Please help me name write chemical formulas for each of the following problems:

1. ammonium sulfantimonite (has "ortho" form and is "trithio").  I'm not sure what "sulfantimonite" is.

2. barium ferrate [color=darkred](has one Fe atom which possesses twice its normal highest valence)[/color].
I thought that it might be BaFeO4, but does the [color=darkred]note[/color] affect the formula?

3. sodium monohydrogen thioarsenite hexahydrate.
I'm not sure how to write these long formulas.

4. potassium ammonium monohydrogen hypothiobismuthite heptahydrate.
Same as above.

5. ferrocobaltinickelous diammonium monohydrogen hypdithiophosphite dodecahydrate.
Same again.

Please help, even if you can only help with one or two.  I'd really appreciate any help.

Thanks

1. $(NH_4)_3SbS_3$

2. $BaFeO_4$

3. $Na_2HAsSO_2\cdot6HOH$

4. $KNH_4HBiSO\cdot7HOH$

5. $FeCo_2Ni(NH_4)_2(HPS_2)_6\cdot12HOH$


Show that every positive integer is a sum of one or more numbers of the form $2^r3^s,$ where $r$ and $s$ are nonnegative integers and no summand divides another.
(For example, $23=9+8+6.)$

A proof, by complete induction. We can work out the first few easily enough. Assume that every number in $\{1,2,\dots,n-1\}$ can be properly represented.

Consider $n.$ If $n$ is even, then $\frac n2$ can be properly represented. Multiply every one of the terms in that representation by $2,$ (increasing the power of $2$ by one in each case), and we have a proper representation of $n.$

Now assume that $n$ is odd. Find $s$ so that $3^s\le n<3^{s+1}.$ If $3^s=n,$ we are done. If $3^s<n,$ then let $m=\frac{n-3^s}2.$ This is an integer because we had assumed that $n$ was odd. Then $m$ can be properly represented. Furthermore, since $m=\frac{n-3^s}2<\frac{3^{s+1}-3^s}2=3^s,$ all of the powers of $3$ in the representation of $m$ are smaller than $3^s.$ Multiply that representation of $m$ by $2$ to get a representation of $2m.$ Every one of the terms of this representation of $2m$ contains a power of $2$, and hence does not divide $3^s$, but all of the powers of $3$ in the representation of $2m$ are smaller than $3^s,$ and hence are not divisible by $3^s.$ Put this representation of $2m$ together with $3^s$ and we have a representation of $n.$

This is a very well known problem, from some recent Tournament of the Towns competition.

Wonder what it is doing on the Putnam paper :huh:

It's not original with any contest- this is a classic, and I think Erdos originally threw it out.

Sometimes it's not worth reinventing the wheel to create an easy problem.

Yeah, but after all, it is problem A1 not like A5  :P

It also happens that $ 6 \equal{} 6$.  :)

[quote="FieryHydra"]I have a related question.

How many pairs of primes $(p, q)$ are there such that  every integer $n$ is a sum of one or more integers of the form $p^r q^s$, where $r$ and $s$ are nonnegative numbers and $n > n_0$ for some integer $n_0$ ?[/quote]

I'm so interested in this question.

I want solution of this question

anybody help me

[quote="FieryHydra"]I have a related question.

How many pairs of primes $(p, q)$ are there such that  every integer $n$ is a sum of one or more integers of the form $p^r q^s$, where $r$ and $s$ are nonnegative numbers and $n > n_0$ for some integer $n_0$ ?[/quote]

really interesting question.

[quote=seynsmartjenny]"where  and  are nonnegative integers and no summand divides another. "  This is not true because $ 6 \equal{} 1 \plus{} 5 \equal{} 2 \plus{} 4 \equal{} 3 \plus{} 3$, and all those sums have at least one summand divide by another.  So if no summand divides another is true, we must not include 1.

but then, if you include 1, in the proof after you are done with the evens, you can just say for the case $ n$ is odd, just add 1.[/quote]

But $6= 6 $.

P.S. : Your username does not permit quoting, so I had to remove the x.

Let $n$ be the number we are trying to represent. We proceed with strong induction on $n$, with the case $n=1$ obvious. Now suppose we're trying to represent $n$, and we know all $m<n$ are representable.

If $2\mid n$, then we're done as we can just multiply the representation of $n/2$ by $2$. The same argument resolves the case when $3\mid n$.

Now suppose $\gcd(n,6)=1$. Let $3^k$ be the largest power of $3$ that is at most $n$. Let $m=\frac{n-3^k}{2}$, which is a positive integer as $n$ is odd. Let
\[m = 2^{r_1}3^{s_1} + \cdots + 2^{r_t}3^{s_t}\]
be a valid representation of $m$. Then, we claim that
\[n = 3^k + 2^{r_1+1}3^{s_1} + \cdots + 2^{r_t+1}3^{s_t}.\]
Indeed, none of the latter $t$ terms divide each other, and since they are even, they don't divide $3^k$. Furthermore, we see that $m<3^k$, so $s_i<k$, so $3^k$ does not divide any of the latter $t$ terms. Thus, we have a produced a valid representation of $n$, which completes the strong induction.

We can also prove through strong induction. Sketch: Base case is $n=1$. Recursive step is for odd $n$, subtract the largest power of $3$ less than $n$ and divide by $2$ and inductively proceed. For even $n$, divide out largest power of $2$ and inductively proceed.


parameterization
logarithms
algebra unsolved
algebra

Find all real values of the real parameter $a$ such that the equation \[ 2x^{2}-6ax+4a^{2}-2a-2+\log_{2}(2x^{2}+2x-6ax+4a^{2})= \] \[ =\log_{2}(x^{2}+2x-3ax+2a^{2}+a+1). \]
has a unique solution.

One little mistake: not for both solutions of $x^{2}-3ax+2a^2-a-1=0$ the equation above necessarily makes sense. One can check that for $a\in(-\frac13;0)$ exactly for one of them the under-logarythm expressions are positive. Hence it is a solution too.

[quote="Gabriel(fr)"]but note that,
if $(S-P) > 0 (ie. S>P) \implies \frac{S+P}{2S}>1 \implies S+P>2S \Longleftrightarrow P>S$ (contratidion!)

if $(S-P) < 0 (ie. S<P) \implies \frac{S+P}{2S}<1 \implies S+P<2S \Longleftrightarrow P<S$ (contratidion!)

[/quote]
But that's only true when $S>0$.

[quote=Gabriel(fr)][hide]If $2x^{2}+2x-6ax+4a^{2}=S$ and $2(x+a+1)=P$ we can rewrite the equation as,

$S-P + log_2({S})= log_2({\frac{S+P}{2}) \Longleftrightarrow S-P=log_2(\frac{S+P}{2S})}$

but note that,
if $(S-P) > 0 (ie. S>P) \implies \frac{S+P}{2S}>1 \implies S+P>2S \Longleftrightarrow P>S$ (contratidion!)

if $(S-P) < 0 (ie. S<P) \implies \frac{S+P}{2S}<1 \implies S+P<2S \Longleftrightarrow P<S$ (contratidion!)

consequently the only possible values of $P$ and $S$ occur when $P=S$ therefore,

$S-P=log_2({1})=0$ which is true. Now we need to find the value of $x$ such that the equation $S=P$  has a unique solution, consequently,

$2x^{2}+2x-6ax+4a^{2}=2(x+a+1) \implies x^{2}-3ax+2a^2-a-1=0$ this second-degree equation has a unique root when its discriminant is zero then,

$\Delta=\sqrt{a^2+4a+4}=0 \implies a=-2$

[/hide][/quote]

Is that full solution?


function
parameterization
trigonometry
inequalities unsolved
inequalities

Let $ A\equal{}cosx\plus{}cosy\plus{}cos(x\plus{}y)$, with $ x, y \in [\minus{}\pi;\pi]$.

Find maximum and minimum of $ A$.

The maximum is $ 3$ when $ x\equal{}y\equal{}0$.

The minimum could be obtained by considering the expression as a function of $ x$ and parameter $ y$. It is $ \minus{}\frac{17}{27}\minus{}\frac{7\sqrt{7}}{27}$ when $ y\equal{}2 \arccos(\minus{}\frac 16 \plus{} \frac{\sqrt{7}}{6})$ and $ x\equal{}\frac{y}{2}$. (I obtained this from Maple.)

What about $ x \equal{} y \equal{} \frac {2\pi}{3}$?

[hide]Suppose $ cosx_0 \plus{} cosy_0 \plus{} cos(x_0 \plus{} y_0)$ is the minimum possible value (there is a minimum because $ [ \minus{} \pi, \pi]^2$ is a [url=http://en.wikipedia.org/wiki/Compact]compact[/url]). To discover something about $ x_0$ and $ y_0$, use the formula to transform a sum of cosines in a product (how is it called in English?). How could you change $ x_0$ and $ y_0$ to make the sum get smaller? In the worst case, you can't do it.[/hide]

Because of $ \cosh\leq 1,\cosh\leq 1, \cos(x+y)\leq 1$ then $ max A=3$ if $ x=y=0$
On the other hand
$ A=2\cos\frac{x+y}{2}.\cos\frac{x-y}{2}+2\cos^2\frac{x+y}{2}-1$
$ 2\left(\cos\frac{x+y}{2}+\frac{1}{2}\cos\frac{x-y}{2}\right)^2-1-\frac{1}{2}.\cos^2 \frac{x-y}{2}$
$ \geq -1-\frac{1}{2}=-\frac{3}{2}$
Hence min A =$ \frac{-3}{2}$ if $ x=y=\frac{2\pi}{3}$


limit
logarithms
calculus
calculus computations

show :D
$ \lim_{x\to\infty}\;\left(e^{2x}\plus{}e^x\plus{}x\right)^{\dfrac{1}{x}}\;\equal{}\;\boxed{e^2}$

Without using L'Hopital can kill this problem. :D

hello, it is $ \lim_{x\to\infty}\;\left(e^{2x}\plus{}e^x\plus{}x\right)^{\frac{1}{x}}\equal{}
e^{\lim_{x\to\infty}\;\frac{\ln\left(e^{2x}\plus{}e^x\plus{}x\right)}{x}}$. Using the L'Hospital rule, we get:
$ e^{\lim_{x\to\infty}\;\frac{\ln\left(e^{2x}\plus{}e^x\plus{}x\right)}{x}}\equal{}
e^{\lim_{x\to\infty}\;\frac{\left(2e^{2x}\plus{}e^x\plus{}1\right)}{e^{2x}\plus{}e^x\plus{}x}}\equal{}e^2$.
Sonnhard.

More simply: write it as

$ \left[e^{2x}\left(1\plus{}e^{\minus{}x}\plus{}xe^{\minus{}2x}\right)\right]^{\frac1x}\equal{}
e^2\left(1\plus{}e^{\minus{}x}\plus{}xe^{\minus{}2x}\right)^{\frac1x}$

That isn't indeterminate: $ 1^0\equal{}1$ with no ambiguity.

$ (a,\infty )\subset \mathcal D_{f}$ , $ \lim_{x\to\infty}\frac {f(x)}{e^{mx}} \equal{} l\in\mathcal R$ $ \implies \lim_{x\to\infty}\left[e^{mx} \plus{} f(x)\right]^{\frac 1x} \equal{} e^m\ .$

[b]Proof.[/b] $ L \equal{} \lim_{x\to\infty}\left[e^{mx} \plus{} f(x)\right]^{\frac 1x} \equal{} e^m\cdot\lim_{x\to\infty}\left(\frac {e^{mx} \plus{} f(x)}{e^{mx}}\right)^{\frac 1x} \equal{} e^m$.

[quote="misan"]show :D
$ \lim_{x\to\infty}\;(e^{2x} + e^x + x)^{\dfrac{1}{x}}\; = \;\boxed{e^2}$[/quote]
or because
$ (e^{2x} + e^x + x)^{\dfrac{1}{x}}$~$ (e^{2x})^{\frac{1}{x}}$ and we are done

Squeeze it between: $ (e^{2x}) ^{\frac {1}{x}} \equal{} e^2$
and $ (e^{2x} \plus{} 2e^{x} \plus{} 1)^{\frac {1}{x}} \equal{} ((e^x \plus{} 1)^{\frac {1}{x}})^2 \to e^2$

Though Prof. Merryfield's is more elementary


Este es el problema 2 del TST Cono Sur Peru 2006:

$AA_{1}$ y $BB_{1}$ son las alturas de un triangulo
acutangulo no isosceles $ABC$. $A_{0}$ y $B_{0}$ son los puntos
medios de $BC$ y $CA$, respectivamente. El segmento $A_{1}B_{1}$ corta
al segmento $A_{0}B_{0}$ en $C'$. Probar que $CC'$ es perpendicular a
la recta que une el ortocentro y el circuncentro de $ABC$.
Pongan su solucion

H: ortocentrro; O: circuncentro; P : inters. de  HC y A1B1; Q: inters. de OC y A0B0.
Tenemos las siguientes semejanzas de triangulos:
HA1C ~ OB0C
PA1C ~ QB0C
las cuales implican la semejanza:
HCO ~ PCQ
En particular, HC es paralela a PQ; y, como C' es el ortocentro de PCQ, obtenemos lo pedido.

El problema es un caso particular de la siguiente configuración:

[b]Teorema:[/b] En un $ \triangle ABC,$ sean $ P,Q$ dos puntos isogonales en él. $ \triangle P_aP_bP_c$ es el triangulo pedal de $ P$ y $ \triangle Q_aQ_bQ_c$ es el triangulo pedal de $ Q.$  Sea $ A' \equal{} P_aP_b \cap Q_aQ_b$ y definimos ciclicamente $ B',C'.$ Entonces $ AA',BB',CC'$ son perpendiculares a $ PQ.$

 $ M \equal{} AQ \cap P_bP_c \ ,\ N \equal{} AP \cap Q_bQ_c \ , \ K \equal{} AQ \cap Q_bQ_c \ , \ L \equal{} AP \cap P_bP_c$

Es claro en vista que $ AM$ y $ AN$ son perpendiculares a $ P_bP_c$ y $ Q_cQ_b$ que $ A'$ es ortocentro de $ \triangle AKL$ y por la semejanza de los triangulos $ \triangle AQQ_c \sim \triangle APP_b$ y $ \triangle AKQ_c \sim \triangle ALP_b$ deducimos que $ KL \parallel PQ$ como consecuencia del teorema de Thales. Por ende, $ AA' \perp PQ$ y analogamente $ BB',CC'$ son perpendiculares a $ PQ.$


I have a bag with blue marbles and yellow marbles in it. At the moment, the ratio of blue marbles to yellow marbles is 8:5. If I remove 12 blue marbles and add 21 yellow marbles, the ratio will be 1:3. How many blue marbles were in the bag before I removed some?


In the addition problem  each letter represents a distinct digit. What is the numerical value of E?
[asy]size(50);
label("G",(0,0)); label("M",(1,0)); label("M",(2,0));
draw((-0.3,0.6)--(2.3,0.6));
label("G",(1,1)); label("M",(2,1));
label("E",(0,2)); label("G",(1,2)); label("M",(2,2));
label("+",(0,1));[/asy]

M=0
G=5
E=4

The answer given is not this. The "correct" answer is 9.

E cannot be 9 under any circumstances

we can take into account two possibilities: 1) there is no carried one after adding the tens column and 2) there is a carried one

if it's case 1, then the number in the hundreds place in the answer must be 9, but that cannot be because G cannot be equal to E
if it's case 2, then we would have 10 hundreds, or 1 thousand, in which case we would need a thousandths place in the answer

mewto's answer works, but i don't see how E could be equal to 9

think about it this way,

EGM
+GM
-------
GMM

or
450+50=500
because   M=0   G=5 and E=4

:) :) :D

We first look at the hundreds place. Since $E\ne G$, it must be that $E+1=G$ in order to get $G$ in the hundreds place. Since a $1$ is carried over, we have $G+G=10+M$. Now we look at the units place. Either $M+M=M$ or $M+M=10+M$. In the second case, $2M=10+M\qquad\Rightarrow M=10$, which is not a possible digit. So it must be that $2M=M$, which is only possible if $M=0$. Now $2G=10\qquad\Rightarrow G=5$ and $E+1=G\qquad\Rightarrow E=4$. The numerical value of $E$ is $\boxed{4}$. We can check that $450+50=500$, which matches the digits in the addition problem.

The correct answer is 4 on FTW, just played this.


1. using  nCr

to study the uity of canada, a 7-person committe is to be selected from 5 public figures from western canada, 5 from central, and 5 from east

how many committees would have at least one representative from each region of canada?

how to solve for this?

4(nC[size=75]2[/size]) = 2 (nC[size=75]1[/size])

one last question, how do i post the solutions like you all do?, with the pictures and special functions you dont usually see? :D

Well, put one of those people from each region into the committee.  You have 12 people left and 4 spots on the committee.  The answer is 12c4.

P.S.
Use the attachment button to post pictures and diagrams.

The special pictures and stuff is called $\LaTeX$ and to use it, all you have to do is put dollar ($\$$) signs around the math stuff you're doing. For example, if you put x^2=3 with dollar signs around it, you get $x^2=3$.


calculus
integration
function
trigonometry
search
derivative
calculus computations

Find the function $f(x)$ which is defined for $\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right]$ such that $f(x)+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin (x-y)\cdot f(y)\ dy=x+1\ \left(-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}\right).$

I got $f(x) = c_{1}\cos x+c_{2}\sin x+x+1$ as a general solution.

Do we also need to find out the constants $c_{1}$ and $c_{2}$?

Before I present my solution, I wish to write a few words with regard to posting of solutions on this forum.

I think penning down initial thoughts/ideas which we come across while working on problems is very beneficial to anyone who did not or could not come up with a way to find a solution himself/herself. Oftentimes, though the solution presented for a problem may be elegant, it is hard to see how the author came to think of it. In such cases, learning about the chain of ideas/thoughts leading to a particular solution may be more fruitful than knowing the final solution itself. I am aware this requires a little bit more effort on the part of the author but I also believe the effort put in doing so is worth it even if for its own sake. Hence, from now on, I would ask anyone posting solutions (to problems) to write a few sentences on how he/she approached or attacked the problems. I am sure this will greatly help others in learning how to solve problems themselves. My sincere apologies if I was too verbose.
-----------------------------------------------------------

Here's my solution to the current problem.

Earlier, when I looked at the equation, I tried to search for a "key" idea or point that would help me reach some kind of a "breakthrough". Computing the integral on the LHS was out of question as it seemed it would make the problem even more complicated than it already was: in fact, there was no way to compute the integral without knowing what $f(x)$ itself was. Then, I looked for sometime at the lower and upper limits of the integral - in this case $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ - to see if I could get anything out of them but I couldn't find anything worthwhile. Then, I wondered what $\sin (x-y)$ was doing inside the integral - may be that was the "key". And, then it struck me that $g(x)=\sin x$ is a special kind of function (much like $h(x)=e^{x}$) because when we differentiate it twice we get back the same function albeit with a negative sign attached to it. So, naturally, I thought, why not differentiate both sides of the given equation w.r.t. $x.$ And, it worked!

When, we differentiate both sides of the equation (w.r.t. $x$) using differentiation under the integral sign, we obtain

$f'(x)+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos (x-y)\cdot f(y)\ dy = 1,$ which on further differentiation leads to

$f''(x)-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin (x-y)\cdot f(y)\ dy = 0.$ Adding this to our original equation helps us get rid of the (bothersome) integral, and we are left with a nice second order differential equation, $f''(x)+f(x) = x+1$ that can, of course, be solved using standard methods, which I won't go into much detail.

I would ask the interested reader to learn about solving differential equations such as the one above himself/herself as I myself did while solving this problem. In the process, I learned quite a few terms such as the Wronskian as well as general methods of  solving such ODEs.

In this case, the homogeneous equation has the linearly independent solutions $u_{1}(x) = \cos x$ and $u_{2}(x) = \sin x.$ Their Wronskian $W(x)$ is equal to $1.$ Thus a particular solution of the non-homogeneous equation is given by $y_{1}= x+1.$ And, so the general solution is $\boxed{f(x) = c_{1}\cos x+c_{2}\sin x+x+1}.$

At this stage, I am not sure if we can compute the constants $c_{1}$ and $c_{2}.$ Perhaps, we could put $x=0$ and $x=\frac{\pi}{2}$ in our original equation, and then evaluate the integral to compute those two constants. Anyone has other ideas?

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

I welcome comments/criticisms on my initial thoughts which I wrote (in this post) right before presenting the above solution.

when you took the derivative of the part inside the integral why did u not take the derivative of $f(y)$?

He differentiated with respect to $x$. There's a theorem which tells you that you can keep $f(y)$ "constant" (where $y$ is the variable of integration), whilst differentiating wrt another variable. Look [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=620407&highlight=written#620407]here[/url] for further details.


I have so far read half of Economics in One Lesson, and I like it a lot.

Essential Economics is a good one to learn the terms. It is featured in the Economist. Read the economist, too.

Here is my suggestions/personal reading list

Hard Heads, Soft Hearts: Tough-Minded Economics for a Just Society (Alan Blinder)
Principles of Political Economics (John Stuart Mill) - a little outdated, but still an awesome text
Stuck in Traffic (and the updated form: Stll Stuck in Traffic) (Anthony Downs) - an economics view of traffic congestion. Interesting read..mostly when one is stuck in traffic

I think those are all classics (then there are Adam Smith, Milton Friedman, Keynes, Say, Mill, and Marx - all really classical texts. But a little dry)

A purely "fun" read would be "Eat the Rich," by P.J. O'Rourke.


For each "exterior, pointy" vertex, there are $ 2$ other non-"pointy" vertices that it touches. Also, there are $ 3$ "long" diagonals. Thus, our total is $ 6\cdot2\plus{}3\equal{}\boxed{15}$.

For every convex vertex, there are three interior diagonals and for every concave vertex, there is seven interior diagonals. However we counted every vertices twice hence:

$\frac{(3)(6)}{2}+\frac{(7)(6)}{2}=\boxed{30}$


Mario and Jose were separately solving an expression that would simplify to a quadratic equation. Mario got the roots $ 15$ and $ \minus{}17$ while Jose got $ 8$ and $ \minus{}3$. Checking each other's solution, Mario noticed that Jose had a mistake in the coefficient of the first degree term. According to Jose, Mario made a mistake in the constant term. Therefore, What are the actual roots of the function??

[hide]
$ (x\minus{}15)(x\plus{}17) \equal{} x^2 \plus{}2x \minus{} 255$
$ (x\plus{}3)(x\minus{}8) \equal{} x^2 \minus{} 3x \minus{} 24$

$ \Rightarrow x^2 \plus{} 2x \minus{}24 \equal{} (x\plus{}6)(x\minus{}4) \Rightarrow x \equal{} \minus{}6,4$[/hide]


geometry
trapezoid
trigonometry
perimeter
analytic geometry
function
calculus

Sorry if I'm a bit vague because I can't remember the details of a problem, but... How would you derive the formula of a trapezoid using trigonometry (identities perhaps?). Given diagram: The trapezoid is drawn on the unit circle with the longer base on the x axis....  :|

What do you mean by the formula of a trapezoid? The perimeter? The area? Please be more specific.

In terms of what? The angles of the trapazoid? Or the coordinates? Or the side length?

[quote="jeez123"]Given diagram: The trapezoid is drawn on the unit circle with the longer base on the x axis....  :|[/quote]

Is it isosceles?  Are you given its height, or just the side lengths?

[hide="hint"]
The area of a triangle is $\frac{1}{2}ab\sin\theta$ where a and b are two sides of the triangle and $\theta$ is the angle between them.
[/hide]

[quote="jeez123"]Sorry, this is the actual problem...

A trapezoid inscribed in the upper half of a unit circle as shown in the figure below.

a. Write the area of the trapezoid as a function of $\theta$.
b. Find the value of $\theta$ that maximizes the area of the trapezoid and the maximum area.[/quote]

[hide="Solution for part a"]

You know that the area of a triangle is $\frac{1}{2}ab\sin\theta$. You also know for this problem that the radius of the circle is $1$. So you have three triangles (draw line from origin to the vertices).

So you apply the above three times to get:

$A$=area of trapazoid

$A=2(\frac{1}{2}\sin\theta)+\frac{1}{2}\sin(180-2\theta)=\sin\theta+\frac{1}{2}\sin\theta\cos\theta$.
[/hide]

[hide="Solution for part b"]

Take the derivative of the above and set it equal to zero to get the $\theta$ that will maximize and minimize the area, then plug into the above to get the actual maximim (or minimum, which is going to be $0$).

[/hide]


vector
trigonometry
function
geometry
geometric transformation
rotation

Suppose we have a disc that the center is centered in a motor.

The disc is mono-color, with a frequency $f$.
The disc is rotating with a constant angular frequency $\omega$.

To make easy see. We can make a vector equation.

$s = ( R. \cos( \omega. t) , R. \sin( \omega. t) , 0 )$

Where $R$ Is a linear function that goes to $( 0, R_{max})$   (Note: $R$ is not in function of time)

Suppose we have a observer in the point

$P = ( r. \cos( \theta) , r. \sin( \theta) , h )$

where $r, \theta , h$ are constants.

A infinitesimal element $dA$ of the disc is rotating with constante speed.
But, the speed of this infinitesimal element relatives to $P$ is changing in time.

So. If $\omega$ is very very large and $R_{max}$ is big. (Dont consider Relativity effects) So, we can see the doppler effect and calculate the new color of the objet

Since a infinitesimal element $dA_{1}$ has a infinitesimal elemento $dA_{2}$ some angle $d\phi$ in front of it, so, we can conclude:

"All infinitesimal elements that passes througth a certain point in a certain time, has the same aparant color for the observer"

And the logical intuition is valid too:

"If the observer is in the same axis of rotation of the disc, he will see the disc with the same frequency"

So, the Question is:

"What is the function Frequency $F$ (Saw by the observer) in function of Radius $R$ of the infinitesimal element of the disc (Considered Static, because this infinitesimal element is different, because we reached a stationary situation) and the angle $\phi$ that this element form with the $x$ axis?

I'm having problem with this, if someone could help

[hide]$\noindent\(f \sqrt{\frac{c \sqrt{d^{2}+r^{2}+R^{2}-2 r R \text{Cos}[\theta-\phi ]}+r R \omega \text{Sin}[\theta-\phi ]}{c \sqrt{d^{2}+r^{2}+R^{2}-2 r R \text{Cos}[\theta-\phi ]}-r R \omega \text{Sin}[\theta-\phi ]}}\)$
[/hide]

I found this equation for the frequency. But. is this rigth?


The diagonals of a convex quadrilateral $ ABCD$ are intersecting at the point $ O$.The circumcircles of the triangles $ AOB$ and $ COD$ are intersecting on a point found on the side $ AD$.Prove that $ AO \ge AB$.

I think, there must be $ AO > AB$.
My solution:
$ \angle{ABO} = 180^{\circ} - \angle{AKO} = \angle{OKD}$.
$ \angle{BOA} = \angle{COD} = \angle{CKD}$.
Since $ \angle{OKD} > \angle{CKD}$, we obtain that $ \angle{ABO} > \angle{BOA}$, so $ AO > AB$, q.e.d.


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inequalities proposed
inequalities

Consider the positive integers $ x_1,x_2,...,x_n$, any two of them distinct. Prove that:

$ 4(x_1\plus{}x_2\plus{}...\plus{}x_n)^3 \le 3(x_1^5\plus{}x_2^5\plus{}...\plus{}x_n^5)\plus{}(x_1^3\plus{}x_2^3\plus{}...\plus{}x_n^3).$


How does one show that $\exp(s \log(1 + z)) = \sum_{n=0}^\infty{s \choose n}z^n$ for all $z,s \in \mathbb{C}, |z| < 1$ where ${s \choose n} = \frac{s(s-1)\cdot \ldots \cdot (s-n+1)}{n!}$? ($\exp$ defined by the usual power series, $\log{z} = \log{|z|} + i \phi$ where $\phi$ is the argument of $z$)

Is the following argument correct?

By real Taylor series expansion, the identity is true for [i]real[/i] $z,s$. For $z$ fixed, $s \in \mathbb{R}$ is an infinite subset with limit point $0$ of $\mathbb{C}$, so by the identity theorem ($\exp(s \log(1 + z))$ is holomorphic), the formula holds for all $s$. Analogously with $s$ fixed and $|z| < 1, z \in \mathbb{C}$.

So it is true for all $z,s \in \mathbb{C}, |z| < 1$ because we can first fix $z \in \mathbb{R}$ and show it for all $s$ and then fix $s \in \mathbb{C}$ and then show it for our $z$.

Yes, that's the way to do it. Every real power series you know from calculus extends in this way.


Is it possible to prove
$ \frac {a}{a^2 \plus{} 3} \plus{} \frac {b}{b^2 \plus{} 3} \plus{} \frac {c}{c^2 \plus{} 3}\le\frac3{4}$ for $ a,b,c >0$ and $ abc \equal{} 1$
with using General Mean Inequality, Schur, Cauchy-Schwartz or Chebyshev. Nothing more than these inequalities and without using pqr. Please answer my question and after that write solutions.

After using Am-Gm, it's becomes:
$ \sum\ \sqrt[]{\frac{a}{a\plus{}b}} \leq\ \frac{3}{\sqrt[]{2}}$  (always trues)
PS: It's an old problem  :)

[quote="nguoivn"]After using Am-Gm, it's becomes:
$ \sum\ \sqrt []{\frac {a}{a \plus{} b}} \leq\ \frac {3}{\sqrt []{2}}$  (always trues)
PS: It's an old problem  :)[/quote]
But why?,why after use AMGM,we have this ineq,sorry because I am too silly,so I cant find how you get it
With the second ineq ,Can hang has a stronger ineq(it is really nice,and quite hard)

can_hang2007 's one
$ \frac{a}{a^3\plus{}1}\plus{}\frac{b}{b^3\plus{}1}\plus{}\frac{c}{c^3\plus{}1} \le\ \frac{3}{2}$ for $ a;b;c>0$ and $ abc\equal{}1$

[quote="SAPOSTO"]Is it possible to prove
$ \frac {a}{a^2 \plus{} 3} \plus{} \frac {b}{b^2 \plus{} 3} \plus{} \frac {c}{c^2 \plus{} 3}\le\frac3{4}$ for $ a,b,c > 0$ and $ abc \equal{} 1$
with using General Mean Inequality, Schur, Cauchy-Schwartz or Chebyshev. Nothing more than these inequalities and without using pqr. Please answer my question and after that write solutions.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1156978#1156978

[quote="SAPOSTO"]Is it possible to prove
$ \frac {a}{a^2 \plus{} 3} \plus{} \frac {b}{b^2 \plus{} 3} \plus{} \frac {c}{c^2 \plus{} 3}\le\frac3{4}$ for $ a,b,c > 0$ and $ abc \equal{} 1$
with using General Mean Inequality, Schur, Cauchy-Schwartz or Chebyshev. Nothing more than these inequalities and without using pqr. Please answer my question and after that write solutions.[/quote]
I found the following interesting solution. :) Hope you will like it :)
Also, the stronger inequality is still true
\[ \frac{a}{(a\plus{}3)^2}\plus{}\frac{b}{(b\plus{}3)^2} \plus{}\frac{c}{(c\plus{}3)^2} \le \frac{3}{16}\]
Will your way still work, quoc anh? :)

[quote="nguoivn"][quote="onlylove_math"]can_hang2007 's one
$ \frac {a}{a^3 \plus{} 1} \plus{} \frac {b}{b^3 \plus{} 1} \plus{} \frac {c}{c^3 \plus{} 1} \le\ \frac {3}{2}$ for $ a;b;c > 0$ and $ abc \equal{} 1$[/quote]
Am-Gm also solved it easily  :)

[quote="can_hang2007"]
I found the following interesting solution. :) Hope you will like it :)
Also, the stronger inequality is still true
\[ \frac {a}{(a \plus{} 3)^2} \plus{} \frac {b}{(b \plus{} 3)^2} \plus{} \frac {c}{(c \plus{} 3)^2} \le \frac {3}{16}
\]
Will your way still work, quoc anh? :)[/quote]

Your solution is nice but hard with me to find.
I 'll try to prove the stronger.  And with the stronger, your way is same to above ?  :maybe:[/quote]
Yes. But there exists another solution with another aproach :)

[quote="nguoivn"][quote="onlylove_math"]can_hang2007 's one
$ \frac {a}{a^3 \plus{} 1} \plus{} \frac {b}{b^3 \plus{} 1} \plus{} \frac {c}{c^3 \plus{} 1} \le\ \frac {3}{2}$ for $ a;b;c > 0$ and $ abc \equal{} 1$[/quote]

Using Am-Gm, we can prove the stronger:
$ (\frac {a}{a^3 \plus{} 1})^{\frac {3}{2}} \plus{} (\frac {b}{b^3 \plus{} 1})^{\frac {3}{2}} \plus{} (\frac {c}{c^3 \plus{} 1})^{\frac {3}{2}} \le\ \frac {3}{2\sqrt []{2}}$ for $ a;b;c > 0$ and $ abc \equal{} 1$[/quote]
Yes, you are right, nguoivn. The inequality can be proved easily with the AM-GM. But how about the following?
\[ \sum \left( \frac{2a}{a^3\plus{}1}\right)^5 \le 3\]
and
\[ \sum \frac{a}{2a^3\plus{}1} \le 1\]
:)

Are the following ineq with the same condition $ abc\equal{}1$?? :maybe:

[quote="Sunkern_sunflora"]Are the following ineq with the same condition $ abc \equal{} 1$?? :maybe:[/quote]
Yes, sunkern. :)

[quote="can_hang2007"]
$ \sum \left( \frac {2a}{a^3 \plus{} 1}\right)^5 \le 3$
[/quote]

Am-Gm also prove easily with $ k \equal{} \frac {9}{2}$, but $ k \equal{} 5$, I can't. I really want to see your proof  :)
Besides, by the following way, we can prove:
$ \sum\ (\frac {8ab^2}{(a \plus{} b)^3})^{\frac{3}{2}} \leq\ 3$

[quote="Sunkern_sunflora"]Yeah.....what I was only able to deduce is that the ineq (second one) is equivalent to:
\[ \frac {1}{2a^2 \plus{} bc} \plus{} \frac {1}{2b^2 \plus{} ca} \plus{} \frac {1}{2c^2 \plus{} ab} \leq 1
\]
Which, I think, every mathlinkers have found...!!!! :([/quote]
and we can use pqr technique here,but we are waiting the proof without it
[quote="SAPOSTO"]with using General Mean Inequality, Schur, Cauchy-Schwartz or Chebyshev. Nothing more than these inequalities and without using pqr. Please answer my question and after that write solutions.
[/quote]

$ \frac {1}{2a^2 \plus{} bc} \plus{} \frac {1}{2b^2 \plus{} ca} \plus{} \frac {1}{2c^2 \plus{} ab} \leq 1$
Indeed, it's not easy althought using pqr. And with:
$ \sum \left( \frac {2a}{a^3 \plus{} 1}\right)^5 \le 3$
$ \sum\ (\frac {8ab^2}{(a \plus{} b)^3})^{\frac{3}{2}} \leq\ 3$
I don't think pqr can solve them  :wink:

Hey nguoivn! Can it be solved by mixing variables $ b$ and $ c$ to $ \sqrt{bc}$ or $ \sqrt{\frac{b^2\plus{}c^2}{2}}$?? :maybe:

[quote="Sunkern_sunflora"]Hey nguoivn! Can it be solved by mixing variables $ b$ and $ c$ to $ \sqrt {bc}$ or $ \sqrt {\frac {b^2 \plus{} c^2}{2}}$?? :maybe:[/quote]
I don't know because I don't like that method.
However:
$ \sum \left( \frac {2a}{a^3 \plus{} 1}\right)^{\frac{9}{2}} \le 3$
$ \sum\ (\frac {8ab^2}{(a \plus{} b)^3})^{\frac{3}{2}} \leq\ 3$
They can be solved easily by Am-Gm.
With $ k\equal{}5$, I think can_hang had a solution with Cauchy-Schwart  :)

[quote="SAPOSTO"]Is it possible to prove
$ \frac {a}{a^2 \plus{} 3} \plus{} \frac {b}{b^2 \plus{} 3} \plus{} \frac {c}{c^2 \plus{} 3}\le\frac3{4}$ for $ a,b,c > 0$ and $ abc \equal{} 1$
with using General Mean Inequality, Schur, Cauchy-Schwartz or Chebyshev. Nothing more than these inequalities and without using pqr. Please answer my question and after that write solutions.[/quote]
I have a very nice solution by Karamata :P
Let $ x \equal{} \ln(a), y \equal{} \ln(b), z \equal{} \ln(c)$. THen we have to prove:
$ f(x) \plus{} f(y) \plus{} f(z) \le \frac{3}{4}$ where $ f(x) \equal{} \frac{e^x}{e^{2x}\plus{}3}$ and $ x\plus{}y\plus{}z \equal{} 0$.
Then $ f''(x) \equal{} \frac{e^{x}(e^{4x} \minus{} 18^{2x}\plus{}9)}{(e^{2x}\plus{}3)^3} \equal{} \frac{e^x}{(e^{2x}\plus{}3)^3}(e^{2x}\minus{}(9\minus{}\sqrt{72}))(e^{2x}\minus{}(9\plus{}\sqrt{72}))$. Hence $ f(x)$ is concave when $ x \in \left [ \frac{1}{2} \ln(9\minus{}\sqrt{72}) ; \frac{1}{2} \ln(9\plus{}\sqrt{72}) \right ]$. From now on let $ I_1 \equal{} (\minus{}\infty;a], I_2 \equal{} [a;b], I_3 \equal{} [b; \plus{}\infty )$ where $ a \equal{} \frac{1}{2} \ln(9\minus{}\sqrt{72}), b \equal{} \frac{1}{2} \ln(9\plus{}\sqrt{72})$.

[b]Case 1[/b]: At least two of $ x,y,z$ is in $ I_2$, assume $ x,y \in I_2$.
Then it follows by karamata that $ f(x) \plus{} f(y) \plus{} f(z) \le f \left ( \frac{x\plus{}y}{2} \right ) \plus{} f \left ( \frac{x\plus{}y}{2} \right ) \plus{} f(z)$. Hence we only consider the case $ x\equal{}y$.
[b]Case 2[/b]: Exactly one of $ x,y,z$ is in $ I_2$ assume $ x \in I_2$.
[b]Case 2.1[/b]: $ y,z \in I_1$. Then by karamata $ f(x) \plus{} f(y) \plus{} f(z) \le f(x) \plus{} f(y\plus{}z\minus{}a) \plus{} f(a)$. Then since $ a,x \in I_2$ we are back to [b]Case 1[/b]: Hence we only consider the case $ y\equal{}z$.
[b]Case 2.2[/b]: $ y,z \in I_3$. Then by karamata $ f(x) \plus{} f(y) \plus{} f(z) \le f(x) \plus{} f(y\plus{}z\minus{}b) \plus{} f(b)$. Then since $ b,x \in I_2$ we are back to [b]Case 1[/b]: Hence we only consider the case $ y\equal{}z$.
[b]Case 2.3.1[/b]: $ y \in I_1$, $ z \in I_3$. Obviously either $ y\plus{}z\minus{}a \in I_3$ or $ y\plus{}z\minus{}b \in I_1$.
[b]Case 2.3.1.1[/b]: $ y\plus{}z\minus{}a \in I_3$. $ f(x) \plus{} f(y) \plus{} f(z) \le f(x) \plus{} f(a) \plus{} f(y\plus{}z\minus{}a)$. Then since $ a,x \in I_2$ we are back to [b]Case 1[/b]: Hence we only consider the case $ y\equal{}z$.
[b]Case 2.3.1.2[/b]: $ y \in I_1$, $ z \in I_3$. $ f(x) \plus{} f(y) \plus{} f(z) \le f(x) \plus{} f(b) \plus{} f(y\plus{}z\minus{}b)$ Then since $ b,x \in I_2$ we are back to [b]Case 1[/b]: Hence we only consider the case $ y\equal{}z$.
[b]Case 2.3.2[/b]: $ z \in I_1$, $ y \in I_3$. The same as [b]Case 2.3.1[/b]
[b]Case 3[/b]: None of $ x,y,z$ is in $ I_2$.
[b]Case 3.1[/b]: Two of $ x,y,z$ is in $ I_1$. Assume $ x,y \in I_1$. Then by Karamata $ f(x)\plus{}f(y)\plus{}f(z) \le f(x\plus{}y\minus{}a) \plus{} f(a) \plus{} f(z)$, and we are back to [b]Case 2[/b], since $ a \in I_2$.
[b]Case 3.2[/b]: Two of $ x,y,z$ is in $ I_3$. Assume $ x,y \in I_3$. Then by Karamata $ f(x)\plus{}f(y)\plus{}f(z) \le f(x\plus{}y\minus{}b) \plus{} f(b) \plus{} f(z)$, and we are back to [b]Case 2[/b], since $ b \in I_2$.

(Note: I did only use that $ f(x)$ is convex, then concave, then convex, so it is a pretty general result :) )

Now we only have to consider the case $ x\equal{}y \iff a\equal{}b$. Let $ a\equal{}b\equal{}t$ then $ c \equal{} \frac{1}{t^2}$. Inserting gives:

$ \frac{2t}{t^2\plus{}3} \plus{} \frac{t^2}{3t^4\plus{}1} \leqslant \frac{3}{4} \iff$

$ 9t^6\minus{}24t^5\plus{}32t^4\minus{}9t^2\minus{}8t\plus{}12 \geqslant 0 \iff$

$ (t\minus{}1)^2((3t^2\minus{}t)^2 \plus{} t^2 \plus{} 10t \plus{} 9) \geqslant 0$. And we are done :)

[quote="Mathias_DK"]
(Note: I did only use that $ f(x)$ is convex, then concave, then convex, so it is a pretty general result :) )

[/quote]

Nice. That looks somewhat similar to the [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=391537]Half concave and half convex theorem[/url] by Zhaobin and Vasc.  :)

[quote="nguoivn"]
$ \sum \left( \frac {2a}{a^3 \plus{} 1}\right)^{\frac {9}{2}} \le 3$
$ \sum\ (\frac {8ab^2}{(a \plus{} b)^3})^{\frac {3}{2}} \leq\ 3$
They can be solved easily by Am-Gm.
[/quote]
And here is my proof which I have mentioned:
Example, I'll prove the result:
$ \sum\ (\frac {8ab^2}{(a \plus{} b)^3})^{\frac {3}{2}} \leq\ 3$
Notice that by AM-GM, easily to see that : $ (a \plus{} b)^3 \ge 4\sqrt [3]{4}.ab\sqrt [3]{a^3 \plus{} b^3}$
=> $ (\frac {8ab^{2}}{(a \plus{} b)^{3}})^{\frac {3}{2}} \le \sqrt {\frac {2b^3}{a^3 \plus{} b^3}}$
The proof'll complete if we can show that:
$ \sqrt {\frac {2x}{x \plus{} z}} \plus{} \sqrt {\frac {2y}{y \plus{} x}} \plus{} \sqrt {\frac {2z}{z \plus{} y}} \le 3$ $ (x \equal{} a^3, y \equal{} b^3,z \equal{} c^3)$
It's a well-known result. So, we have done  :)
Similarly, we can prove the result: $ \sum\ (\frac {2ab^2}{a^3 \plus{} b^3})^{\frac {9}{2}} \leq\ 3$


How many sets of two or more consecutive positive integers have a sum of 15?

$ \textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

[hide] These are the consecutive integers that will add up to 15.

           1+2+3+4+5=15

           4+5+6=15

           7+8=15

          There are three sets, therefore the answer is C.[/hide]

You could also look at the averages (I think this is how the solution manual described it).  If there are two numbers, the mean will be 7.5 -- so you have {7,8}.  If there are 3, the mean is 5 -- so you have {4,5,6}.  Finally, if there are 5 numbers, the mean is 3 --  so you have {1,2,3,4,5}.

[hide="Answer"]This one is simple enough to work out...

$7+8=15$
$4+5+6=15$
$1+2+3+4+5=15$

$\boxed{C}$[/hide]

[hide]
You can solve it systemically, by using the sum of an arithmetic sequence with length k and first term a:

(2a+k-1)(k)/2 = 15

k can be 6 5 3 2, 6 is too big, and there are working cases for lengths of 2, 3 and 5, so the answer is
C[/hide]

who doesn't 0,1,2,3,4,5 work? that would make the answer 4

They specifically say positive integers, not nonnegative integers


This problem was on another thread about three months ago (so I searched  :) ) but the solution was incomplete.

It was this problem:

In $\triangle{ABC}$, AB=425, BC=450, and CA=510. P is an interior point chosen so that the segments DE, FG, and HI are each of length d, contain P, and are parallel to the sides AB, BC, and CA, respectively. Find d.

EDIT... err... wait. I read 4everwise's wonderful solution in the contest section, and after pondering for 10 minutes, I got it!
Mods, can you please lock this (or you can leave for others to answer, but I doubt anyone would want to)?

Actually, I have a question about this question :P. Uh, how do you solve the last part? I got through all of it until the last past by myself, but I gave up because I can't solve the two equations (three if you could that $XY+XZ=d$)

[hide]
do you do something like multiply by the common multiple and then add the equations?
[/hide]

Are the segments supposed to have endpoints on the sides of the triangle? Because that's by no means specified, but it's the only way that there's a definite answer to the problem.

[quote="Zellex"]uh, the link didn't help >.<.[/quote]

Well now it does. THat was the one that was previously unsolved, but now there's finally discussion again.


Here's a problem I just made up.
Simplify:
\[ \cos^2\theta-7\sin^2\theta+\tan^2\theta \]

Enjoy.

Masoud Zargar

Very nice!
My solution is completely different.

$\\\cos^2\theta-7\sin^2\theta+\tan^2\theta\\ =\cos^2\theta-sin^2\theta(7-\sec^2\theta)\\ =\cos^2\theta-sin^2\theta(6-\tan^2\theta)\\ =\cos^2\theta-6sin^2\theta+\frac{sin^4\theta}{\cos^2\theta}$

Let $x=\cos^2\theta-6sin^2\theta+\frac{sin^4\theta}{\cos^2\theta}$.
Hence,
$x\cos^2\theta=\cos^4\theta-6sin^2\theta\cos^2\theta+sin^4\theta=\cos 4\theta$
$x=\frac{\cos 4\theta}{\cos^2\theta}$

Masoud Zargar


Hi,

Is there any way to insert a halfway-page-break?  That is, I am writing a document in which I need each new..."page" if you will, to begin at the half-way point of each page (basically, I need the equivalent of a column break, except horizontal and halfway down the page).  How can I do this?

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=geometry]geometry package[/url] which makes it easy to use a large bottom margin or a smaller textheight or whatever suits your needs. The [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/geometry.pdf]manual[/url] explains what to do.

Sorry, I realize now that my post was unclear.  What I meant was that I need a "break" in the sense that I have text both above and below it on each page, but a constant break (of maybe two lines or so) at the half-way point (and importantly the exact same point) on each page.

\vspace{[i]size[/i]} will give you blank space of whatever size you decide. The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=textpos]textpos package[/url] will allow you to use absolute positioning.


geometry
incenter
circumcircle
trapezoid
geometric transformation
reflection
geometry proposed

$ \mbox{If I is the incircle center of } \triangle{ABC}, \mbox{ then points: } A, B_1, C_1 \mbox{ and } I \mbox{ concyclic!}$
[b]SOLUTION:[/b]
$ \mbox{I is the circumcircle center of }\triangle{B_1MC_1}\Rightarrow$
${ \Rightarrow I\in \mbox{midperpendicular line of the }[B_1C_1] \mbox{ and } (AI \mbox{ bisectrix of }\angle B_1AC_1} \mbox{ of } \triangle{AB_1C_1} \Rightarrow$
$ \Rightarrow I\in \mbox{ circumcircle of }\triangle{AB_1C_1}.$

[b]REMARK:[/b] $ \mbox{The orthocenter H of }\triangle{ABC} \mbox{ not lie on the circumcirle of } \triangle{AB_1C_1}.$

Let $ \mathcal M_B \equiv (M, MB),$  $ \mathcal M_C \equiv (M, MC)$ be 2 concentric circles with common center $ M$ and radii $ MB, BC,$ cutting $ AB, AC$ again at $ B_1, C_1,$ respectively. Rays $ (MB_1, (MC_1$ cut $ \mathcal M_C, \mathcal M_B$ at $ C_2, B_2,$ respectively. $ B_1B_2C_1C_2$ is isosceles cyclic trapezoid with $ B_1B_2 \parallel C_1C_2,$ let $ (P)$ be its circumcircle, $ MP$ bisects the angle $ \angle B_1MC_1.$ Let $ K$ be the midpoint of $ B_1C_2,$ then $ MK \equal{} \frac{_1}{^2} (MB_1 \plus{} MC_2) \equal{} \frac{_1}{^2} (MB \plus{} MC) \equal{} \frac{_1}{^2} BC.$ Angles $ \angle BMC_1 \equal{} 2 \angle C$ and $ \angle CMB_1 \equal{} 2 \angle B$ are fixed, the right $ \triangle MKP$ just translates with the point $ M \in BC,$ $ P$ is on a line $ p$ parallel to $ BC.$

WLOG, assume $ MB \le MC.$ Since $ \angle B_1MC_1 \equal{} 2\angle B \plus{} 2\angle C \minus{} 180^\circ,$ then from the isosceles $ \triangle MC_1C_2,$ $ \angle MC_1C_2 \equal{} 90^\circ \minus{} \frac{_1}{^2} \angle B_1MC_1 \equal{} 180^\circ \minus{} (\angle B \plus{} \angle C) \equal{} \angle A$ $ \Longrightarrow$ $ A \in (P).$ For any $ M \in BC,$ the centers $ P \in p$ of the circles $ (P)$ are collinear and they all go through $ A,$ they all go through reflection of $ A$ in $ p \parallel BC,$ which lies on the A-altitude of the $ \triangle ABC.$ When $ M$ is the midpoint of $ BC,$ the circles $ \mathcal M_B, \mathcal M_C$ coincide and $ B_1, C_1$ become feet of the C-, B-altitudes of the $ \triangle ABC.$ The circumcircle $ (P)$ of the $ \triangle AB_1C_1$ then goes through the orthocenter $ H$ of the $ \triangle ABC$ on account of the right angles $ \angle AB_1H, \angle AC_1H,$ i.e., it cuts the A-altitude at $ H,$ so that the orthocenter $ H$ of the $ \triangle ABC$ is the other fixed point of the circles $ (P).$

As the incenter $ I$ of the $ \triangle ABC$ is different from $ A, H,$ exactly one circle from the pencil $ \left \{(P) | A, H \in (P)\right\}$ goes through $ I,$ for exactly one position of $ M \in BC.$ Therefore, $ I \not\in (P)$ in general.

$ \angle B_1PC_1 \equal{} 2 \angle A$ and $ \angle B_1PC_1 \plus{} \angle B_1MC_1 \equal{} 2 \angle A \plus{} 2 \angle B \plus{} 2 \angle C \minus{} 180^\circ \equal{} 180^\circ$ $ \Longrightarrow$ $ MB_1PC_1$ is cyclic, $ P$ is on circumcircle of the $ \triangle MB_1C_1$ $ \Longrightarrow$ $ (P)$ goes through the incenter $ J$ of this triangle.

$ \mbox{Sorry! I have another problem solved... We design wrong to me: } \\
|B_1B| \equal{} |BM|(M\in[AB]); |C_1C| \equal{} |MC|(C_1\in[AC]).$

[quote="yetti"]$ I$ is incenter of the triangle $ \triangle MB_1C_1,$ isn't it ?[/quote]

Yes problem is now corrected by the moderators.

my solution is similar to yetti's
Notice ∠B1MC1=180-2∠A   C,H,C2 are collinear(∠HCC1=∠C2CC1=90-∠A)
Then the problem is trivial

[quote="Omid Hatami"]Suppose that $ M$ is an arbitrary point on side $ BC$ of triangle $ ABC$. $ B_1,C_1$ are points on $ AB,AC$ such that $ MB = MB_1$ and $ MC = MC_1$.  Suppose that $ H,I$ are orthocenter of triangle $ ABC$ and incenter of triangle $ MB_1C_1$. Prove that $ A,B_1,H,I,C_1$ lie on a circle.[/quote]
Let the altitude from $A$ to $BC$ intersect $BC$ at $Z$. Let the circumcircle of $\triangle AB_1C_1$ intersect $AZ$ at $A,H$. We will prove $H$ is the orthocentre of $\triangle ABC$. Let $AZ$ intersect the circumcircle of $\triangle ABC$ at $T$. Then it suffices to prove $HZ = ZT$.

Let $X,Y$ be the midpoint of $BB_1, CC_1$ respectively, and let $P$ be the second intersection of the circumcircles of $\triangle ABC, \triangle AB_1C_1$. Then $P$ is the center of spiral similarity that takes $B \rightarrow C, B_1 \rightarrow C_1$. Hence it takes the midpoint of $BB_1$, that is, $X$, to the midpoint of $CC_1$, that is, $Y$.

Now, the center of spiral similarity (which we already know is $P$) that takes $X \rightarrow Y$, $B \rightarrow C$ must lie on the circumcircles of $\triangle ABC, \triangle AXY$ (as $A = BX \cap CY$). Thus, $P,A,X,Y$ are concyclic. But $M,X,A,Y,Z$ all lie on the circle with diameter $AM$, hence $P,A,X,Z$ are concyclic.

Now, $P$ is the center of spiral similarity that takes $B \rightarrow B_1$ and $H \rightarrow T$, hence it must take the midpoint of $BB_1$, that is $X$ to the midpoint of $HT$. But that would mean that the midpoint of $HT$ also lies on the circumcircle of $\triangle APX$, but the only intersections with the altitude of this circle are $A,Z$, hence $Z$ is the midpoint of $HT$.


Solve for x
$2\cos x + 2\cos 3x + 2\cos 5x = 1$
$x \in [0,2\pi ]$

[quote="Bruno Bonagura"][hide]$2\cos (x) + 2\cos (3x) + 2\cos (5x) = 1$
$\cos (x) + \cos (3x) + \cos (5x) = \frac12$
$\frac{\cos\left(\frac{x + 5x}{2}\right)\cdot\sin\left(\frac{3\cdot 2x}{2}\right)}{\sin\frac{2x}{2}} = \frac12$
$\frac{\cos(3x)\sin(3x)}{\sin x} = \frac12$
$2\cos(3x)\sin(3x) = \sin x$
$\sin(6x) = \sin x$
$6x = x + 2k\cdot\pi \quad k = 1,2,3,\dotsc$
$x = \frac25 k\cdot\pi$[/hide][/quote]

You are missing some solutions
Think about the domain, K cant be 6,7,...

after you get $\sin 6x=\sin x$ , you will get

$6x = n\pi+(-1)^nx$

When $n=2k$ , $x=\frac{2k\pi}{5}$

When $n=2k+1$ , $x=\frac{(2k+1)\pi}{7}$

 :)


A particle of mass $ m$ moving along a straight line is acted on by a retarding force $ F\equal{}be^{\alpha v}$, where $ b$ and $ \alpha$ are constants and $ v$ is the velocity. At $ t\equal{}0$ it is moving with velocity $ v_0$. Find the velocity at later times.

If $ F=be^{\alpha v}$ be the magnitude of the retarding force, then the velocity after time $ t$ is given by \[ v=v_{0}-\frac{1}{\alpha} {\ln {\left(\frac{abt}{m}\right)}}\]

[hide="solution"]
Let after time $ t$, velocity of the particle be $ v$.
Thus, from Newton's law of motion,

$ -m{\frac{dv}{dt}}=be^{\alpha v}$

$ \Rightarrow - \frac{m}{b} e^{-{\alpha v}} {dv}=dt$

Integrating both sides and putting the limits, we get,

$ \frac{m}{\alpha b} \left[e^{-{\alpha v}} - e^{-{\alpha v_{0}}}\right]=t$

$ \Rightarrow e^{-{\alpha v}}=e^{-{\alpha v_{0}}}+\frac{\alpha bt}{m}$

$ \Rightarrow v=v_{0}-\frac{1}{\alpha} {\ln {\left(\frac{abt}{m}\right)}}$
We are done.....!
 :)[/hide]

Danger: for t->0 your v goes to infinity.

It's because the logharitm of a sum is not rhe sum of the logharitms; seems (for b>0) it should be

\[ v=v_{0}-\frac{1}{\alpha} {\ln {\left(\frac{abt}{m}.e^{\alpha v_{0}}+1\right)}}\]


The Michigan forum has been split into 2 forums: this public one, and a private one. If Michiganders want to join the private one, they can send a private message to tetrahedr0n, the group's moderator. Thanks.

[quote="1234567890"]If Michiganders want to join the private one, they can send a private message to tetrahedr0n, the group's moderator. Thanks.[/quote]

Yup! Thanks 1234567890.

Anyone can feel free to post here in the public one. If you are interested in the private forum, PM me and I can add you to the group.

This is annoyinh.  I keep clicking on this forum instead of the other one.....

[quote="1234567890"]Since I am now the Michigan forum's moderator, anyone who wants to join should send a PM to [b]me [/b]now instead of tetrahedr0n.[/quote]

I am wishing to join this forum.

Hmmm.

Oh yeah, I thought that the forum he is talking about was something different :roll:

[quote="1234567890"]Since I am now the [b][color=darkblue][size=200]Michigan forum[/size][/color][/b]'s moderator, anyone who wants to join should send a PM to [b]me [/b]now instead of tetrahedr0n.[/quote]
Right neil...

[quote="jhredsox"]he has a good excuse, all of this math has gotten to him

-jorian[/quote]

I don't want to say get out of the Michigan Forum, but get out.

And no, I don't do math only.

wow this is a really active forum.... :maybe:

Mostly stuff happens in the Private forum, but that is currently deceased.

deceased cuz i don't post in it anymore  :lol:

Yeah, you accounted for half of the posts.

lol only like 2 ppl who post on the michigan forum are actually from michigan. do the michiganders mind if i pretend to be from michigan and post here like i was posting on the IN forum. don't worry i won't spam it up.   :P

hey can i join the forum?
(guys, ignore the location part of my profile)

I am mad that there isn't an Alaska forum. I'm in Alaska, but I was born in Michigan, so I joined this group.

*BUMP* sorry but I live in Michigan and I haven't been in the Michigan private forum yet so could someone please add me? :)

I believe the group is open so anybody can join without having to get added.

I wasn't been able to get on the private forum today  :(  maybe a technical problem?

Well, you might want to try going to usergroups under "profile" in the top panel and joining the Michigan group - does that work?

Uh, do I have to pm you kyyuan to be added?

How can I join Michgan Private? Please help.

[quote="Aopser101"]I can't join![/quote]
if you dont live in michigan you probably cant join......

The user you are talking to does live in Michigan. To answer most people's question, levans told me that the Michigan private forum was deactivated becaise of inactivity, and they don't plan on bringing it back

Late question: Why was the forum split in the first place?


function
calculus
integration
inequalities
real analysis
real analysis solved

P2

Determine all continous functions f:[0,2^1/2]->R s.t. integral from 0 to 2^1/2 of (f(x))^2 dx = integral from 0 to 2^1/2 of (F(x))^2 dx where F?=f and F(0)=0.

This one is cauchy -schwartz again:

F^2(x)=(F(x)-F(0))^2=( \int 1*f(t) dt)^2<= \int 1^2dt * \int f^2(t)*dt=x*  \int f^2(t) dt<= x* \int f^2(t) dt all the integrals are from 0 to x exept the last one which is from 0 to sqrt(2)

cheers! :D :D

[quote="Lagrangia"]This one is Cauchy-Schwarz again:

F 2(x) \leq  x* \int_[0;\sqrt2/2] f 2(t) dt

[/quote]

We integrate this inequality on [0;\sqrt2/2]

\int_[0;\sqrt2/2]F2(x)dx \leq(\int_[0;\sqrt2/2]xdx)(\int_[0;\sqrt2/2]f2(t)dt)

 \int_[0;\sqrt2/2]F2(x)dx \leq 1/4\int_[0;\sqrt2/2]f2(t)dt

By hypothesis the integrals just above are equal, we must have

\int_[0;\sqrt2/2]f2(t)dt = 0

(f2  is continous positive with integral 0 on the compact
[0;\sqrt2/2] ) => f2=0 => f=0

The function solution of the problem is 0


A sequence of 20 terms is formed by using the rule f(x) = 4x - 1 for x= 1,2,3,...20. What is the sum of the last eight terms?

The sum of the first twenty terms is $ (79\plus{}3)*20/2\equal{}820$
The sum of the first twelve is $ (47\plus{}3)*12/2\equal{}300$
So the sum of the eight terms is $ 820\minus{}300\equal{}520$


新 年 快 乐!  :D

aww if only my labtop could let me type chinese and use chinese programs...  :mad:

woh eeng gai dway may guh zuhn swoh ching chee kwai luh!

!!  :lol:

[quote="Fraxia"]:coolspeak:  :arrow:   [size=150]红包[/size]

祝你们新年快乐！[/quote]

no fair you get money? as in hohng bao? NO FAIR lolz

I got nothing much  :rotfl:

yay lets sing the new years song!!!!

Does anyone want to start?


Choose seven numbers from the set $\{1,2,3\ldots,50\}$ in such a way that every pair in this 7-set differs more than 2. (Example: $\{1,4,7,10,13,16,19\}$. I how many ways is this possible?

[quote="boxedexe"]If we choose the 7-tuple $(a_{1},\hdots,a_{7})\rightarrow (b_{1},b_{2}-2,\hdots, b_{7}-12)$, then $(b_{1},b_{2},\hdots, b_{7})\leftarrow (a_{1},a_{2}+2,\hdots, a_{7}+12)$ gives us a set $(b_{1},b_{2},\hdots, b_{7})$ with a minimum difference of elements two less than that of $(a_{1}\hdots, a_{7})$.  As a result, we have a bijection between the sets of minimum difference of $1$ and $3$.  So the answer is $\binom{50-12}{7}=\binom{38}{7}$.[/quote]

Thanks for your answer. Could you develop your reasoning a bit, please?

Here's a better explanation.

[b]Solution.[/b]

Suppose that the 7-tuple $(a_{1},\hdots, a_{7})$ satisfies the property of a minimum difference of $3$, where $1\leq a_{1}<a_{2}<\hdots<a_{7}\leq 50$.  This implies that there exists a 7-tuple $(b_{1},\hdots, b_{7})$ with a minimum difference of $1$ between each element such that $(a_{1},\hdots,a_{7})=(b_{1},b_{2}+2,b_{3}+4,\hdots, b_{7}+12)$.

Conversely, we are to find $(b_{1},\hdots,b_{7})=(a_{1},a_{2}-2,\hdots, a_{7}-12)$. As a result, we are to find $(b_{1},\hdots, b_{7})$ such that $b_{i}\in\{1,\hdots, 38\}$.  This can be done in $\binom{38}{7}$.$\blacksquare$

[quote="boxedexe"]Here's a better explanation.

[b]Solution.[/b]

Suppose that the 7-tuple $(a_{1},\hdots, a_{7})$ satisfies the property of a minimum difference of $3$, where $1\leq a_{1}<a_{2}<\hdots<a_{7}\leq 50$.  This implies that ...[/quote]
Why does it imply that?

And why are we to find a $b$-tuple, these have a difference more than (or eq. to) one, while the desired is more than 2? Am I wrong somewhere?
Thanks.


ARe the old AMC's (1990 and earlier) still being sold somewhere? I would really like to get my hands on some on those since they offer excellent practice. Also, did AIME's exist before 1983? Since obviously they had AMC's before 1983.

From my knowledge, USAMO selection was based solely on the AHSME prior to 1983. Then they made the AHSME easier (sheesh, it was still too hard) and included another test, the AIME, for people who did exceptionally well on the AHSME. So that's why there are no AIME's before 1983.

yes, but are there books or something from which I can buy them? like all the way back to the 1950's.

Yes, there are 6 books that contain all the AHSME's from 1950 to 1994.  You can find them on the [url=https://enterprise.maa.org/ecomtpro/timssnet/books/NML.cfm]MAA site[/url].  They used to be offered on the AMC site too, but I don't see them there now.


Not all people who do math are geeks though
the reason why one of my friends first started doing math is because he broke up with his girlfriend and had to do math to escape that reality, and later his score on Cayley was like 24 higher than Pascal

According to that definition I would be a class A geek since I spend all my spare time doing math/cs.  But I barely have any spare time because I am always so busy with non-math things (sports/extracuricculars/music/etc at school).  So I don't think I deserve to be class A  :( ...maybe next year lol


It is equivalent to $ (\frac{n}{n\minus{}1})^n>(\frac{n\plus{}1}{n})^{n\plus{}1}$, or $ (1\plus{}\frac{1}{n\minus{}1})^n>(1\plus{}\frac{1}{n})^{n\plus{}1}$. But the sequence $ \{(1\plus{}\frac{1}{n})^{n\plus{}1}\}$ is decreasing (well-known).

[quote="N.T.TUAN"] But the sequence $ \{(1 \plus{} \frac {1}{n})^{n \plus{} 1}\}$ is decreasing (well-known).[/quote]

proof by AM-GM  :wink:


function
group theory
superior algebra
superior algebra unsolved

Is there an easy way to find the maximum order of an element in, say, $S_{18}$? The only ways to do it I can think of involve a lot of checking. I mean, we want to find the maximum of $lcm(a_{1},...a_{r})$ for some positive numbers $a_{i}$ with $a_{1}+\cdots+a_{r}= 18$, but this is not so easy.

Also,  I should note that I only know basic group theory, so I don't understand most advanced solutions.

If $a_{i}$ has two distinct prime factors, you can split this cycle into two and reduce their total length. Thus, each $a_{i}$ is a power of a prime. This leaves 2,3,4,5,7,8,9,11,13,17. Smaller primes are preferable here, so it seems that $2+3+5+7$ is the optimal configuration of cycles, with the order of $210$. But yes, this involves some checking.

Edit: This is called [url=http://mathworld.wolfram.com/LandausFunction.html]Landau's function[/url]; sequence [url=http://www.research.att.com/~njas/sequences/A000793]A000793[/url] in Sloane's Encyclopedia of Integer Sequences.


vector
induction
number theory unsolved
number theory

Special case of http://www.mathlinks.ro/Forum/viewtopic.php?t=27089 .
I know that pbornsztein has posted a solution to this special case in a more elementary way, but can't find it right now.

We can show that as a vector space, the dimension of the field
extension gotten from $ Q$, the field of rational numbers, by adjoining
the square roots of $ s$ distinct integers > 1 which are squarefree is $ 2^s$.  Proceed by induction on the number of such
integers adjoined.

We know it works for $ s \equal{} 1$ and $ s \equal{} 2$,(quite easy)
Assume it works for $ 1, 2, ..., s$ roots.  Let $ n_1, n_2, ... n_s, n_{s\plus{}1}$ be such a set of $ s\plus{}1$ integers.  Let us use the letters $ E$ and $ F$
to represent the following fields:

$ E \equal{} Q(\sqrt{n_1}, ..., \sqrt{n_{s\minus{}1}})$
$ F \equal{} E(\sqrt{n_s})$

By induction, $ \dim(E) \equal{} 2^{s\minus{}1}$ and $ \dim(F) \equal{} 2^s$.  We would be done
with the proof if we could show that $ \sqrt{n_{s\plus{}1}}$ is not an element
of $ F$.

Assume otherwise, that $ \sqrt{n_{s\plus{}1}}$ is in $ F$.  Then for some $ a, b$ in $ E$,
we can write:

$ \sqrt{n_{s\plus{}1}} \equal{} a \plus{} b\sqrt{n_s}$

Squaring both sides gives:

$ n_{s\plus{}1} \equal{} a^2 \plus{} 2ab\sqrt{n_s} \plus{} b^2n_s$ or
$ 2ab\sqrt{n_s} \equal{} n_{s\plus{}1} \minus{} a^2 \minus{} b^2 n_s$

The righthand side lies in $ E$.  Three cases are possible:

1) a = 0
2) b = 0
3) $ \sqrt{n_s}$ is in $ E$.

all the three give contradiction

Thanks to mathmanman for sending me the link: http://www.mathlinks.ro/viewtopic.php?t=4496 .


calculus
integration
trigonometry
calculus computations

show :D

$ \int_{ \minus{} \infty}^\infty\;\frac {b\cos\,kx \plus{} (x \minus{} a)\sin\,kx}{b^2 \plus{} (x \minus{} a)^2}\;\;\textbf dx\;\; \equal{} \;\;\boxed{2\pi\,e^{ \minus{} kb}\cos\,ka}$

do [b]not[/b] use techniques from complex analysis.

by substituting $ x \minus{} a \equal{} z$ we will get integral of the forms: ( 1)+2)+3)+4) ) where
1)$ b_{1} \int_{ \minus{} \infty}^{\infty}\frac {\cos mz}{c^{2} \plus{} z^{2}}dx$
2) $ b_{2}\int_{ \minus{} \infty}^{\infty}\frac {\sin pz}{c^{2} \plus{} z^{2}}dx$

and
3) $ b_{3}\int_{ \minus{} \infty}^{\infty}\frac {z\sin qz}{c^{2} \plus{} z^{2}}dx$
 and
4)$ b_{4}\int_{ \minus{} \infty}^{\infty}\frac {1}{c^{2} \plus{} z^{2}}dx$
all of this can be easy calculated . 4) it's $ arctan$ 3) it's zero . 2),1) it's wellknown Laplace integrals.


Dear AoPS,

My last post asking about Vedic math was deleted, perhaps because I posted a company name and web site? I apologize if I made a gaffe, and will keep this version more generic.  Does anyone have any thoughts on whether teaching the Vedic system of math (from India, I think) to elementary school kids in the U.S. is helpful in having them grasp the concepts, or harmful because it just confuses them? I have a fairly bright kiddo going into kindergarten, and he has the opportunity to learn Vedic math, which seems handy from what I've read here and there on the web, but I am just a mom and not a math teacher...

Thanks in advance for thoughts and opinions posted.

MathMomSD

This should probably be moved to the Round Table and merged with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1968600307&t=152022]this thread[/url].

As to the question itself, tricks for quick multiplication and such is not what mathematics is really about (but the ideas behind them are), so a lot depends on how such things are taught (by the way, this remark applies to almost any math. subject). Memorizing the techniques without understanding what's behind them is rather useless but analyzing why $ 125^2\equal{}(12*13)*100\plus{}25\equal{}15625$ may be quite beneficial for understanding how our decimal number system works.

The forum that this was originally posted in is intended for questions, suggestions, etc. about this website, not for general questions or discussions.

WELL, i dont understand what trick u wrote above , but i have a far simpler trick to square things by vedic maths.and many questions are solved just by the process it is done. i have solved 2 on aops., it is quite helpful
however i am still, analyzing why it works.

vedic maths should not be taught to kids as at that time concepts should built up rather than learning tricks to solve faster without even knowing how they work. working with formulas which we know how to derive makes understanding of numbers better. afterwards they may see through those formulas , try to derive them, but there's no point cramming them
  however, i believe if i prove the formula i have , it would be quite knowledgeable concept clearing for me.

[quote="analytic"]WELL, i dont understand what trick u wrote above.[/quote]
If you want to square a number ending by $ 5$, remove $ 5$, multiply the resulting number $ A$ by $ A\plus{}1$ and add $ 25$ to the end (like $ 75^2\equal{}5625$ because $ 7\cdot 8\equal{}56$). Of course, this particular trick may be not really representative of Vedic math., but it shows the point: memorizing it is rather useless unless you want to impress your friends with your counting skills, but if a small kid understands that $ (10A\plus{}5)^2\equal{}100A^2\plus{}100A\plus{}25\equal{}100A(A\plus{}1)\plus{}25$ (you do not really need algebraic notation in this formula; a few number examples will convey the idea equally well), he will be on his way to building basic algebra skills.

well, yes if a child understands the trick u mentioned it would be quite nice.
however, i am working on squaring trick which gets squares of any no.

but that trick is simple algebra, i think not vedic maths

All vedic math. tricks are "simple algebra" once you figure out how they work  :) The standard multiplication table above $ 5\times 5$ thingy (which is what they often start with) is just the identity $ (10\minus{}a)(10\minus{}b)\equal{}10(10\minus{}a\minus{}b)\plus{}ab$. The way it is usually presented is "take the complement of the second large number to $ 10$ (i.e., $ b$), subtract it from the first number (i.e., get $ 10\minus{}a\minus{}b$), add $ 0$ to the result (i.e., write the first term) and add the product of the complements (add $ ab$)". Take $ 8\times 7$ for instance. The complements are $ 2$ and $ 3$. The difference is $ 8\minus{}3\equal{}5$. Write $ 50$. Add $ 2\times 3\equal{}6$. Get $ 56$.

I would think that until a human is around 10-12, you should not expect much abstract thought. Maybe you could interest them in $ (a\plus{}b)^2\equal{}a^2\plus{}b^2\plus{}2ab$, which can be arrived at by asking what it would take to enlarge a square patio $ a$ tiles per side by an extra $ b$ tiles, keeping it square. There probably are some children under 12 who would think for a while and tell you that the square of a sum is the sum of two sqares plus two rectangles. The rare ones who do so should be recruited for math competitions.
I would just concentrate on getting the younger kids to understand the rationale behind the really common algorithms their culture expects them to memorize and use at their age. Maybe in India it would involve some vedic math tricks. In the US, my biggest concern would be long division and adding fractions. Of course, earlier yet it would be nice for the little tykes to understand place value in our base 10 number system, know why thay have to carry the "1" when adding and why they have to regroup when calculating differences like $ 41\minus{}16$. I am betting that their teachers did not give an explanation that they could understand.

[quote="fedja"]All vedic math. tricks are "simple algebra" once you figure out how they work  :) The standard multiplication table above $ 5\times 5$ thingy (which is what they often start with) is just the identity $ (10 \minus{} a)(10 \minus{} b) \equal{} 10(10 \minus{} a \minus{} b) \plus{} ab$. The way it is usually presented is "take the complement of the second large number to $ 10$ (i.e., $ b$), subtract it from the first number (i.e., get $ 10 \minus{} a \minus{} b$), add $ 0$ to the result (i.e., write the first term) and add the product of the complements (add $ ab$)". Take $ 8\times 7$ for instance. The complements are $ 2$ and $ 3$. The difference is $ 8 \minus{} 3 \equal{} 5$. Write $ 50$. Add $ 2\times 3 \equal{} 6$. Get $ 56$.[/quote]

probably, u havnt come across good tricks.anybody wh doesnt know vedic maths would have applied this, e.g. i did before. this was a simple derivation u did. it wuold be impressive if u come out with a trick for squares with explanation [b]for all numbers.[/b]

@kmst
[quote]Maybe in India it would involve some vedic math tricks.[/quote]
well, it is not common in india. books relating can be found in fairs, but children dont buy such books here, only few people of around 18 buy these. children are made to form concepts and if they have such a trick they must have a explanation or they cannot use it. and this topic was about vedic maths  at 12, i agreee with others that concepts should be built rather.


Hello, my sister is doing MOEMS.  She is in 6th grade and i was wondering if you guys any tips for getting better at these types of problems.  Thanks.

also check your work and do practice problems.

practice and work hard with determination. Also dont get tensioned when doing the problems.[/b]

one thing really important is to solve different types of problems. there are many different types of math problems out there (like time-distance problems, equation problems, etc.).

[quote="Naonao"]My suggestion is to learn to READ THE QUESTION i got a 3 cuz i can't read the question, i always miss 1 or 2 words that are VERY VERY important.[/quote]

SO IMPORTANT!! I always have that problem..

MOEMS gives alot of time per question.  Alot of people at my school turned in their answers with 10-15 minutes left.  I think that checking your work during extra time can really help.

Please don't revive threads from the previous year.


I like these inequalities, each time when I have hunger I take one of them and eat it up  :D  :D

In fact, you can take the first of the three fractions and rewrite its denominator as

  $\left( 1+x+y\right) \left( 1+x+z\right) = 1+x+y+x+z+x^2+xy+xz+yz$
  $= 1 + x + \left( x+y+z\right) + x \left( x+y+z\right) + yz = 1 + x + 1 + x + yz = 2 + 2x + yz$,

since x + y + z = 1. Thus,

  $\displaystyle \frac{yz+2x}{\left( 1+x+y\right) \left( 1+x+z\right)} = \frac{yz+2x}{2+2x+yz} = 1 - \frac{2}{2+2x+yz}$.

Similarly we find

  $\displaystyle \frac{zx+2y}{\left( 1+y+z\right) \left( 1+y+x\right)} = 1 - \frac{2}{2+2y+zx}$;
  $\displaystyle \frac{xy+2z}{\left( 1+z+x\right) \left( 1+z+y\right)} = 1 - \frac{2}{2+2z+xy}$.

Hence, your inequality

  $\displaystyle \frac{yz+2x}{\left( 1+x+y\right) \left( 1+x+z\right)} + \frac{zx+2y}{\left( 1+y+z\right) \left( 1+y+x\right)} + \frac{xy+2z}{\left( 1+z+x\right) \left( 1+z+y\right)} \leq \frac{21}{25}$

is equivalent to

  $\displaystyle \left( 1 - \frac{2}{2+2x+yz} \right) + \left( 1 - \frac{2}{2+2y+zx} \right) + \left( 1 - \frac{2}{2+2z+xy} \right) \leq \frac{21}{25}$.

By reversing the sign, we see that this is the same as

  $\displaystyle \frac{2}{2+2x+yz} + \frac{2}{2+2y+zx} + \frac{2}{2+2z+xy} \geq \frac{54}{25}$.

Dividing by 2, this simplifies to

  $\displaystyle \frac{1}{2+2x+yz} + \frac{1}{2+2y+zx} + \frac{1}{2+2z+xy} \geq \frac{27}{25}$.

But from HM-AM or from Cauchy-Schwarz, you know that

  $\displaystyle \frac{1}{2+2x+yz} + \frac{1}{2+2y+zx} + \frac{1}{2+2z+xy} \geq \frac{9}{\left( 2+2x+yz\right) + \left( 2+2y+zx\right) + \left( 2+2z+xy\right) }$.

Hence, it remains to show that

  $\displaystyle \frac{9}{\left( 2+2x+yz\right) + \left( 2+2y+zx\right) + \left( 2+2z+xy\right) } \geq \frac{27}{25}$.

But

  (2 + 2x + yz) + (2 + 2y + zx) + (2 + 2z + xy) = 6 + 2 (x + y + z) + (yz + zx + xy) = 6 + 2 + (yz + zx + xy) = 8 + (yz + zx + xy)

using x + y + z = 1 again; also, since we have $yz+zx+xy \leq \frac13 \left( x+y+z\right) ^2$ for any three positive numbers x, y, z, we have

  $\displaystyle \left( 2+2x+yz\right) + \left( 2+2y+zx\right) + \left( 2+2z+xy\right) = 8 + \left( yz+zx+xy\right)$
  $\displaystyle \leq 8 + \frac13 \left( x+y+z\right) ^2 = 8 + \frac13$

(remember that x + y + z = 1). Hence,

  $\displaystyle \frac{9}{\left( 2+2x+yz\right) + \left( 2+2y+zx\right) + \left( 2+2z+xy\right) } \geq \frac{9}{8+\frac13} = \frac{27}{25}$,

completing our proof.

  Darij


I'm having a lot of trouble prooving this trig identity. I need some help!

$\frac{sin(x)+tan(x)}{cos(x)+1}=tan(x)$

i dont think thats right..like the equality doesnt hold..

[hide]Using $\sin{x}=\cos{x}\tan{x}$

We see..

$\frac{\sin{x}+\tan{x}}{\cos{x}+1}=\frac{\cos{x}\tan{x}+\tan{x}}{\cos{x}+1}=\frac{\tan{x}(\cos{x}+1)}{\cos{x}+1}=\tan{x}=\boxed{\text{R.S.}}$[/hide]

Actually, it does hold.

[hide]$\frac{\sin x+\tan x}{\cos x+1}= \frac{\sin x+\frac{\sin x}{\cos x}}{\cos x+1}$

$= \frac{\frac{\sin x \cos x+\sin x}{\cos x}}{\cos x+1}$

Factoring $\sin x$ out of the numerator and changing the multiple denominators to a single one:

$= \frac{\sin x(\cos x+1)}{\cos x (\cos x+1)}$

$= \tan x \frac{\cos x+1}{\cos x+1}= \tan x$.[/hide]

Man, that's more complicated than SM4RT's solution. Figures that I was typing that as he did his.

Wow.... i feel stupid lol. Exams do wonders to your mind capabillity.
Thanks guys!!

[quote="Jeremic"]SOrry bout that I think I'm brain dead... I've been trying to solve it for over 15 minutes...[/quote]

Oh no, a whole 15 minutes :!:  :!:

 :| That's not saying too much on a forum where people take tests that are 9 hours long and have 6 problems lol


I have seen another solution. where 1+2x = a. The LHS becomes a function F(x) where you can easily see that F(a) = F(a/3). Nice and elegant!


given sinhx=tan b 0<b<(pi/2), express cosh x and tanh x in terms of B ?

As I say I feel like I'm missing something obvious but I am shattered so... :)

Use [[LaTeX]] -- "\sinh x" gives $ \sinh x$, and so on.

Probably, you should try to remember some identities that relate values of the various hyperbolic trig functions.


[hide="hint"]what is 1+100?  what is 2+99?  what is 3+98?...what is 50+51?[/hide]

[hide="hint"]
use a bit of physics
[/hide]
[hide="hint2"]
In physics,
$\text{Avg velocity}=\frac{u_1+u_2}{2}$
[/hide]

[quote="riddler"] [hide="hint2"]
In physics,
$\text{Avg velocity}=\frac{u_1+u_2}{2}$
[/hide][/quote]
Nope. The avg speed is a HM not an AM and only if all distances ([tex]d_i[/tex]) are equal. Otherwise it's a weighted HM.

By definition: [tex]\overline{v}=\frac{\sum\limits_{i=1}^n d_i}{\sum\limits_{i=1}^n t_i}=\frac{\sum\limits_{i=1}^n d_i}{\sum\limits_{i=1}^n \frac{d_i}{v_i}}[/tex]

So if a car travels from A to B at 60mph and then from B to A at 40mph then it's making 48mph average and not 50mph.

[quote="nikkimarie"]What is the sum of all the integers from 1 to 100 (i.e. 1 + 2 + 3 + . . . + 100)?  How about from 1 to 1000? (Try to think of a way to avoid adding these all up one at a time![/quote]

$\frac{n(n+1)}{2}$. Why is this in intermediate ;) ?

Well, the physics application was interesting.

Hint:Hint:Hint:
[hide]
This will really give it away
[hide]Look under the toolbar of this website  :P [/hide]
[/hide]


analytic geometry
logarithms
function
graphing lines
slope
rate problems
parameterization

Point p(x,y) moves in the XY-plane in such a way that dx/dt=1/(t+1) and dy/dt=2t for t > or = 0.
a) Find the coordinates of P in terms of t if, when t=1, x=ln2 and y=0.
b) Write an equation expressing y in terms of x.
c) Find the average rate of change of y [u]with respect to x[/u] as [u]t[/u] varies from 0 to 4.
d) Find the instantaneous rate of change of y with respect to x when t=1.

ahhhh @_@  :huh:

$ dx \equal{} \frac {dt}{t \plus{} 1} \equal{} > x \equal{} \ln(t \plus{} 1) \plus{} C$. Now x = ln 2 when t = 1 implies $ \ln 2 \equal{} \ln 2 \plus{} C \equal{} > C \equal{} 0$ and so $ x \equal{} \ln(t \plus{} 1)$.

$ dy \equal{} 2tdt \equal{} > y \equal{} t^2 \plus{} C'$. y = 0 when t = 1 implies $ 0 \equal{} 1 \plus{} C' \equal{} > C' \equal{} \minus{} 1$ and so $ y \equal{} t^2 \minus{} 1$.

From $ x \equal{} \ln(t \plus{} 1)$ we get $ t \plus{} 1 \equal{} e^x \equal{} > t \equal{} e^x \minus{} 1$. Now substituting this on the parametric equation for y we have

$ y \equal{} (e^x \minus{} 1)^2 \minus{} 1 \equal{} > y \equal{} e^{2x} \minus{} 2e^x \equal{} > y \equal{} e^x(e^x \minus{} 2)$.

Now try to solve c) and d) by yourself.

ok so the answer to part a (P(x,y)) is p(ln(t+1),t^2-1)? also just checking...

c) dy/dx= 2t/(1/(t+1))=2t*(t+1)=2t^2+2t
dy/dx @ t=0 = 0
dy/dx @ t=4 = 2(4)^2+2(4)=40
soo... dy/dx @ t=4 minus dy/dx @ t=0 equals 40?

d) dy/dx @ t=1 = 2(1)^2+2(1) = 4?

Does that all look right? and btw, thank you very much for helping me out =)

For (c), the [i]average[/i] rate of change is given by $ \frac{y(4)\minus{}y(0)}{x(4)\minus{}x(0)}$. The numerator is the net change in $ y$ between $ t\equal{}0$ and $ t\equal{}4$, and the denominator is the net change in $ x$. (Here $ x(t)$ and $ y(t)$ are functions of $ t$.)

Although the formula $ \frac{dy}{dx}\equal{}\frac{y^\prime (t)}{x^\prime (t)}$ is correct (and can even be used when $ y^\prime (t)$ and $ x^\prime (t)$ are both 0, if the limit of $ \frac{y^\prime (u)}{x^\prime (u)}$ as $ u\rightarrow t$ exists), $ \frac{dy}{dx}$ is the [i]instantaneous[/i] rate of change.

You can also see the difference using a graph of $ y$ versus $ x$: the average rate of change between two points is slope of the line connecting those points, while the instantaneous rate of change at a point is the slope of the tangent line at that point.


let $ a,b,c\geq 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that
$ \sum\frac {1}{\sqrt {a^2 \plus{} ab \plus{} b^2}}\geq 4 \plus{} \frac {2}{\sqrt {3}}$

Reference: [url]http://vnineqmath.hnsv.com/viewtopic.php?f=46&t=90&st=0&sk=t&sd=a&start=10[/url]

This is inelegant but I will post it in the absence of anything else.
It is enough to assume we are in the region R: a>=b>=c>=0.
Call the function under consideration phi(a,b,c).
Then first show that in the region R, d(phi)/da>=d(phi)/db.
Hence the min of phi over R on any line a+b=const actually occurs on a=b.
Then we just have to consider Phi(c)=phi((1-c)/2,(1-c)/2,c) for c in [0,1/3].
We have to show Phi(c)>=Phi(0) and this can be done by interval subdivision,
or I suppose various other methods.
I can type in more details if that would help.


The areas of two squares are in the ratio of 3:1.  Find the ratio of the perimeter of the larger square to the perimeter of the smaller square.  Express your answer in the form of x:1 with x rounded to the nearest tenth.

[hide]The ratio is 3:1.  so the side of the larger is  :sqrt: 3x and the sides of the smaller one is 1x.  Since one side is already 1, you dont need to find anything else.  root three is closest to 1.7[/hide][/hide]

There seem to be serious problems with 10ths here ...

heh, scrambled has the answer; I think churchill needs to rethink the difference between nearest tenth and nearest hundredth.

Actually, Scrambled's first answer was 2, which was also not nearest 10th.  He's fixed it, though.


trigonometry
geometry
circumcircle
geometry solved

Let ABC be a non-iscoceles triangle . Let P be a point s.t. PA=PB and PB is perpendicular to BC. similarly, let Q be a point s.t. QA=QC and QC is perpendicular to BC. Let PQ meet BC at R. Prove AR is tangent to the circumcircle of ABC.

I might be way off, but wasn't this proiposed by Valentin for some BMO?

Anyway, we have to show that $\frac {RB}{RC}=\frac {AB^2}{AC^2}$. This is equivalent to $\frac {AP}{AQ}=\frac {AB^2}{AC^2}$, but a quick angle chase shows that triangles $ABC,\ APO,\ AOQ$ are similar (with this orientation), so $\frac {AP}{AO}=\frac {AB}{AC},\ \frac {AO}{AQ}=\frac {AB}{AC}$. We multiply these and we're done.


agha salam
agha bebakhshid ke daram idatoono kopi mikonam vali man in topic ro baz mikonam ta hame be ziyafate inequalities sarmast beshim...
faghat kheili entezaar nadashte bashin man khodam masale ha ro hal konam chon man koochek tar azin harfa hastam :blush:  :oops:

first problem...
farz konid $a,b,c$ adaade haghighi bashand.be tori ke $abc=-1$.neshan dahid:
$a^4+b^4+c^4+3(a+b+c)\geq \frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{c}+\frac{b^2}{a}+\frac{c^2}{a}+\frac{c^2}{b}$
be inke doroste shak nakonin... :)

salam marhale dovome chand saal pishe
agar ghalame latex yaari kone migam halesho bedrood

salam mojadad :
tarafe chap ra dar $-1$ zarb mikonim va tarafe raast ra dar  $xyz= -1$ zarb mikonim bedast miayad:
$x^4 + y^4 + z^4 + 3(x + y + z) + x^3 .y + x^3 .z + y^3 .x + y^3 .z + z^3 .x + z^3 .y \geq 0$
ke pas az sade saazi:
$(x+y+z). (x^3 + y^3 + z^3 + 3) \geq 0$
hala bejaaye $3$ mizaarim $-3xyz$ va az etehade maroofe $oiler$ estefaade mikonim:
$(x+y+z). (x+y+z).(x^2 + y^2 + z^2 - xy-xz-yz)\geq 0$
ke bargharaare ;)  va haminaa ro barmigar doonim bala ta hokm bedast biyad
omid varam eshtebah naneveshte basham

agar ghabool daari ta maale khodamo befrestam haal konin

az halle zibaye shoma besyar lezzat bordam.khob hala masala ro begin lotfan

[quote="pirhoosh"]secend :
salam in soal ro tooye "nice_ineq_dars_va_masale__610"" maale khodete " didam va benazaram khoob oomad:
$let a , b ,c > 0$
$(\frac{a}{b} + \frac{b}{c}+ \frac{c}{a} )^2 \geq (a+b+c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
inshalah dorost biyad baraatoon[/quote]

Salom aleykum... ba fikri man in nobarobarii shoma fakat baroi abc=1 durust ast  ;)

Davron

na fekonam baraaye har se adade mosbati doroste

khodam halesh kardam
haalaa baghiye nazar bedan taman dobaare baresi konam vali motmaen bash dooste aziz

[quote="amir2"]ghablan hal shode .soale marhale sevome hamin tabestoon boode.
http://www.mathlinks.ro/Forum/viewtopic.php?p=316311#316311[/quote]
 mamnoon amir2
 vali man yetore dige halesh kardam (albate fekr mikonam!) say mikonam halamo benevisam
ta shoma bozorgaan bebinid.

salam javaabe masalaro rooye linke bala post kardam bedrood

in ham az tarafe man. :)
agar $a+b+c=3$ bashad sabet konid:
$\frac{a}{b^2 +1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\geq \frac{3}{2}$

salam dooste aziz soale ghashangiye faghat agar momkene begid $a,b,c$ dar che majmooei hastand

salam,
$a,b,c$ mosbat hastand va ozve $\mathbb{R}$ :)


number theory
relatively prime
Diophantine Equations

We start with the following observation: if $ a^x \equal{} b^y$ with $ a, b, x, y$ positive integers and $ (x, y) \equal{} 1$ ten $ a \equal{} c^y$ for some $ c$. This is almost obvious: take prime $ p|a$ and $ p^{\alpha}\parallel{}a$. Then $ y|\alpha x$, so $ y|\alpha$.

We can write $ r \equal{} \frac {a}{b}$ with $ a, b > 0$ and $ (a,b) \equal{} 1$. Suppose that $ \sqrt [r \minus{} 1]{r} \equal{} \frac {c}{d}$ for some positive, relatively prime integers $ c, d$. Then we of course have $ \frac {a^b}{b^b} \equal{} \frac {c^{a \minus{} b}}{d^{a \minus{} b}}$ which clearly forces $ a^b \equal{} c^{a \minus{} b}$ and $ b^b \equal{} d^{a \minus{} b}$.  We must have $ a > b$ and since $ (a, b) \equal{} 1$ we also have $ (b, a \minus{} b) \equal{} 1$, so it follows from our observation that $ a \equal{} x^{a \minus{} b}$ and $ b \equal{} y^{a \minus{} b}$ for some positive integers $ x, y$ with $ x > y$. But if $ a \minus{} b > 1$:
$ a \minus{} b \equal{} x^{a \minus{} b} \minus{} y^{a \minus{} b} \geq (y \plus{} 1)^{a \minus{} b} \minus{} y^{a \minus{} b} \equal{} y^{a \minus{} b \minus{} 1}(a \minus{} b) \plus{} ... \plus{} 1 > a \minus{} b$
Hence $ a \minus{} b \equal{} 1$ and it's also easy to check that it is the sufficient condition.
So: $ r\equal{}\frac{p\plus{}1}{p}$ with $ p$ positive integer.

??? The order of a root is a NATURAL number, not a fraction...

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=376440& think this is from 1999 rather than 2000...

[quote="Peter"]Determine all positive rational numbers $r \neq 1$ such that $\sqrt[r-1]{r}$ is rational.[/quote]
Do you mean that r-1 can be a rational number that isn't an integer number?

[quote="Peter"]Determine all positive rational numbers $r \neq 1$ such that $\sqrt[r-1]{r}$ is rational.[/quote]
Rewrite $r^{\frac{1}{r-1}}$.
For Gelfond's theorem $ a^b $ with $a$ rational can be rational only if b is an integer or if $ a=c^ \frac{1}{b} $ for some rational c. Under the conditions of the problem, b is an integer only if r=2 or r=1+1/n, n natural number.

[quote="Weeks_behind"][quote="ctumeo"][quote="Peter"]Determine all positive rational numbers $r \neq 1$ such that $\sqrt[r-1]{r}$ is rational.[/quote]
Rewrite $r^{\frac{1}{r-1}}$.
For Gelfond's theorem $ a^b $ with $a$ rational can be rational only if b is an integer or if $ a=c^ \frac{1}{b} $ for some rational c. But this is clearly impossible under the conditions of the problem.
If you don't agree, post a counterexample (i.e. any rational values for r).[/quote]

What about $r=2$?[/quote]
I've edited my post.


inequalities
function
calculus
calculus computations

Can anyone clearly explain the concept of Lagrange multipliers and recommend  good books that treat the matter clearly from the beginning up to an advanced level and cite examples how they can be helpful in Olympiad level inequalities?

$ Thanks$   $ in$   $ advance$

PS:I heard that use of Lagrange multipliers make inequalities cumbersome..... :wink: Is that so?

In a nutshell, [url=http://en.wikipedia.org/wiki/Lagrange_multipliers]Lagrange multipliers[/url] allow you to find the maxima and minima of functions subject to constraints.


function
inequalities
induction
Miscellaneous Problems
pen

The sum of the digits of a natural number $n$ is denoted by $S(n)$. Prove that $S(8n) \ge \frac{1}{8} S(n)$ for each $n$.

Some simple and often used properties of the function $ s(n)$ are: $ s(a \plus{} b)\le s(a) \plus{} s(b)$. This is really easy to prove by analyzing the process of adding $ a$ and $ b$ 'in column'. The second property is $ s(mn)\le s(m)s(n)$. To see this let firstly note that for a digit $ 1\le d\le9$ by the first property we have $ s(md)\le ds(n)$. Now let $ n \equal{} \overline{a_1a_2\ldots a_k}$ be the decimal representation of $ n$. Then $ s(mn) \equal{} s(ma_110^{k \minus{} 1} \plus{} ma_210^{k \minus{} 2} \plus{} \ldots \plus{} ma_{k \minus{} 1}\cdot10 \plus{} ma_k)\le$ $ s(ma_110^{k \minus{} 1}) \plus{} s(ma_210^{k \minus{} 2}) \plus{} \ldots \plus{} s(ma_{k \minus{} 1}\cdot10) \plus{} s(ma_k) \equal{} E$. Further $ E \equal{} s(ma_1) \plus{} s(ma_2) \plus{} \ldots \plus{} s(ma_k)\le s(m)[a_1 \plus{} \ldots \plus{} a_k] \equal{} s(m)s(n)$ and we are done.

Now the desired inequality immediately results from $ s(n) \equal{} s(1000n) \equal{} s(125\cdot8n)\le8s(8n)$.

wow, the above is elegant. I found two proofs based on induction, one of which I will give in full detail. Every letter will denote a positive integer.

Assume we have $S(8n) \ge \frac{1}{8}S(n)$ for all $n \le 1.25 * 10^{k+2}$ for some $k \ge 1$ (it can be easily checked to hold for $k = 1$). Then, for every $a \le 10^k$, we have:

$S(a) = S(1000a) = S(8*(125a)) \ge \frac{1}{8}S(125a) = \frac{1}{8}S(1250a)$

Now, set $n = 1250a + b$ with $a \le 10^k$, $b \le 1249$. We then have:

$\mbox{S(8n) = S(10000a + 8b) = S(a) + S(8b) \ge \frac{1}{8}S(1250a) + \frac{1}{8}S(b) \ge \frac{1}{8}S(1250a + b) = \frac{1}{8}S(n)}$

And we see that our inequality holds for all $n \le 1250*10^k = 1.25*10^{k+3}$. This finishes our induction step.

The second proof, while also based on induction, is completely different; it feels more like a direct computation.
Let $k = k(n)$, $n = \sum_{j=0}^k A_{j,n}*10^j$ and let $n'$ be $n$ without its first digit; $n' = \sum_{j=0}^{k-1} A_{j,n}*10^j$. Now, we have the following (non-trivial) property that a smallest counterexample $> 125$ to the inequality $S(8n) \ge \frac{1}{8}S(n)$ must have: $100*A_{i+2,n} + 10*A_{i+1,n} + A_{i,n} > 125$ for all $0 \le i \le k - 2$. In essence this boils down to the fact that a counterexample that doesn't satisfy that property can be split into two smaller parts, at least one of which also has to be a counterexample. For example: if 9870654 is a counterexample (note that 100*0 + 10*6 + 5 = 65 < 125), either 987 or 654 must be a counterexample too. Now, assume that $S(8n) \ge \frac{1}{8}S(n) + A_{k,8n}$ holds for all $n < 10^m$ that satisfy $100*A_{i+2,n} + 10*A_{i+1,n} + A_{i,n} > 125$ for all $0 \le i \le k - 2$ and set $n$ between $10^m$ and $10^{m+1}$. We can then write $S(8n)$ in terms of $S(8n')$ and some digits of $8n$ and $8n'$. Now, use the induction hypothesis on $S(8n')$, use the fact that $S(n') = S(n) - A_{k,n}$ and you're basically done. If I'm not mistaken, it can be shown that $S(8n) \ge \frac{1}{8}S(n) + A_{k,8n}$ holds for all $n > 1$, which is (albeit very marginally) more than what was asked for. But it should be clear that the extra term $A_{k,8n}$ came with a price: a longer and uglier proof that has no way of being generalised and is nowhere near as powerful as the easy and beautiful inequality of freemind: $s(mn) \le s(m)s(n)$

[hide=Solution][b]Lemma 1:[/b] $S(m)S(n)\ge S(mn)$

Proof: Let $n = \overline{d_kd_{k-1}\cdots d_0}$. Then

$$S(m)S(n)=S(m)\sum_{i=0}^k d_i \ge \sum_{i=0}^k S(md_i) = \sum_{i=0}^k S(md_i\cdot 10^i) \ge S(\sum_{i=0}^k md_i\cdot 10^i)=S(mn).$$

Using the above lemma, $8S(8n)=S(125)S(8n)\geqslant S(1000n)=S(n)$, as desired.[/hide]


Now your horoscope is screwed up because Pluto isn't a planet anymore, congratulations.

Bahah now the planet is Mars. God of war FTW. :lol:

This is a crisis for astrologers!!! Nergal?!?? I hope the ecliptic curve is still accurate...oh and I'm not a scorpio sorry.

Well when I read secrets of the vaulted sky it was all crap anyway.

[quote="alex__ib"]I don't believe in these things because there is NO MATHEMATICAL PROOF for them!  :D[/quote]

So by your standards, you do not believe in math. As there is no mathematical proof for axioms so you don't believe in those. But since everything in math relies on axioms, you do not believe in anything in math.

Yes Nergal can't hold as two planets therefore Pluto is a planet astrologically but not scientifically  :roll:  Really does it ruin the ecliptic line?

let's just say it's all too far away to matter
by the way I am a scorpio...

but it's your destiny *looks like crazy old prof. trelawney*


logarithms
probability
probability and stats

Let $ X_i, \ i\in \mathbb{N}$ be centered iid random variables, $ S_n \equal{} \sum^n X_i$. Let $ T_n \equal{} \frac {S_n}{\sqrt {X_1^2 \plus{} \dots \plus{} X_n^2}}$. What is the possible behaviour of $ T_n$ in the sense of weak limit?

[quote="Yustas"]Let $ X_i, \ i\in \mathbb{N}$ be centered iid random variables, $ S_n = \sum^n X_i$. Let $ T_n = \frac {S_n}{\sqrt {X_1^2 + \dots + X_n^2}}$. What is the possible behaviour of $ T_n$ in the sense of weak limit?[/quote]

If random variables $ X_i$ have finite variance then
\[ T_n = \frac {S_n/\sqrt {n}}{\sqrt {(X_1^2 + \dots + X_n^2)/n}} \Rrightarrow{d} {\mathcal N}(0,1),
\]
since $ S_n/\sqrt {n} \Rrightarrow{d} {\mathcal N}(0,\sigma^2)$ and $ (X_1^2 + \dots + X_n^2)/n \Rrightarrow{a.s.} \sigma^2$.

However, if the variance of $ X_i$ is infinite, the sequence $ T_n$ can diverge (or converge to a non-Gaussian distribution).

The case of finite variance is, of course, trivial. The more interesting question is to give examples for the second part.

I. Here is an example when $ T_n$ converges to a non-Gaussian distribution. Let $ Y_i$ be a r.v uniformly distributed on $ [0,1]$ and $ X_i \equal{} \plus{} e^{1/Y_i}$ with prob. $ 1/2$ and  $ X_i \equal{} \minus{} e^{1/Y_i}$ with prob. $ 1/2$
Let $ Y_{i^*}$ be the smallest among $ Y_i$ (where $ i^*$ is a r.v.). Note that with prob. $ 1 \minus{} o(1)$:
 i) $ Y_{i^*} < 1/n^{0.9}$;
 ii) $ \min_{i\neq i^*} Y_i > Y_{i^*} \plus{} 1/n^{1.1}$.
Therefore, for every $ i\neq i^*$, with prob. $ 1 \minus{} o(1)$
\[ \frac {1}{Y_{i^*}} \minus{} \frac {1}{Y_i} > (1 \minus{} o(1)) n^{0.7}.
\]
Thus
\[ |X_{i^*}/X_i| > e^{(1 \minus{} o(1))n^{0.7}} > n^2
\]
(for suff. large $ n$). Hence $ S_n \equal{} X_{i^*} (1 \plus{} o(1))$ and $ T_n \equal{} \text{sign}\ X_{i^*}$ with prob. $ 1 \minus{} o(1)$. That is,
\[ T_n \xrightarrow{d} \text{Bernoulli distribution}.
\]
II. Define a sequence $ a_i$ by $ a_0 \equal{} 1/2$, $ a_{i \plus{} 1} \equal{} a_i^2$. Let $ A \equal{} (a_1,a_0) \cup (a_3,a_2)\cup (a_5, a_4), \dots$; $ B \equal{} (a_0,1) \cup (a_2,a_1) \cup (a_4,a_3)\cup (a_6, a_5), \dots$.
Define $ X_i$ as follows:
\[ X_i \equal{} \begin{cases} \pm e^{1/Y_i}, & \text{ if } Y_i\in B; \\
\pm e^{1/a_{2k \plus{} 1}}, & \text{ if } Y_i\in (a_{2k \plus{} 1},a_{2k}) \end{cases}
\]
(we choose $ \plus{}$ or $ \minus{}$ with prob. $ 1/2$; $ Y_i$ is the same as in part I)
It is easy to see that for some $ k$ the distribution of $ T_k$ is close to the Bernoulli distribution (as in part I), and for some it is close to the Gaussian distribution. Thus $ T_n$ doesn't converge in distribution.

Looks interesting, but i think in your example the random variables have infinite expectation: indeed, the density is something like $ \frac{1}{t(\ln t)^2}$.

Of course, it has no expectation. But one can construct a similar example with a finite first moment but infinite second moment. Let $ a_n$ be a very fast growing sequence; say, $ a_0 \equal{} 1$, $ a_{i\plus{}1}\equal{}2^{a_i}$. Let $ A\equal{}\{a_i\}_i$. Define $ Y_i \equal{} k$ with probability $ c/k$ if $ k\in A$ (where $ c$ is a normalization constant); and $ X_i \equal{} \pm Y_i^{0.6}$. $ X_i$ has expectation $ 0$.

If $ n\in A$, then with some constant probability exactly one of $ \{Y_{i}\}_{i\equal{}1,\dots,n}$ equals $ n$ (denote it by $ Y_{i^*}$), and all other $ Y_i$ are less than $ n$.  Then with a constant probability sum of all $ \{Y\}_{i\neq i^*}$ is at most $ 10 \sqrt{n} \log n \equal{} o(n^{0.6})$. Therefore, with a constant probability $ T_n$ is very close to $ 1$. Thus the distribution of $ T_n$ is not Gaussian. Now it is easy to see that $ T_i$ has no limit  :D


analytic geometry
superior algebra
superior algebra unsolved

Let $ M \equal{} \prod_{p} \mathbb{Z}/p\mathbb{Z}$. I checked that $ \textrm{Tor}(M) \equal{} \sum_p \mathbb{Z}/p\mathbb{Z}$. Now we have to show that $ M/TM$ is divisible (as $ \mathbb{Z}$-module).

* I don' t think this is correct, it was given to us by our professor though. $ T/TM$ consists of the sequences with infininitely many non-zero entries. But, if it were divisible, then this would imply that $ \mathbb{Z}/p\mathbb{Z}$ is divisible, which is not the case (the problematic case is when one divides by a prime number).

No this is correct. For any element, it suffices to "divide it by $ p$" for each prime $ p$. The only obstruction to this, as you have said is the $ \mathbb{Z}/p\mathbb{Z}$ part. But we have mod out the torsion part - meaning that we can freely modify finitely many entries of an element. In particular this means that for any element in the infinite product, we can replace the coordinate at $ \mathbb{Z}/p\mathbb{Z}$ by 0, and there would be no problem of division at other places.


geometry
rectangle
ratio
ceiling function

Rashida is mowing a rectangular lawn that measures $ 100$' x $ 80$'. She mows by making laps around the rectangle and her lawn mower mows a path that is $ 18$" wide. What is the least number of laps necessary for Rashida to mow the lawn?

I don't think that's correct...in each lap, the sides of the rectangle decrease $ 3'$ per dimension. So it will take at least $ \left\lceil\frac{80}3\right\rceil\equal{}\boxed{27}$ laps?


Here you can download PDF-files with some problems from contest of Ukrainian journal "[i][b]In the World of Mathematics[/b][/i]" ("У світі математики" - "U sviti matematyky").
Official web-site: [url]http://www.mechmat.univ.kiev.ua/probability/Journals/mathw.html[/url]
One more site: http://olympiads.win.tue.nl/imo/matworld.html

Problems 148-159, 166-171
Vol. [b]7[/b] (2001), #1-2,4

[quote="Lovasz"]Hey dmitin, I remember that i saw the pdf files you had posted to forum. But now I don't see them. What's happened?[/quote]

Well I can see and download them right now.

New version of Prob2006.pdf (problems from #3 are added)
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=465391#p465391]Vol. 12 (2006), #1-3[/url]

New version of Prob2006.pdf (problems from #1-4)

I tried to replace previous version [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=465391#p465391]here[/url] but failed :(

Problems (in ukrainian and english): 472-477 (of n.3/2014):
http://www.mechmat.univ.kiev.ua/dload/school/our_konkurs/Problems472-477_N3_14.pdf

Please post the ENGLISH versions of these problems on the appropriate fora then post a list of the links at the National Olympiads forum for the RManagers to add them to the resources page.

The Link is:
http://www.mechmat.univ.kiev.ua/u/node/142

[quote=rogue][quote="dmitin"]Previous problems of this contest may be found here:
http://www.mechmat.univ.kiev.ua/probability/Journals/problems.html
http://www.math.chalmers.se/~volody/itwom.html[/quote]
These links are unattainable now.  :(

Current link is [url]http://www.math.uu.se/~mazor/itwom.html[/url]  (Problems 22-154) :)[/quote]

I have collected using archive.org in 2 pdfs the above mentioned files in English,
missing are problems 1-21 and 29-35, that were not published online
and also problems 160-165 (2001 issue 3, also not online)
thanks to all the contributors above

In case of back up and to avoid possible typos, [url=https://drive.google.com/file/d/1xYMsFp-uOd0yXhcxNgwbbD10WjaRD-p6/view?usp=sharing]here [/url]are most of the problems in Ukrainian and English
(problems contained in the above pdf are  22-28, 36-159, 166-495)

Here is Volume 23 year 2017 and volume 24 year 2018. But they are in Ukranian Language. If you can translate them then it will be great

[url=http://probability.univ.kiev.ua/usm/wp-content/uploads/2018/02/usmvolume1.pdf]Volume 23[/url]
[url=http://probability.univ.kiev.ua/usm/wp-content/uploads/2018/11/usmvolume2.pdf]Volume 24[/url]

I am interested in the previous version of the magazine, the problem column only, i do not speak Ukrainian or Russian, god bless google translate :)

But in which section problem column can be found

there are problems missing in the last numbering, I am asking for the problems missing in that numbering
the magazine restarted a new numbering, I am looking for the old ones[quote=parmenides51]I am looking in order to complete the problem collection [url=https://imogeometry.blogspot.com/p/u-sviti-matematyky-in-world-of.html]here[/url],
in any form (scanned or even photo from the hard copy of a magazine),
the missing problems [b]01-21, 29-35, 160-165[/b] and all that were posted [b]after 531[/b]

the pdfs with the links are on the top of that page[/quote]


geometry
vector
analytic geometry
combinatorics unsolved
combinatorics

We consider a table (n*n) that has n^2 blocks , and every line and every column has a "good" block. a) What is the area of the n-gon determinated by the center of those n blocks ? b) Determine the "G- center of weight" of the n-gon from a).

It's easy to find the G.
For N points in the plane Ai(xi,yi), everybody knoes that G( (x1+..+xn)/n, (y1+...+yn)/n).

Do you have any idea of how can i figure out what the area is ?

There are many $n$-gons with given vertices. So in the current state problem is meaningless.

About the area, here is a thought:
Let the points be $A_i(x_i,y_i)$ and also let $G(x,y)$. Let x denote the vector product that gives the area, I don't know hot it will be in Latex.
Then, by associativity one can find the area..... :)

[quote="DusT"]Then, by associativity one can find the area..... :)[/quote]
Please, explain.

Ok, let's see.....
Let the product of 2 vectors $(x_1,y_1)*(x_2,y_2)=x_1 y_2 - x_2 y_1$ . This is the double area of the triangle formed by this 2 vectors.
Now, the double area will be \[\sum G*A_i=G[\sum X_i + \sum Y_i]\]
where $X_i$ and $Y_i$ are the coordinate vectors of the points $A_i$.
Something like this???? :)

Hmm... My head works badly today.
What if $G=0$?

No, I think your head work well :) Mine, instead, a little worse :)
If G is 0, then the area will be 0 indeed. The sum should have been $\sum A_i*A_{i+1}$. But now, I see no reasonable good way to calculate it, sorry for that thing....

I said earlier that problem is meaningless, because we can draw various $n$-gons with given vertices and they will have distinct areas...
Something is wrong with that problem  :?