function
inequalities
logarithms
inequalities proposed

Let $ x,y,z$ be positive real numbers. Prove that for any $ k > 0$ such that $ k^3 \plus{} 3k^2 \minus{} 6k \minus{} 2 \ge 0,$ then
$ \left( \frac {x^2 \plus{} yz}{x^2 \plus{} kyz}\right)^2 \plus{} \left( \frac {y^2 \plus{} zx}{y^2 \plus{} kzx}\right)^2 \plus{} \left( \frac {z^2 \plus{} xy}{z^2 \plus{} kxy}\right)^2 \ge \frac {12}{(k \plus{} 1)^2}.$
:)

It follows from the convexity of function $ f(t) \equal{} \left( \frac{e^t\plus{}1}{e^t\plus{}k} \right)^2$

If we let $ xyz \equal{} 1$ then the inequality becomes:
$ f(\ln x) \plus{} f(\ln y) \plus{} f(\ln z) \geq 3f(0)$ which is Jensen

Hello, thank you for your reply. But I think the function $ f(t)\equal{}\left( \frac{e^t\plus{}1}{e^t\plus{}k}\right)^2$ is not always convex, so I think you can't use Jensen's Inequality Here. Please try it again. BTW, the condition $ k^3\plus{}3k^2\minus{}6k\minus{}2 \ge 0$ is also the best condition such that the original inequality holds.
Regards


[quote="4865550150"]Let {x_i} be a set of reals such that $ \sum^n x_i \equal{} n$. Prove that

$ \sum^n (n \plus{} x_i \plus{} \frac {1}{x_i})(n \plus{} x_i \plus{} x_i^2) \ge n^3 \plus{} 4n^2 \plus{} 4n$[/quote]

The LHS is equivalent to $ n^3\plus{}2n^2\plus{}2n\plus{}\sum x^{3}_{i}\plus{}(n\plus{}1)\sum x^{2}_{i}\plus{}\sum x_{i}\sum \frac{1}{x_i}\geq n^3\plus{}4n^2\plus{}4n$.

so, it suffices to prove that $ \sum x^{3}_{i}\plus{}(n\plus{}1)\sum x^{2}_{i}\plus{}\sum x_{i}\sum \frac{1}{x_i}\geq 2n^2\plus{}2n$.

But from Chebychev's inequality we get that $ \sum x^{3}_{i}\geq \sum x^{2}_{i}$. So, we need to prove that $ (n\plus{}2)\sum x^{2}_{i}\plus{}\sum x_{i}\sum\frac{1}{x_i}\geq n^2\plus{}n^2\plus{}2n$ which is obviously true from CS and AM-GM


Let a,b,c > 0 .Prove that :

i, $ \frac{a}{a\plus{}2b}\plus{}\frac{b}{b\plus{}2c}\plus{}\frac{c}{c\plus{}2a}\leq \frac{3(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}$

ii, $ \frac{a^2}{a^2\plus{}2b^2}\plus{}\frac{b^2}{b^2\plus{}2c^2}\plus{}\frac{c^2}{c^2\plus{}2a^2}\leq \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}$

[quote="PHONGVAN"]Let a,b,c > 0 .Prove that :

i, $ \frac {a}{a \plus{} 2b} \plus{} \frac {b}{b \plus{} 2c} \plus{} \frac {c}{c \plus{} 2a}\leq \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}$

[/quote]
\[ \Longleftrightarrow \sum{\left(1\minus{}\frac{a}{a\plus{}2b}\right)}\geq 3\minus{}\frac{3(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}\equal{}\frac{6(ab\plus{}bc\plus{}ca)}{(a\plus{}b\plus{}c)^2}\]
\[ \Longleftrightarrow \frac{b}{a\plus{}2b}\plus{}\frac{c}{b\plus{}2c}\plus{}\frac{a}{c\plus{}2a}\geq \frac{3(ab\plus{}bc\plus{}ca)}{(a\plus{}b\plus{}c)^2}\]
By $ Cauchy$ we get
\[ \sum{\frac{b}{a\plus{}2b}}\sum{b(a\plus{}2b)}\geq (a\plus{}b\plus{}c)^2\]
It suffice to show that
\[ (a\plus{}b\plus{}c)^4\geq 3(ab\plus{}bc\plus{}ca)(ab\plus{}bc\plus{}ca\plus{}2a^2\plus{}2b^2\plus{}2c^2)\]
Without loss of generosity,assume that $ ab\plus{}bc\plus{}ca\equal{}3$,then it becomes
\[ \left[(a\plus{}b\plus{}c)^2\minus{}9\right]^2\geq 0\]
which is obvious.


Rohan plans to open a bank for his friends. Some friends will allow him to keep their money in savings, and others will borrow money from him. The charts below describe the amounts that students are saving and borrowing from him. What is the least possible value, in dollars, of the money that Rohan is currently keeping in his bank? Express your answer to the nearest whole number.

\[ \begin{array}{cccc}
\multicolumn{2}{c}{\textrm{\underline{Amount put in for savings}}} & \multicolumn{2}{c}{\textrm{\underline{Amount taken for loans}}} \\
\textrm{\underline{Value of accounts}} & \textrm{\underline{Number of accounts}} &
\textrm{\underline{Value of loan}} & \textrm{\underline{Number of loans}} \\
\$0.01\minus{}\$20.00 & 3 & \$0.01\minus{}\$20.00 & 5\\
\$20.01\minus{}\$40.00 & 4 & \$20.01\minus{}\$40.00 & 3 \\
\$40.01\minus{}\$60.00 & 6 & \$40.01\minus{}\$60.00 & 1 \\
\$60.01\minus{}\$80.00 & 2 & \$60.01\minus{}\$80.00 & 1 \\
\end{array}\]

[quote="BOGTRO"][quote="GameBot"]Rohan plans to open a bank for his friends. Some friends will allow him to keep their money in savings, and others will borrow money from him. The charts below describe the amounts that students are saving and borrowing from him. What is the least possible value, in dollars, of the money that Rohan is currently keeping in his bank? Express your answer to the nearest whole number.
\[ \begin{array}{cccc} \multicolumn{2}{c}{\textrm{\underline{Amount put in for savings}}} & \multicolumn{2}{c}{\textrm{\underline{Amount taken for loans}}} \\
\textrm{\underline{Value of accounts}} & \textrm{\underline{Number of accounts}} & \textrm{\underline{Value of loan}} & \textrm{\underline{Number of loans}} \\
\$0.01 \minus{} \$20.00 & 3 & \$0.01 \minus{} \$20.00 & 5 \\
\$20.01 \minus{} \$40.00 & 4 & \$20.01 \minus{} \$40.00 & 3 \\
\$40.01 \minus{} \$60.00 & 6 & \$40.01 \minus{} \$60.00 & 1 \\
\$60.01 \minus{} \$80.00 & 2 & \$60.01 \minus{} \$80.00 & 1 \\
\end{array}
\]
[/quote]

$ .01 \cdot 3\plus{}20.01 \cdot 4\plus{}40.01 \cdot 6\plus{}60.01 \cdot 2\equal{}440.15$.

$ 440.15\minus{}100\minus{}120\minus{}60\minus{}80\equal{}80.15$. Rounded this is $ 80$.[/quote]

How did you get the second line?


如图，正方形ABCD的边长为1，点M、N分别在BC、CD上，使得△CMN的周长为2。求△MAN面积的最小值。

旋转我知道，可证角MAN=45度，可否告知接下来的推理、运算过程？


For a given natural number $n > 3$, the real numbers $x_1, x_2, \ldots, x_n, x_{n + 1}, x_{n + 2}$ satisfy the conditions $0
< x_1 < x_2 < \cdots < x_n < x_{n + 1} < x_{n + 2}$.  Find the minimum possible value of
\[\frac{(\sum _{i=1}^n \frac{x_{i + 1}}{x_i})(\sum _{j=1}^n \frac{x_{j + 2}}{x_{j +
1}})}{(\sum _{k=1}^n \frac{x_{k + 1} x_{k + 2}}{x_{k + 1}^2 + x_k
x_{k + 2}})(\sum _{l=1}^n \frac{x_{l + 1}^2 + x_l x_{l + 2}}{x_l
x_{l + 1}})}\] and find all $(n + 2)$-tuplets of real numbers $(x_1, x_2, \ldots, x_n, x_{n + 1}, x_{n + 2})$ which gives this value.

If it holds for all $a_i$ equal, shouldn't the minimum be 1?
 And why can we choose $a_1=a_{n+1}$?

Using Rust 's notation,in fact we are looking for the minimum value of $$\dfrac {1}{\left( \sum ^{n}_{k=1}\dfrac {1}{\dfrac {1}{a_{k}}+\dfrac {1}{a_{k+1}}}\right) (\dfrac {1}{\sum ^{n}_{l=1}a_{l}}+\dfrac {1}{\sum ^{n}_{l=1}a_{l+1}})}$$ for all $a_{i} > 1$.
And I think it's not so easy.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

This is not fundamentally different from the Vandermonde case:  the inversion of this matrix decides the solvability of the interpolation problem $ P(a_i) \equal{} y_i$ for a polynomial $ P(x) \equal{} c_0 \plus{} c_1 x^{k_1} \plus{} c_2 x^{k_2} \plus{} ...$, which is clearly well-posed by a dimension argument.  I believe jmerry made comments to this effect in a thread a few weeks ago.


inequalities
AMC
USA(J)MO
USAMO
IMO Shortlist
inequalities solved

For a,b,c positive real numbers such that a+b+c=1 prove

1/(1+a+b)  + 1/(1+b+c) + 1/(1+c+a)  \leq 1

There's definitely sth wrong with this one: what if a=b=c=1/3? The sum on the left becomes 9/5>1...

Sorry very much. The condition is abc=1 and not a+b+c=1.

Well, put x = a3, y = b3, z = c3. Then we have to prove that

\sumcyclicxyz/(x3 + y3 + xyz) \leq 1.

This is exactly USAMO 1997/5. Solution at the kalva site.

This is a well-known trick:

I used the sum Arne wrote. If we consider the fact that x^3+y^3>=xy(x+y) we get left sum <=  \sum z/(x+y+z)=1. This was the official solution of a very similar (if not identical, I can't eally remember      :D) IMO problem.

You probably mean A1 of the IMO shortlist 1996 ?

If x, y, z > 0 and xyz = 1 then

\sumcyclic xy/(x5 + y5 + xy) \leq 1.

x=b+c y=c+a z=a+b
So we have:
 \sum 1/(1+a+b)= \sum 1/(1+x)=(3+2 \sum x+ \sum xy)/(1+ \sum x+ \sum xy+xyz)
2+ \sum x \leq xyz
2+2 \sum a \leq (a+b)(b+c)(c+a)
2 \sum a \leq  \sum (a^2b+a^2c+1)
a^2b+a^c+1 \geq 3a
 \sum(a^2b+a^c+1) \geq 3 \sum a
3 \sum a \geq 2 \sum a+3
 \sum a \geq 3

If $x,y,z$ positive real numbers such that $yz+zx+xy=3$ and $k\geq1$, then

$\frac{1}{k+y+z}+\frac{1}{k+z+x}+\frac{1}{k+x+y}\leq\frac{3}{k+2}$,

with equality if and only if $x=y=z=1$.

use harazi's substitution
a=x/y,b=y/z,c=z/x and we are through! :roll:


search
arithmetic sequence
geometric sequence
number theory open
number theory

Does an arithmetic progression of four distinct triangular numbers exist?

I've done search through all possible quadruples with the largest member smaller than $ T_{32000}$, however I didn't find anything. There exist a lot of triples, e.g. the smallest ones $ (T_1, T_7, T_{10}), (T_3, T_6, T_8)$.

Does anybody know something about this problem or have any ideas?

Note: Similar problem with four triangular numbers in geometric progression (Sierpinski, Szymiczek) was proven to have no solution.

Note that $ T_n \equal{} \frac{n(n\plus{}1)}{2} \equal{} \frac{(2n\plus{}1)^2\minus{}1}{8}$. Therefore, if triangular numbers $ T_{n_1}, T_{n_2}, T_{n_3}, T_{n_4}$ are consecutive term of an arithmetic progression, then such are the numbers $ (2n_1\plus{}1)^2, (2n_2\plus{}1)^2, (2n_3\plus{}1)^2, (2n_4\plus{}1)^2$. But it is well known (see [url=http://www.mathpages.com/home/kmath044/kmath044.htm]this page[/url], for example) that there are no 4 consecutive squares in arithmetic progression.


inequalities
trigonometry
inequalities unsolved

Prove the inequality
\[ \frac{\sqrt{1\plus{}8\cos^{2}A}}{\sin B}\plus{}\frac{\sqrt{1\plus{}8\cos^{2}B}}{\sin C}\plus{}\frac{\sqrt{1\plus{}8\cos^{2}C}}{\sin A}\geq 6,\]
where $ A$, $ B$ and $ C$ are the three angles of a triangle $ ABC$.

The inequality above is the corollary of this one:
 $ (1+8cos^2A)(1+8\cos^2B)(1+8\cos^2C)\geq\ 64sin^2Asin^2Bsin^2C $  (*)
To prove (*)  we replace $ sin^2A=1-cos^2A $ and etc.
Expand the expression the inequality becomes :
$  -63+72(cos^2A+cos^2B+cos^2C)+576cos^2Acos^2Bcos^2C \geq\ 0 $
Use the indentity $ cos^2A+cos^2B+cos^2C+2cosAcosBcosC=1 $ (*) is equivalent to $ 64(cosAcosBcosC)^2 - 16(cosAcosBcosC) + 1 \geq\ 0 $
The last inequality is clearly true. Therefore (*) is proved. Apply the Cauchy's inequality for the LHS of the original inequality we finish.

[quote="Myth"][quote="tranminhhoang"]$ cos^2A+cos^2B+cos^2C+2cosAcosBcosC=1 $[/quote]
It was unknown for me.[/quote]

I use this identity so frequently that suddenly it is difficult to write the proof. But anyway, i have the proof now.
Use the formulae : $ cos^2A=(1+cos2A)/2 $ and etc. All we have to prove now is :
$ P=cos2A+cos2B+2cos^2C+4cosAcosBcosC = 0 $
We have $cos2A+cos2B=2cos(A+B)cos(A-B)=-2cosCcos(A-B) $ Therefore it is equivalent to prove
$ P=2cosC(cosC-cos(A-B)+2cosAcosB)=0 $
But this is clearly true because $ cosC-cos(A-B)=-cos(A+B)-cos(A-B)=-2cosAcosB $


inequalities
combinatorics proposed
combinatorics

Here's a quite easy, but elegant problem:

[i]Let $ A \equal{} (A_{i})_{1\le i\le n}$, $ B \equal{} (B_{i})_{1\le i\le n}$, $ C \equal{} (C_{i})_{1\le i\le n}$ be three partitions of a finite set $ M$.
Prove that if $ |A_{i}\cap B_{j}|\plus{}|B_{j}\cap C_{k}|\plus{}|C_{k}\cap A_{i}|\ge n$ holds for any $ i,j,k$, then $ |M|\ge\frac{n^{3}}{3}$.

Equality can hold iff $ n\equiv0($mod $ 3)$.
[/i]

Summation over all $ i,j,k$ in your inequality proves the first part of your assertion

$ \begin{array}{lcl}n^{4}&\leq &\sum_{i,j,k=1}^{n}\left(|A_{i}\cap B_{j}|+|B_{j}\cap C_{k}|+|C_{k}\cap A_{i}|\right)\\[0.1cm] & =&\sum_{i,j=1}^{n}\left(n|A_{i}\cap B_{j}|+|B_{j}|+|A_{i}|\right)\\[0.1cm] & = &\sum_{i=1}^{n}\left(n|A_{i}|+|M|+n|A_{i}|\right)\\[0.1cm] & = & n|M|+n|M|+n|M|\\[0.1cm] & = & 3n|M|\end{array}$

Equality obviously implies $ n$ divisible by 3. This proves one direction of the "iff"-statement. As for the other direction (the "can hold") let $ n$ be divisible by 3 and choose an arbitrary partition $ \left(D_{r,s}\right)_{r,s=1,\ldots n}$ such that $ |D_{(r,s)}|=\frac{n}{3}$ holds. We interpret the indexes $ (r,s)$ of this partition as all possible pairs of residues $ 1,\ldots, n$ modulo $ n$. With that interpretation in mind define

$ \begin{array}{lcl}A_{i}& = &\bigcup_{t=1}^{n}D_{(i,t)}\\[0.1cm] B_{j}& = &\bigcup_{t=1}^{n}D_{(t,j)}\\[0.1cm] C_{k}& = &\bigcup_{t=1}^{n}D_{(t,t+k)}\end{array}$

and observe for all indexes $ i,j,k$ the relationship

$ |A_{i}\cap B_{j}|+|B_{j}\cap C_{k}|+|C_{k}\cap A_{i}|=|D_{(i,j)}|+|D_{(j-k,j)}|+|D_{(i,i+k)}|=n$ verifying the last bit.


whats wron with my solution

When Bill drives to work at 45 mph, he is 5 minutes late. when bill drives 53 mph to work, he is 5 minutes early. how fast should he drive to work if he wants to be right on time?

[hide]
set the time it takes for him to be right on time x.
$ D \equal{} 45(x \plus{} 5)$
$ D \equal{} 53(x \minus{} 5)$
substitute
$ 53(x \minus{} 5) \equal{} 45(x \plus{} 5)$
i solve and get 61.25
which is clearly not the answer...[/hide]

You solved for $ x$, not the rate.  The rate would be $ \frac {D} {x}$.  So, go redo it, and see if you end up with the correct answer.

[hide="new solution"]
i did what i did before, got $ x\equal{}61.25$
i plugged it back in, and got $ D\equal{}2981.25$
$ \frac{D} {x}\equal{}48.67$(i dont know how to make the aproximate sign)[/hide]

The approximate sign is \approx. $ \approx$


Let $ a,b,c \ge \frac{1}{2}$ be positive real numbers such that $ abc\equal{}1$. Prove that
\[ \frac{1}{a\plus{}2b}\plus{}\frac{1}{b\plus{}2c}\plus{}\frac{1}{c\plus{}2a} \le 1\]

We can prove this inequality using Cauchy Schwarz Inequality. :)

Can, see your problem, I remembered an inequality created by Anh Tran.
[hide]Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {b}{a \plus{} 2b} \plus{} \frac {c}{b \plus{} 2c} \plus{} \frac {a}{c \plus{} 2a}\leq 1$[/hide]

[quote="Sudoku"]Can, see your problem, I remembered an inequality created by Anh Tran.
Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {b}{a \plus{} 2b} \plus{} \frac {c}{b \plus{} 2c} \plus{} \frac {a}{c \plus{} 2a}\leq 1$[/quote]
Yes, you are right. :)
But I think this problem was proposed first time by VasC at Romany team selection test (it is the generalization)
\[ \overset{n}{\underset{i\equal{}1}{\sum}} \frac{1}{a_i\plus{}n\minus{}1} \le 1\]
with $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1a_2\cdots a_n\equal{}1$.
:)


inequalities
topology
triangle inequality
real analysis
real analysis unsolved

Let A be a bounded subset of a metric space X equipped with a metric 'd'. show that A is bounded iff for every $x\in X$ there exists r>0 such that $A\subset B(x,r)$

let A be bounded, then there exists $y\in X$ and R>0, so that $A\subset B(y,R)$

now i am stuck when proving that if A is bounded, then that condn holds. here's what i tried

let us assume the contrary. ie$\exists x\in X$ such that however big u take r,points of A lie outside $B(x,r)$. tkae r=n, then $\exists a_n$ such that $d(a_n,x)\geq n$ . now using this and the bounded condition i tried to arrive at a contradiction,by triangle inequality but i'm getting stuck.is my approach on the right lines?could someone give me a hint? or suggest another approach. i dont want a detailed proof as i want to do it myself

p.s- by $B(x,r)$ i mean the open ball, with radius r and centre x

Let $R_1 = R+\rho(x,y)$, and look at the $B(x,R_1)$


Does anyone know of a really good Partial Differential Equation's Book?

we used evan's, "partial differntial equations" at my school


Take the figure $ 123456789$. Now insert two and only two multiplication signs ($ \times$) anywhere inbetween.

How many even numbers will you be able to generate by this exercise ?

For example,

$ 12\times34\times56789$ and $ 1\times2\times3456789$ are two such exercises which will generate even numbers.

You are not permitted to change the sequence of the digits $ 1$ thru' $ 9$.

[hide="Hint"]Count all the cases (number of ways of havin 2 x and 7 numbers) and substract the odd ones (remembering the only you get and odd number by multipliying odd numbers only)[/hide]

[hide] Total ways = $ \binom{9\minus{}1}{2}\equal{}28$
To make odd numbers, we can only put both "X"s after 1,3,5,7. Therefore, we have $ \binom{4}{2}\equal{}6$ ways
So the there're 22 ways. Right?[/hide]


geometry
incenter
inradius
rectangle
geometry proposed

Given three parallel lines on the plane. Find the locus of incenters of triangles with vertices lying on these lines (a single vertex on each line).

Call $ r$ the parallel line that is contained in the stripe determined by the other two lines $ s,s'$.  Clearly, if the incenter $ I$ is contained in the stripe defined by $ r$ and $ s$, then it must be closer to $ r$ than to $ s$; otherwise, since the incircle cannot touch line $ s$ because the triangle is entirely contained in the stripe between $ s$ and $ s'$, then the inradius is less than half the distance between $ r$ and $ s$, hence it would not cross $ r$, and any ray originating in any point on $ r$ and touching the incircle could not cross $ s'$.  Similarly, if the incenter is contained in the stripe defined by $ r$ and $ s'$, then it is closer to $ r$ than to $ s'$.  We will now show that for any point strictly inside the stripe defined by the parallel lines halfway between $ r$ and $ s$, and halfway between $ r$ and $ s'$, one point may be found on each of the three given lines such that this point is the incircle of the triangle thus formed.  Take $ A\in r$, $ B\in s$ and $ C\in s'$, collinear and such that $ AB\perp r$.  Displace $ C$ along $ s'$ away from $ A,B$.  As $ C$ goes arbitrarily far, the figure tends towards an open rectangle with one side $ AB$, two sides being rays on $ r$ and $ s$, hence the incircle tends towards the circle touching $ AB$, $ r$ and $ s$, ie, the incenter tends towards a point on the line halfway between $ r$ and $ s$, getting arbitrarily close but never reaching it.  From the original position, displace $ A$ along $ s$ away from $ B,C$, the incenter tends towards a point on the line halfway between $ r$ and $ s'$, getting arbitrarily close but never reaching it.  Since $ C$ on the first displacement, and $ A$ on the second displacement, may be moved continuously, and in doing so the incenter moves continuously, for each line $ t$ parallel to $ r$ contained strictly in the stripe defined by the lines halfway from $ r$ and $ s$, and halfway from $ r$ and $ s'$, there is a position of $ A,B,C$ such that the incenter is on $ t$.  Displace now simultaneously $ A,B,C$ along $ s,r,s'$ as needed until the incenter reaches any desired point on $ t$.  The conclusion follows.


Are you actually supposed to do [url=http://planetmath.org/encyclopedia/QuarticFormula.html]this[/url], or is there an easier way?

OF COURSE NOT!!

Hint: try reducing it to something you can solve, such as.......................................................... a quadratic?

Oh yeah, and what are those stars under your name for?? :?:

[hide="Hint"]Let $ X\equal{}x^{2}$.[/hide]

and the stars relate to how many posts you have.

thats what i was thinking. only i used y. so we have y^2+5y=36 so y^2+5y-36=0 (y+9)(y-4). plug in x^2. x^2+9=0 and x^2-4=0.

are there imaginary numbers involved cuz i get [hide="my answer"]$ 2,\minus{}2,3i,\minus{}3i$[/hide]

[quote Kevin K."]Find the value of x. There are 4 solutions.
(x^4) + (5 x^2) = 36[/quote]
Solution (by galois01)
$ x^{4}\plus{}5x^{2}\equal{}36\Longleftrightarrow$
$ x^{4}\plus{}5x^{2}\minus{}36\equal{}0 (1)$
Let:
$ x^{2}\equal{}z$ (2)          $ z\geq 0$
$ (1)\Longrightarrow z^{2}\plus{}5z\minus{}36\equal{}0$
$ D\equal{}5^{2}\minus{}4(\minus{}36)\equal{}169$
$ z_{1,2}\equal{}\frac{\minus{}5\pm\sqrt{169}}{2}$
$ \Longrightarrow$  $ z\equal{}4$and $ z\equal{}\minus{}9$
From (2) we get
$ x^{2}\equal{}4\Longleftrightarrow x\equal{}\pm2$
$ x^{2}\equal{}\minus{}9\Longleftrightarrow x\equal{}\pm3i$
So the solutions are:
$ x\equal{}2,\minus{}2, 3i,\minus{}3i$

Oh. yeah. Oops. This is simple, sorry. I thought...

Here, let me try it on my own.

[hide]
$ x^{4}\plus{}5x^{2}\equal{}36$
$ n\equal{}x^{2}$
$ n^{2}\plus{}5n\equal{}36$
$ n^{2}\plus{}5n\plus{}6.25\equal{}42.25$
$ \pm(n\plus{}2.5)\equal{}6.5$
$ n\equal{}4,\minus{}9$
$ x\equal{}2,\minus{}2,3i,\minus{}3i$[/hide]

[hide]
Replace $ x^{2}$ with $ y$
$ y^{2}\plus{}5y\minus{}36\equal{}0$
Factor: $ (y\plus{}9)(y\minus{}4)\equal{}0$
So, $ y\equal{}\minus{}9 , 4$
The 4 solutions for x are $ \boxed{\minus{}2 , 2 ,\minus{}3i, 3i}$[/hide]

Is there any particular reason that you have to make the substitution $ y\equal{}x^{2}$? You can just directly factor...

\[ x^{4}\plus{}5x^{2}\minus{}36\equal{}(x^{2}\minus{}4)(x^{2}\plus{}9)\equal{}(x\minus{}2)(x\plus{}2)(x\minus{}3i)(x\plus{}3i)\equal{}0\]

you dont have to, it just makes it easier to see what we have to do. but yes you can do it that way to


Solve in natural numbers x,y,z,t the equation: $x^2+y^2+z^2=2^t$

Well, if t equals 0, the solution is trivial : (x,y,z)=(1,0,0) of any permutation of this solution.
Now assume that $t\geq 1$. The value of  $x^2+y^2+z^2$ must be divisible by 2. So x,y and z are all even, or exactly one of those is even. In the first case, we can put $x=2x',y=2y',z=2z'$ and the equality becomes $x'^2+y'^2+z'^2=2^{t-2}$. If t equals 2, we have our trivial case, otherwise we have one more time two cases to see. We can do the substitution again and again to get exactly one of the numbers even. Let WLOG  $x=2x'+1,y=2y'+1,z=2z'$ and the equality becomes now $2(x'^2+x'+y'^2+y'+z'^2)+1=2^{t-1}$. So $2^{t-1}$ is not even and t=1. Then obviously $x'=y'=z'=0$. Finally we have
$(x,y,z,t)\in\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(2^k,2^k,0,2k+1),(2^k,0,2^k,2k+1),(0,2^k,2^k,2k+1)|k\in\mathbb{N}\}$

You omitted $(2^k,0,0,2k)$ and the permutations (since you left the discussion on all the numbers even unfinished), and for $k=0$ you get the trivial solution.


inequalities
ceiling function
number theory unsolved
number theory

Let p(n) be the greatest odd divisor of n. Prove that
                  $\frac{1}{2^n} \sum_{k=1}^{2^n} \frac{p(k)}{k} > \frac{2}{3}$

Solution: Let $ n,k$ be two positive integers.
We define $ v_2: \mathbb{N} \rightarrow \mathbb{N}$, by $ v_2(n) = k$ iff $ 2^k \parallel{} n$ (i.e $ 2^k$ divides $ n$ and $ 2^{k + 1}$ does not divide $ n$)
We have

\begin{eqnarray*} \sum_{k = 1}^{2^n} \frac {p ( k )}{k} & = & \sum_{j = 0}^{\infty} \sum_{\tmscript{\begin{array}{c} k \leq 2^n \\
v_2 ( k ) = j \end{array}}} \frac {p ( k )}{k} \\
& = & \sum_{j = 0}^{\infty} \sum_{\tmscript{\begin{array}{c} k \leq 2^n \\
v_2 ( k ) = j \end{array}}} \frac {\frac {k}{2^j}}{k} \\
& = & \sum_{j = 0}^{\infty} \frac {1}{2^j} \sum_{\tmscript{\begin{array}{c} k \leq 2^n \\
v_2 ( k ) = j \end{array}}} 1 \\
& = & \sum_{j = 0}^n \frac {1}{2^j} \sum_{\tmscript{\begin{array}{c} 2^j k \leq 2^n \\
k \mbox{ odd} \end{array}}} 1 \\
& = & \sum_{j = 0}^n \frac {1}{2^j} \sum_{\tmscript{\begin{array}{c} k \leq 2^{n - j} \\
k \mbox{ odd} \end{array}}} 1 \\
& = & \sum^n_{j = 0} \frac {1}{2^j} \cdot \lceil 2^{n - j - 1} \rceil \\
& = & \sum_{j = 0}^{n - 1} \frac {1}{2^j} \cdot 2^{n - j - 1} + \frac {1}{2^n} \\
& = & 2^{n - 1} \sum_{j = 0}^{n - 1} \left( \frac {1}{4} \right)^j + \frac {1}{2^n} \\
& = & 2^{n + 1} \cdot \left( \frac {1 - \left( \frac {1}{4} \right)^n}{3} \right) + \frac {1}{2^n} \end{eqnarray*}

Hence

\begin{eqnarray*} \frac {1}{2^n} \sum_{k = 1}^{2^n} \frac {p ( k )}{k} & = & 2 \cdot \left( \frac {1 - \frac {1}{4^n}}{3} \right) + \frac {1}{4^n} \\
& = & \frac {2 - \frac {2}{4^n} + \frac {3}{4^n}}{3} \\
& = & \frac {2 + \frac {1}{4^n}}{3} > \frac {2}{3} \end{eqnarray*}

Funny solution, huh ? :)


Five integers have a sum of 92. What is the greatest possible product of these five numbers?

But we want 5 numbers  :roll:  (and I never thought of AM-GM even though it is obvious that you can use it).

EDIT: Deleted post

[quote="236factorial"]But we want 5 numbers  :roll:  (and I never thought of AM-GM even though it is obvious that you can use it).[/quote]

Sorry, I misread the question and caught my mistake before you posted.  I found the answer for 5 numbers, but realised that this is the middle school section, so I'll let someone else post.


Give that three roots of $ f(x)\equal{}x^4\plus{}ax^2\plus{}bx\plus{}c$ are 2, -3, and 5, what is the value of $ a\plus{}b\plus{}c$?

well plugging in

16+4a+2b+c=0
81+9a-3b+c=0
625+25a+5b+c=0

subtracting (2) and (1) you get 5b-5a=65, b-a=13
subtracting (3) and (1) you get 544=16a+8b or 2a+b=-68

substituting b=13+a, you get 3a=-81, a=-27, so b=-14

plugging into any of the three equations, c=120

so -27-14+120=79


$ S_n \equal{} a_1 \plus{}a_2 \plus{} \cdots \plus{} a_n , a_n \plus{}2 S_n S_{n\minus{}1} \equal{}0 (n \geq 2),a_1 \equal{} \frac 12 ,$
$ b_n \equal{} 2 (1\minus{}n) a_n (n \geq 2), \Longrightarrow b_2 ^2 \plus{}b_3^2 \plus{} \cdots \plus{}b_n^2<1.$


How many C's are used when writing down all of the numbers from 1 to 500 using Roman numerals?

I keep getting 720, but the correct answer is 750.  Can someone explain this to me?

Here's a quick chart if you forgot which letters go with which numbers:

I = 1
V = 5
X = 10
L = 50
C = 100
D = 500

I don't know how to express numbers higher than like 10, so could someone explain how to express higher numbers to me?

Here are the rules according to the makers of the problem:

[quote] As many as three of the same marks that represent $10^n$ may be placed consecutively to form other numbers:

    * III is 3
    * CCC is 300

Marks that have the value $5\cdot10^n$ are never used consecutively.

Generally (with the exception of the next rule), marks are connected together and written in descending order to form even more numbers:
# CCLXVIII = 100+100+50+10+5+1+1+1 = 268

Sometimes, a mark that represents 10^n is placed before a mark of one of the two next higher values (I before V or X; X before L or C; etc.). In this case, the value of the smaller mark is SUBTRACTED from the mark it precedes:

    * IV = 4
    * IX = 9
    * XL = 40

This compound mark forms a unit and may not be combined to make another compound mark (e.g., IXL is wrong for 39; XXXIX is correct).

Compound marks like XD, IC, and XM are not legal, since the smaller mark is too much smaller than the larger one. For XD (wrong for 490), one would use CDXC; for IC (wrong for 99), one would use XCIX; for XM (wrong for 990), one would use CMXC. [/quote]

C is first used at 90 = XC
One C is used until 190 = CXC
Two C's are used until 290 = CCXC
Three C's are used until 390 = CCCXC
Four C's are used until 400 = CD
One C is again used until 490 = CDXC
Two C's are used until 500 = D

100 + 200 + 300 + 40 + 90 + 20 = 750

I need help again with something similar to this (It's for a computer program).

How many X's, L's and C's are in 2974.  I need the work, since I already know the answers.

Thanks.

Sorry, I worded it wrong.  How many of those letters do you need to write [b]all[/b] of the number [b]from[/b] 1 [b]to[/b] 2974.

The bolds are just to emphasize, not to act like I'm getting mad at you or anything.

Here's what I have for the number of L's for all of the numbers from 1 to 2974 so far:

Here are all of the multiples of 10 from 40 to 100 that need L's:

XL = 40 to 49
L = 50 to 59
LX = 60 to 69
LXX = 70 to 79
LXXX = 80 to 89

That means for every 100 numbers you wirte, you need 50 L's.  That means we have 28*50 L's from 1 to 2900.  But, we still have 74 numbers left to check.

From 40 to 69, there are 30 L's, and from 70-74 there are 5 L's.

So in total, we have 28*50 + 30 + 5 = 1400+35 = 1435.  But the correct answer is 1485.  This is the one I'm having to most trouble with.

Yeah, I forgot about the 1-100 part.  I though that it should have been 28 since 2974 is in the 2900-3000 range.

Sorry, but I always make stupid mistakes like that.


A house with a square floor plan is to have each of its four outer walls finished with brick, stone, stucco or wood. If no two adjacent outer walls are to have the same finish in how many ways can the house's walls be constructed? (One such way is to brick on the front and back and stone on the left and right.)

If all four walls have the same finish, there are $ 4!\equal{}24$ different ways to finish the walls.
If two walls have the same finish and the other walls have different finishes, there are $ 4\cdot 2\cdot {3\choose 2}\cdot 2\equal{}4\cdot 2\cdot 3\cdot 2\equal{}48$ different ways.
If two walls have the same finish and the other two walls also have the same finish, then there are $ {4\choose 2}\cdot 2\equal{}6\cdot 2\equal{}12$ different ways.
In total, there are $ 24\plus{}12\plus{}48\equal{}\boxed{84}$ ways.

I got this with the casework solution, but why can't we just say that there are 4 choices for the front most wall. Then 3 choices for each of the side and 2 in the back to get $4 \cdot 3^2 \cdot 2 = 72$?

It gets complicated then. What if one of the sides with 3 has the same choice as the other side with 3? Then instead of 2, you have 3 for the back, etc. Those things screw up the easy count.

So it's then 4x3x3 if the two sides are the same and 4x3x2x2 if the two sides are different (Note 3x2 instead of 3x3 as the two sides have to be different) for a total of 84

What's wrong with this [hide=solution?]
Let's call the walls $A$, $B$, $C$, and $D$ and paint them in that order. Wall $A$ has four choices, wall $B$ has three choices (because it cannot be the same as wall $A$, we aren't considering wall $C$ yet), and wall $C$ has three choices (same logic). So the ways to paint these three walls is $4\cdot3\cdot3=36$. Exactly one quarter of the time, $A$ and $C$ will be painted with the same material, so for that one quarter of the time ($36\cdot\frac{1}{4}=9$ possibilities), $D$ has three choices, so it's $9\cdot3=27$. The other three quarters of the time ($36\cdot\frac{3}{4}=27$ possibilities), $A$ and $C$ have different material, so $D$ has two choices, so it's $27\cdot2=54$. Add the two up, and the answer is $81$.

But clearly, the answer isn't 81! I don't understand why.
[/hide]

I hope this isn't considered revival, I just didn't understand, and I know that we are supposed to discuss with Gamebot. Thanks!
hiabc

ｕｍ　ｔｈｉｓ　ｉｓ　ｍｉｓｐｌａｃｅｄ．　Ｉｔ　ｓｈｏｕｌｄ　ｂｅ　ｉｎ　ｃａｓｅｗｏｒｋ，　ｎｏｔ　ｃｏｕｎｔｉｎｇ　ｗｉｔｈ　ｒｅｓｔｒｉｃｔｉｏｎｓ

Hello,

[hide=Alternate Casework]
The north side has 4 possibilities.
[b]Case 1: East and west side have same finishing.[/b] There are 3 ways to choose the shared finishing and 3 ways to choose the south side's finishing.
[b]Case 2: East and west side have different finishings.[/b] There are 3 ways to choose the east side's, then 2 ways to choose the west side's, then 2 ways to choose the south side's.
Altogether, there are a total of $4 \cdot (3 \cdot 3 + 3 \cdot 2 \cdot 2) = \boxed{84}$ choices. $\square$
[/hide]

[b]@hiabc:[/b] [hide=Hopefully this makes sense.]Let your $A, B, C, D$ be the north, east, south, and west walls of the house, respectively. If $A = C$ then there are 4 ways to choose the shared finishing, 3 ways to choose $B,$ and 3 ways to choose $D.$ If $A \neq C$ then there are 2 ways each to choose $B$ and $D$ finishing. This gives us an answer of $4 \cdot 3 \cdot 3 + 4 \cdot 3 \cdot 2 \cdot 2 = \boxed{84}$ ways. $\square$[/hide]

$\varnothing$

[quote=MarcusRashford][quote=SigmaPiE][hide = solution]
Equivalent to the 4th chromatic number, which is simply 3^4+1*3=84[/quote]

What's a "chromatic" number?[/quote]

a chromatic number is the least number of colours needed to colour a graph's vertices such that no two adjacent vertices has the same colour.


number theory
least common multiple
arithmetic sequence

The two arithmetic sequence, 1, 5, 9, 13, $ \ldots$ and 1, 6, 11, 16, $ \ldots$, have infinitely many terms in common. What is the sum of the first three common terms?

The lists are $ 1 \plus{} 4x$ and $ 1 \plus{} 5x$, respectively, where x >= 0.
Subtract 1 from every term in the list, giving lists of $ 4x$ and $ 5x$, x>=0.
Since the lcm of 4 and 5 is 20, we want to find the first 3 terms of $ 20x$, which are 0, 20, and 40.
Adding those up gives 60, but we still need to add the 3 that we subtracted before.
So the sum is $ 60 \plus{} 3 \equal{} \boxed{63}$.


function
limit
LaTeX
real analysis
real analysis unsolved

Let $f\text{: }[0,4]\rightarrow\mathbb{R}$ be a continuous function. such that $f(0)\cdot f(4)<0$. Show that $(\exists) c\in(0,2)$ such that $1-<f(c^{2})+f(2c)<0$.

If we denote $g: [0,2]\to\Re,g(x)=f(x^2)+f(2x)$ then $g(0)g(4)<0$ so let $g(\alpha)=0,\alpha \in (0,2)$. Suppose $g(0)<0$,$g$ is continous so $\lim_{x\to\alpha}g(x)=g(\alpha)=0$ (don`t know how to write this in $LaTeX$ but in the precent limit we consider it from te left) so that means that $\forall \epsilon >0 |g(x)|<\epsilon$ iff $|x-\alpha|<\delta$. If we take $\epsilon=1$ then $-1<g(x)<1$ and from the fact that $g$ is continous and $g(\alpha)=0$ then it is clearly that exists $\beta$ such that $-1<g(\beta)<0$ ... Is this proof ok?

I think it's ok... I found the following statment:

Let $f: [0,2]\to\mathbb{R}$ a continuous funtion that satisfyes $f(a)\cdot f(b)\leq0$. Then there is at least on point $c\in[a,b]$ such that $f(c)=0$.

Can this help?

Well ... I used that.... It`s called Cauchy-Bolzano theorem. :P

I thought so  :blush: . I'm new with this type of problems... need to practice more  :read:


Let ABC be a triangle inscribed in a circle of radius R

and a, b, c be respectly sides of the triangle ABC.

Prove that:

   R/r >= b/c + c/b

Pd.  r: inradio of the triangle ABC.

R/r >= (b^2+c^2)/bc   (1)

(abc)/4(p-a)(p-b)(p-c)  >= (b^2+c^2)/bc    (2)

We denote   x=2(p-a)        y=2(p-b)           z=2(p-c)
so we have  a=(y+z)/2      b=(x+z)/2         c=(y+x)/2

(2) equivalent to

(x+y)(y+z)(z+x)/4xyz   >=   ((x+z)^2 + (y+x)^2)/(x+z)(y+x)

((x+y)(y+z)(z+x)-8xyz)/4xyz   >=   ((x+z)^2 + (y+x)^2)/(x+z)(y+x)

(x(y-z)^2 + y(z-x)^2 +z(x-y)^2 )/4xyz   >=   (y-z)^2/(x+z)(x+y)

We have (y(z-x)^2 + z(x-y)^2)/4xyz = (z-x)^2/4xz  +  (x-y)^2/4xy
>= (z-y)^2/4x(y+z)

Therefore now we have to prove :

1/4yz + 1/4x(y+z)  >=  1/(x+z)(x+y) (3)

Use the lemma for a,b>0 1/a +1/b >=4/(a+b)  (3) is proved.

After all, (1) is proved.  The equality holds iff ABC is the equalateral triangle


Is everyone who posts here actually from the bay area? I was wondering if possibly there are some ppl here who don't actually live here...

I really do by the way...

This forum isn't restricted to residents of the Bay Area. It's just that most people who don't live here wouldn't really have anything to post here, now, would they?

By the way, I live in Millbrae, which is near the SF Airport.

Yeah!!!!! I'm from the east bay. In a town near concord. Go campo!

eh whats the point of this topic
SPAM SPAM

um

doom

um

dio is a better singer than osbourne
yea

666

how many people here from cupertino? or cupertino schools (as in cusd or parts of fuhsd)?  i am.

here

[size=9]This message is too small. Please make the message longer before submitting.[/size]

[quote="xpmath"]I shouldn't be posting here, as I'm from Ohio, but oh well.[/quote]
dude.
stop.

i'm from IN. i don't normally post here but i am a member of the usergroup and i read threads here.

WELCOME TO THE FORUMS!!!

alrighty Bay Area-ers let's introduce myself.

I'm Ubemaya. I suck at math. And I want Meggy-Meg.

OMFG stop posting that you want Meggy-Meg everywhere on the forums! It's hard now to find a place where you have posted that doesn't mention Meggy-Meg anymore...

And you don't #$%@! suck at math!!!

WELL COMPARED TO PEOPLE AT MY SCHOOL I DON'T BUT COMPARED TO YOU I'M LIKE A PRESCHOOLER!!!!!

[quote="person w/o a username"]not that you want me to, austin :P[/quote]

What's your real name?

EDIT:
[quote="xscapezaer"]OMFG stop posting that you want Meggy-Meg everywhere on the forums! It's hard now to find a place where you have posted that doesn't mention Meggy-Meg anymore...

And you don't #%@! suck at math!!![/quote]

Eh, just ignore him.

Besides, you are starting to sound like Ubemaya the II, anyways, so you shouldn't be talking... :rotfl:
You say that you're going to be on the C team, but I think you're going to be on the B team, and if you really work hard, you have a chance of making A team.

wtf?! [b][i]Me[/i][/b] sound like Ubemaya v.2?! And what about you?! Remember those "enlightening words of wisdom" you put on your calculator :roll: or the...well, you know what i'm talking about.

EDIT: and I still don't have a chance of doing well on ARML...so shaddup.

Yeah, that is what the world may become.

[quote="CheeseIsGood"]That is UNPOSSIBLE!

Then the world will suck, there will be no more food or any other products.
Maybe there would, until all that we have is used up, and then there will be nothing left, and everyone will die, since no new things are being made.[/quote]

nah I'll just grow some food and farm like a farmer in those days. It's going to be hard work. I should start now. Lemme save up my apples

[quote="moogra"]
nah I'll just grow some food and farm like a farmer in those days. It's going to be hard work. I should start now. Lemme save up my apples
[/quote]

Awesome, I want a banana.

I'm sorry. I don't own a banana tree. Maybe when I get one, I'll send you a free banana.

[quote="happyme"]Someone I know gave me this account, apparently he didn't need AoPS anymore.

I go to Redwood Middle...probably going to Mitty or Harker for High School though.[/quote]
Oh hi! Were you at Mathcounts last year?

[quote="happyme"]...apparently he didn't need AoPS anymore.[/quote]
I sense arrogance.

Anyway, yes, I'm in the bay area. I go to Hopkins (in Fremont), as well as lifeisacircle, superquark1123, psichilikapati, and another person who's username i dont know (the first two people are people who joined AoPS because i told them about it and pestered them until they made accounts and bought all the books, lol). bowei is from mission high, where im going in 2 years time. aznphatso is the brother of the person who's username i dont know that used to be in mission but is now in MIT. yeah.

Okay.

End of spam message.

[quote="archimedes1"][quote="happyme"]
I go to Redwood Middle...probably going to [b]Mitty[/b] or Harker for High School though.[/quote]
we need more math people![/quote]

you're probably good enough for your whole high school...

Wait... who's happyme now?

Uh my friend Ho (I'm not sure if he wanted people to know his first name o.o) gave me this.

I am currently nobody, and I suck at math.

I was just kidding. But I bet I suck more.

haha tony had an account? lol
and now he quit and gave it to you, wow
how do you know tony?

I don't know who tony is.

o.o

What grade are you in Mission? I know a lot of people that went to Hopkins/Mission.

bowei is a freshman

seeing as i'm not in mission, i don't know a lot of people. but i do have a brother in mission. is tony a junior? i can remember some times when a guy called tony is mentioned.

oopsie lol i thought when you said ho you meant tony ho
my bad but i would have expected him to have an aops account
theres only one more person i know that goes onto aops in hopk/mission
its mathclass, i dunno if he ever posted here


geometry
circumcircle
geometry proposed
moving points

Let a triangle ABC and a given line $ d$. A point M moves on d. The perp line from the midpoint of MB to $ d$ meets AB at P. The perp line from the midpoint of MC to $ d$ meets AC at Q. Prove that the perp line from M to PQ always passes through a fixed point when M moves on the line $ d$.
[hide="Consequences"]Two consequences ( one is Pr 7, VMO 2008 and other is my own problem ) were posted at [url=http://www.mathlinks.ro/viewtopic.php?t=185640]here[/url]. They refer to two particular cases of the above problem.[/hide]

Hello!

Your problem can be generalized:

Let $ ABC$ be a triangle, $ d$, $ g$ are two given lines $ ($ the line $ g$ is not parallel to the lines $ AB$, $ AC$ or $ d$ $ )$ and a point $ M$ moves on the line $ d$.
$ S\in MB$, $ T\in MC$ such that $ ST\parallel{}BC$ and $ \frac {SM}{SB} \equal{} k$  $ ($ $ k\in\Re^{*}_{ \plus{} }$, $ k$ is a fixed number $ )$.
$ X,Y\in d$ such that $ SX\parallel{}TY\parallel{}g$, $ \left\{P\right\} \equal{} SX\cap AB$ and $ \left\{Q\right\} \equal{} TY\cap AC$.
The circumcircles of the triangles $ PXM$ and $ QYM$ intersect again at the point $ N$. Prove that the line $ MN$
always passes through a fixed point when the point $ M$ moves on the line $ d$.  :)

Oh, you have a wonderful generalization, Petry.
But I don't have enough time to check it. If your generalization is true, I hope that you will introduce us about its proof. For my problem, I have a synthetic proof, but maybe your solution to the generalized problem is very nice and I am really curious about it.
Sincerely, mr.danh

[quote="mr.danh"][quote="Petry"]I don't have a proof :(, but I will try  :) ![/quote]
 :?:  Why can you generalize my problem?  :)  Are you sure that your generalization is true?[/quote]
As I can see by the drawing, I think this challenge problem is true.

I will check more carefully before a definite answer (probably, later today).

Kostas Vittas.

I verified the generalization with "Compass and Ruller" program, it seems to be true, so maybe one of the Mathlinkers will prove this problem.
   Sincerely, Petry

@mr.danh: I have seen your solution to your extensive problem in a link. It is very nice  :lol:
@vittasko:
[quote="vittasko"]As I can see by the drawing, I think this challenge problem is true.
I will check more carefully before a definite answer (probably, later today).
Kostas Vittas. [/quote]
Did you solve the generalized problem of Petry? Why don't you introduce to your proof ?

[quote="thaithuan_GC"][quote="vittasko"]As I can see by the drawing, I think this challenge problem is true.
I will check more carefully before a definite answer (probably, later today).
Kostas Vittas. [/quote]
Did you solve the generalized problem of Petry? Why don't you introduce to your proof ?[/quote]
I can only say by the drawing that it is true the problem in generalization of [b][size=100]Petry[/size][/b], but unforunately, I have not in mind any proof of this nice result dear [b][size=100]thaithan_GC[/size][/b] and sorry for the late answer.

Kostas Vittas.

Hello!

Finally, I proved the generalization of the proposed problem of mr.danh.
First we prove a generalization of the Carnot's theorem, then a aplication of this generalization and at last the generalization of the proposed problem.

The generalization of the Carnot's theorem:

Let $ ABC$ be a triangle and the points $ X$, $ Y$, $ Z$ are the feet of the lines equally inclined to the sidelines $ BC$, $ CA$, $ AB$ from a point $ P$
($ \angle\left(PX;BC\right) \equal{} \angle\left(PY;CA\right) \equal{} \angle\left(PZ;AB\right) \equal{} \alpha$; this angles have the same orientation). Prove that
$ \left|BX^{2} \minus{} CX^{2} \plus{} CY^{2} \minus{} AY^{2} \plus{} AZ^{2} \minus{} BZ^{2}\right| \equal{} \left|4\cdot ctg\alpha\cdot\left[ABC\right]\right|$.
The reciprocal is also true.

$ \left[...\right]$ - the area of the triangle $ ...$

Solution:
$ PB^{2} \equal{} PX^{2} \plus{} BX^{2}\mp 2\cdot PX\cdot BX\cdot cos\alpha\Rightarrow PB^{2} \equal{} PX^{2} \plus{} BX^{2}\mp 4\cdot ctg\alpha\cdot\left[PXB\right]$    (*)
$ PC^{2} \equal{} PX^{2} \plus{} CX^{2}\pm 2\cdot PX\cdot CX\cdot cos\alpha\Rightarrow PC^{2} \equal{} PX^{2} \plus{} CX^{2}\pm 4\cdot ctg\alpha\cdot\left[PXC\right]$     (**)
(*), (**) $ \Rightarrow PB^{2} \minus{} PC^{2} \equal{} BX^{2} \minus{} CX^{2}\mp 4\cdot ctg\alpha\cdot\left[PBC\right]\Rightarrow$ (1)
Similarly $ PC^{2} \minus{} PA^{2} \equal{} CY^{2} \minus{} AY^{2}\mp 4\cdot ctg\alpha\cdot\left[PCA\right]$(2)
and       $ PA^{2} \minus{} PB^{2} \equal{} AZ^{2} \minus{} BZ^{2}\mp 4\cdot ctg\alpha\cdot\left[PAB\right]$(3)
(1), (2), (3) $ \Rightarrow BX^{2} \minus{} CX^{2} \plus{} CY^{2} \minus{} AY^{2} \plus{} AZ^{2} \minus{} BZ^{2} \equal{} \pm 4\cdot ctg\alpha\cdot\left[ABC\right]\Rightarrow$
$ \Rightarrow\left|BX^{2} \minus{} CX^{2} \plus{} CY^{2} \minus{} AY^{2} \plus{} AZ^{2} \minus{} BZ^{2}\right| \equal{} \left|4\cdot ctg\alpha\cdot\left[ABC\right]\right|$.

A aplication of the generalization of the Carnot's theorem:

Let $ ABC$ be a triangle and $ d$ a line. The points $ D$, $ E$, $ F$ are the feet of the lines equally
inclined to the line $ d$ from the points $ A$, $ B$, $ C$ ($ \angle\left(AD;d\right) \equal{} \angle\left(BE;d\right) \equal{} \angle\left(CF;d\right) \equal{} \alpha$;
this angles have the same orientation, so $ AD\parallel{}BE\parallel{}CF$).
The points $ X$, $ Y$, $ Z$ are the feet of the lines equally inclined to the lines $ BC$, $ CA$, $ AB$ from the points $ D$, $ E$, $ F$ respectively
($ \angle\left(DX;BC\right) \equal{} \angle\left(EY;CA\right) \equal{} \angle\left(FZ;AB\right) \equal{} \minus{} \alpha$; this angles have the same orientation).
The angles $ \angle\left(AD;d\right)$ and $ \angle\left(DX;BC\right)$ have different orientation. Then prove that the lines $ DX$, $ EY$, $ FZ$ are concurrent.

$ \left[...\right]$ - the area of the triangle or quadrilateral $ ...$

Solution:
$ DB^{2} \equal{} DX^{2} \plus{} BX^{2}\mp 2\cdot DX\cdot BX\cdot cos\alpha\Rightarrow DB^{2} \equal{} DX^{2} \plus{} BX^{2}\mp 4\cdot ctg\alpha\cdot\left[DXB\right]$     (*)
$ DC^{2} \equal{} DX^{2} \plus{} CX^{2}\pm 2\cdot DX\cdot CX\cdot cos\alpha\Rightarrow DC^{2} \equal{} DX^{2} \plus{} CX^{2}\pm 4\cdot ctg\alpha\cdot\left[DXC\right]$    (**)
(*), (**) $ \Rightarrow DB^{2} \minus{} DC^{2} \equal{} BX^{2} \minus{} CX^{2}\mp 4\cdot ctg\alpha\cdot\left[DBC\right]\Rightarrow$
$ \Rightarrow BX^{2} \minus{} CX^{2} \equal{} DB^{2} \minus{} DC^{2}\pm 4\cdot ctg\alpha\cdot\left[DBC\right]$   (1)

$ DB^{2} \equal{} DE^{2} \plus{} BE^{2}\pm 2\cdot DE\cdot BE\cdot cos\alpha\Rightarrow DB^{2} \equal{} DE^{2} \plus{} BE^{2}\pm 4\cdot ctg\alpha\cdot\left[DEB\right]$ ($ \bullet$)
$ DC^{2} \equal{} DF^{2} \plus{} CF^{2}\mp 2\cdot DF\cdot CF\cdot cos\alpha\Rightarrow DC^{2} \equal{} DF^{2} \plus{} CF^{2}\mp 4\cdot ctg\alpha\cdot\left[DFC\right]$  ($ \bullet\bullet$)
($ \bullet$), ($ \bullet\bullet$) $ \Rightarrow DB^{2} \minus{} DC^{2} \equal{} DE^{2} \minus{} DF^{2} \plus{} BE^{2} \minus{} CF^{2}\pm 4\cdot ctg\alpha\cdot\left(\left[DEB\right] \plus{} \left[DFC\right]\right)$   (2)

(1),(2)$ \Rightarrow BX^{2} \minus{} CX^{2} \equal{} DE^{2} \minus{} DF^{2} \plus{} BE^{2} \minus{} CF^{2}\pm 4\cdot ctg\alpha\cdot\left[BEFC\right]$   (3)
Similarly $ CY^{2} \minus{} AY^{2} \equal{} EF^{2} \minus{} ED^{2} \plus{} CF^{2} \minus{} AD^{2}\mp 4\cdot ctg\alpha\cdot\left[CFDA\right]$   (4)
and $ AZ^{2} \minus{} BZ^{2} \equal{} FD^{2} \minus{} FE^{2} \plus{} AD^{2} \minus{} BE^{2}\mp 4\cdot ctg\alpha\cdot\left[ADEB\right]$   (5)

(3), (4), (5) $ \Rightarrow BX^{2} \minus{} CX^{2} \plus{} CY^{2} \minus{} AY^{2} \plus{} AZ^{2} \minus{} BZ^{2} \equal{} \mp 4\cdot ctg\alpha\cdot\left[ABC\right]\Rightarrow$
$ \Rightarrow\left|BX^{2} \minus{} CX^{2} \plus{} CY^{2} \minus{} AY^{2} \plus{} AZ^{2} \minus{} BZ^{2}\right| \equal{} \left|4\cdot ctg\alpha\cdot\left[ABC\right]\right|$.
From the reciprocal of the generalization of the Carnot's theorem we have that
the lines $ DX$, $ EY$, $ FZ$ are concurrent.

The generalization of the proposed problem:

Let $ ABC$ be a triangle, $ d$, $ g$ are two given lines (the line $ g$ is not parallel to the lines $ AB$, $ AC$ or $ d$) and a point $ M$ moves on the line $ d$.
$ S\in MB$, $ T\in MC$ such that $ ST\parallel{}BC$ and $ \frac {SM}{SB} \equal{} k$, ($ k\in\Re^{*}_{ \plus{} }$, $ k$ is a fixed number).
$ X,Y\in d$ such that $ SX\parallel{}TY\parallel{}g$, $ \left\{P\right\} \equal{} SX\cap AB$ and $ \left\{Q\right\} \equal{} TY\cap AC$.
The circumcircles of the triangles $ PXM$ and $ QYM$ intersect again at the point $ N$. Prove that the line $ MN$
always passes through a fixed point when the point $ M$ moves on the line $ d$.

Solution:
$ \angle\left(g;d\right) \equal{} \alpha$ and it's easy to prove that $ N\in PQ$.

Construct $ AA'\parallel{}g$, $ A'\in d$, so $ AA'\parallel{}PX\parallel{}QY$.
The points $ D$, $ E$, $ F$ are the feet of the lines equally inclined to the lines $ PQ$, $ QA$, $ AP$ from the points $ A'$, $ X$, $ Y$ respectively
($ \angle\left(A'D;PQ\right) \equal{} \angle\left(XE;QA\right) \equal{} \angle\left(YF;AP\right) \equal{} \minus{} \alpha$; this angles have the same orientation).
The angles $ \angle\left(AA';d\right)$ and $ \angle\left(A'D;PQ\right)$ have different orientation.
From the aplication of the generalization of the Carnot's theorem we have that the lines $ A'D$, $ XE$, $ YF$ are concurrent. $ \left\{W\right\} \equal{} A'D\cap XE\cap YF$.
It's easy to prove that $ MN\parallel{}A'D$.

Construct $ BB'\parallel{}CC'\parallel{}g$, $ B',C'\in d$. Through the points $ B'$, $ C'$ construct parallels to the lines $ XE$, $ YF$ respectively
and let $ V'$ be their point of intersection.
It's easy to prove that the points $ M$, $ W$, $ V'$ are collinear and  $ \frac {WM}{WV'} \equal{} \frac {XM}{XB'} \equal{} \frac {SM}{SB} \equal{} k$.

$ \left\{V\right\} \equal{} V'A'\cap MN$
Construct $ VB"\parallel{}V'B'$, $ B"\in d$ and $ VC"\parallel{}V'C'$, $ C"\in d$.
$ \frac {A'B"}{A'B'} \equal{} \frac {A'V}{A'V'} \equal{} \frac {WM}{WV'} \equal{} k$. Similarly $ \frac {A'C"}{A'C'} \equal{} \frac {A'V}{A'V'} \equal{} \frac {WM}{WV'} \equal{} k$

The points $ A'$, $ B'$, $ C'$ are fixed, $ k$ is a fixed number, so the points $ B"$, $ C"$ are fixed. (1)
$ VB"\parallel{}V'B'\parallel{}XE\Rightarrow\angle\left(VB";AC\right) \equal{} \alpha$ (2)
$ VC"\parallel{}V'C'\parallel{}YF\Rightarrow\angle\left(VC";AB\right) \equal{} \alpha$ (3)

(1),(2),(3)$ \Rightarrow V$ is a fixed point.

Wow! Thanks Petry for your nice complete proof of the generalized problem. And I will be very glad if someone have a synthetic proof.  :)

Here's a proof using moving points for this beautiful problem: Let $X,Y,Z$ be the feet of perpendicular from the points $A,B,C$ respectively to $d$. Suppose the line through $Y$ perpendicular to $AC$ meets the line through $Z$ perpendicular to $AB$ at $T$, and let $T'$ be the reflection of $T$ in the point $X$. We claim that $T$ is the desired fixed point. Fix the line $d$, and the points $A,B,M,Z$ and animate $C$ on the line through $Z$ perpendicular to $d$ (call this line $\ell$). Note that $X,Y,P$ automatically get fixed, and that there is no loss of degrees of freedom in doing this. Let $U$ be the midpoint of $CM$. Then the line $UQ$ is simply the perpendicular bisector of $MZ$. As $Z$ is fixed, so this means that $UQ$ is also fixed. Then $$C \mapsto AC \mapsto AQ \mapsto Q \mapsto PQ \mapsto P\infty_{\perp_{PQ}} \mapsto \infty_{\perp_{PQ}}$$ is a projective map (where the last two maps follow from rotation of angle $90^{\circ}$ about $P$). Now, as both $AB$ and $Z$ are fixed, so the line through $Z$ perpendicular to $AB$ (i.e. $ZT$) is also fixed. This gives $$C \mapsto AC \mapsto \infty_{\perp_{AC}} \mapsto Y\infty_{\perp_{AC}} \mapsto YT \mapsto T \mapsto T'$$
is also projective (where the last map is reflection in $X$). Thus, it suffices to prove that $M,T',\infty_{\perp_{PQ}}$ are collinear for three positions of $C$. Take $C=\infty_{\ell}$ and $C=AB \cap \ell$ for the first two positions. One can easily check that these work. Finally, take $C$ as the foot of perpendicular from $B$ to $\ell$ (i.e. $BC \parallel d$). As $T$ is the orthopole of $\triangle ABC$ with respect to $d$ (by the definition of the point $T$), so the line $TX$ must be perpendicular to $BC$, or in this case, to $d$ as well. So this means that $X'$ is simply the reflection of $X$ in $d$. Now, notice that if we move the line $d$ with constant velocity, keeping it parallel to $BC$, and without changing the distances $MY$ and $MZ$, then the points $P,Q$ remain fixed (as they lie on perpendicular bisectors of $MY$ and $MZ$ respectively), while the line $MT'$ simply moves parallel to its original position. So it suffices to prove the desired conclusion for just one position of the line $d$. Take $d=BC$ (obviously :D). Then $T$ becomes the orthocenter of $\triangle ABC$, while $T'$ is the point where the $A$-altitude meets $\odot (ABC)$ again. Suppose $T'M$ meets $\odot (ABC)$ again at $N$. Then $$2\angle MNB=2\angle T'NB=2\angle T'AB=\angle MPB \Rightarrow \text{As PM=PB, so } N \in \odot (P,PM)$$
Similarly, $N$ lies on $\odot (Q,QM)$, which in turn gives that $T'$ lies on the radical axis of $\odot (P,PM)$ and $\odot (Q,QM)$. Thus, $T'M \perp PQ$, as desired. Hence, done. $\blacksquare$


I was wondering how to do this without guess and check.

#15
15. The integers from 1 through 15 are written in numerical order in pencil going clockwise around a circle. A student begins moving clockwise around the circle erasing every third integer that has not yet been erased until only the integer 11 remains. Which integer did the student erase first?

#27
A rectangular band formation is a formation with m band members in each of r rows, where m and r are integers. A particular band has less than 100 band members. The director arranges them in a rectangular formation and finds that he has two members left over. If he increases the number of members in each row by 1 and reduces the number of rows by 2, there are exactly enough places in the new formation for each band member. What is the largest number of members the band could have?

for #27, (m+1)(r-2)=mr+2
r-2m=4
find the largest m such that mr+2<100 which happens to be m=6, r=16
so the answer 6*16+2=98

[quote="math154"]Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=32450337&t=249626]here[/url] for number $ 15$.

[hide="27"]
Let there be some $ a$ and $ b$ so that $ ab\plus{}2\equal{}mr$, which satisfies the first formation. Then $ mr\equal{}(a\plus{}1)(b\minus{}2)\equal{}ab\minus{}2a\plus{}b\minus{}2\implies mr\minus{}ab\equal{}2\equal{}\minus{}2a\plus{}b\minus{}2\implies b\minus{}2a\equal{}4$ as well. $ mr\equal{}(a\plus{}1)(b\minus{}2)$ is maximized when $ b$ is maximized, since both $ a$ and $ b$ increase as $ b$ increases. Testing, $ (a,b)\equal{}(6,16)$ is the largest pair that works, so the answer is $ mr\equal{}(a\plus{}1)(b\minus{}2)\equal{}7\cdot14\equal{}\boxed{98}$.
[/hide][/quote]
$ mr=(a+1)(b-2)=ab-2a+b-2\implies mr-ab=2=-2a+b-2\implies b-2a=4 $

What's the point of the above step if in the end you're looking at $(a+1)(b-2)=mr$?

[quote=Aequilipse]Any way to solve #27 using mods?[/quote]

Sorry for the huge bump (oops 4 year bump)

Can someone explain how you would do #27 with mods?

I started off with something like this. We would $mr + 2 = (m + 1)(r - 2)$. Which would mean that (m + 1)(r - 2) is either $1 \pmod{2}$ (this is when $mr$ is even, and $0 \pmod{2}$ when $mr$ is odd. I am not sure on how to go from here, can someone explain?

I remember doing that sprint, #15 was a pain in the neck


trigonometry
logarithms
complex analysis
calculus
calculus computations

Find the Taylor series centered at $1$ for $f(x)=\tan^{-1}(x).$ Determine the radius of convergence of that series.

[hide="Hint"]Somwhere in this forum.

[hide="OK I give up"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42805]Here.[/url] [/hide][/hide]

Instead of explictly calculating the series, I give another way to find the radius of convergence, using Complex Analysis
[hide="Radius"]For the series $\arctan(z)$ with $z \in \mathbb{C}$ we can calculate the radius of convergence as the shortest distance from the centre, to a singularity (Singularity or pole...I always get confused). Anyway this is the distance from 1 (the centre) to $i$, or $\sqrt{2}$.
Hope its correct![/hide]

[quote="Kent Merryfield"]That is correct, as far as it goes. We're still looking for the series itself.

[hide]The arctangent has singularities at both $i$ and $-i$ and your argument should acknowledge both of them. Of course, the distance from $1$ to either one is the same, $\sqrt{2}.$

The singularities of the arctangent at $\pm i$ are not poles. They are branch points. They're the same kind of singularities that the logarithm has at zero. (One can even express the arctangent in terms of logarithms.)[/hide][/quote]

Thank you kent. How can we tell we we have a pole or a branch point?

BTW: The answer to the original problem is
[hide]$\frac{\pi}{4}+\frac{1}{2}(x-1)-\frac{1}{4}(x-1)^2+\frac{1}{12}(x-1)^3+ \cdots $[/hide]


Which program(s) is/are the best for drawing geometric figures?


The mean of 20 numbers is 30, and the average of 30 other numbers is 20.  What is the average of all 50 numbers?

$a.) 23 b.) 24 c.) 25 d.) 26 e.) 27$

[hide]
$\frac{30 \cdot 20 + 20 \cdot 30}{50}=\frac{1200}{50}=24$, so B
[/hide]
I just took the AMC-12 2005 and the same problem was on it, word for word. :o  :o  :o

[quote="mathnerd314"]I just took the AMC-12 2005 and the same problem was on it, word for word. :o  :o  :o[/quote]
I think the 15 or so problems are the same on the AMC-10 and AMC-12.


Find a series $ \sum_{n}(\minus{}1)^na_n$ with these properties:

1) $ a_n>0$ for all sufficiently large $ n.$

2) $ \lim_{n\to\infty}na_n\equal{}1.$ (That is, the terms "look like" $ \frac1n.$)

3) The alternating series diverges.

[hide]
how about $ \frac {ln(n) \plus{} ( \minus{} 1)^n}{nln(n)}$

also, doesn't requirement 2 imply requirement 1?
[/hide]

[quote="0714446459923"]also, doesn't requirement 2 imply requirement 1?[/quote]
True, that was redundant. I think I changed my mind about the exact phrasing in the middle of writing down the problem, and accidentally introduced the redundancy.

Nice example.

Of course, what I was pushing here is that the requirement that the absolute values decrease monotonically cannot be dispensed with as a hypothesis of the alternating series test.


function
inequalities
trigonometry
calculus
integration
real analysis
complex numbers

Prove the following inequality.
$\sum_{k=1}^{1001}\sin\frac{k\pi}{1002}<\frac{2004}{\pi}$

It's not very hard. The left side is less than 2 times the integral from 1 to 501 of the function sin(xPi/1002) plus 1. This expression is 2*1002*cos(Pi/1002)/Pi+1 which is more than 2004/Pi which can be proved with a calculator.

Misha

This is a very well kown trick for summing trigs.

We have\begin{eqnarray*}
&& 2\sin\frac{\pi}{2004} \sum_{k=1}^{1001} \sin\frac{k\pi}{1002} \\
&=& \sum_{k=1}^{1001} \cos\frac{(2k-1)\pi}{2004} - \cos\frac{(2k+1)\pi}{2004} \\
&=& \cos\frac{\pi}{2004} - \cos\frac{2003\pi}{2004}
&=& 2\cos\frac{\pi}{2004}
\end{eqnarray*}

So \[
\sum_{k=1}^{1001} \sin\frac{k\pi}{1002} = \cot\frac{\pi}{2004}
\]

And the result come immediately from the fact that $\tan x > x$ in $(0, \pi/2)$.

Two comments:

The first is a variation on what Misha123 said.

$\frac{\pi}{1002}\sum_{k=1}^{1001}\sin\left(\frac{k\pi}{2002}\right)$ is both the right-hand endpoint Riemann sum and the left-hand endpoint Riemann sum (with 1002 subintervals of equal width) for the integral $\int_0^{\pi}\sin x dx$. The average of the right and left-hand Riemann sums is the trapeziodal rule estimate. Since the function is concave downward on the whole interval, the trapezoidal rule estimate is less than the actual integral. Hence,
$\frac{\pi}{1002}\sum_{k=1}^{1001}\sin\left(\frac{k\pi}{1002}\right) < \int_0^{\pi}\sin x dx = 2$, which proves the inequality quite directly. You could even use the error estimate for the trapezoidal rule to determine how close together the two sides must be.

The second is a comment on what billzhao said. This exact summation is both familiar and important in Fourier analysis. The exact formula can be recovered without remembering as many trigonometric identities by employing complex numbers and the formula for the sum of a finite geometric series.
$\sum_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)= \sum_{k=0}^{n-1}\sin\left(\frac{k\pi}{n}\right)$
$=Imag\left[\sum_{k=0}^{n-1}e^{ik\pi/n}\right] = Imag\left[\frac{1-e^{i\pi}}{1-e^{i\pi/n}}\right] = Imag\left[\frac{2}{1-e^{i\pi/n}}\right]$
$=Imag\left[\frac{2e^{\frac{-i\pi}{2n}}} {e^{\frac{-i\pi}{2n}}-e^{\frac{i\pi}{2n}}}\right] = \frac{\cos\left(\frac{\pi}{2n}\right)}{\sin\left(\frac{\pi}{2n}\right)}$, which is the same formula billzhao found.

Thank you for your comment. I used the trapezoidel rule several times indirectly but I never realized the real matture. Your proof is of course the 'perfect' proof for this problem.

Misha

Thank you for your replies!  Sorry for my delaying reply.
Indeed, my solution is the same as that of Kent Merryfield.

kunny


abstract algebra
geometry
calculus
integration

Post your jokes here! Here's two to get started.

Old maid's laughter--He! he! he!

A flea and a fly in a flue,
Were improsoned, so what could they do?
Said the fly, "let us flee!"
"Let us fly!" said the flea,
And they flew through a flaw in the flue.

Why are they called apartements when they are all stuck together?

What's nice and works for the phone company? A deferential operator!!

What's purple and commutes? An abelian grape!!

Two wrongs don't make a right... but two Wrights make an airplane...

A mathematician decides that his job as a professor isn't paying enough, and decides to be a plumber instead. However, he lies about his educational experience, saying that he dropped out of seventh grade, because he thinks that the plumbing company doesn't like educated people. Well, one evening, the company decides to round up all the plumbers and enroll them in an eighth grade level class. It just happens that the first class of the evening is math, the first formula is to find the area of a circle, and the mathematician is called up to the board to demonstrate his knowledge. The mathematician suddenly has a mental blank and forgets, so he fills the board with equations trying to derive the formula. But he keeps getting $A=-\pi r^{2}$. Thoroughly confused, he desperately looks to the other plumbers for help, only to find them all mouthing, "Switch the limits of the integral!" (No offense to plumbers :D)

:roll:

[quote="K81o7"]One off the top of my head, and not so good

A man goes into a bar.  He notices a woman sitting alone at a table, so he goes to her.

Man: Can I buy you a drink?

She replies at the top of her lungs "NO I WILL NOT SLEEP WITH YOU TONIGHT!"  Everyone in the bar stares as the man crawls away to a table in the corner and sits dejected.  A few minutes later, the woman goes to the man.

Woman: I'm sorry about earlier.  I'm a psychologist, and I wanted to see how you'd react to that situation.

Man: WHAT DO YOU MEAN TWO HUNDRED DOLLARS?![/quote]

:rotfl:

He: I'm a photographer. I've been looking for a face like yours.
She: I'm a plastic surgeon. I've been looking for a face like yours.

-There are two groups of people in the world; those who believe that the world can be divided into two groups of people, and those who don't.

-The less you know, the more you make.
Proof:
Postulate 1: Knowledge is Power.
Postulate 2: Time is Money.
As every engineer knows: Power = Work / Time
And since Knowledge = Power and Time = Money
It is therefore true that Knowledge = Work / Money .
Solving for Money, we get:
Money = Work / Knowledge
Thus, as Knowledge approaches zero, Money approaches infinity, regardless of the amount of Work done.

-Teacher: "Who can tell me what 7 times 6 is?"
Student: "It's 42!"
Teacher: "Very good! - And who can tell me what 6 times 7 is?"
Same student: "It's 24!"

What do vampires say when they kiss each other?
Ouch!

Why did the monster take six hours to finish the small book?
He was full!

[quote="lingomaniac88"]Why is the alphabet in that order?  Is it because of that song?[/quote]

Goooooooooood question...

Philosophy Explained in Two-Cow Terms!!

Socialism: You have two cows. You give one to your neighbor.
Communism: You have two cows. You give them to the government and the government gives you some milk.
Fascism: You have two cows. The government confiscates them and sells you some milk.
Totalitarianism: You have two cows. The government shoots you and keeps the cows.
Democracy: You have two cows. They outvote you 2-1 to ban all meat and dairy products.


You must have the condition $a,b,c\ge 0$
Then In fact you can let:
$a=2\cos A,b=2\cos B,c=2\cosh$
$A,B,C$ is three angles of an triangle.
The rest is easy.

[quote="Stuart"]no one nows how to solve it[/quote]

Don't generalize from yourself to others!

http://www.mathlinks.ro/Forum/viewtopic.php?t=25977

  darij

[quote="arqady"][quote="Stuart"]if a²+b²+c²+abc=4, a,b,c>=0 prove a+b+c<=3[/quote]
The following inequality is stronger:
let $a,$ $b$ and $c$ are non-negative numbers such that $a^{2}+b^{2}+c^{2}+1.5abc=4.5.$
Prove that $a+b+c\leq3.$ :wink:[/quote]

Dear [b]arqady[/b], can you give a solution to your problem? :roll:

[quote="cefer"]
Dear [b]arqady[/b], can you give a solution to your problem? :roll:[/quote]
For you, cefer :lol:
Let $a+b+c>3,$  $a=kx,$ $b=ky$ and $c=kz$ such that $x+y+z=3.$
Hence, $k>1$ and $4.5=a^{2}+b^{2}+c^{2}+1.5abc=k^{2}(x^{2}+y^{2}+z^{2})+1.5k^{3}xyz>x^{2}+y^{2}+z^{2}+1.5xyz.$
Thus, $4.5>x^{2}+y^{2}+z^{2}+1.5xyz.$ It's contradiction since, $4.5\leq x^{2}+y^{2}+z^{2}+1.5xyz\Leftrightarrow\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)\geq0.$ :)
Also see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123877

[quote="arqady"]For you, cefer :lol:
[/quote]

Thanks, very much


geometry
integration
calculus
calculus computations

I need help in solving this..... Please and Thank You
Find the area of the region in the first quadrant bounded by the line y=8x, the line x=1 , the curve y=1/sqaureroot(x), and the x-axis.

Hi :) !!!

are you want to say thatt?????

calcul: $ \int_1^x \left|8t \minus{} \frac{1}{\sqrt{t}} \right|dt$

and thanks :)

How to start: draw all the curves given to you.  Shade in the region described by the problem -- there's only one portion of the plane that has all four of those curves as part of its boundary.  Then figure out an integral (or sum of integrals) that gives you this area.  (mathema*'s answer isn't right -- he or she made false assumptions about what the region looks like.  Draw the picture and you won't make the same mistake.)

[quote="JBL"]How to start: draw all the curves given to you.  Shade in the region described by the problem -- there's only one portion of the plane that has all four of those curves as part of its boundary.  Then figure out an integral (or sum of integrals) that gives you this area.  [b](mathema*'s answer isn't right -- he or she made false assumptions about what the region looks like.  Draw the picture and you won't make the same mistake.[/b])[/quote]

HI JBL

I didn't give any answer and i didn't understood the questions of "dizizviet". than i just asked (he or she) !!!!

and thanks

PS: i'm a man ((he))

Instead of "Are you say thatt?" use "Is this what you mean?" for the question you intended.  Also, "calcul" is not an English word -- the verb you want is "calculate."

dizizviet gave four curves, $ y \equal{} 8x$, $ x \equal{} 1$, $ y \equal{} 0$ and $ y \equal{} \frac{1}{\sqrt{x}}$.  These curves cut the plane into several regions.  One of these regions is bounded by all four curves.  dizizviet is asking for the area of this region.

Find where:
$ 8x\equal{}\frac{1}{\sqrt{x}}$
$ x^\frac{3}{2}\equal{}\frac{1}{8}$
$ \sqrt{x}\equal{}\frac{1}{2}$
$ x\equal{}\frac{1}{4}$

Thus, the area under the curves is equal to:

$ \int^{\frac{1}{4}}_08x \plus{} \int_{\frac{1}{4}}^1 \frac{1}{\sqrt x}$
$ \equal{}4 \left (\frac{1}{4} \right )^2  \minus{}4(0)^2\plus{}2 \sqrt{1} \minus{} 2 \sqrt{\frac{1}{4}}$
$ \equal{}\frac{5}{4}$


hi,
we consider the polynominal with integer coeffitients
\[ P(x) \equal{} a_{0}x^{n} \plus{} a_{1}x^{n\minus{}1} \plus{} ... \plus{} a_{n},\]
and a prime $ p$ s.t. $ a_{n} \not\equiv 0 \bmod{p}$. If there exists $ n\plus{}1$ integer numbers $ \alpha_{1}, ..., \alpha_{n\plus{}1}$ with $ P(\alpha_{r}) \equiv 0 \bmod{p}, \forall r \equal{} 1, ..., n\plus{}1$, prove that there exists two of the $ \alpha_{i}, \alpha_{j}$ s.t. $ \alpha_{i} \equiv \alpha_{j} \bmod{p}$.

[quote="doicu"]hi,
we consider the polynominal with integer coeffitients
\[ P(x) \equal{} a_{0}x^{n} \plus{} a_{1}x^{n \minus{} 1} \plus{} ... \plus{} a_{n},
\]
and a prime $ p$ s.t. $ a_{n} \not\equiv 0 \bmod{p}$. If there exists $ n \plus{} 1$ integer numbers $ \alpha_{1}, ..., \alpha_{n \plus{} 1}$ with $ P(\alpha_{r}) \equiv 0 \bmod{p}, \forall r \equal{} 1, ..., n \plus{} 1$, prove that there exists two of the $ \alpha_{i}, \alpha_{j}$ s.t. $ \alpha_{i} \equiv \alpha_{j} \bmod{p}$.[/quote]
We will prove $ P(x)$ has max $ n$ root $ mod p$(*)
$ n \equal{} 1$,easy
Suppose that we have (*) for $ n \equal{} k$
We will prove that this is true for $ n \equal{} k \plus{} 1$
Consider $ degP(x) \equal{} k \plus{} 1$ and
$ P(x) \equiv 0(mod p)$(**)
If (**) hasn't root,we done
If (**) has a root is $ x_o$ then we have
$ P(x) \equiv (x \minus{} x_o)Q(x) \plus{} R(x)(mod p)$
$ degR(x) < 1$ so $ R(x) \equiv const \equiv c$=>$ degQ(x) \equal{} k$
$ P(x_o) \equiv R(x_o) \equiv c(mod p)$ so $ p|c$
=>$ P(x) \equiv (x \minus{} x_o)Q(x)(mod p)$
Then $ P(x) \equiv 0(mod p)$ <=>$ x \equiv x_o(mod p)$ or $ Q(x) \equiv 0(mod p)$
On other hand $ degQ(x) \equal{} k$ so $ Q(x)$ has max $ k$ root $ mod p$
So $ P(x)$ has max $ k \plus{} 1$ root $ mod p$
 :P
Sorry beacuse my English isn't good!


In $\triangle ABC$, $\angle A=60^\circ$. The incentre is $I$. The inscribed circle of $\triangle ABC$ meets $AB$, $AC$ at $D$, $E$ respectively. Line $DE$ meets line $BI$ and line $CI$ at $F$, $G$ respectively. Show that $FG=\frac12 BC$.

see http://www.mathlinks.ro/Forum/viewtopic.php?p=224997#p224997 , it's almost the same problem.

 this is my solution:
easy to see that $\angle ADE=\angle AED=60$ ,$\angle BIC=120$

then $\angle CGE=60-\frac{c}{2}$ , $\angle BFD=\frac{c}{2}$
and $\angle GDI=30$
sin theo. in $\triangle DGI$ :

$\frac{DG}{sin(\frac{B-60}{2})}=\frac{DI}{sin(\frac{B}{2})}$ then

$DG=\frac{rsin(\frac{B-60}{2})}{sin(\frac{B}{2})}$

if we use the sin theo. in $\triangle DIF$

we obtain $DF=\frac{rcos(\frac{B}{2})}{sin(\frac{C}{2})}$

$\Rightarrow$ $DF-DG=GF=r(\frac{sin(\frac{B}{2})cos(\frac{B}{2})+sin(\frac{B-60}{2})sin(\frac{C}{2})}{sin(\frac{C}{2})sin(\frac{B}{2})})$
we use the trigonometrical changes we obtain:
$GF=\frac{1}{2}\frac{rsin(60)}{sin(\frac{C}{2})sin(\frac{B}{2})}$ we must prove $GF=\frac{1}{2}BC=Rsin(60)$
LEMMA: $r=4Rsin(\frac{C}{2})sin(\frac{B}{2})sin(\frac{A}{2})$
(this is a famous lemma this proves easy)
if we use the lemma we obtain
$GF=4Rsin(\frac{60}{2})sin(60)\frac{1}{2}=Rsin60=BC\frac{1}{2}$

and that is the end of the solution :lol:

[quote="modeler"]can anyone give a pure geom soln?[/quote]
See the link at Post 2.


ceiling function
algebra proposed
algebra

Find all possible triplets that satisfy:

$ xyz\equal{}4104$

and

$ x\plus{}y\plus{}z\equal{}77$

 :)

$ 4104|n \ \text{for} \ \left\lceil \frac{77\minus{}2}{2}\right\rceil<n<77\minus{}2 \Rightarrow n\equal{}38, \ 54, \ 57, \ 72$.

Case $ n\equal{}38$
$ \frac{4104}{38}|m \ \text{for} \ \left\lceil \frac{77\minus{}38\minus{}1}{2}\right\rceil<m<77\minus{}38\minus{}1 \Rightarrow m\equal{}27, \ 36$.

Subcase $ n\equal{}38, \ m\equal{}27$
$ 38\plus{}27\plus{}\frac{4104}{38\times27}\equal{}38\plus{}27\plus{}4\equal{}69 \neq 77$

Subcase $ n\equal{}38, \ m\equal{}36$
$ \mathbf{38\plus{}36\plus{}\frac{4104}{38\times36}\equal{}38\plus{}36\plus{}3\equal{}77}$

Case $ n\equal{}54$
$ \frac{4104}{54}|m \ \text{for} \ \left\lceil \frac{77\minus{}54\minus{}1}{2}\right\rceil<m<77\minus{}54\minus{}1 \Rightarrow m\equal{}19$.

Subcase $ n\equal{}54, \ m\equal{}19$
$ \mathbf{54\plus{}19\plus{}\frac{4104}{54\times19}\equal{}54\plus{}19\plus{}4\equal{}77}$

Case $ n\equal{}57$
$ \frac{4104}{57}|m \ \text{for} \ \left\lceil \frac{77\minus{}57\minus{}1}{2}\right\rceil<m<77\minus{}57\minus{}1 \Rightarrow m\equal{}12, \ 18$.

Subcase $ n\equal{}57, \ m\equal{}12$
$ 57\plus{}12\plus{}\frac{4104}{57\times12}\equal{}57\plus{}12\plus{}6\equal{}75 \neq 77$

Subcase $ n\equal{}57, \ m\equal{}18$
$ 57\plus{}18\plus{}\frac{4104}{57\times18}\equal{}57\plus{}18\plus{}4\equal{}79 \neq 77$

Case $ n\equal{}72$
$ \frac{4104}{72}|m \ \text{for} \ \left\lceil \frac{77\minus{}72\minus{}1}{2}\right\rceil<m<77\minus{}72\minus{}1 \Rightarrow m\equal{}3$.

Subcase $ n\equal{}72, \ m\equal{}3$
$ 72\plus{}3\plus{}\frac{4104}{72\times3}\equal{}72\plus{}3\plus{}19\equal{}94 \neq 77$

Thus, $ (x, \ y, \ z)\equal{}(3, \ 36, \ 38), \ (4, \ 19, \ 54)$


Putnam calculus AMC USA(J)MO USAMO MIT college

I'm pretty sure that this is the wrong forum to post this, and I apologize in advance, so someone can move this if they are disturbed  :P

What is the Putnam, and what is it's difficulty level? Is it much harder than the IMO? How can you prepare for this? I'm still in high school, but I guess i could start knowing what it is so when it comes, I can be prepared  :)

The Putnam is the most prominent under grad math competition. It includes topics that wouldn't be normally learned in a high school curriculum. I wouldn't say "more difficult" than the IMO, but the Putnam is with no doubt at a much faster pace. Someone in college who has taken it probably has a lot more to say...

As krustyteklown points out, the Putnam is an undergraduate level competition.  The Putnam tests the same problem solving skills as any olympiad such as IMO, but the topics also include differential equations, linear algebra, and multivariable calculus, since they are covered in undergraduate mathematics.

You can find previous Putnam problems at these sites:
[url=http://www.kalva.demon.co.uk/putnam.html]www.kalva.demon.co.uk/putnam.html[/url]
[url=http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pindex.html]www.unl.edu/amc/a-activities/a7-problems/putnam/-pindex.html[/url]

The AoPS bookstore also sells compilations of Putnam problems:
[url=http://www.artofproblemsolving.com/Books/AoPS_B_CP_Putnam.php]www.artofproblemsolving.com/Books/AoPS_B_CP_Putnam.php[/url]

Look in the "College Playground" section here at AoPS. We do have a whole forum devoted to the Putnam. It's briefly active in two spurts - December (the exam itself) and March (the results are announced).

The structure of the Putnam: no qualifying rounds - whoever wants to take it can. Many colleges have study groups or even courses devoted to preparing for it. Held on the first Saturday in December. Two three-hour sessions on the same day, morning and afternoon. 6 problems each session, 12 total. (When krustyteklown said "at a much faster pace" he was probably referring to this 30 minute per problem time allowance, in contrast to the USAMO or IMO's hour and a half.) Usually a considerable range of difficulty. Scored 0-10 per problem but the numbers 3 through 7 are not used. About 3500 participants. Median score (out of 120 possible) usually low single digits, occasionally zero.

There is "college-level" material on the Putnam, but it's usually not highly technical. Students who bring a strong problem-solving and competition background with them from high school have a substantial advantage over those trying to pick it up for the first time at the college level. College freshmen can and do do well on it.

Look at the forum here to see discussion of the 2005 and 2004 versions.

[quote="sapphyre571"]yes over 2/3 of the participants this year got a 0 out of 120
highest score was a 92[/quote]
Are you sure about the highest score?

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-html/putnam2005stats.html

Just wondering, do you have to be math major to take Putnam? I know it's hard to answer since it might differ from college to college but this is something I always wondered.

[quote="amcavoy"][quote="sapphyre571"]yes over 2/3 of the participants this year got a 0 out of 120
highest score was a 92[/quote]
Are you sure about the highest score?

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-html/putnam[b]2005[/b]stats.html[/quote]

;)

Actually, in 2005, 46.7% (1657 out of 3545) of the participants got a score of zero. The high score was 100. The median score was 1.

The percentages of particpants scoring zero:

2005: 46.7%
2004: 53.6%
2003: 27.8%
2002: 34.7%
2001: 44.9%
2000: 57.7%
1999: 60.2%
1998: 30.2%

I didn't find a year in there with a top score of 92 and a 2/3 rate of zeros, although 1999 featured a top score of 74.

Do you have to be a math major to take the Putnam? Almost certainly not. However, it will probably be run by a Math Dept. faculty member, and information about it will come through the Math Dept. If you're not a math major but are interested, you may need to actively seek out the information about it at your college. The entry form is due in mid-October, and each college must at that time name it's three-member "team" - the three individuals whose scores will count for the team score. Note that schools with a huge number of students taking the Putnam and scoring well on it - MIT and Harvard, in particular - almost never get that naming of the team "right" in hindsight.

There may or may not be some kind of qualifying test; if so, that's likely to be in early October so that the results could be used to name the team. Even if there is a "qualifying" test, any student who truly wants to take the Putnam will probably be allowed to do so.

I've run the Putnam for 7 years at my school, and usually get 6 to 8 students to take it, almost all of whom take (or at least sit in on) the special problem-solving course. In those years I've had maybe 2 who weren't math majors (or at least double majors - math with something else), and one of those had been on an ARML team when he was in high school.

For future reference, what is the normal procedure for high school students taking the Putnam (if there is a "normal procedure")?

[quote="Ravi B"]By the way, I think you get at most 4 shots to take the Putnam.  So if you take it as a high-school student, you use up one of your shots.[/quote]It seems this is true :  [url]http://www.cs.berkeley.edu/~wkahan/MathH90/PutnamRegs.pdf[/url]
[quote="Putnam Regulations"]This contest is open only to registered students who have not received a college degree, and who have competed fewer than four times in previous Putnam examinations.[/quote]


function
vector
analytic geometry
group theory
abstract algebra
absolute value
graph theory

I have a very ambigous and ugly solution. Vess, please show us yours!
We prove the following lemma:
Let $u_1,\cdots,u_m$ be pairwise distinct vectors in $Z^n$.
Then there is a function $f:Z^n \rightarrow Z*$ with
$f(x+u_1)+\cdots +f(x+u_n)=0$ for any $x$.
Proof.
Define $S_x=\{x+u_1,\cdots,x+u_n\}$.
Sublemma. The vectors in $Z^n$ can be enumerated
$v_1,v_2,\cdots,v_n,\cdots$ in such a way that for any $k$
there is at most one $X$ with $v_k \in S_{X}$ and $S_{X} \subseteq \{v_1,v_2,\cdots,v_k\}$.
Proof. Consider a function $g: Z^n \rightarrow Z$ defined by
 $g(x_1,\cdots,x_n)=\sum_{i=1}^n m_i x_i$
where $m_i \in N / \{0\}$ are choosen in such a way that
$g(u_i)$ are all distinct and non-zero.
Note that $g$ is additive.
Now define $U_k$ be the set of all vectors $v$ with $|g(v)|=k$.
Let's enumerate the vectors in $Z^n$ as follows:
Choose a sufficiently large number $N$.
Then begin:
Step $1$: Enumerate all vectors in $U_0$ having all coordinates
at most $N$ by absolute value (there are finitely many of them)
Step $2$: Enumerate all vectors in $U_0$ having all coordinates
at most $N^2$ by absolute value and not already enumerated.
Step $3$: Enumerate all vectors in $U_0$ having all coordinates
at most $N^3$ by absolute value and not already enumerated,
then enumerate all vectors in $U_1$ having all coordinates
at most $N$ by absolute value.
$\cdots$
Step $k$: Enumerate all vectors in $U_0$ having all coordinates
at most $N^k$ by absolute value and not already enumerated,
then enumerate all vectors in $U_1$ having all components
at most $N^{k-2}$ by absolute value and so on.

Of course every vector will be enumerated at some time.
Now let's prove the conclusion of the sublemma:
Suppose $v_k \in S_X,S_Y$. WLOG $g(v_k) \geq 0$.
We note that we must have $X=v_k-u_i$, $Y=v_k-u_j$.
 WLOG $g(u_i)>g(u_j)$.
Then $V=v_k-u_j+u_i \in S_Y$ and hence was alredy enumerated.
But $g(V)=g(v_k)-g(u_j)+g(u_i)>g(v_k)$ so $|g(V)|>|g(v_k)|$.
Now suppose $|g(v_k)|=s,|g(v)|=t>s$ and
$v_k$ was enumerated at step $r$, $V$ at step $q\leq r$.
Then by the properties of our enumeration
 $v_k$ has at least one component greater that $N^{r-2s-1}$
by absolute value
(or else it would have been  enumerated at a previous step).
But then if $N$ is sufficiently large
then $V$ must have at least one component
greater than $N^{r-2s-2}$ by absolute value.
But this contradicts the fact that $V$ has all components
 smaller than $N^{q-2s}\leq N^{r-2s-2}$ by absolute value.
The sublemma is proven.

To prove our lemma,
consider $\left(v_i\right)$ our enumeration as in the sublemma.
Now we can define $f(v_i)$ inductively on $i$.
We can choose $f(v_i)$ at random unless
$v_i \in S_X$ with $S_{X} \subseteq \{v_1,v_2,\cdots,v_i\}$.
But then by sublemma, this $X$ is unique and then
we can choose an unique value of $f(v_i)$ to satisfy
 $\sum_{v \in S_x} f(v)=0$.
Moreover, we can ensure
$f(v_i)=1$ $(mod$ $n)$ at each step so $f(v_i) \neq 0$.

the lemma is proven.

Now to pass to the solution of the problem.
Let $S={x_1,\cdots,x_n}$ and
wlog suppose $S$ doesn't contain $0$ (it doesn't matter).
We define $T=\{\prod x_i^{a_i}|a_i \in Z\}$.
Let $u_1,u_2,u_3,\cdots,u_m$
be the basis of $T$ over $Q$(it has dimension at most $n$).
Then every element $x$ of $T$ can be written as
$\prod u_i^{a_i}$ for $a_i \in Q$.
But multiplying by a suitable constant
we can suppose that $a_i \in Z$ for $x=x_i$.
Then as easily seen, $a_i \in Z$ for any $X$.
Now extend $T$ to become subgroup of $(R,\times)$
 (still having dimension $m$).
So we can take the group $R\T$.
By axiom of choice  we can choose from every class $C$
 an element $x \in C$.
Now every element in $C$ can be expressed uniquely as
$x \prod u_i^{a_i}, a_i \in Z$
and hence can be associated with the vector
$(u_1,\cdots,u_n)$.
Now taking $w_i$ be the vector associated with $x_i$
(in $T'$ this time, not in $C$)
and using the lemma we get the result.

Vess, am I right? I'm not sure..

  [Edit: I have edited my proof to correct one possible flaw]

[quote="vess"][b](i)[/b] Let $S$ be a finite set with at least three elements. Show that there exists a function $f : \mathbb{R} \to \mathbb{Q}$ with the following properties: 1) $f(x) = 0$ if and only if $x=0$; and 2) for any $x \in \mathbb{R}$, $\sum_{a \in S} f(ax) = 0$.
[b](ii)[/b] Does there necessarily exist such function $: \mathbb{R} \to \mathbb{Z}$?[/quote]

To avoid irrelevant issues about Zorn's lemma let me rephrase the meat of the problem:
You have a finitely generated group, generated by the nonzero elements of S.
Does its Cayley graph have a labelling where the sum of labels of the neighbors
of any point = 0, and all labels are nonzero?
 For R or Q as the labels maybe this is true for any f.g group.  For Z, I don't know.


limit
algebra
polynomial
calculus
calculus computations

What is the limit of

(10^8 * x^5 + 10^6 * x^4 + 10^4*x^2) / (10^9 * x^6 + 10^7 * x^5 + 10^5*x^3)

as x approaches positive infinity

thanks

0

[hide="."]
look at the degree of the denominator
[/hide]

Let $ L \equal{} \lim_{x \to \infty} \frac {P(x)}{Q(x)}$, where P(x) and Q(x) are two polynomials in x with degrees m and n, respectively. Then we have:

L = 0, if m < n;
L = a/b, if m = n (whera a and b are the coefficients of the terms with degrees m and n);
L = infinity, if m > n.

To see this write P(x)/Q(x) in following way:

$ \frac {P(x)}{Q(x)} \equal{} \frac {ax^m \plus{} a_1x^{m \minus{} 1} \plus{} a_2x^{m \minus{} 2} \plus{} ... \plus{} a_m}{bx^n \plus{} b_1x^{n \minus{} 1} \plus{} b_2x^{n \minus{} 2} \plus{} ... \plus{} b_n} \equal{} \frac {x^m\left(a \plus{} \frac {a_1}{x} \plus{} \frac {a_2}{x^2} \plus{} ... \plus{} \frac {a_m}{x^m}\right)}{x^n\left(b \plus{} \frac {b_1}{x} \plus{} \frac {b_2}{x^2} \plus{} ... \plus{} \frac {b_n}{x^n}\right)}$

$ \equal{} x^{m \minus{} n} \underbrace{\frac {a \plus{} \frac {a_1}{x} \plus{} \frac {a_2}{x^2} \plus{} ... \plus{} \frac {a_m}{x^m}}{b \plus{} \frac {b_1}{x} \plus{} \frac {b_2}{x^2} \plus{} ... \plus{} \frac {b_n}{x^n}}}_{ \equal{} T}$.

Now letting x -> +oo, the limit of the fraction T is always a/b, but the limit of the term $ x^{m \minus{} n}$ is zero if m < n (and so the total limit is also zero), is 1 if m = n (and so the total limit is a/b), and is infinite if m > n (and so the total limit is also infinite).


arithmetic sequence
number theory unsolved
number theory

I have easy problem
Prove that if$a,b$be two realnumbers $\geq2$.Provethat ${a^n-b^n}$ doesnot contain a arithmetic progression contains three terms

I split it to start a new topic... ;)

Pierre.

Assume that $0<b<a$ and $0 \leq n<q<p$ are integers such that
$(a^n-b^n) + (a^p-b^p) = 2(a^q-b^q)$ (which means that $a^n-b^n,a^q-b^q,a^p-b^p$ form a $3$-arithmetical progression).
Then $a^n+a^p-2a^q = b^n+b^p - 2b^q .$

Let $f(x) =x^n+x^p-2x^q$ defined for $x \geq 2.$
Then $f'(x) = \frac 1 x (nx^n + px^p - 2qx^q)$.
But, since $x \geq 2$ and $q \leq p-1$, we have :
$nx^n + px^p - 2qx^q \geq px^p - 2qx^q > qx^p - 2qx^q \geq 2qx^{p-1}-2qx^q \geq 0$.
Thus $f'(x) >0$ and $f$ is increasing on $(2; + \infty)$, which contradicts that $f(a)=f(b)$ with $a<b.$

Pierre.

Yes,my solution is above.I said that this problem is very easy.


Hmm check out
[url=http://www.trendio.com]www.trendio.com[/url].  It's a pretty interesting site, where you "buy" popular words that appear often in the news - kind of like the stock market.  The price of the words fluctuates based on how often the word appears in the news, and the object is to get as much money as possible.

JESUS CHRIST JOSE, STOP PISSING PEOPLE OFF AND SPAMMING THE ROUND TABLE. u posted like 10 meaningless posts in the last 2 mins. WTF is with that?If u're gonna spam something, go to G&FF.  here are two examples:

[quote="José"][hide="lol"]superb signature![/hide][/quote]

[quote="José"][hide="for me is"]200% waste of time[/hide][/quote]

[quote="José"]There's a huge problem here in MathLinks: Everything is spam.... you can't even post your opinion! come on! what's this then??? a military service??? ONLY MATH ONLY MATH... come on![/quote]

Jose, look around.  No one else posts as much as you.  And yet your posts contain very little content.  They are short and devoid of insight.  You do not have to post in every single thread you read.

Of course you can post your opinion.  But posting an opinion without any justification is spam (not to mention rude).

Regards


Let $ a,b,c>0$ so that $ a\plus{}b\plus{}c\equal{}1$. Prove that:
$ \frac{\sqrt{a^2\plus{}abc}}{c\plus{}ab}\plus{}\frac{\sqrt{b^2\plus{}abc}}{a\plus{}bc}\plus{}\frac{\sqrt{c^2\plus{}abc}}{b\plus{}ca}\leq\frac{1}{2\sqrt{abc}}$

I failed to find a less uglier solution  :(

Note that $ \sum\frac{\sqrt{a^2+abc}}{c+ab}=\sum\frac{\sqrt{a(c+a)(a+b)}}{(b+c)(c+a)}$.

Therefore our inequality is equivalent to
\begin{align*}&\sum\frac{\sqrt{a(c+a)(a+b)}}{(b+c)(c+a)}\le\frac{a+b+c}{2\sqrt{abc}}\\
\iff &\sum a(a+b)\sqrt{bc(c+a)(a+b)}\le \frac 12(a+b+c)(a+b)(b+c)(c+a)\end{align*}
By AM-GM,
\begin{align*}\sum a(a+b)\cdot 2\sqrt{bc(c+a)(a+b)}&\le\sum a(a+b)(b(c+a)+c(a+b))\\
&=\sum a(a+b)(ab+2bc+ca)\end{align*}
Now
\begin{align*}\sum a(a+b)(ab+2bc+ca)&=\sum a^2(ab+bc+ca)+\sum a^2bc+\sum ab(ab+bc+ca)+\sum ab^2c\\
&=(a^2+b^2+c^2+ab+bc+ca)(ab+bc+ca)+2abc(a+b+c)\\
&=(a+b+c)^2(ab+bc+ca)-(ab+bc+ca)^2+2abc(a+b+c)\\
&=(a+b+c)^2(ab+bc+ca)-(a^2b^2+b^2c^2+c^2a^2)\\
&\le (a+b+c)^2(ab+bc+ca)-abc(a+b+c)\\
&=(a+b+c)(a+b)(b+c)(c+a)\end{align*}
which was what we wanted.

@nayel:

In your proof, you used AM-GM to give an upper bound for the square root. In particular, you did

$ 2\sqrt{bc(c\plus{}a)(a\plus{}b)}\le b(c\plus{}a)\plus{}c(a\plus{}b)$

I am wondering, why did you choose that particular way of breaking up the square root. For example, one could group $ bc$ and $ (c\plus{}a)(a\plus{}b)$ instead. What is your thinking on this?

Actually when using AM-GM or Cauchy-Schwarz in this sort of cases, I always try to end up with the form where the degree minorizes as much as possible. For example, if I used AM-GM for $ 4bc$ and $ (c \plus{} a)(a \plus{} b)$ I would get $ [2,0,0]$ and $ [1,1,0]$ terms. But if I use it on $ b(c \plus{} a)$ and $ c(a \plus{} b)$ I get all $ [1,1,0]$ terms, which in the long run could possibly be useful.

EDIT: Thanks arqady! :)

Geat proof, but werent you suposed to show it was less than half of what you showed?

But I already multiplied both sides by $ 2$ before applying AM-GM.


I'm not sure how to use the math symbols with a post like I've seen in other posts; but let me give it a shot.

I'm asked to write the equation of a circle in standard form and sketch its graph.  I thought: Simple enough; I'll just work backwards and put them back in their respective places for the standard equation of a circle.  Then I realized there was no way I could tell how the orignal two were split up (because in the end they were combined)... how else can I approach this?  Any suggestions?

x²+y²-2x+6y+6=0

Thanks,

BTW; The standard equation of a circle is (x-h)²+(y-k)²=r².

r = radius
(x,y) = point on circle
(h,k) = center

This forum makes me feel immensely stupid in all honesty...  maybe I should find a group of less intelligent people.  :P

Thank you for the quick response; I haven't done any of this stuff since Algebra II.  We are reviewing an incredible amount of old material for the first week in A.P. Calculus AB and these topics seem almost foreign to me.  However it all seems to be coming back.

Thanks (once again),

[hide]
$x^2+y^2-2x+6y+6=0$
$(x^2-2x)+(y^2+6y)+6=0$
$(x^2-2x+1)-1+(y^2+6y+9)-9+6=0$
$(x-1)^2+(y+3)^2=4$

So the center is $(1,-3)$ and the radius is $2$.
[/hide]

[quote="shyce"]This forum makes me feel immensely stupid in all honesty...  maybe I should find a group of less intelligent people.  :P
[/quote]
You can use [url=http://www.artofproblemsolving.com/Forum/index.php?f=149]Getting Started[/url] forum ;)

Also, for "math symbols" ($\LaTeX$), all you have to do is to put dollar signs before and after the expression.


Prove that $ x^2-(m^2+3m+1)x+(3n^2+5n+7)=0$ has no integer root $ \forall m,n \in \mathbb{Z}$

[size=150]It is important to use the fact  that $ x^2\plus{}x$ is even (because $ x^2\plus{}x\equal{}x(x\plus{}1)$, and one of $ x$ or $ x\plus{}1$ is even).[/size]
$ m^2\plus{}3m\plus{}1\equal{}(m^2\plus{}m)\plus{}(2m\plus{}1)$, which is always odd.
$ 3n^2\plus{}5n\plus{}7\equal{}3(n^2\plus{}n)\plus{}(2n\plus{}7)$, which is always odd.
If $ x$ is even or odd, $ x^2\minus{}x(m^2\plus{}3m\plus{}1)\plus{}3n^2\plus{}5n\plus{}7$ is always odd, while $ 0$ is even, so it is impossible the equality.

[quote="erudito"][size=150]It is important to use the fact  that $ x^2 \plus{} x$ is even (because $ x^2 \plus{} x \equal{} x(x \plus{} 1)$, and one of $ x$ or $ x \plus{} 1$ is even).[/size]
$ m^2 \plus{} 3m \plus{} 1 \equal{} (m^2 \plus{} m) \plus{} (2m \plus{} 1)$, which is always odd.
$ 3n^2 \plus{} 5n \plus{} 7 \equal{} 3(n^2 \plus{} n) \plus{} (2n \plus{} 7)$, which is always odd.
If $ x$ is even or odd, $ x^2 \minus{} x(m^2 \plus{} 3m \plus{} 1) \plus{} 3n^2 \plus{} 5n \plus{} 7$ is always odd, while $ 0$ is even, so it is impossible the equality.[/quote]

Correct.

It is interesting how people sometimes are only able to see the most difficult solution, trying to calculate the both roots of the equation:
$ x \equal{} \dfrac{m^2 \plus{} 3m \plus{} 1 \plus{} \sqrt {( \minus{} (m^2 \plus{} 3m \plus{} 1))^2 \minus{} 4(3n^2 \plus{} 5n \plus{} 7)}}{2}$ and $ \dfrac{m^2 \plus{} 3m \plus{} 1 \minus{} \sqrt {( \minus{} (m^2 \plus{} 3m \plus{} 1))^2 \minus{} 4(3n^2 \plus{} 5n \plus{} 7)}}{2}$, trying to see when the term $ ( \minus{} (m^2 \plus{} 3m \plus{} 1))^2 \minus{} 4(3n^2 \plus{} 5n \plus{} 7)$ is a perfect square.
There is a problem (actually easier then this one) in which people normally procced by the most difficult solution:
Two trains in a straight line are going to crash, and both run with speed $ 60$ against the other. A bug, which flies with speed $ 80$, always keep his way in this straight line, and when arrives in the front side of a train, begin to fly exactly in the opposite way, same direction, without stoping, against the other train, and keep itself like this until both hit, then stop. Initially, they are at a distance $ 300$ one from other. How much is the bug going to fly ?
People try to sum the progression of distances that the bug flies from one to other train each time, but realise that they are going to hit in $ 2.5$ "hours", and the speed of the bug is constant, so it flies $ 200$.


[quote="LaughingMan42"]The following numbers are the first seven numbers obtained through a few logical tranformations involving a non-Cardinal sequnce of numbers.
All i can realy say about this term "logical sequence" is that i'm not sure if i can write an equation for it, but i can deffinatly write a program that outputs numbers in the sequnce, without useing conditional statments.

Here are the first seven numbers produced by the formula:
7,15,36,76,144,220,324

Thre may be more then one way to get this sequnce of numbers... i'll provide more numbers if necisary.[/quote]
$a_{n}=\frac{(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)}{(1-2)(1-3)(1-4)(1-5)(1-6)(1-7)}\cdot7+$
$+\frac{(n-1)(n-3)(n-4)(n-5)(n-6)(n-7)}{(2-1)(2-3)(2-4)(2-5)(2-6)(2-7)}\cdot15+$
$+\frac{(n-1)(n-2)(n-4)(n-5)(n-6)(n-7)}{(3-1)(3-2)(3-4)(3-5)(3-6)(3-7)}\cdot36+$
$+\frac{(n-1)(n-2)(n-3)(n-5)(n-6)(n-7)}{(4-1)(4-2)(4-3)(4-5)(4-6)(4-7)}\cdot76+$
$+\frac{(n-1)(n-2)(n-3)(n-4)(n-6)(n-7)}{(5-1)(5-2)(5-3)(5-4)(5-6)(5-7)}\cdot144+$
$+\frac{(n-1)(n-2)(n-3)(n-4)(n-5)(n-7)}{(6-1)(6-2)(6-3)(6-4)(6-5)(6-7)}\cdot220+$
$+\frac{(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{(7-1)(7-2)(7-3)(7-4)(7-5)(7-6)}\cdot324.$ :D


geometry
geometric transformation
reflection
incenter
parallelogram
circumcircle
ratio

Let $ABC$ be a triangle in which no angle is $90^{\circ}$. For any point $P$ in the plane of the triangle, let $A_1, B_1, C_1$ denote the reflections of $P$ in the sides $BC,CA,AB$ respectively. Prove that

(i) If $P$ is the incenter or an excentre of $ABC$, then $P$ is the circumenter of $A_1B_1C_1$;

(ii)  If $P$ is the circumcentre of $ABC$, then $P$ is the orthocentre of $A_1B_1C_1$;

(iii) If $P$ is the orthocentre of $ABC$, then $P$ is either the incentre or an excentre of $A_1B_1C_1$.

(1) The distance from $I$ to the sides of the triangles are all equal. Therefore we have $IA_1=IB_1=IC_1$, Hence $I$ is the circumcentre of $\triangle{A_1B_1C_1}$.

(2) Since $O$ is the circumcentre, so $BC_1 || AO || CB_1$ and $BC_1=AO=CB_1$, therefore $BCB_1C_1$ is a paralellogram hence $B_1C_1 || BC$. Hence $A_1O \perp B_1C_1$. Similarly, we can get that $O$ is the orthocentre of $\triangle{A_1B_1C_1}$.

(3) The reflections of the orthocentre of the triangle is on the circumcircle of the triangle. Also, since $H$ is the intersection of the altitudes, we have $A, H, A_1$ collinear. Thus: \[ \angle{B_1A_1A}=\angle{B_1BA}=\angle{C_1CA}=\angle{C_1A_1A} \]
So $AHA_1$ is the bisector of $\angle{B_1A_1C_1}$. Similarly, we will have other two results. Therefore, $H$ is the incenter of $\triangle{A_1B_1C_1}$.

The proof of excentre is basically the same as above.

Well for the incenter and circumcenter, it is very easy as shown by shobber.
For the part 3) another proof:
COnsider the orthocenter $ H$ of $ ABC$ and its reflections $ A_1,B_1,C_1$ . We know and can prove easily that reflections are concyclic with the the points $ A,B,C$ by proving specifically the concyclity for $ A_1, A,B,C$, $ B_1,A,B,C$, $ C_1,A,B,C$. So now consider triangle $ A_1B_1C_1$ with the same circumcircle as $ ABC$. Draw the pedal triangle of $ ABC$. Let it be $ DEF$. Obviosuly, $ DEF$ and $ A_1B_1C_1$ are homothetic with ratio $ \frac{1}{2}$ center of homothety $ H$. So, now the the $ H$ of the triangle $ ABC$ is the incenter of the podar triangle and done. Similarly, we can do for excenter in the case of the original triangle being obtuse as the prthocenter of the original triangle will be the excenter of the podar triangle in that case.

What is reflection? How can we construct it here and what point is P ? Pls can someone explain.

My sol
1use incentre is circumcentre of pedal triangle and homothety about p with ratio 1/2
2use circumcentre is orthocentre of medial triangle and homothety
3use orthocentre is incentre of pedal triangle and homthety


What is the value of $ (\minus{}3) \plus{} (\minus{}6) \div (\minus{}6) \times (\minus{}3) \minus{} (\minus{}6)$?

remember order of operations
\[ ( \minus{} 3) \plus{} ( \minus{} 6) \div ( \minus{} 6) \times ( \minus{} 3) \minus{} ( \minus{} 6)\]
\[ ( \minus{} 3) \plus{} 1 \times ( \minus{} 3) \minus{} ( \minus{} 6)\]
\[ ( \minus{} 3) \plus{} ( \minus{} 3) \minus{} ( \minus{} 6)\]
\[ \minus{} 3 \minus{} 3 \plus{} 6\]
\[ 0\]


quadratics
inequalities proposed
inequalities


function
trigonometry
limit
inequalities
logarithms
real analysis
real analysis unsolved

Prove that a function $f\in C^{1}\bigl((0,+\infty)\bigr)$ which satisfy $\ds f'(x)=\frac{1}{1+x^{4}+\cos f(x)},\, x>0,$ is bounded at $(0,+\infty).$ (O. Nesterenko)

See all the problems from [color=orange]Kyiv Mechmat Competition, 2007[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=135749]here[/url]

This is an idea:

Observe that $f'(x)>0$ so $f$ increases, therefore you only need to show that $\lim_{x\rightarrow \infty}f(x)<\infty$.

We have

$-1\leq \cos f =\frac{1}{f'}-1-x^{4}\leq 1$, it follows that

$\frac{1}{2+x^{4}}\leq f'\leq \frac{1}{x^{4}}$

$(f+\frac{x^{-3}}{3})'\leq 0$ for all $x\geq 1$. Thus the function

$x\rightarrow f(x)+\frac{x^{-3}}{3}$ decreases on $(1,\infty)$ hence

$f(x)\leq f(1)+\frac{1}{3}-\frac{x^{-3}}{3}$, for all $x\geq 1$ and the desired result follows.

[quote="didilica"]This is an idea:
Observe that $f'(x)>0$ so $f$ increases, therefore you only need to show that $\lim_{x\rightarrow \infty}f(x)<\infty$.
[/quote]
And what about $\lim_{x\to 0+}f(x)>-\infty$? :maybe:

Use the other inequality:

Let $F$ be a primitive of $\frac{1}{2+x^{4}}$. Then

$(f-F)'\geq 0$ implies that $x\rightarrow f(x)-F(x)$ increases on $(0,\infty)$ hence

$f(x)\geq F(x)+f(0)-F(0)$ for all $x\in (0,\infty)$.

A calculation shows that
 $F(x)=\frac{1}{8\sqrt[4]{2}}(-2\arctan (1-\sqrt[4]{2}x)+2\arctan(1+\sqrt[4]{x})+\ln \frac{\sqrt{2}x^{2}+2\sqrt[4]{2}x+2}{\sqrt{2}x^{2}-2\sqrt[4]{x}+2})$

Now, let $x\rightarrow 0$ and you get that the function is bounded. I hope that there are no computational mistakes.

[quote="didilica"]Use the other inequality:

Let $F$ be a primitive of $\frac{1}{2+x^{4}}$. Then

$(f-F)'\geq 0$ implies that $x\rightarrow f(x)-F(x)$ increases on $(0,\infty)$ hence

$f(x)\geq F(x)+f(0)-F(0)$ for all $x\in (0,\infty)$.

[/quote]

 $f(0)$ ?
you don't know if it is defined


http://www.dailymail.co.uk/sciencetech/article-1031438/Pictured-The-floating-cities-day-house-climate-change-refugees.html

I have a few questions: There are bound to be sea storms. If these are lilypads, then what's to keep them from floating into hurricanes??

Also, what keeps them from floating into each other, or onto debris, like reefs or rocks?

what exactl do u mean or???? and how is this related to aops??? :huh:

I completely support innovation like this, but there is so much that needs to go into this like proper safety precautions which with 50,000 inhabitants it sounds over the top. So I am somewhat against it.

rocket boosters.

jk, but they could have GPS systems and propulsion systems (probably underwater) that prevent collisions. Maybe a wide buffer zone around each lilypad? The propulsion systems won't have to be very powerful then.