modular arithmetic
probability
function
geometry
calculus
integration
number theory

Satellite X and Satellite Y orbit around Earth's equator at 4km/sec and 9km/sec respectively. Both satellites have circular orbits such that the radius of Satellite X's orbit is twice the length of Satellite Y's orbit. If Satellite X and Y are lined up exactly 24 times a year, what is the length of Satellite X's orbit in kilometers?

Can someone give a solution?

#1: How many solutions are there to $11x+12y=85$ if $x$ and $y$ are natural numbers?

[hide]
Take $11x+12y=85$ $\pmod{11}$. Then $11x \pmod{11}+12y \pmod{11}=85 \pmod{11}$ $0 \pmod{11}+ y \pmod{11}=8 \pmod{11} \Rightarrow y \pmod{11}=8 \pmod{11}$, so $y=8 \Rightarrow x=-1$. $-1$ is not a natural number, so the answer is 0.
[/hide]

I think
[hide]Satellite X travels $\frac 49$ slower than Satellite Y but travels on an orbit that's twice as long so Satellite Y is catching up on X at 7 kilometers a second. Since there are $31536000$ seconds in a year they are lined up every $1314000$ seconds. The amount of time it takes Satellite Y to line up with X is equal to $\frac {2r \pi}{7}$ where $r$ is the radius of Y's orbit because time is equal to distance over rate. So $\frac {2r \pi}{7} = 1314000$. X's orbit is twice the length of Y's orbit so we are looking for $4 r \pi$.

$1314000 \cdot 7 \cdot 2 = 18396000$[/hide]

#1: you are right

#2:heads or tails is 1/2 so the probability of one outcome is 1/(2^7) = 1/128
the favorable outcomes are hththth and thththt = two of them h=heads t=tails
1/128 * 2 = 2/128 = 1/64

#3: you are right

4. Find the maximum height of a truck of width 6 passing through a parabolic tunnel of height 16 and width 10.

[hide]
We have the points $(0,16)$ and $(5,0)$ on the tunnel represented by the parabola. The y-intercept is 16, so the function is the form $y=ax^2+16$. We know that $0=a \cdot 5^2+16$, so $a=-\frac{16}{25}$. If the width of the truck is 6, the x-value is 3. Then the max height is $-\frac{16}{25} \cdot 3^2+16$, or $\frac{256}{25} = 10.24$.
[/hide]

5. $ABCD$ is a square where $\overline{AB}=10$ and $\overline{AE}=6$. $E$ is a point on line $\overline{AD}$ and $F$ is a point on line $\overline{BC}$ such that $\overline{BF}=9$. Find the area of $\triangle ABG$, where $G$ is the intersection of $\overline{EB}$ and $\overline{FA}$.

[hide]
The line $EB$ can be represented by the equation $y=\frac{6x}{10} - 6$. The line $FA$ can be represented by the equation $y=\frac{9x}{10}$. The intersection is $\frac{6x}{10} - 6 = \frac{9x}{10}$. Solving for $x$, we see that $x=4 \Rightarrow y=-\frac{18}{5}$. $-y$ is the height of the triangle, so $A = \frac{-y \cdot 10}{2} = 18$.
[/hide]

#6: Find the smallest number with 24 factors that are perfect integral squares.

[hide]I don't know.[/hide]

#6:[hide]
We want the smallest number with 24 square divisors. The number of square divisors for
2^n*3^m, etc. is (n/2 + 1)(m/2 + 1) etc.
To achieve the smallest number, we have to minimize the values of n, m, etc, and (n/2 + 1), etc. as well. For 24, we have 3*2*2*2 = 24, so
we find n/2 + 1 = 3, n = 4 for base 2
m/2 + 1 = 2, n = 4, for base 3, 5, etc.

Then the number is 2^4 * 3^2 * 5^2 * 7^2 = [b]176400[/b].[/hide]

[hide="4"]Set the vertex to be $(0,16)$ and you knnow a point $(5,0)$ and $(-5,0)$, so the equation is $y=\displaystyle \frac{-16x^2}{25}+16$.  Then since the width is $6$, plug in $3$.

$y=\displaystyle \frac{-16\cdot{3^2}}{25}+16$

$y=\displaystyle \frac{-16\cdot{9}}{25}+\displaystyle \frac{400}{25}$

$y=\displaystyle \frac{-144}{25}+\displaystyle \frac{400}{25}$

$y=\displaystyle \frac{256}{25}=10.24$[/hide]

I believe that 176400 isn't the solution to number 6, because it isn't divisible by 162. One is enough to prove that it is wrong.

I don't know the answer though. I don't have the solutions with me now.

#8: The planet Phaesor has only 5-cent and 8-cent pieces as its currency. For example, one can make 51 cents with seven 5-cent pieces and two 8-cent pieces. What is the largest value that cannot be obtained using these 5-cent and 8-cent pieces?

[hide]
Chicken McNugget Formula:
If $m$ and $n$ are relatively prime, then the answer to this type of question is $mn-m-n$, or $(5)(8)-5-8=27$.
[/hide]

#9: See topic "Area of a Circle based on Chords" or whatever.

[quote="chess64"]#7: If Antoine rolls a standard 6-sided die three times, such that the first roll is given the quantity of $A$, the second roll is given the quantity of $B$, and third roll is given the quantity of $C$, what are the two most probable outcomes of $A \cdot B \cdot C$?

[hide]I don't know.[/hide][/quote]

I would guess 18 (that's 2 x 3 x 3) and 12 (2 x 2 x 3), but that's based on thirty seconds of thinking through a haze of sugar high.

The answer is 36 (and don't ask me why) to number 7.

[url]http://www.geocities.com/rocket_city_math_league/[/url]contains the answers to these apollo division problems.

[quote="chess64"]#6: Find the smallest number with 24 factors that are perfect integral squares.[/quote]

Find the smallest number with 24 factors. Then square that.

Ehrm...

Three sixty squared?

EDIT: I believe that's [b]129600[/b].

[quote="chess64"]#2: If Jason flips a coin 7 times, what is the probability that at no time he gets two consecutive heads or two consecutive tails?

[hide]I don't know the solution.[/hide][/quote]

The first flip doesn't matter. All of the other six require the flip that wasn't flipped before.

(1)(1/2)^6

I think the two most probable products are 12 and 24.

Here's why:

Let's just assume (for common sense's sake) that the #s which are most probable have 6 or more factors and are less than or equal to 64 (why? Because 64=4x4x4, and there is no way to further interchange the factors without having one of them greater than 6). Out of those numbers, the smaller ones seem most probable.

Testing out #s shows that 12=(4,3,1)=6 combos, (2,3,2)=3, and 6,2,1)=6 and 24=(6,2,2)=3, (6,4,1)=6, and (2,3,4)=6 with a total of 15 combos each. Close seconds are 60, 42, 36, and 30 with 12 combos each.

[b]Answer(s):12 and 24[/b]

[quote="chess64"]10. John is in charge of school text books. There are four subjects and four years for each subject so he has 16 books in all. How many ways can he rearrange these books if he wants to keep each subject together, but does not care in which order the books for each subject are in.

[hide]
Rearrangements for the books in each subject are $4!=24$. Rearrangements for the different subjects are $4!=24$. That's all I know. Does anyone know how to get the answer?
[/hide][/quote]

alright, you have that in each subject, you can arrange the books 24 ways, so you'll have that for each subject, in other words, 24*24*24*24, or 24^4, then you know that you will be able to arrange these subjects in 24 different ways, so (24^4)(24)=
24^5=7962624 is the answer i got

wow I just looked at the results of this and I got tied for 4th place with my pathetic 34... ROFL

11. Find the 10th term in the following recursive sequence of numbers where 8 is the 3rd term, which is defined by the 1st and 2nd terms. $\{1, 3, 8, 22, 60, 164, ...\}$

[hide]
$\{1, 3, 8, 22, 60, 164, 448, 1224, 3344, 9136\}$
Is there a faster way to do this?
[/hide]

#12: The answer key says that you're wrong, tarquin. Anyone else wants to try?

[quote="chess64"][quote="biffanddoc"][hide="number 6"]Find the least integer with 24 factors = 360. 3602 = 129600[/hide][/quote]

how do you find the smallest integer w/ 24 factors?[/quote]

With the power of roffle waffles.

[quote="KhL1392"][quote="chess64"][quote="biffanddoc"][hide="number 6"]Find the least integer with 24 factors = 360. 3602 = 129600[/hide][/quote]

how do you find the smallest integer w/ 24 factors?[/quote]

well, first, you put 24 into prime factorization, 2*2*2*3, since you add one to the powers and multiply them in order to find the number of factors, we subtract one, so the obvious answer would be $(2^2)*(3^1)*(5^1)*(7^1)$ (which is a mixture of  the powers [3-1]*[2-1]*[2-1]*[2-1]) that is the obvious answer is it not, but notice that 2 2  is less than 71 , so instead of the powers being  2, 1,  1, 1 it would be... some order of 3,2,1,(which is a mixture of the powers [4-1]*[3-1]*[2-1]) so we arrange with the least value, $(2^3)*(3^2)*(5^1)$ which gives us $360$ then, we square that for our answer of $129600$
sry if this is a little confusing[/quote]

what?

I don't know if it's as good as your solution, but it's easier to understand (for me):

The number that we are looking for, $N$, can be written as the product of prime numbers as follows: $N=p_1^{a_1} \cdot p_2^{a_2} \cdot p_3^{a_3} \cdot ... \cdot p_n^{a_n}$, where $p_1$ through $p_n$ are the prime numbers, least to greatest. Since we are trying to minimize the value of $N$, $a_1 > a_2 > a_3 > \cdots > a_n$.

Also notice that $(a_1 +1)(a_2 + 1)(a_3 + 1) \cdots (a_n +1)=24$. $24 = 2^3 \cdot 3$.

We need to keep the exponents as close to each other as possible, so if we have 2 prime factors, the exponents must be 5 and 3. Then we have $2^5 \cdot 3^3 = 864$.

If we have three prime factors, the exponents must be 3, 2, and 1. Then our number is $2^3 \cdot 3^2 \cdot 5^1 = 360$.

If we have four prime factors, ... well, we can't have 4 prime factors for 24 (I don't think) so 360 is the smallest.

EDIT: actually, you can but whatever

Anyway, you probably like your solution better because mine seems so long. I probably would like yours better too, if I could understand it  ;)


Source: Massachusetts Mathematics Leagues States 2005

Determine all values of $k$ such that the solution set of $|x-k|<2$ is a subset of the solution set of $\frac{2x-1}{x+2}<1$. Use inequality notation to express your answer.

$\frac{2x-1}{x+2}<1$ works for all negative $x>-2$, and for positive x, by cross multiplying, $2x-1<x+2 \Rightarrow x<3$. So therefore all solutions for x in $|x-k|<2$ must be less than 3, so $k<1$. Lastly, to satisfy the $x >-2$ we get $k>0$ so $\boxed{1>k>0}$ is our final solution.

[quote="matt276eagles"]Right idea, but the answer is actually $0\le k\le 1$.[/quote]

why????

You can tell by looking at the original that it can be equal to the values.


function
integration
logarithms
limit
calculus
calculus computations

Let $ f$ be a nondecreasing function on $ \left[ 1,\infty\right)$, and define

               $ F\left( x\right) \equal{} \int_{1}^{x}\frac {f\left( t\right) }{t}dt,$          $ x\geq1.$

Prove that if  $ \frac {F}{\log}$ is bounded on $ \left[ 1,\infty\right)$ then $ f$ is bounded on  $ \left[ 1,\infty\right)$

what does $ \frac{F}{\log}$ stand for? $ \frac{F(x)}{\ln(x)}$ ? $ \frac{F(x)}{\log_{a}(x)}$ ???

That is $ \frac {F(x)}{\log(x)}$ =$ \frac {F(x)}{\ln(x)}$ . :lol:

For any $ c\ge 1$, and any $ x > c$, $ F(x)\ge \int_c^x \frac{f(t)}{t}\,dt\ge f(c)(\ln x \minus{} \ln c)$.
What happens as $ x\to\infty$?

For any $ c\geq1$, $ M\equal{}\underset{x\rightarrow\infty}{\lim}M\geq\underset
{x\rightarrow\infty}{\lim}\frac{F\left(  x\right)  }{\ln x}\geq f\left(
c\right)  \underset{x\rightarrow\infty}{\lim}\frac{\ln x\minus{}\ln c}{\ln
x}\equal{}f\left(  c\right)$

 :lol:

[quote="jmerry"]For any $ c\ge 1$, and any $ x > c$, $ F(x)\ge \int_c^x \frac {f(t)}{t}\,dt\ge f(c)(\ln x \minus{} \ln c)$.
What happens as $ x\to\infty$?[/quote]

 $ F(x)\ge \int_c^x \frac {f(t)}{t}\,dt$ is not really correct because if f(t)<0 for c>t $ \ge$1

exist M'=M+Mlnc/lnx

F（ｘ）－F（ｃ）＜　M’

the result is the same.　Q.E.D.


calculus
integration
trigonometry
function
logarithms
calculus computations

Is it possible to find the integral of exp(x^3) analytically?

Sure, use power series:

Since $e^{x^{3}}=\sum_{k=0}^{\infty}\frac{x^{3k}}{k!}$ we have

$\int e^{x^{3}}dx = \int\sum_{k=0}^{\infty}\frac{x^{3k}}{k!}dx = \sum_{k=0}^{\infty}\frac{x^{3k+1}}{(3k+1)k!}+C$

[quote="Kent Merryfield"]Note the equivalence to $\int u^{-2/3}e^{u}\,du$ and hence $\Gamma\left(\frac13\right).$

I'm not familiar enough with the esoteric algebraic topic of elementary representation of integrals to make any definitive pronouncements, but I will say this: there is no good reason to assume that it can be represented in elementary terms.

["Elementary terms" means starting with powers, trig and inverse trig functions, exponentials and logarithms, and building the function in question in a finite number of arithmetic or function-composition steps. A power series involves an infinite number of steps. Of course, benorin is right; this has an easily found power series which is useful for many - but not all - purposes.]

And that's the usual response to many similar questions: there's no good reason to assume that it can. One should really get in the habit of thinking of elementary representation of integrals as the exception rather than the rule.[/quote]

This only holds if the integral is over $(0,\infty )$!


Set $ S \equal{} (1, 2,... , 2004).$ We denote by $ d_1$ the number of subset
of S such that the sum of elements of the subset has remainder
7 when divided by 32. We denote by $ d_2$ the number of subset of
S such that the sum of elements of the subset has remainder 14
when divided by 16. Compute $ d_1/d_2$.

[quote="pier1"]Set $ S \equal{} (1, 2,... , 2004).$ We denote by $ d_1$ the number of subset
of S ...[/quote]
$ S$ is not a set.  :P

See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1272872#1272872 -- this is a link to the last post on that thread, by kevinatcausa, which contains the solution.  The identical method applies here, giving an answer of $ \frac{1}{2}$.  (Note that this is independent of the numbers "7" and "14" but very dependent on the numbers "16" and "32.")


function
algebra
functional equation
algebra unsolved

Find all function $f:R \rightarrow R$ satisfy:
$f((x-y)^2)=(f(x))^2-2xf(y)+y^2;\forall x,y \in R$.

$x=y=0$: $f(0) = f^2 (0)$, $f(0)=0, 1$.

Case 1: $f(0)=0$.  Then setting $x = y$ gives $0=(f(y)-y)^2$ or $f(y)=y$ for all $y$.  This holds in the original functional equation.

Case 2: $f(0) = 1$.  Setting $x=y$ gives $1 = (f(y) - y)^2$ or $f(u)=u+-1$ for any particular $u$.  Suppose $f(y)=y-1$ and $f(x)=x+1$.  Then $(x-y)^2+1-4y = f((y-x)^2) = f((x-y)^2) = (x-y)^2+1+4x$.  Therefore, $x=-y$ and since $(x-y)^2+1+4x = (x-y)^2 + 1 - 4y = f((x-y)^2) = (x-y)^2 +- 1$, $x=y=0$.  But this is a contradiction since it implies $-1 = y-1 = f(y) = f(0) = f(x) = x+1 = 1$.  Therefore, $f(x)=x+1$ for all $x$ or $f(x)=x-1$ for all $x$ provide two additional solutions as per the original equation.

[quote="Michael Lipnowski"]Then $(x-y)^2+1-4y = f((y-x)^2) = f((x-y)^2) = (x-y)^2+1+4x$.[/quote]

Michael, I don't understand how you get this line.  Anyway, $f(x) = x -1 \forall x$ isn't a solution to the functional equation; it doesn't work if you substitute it back into the original equation.

My solution for Case 2 is as follows:
Setting $x = 0, y = 1$ we obtain $f(1) = 2$, and setting $x = y$ gives us $f(x) - x = \pm 1$.  Assume that there exists $k$ such that $f(k) - k = -1$.  By setting $x = k, y = 0$ we find that the only possible values of $k$ are $\frac{1}{2}$ or 0 which is impossible.

Now, setting $x = 1, y = \frac{1}{2}$, we obtain a contradiction.  Hence $f(x) = x+1 \forall x$, and we can easily verify that this satisfies the given equation.

[quote="Michael Lipnowski"]Therefore, $x=-y$ and since $(x-y)^{2}+1+4x = (x-y)^{2}+1-4y = f((x-y)^{2}) = (x-y)^{2}+-1$, $x=y=0$.[/quote]
I don't see how you got it. Here you can set $x=0$ since $f(0)=0+1$, so you will really obtain $y=0$, and then contradiction: $f(0)=f(y)=y-1=0-1=-1$.

[quote="Valiowk"]Assume that there exists $k$ such that $f(k)-k =-1$.  By setting $x = k, y = 0$ we find that the only possible values of $k$ are $\frac{1}{2}$ or 0 which is impossible.[/quote]
How did you find it?


geometry
circumcircle
superior algebra
superior algebra unsolved

Let P be a regular n-gon, vertices A1, A2.....An. Define the centre P to be the point C such that the distance

CAi=CAj for all i,j

a) Prove that if a regular n-gon can be constructed then a regular (2n)-gon can be constructed.

construct the bisectors of all the angles $ \angle A_{i}CA_{i\plus{}1}$, and let these bisectors intersect the circumcircle of the $ n$-gon. these intersection points together with the vertices of the regular $ n$-gon form a regular $ 2n$-gon.


[quote="firecricket91"]Does anyone know the 2003 ARML individual answers?

I have the test but I have no answers for it.[/quote]

See if they're on the website... URL manipulation is your friend.

Yea I did that, it didnt work for 2003.

Works for 2004,2005,2006 though...

Thanks but do you by anychance have like a solution for #3?

I cant seem to understand why it isnt

1/3(3/9+4/10+4/10)=17/45

which is like .37777777777

The original question was

Let N be a three-digit number that is divisible by 3.  One of N's digits is chosen at random and removed.  COmpute the porbablilty that the remaining number is divisible by 3 (Note if we have like 300 we can form either 30,30 or 00=0.  Or if we have 207 we can have 20, 07=7 or 27)

Your calculation is based on the assumption that each possible hundreds (or tens, or units) digit occurs equally frequently in three-digit multiples of 3.  There's no reason to suppose that this is true.

[hide="My solution"]The remaining number is divisible by three if and only if the digit removed is divisible by three. We have three cases:
1) All three of the numbers digits are divisible by three (note that 0 is divisible by three): there are four possible digits to use: 0, 3, 6, and 9. 0 cannot be the first digit, so there are 3*4*4=48 such numbers. If a digit is removed, there is a 1/1 probability that it is divisible by three.
2) None of the digits are divisible by three: either all of the digits are congruent to 1 (mod 3) or all of them are congruent to 2 (mod 3). There are 27 cases for the former and 27 cases for the latter for a total of 54 digits. If a digit is removed, there is a 0/1 chance that it is divisible by three.
3) One of the digits is divisible by three: This case is true for all remaining three digit multiples of three - there are 300 total multiples, so 300-54-48=198 numbers have one digit divisible by three. If a digit is removed from one of these numbers, there is a 1/3 chance that it is divisible by three.
Thus, the total probability is (1*48+0*54+1/3*198)/300=19/50
Sorry I did not LaTeX this. I'm feeling lazy today. :wink: [/hide]


In your opinion, who's gonna win this year the Davis Cup????

(Argentina is 2-1 up versus Croatia!!!)

If Ancic was playing, that tie might have had a different ending.
Argentina took it in the final fifth rubber, 3-2 over Croatia.

But, I would agree that Argentina has one of the strongest teams.  Russia does as well though - they defeated France 4-1.  If I had to pick, I think one of those two teams will win it.  Argentina or Russia, I don't know which.

Supposing that Argentina defeat Australia, we would play against Russia or the USA...

Where will we play if we play against Russia? And against the USA?????

that's a good question, you kno.. let me check

i believe, if i'm not mistaken, if argentina plays russia, it'll be russia, and russia's choice of surface.
if argentina plays the states, it'll be in argentina, and argentina will have choice of surface.. n that means clay.

so i guess that means you are routing for USA, right?

Yeah!!! I just want to play in Argentina ( I think that if we are locals at the final we can win!!!!) jajaja lol.


Each one of three consecutive integers is a nower of some prime .find all triples integer
with this property

Look modulo 2 and modulo 3. You'll get two cases to check:

1. $n-1$ and $n+1$ are powers of 2
2. $n$ is a power of 2 and $n+1$ or $n-1$ is a power of 3 - this is special case of Catalan's theorem so you'll get  easy a solution (look if you need here for details: http://mathworld.wolfram.com/CatalansConjecture.html ) - but I guess you don't need to use such an atom bomb, the special case $3^a-2^b=1$ or $3^a-2^b=-1$ is easy to handle with some congruences if I remember correctly.


Hey guys, can anyone suggest good chapters to do in each book for good Euclid preparation

i think you can probably try the worksop problems and decide what you need to learn more about.


(this is for summer work, and I am not sure if this answer is correct)

I calculated that the problem:
[quote]
4
---------
5 - 2√3
[/quote]
Comes out to be:
[quote]
20 + 8√3
-----------
13
[/quote]
Or:
[quote]
(20/13) + (8/13)√3
[/quote]
And I sort of have a feeling this is wrong... can someone verify the answer, and maybe explain how to do it correctly if it is incorrect?

The correct $ \text{\LaTeX}$ syntax for what you wrote is:[code]$\frac{4}{5-2\sqrt{3}}$[/code]
Rationalizing the denominator leaves us with
$ \frac{4}{5-2\sqrt{3}}\cdot\frac{5+2\sqrt{3}}{5+2\sqrt{3}}=\frac{20+8\sqrt{3}}{25-12}=\frac{20+8\sqrt{3}}{13}$.

Thank you. To "simplify" it more, should I put it in fractions (second example of answer) or keep it in that form?

If those lines are meant to be fraction bars, then see [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]this[/url]

We rationalize a denominator by multiplying by $ 1$ in the form of $ \frac {\text{its radical conjugate}}{\text{its radical conjugate}}.$

[hide="see this if you do not know what a radical conjugate is"]
If a number is in the form $ a\pm\sqrt {b},$ then its radical conjugate is $ a\mp\sqrt {b}.$[/hide]

Why do we multiply by the radical conjugate? Recall the identity $ (a + b)(a - b) = a^2 - b^2.$ We want the radicals to go away, and the best way to make them do that is to have them squared. Multiplying by the radical conjugate gives us $ (a + b)$ and $ (a - b).$

Doing so,
\[ \frac {4}{5 - 2\sqrt {3}} = \left(\frac {4}{5 - 2\sqrt {3}}\right)\left(\frac {5 + 2\sqrt {5}}{5 + 2\sqrt {5}}\right).
\]
Now we use our identity in reverse to get
\[ \frac {20 + 8\sqrt {5}}{5^2 - \left(2\sqrt {3}\right)^2\right)} = \frac {20 + 8\sqrt {5}}{25 - 12} = \boxed{\frac {20 + 8\sqrt {5}}{13}}.
\]
[hide="concerning JRavs post"]I felt it might be helpful to explain why you multiplied the original fraction by what you did.[/hide]

[hide="concerning StephenBs question"]It isn't possible to simplify it any more, unless you want it to be in the form $ \frac {20}{13} + \frac {8\sqrt {5}}{13}.$ This form is not necessarily more simplified than the other form, it's merely another way of presenting an answer.[/hide]

What he's asking is if the 2nd form is more simplified than the first.

I'm pretty sure you should keep it in the first form.

i'm not sure what 1st/2nd are, but i'm pretty sure roflcopt's form is more universally accepted.

In the case like this in many math competitions, the standard is to keep it as one unless it can be reduced (which is not the case here).
If this problem comes up on AIME, they would probably say, "The value is in the form $ \frac{a \plus{} b \sqrt{c}}{d}$ in a reduced form and $ c$ is a square free. Find the value of $ a\plus{}b\plus{}c\plus{}d$).

How can this be simplified even more. He already rationalized the denominator.


Express $(a_1+a_2+\cdots+a_x)^n$ in $\Sigma$ and $\Pi$ notation

[quote="Altheman"]$\prod_{k=1}^{n}(\sum^{x}_{i=1} a_{i})^{k}$[/quote]

I dont think thats right. But I do not have one either.

$\prod_{k=1}^{n}(\sum^{x}_{i=1} a_{i})$

there just should not have the k there, but it seems pretty trivial, and the product notation is useless, maybe there is something else to the problem

yes, theres something else to the probelm, so thats not right

well, it is not clear what else is contained in the problem, the way that i represented it answers the question, do you want the expansion? i.e. like binomial expansion just not a binomial?

He basically want's the extended binomial theorem.

the best way that i can think of it would be to have $A_n$ =  the binomial expansion of $A_{n-1}$ and $a_n$
$A_n = \sum^{n}_{k=0}\binom{n}{k}(A_{n-1})^{k}a_{n}^{n-k}$
then repeat until $A_1$, i think that there is something that is not clear here, but i don't see what exactly
i wonder if there is a better way, since this gets complex fast


linear algebra
matrix
algebra
polynomial
integration
induction
complex numbers

Let A,B be two matrices from M 3 ([b]C[/b]) so that AB(AB-BA)=O3 . Prove that (AB-BA) 3=O3 . ( [b]C[/b] is the set of complex numbers)

Nice (not the town in south of France  :D ) problem pcalin,

Three ingredients for solving this nice problem:

1)A caracterisation of nilpotent matrix in  M_3(C)
Let N \in M_3(C)
trN=trN=trN^3=0 if and only if N is nilpotent
where tr is the trace
2) formula trMN=trNM
[b]3) Fadeev formula : let M a matrix in M_n(C), P(x)=x^n+a_1.x^(n-1)+...+
a_(n-1)x+a_n
we have
trM i + a1.trMi-1+...+ai-1.trM + i.ai = 0 for any i\in {1;2;...;n}[/b]
The hypotethis => (i) ABAB=ABA and
(ii) trABAB=trABA=trBAB
--------------------------------------------------
tr(AB-BA)=tr(AB)-tr(BA)= 0
-------------------------------------------------
(AB-BA)=ABAB-ABA-BAB+BABA
(AB-BA)=BABA-BAB with (i)

tr(AB-BA)
= trBABA-tr((BA)(AB))
=tr(BAB)A-trABA
=trABAB-trABA=0
--------------------------------------------------
(AB-BA)^3=(BABA-BAB)(AB-BA)
=BABA(AB-BA)-BA[AB(AB-BA)] the term in [ ] is 0 with (i)
=BABAAB-BABABA

tr(AB-BA)^3=
tr(BABAA)B - tr(BABABA)
=tr(BABA)A - tr(BABABA)
=tr((ABA)BA) - tr(BABABA)
=tr(ABAB)BA - tr(BABABA)
=trAB(ABA) - tr(BABABA)
=trABABAB - tr(BABABA) = 0

Let P(x)=x^3+ax+bx+c  the polynomial caracteristic of AB-BA
with Cayley-Hamilton theorem
P(AB-BA)=(AB-BA)^3 + a(AB-BA)+b(AB-BA)+c.I_3=0 (*)

a=tr(AB-BA)=0
in (*) take the trace 3c=0=>c=0
Fadeev formula give b=0

So P(AB-BA)=(AB-BA)^3 = 0

I hope pcalin you have a nicer solution , can you post it please ?

thx
 :D  \cup \beta  \int

I didn't have any solution. I tried the same tricks but I couldn't prove that b=0. (I didn't knew Fadeev's formula). By the way, can you post an idea of the proof of Fadeev's formula?

Is this fadeev?

Fadeev formula : let M a matrix in M_n(C), P(x)=x^n+a_1.x^(n-1)+...+
a_(n-1)x+a_n
we have
trM^i + a_1.trM^(i-1)+...+a_(i-1).trM + i.a_i = 0 for any i in {1;2;...;n}

I mean do you have a proof for it? and is it that the coeficient of a_i is i?

bye!

For the Fadeev Formula:
Let a1,...,an be the eigenvalues of M. Then ai k are the eigenvalues of M i.
We have tr(M i)= \sum {j in 1..n} aj i.

Fadaev formula really comes from the Newton's sums, and we have some formula for i>n also:
tr(M n+k)+...+ak*tr(M k)=0.
To prove the last, just isolate the terms of the sum in which ai appears, and use the fact that ai is a root of P.

For the first part do smth similar using Viete's formulas.
 \sum (a1 i)- (\sum (a1 i*a2)-\sum(a1 i-1a2a3))+... In the last sum, the sums are symetrical.
Or you could use induction, but I can't remember how.


on the AIME geometry problems, do they give you pictures for the geometry problems? or is it mixed, like on the amc's. or do they just not have them at all?

thanks.

The great majority of the time, you don't get pictures. Rarely, especially for late geo problems, there is a diagram if it would be complicated to explain.

Yeah, there was a 3-D geometry problem with some irregular polyhedron, and no diagram was provided.


Look what I found!!!!!!

         Solve the puzzles by saying them out loud, over and over, faster and faster, repeating the phrase, until you "hear" the answer.

Example: LAWN SAND JEALOUS (place) Answer: Los Angeles

1. SHOCK CUSSED TOE (person)

2. SAND TACKLE LAWS (fictional character)

3. MY GULCH HOARD UN (person)

4. MOW BEAD HICK (book)

5. TALL MISCHIEF HER SUN (person)

6. CHICK HE TUB AN AN US (product)

7. THOUGH TIGHT AN HICK (thing)

8. AISLE OH VIEW (phrase)

9. TUB RAID HEAP HUNCH (TV show)

10. CARESS TROUGHER CLUMP US (person)

11. DOCKED HEARSE WHOSE (person)

12. THUMB ILL KEY WAKE OWL LICKS HE (place)

13. AGE ANT HUB BLOWS HEAVEN (fictional character)

14. THESE HOUND DOVE MOO SICK (movie)

15. BUCK SPUN HE (fictional character)

[hide="I think"]1. Jacques Cousteau
2. Santa Clause
3. Michael Jordan
4. Moby Dick
5. Thomas Jefferson
6. Chiquita Bananas
7. The Titanic
8. I love you
9. The Brady Bunch
10. Christopher Colombus
11. Dr. Seuss
12. The Milky Way Galaxy
13. Agent 007
14. The Sound of Music
15. Bugs Bunny[/hide]

Well that didn't last long. Yippee for perfect. I can't believe you knew #1.

he takes french, he has to know jacques cousteau


A certain number of oranges are being divided among three people. One third of the oranges are given to person A. One fifth of the remaining oranges is given to person B and then all of the remaining, 56 oranges, are given to person C. How many oranges were there to begin with?

[hide]Alright, let us keep x as the number of oranges.
1/3x= given to person A
1/15x=Given to B
and the rest is given to c, and that is 56...
Well, 2/3x=given to person C
Cause, you add 1/3+1/15, and you get 5/15=1/3
Then you subtract 3/3-1/3=2/3
So 2/3x=56,
Then you solve the equation for x, and you get the answer as 84.[/hide]

[hide]Let us seee\

A:1/3
B:1/5(2/3)
c:56

soo

if we add a and b we get 7/15 so

that means 8/15 of the total lets say x is 56 so therefore x equals [b]105[/b] or (56x15)/8[/hide]

K001 1/15 isnt how much person B gets because it is the remaining of what A takes which is 2/3

... I think... Plz let me know if im wrong


What is the units digit in the product $ 3^1 \cdot 3^2 \cdot 3^3 \cdot 3^4 \cdots 3^{99} \cdot 3^{100}$ where all the whole number powers of $ 3$ from $ 3^1$ to $ 3^{100}$ inclusive are multiplied?

it will be the units digit of $ (3\cdot9\cdot7\cdot1)^25$, which is the units digit of $ 9^25$. knowing that powers of $ 9$ go in cycles like $ 9, 1, 9, 1, 9, 1, \cdots$, we see that $ 25$ is odd, and the units digit will be $ \boxed9$

Another way to approach this would be to add the exponents (this is one of the rules of exponents: when the base is the same, add the exponent).
Adding, we get $ 3^{\frac{100(1\plus{}100)}{2}}\equal{}3^{5050}$. Since the powers of $ 3$ repeat in cycles of $ 4$ ($ 3,9,7,1,3,9,7,1\ldots$), take $ 5050 (\text{mod }4)\equiv2 (\text{mod }4)$.
This gives an answer of $ 3^2\equal{}\boxed{9}$, same as nike's :)


geometry trigonometry ratio incenter inradius circumcircle parallelogram

[hide]Consider $ \Delta AHF$ and $ \Delta AHE$, by using sine law yields $ \frac {FH}{HE} \equal{} \frac {sin (90 \minus{} C)}{sin (90 \minus{} B)} \equal{} \frac {cos C}{cos B}$.
By some angle tracing we can get $ \angle HDF \equal{} \frac {1}{2} \angle C$ and $ \angle HDE \equal{} \frac {1}{2} \angle B$
Consider $ \Delta DHF$ and $ \Delta DHE$, we get the ratio $ \frac {FH}{HE} \equal{} \frac {tan \angle HDF}{tan \angle HDE} \equal{} \frac {tan \frac {B}{2}}{tan \frac {C}{2}}$
So $ \frac {cos C}{cos B} \equal{} \frac {tan \frac {B}{2}}{tan \frac {C}{2}}$.
Denote the incenter as I and the inradius as r.
Then consider $ \Delta BHF$ and $ \Delta CHE$ yields $ \frac {BE}{CF} \equal{} \frac {tan \frac {B}{2}}{tan \frac {C}{2}} \equal{} \frac {cos C}{cos B} \equal{} \frac {FH}{HE}.$
As $ \angle HFB \equal{} \angle HEC$ so $ \Delta HFB \sim \Delta HEC$.
So $ \frac {sin \angle HBD}{sin \angle HCD} \equal{} \frac {HB}{HC} \equal{} \frac {cos C}{cos B} \equal{} \frac {sin (90 \minus{} C)}{sin (90 \minus{} B)}$ and $ \angle HBD \minus{} \angle HCD \equal{} \angle B \minus{} \angle C \equal{} (90 \minus{} C) \minus{} (90 \minus{} B)$.
Let $ k \equal{} \angle B \minus{} \angle C, a \equal{} \angle HCD, b \equal{} 90 \minus{} \angle B$
Then $ \angle HBD \equal{} a \plus{} k,90 \minus{} \angle C \equal{} b \plus{} k$.
So $ \frac {\sin (a \plus{} k)}{\sin {a}} \equal{} \frac {\sin(b \plus{} k)}{\sin {b}}$, which yields $ \sin {k} \sin (a \minus{} b) \equal{} 0$ As $ AB \neq AC$ so $ k \neq 0$ so $ a \equal{} b$ i.e. $ \angle HCD \equal{} 90 \minus{} \angle C$ so $ BH \bot AC$ thus H is the orthocenter.
[/hide]
Remark: can anybody provide a nicer proof for $ \angle HCD \equal{} 90 \minus{} \angle C$?

[quote="brianchung11"]$ \Delta ABC$ is a triangle such that $ AB \neq AC$. The incircle of $ \Delta ABC$ touches $ BC, CA, AB$ at $ D, E, F$ respectively. $ H$ is a point on the segment $ EF$ such that $ DH \bot EF$. Suppose $ AH \bot BC$, prove that $ H$ is the orthocentre of $ \Delta ABC$.

Remark: the original question has missed the condition $ AB \neq AC$[/quote]
An other solution . Let $ EF$ cuts $ BC$ at point T then $ (BCDT) \equal{} \minus{} 1$ .Because $ RH\perp EF$ so it is the angle bisector of $ \angle{AHB}$ . Therefore $ \angle{BHP} \equal{} \angle{AHQ}$ . From this statement , it is easy to check that triangle  AHQ is similarity to triangle BHP . So $ BH\perp AC$ , and H is the orthocentre of triangle ABC .

http://www.mathlinks.ro/viewtopic.php?p=1321704&search_id=509318415#1321704
An almost equivalent problem here  :)

[quote="brianchung11"] [color=darkred]Let $ ABC$ be an acute triangle with the orthocenter $ H$ . The its incircle $ C(I,r)$ touches

its sides in $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ respectively. Suppose $ AB\ne AC$

and denote $ R\in AH\cap EF$ . Prove that if $ DR\perp EF$ , then $ R\equiv H$ .[/color] [/quote]
[color=darkblue]Prove easily that (or is well-known) : $ m(\angle EDF) \equal{} \frac {B \plus{} C}{2} \equal{} 90^{\circ} \minus{} \frac A2$ a.s.o. ;  $ AH \equal{} 2R\cos A$ ;

$ r \equal{} (p \minus{} a)\tan\frac A2 \equal{} (p \minus{} b)\tan\frac B2 \equal{} (p \minus{} c)\tan\frac C2$ ; $ \boxed {\cos A \plus{} \cos B \plus{} \cos C \equal{} 1 \plus{} \frac rR}\ \ (1)$ .

$ \frac {RF}{RE} \equal{} \frac {AF}{AE}\cdot\frac {\sin\widehat {RAF}}{\sin\widehat {RAE}} \equal{} \frac {\sin\left(90^{\circ} \minus{} B\right)}{\sin\left(90^{\circ} \minus{} C\right)} \equal{} \frac {\cos B}{\cos C}$ $ \implies$ $ \boxed {\ \frac {RF}{RE} \equal{} \frac {\cos B}{\cos C}\ }\ \ (2)$ .

$ \frac {AR}{\sin \widehat {AFR}} \equal{} \frac {AF}{\sin\widehat {ARF}}\implies\frac {AR}{\cos\frac A2} \equal{} \frac {p \minus{} a}{\sin\left(B \plus{} \frac A2\right)}\implies$ $ AR \equal{} \frac {(p \minus{} a)\cos\frac A2}{\sin\left(B \plus{} \frac A2\right)} \equal{}$

$ \frac {r}{\tan\frac A2}\cdot \frac {\cos\frac A2}{\sin\left(B \plus{} \frac A2\right)} \equal{}$ $ r\cdot\frac {\cos^2\frac A2}{\sin\left(B \plus{} \frac A2\right)\sin\frac A2}$ $ \implies$ $ \boxed {\ AR \equal{} r\cdot\frac {1 \plus{} \cos A}{\cos B \plus{} \cos C}\ }\ \ (3)$ .

$ \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{}$

$ \blacktriangleright\ DR\perp EF\implies$ $ \frac {RF}{RE} \equal{} \frac {DF}{DE}\cdot\frac {\sin\widehat {RDF}}{\sin\widehat {RDE}}\stackrel {(2)}{\ \implies\ }\frac {\cos B}{\cos C} \equal{} \frac {2(p \minus{} b)\sin\frac B2}{2(p \minus{} c)\sin\frac C2}\cdot \frac {\sin\frac C2}{\sin \frac B2}$ $ \implies$

$ \frac {\cos B}{\cos C} \equal{} \frac {p \minus{} b}{p \minus{} c} \equal{} \frac {\tan\frac C2}{\tan\frac B2}$ $ \implies$ $ \cos B\tan\frac B2 \equal{} \cos C\tan\frac C2\stackrel {(B\ne C)}{\ \implies\ }\underline {1 \plus{} \cos A \equal{} \cos B \plus{} \cos C}$ .

Thus, $ \left\|\begin{array}{c} (3)\implies\underline {AR \equal{} r} \\
 \\
(1)\implies 2R\cos A \equal{} r\implies \underline {AH \equal{} r}\end{array}\right\|\ \implies\ AR \equal{} AH \equal{} r\ \implies\ R\equiv H$ .

Hence $ \boxed {\ DR\perp EF\ \implies\ R\equiv H\ }$ . Observe that if $ M$ is the midpoint of $ [BC]$ , then $ H\in MI$ .

[b][u]Remark.[/u][/b] If $ B\ne C$ , then $ 1\plus{}\cos A\equal{}\cos B\plus{}\cos C\Longleftrightarrow$ $ 1\plus{}\sin B\sin C\equal{}\frac {\sin (B\minus{}C)}{\sin B\minus{}\sin C}\Longleftrightarrow$

$ \sin\frac A2\cos\frac B2\cos\frac C2\equal{}\frac 12\Longleftrightarrow$ $ (p\minus{}b)(p\minus{}c)\equal{}2Rr\Longleftrightarrow$ $ OI\equal{}OD$ , where $ O$ is the circumcenter of $ \triangle ABC$ . [/color]

We have:
*$ \angle{HED}\equal{}\angle{DIB}\equal{}90\minus{}\frac{\angle{B}}{2}$
$ \rightarrow \triangle{DHE} \sim \triangle{BDI} \rightarrow ID.HD\equal{}BD.HE$(1)
*$ \angle{HFD}\equal{}\angle{DIC}\equal{}90\minus{}\frac{\angle{B}}{2}$
$ \rightarrow \triangle{DHF} \sim \triangle{CDI} \rightarrow ID.HD\equal{}CD.HF$(2)
From (1) ,(2) we get : $ \triangle{HEC} \sim \triangle{HFB} \rightarrow \angle{ABH}\equal{}\angle{ACH}$ (3)
$ CH$ intersect $ AB$;$ BH$ intersect $ AC$ at $ P,Q$ ,respectively
$ \rightarrow APQC$ is inscribed a circle.
We have:
*$ \angle{HAB}\equal{}90\minus{}\angle{ABC}\equal{}90\minus{}\angle{AQP}$
$ \rightarrow AH$  pass through circumcenter of $ \triangle APQ$
*$ \angle{AHP}\equal{}180\minus{}\angle{HAB}\minus{}\angle{APH}\equal{}90\minus{}\angle{PCB}\equal{}90\minus{}\angle{PQH}$
$ \rightarrow AH$  pass through circumcenter of $ \triangle HPQ$
But, there is only one point $ T$ lie on $ AH$ such that $ TQ\equal{}TP$($ AB\neq AC$)
$ \rightarrow APHQ$ is inscribed a circle $ \rightarrow \angle{APH}\equal{}\angle{AQH}\equal{}90$
Therefore, $ H$ is orthocenter of triangle$ ABC$

why there is only one point T lie on AH
there are two points  	symmetrical

Another approach: Let $ T$ be the intersection of $ EF$ with $ BC$. We have that $ (T, B, D, C)\equal{}\minus{}1$, and since $ DH \perp TH$, the line $ HD$ is the $ H$-angle bisector of triangle $ BHC$. Now, $ AI \perp EF$ and so $ AI \| DH$. Combining this with $ AH \perp BC$ and $ ID \perp BC$, we get that $ AIDR$ is a parallelogram (where $ I$ is the incenter of $ ABC$). Consider now the homothety which maps $ A$ to $ R$ and $ I$ to $ D$. Since in a triangle the circumcenter and orthocenter are isogonal conjugates, the $ A$, $ R$-circumcenter cevians of $ ABC$ and $ RBC$ are parallel and so the circumcenter $ O$ of $ ABC$ is mapped into the circumcenter $ O'$ of $ BRC$. It follows that $ AH \equal{} 2OM$, where $ M$ is the midpoint of $ BC$, and so $ H$ is the orthocenter of $ ABC$.

Once we have proven that $ \angle{BHD}\equal{}\angle{CHD}$, as $ \angle{HFB}\equal{}\angle{HEC}$, we have $ \angle{ABH}\equal{}\angle{ACH}$. Then reflect $ C$ with respect to $ AH$, and call this new point $ C'$. Therefore $ AHBC'$ is cyclic quadrilateral and $ \angle{HBC}\equal{}\angle{HAC'}\equal{}\angle{HAC}\equal{}90\minus{}\angle{C}$.  :)

Let's assume without loss of generality that AB < AC, Sea I incenter ABC then it is easy to see that AI is perpendicular to EF that AI / / DH also as ID is perpendicular to BD have ID / / AH so as DIAH is a parallelogram, then AH = ID, AI Pack = DH, < DBF = 2 < IBD = < EHL besides < EHD = 90 = < IDB then IBD and DEH are similar so EH/HD = ID / DB = AH/DB from here
EH/HD = AH / DB i.e. EH / AH = HD /DB and < EHA = < HDB, DHB and EHA are therefore similar < DHB = <AEH= < AFH = x then < EAF = 180 - 2 x now prove that < CHD = x continuing EOF until that Court to CB in M luiego is known that C, D, B and M are in Tetrad harmony and as < DHM = 90 get < DHB = < CHD = x this means that H meets that < CAB + < CHB = 180 and H is the height so as the orthocenter only conclude that H is the orthocenter. THIS COMPLETES THE SOLUTION


geometry
trigonometry
inequalities
geometric transformation
reflection
perpendicular bisector
inequalities unsolved

Prove that the area of a quadrilaterial with sides a,b,c,d is $\leq \frac{1}{4}((a+c)^2+bd)$

But equality [b]A = sqrt ( (p-a)(p-b)(p-c)(p-d) )[/b] isn't  right for all quadrilaterals.
Inequlity   [b]A <= sqrt ( (p-a)(p-b)(p-c)(p-d) )[/b] is right, and helps for solving this problem too.

A nice problem. Here is my solution:

We show a lemma about triangles:

[color=blue][b]Lemma 1.[/b] Let ABC be a triangle with area $\Delta$. Then,

[b](a)[/b] we have $\left(a+b\right)^{2}-\frac34c^{2}\geq 4\Delta$;
[b](b)[/b] we have $ab+\frac34c^{2}\geq 4\Delta$.[/color]

[i]Proof of Lemma 1.[/i] [b](a)[/b] Let $h_{c}$ be the altitude of triangle ABC from the vertex C. Then, on the one hand, by a very well-known formula, the area $\Delta$ of triangle ABC equals $\Delta=\frac12 ch_{c}$. On the other hand, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=27965 , we have $\left(a+b\right)^{2}\geq c^{2}+4h_{c}^{2}$. By AM-GM, we have $\left(\frac12 c\right)^{2}+\left(2h_{c}\right)^{2}\geq 2\cdot \frac12c\cdot 2h_{c}=4\cdot\frac12 ch_{c}=4\Delta$. Hence,

$\left(a+b\right)^{2}-\frac34c^{2}\geq\left(c^{2}+4h_{c}^{2}\right)-\frac34c^{2}=\left(\frac12 c\right)^{2}+\left(2h_{c}\right)^{2}\geq 4\Delta$,

and Lemma 1 [b](a)[/b] is proven.

[b](b)[/b] First we use the cosine law for rewriting $c^{2}$ as $c^{2}=a^{2}+b^{2}-2ab\cos C$. Since $a^{2}+b^{2}\geq 2ab$, we thus have $c^{2}\geq 2ab-2ab\cos C=2ab\left(1-\cos C\right)$. On the other hand, the area $\Delta$ of triangle ABC equals $\Delta=\frac12 ab\sin C$, so that $4\Delta=4\cdot\frac12 ab\sin C=2ab\sin C$. Hence, in order to prove the inequality $ab+\frac34c^{2}\geq 4\Delta$, it is enough to show the stronger inequality

$ab+\frac34\cdot 2ab\left(1-\cos C\right)\geq 2ab\sin C$.

Division by $\frac{ab}{2}$ transforms this into $2+3\left(1-\cos C\right)\geq 4\sin C$. This rewrites as $5-3\cos C\geq 4\sin C$ or $5\geq 3\cos C+4\sin C$. But this is rather obvious: By Cauchy-Schwarz,

$\left(\underbrace{3^{2}+4^{2}}_{=5^{2}}\right)\cdot\left(\underbrace{\cos^{2}C+\sin^{2}C}_{=1}\right)\geq\left(3\cos C+4\sin C\right)^{2}$,

so that $5^{2}\geq \left(3\cos C+4\sin C\right)^{2}$, or, after taking the square root, $5\geq\left|3\cos C+4\sin C\right|$, so that automatically $5\geq 3\cos C+4\sin C$. Thus, Lemma 1 [b](b)[/b] is proven.

Now, we can proceed to solve the actual problem:

[color=blue][b]Problem.[/b] Prove that the area of any quadrilateral $ABCD$ is $\leq\frac14\left(\left(AB+CD\right)^{2}+BC\cdot DA\right)$.[/color]

[i]Solution.[/i] Let $\left|P_{1}P_{2}...P_{n}\right|$ denote the area of an arbitrary n-gon $P_{1}P_{2}...P_{n}$.

We have to prove that, for any quadrilateral ABCD, we have $\left|ABCD\right|\leq\frac14\left(\left(AB+CD\right)^{2}+BC\cdot DA\right)$.

Obviously, reflection in the perpendicular bisector of the segment AC maps the point A to the point C and vice versa. Let this reflection map the point B to a point E. Then, since reflections are congruence transformations, the triangles ACB and CAE are congruent, so they have the same area: |CAE| = |ACB|, and corresponding sides are equal, so that AB = CE and BC = EA.

Now, by Lemma 1 [b](a)[/b], applied to the triangle ECD, we have $\left(CE+CD\right)^{2}-\frac34\cdot ED^{2}\geq 4\cdot\left|ECD\right|$, and by Lemma 1 [b](b)[/b], applied to the triangle EAD, we have $EA\cdot DA+\frac34\cdot ED^{2}\geq 4\cdot\left|EAD\right|$. Thus,

$\left(AB+CD\right)^{2}+BC\cdot DA=\left(CE+CD\right)^{2}+EA\cdot DA$
$=\left(\left(CE+CD\right)^{2}-\frac34\cdot ED^{2}\right)+\left(EA\cdot DA+\frac34\cdot ED^{2}\right)$
$\geq 4\cdot\left|ECD\right|+4\cdot\left|EAD\right|=4\cdot\left(\left|ECD\right|+\left|EAD\right|\right)=4\cdot\left|AECD\right|$
$=4\cdot\left(\left|CAE\right|+\left|ACD\right|\right)=4\cdot\left(\left|ACB\right|+\left|ACD\right|\right)=4\cdot\left|ABCD\right|$,

so that $\left|ABCD\right|\leq\frac14\left(\left(AB+CD\right)^{2}+BC\cdot DA\right)$, and the problem is solved.

  darij

[quote="mecrazywong"]The area of a convex quadrilateral can be expressed as\[ S=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\frac{B+D}{2}}=\frac{1}{4}\sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}  \]where $p,q$ are the lengths of the two diagonals.[/quote]

How to prove it? (I have really no idea  :D )

[quote="Kalimdor"]I think apply [b]Ptolemy[/b] is a good idea here, but is this theory suitable for all quadralateral?
And what if we don't have the condition "convex", will the inequality still holds? :huh:[/quote]
But the quadrilateral may not be cyclic ;)
Here is the generalized one, $AC^2\cdot BD^2=AB^2\cdot CD^2+BC^2\cdot DA^2-2AB\cdot BC\cdot CD\cdot DA\cos(B+D)$ :)

What about the area formula? Can anyone show?


ceiling function
irrational number
absolute value
number theory proposed
number theory

assume that the interval $ [n\phi\minus{}\frac{1}{n}, n\phi\plus{}\frac{1}{n}]$ where $ \phi\equal{}(1\plus{}\sqrt{5})/2$,
contains an integer. Prove that $ n$ is a Fibonacci number.

The condition $ s\in [n\phi\minus{}\frac{1}{n},n\phi\plus{}\frac{1}{n}]$ can be rewritten

$ |n\phi \minus{} s|<\frac{1}{n}$ or

$ |\phi \minus{} \frac{s}{n}|<\frac{1}{n^2}$.

Now if $ \tau$ is any positive irrational number, the condition

$ |\tau \minus{} \frac{s}{n}|<\frac{1}{2n^2}$

does imply that $ \frac{s}{n}$ is a convergent to $ \tau$, and that will be enough, but I don't see how to overcome that factor of 2 in the denominator. Maybe we can use the fact that successive convergents to an irrational number are alternately > and < that number?

My incomplete thought was to use the fact that $ n$ can be represented (uniquely) as the sum of non-adjacent Fibonacci numbers.  For example, suppose that $ n$ is $ F_{10} \plus{} F_8 \plus{} F_5 \plus{} F_2$.  Then we can compare $ n \phi$ to $ F_{11} \plus{} F_9 \plus{} F_6 \plus{} F_3$.  The error $ F_{k \plus{} 1} \minus{} \phi F_k$ is a decreasing (in absolute value) geometric series.  So it is biggest for the final term in the representation ($ k \equal{} 2$ in my example), and we can hopefully show that the sum of the error terms is sufficiently smaller than 1 that we don't have to worry about wraparound to the next integer.

Just a thought...

It may be hard to expand Larry's idea, however it would be very elegant proof.

using Ravi's idea:

Let $ n \equal{} f_{i_N} \plus{} f_{i_{N \minus{} 1}} \plus{} \dots \plus{} f_{i_1}$ - Zeckendorf representation of $ n$.

I can prove that $ 1 \plus{} f_{i_N \plus{} 1} \plus{} f_{i_{N \minus{} 1} \plus{} 1} \plus{} \dots \plus{} f_{i_1 \plus{} 1}$ is the least integer bigger then $ n\phi$, that is $ \lceil n\phi\rceil$.

but what now ?


Cocamide DEA is cocamide diethanolamine. It is used to thicken the shampoo, a body wash, or a facial cleanser and give it a nice goopy consistency.

It is made by reacting fatty acids in coconut oils with diethanolamine. It is a viscous liquid and is used as a foaming agent in bath products like shampoos and hand soaps, and in cosmetics as an emulsifying agent.

Product manufacturers believe that the thicker is the product, the more appealing it seems to the customers. Maybe they think the product is more “rich” or “nutritious” or “natural”. But there’s nothing natural about [url=http://www.useful-chemicals.com/?p=39]cocamide DEA[/url].

DEA and its variants are suspected of increasing the risk of cancer. DEA can combine with amines present in cosmetic formulations to form nitrosamines (N-nitrosodiethanolamine), which are known to be highly carcinogenic.

Is there a point to this thread?  You didn't ask any questions or express an opinion.  Do you want people to discuss this?  If so, it's better off in the Round Table.

Obviously, a possible question is:

Provide a synthesis of Cocamide DEA (look structure [url=http://en.wikipedia.org/wiki/Cocamide_DEA]here[/url]).


Please try the question:

A set of numbers is called [i]sum-free set[/i] if no two of them add up to a member of the same set and if no member of the set is double another member. How big could be a sum free subset of {1,2,3,...,2n+1)?

Nice problem (probably very well known).

Taking $1,3, ..., 2n+1$ we see that the number is $\geq n+1$

Let's we take a subset with (strictly) more then $n+1$ elements : $a_1 < ... < a_k$
The difference sequenc $a_k-a_{k-1}, ..., a_k-a_1$ takes at least $n+1$ different values, which, by pigeon-hole, can't be all different from the $a_1, .., a_{k-1}$.

So max is $n+1$.


If $ a,b,c   \ge 0$  then prove:

$ \frac{1\plus{}a}{\sqrt{1\plus{}a\plus{}a^2}}\plus{} \frac{1\plus{}b}{\sqrt{1\plus{}b\plus{}b^2}}\plus{} \frac{1\plus{}c}{\sqrt{1\plus{}c\plus{}c^2}} \le 2\sqrt 3$.

It's easy to show that for all $ x$ we have (*) $ \frac {1 \plus{} x}{\sqrt {1 \plus{} x \plus{} x^2}}\leq\frac {2\sqrt {3}}{3}$, there only sum the inequality (*) for $ a,b,c$, and done.


Kim's flight took off from Newark at 10:34 AM and landed in Miami at 1:18 PM. Both cities are in the same time zone. If her flight took $ h$ hours and $ m$ minutes, with $ 0<m<60$, what is $ h\plus{}m$?

$ \textbf{(A)}\ 46 \qquad
\textbf{(B)}\ 47 \qquad
\textbf{(C)}\ 50 \qquad
\textbf{(D)}\ 53 \qquad
\textbf{(E)}\ 54$

[hide=Solution]Comparing $12:00-10:34$ is $1$ hour and $26$ minutes. Comparing $1:18-12:00$ is $1$ hour and $18$ minutes. Adding, we get $2$ hours and $44$ minutes, so $2+44=\boxed{46}.$

We can rewrite 1:18 as 13:18 (military time). So the time would be 3 hours minus 16 minutes, which is 2 hours and 44 minutes.

Thus, the answer is $\boxed{(A)}$.


[quote="Tiks"]9. Let $ABC$ is a triangle and $A'\in{[BC]}$,$B'\in{[AC]}$ and $C'\in{[AB]}$.
Prove that \[ S_{ABC}S_{A'B'C'}^2\geq 4S_{AB'C'}S_{A'BC'}S_{A'B'C}  \]
and equality holds only when the lines $AA',BB'$ and $CC'$ intersecting in the same point. $\clubsuit$[/quote]
Hey guys,why have you forgotten to solve this easy problem?

[quote="Tiks"]Hey guys,why have you forgotten to solve this easy problem?[/quote]

I guess geometry people don't like this kind of problems. :-)

Let $x=\frac{BA'}{A'C}, y= \frac{CB'}{B'A}, z=\frac{AC'}{C'B}$, Then

$\frac{S_{AB'C'}}{S_{ABC}} = \frac{z}{(1+z)(1+y)}$,

$\frac{S_{A'BC'}}{S_{ABC}} = \frac{x}{(1+z)(1+x)}$,

$\frac{S_{A'B'C}}{S_{ABC}} = \frac{y}{(1+x)(1+y)}$,

With a little calculation, we have

$\frac{S_{A'B'C'}}{S_{ABC}} = 1- \frac{S_{AB'C'}}{S_{ABC}} -\frac{S_{A'BC'}}{S_{ABC}} -\frac{S_{A'B'C}}{S_{ABC}} =\frac{1+xyz}{(1+z)(1+y)(1+x)}$

Therefore, the original inequality is equivalent to:

$(1+xyz)^2 \ge 4xyz$ or $(1-xyz)^2\ge0$.

The equality holds iff $xyz=1$ iff three lines concur at the same point by Ceva.

[quote="fuzzylogic"][quote="Tiks"]Hey guys,why have you forgotten to solve this easy problem?[/quote]

I guess geometry people don't like this kind of problems. :-)

[/quote]
Maybe you are right,'cous that is really strange that this easy problem was unsoved so long time :(  :) .
Thanks for solution,my solution is same with yours ;) .

[quote="No Reason"]Sorry for a stupid question,but are there a "Geometry Olympiad" in Russia?Can I have more information about that?[/quote]
This section is irrelevant place to ask such things.
 Yes, there is, so called Шарыгин(Sharygin) Geometrical Olympiad, it is held every year in the end of january(approximately), and if I am not mistaken, everyone can participate in it, the whole thing you should do is to send them letter with solutions.
  Several Past Olympiads are available here:

http://www.alleng.ru/d/math/math101.htm

   Click on the Скачать button.

[quote="April"]You also can find more information, also problems from some previous Olympiads here: http://geometry.ru/olimp.htm .

@Erken: I see that you are very good at Russian, so could you please translate and post some nice problems from this Competition on MathLinks? Thanks.[/quote]
Sure.

if I am not mistaken DoctorA has already posted some problems...so I'll translate and post what's left.


Let $ a \in (0,1) \cup (1, \infty)$ and $ A\equal{} a^{x\plus{}y} \plus{} a^{y\plus{}z}\plus{}a^{z\plus{}x}$ , $ \forall x,y,z \in \mathbb{R}$ . Find $ x,y,z$ if :

$ \frac{a^x}{\sqrt{A \plus{} a^{2x}}} \plus{} \frac{a^y}{\sqrt{A \plus{} a^{2y}}} \plus{} \frac{a^z}{\sqrt{A \plus{} a^{2z}}} \equal{} \frac{3}{2}$ .


Consider all 6 digit numbers from 000000 to 999999. A number is good if (1) none of its digits are 9 (2) the sum of the first 3 digits = sum of last three digits.
(i) show the sum of all good numbers is divisible by 2002
(ii) how many good numbers are there?

(i) because of (2) the sum will be a multiple of 1001, and the sum of all 3 digit numbers not containing 9 is even.

(ii) well you only need to make a 3-digit number that doesn't contain 9, so 8.9.9 options.

(ii) is incorrect ...
bcoz for each digit sum k, what we are actually doing is pairing up 2 three digit numbers with digit sum k ... so there are a *lot* more good numbers ...

Maybe I'm just misreading, but is a "lucky" number the same as a "good" number? I can't see the definition of "lucky" anywhere in the text of the problem (only the definition of "good").

oh now I see :D yes that's a little mistake of mine :)

Besides, "all 6 digit numbers from 000000 to 999999" --> this is really the first time I see 000000 considered a 6-digit integer :P anyway it simplifies stuff a lot here so I don't mind. ;-)

Ok now my second attempt:

we build up numbers with sum n like this: a permutation of 2 "|" and n "." where "."means +1 and "|" means digit separation. We act like in a 9-digit system instead of a 10-digit.

For n=<8 we got the simple combi giving us the amount of numbers.
For n>=16 we have the same (n vs. 24-n) since we can say the #. = 8- the digit value.
Now for n=9/15 we have 52; 10/14: 57; 11/13: 60; 12: 61

Counting all together now, we get $32661$ good numbers.


number theory
least common multiple
induction
number theory unsolved

$ T(n)$ is defined by $ T(1) \equal{} 2$,and for $ T(n) \equal{} 2^{T(n \minus{} 1)}$ . Prove that  $ T(n) \minus{} T(n \minus{} 1)$ is divisible by $ n!$  for $ n\geq 2$  .

Let $ c(n) \equal{} lcm_{p^k|n}( p^{k \minus{} 1}(p \minus{} 1))|\phi(n).$
If $ c(M_n)|M_{n \minus{} 1}|T(n \minus{} 1) \minus{} T(n \minus{} 2)$, then $ M_n|T(n) \minus{} T(n \minus{} 1)$.
Obviosly $ 2! \equal{} 2|T(2) \minus{} T(1)$ and $ c(n!)|(n \minus{} 1)!$, therefore, by induction $ n!|T(n) \minus{} T(n \minus{} 1) \equal{} T(n \minus{} 1)(2^{T(n \minus{} 1) \minus{} T(n \minus{} 2)} \minus{} 1).$


Should be USAMTS although I've only seen 5 problems for USAMTS and none for USAMO so I'm not sure (the reason why I didn't vote)...moved it to "Other Contests & Programs"

definitely usamo. i have looked at a number of recent usamo problems of which i can only solve a small number. on the other hand, ive had solutions to all of this year's usamts problems. of course, i have lost points for stupidity. still. usamo is much more difficult.

Well, many of the USAMTS questions would be much harder if we had to do them within a specified time limit without using other resources, and the graders wanted rigorous proofs like they do in USAMO. On the other hand, USAMO problems would be many times easier if it was graded under USAMTS guidelines, so I'm not sure the two are comparable.

The USAMTS questions are unquestionably easier.  (Sorry, Jen :) )  I can look at the USAMTS and immediately get 1 of the questions without writing anything down.  If I devoted myself slavishly (as one does on the USAMO), I could have full solutions to the set of 5 easily within the time constraints of the USAMO, with occasional exceptions for questions that are best solved with massive case-examination or computers.

I've only had a "oh, this isn't hard" experience on one or maybe two USAMO questions in competition.

USAMTS is much easier.  You can use calculators, computers, books, etc.; there is virtually no time limit; the proofs don't have to be all that rigorous; and the problems are just plain easier.

I think more of a question would be something like if you were given USAMTS under USAMO conditions which one would u do better on.

...and for that I still say USAMO is much harder.


function
number theory
least common multiple

Is there a way to find values for the independent variable to make the dependent variable an integer, if one is given a function?

For example, if I were to say, find all x values such that $f(x)$ is an integer, if $f(x)=(123/456)x+789/12345$, could you tell me what values?

To clarify, I mean is there a quick way to do so.

[i]Obviously there are an infinite amount, but is there a quick way to find them within a certain range?[/i]

well for that one you should probably clear the denominators first
and then it should be easy

But if you multiply both sides by LCM to clear the denominators (tell me if this is not what you meant), then what are you going to write on the right hand side?


[hide] Say that x pounds of cement added will make the bag have 40% cement.  Since 25% of a 20 pound bag has cement, we have that 4 pounds is cement.  This implies that $\frac{4+x}{20+x}=\frac{40}{100}$ because when you add x pounds of cement, you are also adding x to the total (that is where the 20+x comes from) as well as the cement (that is where the 4+x) comes from.  Solving for x, we have that $x=\frac{20}{3}$, which is our answer. [/hide]

I would recommend using systems of equations to solve mixture problems.

[hide="hint"]Let $x$ be the number of pounds of pure cement that is added.  What equation(s) can you write?[/hide]

[hide="bigger hint"]There are 4 pounds of cement in the mixture at the start.  When pure cement is added, there will be $4+x$ poundso of cement.  There will be $20+x$ total pounds of cement.  Now what equation can you write?[/hide]

b-flat,

I have a hard time setting up equations from the given data in word problems.

[hide]We know that the ratio of the amount of cement to the total amount of mixture is $\frac{4+x}{20+x}$.  If this is $40\%$, then it will be equal to $\frac25$.  Thus, the equation is $\frac{5+x}{20+x}=\frac25$[/hide]

[quote="Interval"]A 20 pound bag of cement mix has 25% cement and 75% sand. How much pure cement must be added to produce a cement mix that is 40% cement?

Another mixture question in my state exam study booklet.

What is the table or chart that will lead to the right answer?

What's the answer?

What is the best way to set up mixture tables and/or chart?[/quote]
[hide]There are currently 5 pound of cement in the bag.

If you add $x$ pounds of cement to the bag, the persentage of cement in the bag will be $\frac{5+x}{20+x}$.

$\frac{5+x}{20+x}=\frac{40}{100}$

$x=\boxed{5\text{ pounds}}$[/hide]

There are two fractions given.

One is:

(4 + x)/(20 + x)

and the other given is:

(5 + x)/(20 + x)

Which is it?

$25\%$ of 20 is $\frac14\cdot20=5$, so the correct fraction is $\frac{5+x}{20+x}$.  Sorry about that (will edit).

:blush: Oh!  I made the same mistake!  It should be (5+x)/(20+x)=2/5, as b-flat and i_like_pie said.


algebra
polynomial
vector
combinatorics unsolved
combinatorics

Let $ p$ be a prime and $ V \equal{} {1,2,...,4p}$. Let $ F$ be a family of subsets of $ V$ such that $ |A| \equal{} 2p$ for all $ A\in F$ and $ |A\cap B|\neq p$ for all $ A, B\in F$. Prove that
\[ |F|\leq 2 {4p \minus{} 1 \choose p \minus{} 1}.\]
[hide]I know it can be proved using multilinear polynomials $ f_A^*$ where $ f_A \equal{} (\sum_{i\in A}x_i)^{p \minus{} 1} \minus{} 1$ (where $ x \equal{} (x_1, x_2,...,x_{4p})$ is incidence vector) in $ Z_p$. I know, that they are linearly independent (because $ f_A^*(B)\neq0\iff B \equal{} A$) but I don't see how to get the RHS in the inequality...[/hide]

I was wondering, what significance would $ p$ being a prime makes here as I am not able to see how its primality can be used. I hope somebody helps me with this gap here.


modular arithmetic
Euler
algebra
polynomial

let integer n >=2. If for all integer k, satisfying 0 <= k <= sqrt. (n/3). k, k^2+ k +n  are all prime numbers. Prove: for all integer k, satifying 0<= k <= n-2. k, k^2 +k +n are all prime numbers.

Suppose $ k^2 \plus{} k \plus{} n$ is not prime for some $ 0\leq k\leq n \minus{} 2$. Then the least nonnegative $ j$ for which $ j^2 \plus{} j \plus{} n$ is not prime is at most $ n \minus{} 2$.

$ j^2\plus{}j\plus{}n$ cannot equal 1 since $ j^2\plus{}j\plus{}n\equal{}\left( j\plus{}\frac{1}{2}\right)^2\plus{}n\minus{}\frac{1}{4}\geq \frac{7}{4}$, so $ j^2\plus{}j\plus{}n$ is composite. Let $ p$ be the smallest prime divisor of $ j^2 \plus{} j \plus{} n$. Then $ j^2 \plus{} j \plus{} n\geq p^2$, but also
\[ 0\equiv j^2 \plus{} j \plus{} n\equiv j(j \plus{} 1) \plus{} n\equiv (p \minus{} j)(p \minus{} (j \plus{} 1)) \plus{} n
\]

\[ \equiv (p \minus{} 1 \minus{} j)^2 \plus{} (p \minus{} 1 \minus{} j) \plus{} n\pmod{p}
\]
So $ p$ divides $ (p \minus{} 1 \minus{} j)^2 \plus{} (p \minus{} 1 \minus{} j) \plus{} n \equal{} (p \minus{} 1 \minus{} j)^2 \plus{} (n \minus{} 1 \minus{} j) \plus{} p\geq p \plus{} 1$, so also $ (p \minus{} 1 \minus{} j)^2 \plus{} (p \minus{} 1 \minus{} j) \plus{} n$ is not prime.

Note also that the above expression is equal to $ (j \minus{} p)^2 \plus{} (j \minus{} p) \plus{} n$. As $ j \minus{} p < j$, $ j \minus{} p$ must be negative, so $ j\leq p \minus{} 1$ making $ p \minus{} 1 \minus{} j$ nonnegative.

Therefore $ p \minus{} 1 \minus{} j\geq j\Rightarrow j\leq \frac {p \minus{} 1}{2}$, giving
\[ p^2\leq j^2 \plus{} j \plus{} n\leq \frac {p^2 \minus{} 1}{4} \plus{} n\Rightarrow p\leq \sqrt {\frac {4n \minus{} 1}{3}}
\]

\[ \Rightarrow j\leq \frac {\sqrt {\frac {4n \minus{} 1}{3}} \minus{} 1}{2} < \sqrt {\frac {n}{3}}
\]
So $ 0\leq j\leq \sqrt {\frac {n}{3}}$ and $ j^2 \plus{} j \plus{} n$ is not prime. So if $ k^2 \plus{} k \plus{} n$ are all prime for $ 0\leq k\leq \sqrt {\frac {n}{3}}$, then they are all prime for $ 0\leq k\leq n \minus{} 2$, as desired.

This is Euler polynomial problem. You can find the answer in many numbertheory books.


How many positive integers from 1 to 1000 are there that contains a 7.

EDIT: Well, I think the three forms of the numbers will be $xx7,7xx,x7x$ and it doesn't matter which digit is for the $x$ so for each one, you'll have the combinations $(1)(10)(9)$ times 3 so you'll have $270$?

[hide]first, i would find all the numbers between 1 and 100 that contained a seven: they include 7, 17, 27,...97 + 70, 71,...,79. that's two sets of 10 with 1 overlap so 19 numbers between 1 and 100 have a 7. there are also 19 numbers in the 100's, 200's...900's. except don't include the 700's in this. so there are 19x9=171 sevens so far. then you add all the numbers between 700-799. that is another 100. thus the toal is [b]271[/b][/hide]

[quote="Iversonfan2005"][quote="yif man12"][quote="Iversonfan2005"]$101$? Since there are $10$ numbers in the first $100$ and those digits keep repeating, so $10(10)=100$ plus the digit $7$ in $700$ so $100+1=101$. I think :?[/quote]

Think about the question a bit more. Is 700 the only 3 digit number that starts with 7? How many more are there?[/quote]

I did, I got $270$, don't know if it's right though... :?[/quote]
270 is wrong. Very close. furious got answer by long way.
1000 numbers, take out all numbers without number 7 in it. So there 9x9x9=729 numbers without number 7 in it. then 1000-729=271

But you cannot do the same to the following problem: How many positive integers from 1 to 1000 are there that contains a 0?

[hide]i think its $182$
because we first consider the numbers from 1 to 99, where there are $9$ numbers
and then we consider the numbers from 100 to 999, and do the same only that we know that there wont be a zero in th first place so 900-9*9*9=$171$
and the thousand is an additional $1$
so we add up to $182$

but we can solve it also diffrently.
coming from the solution for seven, considering all numbers where the numbers counted for seven but not fro zero: when zero is in the first place.
thats the case from 1 to 99
but in 9 cases the zero is counted anyways (10,20,...,90)
and also the thousand is counting extra
so $271-99+9+1=\boxed{182}$[/hide]

One digit numbers: there is [b]one[/b] that contains a 7.

Two digit numbers: there are [b]ten[/b] that have 7 as a first digit, and [b]nine[/b] that have it as a last digit (don't count 77 twice.)

Three digit numbers: there are [b]100[/b] that have 7 as a first digit, [b]100[/b] that have 7 as a middle, and [b]100[/b] that have 7 as a last. But we counted [b]-60[/b] of those twice. Then we add [b]2[/b] more, because we [i]dis[/i]counted 777 twice.

That brings the total to 1 + 10 + 9 + 100 + 100 + 100 - 60 + 1 = [b]271[/b]

For the number of them which have 0...
First we find the numbers which end in 0: 100 of them.
Now find the numbers which have 0 as the second digit and neither one of the other two is 0.  9*9=81 of those.
So we get 181...

well, if u would add $1$ for the thousand, u'd get the same as i have $ \rightarrow $ $182$

maybe we can solve the problem for $1$ to $10^n;\;n\in\mathbb N$ and the digets a) $7$ b) $0$

oops, i cant add: $9+171+1=182$ haha
of course its $181$

and for my second solution, i should substract $100$ instead of $99$ because for 7 its from 700 till 799. or do i still think wrong? i guess im going to write all the numbers down which contain a seven and then a zero ;)

All numbers from 700-799 include at least one 7, giving us 100. Each of the other nine hundreds has a set of ten in it, when the tens digit is 7, giving a total of 190 so far, and each of the nine hundreds also has 9 other tens that each have one 7, giving an additional 81, so the total is $271$.

I don't know where you guys are getting 181 from...

$181$ is the answer to the alternate problem posted by 10000th User, not the original problem.

[hide="Solution"]
1 Digit: $1$ which is 7
2 Digits: $8+10$ which are 17, 27, 37... not including 77; then adding the 7x's
3 Digits: $10(8)+10(8)+100$ which are 170, 171, 270... not including 770's; then 107, 117, 207... not including 7x7's; then the 700's

Subtract: $8$ which are 177, 277, 377... that have been counted twice

Total: $271$
[/hide]

For the problem by 10000thUser, here's how I did it, between $1$ and $100$, you have $10$ numbers with $0, (10,20,30...90,00)$ Now times that by 10 and you get $100$. Now you have also numbers from $200$ onwards such as $201,202....208,209$ etc, so multiply those $9$ by $9$ to get $81$. So in total you have $181$.

The answer for number 1 is $271$ and number 2(the zero problem) is $181$. I don't know why people are getting $182$? And hexaditom, you do realize that there are two problems, right? ;)

following Hexaditidom's madness, let's list all numbers from 1 to 1000 containing 0

those that ends in zero
010 020 030 040 050 060 070 080 090 100
110 120 130 140 150 160 170 180 190 200
210 220 230 240 250 260 270 280 290 300
310 320 and so forth ...
410
510
610
710
810
910 920 930 940 950 960 970 980 990 1000

this is a block of 10x10

what about those that have middle digit zero except divisible by 100?
101 102 103 104 105 106 107 108 109
201 202 203 and so forth ...
301
401
501
601
701
801
901 902 903 904 905 906 907 908 909

this is a block of 9x9

total is $10^2+9^2=\boxed{181}$

seriously, this problem can easily be solved with simple counting:

there are 1000 numbers from 000 to 999 (same thing because we just add 000 and take out 1000, nothing is lost and nothing is else changed) and yet $9\cdot 9\cdot 9=729$ of these 3-digit numbers do not contain the digit 7 (considering 020 as a 3-digit number). so answer is $1000-729=\boxed{271}$

same thing with zeroes, i dunno why my cousin said it doesn't work but it does work for zeroes in a slight change. there are 271 numbers containing digit 0. now we must subtract those that has a zero as first digit and no zero in the third digit. there are $9\cdot 10$ of those numbers. so answer is $271-90=\boxed{181}$

and HOPEFULLY, this stops your doubts. if you still do have doubts, i recommend you write out all the numbers by brute force.

[quote="Palytoxin"]and HOPEFULLY, this stops your doubts. if you still do have doubts, i recommend you write out all the numbers by brute force.[/quote]

Well, hopefully you don't have to use brute force to write.

[quote="Pork_Chop8"]Well, hopefully you don't have to use brute force to write.[/quote]
hah, it really did for my last post since so many aren't convinced which is the right answer, like lost in the sea :)


Prove that for every $n \in \mathbb{N}$ the following proposition holds: $7|3^n +n^3$ if and only if $7|3^{n} n^3 +1$.

[quote="Peter"]Prove that for every $ n\in\mathbb{N}$ the following proposition holds: $ 7|3^{n}\plus{}n^{3}$ if and only if $ 7|3^{n}n^{3}\plus{}1$.[/quote]

One may generalize the result slightly.

Proposition. Let $ p$ be an odd prime. Suppose that $ m$ and $ n$ are positive integers with $ p\not\vert m$. Then, $ p\;\vert\; n^{\frac{p\minus{}1}{2}}\plus{}m$ if and only if $ p\;\vert\; n^{\frac{p\minus{}1}{2}}m\plus{}1$.

The proof is almost same with the above solution. One may also check that

Propostion. Suppose that $ a^{2}\equiv 1\; (mod\; n)$. Then, $ n\;\vert\; a\plus{}b$ if and only if $ n\;\vert\; ab\plus{}1$.

Outline of Proof. Since $ a^{2}\equiv 1\; (mod\; n)$, it is clear that $ gcd(a, n) \equal{} 1$. Now, use the identity $ ab\plus{}1 \equal{} a(a\plus{}b)\minus{}(a^{2}\minus{}1)$.

Simply use that $n^3\equiv \pm 1 (mod~7)$. This works for the generalisation as well.


Which one is used most popular in the university system of US?

Each of them is widely used, for different purposes.  My university makes Maple, Mathematica and Matlab available to its students, as well as SPSS and Stata (statistical software) but not MathCAD.

you mean MATLAB right? or am i in a dream world...

Well I use Maxima; it's not as pretty, but it's easier imo, simpler, free, opensource, powerful, ...

[quote="jarod"]Is it a newest version? In your personal opinion, which one is strongest and easiest to use?[/quote]

Actually, I was wrong; I am using version 5.1, which I downloaded maybe a year or two ago from my university, and for which I need to get a new password at the beginning of every academic year.  I did not bother to download the newer version.

I have no experience with the other software you named, so I'm not a good person to ask.  Which is best probably depends a great deal on what sort of tasks you want to perform.

Maple and Mathematica essentially try to do the same thing.  Matlab is aimed more at highly numerical applied math (the Mat in Matlab is for matrices, not mathematics) computations.  I think Mathcad is more for engineering-ish stuff but I am not entirely sure.  It is best to become proficient in one of Maple or Mathematica, and in Matlab.

I've used Mathcad 8 for some time (it really helps you for some non-elementary stuff), but then I got Mathematica 5.2 and deinstalled Mathcad... I mean, Mathematica is great :D !!!

No idea about US, but in India, Mathematica and MATLAB are both extensively popular.

I have mathematica and it guided me to the answer on a midterm (from AoPS) once... :D


If i am given 2 roots. From these i need to get the equation of the quadratic function. The roots are 3 + root 2 and 3 - root 2. Can some one explain to me how to go about these problems

for quadratic equation: $ax^2+bx+c=0$, we have:
$x_1+x_2=-\frac{b}{a}$ and $x_1x_2=\frac{c}{a}$.


Let a triangle ABC. The incircle (I) touches BC,CA,AB at D,E,F. The bisector of angle EDF meets EF at K. BK meets AC at P, CK meets AB at Q. Prove that AP=AQ and PQ is tangent to the circle (I).

Probably it is well known as a direct application of the [b][size=100]Newton theorem[/size][/b], which states that in every circumscribed quadrilateral, the lines connecting the tangency points of the incircle to its opposide side-segments, pass through the intersection points of its diagonals.

Kostas Vittas.

Oh yes, sorry, it is really well know  :P . My purpose is to aim another problem and this problem is a part as a lemma  :)

Can you show to everyone other problems you said?
Thank you.


Before taking his last test in a class, the arithmetic mean of
Brian's  test scores is 91. He has determined that if
he scores 98 on his last test, the arithmetic mean
of all his test scores will be exactly 92. How
many tests, including the last test, does Brian
take for this class?

We simply set up an equation, where x is the number of tests he has taken (including the last one). $ 91(x\minus{}1)\plus{}98\equal{}92(x);91x\plus{}7\equal{}92x;x\equal{}\boxed{7}$.


geometry AMC USA(J)MO USAMO trigonometry AIME HMMT

well its that time of year guys and gals. How do you think you will do and how have you done in the past. please specify which contest you are taking this year.

for me:

7th grade: Pre-Algebra 39/40 (1st)
8th grade: taking Geometry, i think ill get either 31/32, or 30/32 (due to stupid mistakes) , i make alot of bad mistakes so probly lower.
9th grade: comprehensive.

whoa cognos, that's like way better than I did..

I don't remember my scores but they were nottttt good :(

I'll take the comprehensive one again when it's posted on the site, but estimating maybe 3 wrong. or two.

Lol I got a perfect score last year...

I didnt even sign up this year...

dont care much about it (umm actually I would probably end up going if I dindt have USAMO and 3 AP tests the week/2 weeks after)...but i wonder if i can just show up and take the comprehensive test if Im not busy or something...(didnt sign up for my school)

hmm but I dont like these kinds of tests because they are calculator based and many of the problems require no skill, for example, if im giving a trig problem that might require some factoring/algebraic manipulations, now i can just graph it and find the points...which totally makes it trivial.

same with problems like forexample, that AIME problem find tan(1/3)+tan(1/4)...blah blah, you can now just plug in the 5 answer choices and see which is right...
or like a problem like find x^(2z)+y^(2z)  (these problems are quite famous, seen them in AMCs and HMMTs), (sub in any x and y and z) largest  prime factor instead of factoring it like (x^z+y^z)^2 -2(xy)^z or something and manipluating it, its called plug and chug

plus most of the graph problems that tell you to find some area or something can easily be done on a calc with integrals or something...

and same with geometry problems, a problem that may require some thinking (like laws of cosines and then plug that in to laws of sines), like this old ARML probelm like the find the area  of an isoscles triangle whos sides are sqrt(sinx) ,sqrt(sinx) ,and sin(x) while the vertex angle is x....of course the real way to do it is to find sin x using laws of cosines ro half angle, but the way to do it on this test is to go backwards, look at the choices and see which sinx fits the eqution such that vertex angle eis also x (its called using sin-1 with a calc)

but idk, maybe they want it that way because its more inteneded for people that pay attention in their math class than for people that are actually good and go outside of the box...

Hmm i guess the only "tests" outside of school that I do now are probably AMC/AIME/USAMO, ARML, Rose-Hulman, and if im lucky, USNCO or USAPHO...and then of course SAT/SATII/ACT/PSAT and AP tests...

Ive given up on the "smaller tests"

does anyone know if its too late to sign up? i'm currently not in a math class and i didn't hear about it until a couple days ago...........

and do you have to be take the test that corresponds with the math class you're in? i'm thinking about taking comprehensive, but my school might not let me. if theres no rule that says you can't take a test that is higher than your math class, then i may fight for it.

lol just like you fought for the mathcounts thing in sixth grade :wink:

hmm unfortunently i dont think you can sign up...

apparently your school had to put your name down on a list and pay money or something like that...

oh well i probably cant go anyways so it doesnt matter for me...

if you really want, dont talk to your school, talk to the ICTM people about it...

so what did you ppl get???? actually you know what, wait til 4/27 to talk. and does anyone know any good ppl who did geometry. thank god lyndon and youkow didn't do geo. this year.


¿Y si acá ponemos los equipos de la IMO?

En 1 semana va a ser la selección argentina, cuando estén los resultados los comunico..........

Equipo Argentino:

Julián  EISENSCHLOS, Escuela Técnica ORT N° 1 – Ciudad de Buenos Aires

Ignacio ROSSI, Instituto  San Vicente de Paul – Escobar – Bs. As.

Roberto MORALES, Colegio  Nacional “Rafael  Hernández” – La Plata – Bs. As.

Rodrigo LUGONES, Colegio  Nacional de Buenos Aires - Ciudad de Buenos Aires

Fernando Martín VIDAL,  E. E. M. P. I. N° 3004 – Villa Cañás – Santa Fe

Pablo Nicolás ZIMMERMANN, Inst. Politécnico Superior – Rosario – Santa Fe

Felicitaciones y Suerte!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[color=blue]Muy buena idea José........

Haber si vamos poniendo aqui la informacion
de los procesos de Seleccion de los equipos para la IMO,
tanto examenes, pre-clasificados, clasificados,
informacion sobre las evaluaciones en cada pais
y toda la informacion de estos procesos selectivos, anecdotas, etc...

Aki la info del selectivo de PERU :
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=78141]HERE[/url]

CARLOS  BRAVO:)
LIMA - PERU[/color]

Buena informacion Marko,
podes conseguir los TST en Mexico???

saludos . ..

Carlos bravo ;)
Lima -PERU

[quote="marko avila"]equipo  mexicano

1.- manuel angel  guevara guevara
2.- joshua
3.- pablo soberón
4.- isaac  buenrostro
5.- marco avila (yoooo)
6.-aldo pachiano

  :)  :P[/quote]

Felicitaciones!!!!!! En un posteo del 2005 pusiste que estabas entrenando duro, me acuerdo de eso..... se te dio!!! a disfrutarlo!!!!

[quote="js131090"]Equipo peruano:

PER1:Soncco Huarsaya, Chen
PER2:Alva Gutierrez, Jossy (yo!)
PER3:Vargas Pallete, Franco
PER4:Huanuco Candia, Maria
PER5:Melgar Povez, Paolo
PER6:Quillas Paredes, Aldo

Buena suerte para nosotros !!!!!!!!!!!!! :coolspeak:  :lol:[/quote]

SI realmente deseamos exitos a las selecciones Iberoamericanos en este importante evento ;)
....
alguien ha visto el equipo de Brasil???
.....

Saludos,

Carlos Bravo ;)
Lima. - PERU

Revisar .. equipos para IMO 2006..

http://www.mathlinks.ro/Forum/index.php?f=387

SALUDOS ,

CARLOS BRAVO ;)
LIMA .PERU

josè :

   pues  muchas  gracias,  si  la verdad  trabajèmucho y  afortunadamente se me dio!!!  espero  que algun dia nos veamos  por aia en una imo!!!

 carlos :

pues  los  tsts te los  podria teclear  mas adelante el problema es que ahorita ando  medio ocupado.

  saludos a todos  y  suerte  a todos  en la imo!!!

 :)

Hola a todos soy nuevo aqui, apenas dias atras hice mi primer post jeje!
El equipo venezolano de la IMO es
Ven 01 -  Sofía Taylor (Bronce OMCC 05)
Ven 02 - Rafael Guédez (yo :D ) (IBERO 05)
Ven 03 - Andrés Guzmán (Mención OMCC 04; Mención IBERO 04)[/color]

Los 3 estamos ya inscritos en la pag oficial de la IMO sin embargo como las inscripciones son hasta el 30 inclusive es probable se nos una uno mas este sabado que tendremos la olimpiada nacional que seria Ven 04 - Victor Villamizar (IBERO 05) y bueno quien sabe si otro..
Espero y seguro msi compañeros también, verlos pronto allá, suerte a todos!

Bueno pues olas a todos,

solo quiero reiterar mis felicitaciones a cada uno de los chiccos que conforman las selecciones de sus paises,
y desearles noo suerte, sino exitos en este evento.... que es la IMO 2006.

---
Acabo de recibir uan informacion de Soledad Villar (URU):
Seleccion de Uruguay:

URU1: Daniel Wahrmann
URU2: Daniel Komornik
-----

Saludos,

Carlos Brvo ;)
Lima  - PERU

[quote="js131090"]Equipo peruano:

PER1:Soncco Huarsaya, Chen
PER2:Alva Gutierrez, Jossy (yo!)
PER3:Vargas Pallete, Franco
PER4:Huanuco Candia, Maria
PER5:Melgar Povez, Paolo
PER6:Quillas Paredes, Aldo

Buena suerte para nosotros !!!!!!!!!!!!! :coolspeak:  :lol:[/quote]
hola amiga... mis más sinceras felicitaciones para ti, y en general, a los 6 peruanos que defenderán a su país en la IMO de este año...

me gustaría poner algo del equipo chileno... pero  :o ... no vamos a la IMO  :rotfl:

bueno... la sociedad matemática de Chile no nos da el mérito, y se justifican en nuestros pobres resultados en la APMO...

quiero taparles la boca a esos... ojalá vayamos a la del 2007...  :D
la esperanza es lo último que se pierde

saludos

1) ¿Por qué van solamente 2 uruguayos? (pueden ir hasta 6 no???, o es depende del país??)

2) ¿Por qué no van los chilenos?

alguien me explica lo del post anterior?

hola, hasta donde se los chilenos por su actuación en la apmo, no fue muy destacada. y solo van dos uruguayos porque ellos solo envían a todo aquel que luego de los selectivos sume mas de 30 y en la página web de uruguay se ve que sólo ellos superaron eso.

Al fin salio el equipo Colombiano para la IMO!!! Casi no.

1. SANTIAGO CUELLAR, BOGOTA(OCCM 2005 ORO)
2. JOSE ALEJANDRO SAMPER, BOGOTA(OIM 2005, APMO 2006 PLATA)(yo!!)
3. FABIAN PRADA, BUCARAMANGA
4. SANTIAGO CAMACHO, BOGOTA
5. MIGUEL ACOSTA, BOGOTA (OCCM 2004 BRONCE)
6. DIEGO CIFUENTES, BOGOTA

Es un equipo de gente nueva.. nadie que haya ido a IMO antes. Sin embargo 5 de nosotros podemos repetir el proximo año.

felicidades a todos samper, a todos los vi en enero en Bogota, me imagino que van a estar en el entrenamiento de ahora otra vez, de aqui de vzla vamos el equipo IMO y creo que puede que el de la OMCC pero de ese no estoy seguro, sabes algo de ese entrenamiento?, va aser igual que el anterior?

no se como va a ser para uds. Nostros ya empezamos a entrenar desde esta semana y tenemos entrenamiento continuo hasta el dia anterior al viaje. ¿Uds cuanto tiempo se van a queder?? ¿cuando vienen??? No se como repartan los niveles.
En este entrenamiento vienen profesores que estan en enero y el objetivo es entrenar a la gente que va a IMO principalmente(nivel superior tiene un grupo mucho mas selecto de gente). Hay 3 niveles... no se donde los vallan a meter a uds. Supongo que haran una prueba o algo asi. Bueno mucha suerte.

IMO Brasil team 2006:

André Linhares Rodrigues
Guilherme Rodrigues Nogueira de Souza
Leandro Farias Maia
Leonardo Ribeiro de Castro Carvalho
Rafael Mendes de Oliveira
Régis Prado Barbosa

[]´s!

vamos del 14 al 24 de junio, salimos en la noche el miercoles por avianca, nosotros también vamos a entrenar hasta antes de la IMO pero en bogot}a sólo ese tiempo, uds probablemente tengan mejor nivel que nosotros como todos los años jeje, pero bueno somos un equipo de "nuevos" en la IMO pero todos hmos ido a una olimpiada anteriormente, vamos a entrenar todos a ver que tal, suerte a uds tambien, nos vemos... por cierto uds saben algo de la bolivariana, con esa prueba iban a terminar de seleccionar a los de la OMCC y un amigo queria saber, gracias

yo se de los resultados de la bolivariana en colombia (pues es nuestra misma ronda final), pero no se si ya corrigieron todos los paises. en todo caso, no discutan esa prueba todavia... a nosotros nos dijeron que no la discutieramos hasta que nos avisen.

ah no yo se, lo ke yo keria saber era algo con respcto a los cortes o algo asi...no nada de la prueba, aqui igual se prohibe discutirla.

Equipo español: (En principio creo que son estos, ya que son los que tuvieron medalla de oro en la olimpiada nacional)

--Hugo Fernández Hervás  	       Madrid
--Javier de la Nuez González 	        Madrid
--Diego Gimeno Sanz 	                   Valladolid
--Ángel David Martínez Martínez       Castilla-La Mancha
--Xavier Ros Otón 	                      Cataluña
--Marc Viñals Pérez 	                      Cataluña


inequalities
trigonometry
geometry proposed
geometry

Prove that in an acute triangle $ ABC$ there is the relation

$ \boxed {\ \frac {a}{b^2 \plus{} c^2 \minus{} a^2} \plus{} \frac {b}{c^2 \plus{} a^2 \minus{} c^2} \plus{} \frac {c}{a^2 \plus{} b^2 \minus{} c^2}\ \ge\ \frac {(a\plus{}b\plus{}c)^2}{3abc}\ \ge\ \frac {3(a \plus{} b \plus{} c)}{ab \plus{} bc \plus{} ca}\ \ge\ \frac {9}{a \plus{} b \plus{} c}\ \ge\ \frac {3(a \plus{} b \plus{} c)}{a^2 \plus{} ^2 \plus{} c^2}\ }$ .

[quote="Virgil Nicula"]Prove that in an acute triangle $ ABC$ there is the relation

$ \boxed {\ \frac {a}{b^2 + c^2 - a^2} + \frac {b}{c^2 + a^2 - c^2} + \frac {c}{a^2 + b^2 - c^2}\ \ge\ \frac {9}{a + b + c}}\ \ge\ \frac {3(a + b + c)}{a^2 + ^2 + c^2}$ .[/quote]

$ a^2 + b^2 - c^2=2ab \cos \gamma$ , $ b^2 + c^2 - a^2=2bc \cos \alpha$ , $ c^2 + a^2 - b^2=2ca \cos \beta$

By Chebyshev : $ LHS =  \frac {a}{2bc \cos \alpha} + \frac {b}{2ca \cos \beta} + \frac {c}{2ab \cos \gamma}$ $ \ge \frac{1}{18}(a+b+c)(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})(\frac{1}{\cos \alpha}+\frac{1}{\cos \beta}+\frac{1}{\cos \gamma}) \ge$ $ \frac{1}{3}(a+b+c)(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})$

and ${ \frac{1}{3}(a+b+c)(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}) \ge  \frac {3(a + b + c)}{ab + bc + ca} \ge \frac {9}{a + b + c}}\ \ge\ \frac {3(a + b + c)}{a^2 + b^2 + c^2}$


floor function
number theory open
number theory

do you know the proof of the converse of Beatty's theorem

If $\alpha$ and $\beta$ are irrational and $\lfloor{\alpha}\rfloor,\lfloor{2\alpha}\rfloor,\lfloor{3\alpha}\rfloor,\ldots;\lfloor{\beta}\rfloor,\lfloor{2\beta}\rfloor,\lfloor{3\beta}\rfloor,\ldots$  is the set of integers (there is no repeated integer) then $\frac{1}{\alpha}+\frac{1}{\beta}=1$ ?

[b]Converse of Beatty's theorem:[/b] Let $a$ and $b$ be real numbers. Then the sequence $[a],[b],[2a],[2b],\ldots$ contains all positive integers exactly once iff $a$ and $b$ are irrational conjugates.

It's clear that at least one of $a,b$ is irrational, because if $a=p_{1}/q_{1}$ and $b=p_{2}/q_{2}$ we have $(q_{1}p_{2})a=(q_{2}p_{1})b.$ It's also clear that $a>1,b>1.$ Suppose $a$ is irrational and $b'$ is its conjugate. Then by Beatty's theorem, $[a],[b'],[2a],[2b']$ also covers all $\mathbb{N},$ thus $[nb]=[nb']$ for all $n\in\mathbb{N}.$ But if $|b'-b|=\delta>0,$ then for sufficently large $n, n\delta>1,$ thus $[nb]\neq[nb'].$ So $b=b'.$


Pas aaghaayoone enekaas baaz kojaayand? chi shod? :lol:

albateh kheily mazerat mikham fatehi jan...

chon mesle in ke marakezeshon rooye ye khat nistan...

chon hanooz nemidoonim ke aya in se dayere noghteye moshtaraki darand ya na...

[quote="ashegh"]hanooz nemidoonim ke aya in se dayere noghteye moshtaraki darand ya na...[/quote]
mage har setaa az vasate $OH$ nemigzarand? :huh:

[quote="Ramtin Madani"]saabet konid  agar $H$ markaze ertefaai va $O$ markaze daayereye mohiti $ABC$ baashand, daayerehaaye noh noghteye 3 mosalase $AOH, BOH, COH$ ham mehvarand.[/quote]

Ramtin jan , enekas bezan be markaze vasate $AH,... .$  ;)

[quote="mohamadhosein"]salam
enekas nesbat be dayereye noh noghte masalaro tabdil be yek
masaleye maroof be name\[ tebolet  \]mikone ke mige
agar \[ ad ,be ,cf  \] ertefahaye yek mosalas bashand sabet konid
khotote oylere se mosalase\[ afe ,bfd ,cde  \] hamresand.
ke esbatesh ba ghaziyeye menelaous va seva ziyad sakht nist.[/quote]
yekami bishtae tozih midi?
man ham enekaas zadam vali be injaa khatm nashod :|

[quote="Ramtin Madani"]saabet konid  agar $H$ markaze ertefaai va $O$ markaze daayereye mohiti $ABC$ baashand, daayerehaaye noh noghteye 3 mosalase $AOH, BOH, COH$ ham mehvarand.[/quote]

Ramtin jan agar be markaze dayereye noh noghte va majzore shoaa e dayereye nohnoghte enekas bezani masale be zibaii hal mishe  ;)

[quote="mohamadhosein"]fekr konam manam hamino goftam...
 :rotfl:[/quote]

na azizam fekr nakon ke to inogofti agar khoob bekhooni mibini ke shooa e enekas o nagofti  ;)

man fekr konam goftam enekas nesbat be dayereye noh noghte yani ham markazo dadam va ham shoao.be har hal hala to badesh chi kar kardi?

eyne mas alaro man ghablan to kort dide boodam

kafie sabet kin ke noghate oyler 3 mosallas ke be rooose A,B,C mian hamkhat and


given the array of integers $ a_{n}$ that:
$ a_{0}\equal{}0,a_{k}<1$ for every $ k \leq 0$
$ a_{k\plus{}1}\equal{}a_{k}\plus{}\frac{\sqrt{1\minus{}{a_{k}}^2}}{30000}$ for every $ k \leq 0$
prove that this array have less than 50000 numbers

I think we have to prove that: $ a_{n}\leq \frac{4}{5}$ but I don't know how to prove that.
any ideas?please help :D

I think that we must to prove that:
$ a_{50000}\geq 1$
 :idea:
anybody>>>>


ratio
geometry
rhombus
trigonometry
rectangle
trig identities
Law of Cosines

If $ABCDEF$ is a regular hexagon , find the ratio of the area to $ABCDEF$ to the area of $\Delta ABC$

[hide]
If I get it right, you mean: \[\frac{area ABCDEF}{areaABC}\]
Then, the solution is $6$.

Let's rename the points of the hexagon from $ABCDEF$ to $KLMNOP$ (I'm really sorry, it's because of the picture, it was the best i could find on google...)
[img]http://www-cchs.ccsd.k12.wy.us/cchs_web/wycas8/math/2002/item5.gif[/img]
So we have to seek $\frac{area KLMNOP}{area KLM}$
We know, that the angles of a regular hexagon are 120° ($\frac{2\pi}{3}rad$)
If we name the point in the middle of the hexagon $Q$, we can state that:
\[area KLM = \frac{area KLMQ}{2} = area KLQ\] (We can say that because $KL = KQ = LM = LQ$)
Then, because area $KLQ$ = $\frac {area KLMNOP}{6}, we know that
\[\frac {area KLMNOP}{area KLM} = 6\]
[/hide]

The ratio would be $6/1$ because the triangle $ABC$ is exactly $1/2$ of the rhombus it is contained in. Since it is a regular hexagon, we know that the three rhombuses that make up the hexagon are congruent. Therefore, the ratio of the triangle $ABC$ is $1/6$ to that of the entire hexagon. Therefore, the answer is $6/1$ or just $6$.

[quote="furious"][quote="Iversonfan2005"]some people might have thought it was tough.. ;)[/quote]
some people would like you to hide your answer so they don't see it before they look at the problem.[/quote]

sorry  :oops:

I have another much less ellegant solution to the same problem.

Angle B of triangle ABC must be 120o because it's in a regular hexagon, and the other angles must be the same, and all the angles add up to 180o, so both the other angles must be 30o.  Let $x$ represent the length of one side of the hexagon.  Then using the law of cosines, the side of the triangle on the inside must have a length of

$\sqrt{2x^2 - 2x^2\cos{120}} = x\sqrt3$

Sorry I don't have a picture, but if the hexagon is broken into two triangles (ABC, DEF) and a rectangle (ACDF), the area of the rectangle will be the inside length ($x\sqrt3$) multiplied by the outside length, $x$, to get an area of $x^2\sqrt3$.  The area of the two triangles can easily be found after calculating the height, and they are each $\frac{x^2\sqrt3}4$.

The ratio can be found by taking the total area of all the parts and dividing it by the area of triangle ABC:

$\frac{x^2\sqrt3 + 2\frac{x^2\sqrt3}4}{\frac{x^2\sqrt3}4} = 6$

[quote="riddler"][quote="Iversonfan2005"][quote="riddler"]Now isnt that simple?????
Its just obvious that it is 6:1 from Iversonfan2005's method.
Should this problem be here?
In middle school, maybe?
 :D[/quote]

some people might have thought it was tough.. ;)[/quote]
sorry :D[/quote]
What happened to your quote thingy, it didn't work

[quote="matt276eagles"]The problem is easy if you realize that a hexagon is made up of three congruent rhombuses (rhombi? :lol: ).  However, it is much more difficult if you don't see that immediately.[/quote]

Either way it's pretty easy, I wouldn't use trig though....

Easiest method?

Look at the drawing. Look at the triangle. Use your counting skills to see how many triangles there are.

6 triangles. So... 6/1 ratio right?

[quote="nat mc"][quote="riddler"]sorry :D[/quote]
What happened to your quote thingy, it didn't work[/quote]
its called "Disable BBCode"  :yoda:
and stop making zigurats, admins dont like it

[quote="Iversonfan2005"]
Either way it's pretty easy, I wouldn't use trig though....[/quote]

The very purpose of me posting this problem in the high school forum was to get an answer that invovled trig...
Anyway , calculuslover800 got the solution , the way I wanted it to be......  :lol:


linear algebra
matrix
algebra
polynomial
abstract algebra
linear algebra unsolved

Hey,

I want to compute exp(A), with A=( (2,0,0)', (1,4,2)', (-1,-2,0)' ).

The characteristic polynomial is P(x)=(x-2)^3. De Jordan normal form I have found is J=( (2,0,0)', (1,2,0)', (0,1,2)' ). How can I find the basis of J such that it is similar to A? I have already found (1,0,0)' must be part of the basis.

thnx

$ A\equal{}2I_3\plus{}N$ where $ N^3\equal{}0$. Then $ e^A\equal{}e^2e^N\equal{}e^2(I_3\plus{}N\plus{}N^2/2)$.

Ok, but you have to multiply this result whit the basis of J and the inverse of J? How do you find the basis?

With Lou le Blanc's approach there is no need any more to chase $ J$. Calculate $ N$ and plug it into the result - that's it. ;)

ok, but I don't own that method yet. I want to find the solution with basis multiplication.

Well, for calculating such a basis you must consider the kernels of $ (A \minus{} 2E_3)^m$ for $ m \equal{} 1,2,3$.

But actually this is not needed with what you know already: Calculate $ (J \minus{} 2E_3)^3$. Can you then use Lou blanc's argument for $ J$? If so, then you can copy that for $ A$ - just by using the similarity of $ A$ and $ J$, but [b]without[/b] explicitly knowing the matrix giving the similarity.


function
algebra
floor function
number theory unsolved
number theory

In a sequence $ (x_n)_{n\equal{}1}^{\infty}$, the first term $ x_1$ is a rational number greater than $ 1$. For all $ n\geq 1$, $ x_{n\plus{}1}\equal{}x_n\plus{}\frac{1}{x'_n}$, where $ x'_n$ is the floor function of $ x_n$. Prove that for some $ i$, $ x_i$ is an integer.

OK, maybe it posted before, but I think no complete solution was posted.

Regardless of being solved or not, every problem should only be posted once.


calculus
integration
function
real analysis
real analysis solved

The function $f(x)$ which is defined at $x\geqq0$ has the inverse of function $g(x)$ which is differentiable.
Find $f(x)$ which satisfies the following equation.

\[\int^{g(x)}_0 f(t)dt=x^2\]

Differentiating leads to $g'(x) \cdot f(g(x)) = 2x$ that is $g'(x) = 2$, so that $g(x) = 2x+c$ for some constant $c$. But the initial relation shows that $c=0$.
Therefore $f(x) = \frac 1 2 x$.

Did I miss something?

Pierre.

Hello,Pierre!

How did you find the value of $c$?

kunny

We have $g(x)=2x+v$ so that $f(x)=\frac 1 2 (x-c)$.
Now, let $x=0$ in the initial relation, we have $ \int_0^c f(t)dt = 0$.
But $ \int_0^c f(t)dt =  \int_0^c \frac 1 2 (t-c)dt = - \frac 1 2 c^2$, so that $c=0$, as claimed.

Pierre.

You are right.
Another approach:

After obtaining $g(x)=2x+c$,as $g\left(-\frac{c}{2}\right)=0$,let $\ x=-\frac{c}{2}\ $ into the initial relation,we have $\frac{1}{4}c^2=0$,yielding $c=0$

kunny


trigonometry
calculus
algebra unsolved
algebra

I read in some physic books saying that  f(t) = -kf'(t). And they say that f(t)=Asin(at+b) without proving that. Can you help me prove it?

The statement is false as it is currently worded; you should replace f'(t) with f''(t). To derive the result, simply solve the second order differential equation and combine the resulting sinosoid into a single sine function.

Note that, in the future, you should post physics and calculus problems in the college playground section.


geometry
calculus
integration
function
calculus computations

Hello, this is my first post. I'm currently taken Calculus and I'm encountering some problems with cross-sections, and finding area between two curves.

several questions that are giving me a hard time are these:

1)  Find the area of the region bounded by y2 = 4 – x and y = x + 2

what area or integral is it bounded by? what to what? i dont get that part.

2) Find the area of the region bounded by
f(x) = x2 + 2x + 1 and g(x) = 3x + 3

how should the integer be because im getting confuse if its a washer  or a disk method. also i dont understad how to write the integral.

3)Find the volume of the solid bounded by y2 = x – 1 and y = x – 3 revolved about the y-axis.

i don't get if this is similar to #1 or not because it says find volume but the graph looks kind of similar

4)Find the volume of the solid whose base is bounded by y = x2, x = 0, and y = 4 by taking square cross sections perpendicular to the x-axis.

am i suppose to find the volume for -2 and 2 or just one of them?

Thanks for the all suggestion and advice i could get.

"integer" -> "integral"

Disks and washers are for volumes, not areas.

In 1) and 2) you must find the points where two curves intersect. For example, in 1) you have $x=4-y^{2}$ and $x=2-y$. The curves intersect when these two formulas give the same value of $x$, that is, $4-y^{2}=2-y$. Solutions: $y=-1$ and $y=3$. Notice that I treat $x$ as a function of $y$: this is easier than solving for $y$. The area is $\int_{-1}^{3}((4-y^{2})-(2-y))\,dy$. The order in which you subtract the equations should be such that the integrand is positive for $y$ between $-1$ and $3$.

2) is very similar.

3) First consider the region bounded by $x=y^{2}+1$ and $x=y+3$. Find the intersection points: $y^{2}+1=y+3$ when $y=-1$ or $y=2$. Rotating a horizontal segment $[y^{2}+1,y+3]$ about the $y$-axis, you get a washer with inner radius $y^{2}+1$ and outer radius $y+3$. Find its area and integrate: $\int_{-1}^{2}\pi((y+3)^{2}-(y^{2}+1)^{2})dy$.

[hide="IMHO"]Teaching Calc 2 in HS is a moronic idea[/hide]

i am getting confuse as how you find the solution for y. i think its the way im so use to y and not x but i cant seem to get -1 and 3 when i subtract it. i might be doing something wrong because im getting -1 and 2 but can you explain to me y -1 and 3?

[quote="mlok"]

[hide="IMHO"]Teaching Calc 2 in HS is a moronic idea[/hide][/quote]
Why? Please explain... I think for some teaching Calc 2 any time [b]after[/b] HS is moronic, the concepts are so basic and just being exposed to them are necessary to understand physics, chemistry, and most sciences at the high school level.
Of course, the day when this is reasonable will be a day when school administrators actually understand mathematics and the whole attitude of the United States changes...lol


topology
real analysis
real analysis unsolved

Let $f: \mathbb{R}^{n}\to\mathbb{R}^{n}$ be a homeomorphism such that there is $k\in\mathbb{N}$ such that
     $f^{k}(x)=1_{\mathbb{R}^{n}}$. Prove that $f$ has at least one fixed point.

Discussed several times before. One of the more recent topics is [url=http://www.mathlinks.ro/viewtopic.php?t=151219]this one[/url].


Let a,b,c be positive numbers.Prove that:
    $\sum \frac{1}{a^{2}+bc}\leq 2\sum \frac{1}{(a+b)^{2}}$.

Let Cauchy do it :)
We have $(a^{2}+bc)(b+c)\geq b(a+c)^{2}\Longrightarrow \frac{1}{(a+c)^{2}}\geq\frac{b}{(b+c)(a^{2}+bc)}(\star)$
By the same Cauchy trick you get $\frac{1}{(a+b)^{2}}\geq\frac{c}{(b+c)(bc+a^{2})}(\star\star)$
By combining the last two relations you get $\frac{1}{(a+c)^{2}}+\frac{1}{(a+b)^{2}}\geq\frac{1}{a^{2}+bc}.$ Sum it all up and we`re done ;)

Is there any algebraic way of finding such "Cauchy Tricks"? It seems to me like there is...