_{k\equal{}1}^n \frac{z^k}{k}  \right\vert \leq    \minus{}C  \ln(|1\minus{}z|)$

forall z with |z|=1, |z-1|<1 ,$ z\neq 1$

thanks

If I undestood your question correctly then the answer is again no :) .In the neighbourhood of the point $ |z \minus{} 1| \equal{} 1$, $ |z| \equal{} 1$ the RHS of your inequality breaks. The LHS is absolutely correct. You can  prove thist by observing that RHS greater then $ 1$ (take $ n \equal{} 1$) and for fixed $ z\ne 1$ you can find such $ n$ that
$ |P_n (z) \minus{} ln(\frac {1}{1 \minus{} z})| < \varepsilon$. Thus $ |P_n (z)| > |ln(\frac {1}{1 \minus{} z})| \minus{} \varepsilon\ge C ln\frac {1}{|1 \minus{} z|}$ for $ C \equal{} \frac {1}{2009}$  :)

Well, I surmise that the OP didn't care much whether it was $ |z\minus{}1|<1$ or $ |z\minus{}1|<\frac 12$. In the first case, there is trouble, indeed, but in the second case, everything is fine. :)

I'm sorry leshik but I haven't understood your solution for the RHS.

Ok, I'll try to give a brief explanation . Our aim is to show that $ c\ ln\frac{1}{|z\minus{}1|}<sup|\sum \frac{z^k}{k}|$. Consider the polynomials $ P_n (z)\equal{} \sum\limits_{k\equal{}1}^{n}\frac{z^k}{k}$. Fix point $ z_0$. We know that the sequense $ P_n(z)$ converges uniformly on compact subsets of unit disc that don't intresect real line to the function $ ln\frac{1}{1\minus{}z}$(we can choose such branch of comlex function $ Lnz$). Let us observe that    if we cut out a small open neighbourhoood of the point $ z\equal{}1$ sequence $ P_n(z)$ convergers uniformly on the rest circle (this is an easy exersise,it could be proved by using Abels criterium). Choose compact subset of closed unit disk which contains the point $ z_0$ and does not contain the point $ z\equal{}1$ and observe that our sequense $ P_n$ converges uniformly to the function $ ln\frac{1}{1\minus{}z}$ on this set.Is the rest clear now?

If what Fedja has said is correct (I mean there must be $ |z \minus{} 1| < \frac {1}{2}$ ), then the constastant seems  realy exist. :)

I thank you leshik. This part was clear for me after your explaination.

In fact, i haven't understood when you wrote "the RHS of your inequality breaks"

Did you mean that there's no C such that forall z, |z|=1, z\neq 1

$ \sup\limits_{n\geq 1} \left\vert \sum_{k\equal{} 1}^n \frac{z^k}{k} \right\vert    \leq C |  \ln(1\minus{}z)   |$

thanks

Yes,I did. One  can take the point $ z_0$ with $ |z|\equal{}1$ in the small neighbourhood of point $ |z_1\minus{}1|\equal{}1$, $ |z_1|\equal{}1$ and $ n\equal{}1$ and get a cotradiction :wink:

I'm probably very tired but i still don't understand.

assume we have the RHS, for $ n\equal{}1$ we sould have

$ |z| \leq C|\ln(1\minus{}z)|$

But $ |z|\equal{}1$, so it means that

$ 1 \leq C |\ln(1\minus{}z)|$

i don't see any contradiction when z is near 1.

So you made a typo in the condition of the problem :)  You have written $ sup |P_n (z)| < \minus{} C ln(|1 \minus{} z|)$ :wink:. But, actually, you don't need to have a contradiction in the point $ z\equal{}1$ but in the point where $ |z\minus{}1|\equal{}|z|\equal{}1$. In the small neighbourhood $ ln|(1\minus{}z)|$ is almost $ 0$.

I'm sorry, i've juste understood.

you're looking $ z\equal{}e^{ i\pi /3}$ !!!!

But my condition was |1-z|<1, so i still hope that there's a C>0 such that

$ |z|\equal{}1,|z\minus{}1|<1/2 \quad \Rightarrow \sup\limits_{n\geq 1} \quad \left\vert \sum_{k\equal{}1}^n \frac{z^k}{k} \right\vert \leq \minus{}C\ln |1\minus{}z|$

i've choosen |z-1|<1/2 like fedja said.

thank you very much leshik

Why are you not sure? Let $ z_1 \equal{} e^{i\frac {\pi}{3}}$. Then $ |z_1| \equal{} |1 \minus{} z_1| \equal{} 1$. Thus $ ln|1 \minus{} z_1| \equal{} 0$ and the function $ ln|(1 \minus{} e^{ix})|$ is continous on the interval $ (0,\pi)$. Hence if $ x$ lies "near" $ \frac {\pi}{3}$ then the difference $ |ln|1 \minus{} e^{ix}| \minus{} ln|1 \minus{} e^{i\frac {\pi}{3}|}| \equal{} |ln|1 \minus{} z\parallel{} < \varepsilon$ where $ z \equal{} e^{ix}$[url][/url].

Now, i'm convinced that we have

$ \sup_{n\geq 1} \left\vert \sum_{k\equal{}1}^n \frac{z^k}{k} \right\vert \leq A\plus{}B|\ln|1\minus{}z\parallel{}$

for some $ A,B>0$ and forall $ z$ such that $ |z|\leq 1$ and $ z\neq 1$.

Sketch of the proof. Let $ P_n \equal{}\sum_{k\equal{}1}^n \frac{z^k}{k}$, we know that $ P_n$ converge uniformly on $ |z|\leq 1$ and $ |z\minus{}1|>1/2$.
So $ P_n$ are uniformly bounded on this domain.

Let's study the situation $ |z|\leq 1$ and $ |z\minus{}1| \leq 1/2$. We write

$ P_n (z)\equal{}\int_{0}^1  \frac{z(1\minus{}t^n z^n)}{1\minus{}tz} dt$ and then if $ z\equal{}x\plus{}iy$

$ |P_n(z)|\leq \int_{0}^1 \frac{1\plus{}t^n}{|1\minus{}tz|} dt \leq \int_{0}^1 \frac{1\plus{}t^n}{1\minus{}tx}dt \leq 2 \int_0^1  \frac{dt}{1\minus{}tx}\equal{}\minus{}\frac{2}{x} \ln(1\minus{}x)$

Now, just remark that   $ \frac{1}{2} \leq x$ and  $ |1\minus{}z|^2\equal{}1\plus{}|z|^2\minus{}2x  \leq 2(1\minus{}x)$, and finaly
$ |P_n(z)| \leq A\plus{}B| \ln|1\minus{}z|      |$

Now, i've got two naturals questions :

Let E the set of real B>0 such that there's some A>0 that for all $ z$, $ |z|\leq 1$,$ z\neq 1$  we have

$ \sup\limits_{n\geq 1} \left\vert  \sum_{k\equal{}1}^n \frac{z^k}{k} \right\vert \leq A\plus{} B | \ln|1\minus{}z\parallel{}$

Do we have  $ 1 \in E$ ? or $ 1\equal{} \inf E$ ?

I don't know  if it hepls you, but try to estimate your integral $ \int\limits_{0}^{1} \frac {1 \plus{} t^n}{1 \minus{} xt}$  more "accurately" to obtain something like $ \int\limits_{0}^{1} \frac {1 \plus{} t^n}{1 \minus{} xt}\le \frac {c_n |ln(1 \minus{} x)}{x}|$ where $ c_n\to 1$ uniformly with respect to $ x$. I'll think about that too :)
P.S It seems to me , that it'll be more useful to move these topics to Complex Analysis section to attract attension of other people :)

I don't know how to move all these messages to complex analysis

how can i do ?

I don't know too :) Try to create a new post with your last question and mention the link with this page there :)

that's a good idea!!

the following is at

http://www.mathlinks.ro/Forum/viewtopic.php?t=294098


function
trigonometry
real analysis
real analysis unsolved

It is not difficult to prove that: [b]a nonconstant periodic function $f$ has the least period, if $f$ is continuous.[/b]

But, continuity is NOT a NECESSARY condition. Let us consider the following three functions:

(1) \[D(x)=\begin{cases}0, & \text{for irrational }x,\\ 1, & \text{for rational }x. \end{cases}\] (2) \[R(x)=\begin{cases}0, & \text{for irrational }x,\\ 1/q, & \text{for reduced fraction }x=p/q. \end{cases}\] (3) \[g(x)=x-[x].\] Note that, $D$ is continuous nowhere, and $R, g$ is continuous almost everywhere (a.e.). And, $D$ does not have the least period while $R,g$ both have the least period $T_{min}=1$.

Therefore, I guess that: [b]a nonconstant periodic function $f$ has the least period, if and only if $f$ is continuous ALMOST EVERYWHERE.[/b]

Am I right?????

[quote="mlok"]The function \[F(x)=\begin{cases}2, &\text{for irrational }x,\\ \sin x, &\text{for rational }x\end{cases}\] has the least period $2\pi$ and is not continuous at any point.[/quote]

No, your counterexample is actually wrong. $2\pi$ is NOT a period of your $F$, since $F(0)=\sin 0=0$ while $F(2\pi)=2$.

[quote="mlok"]Thanks. Put $\sin(2\pi x)$ to make period $1$.[/quote]
Good idea! Thank you.

Then what about the other implication?
"A nonconstant periodic function, which is continuous almost everywhere, has a least period."
Is that true?

LiMa

I think even more is true: a nonconstant periodic function that is continuous [b]at least at one point[/b] has a least period.

You are right. After half a page of writing I found the two-line-proof. :-)

Proposition: Let $f$ be continuous at one point and periodic, but without least period. Then f ist constant.
[hide="Proof"]
Assume wlog. that f is continuous at $0$ and $f(0)=0$. Let $\varepsilon>0$. Then there is $\delta>0$ s.th. $0<x<\delta$ implies $|f(x)|<\varepsilon$.
Let $0<p<\delta$ be a period of $f$. For every $x$ there is $k \in \mathbb Z$ s.th. $0\le x-kp<p<\delta$ and so $|f(x)|=|f(x-kp)|<\varepsilon$. Therefore $f=0$.
[/hide]

LiMa

Thank you for all!
I find an essay giving a proof of the following theorem:
[quote]A (Lebesgue) measurable function f: R--->R having arbitrary small periods is a constant a.e. [/quote]


calculus
integration
function
calculus computations

For $a\geq 0,$ find the minimum value of $\int_{-2}^1 |x^2+2ax| dx.$

First we determine where $x^2+2ax \leq 0$ that is $-2a \leq x \leq 0$, other wise it is positif.
If $a \geq 1$ then
\begin{eqnarray*}\int_{-2}^{1} |x^2+2ax| \, dx &=& \int_{-2}^{0} -(x^2+2ax) \, dx + \int_{0}^{1} x^2+2ax \, dx \\ &=& 4a - \frac{8}{3}+ \frac{1}{3}+a \\ &=& 5a - \frac{7}{3} \\ &\geq& \frac{8}{3}\end{eqnarray*}

if $a < 1$ then
\begin{eqnarray*}\int_{-2}^{1} |x^2+2ax| \, dx &=& \int_{-2}^{-2a} (x^2+2ax) \, dx + \int_{-2a}^{0} -(x^2+2ax) \, dx + \int_{0}^{1} x^2+2ax \, dx \\ &=& \frac{8}{3}a^3 - 3a +3 \end{eqnarray*}

The minimum of the function $f(a)=\frac{8}{3}a^3 - 3a +3$, then $f''(a) = 8a^2 -3$.
The the stationery points are $a= \frac{\sqrt{6}}{4}$ and $a= - \frac{\sqrt{6}}{4}$ (both of them less than $1$). But $a \geq 0$ . So we just have $a= \frac{\sqrt{6}}{4}$. Subtituting this to $f(a)$. we have $f \left( \frac{\sqrt{6}}{4} \right) = - \frac{\sqrt{6}}{2}+3$.
This value is less than $\frac{8}{3}$ so we have $- \frac{\sqrt{6}}{2}+3$ as a minimum value.

Yes, you are right. :) Please correct the final result, Beginner.

kunny


A two person game is played in which each player, in turn, removes $1, 2, 3, 4, 5, 6$ or $7$ toothpicks from a common pile, until the pile is exhausted. The player taking the last toothpick LOSES. If the starting pile contains 1000 toothpicks, how many toothpicks must the first player take on his first turn to guarentee a win with perfect subsequent play?

Make sets of 8: (1+7)=8. 1000=0mod8 and after you pick 7 cards, after so many rounds the opponent would be left with one card. Thus the answer is 7.


you might want to wait more than a day before bumping up problems

hmm, that's a really tough one. Anyone know?

[quote="Joey"]I have two problems that i do not entirely understand. They are worded as followed:

1)Find the term free of y in the expansion of (3xy²-1/6x²y³)²°.  Simplify the term

2)Solve the following equation for x. GIve your answer to 4 decimal places if exact values are not possible.
            sqrt2 cosx-sinx=1

3) solve for x     -6=6sin[1/12(x-6π)]-3[/quote]

[hide="1"]
I'm reading the expression as $(3xy^2 - \frac{1}{6x^2y^3})^{20}$ because otherwise there isn't a term free of y.

The term free of $y$ must have $(y^2)^{a} = (y^3)^{20-a} \implies 2a = 3(20-a) \implies 5a = 60 \implies a = 12$.  This term is ${20 \choose 12} 3^{12} x^4 \frac{1}{6^8}$ by an application of the binomial theorem.
[/hide]

[hide="2"]
$\sqrt{2} \cos x = 1 + \sin x$
$\sqrt{2} = \frac{1 + \sin x}{\cos x} = \frac{1 + \cos (\frac{\pi}{2} - x)}{\sin (\frac{\pi}{2} - x)} = \cot (\frac{\pi - 2x}{4} )$
$\tan (\frac{\pi - 2x}{4}) = \frac{1}{ \sqrt{2} }$
$\frac{\pi - 2x}{4} \approx 0.6154797$
$x \approx 0.3398369$
[/hide]

[hide="3"]
$-3 = 6 \sin \frac{x-6\pi}{12}$
$\sin \frac{x-6\pi}{12} = - \frac{1}{2}$
$\frac{x-6\pi}{12} = \frac{7\pi}{6}, \frac{11\pi}{6}$
$x - 6\pi = 14\pi, 22\pi$
$x = 20\pi, 28\pi$
[/hide]

This looks like homework...


graph theory
combinatorics proposed
combinatorics

Suppose $k$ is a natural number. Imagine a graph that its verteces are natural numbers. Suppose this graph doesn't have any complete bipartite $k \times k$ subgraph. Prove that for every $n$ there is an arithmatic progression that no 2 verteces of this progression are connected together.

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=bipartite&t=6201[/url].

P.S. You meant to say that there's no complete $k\times k$ bipartite graph, right?

I saw it in Unsolved Problems and I really thought it is still unsolved :(

Welcome in the Alzheimer club, Myth... :D

Pierre.

No  :mad:
I can give you link, where sam-n posted it for first time. There is no any replies.

and that was the 2005 tourney... while sam-n had posted them twice yet in 2004 :D


A 2 x 3 rectangle and a 3 x 4 rectangle are contained within a square without overlapping at any interior point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?

$ \textbf{(A) } 16 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 49 \qquad \textbf{(E) } 64$

[hide]Balancing the lengths out as much as possible, we see that the smallest length that contains the lengths of the rectangles are 5 when there are the 2x3 and the length of 4 meets the 3 and the 3 forms a line of length 5 with the side length 2. $5^2=25.$[/hide]

[hide]By experimenting, we find that the smallest side length $s$ of a square into which we can fit rectangles with dimensions $a_1\times a_2$ and $b_1\times b_2$ (with $a_1\le a_2$ and $b_1\le b_2$) is $\max(a_2,b_2,a_1+b_1)$, since one of the pairs of dimensions will have to be stacked together. In this case, $s=\max(3,4,2+3)=5$, so $s^2=\fbox{25}$. $\text{(B)}$.[/hide]


linear algebra
matrix
LaTeX
linear algebra unsolved

Suppose $m<n$ and $A$ is a $m\times n$ matrix and $B$ is a $n\times m$ matrix such that $\det(AB) \neq 0$ .prove that $A$ has $m$ column which are linear independent.

What is the definition of linear independence?

You want to show that $\det(A) \neq 0$

If $n>m$, $A$ doesn't have a determinant.

A pair of useful facts:
The column space of $AB$ is contained in the column space of $A$. The row space of $AB$ is contained in the row space of $B$. Try to prove these, then apply to the problem.

$\LaTeX$ tips: use \times to say $m\times n$ matrix - I think that looks better than $m*n.$ Also, \det to say $\det(AB).$

The concept of "rank" is floating around in the air here, but this also feels like the kind of problem you would do just before you define that term. (The only hint I'll give is that you should listen to jmerry.)


When I click the drop down menu in the mood set in blog options, I see [i]default set[/i]. Is there any other set than this one?


calculus
integration
trigonometry
calculus computations

(1) For integers $m,n,$ evaluate $I_{m,n}=\int_0^{2\pi} \cos mx\cos nx\ dx.$

(2) For positive integer $n,$ evaluate $J_n=\int_0^{2\pi} (\sum_{k=1}^n \sqrt{k}\cos kx)^2\ dx.$

A lot of kunny's problems are not solved and they are very good problems!

(1) \[ \begin{aligned} I_{m,n}&=\int_0^{2\pi}\cos mx\cos nx\ dx\\ &=\int_0^{2\pi}\frac12(\cos(m+n)x+\cos(m-n)x)\ dx \end{aligned} \]
If $m=n$, $I_{m,n}=\left.\frac12\left(\frac{\sin(m+n)x}{m+n}+x\right)\right|_0^{2\pi}=\pi$

If $m\ne n$, $I_{m,n}=\left.\frac12\left(\frac{\sin(m+n)x}{m+n}+\frac{\sin(m-n)x}{m-n}\right)\right|_0^{2\pi}=0$

(2) \[ \begin{aligned} J_n&=\int_0^{2\pi}\left(\sum_{k=1}^n\sqrt{k}\cos kx\right)^2\ dx\\ &=\int_0^{2\pi}\left(\sum_{k=1}^nk\cos^2kx+2\sum_{i<j}^n\sqrt{ij}\cos ix\cos jx\right)\ dx\\ &=\sum_{k=1}^nkI_{k,k}+2\sum_{i<j}^n\sqrt{ij}I_{i,j}\\ &=\sum_{k=1}^nk\pi+0=\frac{n(n+1)}2\pi \end{aligned} \]

[quote="10000th User"]A lot of kunny's problems are not solved and they are very good problems![/quote]

Thank you very much, 10000th User :)  From Feb 25 or 26, each national university  has been held university entrance examination. I will post these problems. Have fun!

kunny


A Classic Factorial Counting Problem

We count the zeroes at the end of 320!

that dude has got to get a life...
good problem though!

You know whonn you just insulted, right? :wink:

Edit: Sorry, my 'nn' key was nnising.

It's whom.

If you were referring to the guy talking, that's Mr. Rusczyk. He's pretty awesome. Actually make that super-duper awesome. He runs this site and is the CEO of AOPS. Not the guy to insult.  :rotfl:

mewto, math154 said his "m" key was missing.
And apart from the fact that he's the owner of AoPS, Mr. Rusczyk also has the power to take away your account (He also can probably take you to court and spam you because your email address is in the server, but those are fringe benefits).

And for the zeros: How we solve this is that one zero appears for every multiple of 5 in the number x! So 320/5=64 zeroes. But that's not it: there are also zeroes for multiples of 25 ,and any powers of 5. There are 24 25's in 320, and 2 125's. Thus, there are 64+24+2=90 zeroes.
Did I do this right? I'm more accustomed to problems like "what is the greatest number x so that x! has y zeroes at the end."
Oh, and by the way, I didn't watch the video. So don't be surprised if I'm wrong.

[quote="evilhamster"]mewto, math154 said his "m" key was missing.[/quote]

It was sarcas[u][i][b]m[/b][/i][/u]... :P

[quote]It was sarcasm... [/quote]

He/she was just kidding around.

[quote="yunhua98"][quote="math154"]You know whonn you just insulted, right?

Edit: Sorry, my 'nn' key was nnising.[/quote]

How did you type MY then?
 By the way, the video wont work on my computer, anyone know why? :huh:  :huh: :arrow:   :huh:[/quote]

You might want to post that in the support forum. Does it just come blank? Then your browser probably isn't updated, or your flash. Does it stop in the middle of the lesson (like me)? Then your internet is probably too slow (you can still watch it, but you have to wait for the green thing at the bottom to get to the end first).

The faster way is pretty neat...

320/5 = 64

64/5 = 12 (which is the same thing as 320/25, since we're dividing 320 by 5 and then by 5 again)

12/5 = 2 (which is the same thing as 320/125, since we're dividing 320 by 5 3 times)

[quote="evilhamster"]
And for the zeros: How we solve this is that one zero appears for every multiple of 5 in the number x! So 320/5=64 zeroes. But that's not it: there are also zeroes for multiples of 25 ,and any powers of 5. There are 24 25's in 320, and 2 125's. Thus, there are 64+24+2=90 zeroes.
Did I do this right? I'm more accustomed to problems like "what is the greatest number x so that x! has y zeroes at the end."
Oh, and by the way, I didn't watch the video. So don't be surprised if I'm wrong.[/quote]
Mr. Rusczyk gets 12 multiples of 25 in 300 (so do I). So the total comes out to 78.

A problem for fun that goes with the learning in this video:
Find how many 5's divide into $125!$
[hide="My solution"]
Way 1:
Casework:
[color=#FF0000]$5^3$ multiple:[/color] $\mathbb{P}=125$ or $\boxed{3*1}$ five's.
[color=#FF0000]$5^2$ multiple:[/color] $\mathbb{P}=25, 50, 75, 100$ or $\boxed{4*2=8}$ five's.  Before we move on to $5^1$ we have to note, that the way we can find the number of multiple's of 25, that aren't a multiple of 25, is going to be $\frac{25}{5}-1$ because there was one $5^3$ multiple.
[color=#FF0000]$5^1$ multiple:[/color] We could go out and list them all $\mathbb{P}={5,10,15....}$ but that would take too long.  One way to shorten it, is to find the number of multiple's of $5$ from $1-125$ and subtract the number of multiples of $25$ and $125$.
$\mathbb{P}={5, 10, 15, 20,.......100, 105, 110, 115, 120, 125}$  We know how to count the number of number's in that list!  We divide it all by 5.
$\mathbb{P}=\frac{5, 10, 15, 20,.......100, 105, 110, 115, 120, 125}{5}\implies {1,2,3,4,5,.....20,21,22,23,24,25}$

But we have some over counting here.  The numbers that we over counted where all the numbers in the last two lists combined, or 5 numbers.  We could count those out as:  $\mathbb{P}={25, 50, 75, 100, 125}$

There is $25-5=20$ numbers that have one 5 or a total of $\boxed{20*1}$ five's.

Adding up $\boxed{20*1+3*1+4*2\implies 20+3+8\implies 23+8\implies \boxed{31}}$

Let's try Mr. Ruszcky's method now!

$\frac{125}{5}=25 \cancel{r. 0}$
$\frac{25}{5}=5\cancel{r. 0}$
$\frac{5}{5}=1\cancel{r. 0}$
$\frac{1}{5}=0\cancel{r. 1}$

$\boxed{25+5+1+0\implies 30+1+\cancel{0}\implies \boxed{\boxed{31}}}$ [/hide]


Two circles and three straight lines lie in the same plane. If neither the circles nor the lines are coincident, what is the maximum possible number of points of intersection?

First know that 2 distinct circles can intersect at a maximum of 2 points. THis can be found by trial and error.
[hide="sure-fire but more advanced way"]
3 points define a circle, so if 2 circles share 3 points, then they are the same circle, but we have that they are distinct, so they intersect at a maximum of 2 points.
[/hide]
All 3 lines can intersect the 2 circles a maximum of 4 times each, and the 2 lines can intersect eachother at a maximum of 3 points.
the total is $ 2 \plus{} 3 \cdot 4 \plus{} 3 \equal{} 17$

see diagram if this is unclear

p.s.
the word "all" is only used with 3 or more items
since i used all, you can infer that i meant 3 lines (for 2 you must use "both").

line-to-circle: 3*2*2=12
line-to-line: 3C2=3
circle-to-circle: 2
Total: 17

stevenmeow, check your description... there are a couple of numerical errors.


In Istanbul, there are 40 days which the weather has measured as 18 degrees.
On the other side, Paris has 73 days on same condition.
Along 87 days, the hotter of these two cities has a temperature with 18 degrees.
Along how many days, the colder of these two cities has a temperature with 18 degrees.

Define the three cases for a temperature in a city:
$ 1$ lower than 18 degrees
$ 2$ 18 degrees
$ 3$ more than 18 degrees
$ a_{i,j} | i,j \in \{1,2,3\}$ the combination of events $ i$ for Istanbul and $ j$ for Paris.
$ a_{2,1}\plus{}a_{2,2}\plus{}a_{2,3}\equal{}40$  $ (1)$
$ a_{1,2}\plus{}a_{2,2}\plus{}a_{3,2}\equal{}73$  $ (2)$
$ a_{1,2}\plus{}a_{2,2}\plus{}a_{2,1}\equal{}87$  $ (3)$

Then $ a_{2,3}\plus{}a_{2,2}\plus{}a_{3,2}\equal{}26$  $ (1)\plus{}(2)\minus{}(3)$

With less notation: the number of times the hotter city is 18 degrees plus the number of times the colder city is 18 degrees is the same as the number of times either city is 18 degrees (with multiplicity, since if they are both 18 degrees we can wlog choose one to be "hotter" and one to be "colder").  This question should have been placed in the Intermediate section.


Given positive numbers $ a,b,c,d$ and given that $ cd\equal{}1$, prove that the closed interval $ [ab, (a\plus{}c)(b\plus{}d)]$contains at least one square

[hide="solution"]
$ \lfloor \sqrt {ab} \rfloor \plus{} 1 \ge \sqrt {ab} \Rightarrow (\lfloor \sqrt {ab} \rfloor \plus{} 1)^2 \ge ab$

Also by AM-GM

$ ab \plus{} ad \plus{} bc \plus{} 1 \ge ab \plus{} 2\sqrt {abcd} \plus{} 1 \equal{} (\sqrt {ab} \plus{} 1)^2 \ge (\lfloor \sqrt {ab} \rfloor \plus{} 1)^2$

So $ (\lfloor \sqrt {ab} \rfloor \plus{} 1)^2 \in [ab, (a \plus{} c)(b \plus{} d)]$[/hide]


function
calculus
derivative
real analysis
real analysis unsolved

can any1 pls explain/prove why the following statement is always true?

"although a function f may have derivatives of all orders, the series $ \sum{c_n x^n}$, where $ c_n$ is computed by
$ c_n \equal{} f^{k} (0) / k!$, need not converge to f(x) for any x except for x =0. [u]In this case, f cannot be expanded in a power series about x =0. For if we had f(x) = $ \sum {a_n x^n}$, we should have
n! $ a_n$ = $ f^{n} (0)$, hence $ a_n \equal{} c_n$[/u]

I tried to use the root test but didn't work out. >_<
especially, what does the underlined portion mean?

There is a classical  example  of such  function. It is $ f(x)=\begin{cases} e^{-\frac{1}{x^2}}&x\neq 0&0&x=0\end{cases}$


In an acute triangle $ABC$, $\angle C=45$. $O$ is the circumcenter, $H$ is the orthocenter. The line through $O$ perpendicular to $CO$ meets $AC$,$BC$ at $K,L$ respectively. Prove: $OK+KH=OL+LH$

This is an easy problem.

If we denote as $K',$ $L',$ the intersection points of the circumcircle of $ABC,$ from the segment line $KL,$ then we can easy to prove that the altidudes of $ABC$ through vertices $B,$ $A,$ pass through these points resprctively $($ From the right triangles $\bigtriangleup ADC,$ $\bigtriangleup BEC,$ where $\angle ADC = 90^{o},$ $\angle BEC = 90^{o},$ $\Longrightarrow$ $\angle DAC = 45^{o}$ and $\angle EBC = 45^{o}$ $).$ So, we have that:

$OK+KH = OK+KK' = OK' = OL' = OL+LL' = OL+LH$

Kostas Vittas.


Let $ a$, $ b$ and $ c$ be real numbers such that $ abc\equal{}1$.

Prove that $ \frac a {a^2\plus{}3} \plus{} \frac b {b^2\plus{}3} \plus{} \frac c {c^2\plus{}3} \leq \frac 3 4$.

It seems to me to permit a generalization to other numbers than 3 in the denominator, but I am not sure - I haven't been able to prove anything for other values than $ n\equal{}1$, and the problem is, of course, $ n\equal{}3$.  :blush:

see [url=http://www.mathlinks.ro/viewtopic.php?t=39571]here[/url] for $ n\equal{}2,3,4,5$

I can not see there a solution for $ n\equal{}3$ , but anyway the solution for $ n\equal{}2$ and $ n\equal{}5$ may help you


Consider the polynomials \[f(x) =\sum_{k=1}^{n}a_{k}x^{k}\quad\text{and}\quad g(x) =\sum_{k=1}^{n}\frac{a_{k}}{2^{k}-1}x^{k},\] where $a_{1},a_{2},\ldots,a_{n}$ are real numbers and $n$ is a positive integer. Show that if 1 and $2^{n+1}$ are zeros of $g$ then $f$ has a positive zero less than $2^{n}$.

It's easy to see that $f(x)=g(2x)-g(x).$
Suppose that for any $x\in(0;2^{n})$ f(x)>0. Then
$g(2^{n+1})\ge g(2^{n})>g(2^{n-1})>g(2^{n-2})>\ldots$, i.e.
$g(2^{n+1})>g(0)$.Contradiction.
Similarly if f(x)<0.

But $f(a)$ can be $>0$ and $f(b)$ can be $<0$ for $a,b\in (0,2^{n})$

Nice problem.
For convenience let's rewrite it: let $ f(x) \equal{} \sum_{1\le i \le n} (2^k \minus{} 1)a_{k}x^k, g(x) \equal{} \sum_{1 \le i \le n} a_{k}x^k$.

Lemma: $ \sum_{0 \le i \le n} f(2^i) \equal{} g(2^{n \plus{} 1}) \minus{} g(1)$.
Proof: Since $ f(2^i) \equal{} \sum_{1 \le j \le n} a_{j}(2^j \minus{} 1)(2^{ij})$,
the coefficient of $ a_j$ in $ \sum_{0 \le i \le n} f(2^i)$, is equal to: $ (2^j \minus{} 1)* \sum_{1\le k \le n} (2^{j})^k \equal{} (2^j \minus{} 1) * \frac { (2^j)^{n \plus{} 1} \minus{} 1}{2^j \minus{} 1} \equal{} (2^j)^{n \plus{} 1} \minus{} 1$.
Therefore, $ \sum_{0 \le i \le n} f(2^i) \equal{} \sum_{0 \le i \le n} a_{i} (2^{i(n \plus{} 1)} \minus{} 1) \equal{} \sum_{0 \le i \le n} a_{i} (2^{n \plus{} 1})^i \minus{} \sum_{0 \le i \le n} a_{i} \equal{} g(2^{n \plus{} 1}) \minus{} g(1)$.

Since $ g(2^{n \plus{} 1}) \equal{} g(1) \equal{} 0$, using Lemma: $ f(2^0) \plus{} f(2^1) \plus{} ... \plus{} f(2^n) \equal{} 0$ (*). Suppose that $ f(x)$ does not have any zero between $ 0$ and $ 2^{n}$. If, for some $ i$ ,$ 1 \le i \le n$, $ f(2^{i \minus{} 1})$ and $ f(2^{i})$ have different signs, then $ f$ has a zero between $ 2^{i \minus{} 1}$ and $ 2^{i}$, a contradiction. So $ f(1), f(2) ... f(2^n)$ all have the same sign, and apart from $ f(2^n)$, they are non-zero. But then $ f(1) \plus{} f(2) \plus{} ... \plus{} f(2^n)$ must also be non-zero, which is in direct contradiction with (*).

@ OHO: 'RIP's solution is still correct. Since $ f(x)$ is continuous, if $ f(a) < 0, f(b) >0$ or if $ f(a) < 0, f(b) > 0$, then there is a zero between $ a,b$, which also gives the desired contradiction.

[hide="Solution"]
So $f(x)+g(x)=g(2x)$. This is rather trivial to prove.

Now note that $f(1)+g(1)=f(1)=g(2)$, $f(2)+g(2)=f(1)+f(2)=g(4)$, ... $f(2^n)+g(2^n)=f(1)+f(2)+f(4)+\cdots+f(2^n)=g(2^{n+1})=0$. Since the sum, and thus the average, of all the $f$ values is $0$, either they are all $0$ -- in which case $f(1)=0$ and we are done -- or one of them is positive and another is negative. However, by the intermediate value theorem, that implies that there is some number $c$ which lies strictly between the two x-values such that $f(c) = 0$, so since any possible bounding x-values are at most $2^n$, we know that $c < 2^n$, so we have a root of $f$, as desired.
[/hide]

Note that
\begin{align*}
0=g(2^{n+1})-g(1)&=\sum_{k=1}^{n}\frac{a_{k}}{2^{k}-1}(2^{k(n+1)}-1) \\
&=\sum_{k=1}^{n}\frac{a_{k}}{2^{k}-1}(2^{k}-1)(1+2^{k}+2^{2k}+\dots+ 2^{nk}) \\
&=\sum_{k=1}^n a_k (1+2^k+2^{2k}+\dots +2^{nk}) \\
&=f(1)+f(2)+f(4)+\dots+f(2^n)
\end{align*}

Either all of the $n+1$ terms of $f(1)+f(2)+f(4)+\dots+f(2^n)$ are zero, in which case we can just choose 1 as the root, or there exist two terms which differ in sign. Then, since $f$ is a polynomial, which is continuous, by the Intermediate Value Theorem, there must exist a number strictly between those two terms which is a root of $f$. Since the terms are in the interval $[1, 2^n]$, a number strictly between two terms is in $(1, 2^n)\subset (0, 2^n)$, which is the desired interval, so we're done.


trigonometry
calculus
calculus computations

show $ \sum_{i\equal{}1}^{\infty} 1\minus{}\cos{\frac{1}{i}}$ converges.

You may use limit comparison test with $ 1/i^2$.


Is it possible to purchase an AoPS gift card or something similar?

http://www.artofproblemsolving.com/Books/AoPS_GC_GiftCert.php

 :wink:

Some Mathcounts competitions award those for top scorers. Like mine did in 2008 for 1st place finisher at chapter contest.

Not just MathCounts. BCA Competition did too.  :)


conics
hyperbola
ratio
linear algebra
matrix

Hi guys could somebody give me a procedure to find the equation of the conic that passes through the points
$ (0,0), (0,5), (1,5), ( \minus{} 1, \minus{} 3), ( \minus{} 3, \minus{} 2)$  it is a hyperbola i guess

[hide]

You get the equations (x-h)^2/a+(y-k)^2/b=1 Now input the x and y values to find it.[/hide]

hello im afraid that this form of the equation of the conic doesnt help because it only holds when the focal axis is parallel to the x or y axis , in this case we have a general second  degree equation , what i dont dont is how do i obtain the equation of the conic by 5 points

[hide="Hint"]The standard form is $ Ax^2\plus{}Bxy\plus{}Cy^2\plus{}Dx\plus{}Ey\plus{}F\equal{}0$.

First, plug in $ (0,0)$ to get that $ F\equal{}0$. After that, note that only the [i]ratio[/i] $ A: B: C: D: E$ matters...so keep that in mind when plugging in values and solving the resulting equations.[/hide]

[hide="Alternatively"]This can be expressed as $ \begin{vmatrix}x^2&xy&y^2&x&y&1\\0&0&0&0&0&1\\0&0&25&0&5&1\\1&5&25&1&5&1\\1&3&9&\minus{}1&\minus{}3&1\\9&6&4&\minus{}3&\minus{}2&1\end{vmatrix}\equal{}0$. (Why?) I'm not sure how nicely this will work out, but it's worth a shot (it's probably equivalent to the first way, but whatever...).[/hide]


The numbers $ \log(a^3b^7)$, $ \log(a^5b^{12})$, and $ \log(a^8b^{15})$ are the first three terms of an arithmetic sequence, and the $ 12^\text{th}$ term of the sequence is $ \log{b^n}$. What is $ n$?

$ \textbf{(A)}\ 40 \qquad
\textbf{(B)}\ 56 \qquad
\textbf{(C)}\ 76 \qquad
\textbf{(D)}\ 112 \qquad
\textbf{(E)}\ 143$

[hide]
Using log properties, we have
$ a_1\equal{}3\log{a}\plus{}7\log{b}$,

$ a_2\equal{}5\log{a}\plus{}12\log{b}$, and

$ a_3\equal{}8\log{a}\plus{}15\log{b}$.

Taking differences, we get

$ 2\log{a}\plus{}5\log{b}\equal{}3\log{a}\plus{}3\log{b}$, so

$ 2\log{b}\equal{}\log{a}$ and $ b^2\equal{}a$.

Thus the sequence is $ 13\log{b}$, $ 22\log{b}$, $ 31\log{b}$, $ \ldots$, and the $ 12$th term is $ 112\log{b}$, so $ n\equal{}112$.  Answer is $ D$.
[/hide]

[hide]
Let $ A\equal{}\log(a)$ and $ B\equal{}\log(b)$.
Since the numbers $ 3A\plus{}7B$, $ 5A\plus{}12B$, and $ 8A\plus{}15B$ are the first three terms in an arithmetic sequence, $ 2(5A\plus{}12B)\equal{}(3A\plus{}7B)\plus{}(8A\plus{}15B) \Rightarrow A\equal{}2B$.
Since the first three terms in the sequence are $ 13B$, $ 22B$, and $ 31B$, the $ k$th term is $ (9k\plus{}4)B$.
Thus the 12th term is $ (9\cdot12\plus{}4)B\equal{}112B\equal{}nB \Rightarrow n\equal{}\boxed{112\Rightarrow D}$[/hide]

I thought this was a very nice problem ( wish i had solved it correctly on test). i missed one step that i realized when i got home:
[hide="solution"]Because it's an arithmetic sequence we know that $ log(a^5b^{12})\equal{}log(a^3b^7)\plus{}d$ where d is some common difference. Isolating d we get $ d\equal{}log(a^5b^{12})\minus{}log(a^3b^7)$ which we can simplify using the properties of logarithms that we know and love. Thus we get $ d\equal{}log(a^2b^5)$ This is still not enough information to solve our problem. We thus use the same tactic with the 3rd and 2nd terms of the sequence. From there we get that $ d\equal{}log(a^3b^3)$ Here is where I got frustrated and spent 10 min. thinking the test-writers had made a mistake. How could it be an arithmetic sequence and the d's are different. Instead of thinking this I should have done the simple step of equating the two d's and finding that $ a\equal{}b^2$ From here the problem is very straightforward, and we get our solution as D: 112[/hide]

[quote="Xantos C. Guin"][hide]
Let $ A \equal{} \log(a)$ and $ B \equal{} \log(b)$.
Since the numbers $ 3A \plus{} 7B$, $ 5A \plus{} 12B$, and $ 8A \plus{} 15B$ are the first three terms in an arithmetic sequence, $ 2(5A \plus{} 12B) \equal{} (3A \plus{} 7B) \plus{} (8A \plus{} 15B) \Rightarrow A \equal{} 2B$.
Since the first three terms in the sequence are $ 13B$, $ 22B$, and $ 31B$, the $ k$th term is $ (9k \plus{} 4)B$.
Thus the 12th term is $ (9\cdot12 \plus{} 4)B \equal{} 112B \equal{} nB \Rightarrow n \equal{} \boxed{112\Rightarrow D}$[/hide][/quote]

I don't understand how $ 3A \plus{} 7B$, $ 5A \plus{} 12B$, and $ 8A \plus{} 15B$ was gotten.

Thanks!

[quote="314.math"]I don't understand how $ 3A \plus{} 7B$, $ 5A \plus{} 12B$, and $ 8A \plus{} 15B$ was gotten.Thanks![/quote]

Using the log property $ x\log(y)\equal{}\log(y^x)$,
$ 3A\plus{}7B\equal{}3\log(a)\plus{}7\log(b)\equal{}\log(a^3)\plus{}\log(b^7)$. Similar process for the others.

[hide=solution]Let $A=\log a$ and $B=\log b$. Then the given three numbers rewrite themselves as $3A+7B, 5A+12B,$ and $8A+15B,$ which are the first three terms of an arithmetic sequence. Since the difference between any two consecutive terms of an arithmetic sequence is constant, $(5A+12B)-(3A+7B)=(8A+15B)-(5A+12B)\rightarrow 2A+5B=3A+3B\rightarrow A=2B$. The $12^{\text{th}}$ term of the sequence is the first term plus $11$ times the common difference, which is $3A+7B+11(2A+5B)=3A+7B+22A+55B=25A+62B$. Since $A=2B,$ this is equivalent to $50B+62B=112B$. The answer is $\boxed{\textbf{(D)} ~112}$.[/hide]


I know this is probably very basic, but somehow I forgot how to do it. Maybe it's just that my mind isn't clear.

How do you simplify $ x^{(x^2)}$?

You don't.  That's as simple as it gets.

On a similar topic, is x^y^z x^(y^z) or (x^y)^z

and is a^b^c^d=a^(b^(c^d)) or ((a^b)^c)^d?

Most likely this was not taken from any contest because it is already simplified


Find the maximum value of:
\[ f(x) \equal{} \sqrt{8x\minus{}x^2} \minus{} \sqrt{14x\minus{}x^2\minus{}48}\]

I think you would want to graph this one because it seems pretty hard to get rid of the radicals.

Graphing it would be pretty hard as well... I don't think you can graph it with enough accuracy to get the correct answer, unless you eyeball it and make a guess.

There is probably a solution involving a substitution of some kind and then AM-GM or another simple inequality. Also, completing the square might help a little?

$ f(x)\equal{}\sqrt{16\minus{}(x\minus{}4)^2}\minus{}\sqrt{1\minus{}(x\minus{}7)^2}$

$ \equal{}\sqrt{x(8\minus{}x)}\minus{}\sqrt{(8\minus{}x)(x\minus{}6)}$

$ \equal{}(\sqrt{8\minus{}x})(\sqrt{x}\minus{}\sqrt{x\minus{}6})$.

From the radical in the second one we get $ x\geq 6$. Also, $ 8\geq x$, so $ x$ is bounded by those values.

I believe the answer is $ 2\sqrt{3}$ when $ x\equal{}6$, but I can't find a good way to justify that.

And Since it is AHSME aren't there choices. So you could make a pretty good guess.

$ f(x) \equal{} \sqrt{16\minus{}(x\minus{}4)^2} \minus{} \sqrt{\minus{}(x\minus{}6)(x\minus{}8)}$

the argument of sqrt has to be +ve

possible when $ 6 \leq x \leq 8$

for $ f(x)_{max}$, $ \sqrt{\minus{}(x\minus{}6)(x\minus{}8)}$ has to be min

Besides

$ 16\minus{}(x\minus{}4)^2$ is a decresing function.

So the max will be when x = 6

and its value is $ f(6) \equal{} \sqrt{12}$.


hello,mathlinkers.I have a problem on Inequality and i have solved it but very ugly.so, some one can help me to find another solution
Let $ a_1,a_2,...,a_n$ are positive real numbers.Prove that:
$ \frac{a_1}{a_2\plus{}a_3}\plus{}\frac{a_2}{a_3\plus{}a_4}\plus{}...\plus{}\frac{a_{n\minus{}2}}{a_{n\minus{}1}\plus{}a_n}\plus{}\frac{a_{n\minus{}1}}{a_n\plus{}a_1}\plus{}\frac{a_n}{a_1\plus{}a_2}>\frac{n}{4}$

This is Shapiro's cyclic inequality.

[hide="http://www.mathlinks.ro/Forum/viewtopic.php?t=589&search_id=612167263"]See here.[/hide]

let $ a_{i_1}$ max of $ a_i$ and $ a_{i_2}$ max of $ a_{i_1\plus{}1},a_{i_1\plus{}2}$ . . .  (we say that $ a_{n\plus{}1}\equal{}a_1$) then we will have $ a_{i_{r\plus{}1}}\equal{}a_{i_1}$ if numbers 1,2...n we arrange on circle so numbers $ i_k$ and $ i_{k\plus{}1}$ will  be neighbours or next nearest so $ r\ge n/2$ let S our LHS so $ S\ge \frac{a_{i_1}}{2a_{i_2}}\plus{}\frac{a_{i_2}}{2a_{i_3}}\plus{}. . .\plus{}\frac{a_{i_r}}{2a_{i_1}}\ge\frac{1}{2}r\equal{}n/4$ (last ineq by AM-GM)


limit
trigonometry
real analysis
real analysis unsolved

$u = \frac{1}{x}$

The problem becomes

$\lim_{u \to \infty} \frac{\cos u}{u} = 0$.  :)

You should write as $u\rightarrow\pm\infty.$

[quote="pinky"]lim    x cos (1/x)
x->0[/quote]
$|x \cos \frac{1}{x}| \le |x|$.


A quadratic trinomial $f(x) = x^2+ax+b$ is given. The equation $f(f(x)) = 0$ has four real roots, two of which sum up to $-1$. Prove that $b \le -\frac{1}{4}.$

The problem posted again by  the great Darij [url=http://www.mathlinks.ro/Forum/post-506203.html]here[/url]


I was wonderring if you are considering to either become a mathematician or physicist, which one would you be?

Masoud Zargar

What's the difference, Mathematics is just Physics without puopose. :D (Guess you know what am I )
(Yes I see there is only 1 vote for Physicist - at present - and  I know who  voted for that )

mathematician because i'd like to study pure maths, not applied...i think the second one is more useful but i'm better in abstract things :lol:

In general, if I can point at it, it isn't abstract enough for me.

Hmm... maybe that's why I'm so bad at geometry...

I think if you ask this question here obviously you will get that most people will vote for Mathematician.
I chose Mathematician because I like to work with pure maths... it is like the queen of all sciences to me :)

I study both maths and physics, but I feel more like a mathematician.

However, that might change in the future :)


John ordered 4 pairs of black socks and some additional pairs of blue socks.  The price of the black socks per pair was twice that of the blue.  When the order was filled, it was found that the number of pairs of the two colors had been interchanged.  This increased the bill by 50%.  The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:
\[ \text{(A)}\ 4:1 \qquad \text{(B)}\ 2:1 \qquad \text{(C)}\ 1:4 \qquad \text{(D)}\ 1:2 \qquad \text{(E)}\ 1:8 \]

Let $b$ represent the number of blue socks.
$2b+4=1.5(b+8)$
$2b+4=1.5b+12$
$.5b=8$
$b=16$
$\boxed{\bold{C}}$

[hide]
Let B be the number of black socks, $C_B$ be the cost of the black socks, L be the number of blue socks, and $C_L$ the cost of the blue socks.
$4C_B + LC_L = C$
$LC_B + 4C_L = (3/2)C$

$(2/3)LC_B + (8/3)C_L = C$
$4C_B + LC_L = C$

$(4/3)LC_L + (8/3)C_L = C$
$8C_L + LC_L = C$

$((4/3)L + (8/3))C_L = C$
$(8 + L)C_L = C$

$(4/3)L + (8/3) = 8 + L$
$L = 16$
$B = 4$
$C$
[/hide]


Find the ordered pair $(x,y)$ that is the solution of the system:

$(x+2y)/xy = 11/12$
$(2x-3y)/xy = 2/3$

source: MAO 1990

[hide]Just crossmultiply and substitute:
$12(x+2y)=11xy$                                  $3(2x-3y)=2xy$
$12x+24y=11xy$                                   $12x-18y=4xy$
$12x+24y=11xy$                                   $12x=4xy+18y$
$4xy+42y=11xy$
$7xy=42y$
$x=6$                                                     $72=42y$
                                                               $y=\frac{12}{7}$

$\boxed{\boxed{\left(6,\frac{12}{7}\right)}}$[/hide]

How do you put in spaces so they will show up after you post?  My equations ran together.

cross multiply and so some clever things
it turns out $x=6$ and substituting that into one of the original equations gives $y=\frac{12}{7}$

$(6,\frac{12}{7})$

[hide]
First, rewrite the equations as $\frac{1}{y}+\frac{2}{x}=\frac{11}{12}$ and $\frac{2}{y}-\frac{3}{x}=\frac{2}{3}$.
Then let $Y=\frac{1}{y}$ and $X=\frac{1}{x}$.
The equations become $Y+2X=\frac{11}{12}$ and $2Y-3X=\frac{2}{3}$.

Solving the system gives $(X,Y)=\left(\frac{1}{6},\frac{7}{12}\right)$, so $(x,y)=\left(6,\frac{12}{7}\right)$
[/hide]

bflat, keeyounglee, frt, your answers are correct.


Suppose there are n arbitrary points in the plane.  Let $D$ be the max distance between them and $d$ be the min distance between them.
Show $D>\frac{d\sqrt{3}}{2}(\sqrt{n}-1)$.

[i]Edited by Myth[/i]

This is an IMO 1988 shortlist/longlist problem (or both :D).

[color=red][Moderator edit: It's not on Kalva's Shortlist page, hence it should be a longlist problem.][/color]

I can't believe! My very first dumbest estimate gave the answer! And this problem have ain here for more than 1 year!
Show that $\left(\frac{D}{\sqrt{3}}+\frac{d}{2}\right)^2\geq n\cdot\frac{d^2}{4}$.
Try to guess...  :D


calculus
factorial
algorithm
number theory
Euclidean algorithm

Given any positive integer $ n$, there exists $ n$ consecutive composite numbers? Also does $ \gcd(x, x \plus{} 2) \equal{} 1 \ \text{or} \ 2$ (if $ x$ is odd and even respectively)?

This doesn't really belong [\color=green][in the calculus forum][/color].  The first problem is usually solved using factorials, but it belongs to a somewhat more general class of problems that are easily solved using the Chinese Remainder Theorem:  consider the set of solutions to

$ x \plus{} 1 \equiv 0 \bmod p_1$
$ x \plus{} 2 \equiv 0 \bmod p_2$
..
$ x \plus{} n \equiv 0 \bmod p_n$

where $ p_i$ is the $ i^{th}$ prime.

As for your second question, yes.  The Euclidean algorithm gives $ \gcd(x, x \plus{} 2) \equal{} \gcd(x, 2)$.


How do you make a "<" or a ">" with LaTeX?

I was wondering this myself.  I believe you just use < though, the ordinary less than character.

When you are reading other people's posts in the forum, you can hover your mouse over their Latex to see what Latex code they used.  Pretty useful.

I think you're correct. "$ <$" and "$ \le$" seem to be in the same font.


Find all integer solutions for the equation:

$3(x+y)^{4}-8xy(x^{2}+y^{2})-528=0.$

let me write
$ 3(x+y)^{4}-8xy(x^{2}+y^{2}) = 2(x+y)^{4} + (x-y)^{4} = 528. $
considering that $x-y$ has to be even (because $528$ is) and for the same reason $x+y$ has to me, we can set $2s = x+y$, $2d = x-y$, and we have
$ 2s^4 + d^4 = 33 $
which is $s = 2$, $d = 1$, so $x = 3$, $y = 1$.[/code]


Wikipedia says that Guass-Jordan elimination is "considerably less efficient" then plain Guassian Elimination followed by back substitution.

Why is this? It seems to me that both methods involve the exact same set of steps. Plugging a variable into an equation is equivalent to subtracting that row of the matrix. What am I missing?

The distinction has to do with how you treat the intermediate steps of the original row elimination. Do you zero out the entire rest of the column around each pivot element, or do you only zero out the part of the column below the pivot element? If the former - and your source is probably using "Gauss-Jordan" to refer to the practice of cleaning out above and below both - then you arrive at an identity matrix in one pass but you had to do more work along the way. If you only clean out below, the eliminating left to right leaves you with an upper triangular system. Once you have an upper triangular system, then back substituting is essentially the same thing as row eliminating from right to left to clean out the upper triangle.


topology
real analysis
real analysis unsolved

Show that an open set in $\mathbb{R}^{n}$ cannot be homeomorphic to an open set in $\mathbb{R}^{m}$ unless n=m. this seems to be quite clear intuitively, but seems tough to prove for me. Can anyone give me hints? ( hints appreciated better than solutions)  :)

There's a proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=homology&t=75429]here[/url], using singular homology.

Even though intuitively clear, this is more difficult than it seems.


If $ x \equal{} 1$ satisfies the equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$ with $ a \neq 0,$ find all other solutiosn for $ x$.

[hide="My answer"]
Well since it's quadratic, there's only one other answer if it's distinct. Now, using $ (x\minus{}1)$ as one factor, we see that the other factor must be $ (ax\minus{}c)$ to have $ (x\minus{}1)(ax \minus{} c) \equal{} ax^2 \minus{} (a\plus{}c)x \plus{} c$ so the other root is $ \frac{c}{a}$. [/hide]

It seems to be too obvious..?  :|

[hide="my solution"]
Using synthetic division, we have $ ax^2\plus{}bx\plus{}c\equal{}(x\minus{}1)(ax\plus{}a\plus{}b)\plus{}a\plus{}b\plus{}c\equal{}0$. Plugging in $ x\equal{}1$ into the original polynomial, we have $ a\plus{}b\plus{}c\equal{}0$, so $ (x\minus{}1)(ax\plus{}a\plus{}b)\equal{}0$. Setting $ ax\plus{}a\plus{}b\equal{}0$, we find that $ x\equal{}\minus{}\frac{a\plus{}b}{a}$ is a solution. This is also equivalent to $ \frac{c}{a}$, since $ a\plus{}b\plus{}c\equal{}0$.
[/hide]

[quote="Silverfalcon"]If $ x \equal{} 1$ satisfies the equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$ with $ a \neq 0,$ find all other solutiosn for $ x$.

[hide="My answer"]
Well since it's quadratic, there's only one other answer if it's distinct. Now, using $ (x \minus{} 1)$ as one factor, we see that the other factor must be $ (ax \minus{} c)$ to have $ (x \minus{} 1)(ax \minus{} c) \equal{} ax^2 \minus{} (a \plus{} c)x \plus{} c$ so the other root is $ \frac {c}{a}$. [/hide]

It seems to be too obvious..?  :|[/quote]

[hide]Divide through by $ a$ so the constant is $ \frac{c}{a}$.  The product of the roots is $ \frac{c}{a}$ so the other root is $ \frac{c}{a}$.[/hide]


inequalities
function
inequalities proposed

Here is the corresponding inequality for 4 variables. For all $a,b,c>0 $ we have
  $\sum\frac{1}{4a+1}+\frac{2}{a+b+c+d+1}\geq\frac{1}{2}\cdot\sum\frac{1}{3a+b+1} $ where the second sum has 12 terms.

I think that I was wrong somewhere or the inequality is not too difficult.
We will prove that:
$ \sum \frac{1}{4a}+\frac{2}{a+b+c+d} \geq \frac{1}{2} \sum \frac{1}{3a+b} $.
This one is equal to:
$ C=\sum (a-b)^2[\frac{3(a+b)(c+d)-4ab}{ab(3a+b)(3b+a)}] \geq 0 .$
or $ \sum(a-b)^2 \cdot \frac{A(a,b)}{B(a,b)} \geq 0 $.
Assume that $ a\geq b\geq c\geq d $
It is easy to see that:
$ i) A(a,b) \leq A(a,c); A(a,d),A(b,c),A(b,d),A(c,d) \geq 0 $
$ ii)B(a,b) \geq B(a,c) \geq max[B(a,d),B(b,c),B(b,d),B(c,d)] $
Now we just need to consider the case that $ A(a,b) \leq 0 $.And we have:
$ C \geq \frac{ \sum A(a,b)(a-b)^2}{B(a,c)} $.
However, we can check that $ \sum A(a,b)(a-b)^2 \geq 0 $ by Muirhead.
So the proof is end now. :)

It is only my opinion, but I don't like proofs like this, since it is out of aesthetics.

Such kind inequalities cannot be hard, because it is Caramata Inequality, so we have only to show that first set majorize second set, and it can be done by considering finite number of cases :) Though I am sure there are general method (or whole theorem) for sets like in our inequality.

Dear Myth,

can you please explain better your method based on Caramata inequality.

I am very interested in it.

Thank you very much.

Hi Manlio!
Caramata Inequlity were discussed on forum many times. Recently I saw appropriate topic, for example.
Consider convex function $f(t)=\frac{1}{t+1}$. Due to Caramata Inequality conditions we need to show that set $(4a,4a,4b,...,4d,a+b+c+d,a+b+c+d)$ (12 terms) majorize set $(3a+b,3a+c,...,3d+c)$ (12 terms). Such checking reduces to finite number of cases. But I can repeat: I am sure there are general method for such kind of "symmetric" sets.
Do you need further explanation? Actually, I am going to bed... :)
Good night!

[quote="Myth"]But I can repeat: I am sure there are general method for such kind of "symmetric" sets.[/quote]

Hmm, you do have seen the thread

http://www.mathlinks.ro/Forum/viewtopic.php?p=122493 .

Especially the posting by flip2004 and my reply to that posting. Isn't it a nice symmetric criterion for majorization?

  Darij

[quote="Myth"]
Caramata Inequlity[/quote]

sorry for interrupt but isn't it [b]K[/b]aramata Inequality ? just a detail ...

Of course, it's Karamata. But it's MathLinks, so don't wonder if every second word of a posting has a typo  :)

  Darij

[quote="darij grinberg"]Of course, it's Karamata. But it's MathLinks, so don't wonder if every second word of a posting has a typo[/quote]
Such sentence can offence some people. :?

People's names can have variant spelling of word. It is difficult to remember all canonical english names.

Dear Myth,

if we consider the Harazi ineq previously posted

$\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}+\frac{1}{a+b+c}+\frac{1}{a+b+c}+\frac{1}{a+b+c} \geq \frac{1}{2a+b}+\frac{1}{2b+a}+\frac{1}{2b+c}+\frac{1}{2c+b}+\frac{1}{2c+a}+\frac{1}{2a+c}$

and the convex function $f(x)=\frac{1}{x}$ we can try to apply Karamata ineq considering WLOG that $a\geq b\geq c $ and the two cases

$1)$--->  $b \geq \frac{a+c}{2}$

$2)$--->  $b \leq \frac{a+c}{2}$

For the first case I have put

$x_1=3a,x_2=3b,x_3=x_4=x_5=a+b+c,x_6=3c$ so that $x_1 \geq x_2 \geq x_3 \geq x_4 \geq x_5 \geq x_6 $ and $y_1=2a+b,y_2=2b+a,y_3=2a+c,y_4=2b+c,y_5=a+2c,y_6=b+2c$ so that $y_1 \geq y_2 \geq y_3 \geq y_4 \geq y_5 \geq y_6 $

When I have tried to verify that the number array $(x_1,x_2,x_3,x_4,x_5,x_6)$  majorizes the number arrays $(y_1,y_2,y_3,y_4,y_5,y_6)$  I have found that

$x_1 \geq y_1$
$x_1+x_2 \geq y_1+y_2$

but

[quote]$x_1+x_2+x_3 \leq y_1+y_2+y_3$[/quote]

so I cannot apply Karamata, but probably I am wrong ( I cannot see my error :( ).

Thank you very much.

[quote="Myth"]People's names can have variant spelling of word. It is difficult to remember all canonical english names.[/quote]

Yes, but I don't think this is the case with Karamata. At least, I have always seen him spelt Karamata and never Caramata. But that's not really a matter of discussion. As for typos, I just mentioned them since there are sometimes typos that really hinder people to understand. Of course, I am speaking about typos in formulas, not typos or misspellings like your one (anyway, your posts have almost no typos).

  Darij

Dear Manlio!
[quote="manlio"]so I cannot apply Karamata, but probably I am wrong ( I cannot see my error :( ).[/quote]
You are totally right: there are no error.
I was blinded by article in Kvant magazine where author stated that 3a, 3b, 3c, a+b+c, a+b+c, a+b+c majorize 2a+b, ..., 2c+b.
I am very sorry.

Thus, in the end, it works or not using Karamata? Because I just hate the inequalities that can be proved just applying a majorization. Probably I was blind. Anyway, I have a very estetic proof for this inequality, not just cases and cases.

[quote="harazi"]Thus, in the end, it works or not using Karamata?[/quote]
No (until anyone will check majorization).

[quote="harazi"]Because I just hate the inequalities that can be proved just applying a majorization.[/quote]
I just hate to invent new methods for each inequality of the same kind.

[quote="harazi"]Anyway, I have a very estetic proof for this inequality, not just cases and cases.[/quote]
Estetic is a quite personal sense.


A point P is situated in system (S) (with origin O).If there's system (S')(with origin O') and S' is mooving with Vo along Oy then fiind vector(OP')=f(vector(OP)) .

What do you mean by "(OP')=f(vector(OP))" ?

Oups sorry.I meant O'P=f(OP) .fiind the position vector in S' depending of position vector in S,Vo and time in S (t)

Isn't this a bit trivial? Or what am I missing here?  :maybe:

well it's not a hard problem.
Give the answer after which fiind  the acceleration components in S' depending of the components of acceleration in S


There is gas inside of horizontal  cylinder withporshen of mass  $M$.We give heat to that gas.Then the porshen is moving with constant acceleration and aftyer some time it reaches the speed $u$.The internal energy of gas is changing as like this:$U=CT$Wha the amount of heat given to the system.

Sorry for mistake.I mixed the word 'piston' with the meaning of this word porshen in my language :rotfl:


This isnt a school related question but I thought this would be a good place to post.

I have a total cost of something being 2100 dollars. The store I'm buying it from says the tax is included. I live in Ontario so the tax is 15 percent.
How would I calculate the base price before tax?

[hide="Easy formula"]${2100\over 1.15}={\text{price.after.taxes}\over {1+{\text{tax.rate}\over 100}}}$[/hide]


If the St. Louis Rams field goal kicker, Jeff Wilkins, makes 28 field goals in 32 attempts this year, then he must have a streak of making at least N consecutive field goals.  Under these conditions, what is the longest streak N of consecutive field goals that Jeff is guaranteed to have?

A.   5	B.   6	C.   7	D.   8	E.   9

[hide] He made $ \frac78$ of his field goals. Therefore, he is guaranteed of at least a streak of $ \boxed{7}$ consecutive field goals (make $ 7$, miss $ 1$, make $ 7$, etc.)[/hide]

I disagree. $ 6$, miss, $ 6$, miss, $ 6$, miss, $ 6$, miss, $ 4$.

[hide="Solution"]Let there be $ 5$ boxes in which we will distribute the $ 28$ field goals. The first box corresponds to the field goals earned before the first miss, the second corresponds to the goals after the first miss but before the second... so the fifth box corresponds to the field goals earned after that fourth miss. By the pigeonhole principle, one box must have at least $ \boxed{6}$ goals. Note that $ 5\times 5 < 28$, so $ A$ cannot be the answer. The answer: $ \boxed{B}$.

[/hide]

darkdieugerre is correct, what you must do is subtract 4 from 32. then divide 28/5 to get 5.6. However we need whole numbers so we can have 5,miss,5,miss,5,miss,5 miss, 8 but there 8 is the longest so we try 6, miss, 6, miss, 6, miss, 6, miss, 4, and we have a solution.


[quote="GameBot"]Robin bought a four-scoop ice cream cone having a scoop each
of vanilla, chocolate, strawberry and cherry. In how many
orders can the four scoops be stacked on the cone if they
are stacked one on top of the other?[/quote]

$ 4^4\equal{}256$.

I thought the same thing, but the answer says 24... I'm guessing you just do 4!

Yes, BOGTRO's mistake was that he though that you could as many scoops of each you want. But you can only have one scoop of each flavor. So it would be $4!=4 \cdot 3 \cdot 2 \cdot 1 = \boxed{24}.$


algebra
polynomial
partial fractions
calculus
calculus computations

Is this the correct way of writing LHS=RHS

$ \frac{1}{\left[(\sqrt{3}x\plus{}1)(\sqrt{3}x\minus{}1)\right]^2(1\plus{}x)(1\minus{}x)}\equal{}\frac{A}{\sqrt{3}x\minus{}1}\plus{}\frac{B}{(\sqrt{3}x\minus{}1)^2}\plus{}\frac{C}{\sqrt{3}x\plus{}1}$ $ \plus{}\frac{D}{(\sqrt{3}x\plus{}1)^2}\plus{}\frac{E}{1\plus{}x}\plus{}\frac{F}{1\minus{}x}$

I can only find E,F,D AND B. how do you find A and C.

thanks

Yes, that's the correct generic partial fractions decomposition.  One method that always works is to clear the denominator; you'll get an equality of polynomials, whose corresponding coefficients must be equal.  If you've solved for four of the variables, this means you'll just have to solve two linear equations, which isn't too bad.  (In fact, if you choose coefficients carefully, you won't even have to expand everything out.)  An equivalent way to generate some equations is to plug in selected values for $ x$.  For example, putting $ x \equal{} 0$ will get you one equation involving $ A$ and $ C$; then either choose a coefficient (for example, the coefficient of $ x^5$ lets you avoid multiplying out too much) or choose another convenient value of $ x$ to get another equation.  (Note that putting $ x \equal{} 0$ is actually the same thing as looking at the constant coefficient of the two polynomials.)

you are right thanks.

will use x=0 and x=sqrt3.

Once you're done, [hide="check your answer:"]$ A\equal{}C\equal{}0$, $ B\equal{}D\equal{}\frac{3}{8}$, $ E\equal{}F\equal{}\frac{1}{8}$[/hide]

i get A=1/12 and C=-1/6. others same as yours.

I spent the whole day on this damn question but still can't get the correct answer. i put x=0 and x=sqrt3 to find A and C.

While I'm not usually that much of a fan of computer algebra systems (Maple, Mathematica, TI-89 calculators, etc.), this is precisely the kind of problem for which they are well-suited.


I've been playing around with this for a while. Anyone can help me out?

Over the region, we want to integrate:

$\int^5_0\int^{\sqrt {25-y^2}}_0 xe^ydxdy$

The innermost integral evaluates to:

$\frac {25-y^2}{2}e^y$

You can do the rest (it's straightforward integration by parts).  It might be faster the other way, but I stopped as soon as I saw that this would work.

Oh, and the answer's $\frac{8e^5-23}{2}$

hmm.
Polar transformation:
$x=r\sin\theta$ and $y=r\cos\theta$, and $r\in[0,5], \theta\in[0,\pi/2]$.

$\int_D y e^x dx dy = \int_0^5 \int_0^{\pi/2} r\cos\theta e^{r\sin\theta} rd\theta dr =
\int_0^5 r \left.e^{r\sin\theta}\right|_{\theta=0}^{\pi/2} dx = \int_0^5 r (e^r-1) dr = 4e^5-\frac{23}{2}$


As a moderator, I can see everyone's rating. I happened across a user with a rating of 5.1 who posted spam. I rated the post accordingly(one cube) and was startled to see that the user's rating increased to 5.2!

Is this some sort of bug?

I had a rating of 2.5, and someone rated one of my posts a 3 or 4, and my rating shot down all the way to 1.7!
Don't know if it's related, though

Maybe someone gave you a bad rating on another post.

Perhaps the problem here is because while I was rating that person, another person gave him/her a good rating simultaneously, resulting in an increase of 0.1?


Are the free response questions for the first round always on mechanics (no thermodynamics, electromagnetism, nuclear, relativity, etc.)? Thanks

yea mechanics could also mean gravitation, springs etc. but no magnetism or electricity


Any idea ? Please !  :D
When does the inequality hold ?

$a^{4}+a^{2}b^{2}+b^{4}=(a^{2}+ab+b^{2})(a^{2}-ab+b^{2})$
does this help? :P
and equality holds when $a=b=c$ of course!