Hi,

I have been working through the AoPS Introduction to Number Theory textbook and i just have a question about their solution to one of the problems.

The Problem is 8.41

A faulty car odometer proceeds from 3 to 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005, how many miles has the car actually traveled?

 I though of the correspondence but then the solution says

$ 2005$ is the same as $ 2004_{9}$.

I dont really see how? I said $ 2005_{10} \equal{} 2005_{9}$

I would really appreciate all help

And also, just a general question. For the AoPS Intro to Number Theory, i cant really solve ALL the Challenge problems on the first go, is this alright?

Thankyou![/i]

[quote="Ihatepie"]Imagine that instead of always skipping the digit 4, it always skipped the digit 9. So there would be no 9s, just 0,1,2,3,4,5,6,7, and 8. So it is basically a base 9 number. Now because it skips 4 and it is already on 2005, remember to subtract 1 from your final answer.

@your question: I have the Intro to C&P book and I remember it saying that you aren't expected to solve all challenges on your first try. I assume it's similar for the Number Theory Book so it should be fine.[/quote]

But i considered the first few base 9 digits:

$ 1\,\,\,\, 2 \,\,\,\,3\,\,\,\, 4\,\,\,\, 5\,\,\,\, 6\,\,\,\, 7\,\,\,\, 8\,\,\,\, 10$
$ 11\, 12\, 13\, 14\, 15\, 16\, 17\, 18\, 20$
$ 21\, 22\, 23\, 24\, 25\, 26\, 27\, 28\, 30$

and then base 10:

$ 1\,\,\,\, 2 \,\,\,\,3\,\,\,\, 4\,\,\,\, 5\,\,\,\, 6\,\,\,\, 7\,\,\,\, 8\,\,\,\,9\,\,\,\, 10$
$ 11\, 12\, 13\, 14\, 15\, 16\, 17\, 18\, 19\, 20$
$ 21\, 22\, 23\, 24\, 25\, 26\, 27\, 28\, 29\, 30$

And they all match up for the first 8, but then 10 = 9, 11 = 10 and so on, but when you reach 20 in base 9 it must go with 18, so 20 = 18, 21 = 19 and so on.

So i dont really see how $ 2005_{10} \equal{} 2004_{9}$ Please help  :(

[quote="Ihatepie"]The odometer isn't actually in base 10. It is in a warped form of base 9 with the 4 instead of the 9 gone.[/quote]

Okay, so i understand that, so if its just a base 9 number with the 4 gone instead of 9, then wouldn't the answer be $ 2005_{9} \equal{} 2 \times 9^{3} \plus{} 5$?

I dont see why we would have to subtract 1?

[quote="Ihatepie"]With regular base 9, it would go 2000, 2001, 2002, 2003, 2004...

But with this "base 9" it goes 2000, 2001, 2002, 2003, 2005. So the 2005 here actually corresponds to a 2004 in base 9, not 2005.[/quote]

Ah! I get it  :lol:  :)

Thank you soooooooo much!

Explaining it actually helped me to understand it better. So Thank you!  :lol:


Let $ a;b;c>0$;$ abc\equal{}1$
proove that
$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{a}\plus{}\frac{6}{a\plus{}b\plus{}c}\ge5$

[quote="tranvanluan"]Let $ a;b;c > 0$;$ abc \equal{} 1$
proove that
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{a} \plus{} \frac {6}{a \plus{} b \plus{} c}\ge5$[/quote]
Maybe $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {6}{a \plus{} b \plus{} c}\ge5$ $ ?$
If so see here the stronger inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172820


function
algebra
polynomial
quadratics
Vieta
floor function
quadratic formula

If the function $ f(x)\equal{}x^2\minus{}bx\plus{}9999$ has positive integer roots, what is the difference between the minimum and maximum possible values of $ b$?

The maximum is if the roots are 9999 and 1...the minimum is if the roots are both $ \sqrt{9999}$.

So the answer is...

$ 10000\minus{}2\sqrt{9999}$

Carbon57, the polynomial must have integer roots.

Maximum value of b is when roots are 9999 and 1, and minimum value is when roots are 99 and 101.
10000-200=9980

Oh yeah...my non-integer answer didn't seem correct

i don't see why "Maximum value of b is when roots are 9999 and 1, and minimum value is when roots are 99 and 101."
using the quadratic formula, the roots should be [b+/- sqrt(b²-2*999)]/2
the numerator must be a multiple of 2 in order for the roots to be integer

By vieta's formulas, the product of the roots, call them $ p$, and $ q$, is equal to $ 9999$. We also have that $ b\equal{}p\plus{}q$. $ b^2\equal{}(p\plus{}q)^2\equal{}p^2\plus{}q^2\minus{}2pq\plus{}4pq\equal{}(p\minus{}q)^2\plus{}4*9999$. Now it is equivalent to minimize and maximize the difference between the roots. Suppose that $ p>q$. Then $ p\le 9999$, and $ q\ge 1$, so $ p\minus{}q\le 9998$, equality occurs with $ p\equal{}9999$, $ q\equal{}1$.

For the smallest difference, observe that $ 9999\equal{}pq>q^2$ so $ q<\sqrt{9999}$ but since $ q$ is an integer, $ q\le \lfloor \sqrt{9999}\rfloor\equal{}99$. A similar argument prove that $ p\ge 101$. Then $ p\minus{}q\ge 2$, and equalty occurs when $ p\equal{}101$, $ q\equal{}99$.

I didn't know the formula that for a quadratic $ ax^2\plus{}bx\plus{}c\equal{}0$, the sum of roots is $ \minus{}b/a$ and product of roots is $ c/a$ was called Vieta's formulas. I only thought there existed these formulas.
thanks.


calculus
integration
logarithms
trigonometry
calculus computations

Find \[ \int_{0}^{\dfrac{\pi}{2}} \log(\sin(x)) dx \]. What is the indefinite integration of $\log(\sin(x))$ wrt $x$??

Er... No.. $\int\ln\sin x\ dx$ cannot be expressed in an elementary way. But the definite integral can be computed.
Let $I = \int_0^{\tfrac{\pi}{2}} \ln \sin x \ dx$ and $J = \int_0^{\tfrac{\pi}{2}} \ln \cos x \ dx$
Then
$I+J = \int_0^{\tfrac{\pi}{2}}\ln(\sin x\cos x) \ dx = \int_0^{\tfrac{\pi}{2}} \ln \dfrac{\sin 2x}{2}dx = \int_0^{\tfrac{\pi}{2}} \ln \sin 2x \ dx - \dfrac{\pi}{2}\ln 2 = \dfrac{1}{2}\int_0^{\pi} \ln \sin x \ dx - \dfrac{\pi}{2}\ln 2 = I - \dfrac{\pi}{2}\ln 2$
Hence $J = -\dfrac{\pi}{2}\ln 2$
And it is clear that $I=J$ from the subsitition of variable $x=\dfrac{\pi}{2}-t$.
Hence $I=J=-\dfrac{\pi}{2}\ln 2$.


The number A = 200420052006....2040.Find the remainder when A is divided by 9...
Any formulas for this kind of sums??

sum up the individual digits, then find the remainder when the sum is divided 9. If the sum is still big, take the sum of the individual digits again..  do the same thing

[quote="joml88"][hide]Actually you can just add up $2004+2005+...+2040$ and take it mod 9.[/hide][/quote]

you mean their digits?

No, I meant add up 2004+2005+....+2040.  You [i]can[/i] add up each digit but I believe adding up the arithmetic series 2004+2005+...+2040 is much easier and more accurate.

You can do this because the sum of the digits of a number is congruent to the number modulo 9.

For example, take a smaller number like 121314.  Adding up the digits we get 1+2+1+3+1+4=12=3 mod 9.  Or you can do 12+13+14=39= 3 mod 9.  See, they are the same.


Let $n$ be a natural number. Let $A_n$ denote the set of natural numbers which are not relatively prime to $n$. Find all $n$ such that if $x$ and $y$ are members of $A_n$, then $x+y$ is also a member of $A_n$

Rewrite: find all n such that if x and y each share a common divisor with n, so must x+y.

For this to [b]always[/b] be true, it must be true that x and y have to share a common divisor if they share a common divisor with n. Then n must be divisible by only a single prime--itself. n must be prime.

Could you explain a bit more how your conclusuion follows!


probability
algebra
function
domain
geometry
inequalities
conditional probability

Given

f(x,y)=3/16 (4-2x-y)  ;x>0,y>0 and 2x-y<4

and f(x,y)=0 at elsewhere.

Find P(Y<1|X=5/2)

I wish to tackle this question by first finding the conditional probability,h(Y|x)= f(x,y)/f(x).
but i encounter a problem in finding f(x),that is ,when i want to find f(x) by integrating with respect to y, i am confused with the range of y that should be taken.

Or is there any other way to solve this question?
Can anyone help me?

Isn't that the domain is the area bounded by x>0,y>0 and 2x-y<4?

How to view it as a triangle?

Draw the lines $ x \equal{} 0$, $ y \equal{} 0$ and $ 2x \minus{} y \equal{} 4$.  Shade in the region bounded by these three lines.  What is it?

I still can't get a triangle!

I get the line x=0,y=0 and 2x-y=4 and also shaded the area bounded by x>0,y>0 and 2x-y<4.
x>0 and y>0 ensure that the domain must be in first quad.So,..it's the inequality 2x-y<4 that make me not getting triangle.

Can any one point out where I had gone wrong??

You are completely right in that you are not getting a triangle :). Where you went wrong was when copying the problem from the blackboard or from your textbook. The last condition should be $ 2x\color{red}\plus{}\color{black}y<4$.

I saw $ 4\minus{}2x\minus{}y$ and translated the domain from that. The other ign- well, I saw it and dismissed it unconsciously.

I had double-checked the question in the book.It is indeed true that 2x-y<4.

I just want to ask,if the domain is as what mentioned,is it impossible for us to solve P(Y<1|x=5/2)?Or do we have another way to solve?

The whole question is broken. At $ x\equal{}\frac52$ and $ y\ge 0$, the probability density is negative. What do you get if you condition on something impossible? Nonsense.
The given density does define a probability distribution on the triangle with vertices $ (0,0),(0,4),$ and $ (2,0)$. Note the absence of any values greater than $ 2$.

ok..i wish to ask,can we get E(Y|x) from the domain given?

The given domain does not define a probability distribution.
Try the fixed version ($ 2x\plus{}y<4$). For $ 0\le c<2$, can you find the conditional distribution $ y|x\equal{}c$?

Using the fixed version,  i can find the conditional distribution y|x by first finding f(x).Then divide f(x,y) by f(x) to get the conditional distribution. :)


Don't worry, you'll have all of your scores before the end of your life.  Just... relax.... :D

[quote="13375P34K43V312"]When do we recieve our scores for round 1? Does everyone get it at the same time?[/quote]If all goes well, Monday (11/6).

Having said this: [i]please do not complain if the scores are not available on Monday![/i]  We will do our best to get them all in on Monday, but it may roll over to Tuesday or Wednesday.

I will make an announcement when the scores are posted.

[quote="ehehheehee"]It feels better to not have to wonder...

How do they grade the solutions anyways?[/quote]

I think there's a rubric for each question

[quote="13375P34K43V312"]Don't the precise rubrics get posted in this forum somewhere?[/quote]
They're posted in the "grading issues" forum after scores are released.


a 5-person committee has 2 red people and 3 blue poeple.  if the reds vote for red bills $75\%$ of the time and the blues vote $25\%$ of the time, find the probability that a red measuer will recdieve a majority vote


A quadrilateral is inscribed in a circle.  If angles are inscribed in the four arcs cut off by the sides of the quadrilateral, their sum will be:

$ \textbf{(A)}\ 180^\circ \qquad
\textbf{(B)}\ 540^\circ \qquad
\textbf{(C)}\ 360^\circ \qquad
\textbf{(D)}\ 450^\circ \qquad
\textbf{(E)}\  1080^\circ$

I don't quite understand. Wouldn't it be the case that each angle is half the arc they intercept, and since the entire circle is intercepted, the sum would be $\frac{360^\circ}{2}=\boxed{180^\circ}$?

:o

Math terminology in the 1950s was indeed specific.

What I mean is, exactly how do you know not to inscribe on the other side?

@Phire: I think the aops version of this problem is worded wrong, or in my problem book the editors modified the question. Anyways, in the Contest Problem Book, it states "A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off by the sides of the quadrilateral,[b] without intersecting the sides between vertices, [/b]their sum will be:..."

The original problem on the test might not have contained the added fragment though, so you are right about math terminology in the 1950s being non-specific.

Hope this helps!

[quote="AIME15"][hide="Click for solution"]
$ \angle x\equal{}\frac{1}{2}(a\plus{}b\plus{}c)$.  Thus, the sum of the four angles is:

$ S\equal{}\frac{1}{2}(a\plus{}b\plus{}c\plus{}b\plus{}c\plus{}d\plus{}c\plus{}d\plus{}a\plus{}d\plus{}a\plus{}b)$

which simplifies to

$ \frac{3}{2}(a\plus{}b\plus{}c\plus{}d)\equal{}\frac{3}{2} \cdot 360\equal{}540$, or $ \boxed{\textbf{(B)}}$.
[/hide][/quote]

Can you explain the (1/2)(a+b+c) please?


function
logarithms
trigonometry
real analysis
real analysis unsolved

I have heard in a couple of different courses that $\sum_{1}^\infty \frac{z^{n}}{n}$ converges everywhere on $|z|=1$ except for $z=1$. How can this be shown?

We can show the sum converges for all $z\neq1$ on the unit circle using summation by parts:

We have $\sum_{k=1}^{n}z^{k}=\frac{z-z^{n+1}}{1-z}$ for $z\neq 1$, so
$\sum_{n=1}^{N}z^{n}\cdot\frac1n=\frac1{N+1}\left(\sum_{k=1}^{N}z^{k}\right)+\sum_{n=1}^{N}\left(\sum_{k=1}^{n}z^{k}\right)\cdot \left(\frac1n-\frac1{n+1}\right)$
$=\frac{z-z^{N+1}}{(N+1)(1-z)}+\sum_{n=1}^{N}\frac1{n(n+1)}\cdot\frac{z-z^{n+1}}{1-z}$

As $N\to\infty$, the first term goes to zero; it's at most $\frac2{(1-z)(N+1)}$ in absolute value. The series $\sum_{n=1}^\infty \frac1{n(n+1)}\cdot\frac{z-z^{n+1}}{1-z}$ converges absolutely as well, since its terms are at most $\frac{2}{n(n+1)(1-z)}$ in absolute value.
This proves convergence, and we are done.

Five comments:

1. $\frac1n$ could be replaced in this proof by any sequence $b_{n}$ that is positive, monotone decreasing, and tends to zero. The argument would still work; the series would converge for $|z|=1,z\ne1.$

2. Consider the series $\sum_{n=1}^{\infty}b_{n}z^{n}$ and re-do this proof for the remainder $\sum_{n=N+1}^{\infty}b_{n}z^{n}.$ We find that

$\left|\sum_{n=N+1}^{\infty}b_{n}z^{n}\right|\le \left(\frac{2}{|1-z|}\right)b_{N+1}|z|^{N+1}.$

In other word, the error is no larger than the next term times $\frac{2}{|1-z|}$

3. This then shows that the series converges uniformly on $\{z: |z|=1,|z-1|\ge\delta\}$ for each $\delta>0$ and that the limit function is continuous on the circle except at the point $z=1.$

4. If you let $z=-1,$ what I've said in #1 and #2 would reduce to precisely the alternating series test, including the error estimate.

5. For the particular case of $\sum_{n=1}^{\infty}\frac{z^{n}}{n},$ once we know that it converges, Abel's Theorem tells us what it converges to: $-\log(1-z),$ evaluated at $|z|=1,z\ne1,$ using a branch of the logarithm such that $\log 1=0.$

An interesting collection of trigonmetric identities and purely geometric arguments then show us that for $0<\theta<2\pi,$

$\sum_{n=1}^{\infty}\frac{e^{in\theta}}{n}=-\log(1-e^{i\theta})=-\ln\left(2\sin\frac{\theta}{2}\right)+i\left(\frac{\pi-\theta}{2}\right).$

[color=green][I think jmerry learned this particular argument from me. In any case, it looks familiar.][/color]

That's really nice. I thought the statement was much stronger and harder to show. I haven't seen summation by parts before during my 3 semesters at university. I wonder why not. Seems pretty basic.

This argument is also easier to me than the proofs of Leibniz' alternating series criterion I've seen before, so I wonder a little why it hasn't been shown before!


1.	Ten 10-peso coins are initially lined up side by side in a row. Every second 10 peso coin is replaced by a 5peso coin. Finally, every third coin, is replaced by a 1 peso coin, What is the final value of the money in the row?

2.	A boy arranges three kinds of books which are 36mm, 20mm, and 24mm thick. He places only books of the same thickness into 3 stacks of equal height, and wants to make the height as small as possible. How many books would be used in this arrangement?

3.	Some people interpret 3/7 as March 7 while others interpret it as July 3. How many dates on a year can cause this confusion?

4.	Two identical jars are filled with mixtures of water and vinegar in the ratio 3: 2 and 5:2. If both jars are emptied into a new container, what is the new ratio of water to vinegar in the third container?

Well, this problem doesn't deal with a lot of coins, so it isn't too hard just to do it manually. You get 18 pesos.

For this problem, simply find the least common multiple of 36,20,24, which is 360. That means that each stack must be 360 mm, that means that there will be 10+18+15=33 books.

This confusion can happen in any month, and on any day less than 12, so that means that there will be 12 days in each month that will cause confusion, leading to 144 days in a year with confusion. But we can see that we have overcounted, as 1/1 is january 1st no matter which way you see it, so we have to subtract 12 days from the entire number of days, to get 132 days.

Let's just say that there is 5 units in the 1st jar and 7 in the second. We see that if we pour the two together, we get 12 units, with 4 units of vinegar. Thus, the ratio of water to vinegar in the third jar would be 8:4=2:1

[quote="orangefronted"]
3.	Some people interpret 3/7 as March 7 while others interpret it as July 3. How many dates on a year can cause this confusion?
 [/quote]
Woud it matter if the year is a leap year?

Er, no...there is no 29th month :wink:

By the way, the solution provided for the problem is incorrect:

[quote="eli140"]This confusion can happen in any month, and on any day less than 12, so that means that there will be 12 days in each month that will cause confusion, leading to 144 days in a year with confusion. But we can see that we have overcounted, as 1/1 is january 1st no matter which way you see it, so we have to subtract 12 days from the entire number of days, to get 132 days. [/quote]

We must subtract out $ 12$ [b]two[/b] times, because we want to count it 0 times, and it has been counted twice.  Thus, the answer is $ 144\minus{}24\equal{}\boxed{120}$.

[quote="AIME15"]
We must subtract out $ 12$ [b]two[/b] times, because we want to count it 0 times, and it has been counted twice.  Thus, the answer is $ 144 \minus{} 24 \equal{} \boxed{120}$.[/quote]

[hide]Do you mean the possibilities like 7/7,2/2 etc. ?

If so,how are they counted twice ?

I mean to say 7(month)/7(date) and 7(date)/7(month) don't mean the same.

So,I think,24 need not be subtracted.Please clarify!
[/hide]

[quote="AIME15"]In the original count of 144, we count 7/7 twice.  However, there is [b]no confusion[/b] for 7/7 - either interpretation yields it to be July 7.  Thus, we wish to count it 0 times.  There are 12 7/7 "types" (1/1, 2/2, ..., 12/12), so we subtract 12 * 2 = 24, for a answer of 144 - 24 = 120.[/quote]

Thank you for clarifying! :)


Solve in positive integers $x,y$: \[\frac{x^{2}+y^{2}+8}{xy}\] [hide="Note."]The well-known problem 1 from Estonia 95/96 is $\frac{x^{2}+y^{2}+6}{xy}$, and is in fact very different in solution to the one above.[/hide]

By "solve," is it safe to assume that you mean, "find the values for $x$ and $y$ which make said expression an integer"?

Yes. Sorry if it was a little ambiguous...

This might be the wrong place... as a hint for the basic outline of a solution, look at one of these posts:
[quote="me@home"]See http://www.mathlinks.ro/Forum/viewtopic.php?t=77730
and http://www.mathlinks.ro/Forum/viewtopic.php?t=18491
for the method.[/quote]

But its more difficult since $8$ isn't square free...

Took me forever to solve this!

[hide]
To begin, we have the bounds, $x \geq 1, y \geq 1, k \geq 1$.
For our first case, let's assume that $x=y$.
Hence we have:

$8=x^{2}(k-2)$, thus we derive the solutions $(1,1)$ and $(2,2)$.

Now, let's assume without a loss of generality that $x > y$.

$\frac{x^{2}+y^{2}+8}{xy}=k$

$(1)x^{2}+(-ky)x+(y^{2}+8)=0$

Plugging this into the quadratic formula:

$\frac{ky+\sqrt{k^{2}y^{2}-4(1)(y^{2}+8)}}{2}=x$

Therefore, we are looking for a square determinant, which we will call $z^{2}$, thus $z=2x-ky$ by rearranging the equation.

So we have:

$y^{2}(k^{2}-4)-32=z^{2}$

Let's introduce an $a=k-2$.

$(ya+z)(ya-z)=4(8-ay^{2})$.

Which means that since the right hand side is positive, we have only two values of y to test for, 1 and 2.

Let $y=1$:

$a^{2}-z^{2}=32-4a$

Substituting back in k,

$(k+z)(k-z)=36$

Thus the only solution from this is $(9,1)$.

Similarly, by letting $y=2$:

$(k+\frac{x}{2})(k-\frac{x}{2})=12$

yields $(6,2)$.

So if I haven't made a mistake, I believe the only solutions are:

$(1,1)$, $(2,2)$, $(9,1)$, $(6,2)$, [/hide]

[quote="me@home"][quote="Potato Theory"]
Let's introduce an $a=k-2$.

$(ya+z)(ya-z)=4(8-ay^{2})$.

Which means that since the right hand side is positive, we have only two values of y to test for, 1 and 2.
[/quote]

This might be my mistake, but I don't see why the rhs is positive... don't get me wrong, your proof looks very insightful and nice, but I'm not too sure about this step. I can prove: $ya+z\geq 0$
(since $ya+z=2x-2y\geq 0$)
but the other isn't so sure, in fact I believe it is neither always right or wrong that $ya\geq z$...

I hope your wonderful proof isn't hurt by this :([/quote]

Unfortunately, it [b]is[/b]. There are infinitely many solutions just for
\[\frac{x^{2}+y^{2}+8}{xy}=4\]
namely, we can define $a_{n}$ by $a_{0}=a_{1}=2$ and $a_{n+2}=4a_{n+1}-a_{n}$, and then $(a_{n},a_{n+1})$ will always satisfy the above. What's more, this does not prove that these are [b]all[/b] of the solutions to the above, and then we have to worry about the cases
\[\frac{x^{2}+y^{2}+8}{xy}=6\]
and
\[\frac{x^{2}+y^{2}+8}{xy}=10\]
which also have infinitely many solutions each. Proving that no other integer values of $\frac{x^{2}+y^{2}+8}{xy}$ are possible is similar to (and somewhat easier than) #6 on the 1988 IMO.


We define low-density set as following,

<It contains less than one integer for every successive 3 integers>

Find all subsets of {1,2, $ \cdots$15} whice is a low-density set.(Here we count the empty set)

Thanks.

Well, this is well-suited to one of my favorite methods: the recursion bash.  Let $ a_n$ be the number of such subsets of $ [n]\equal{}\{1, \ldots, n\}$ that contain neither $ n \minus{} 1$ nor $ n$, let $ b_n$ be the number that contain $ n \minus{} 1$ but not $ n$, and let $ c_n$ be the number that contain $ n$ but not $ n \minus{} 1$.  Then we have $ a_{n \plus{} 1} \equal{} a_n \plus{} b_n$, $ b_{n \plus{} 1} \equal{} c_n$ and $ c_{n \plus{} 1} \equal{} a_{n}$.  Then each of the sequences (and so also their sum, the number we're looking for) satisfies the recursion $ x_{n \plus{} 1} \equal{} x_n \plus{} x_{n \minus{} 2}$ (and in fact it's easy to justify this recursion directly) with initial values 1, 2, 3.  Then we can quickly calculate the first few terms as 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, which is sequence [url=http://www.research.att.com/~njas/sequences/A000930]A000930[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


Hi all ,

Can anyone tell me how is math applied in economy .I mean not easy problems , I'd like to see interesting problems (e.g Game theory ),

Also any theoretical hint to solve the problems would be appreciated

Thanks !

[url=http://en.wikipedia.org/wiki/Mathematical_economics]Here[/url] is some information.  No problems, but it explains how math is applied to economics.

Math deals with numbers and variables, so does economics. If you are wondering what subjects within math are applicable to economics, well, the main ones are statistics, probability, algebra, and calculus. Stochastic calculus is used particularly in financial economics.

Game theory is one of my favorite subjects. An application of game theory would  be a strategy for making the most prominent move in an oligopoly that deals with price setting.
(I'm assuming you understand the basic concept. If not: [url]http://en.wikipedia.org/wiki/Game_theory[/url])

For example, airline companies are an example of an oligopoly, or a small group controlling the market, basically it's the plural form of monopoly. A common effect of oligopolies is price fixing, were the companies choose the prices of their service due to lack of competition. This way the competitors all make the most profit. However, sometimes an airline company might break that agreement by lowering their price on certain flights in order to increase demand. So how would the competition respond?

Game theory answers the question using the tit for tat strategy. Basically you respond by doing whatever the other person did. This can be demonstrated with the classic prisoner's dilemma : Two criminals got caught stealing and are being questioned. Criminal 1 (C1) and Criminal 2 (C2) each have two options, confess or deny. You can represent the results with a payoff matrix.

                                             C2
.......................Confess.....................Deny

........Confess.....(3, 3)......................(1, 6)

C1

........Deny..........(6, 1).....................(0, 0)

For each $(a, b)$ $a$ denotes the number of months in jail for C1 and $b$ the the number of months in jail for C2. If one of the criminals intended to never rob again then confessing would be a smart move since he wouldn't risk going to jail for 6 months. It gets interesting when one tries to analyze the best strategy if someone was going to play $N$ games, since then losing 3 every time would certainly not be the wisest move. In order to determine a winning strategy, various strategies were plugged into a computer and pitted against each other numerous times to find the most successful one. The tit for tat strategy returned the best results (Can you see why?).

The strategy is dependent on these four conditions:

1. Unless provoked, the agent will always cooperate.
2. If provoked, the agent will retaliate.
3. The agent is quick to forgive.
4. The agent must have a good chance of competing against the opponent more than once.

That is why it is so effect in the iterated prisoner's dilemma

Overall, game theory is useful in economics when determining the most rewarding decision keeping in mind other participants or "players" ability to affect the final result.


I'm wondering... Is it possible to have double loops in asymptote, or even multiple loops? By double loop, I mean something like:
[code]for(int i=1; i<10, ++i) for(int j=1; j<10, ++j) dot(i,j);[/code]

I want this to make a 10 by 10 grid. But, it doesn't work. Single loops work just fine, but I'd have to make 10 of them. This would also be very useful for making $ m\times n$ grids.

Also, I don't understand how loops work. I tried the Asymptote Manual at sourceforge, but I really don't understand what they are saying. I try to see other peoples' codes and try to fashion my code off of theirs. It works sometimes, and it doesn't at other times.

In particular, I understand the first two arguments, but I don't understand what the third argument is for. In the bottom code, I would like to understand what "??" means:
[code]for(int i=(begin); i<(end), ??) (what it's supposed to do)[/code]

As a final query, I am trying to make a grid, using the code from the Asymptote Manual. I don't know why, but I can't get it quite to work.

It should be dot((i,j)), not dot(i,j)! Also all three parts in for() should be separated by ; and not by ,.
[asy]for(int i=1; i<10; ++i) for(int j=1; j<10; ++j) dot((i,j));[/asy]
[quote]
In particular, I understand the first two arguments, but I don't understand what the third argument is for. In the bottom code, I would like to understand what "??" means:
[code]for(int i=(begin); i<(end), ??) (what it's supposed to do)[/code][/quote]
The third part is very important: it changes the value of $ i$. Without it you get the same thing repeated infinitely many times. The usual ++i is just an abbreviation for i=i+1.


I've run into a few cases in which forum code, either html or emoticons, interferes with the interpretation of $\LaTeX$ with unpredictable results. I've managed, using my moderator editing powers, to clean up a few of the resulting messes. Here are a couple of things to look out for:

The colon ( : ) - if you want a colon anywhere, you need to put a space after it. With no space after it, it gets interpreted as a smiley. (Just try editing out the space I included after the colon in this paragraph for a demonstration.) But if that's inside a TeX passage, look out.

The code \{x: x^2<1\} works fine: $\{x: x^2<1\}$. Try leaving out the space after the colon: \{x:x^2<1\}. Do that inside the Tex dollar signs - well, I won't do that deliberately. Try it yourself in preview to see what happens.

The one I just ran into was "< = >" typed inside TeX dollar signs. Among the side effects was a window approximately tripled in width, containing an alleged link that was complete garbage. Actually, that was typed without spaces - I had to type it here with spaces just to show it.

In doubtful situations, try adding spaces.

Actually it's not the forum code, it's the smillies :) I will deal with this bug at a later date in the future. Untill then, this remains a sticky in here.

There's always the "Disable Smililes in this post" checkbox you can just click on. A heavily LaTeX-laden post is unlikely to have many smileys in it anyway.

Indeed, that too is a simple solution :D

$\{x: x^2<1\}$

$\leq >$

This seems to have been fixed?

Inteluser can you rephrase your question please? If it is was a test, I suggest you do tests in the Test forum in the future ;)

:blush: I was noting that the problem above doesn't seem to exist anymore (those were the two examples Kent Merryfield cited).

I don't know if anyone else gets this problem.
$\binom{k}{n}$
$\binom{n}{k}$

Coded exactly the same way, but one is big and one small...
Is this meant to be in Latex, or is it just on the forum?

Hehe, some of the images are small because they are cached from almost 2 years ago (wow! :) ) when I didn't think about putting \displaystyle in the command.

I am slowly manually removing the annoying cached images like this as they are reported here ;)

[quote="jmerry"]If the original problem doesn't exist anymore, shouldn't the thread be unstickied?[/quote]Good point :P

[quote="Peter VDD"][quote="Valentin Vornicu"]I am slowly manually removing the annoying cached images like this as they are reported here ;)[/quote]You have a function sort on date in your ftp manager... so you can basicly delete all the old ones with 1 click. ;)[/quote]I don't use a ftp manager but you gave me a good idea :)


algebra
polynomial
geometry
3D geometry
complex numbers
superior algebra
superior algebra unsolved

prove that $ \sqrt[3]{5} \notin \mathbb{Q}[\sqrt[3]{2}]$.

i know you can probably brute force by explicitly writing a polynomial in $ \sqrt[3]{2}$ and cubing it.  i am looking for a purely field theoretic approach.

Apply the Rational Root Theorem to $ x^3 \minus{} 5$.  (At least, I think it still holds.  I'm not sure.)

This solution may seem like a lot of brute force, but some of it might be necessary: after all, it's pretty hard to "prove" something like $ \sqrt[3]{54}\not\in \mathbb{Q}(\sqrt[3]{2})$.

[hide]
If $ \sqrt[3]{5}$ were of the form $ a\plus{}b\sqrt[3]{2}\plus{}c\sqrt[3]{4}$ for some $ a,b,c\in \mathbb{Q}$, then $ (a\plus{}b\sqrt[3]{2}\plus{}c\sqrt[3]{4})^3\equal{}5$. Now let $ \omega\equal{}\frac{\minus{}1\plus{}i\sqrt{3}}{2}$ and consider the automorphism $ \sigma$ on $ \mathbb{Q}(\sqrt[3]{2},\omega)$ fixing $ \mathbb{Q}(\omega)$ and mapping $ \sqrt[3]{2}$ to $ \omega\sqrt[3]{2}$. Applying $ \sigma$ gives the equation
\[ (a\plus{}b\omega \sqrt[3]{2}\plus{}c\omega^2\sqrt[3]{4})^3\equal{}5\]
If $ a\plus{}b\omega \sqrt[3]{2}\plus{}c\omega^2\sqrt[3]{4}$ were to equal $ \sqrt[3]{5}$, then equating imaginary parts gives $ \frac{\sqrt{3}}{2}(b\sqrt[3]{2}\minus{}c\sqrt[3]{4})\equal{}0$. So $ b\equal{}c\equal{}0$ forcing $ a\equal{}\sqrt[3]{5}$, a contradiction.

So it must be that $ a\plus{}b\omega \sqrt[3]{2}\plus{}c\omega^2 \sqrt[3]{4}$ is one of the nonreal cube roots of 5 ($ \omega \sqrt[3]{5}$ or $ \omega^2\sqrt[3]{5}$), and so $ a\plus{}b\omega^2\sqrt[3]{2}\plus{}c\omega \sqrt[3]{4}$ is its complex conjugate (the other one). Adding these two equations with the original, we find $ a\equal{}0$.

Now $ b\plus{}c\sqrt[3]{2}\equal{}\sqrt[3]{\frac{5}{2}}$ which is an irrational real cube root. Use the same argument as above to get $ b\equal{}0$, and then $ c\equal{}\sqrt[3]{\frac{5}{4}}$, a contradiction.

This should easily generalize to $ \sqrt[3]{y}\not\in \mathbb{Q}(\sqrt[3]{x})$ if $ x,y$ are integers and none of $ x,y,xy,x^2y$ are perfect cubes.
[/hide]

Here's a field-theoretic solution, sort of: calculate a degree $ 9$ polynomial with integer coefficients and $ \sqrt[3]{2}\plus{}\sqrt[3]{5}$ as a root, and show that it is irreducible.

I did not say that it would not be a little ugly...


quadratics
Vieta
algebra
number theory
prime numbers
special factorizations

[hide] THe sum of the roots is 63, so the primes must be 2 and 61. There is only 1 value of k, so B.[/hide]

Sorry for the revive, but is it possible to instantly know that 61 and 2 are the only two prime numbers whose sum is 63? Or would you have to try out a whole bunch of numbers first?

Well, 63 is odd, so you know it must be the sum of an even and an odd. The only even prime is 2, so the other prime is 61.

Two solutions
[hide] We use Vieta's formulas. We know that the sum of the roots is $-b/a$, so the sum of the roots of this quadratic is $-(-63/1)=63$
We also know that both roots are prime. $63$ is odd, so there must be one even and one odd root. The roots are also both prime, and the only even prime is $2$. So, the roots are $2$ and $61$. So, there is one k, as there is one distinct product of $2$ and $61$, which is $122$, so the answer is $B$.
[/hide]
[hide]We use the quadratic formula. We know there are two distinct roots, from graphing, completing the square, or Vieta's. So, since the plus and minus signs already assure that there are two roots, the $\sqrt{b^2-4*a*c}$ has one value. Since $a=1$ and $b=-63$, we know that c has one distinct value, so the answer is $B$.[/hide]

By Vieta's Formula, the sum of the roots of this quadratic equation is just:
$$\frac{-(-63)}{1} = 63$$
Let $a$ and $b$ denote the prime roots to this quadratic equation.

Clearly, since $a$ and $b$ are prime, $a \in \mathbb{Z}^{+}$ and $b \in \mathbb{Z}^{+}$.

We could actually see the pairs of all prime numbers which add up to $63$ or we could use parity in our favor. We know that:
$$a + b = 63$$
If either $a$ or $b$ are odd, then the other one will be even, and we can't have that unless one equals $2$. Hence:
$$(a,b) = (2,61)$$OR
$$(b,a) = (61,2)$$
By Vieta's Formula, we know that the product of the roots of this quadratic equation is just:
$$\frac{k}{1} = k$$
We know that $ab$ will have a fixed value which is:
$$2\cdot61 = 122$$.
Hence there is only one value of $k$ that satisfies this which is $k = 122$.

Hence our answer is $\boxed{1}$.

[quote=worthawholebean]Both roots of the quadratic equation $ x^2 \minus{} 63x \plus{} k \equal{} 0$ are prime numbers. The number of possible values of $ k$ is

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \textbf{more than four}$[/quote]

Can primes be negative?

No. Primes are not negative. The smallest prime is $2$.

This is a very interesting problem because it connects algebra with number theory!

For the equation $x^{2} - 63x + k = 0$, we need to find two prime numbers, whose product is $k$, and whose sum is $63$ (one will be positive, and one will be negative). Since $63$ is an odd number, and given that there are prime roots, the only two numbers I can think of are $2$ and $61$ (both are negative). If we multiply $2$ and $61$, we get $122$, so there is only one possible value of $k$. The correct answer is B.

$p_1+p_2=63$.By taking parities we need one of them to be $2$, and the other one is $63-2=61$ which is a prime.

The answer is B.

[hide=Sol]Their sum is $63$ and they must be prime, so the only value of $k$ that we have is $2 \cdot 61 = 122$. So the answer is $\boxed{\textbf{(B)}\ 1}$.[/hide]


finaf all functions $ f$ on $ R\minus{}${$ \minus{}1,1$} such that :
         $ f(\frac{x\minus{}3}{x\plus{}1})\plus{}f(\frac{3\plus{}x}{1\minus{}x})\equal{}x$ for all x $ \in$$ R\minus{}${$ \minus{}1,1$}

replace $ x$ by $ \frac{x\minus{}3}{x\plus{}1}$ ,after that replace again in the first equation x by $ \frac{x\plus{}3}{1\minus{}x}$


calculus
integration
geometry
derivative
logarithms
calculus computations

Find the value of $ a$ for which the area of the figure surrounded by $ y \equal{} e^{ \minus{} x}$ and $ y \equal{} ax \plus{} 3\ (a < 0)$ is minimized.

For lack of my own ability to arrive at an analytical solution, I resorted to using Lambert W to solve for the points of intersection between the $ y \equal{} e^{ \minus{} x}$ and $ y \equal{} ax \plus{} 3$ with  $ a < 0$. I will admit some good pointers from other "posters" in the forum.

The integration of the area between the curves seemed easy enough.  The problem, as I see it, was determining the x limits of integration. If I haven't messed it up to badly I get appox. value of $ a \equal{} \minus{} 1.49$ resulting in an area of about 4.92 sq units.

Well...   How bad is it?

Rather bad. I would say 1 point out of 10 if I were grading the exam :P. Think graphically!

[hide]
For each $ a$, let $ p,q$ be the points of intersection with $ p\leq q$. We have that the area is (since the line is above the exponential curve in the "piece")
\[ \int_p^q (ax\plus{}3\minus{}e^{\minus{}x})\ dx\equal{}\frac{1}{2}a(q^2\minus{}p^2)\plus{}\int_p^q (3\minus{}e^{\minus{}x})\ dx\]
Set the derivative with respect to $ a$ equal to 0 to get
\[ 0\equal{}\frac{1}{2}(q^2\minus{}p^2)\plus{}a(qq^\prime \minus{}pp^\prime )\plus{}q^\prime (3\minus{}e^{\minus{}q})\minus{}p^\prime (3\minus{}e^{\minus{}p})\]
\[ \equal{}\frac{1}{2}(q^2\minus{}p^2)\plus{}a(qq^\prime \minus{}pp^\prime)\minus{}q^\prime (aq)\plus{}p^\prime (ap)\]
(using the equations for $ p,q$).

But look! The $ qq^\prime$ and $ pp^\prime$ terms cancel. Therfore $ q^2\equal{}p^2$. But $ q\neq p$ because $ ax\plus{}3\minus{}e^{\minus{}x}$ is continuous, positive at $ x\equal{}0$, and integrated over the interval where it's nonnegative. This forces $ p<0<q$. So we get $ p\equal{}\minus{}q$.

Now we have the equations $ e^{\minus{}q}\equal{}aq\plus{}3,e^q\equal{}\minus{}aq\plus{}3$. Add the two to get $ e^q\plus{}e^{\minus{}q}\equal{}6$, so that $ e^q\equal{}3\plus{}2\sqrt{2}$ and $ q\equal{}\ln (3\plus{}2\sqrt{2})$ (using $ q>0$). Finally,
\[ a\equal{}\frac{e^{\minus{}q}\minus{}e^q}{2q}\equal{}\boxed{\minus{}\frac{2\sqrt{2}}{\ln (3\plus{}2\sqrt{2})}}\]
[/hide]

scorpius119:   Very nice solution.

fedja:	Glad I've never had to be in your class... :)

[quote="kunny"]That's right. :)
This problem is difficlut for Japanese high school students. But the simillar problem has been posed at least 3 times with hint.
I think that university students would solve the problem in using calculus of variations.As pointed by fedja, we would solve in thinking two graphs.
The key point is the point $ (0,\ 3)$ which is the mid point of the intersection points between 2 graphs.[/quote]

The problem may be difficult for Japanese student. And university students may use variational calculus but some of us are just self motivated learners trying to learn somethings.

I learned a number of things. Thanks for posting it!!


superior algebra
superior algebra theorems

[quote="queue_zniv16"]Show that the center of the group $ D_{n}$ is $ \langle e\rangle$ if $ n$ is odd and isomorphic to $ Z_{2}$ if $ n$ is even.[/quote]

Say $ n=2$ then $ Z(D_{n}) = D_{n}$. So, it wrong. Maybe it means that $ n\geq 3$?


I am new user in this site, I want to participate to the contest Round7. How can we post to my answers?

Can you give me more information about participating contests?

Thanks.

You are pretty much late :)

Now the Contest is down, so just wait or look at http://www.mathlinks.ro/viewtopic.php?t=302609


geometry
perimeter
function
monovariant
combinatorics proposed
combinatorics

Here is a nice problem (not very difficult). I'm not sure where it originated ... .

Initially, some configuration $S$ of cells of a given $n \times n$ chessboard are infected. Then, the infection spreads as follows: a cell becomes infected iff at least two of its neighbors (not including diagonal ones) are infected. If the entire board eventually becomes infected, prove that $|S| \geq n$ (that is, at least $n$ of the cells were infected initially).

Enjoy!  ;)

This is indeed very famous :). You probably have a generalization up your sleeve :D.

Anyway, we look at the perimeter of the infected zone. When a cell becomes infected, the perimeter can only decrease (it's a monovariant). Since in the end the perimeter is $4n$, it must have been at least $4n$ at the beginning, so that's that :).

As an aside, t's also clear that $n$ infected cells suffice to infect all the others (take the ones on a diagonal, for instance).

[quote="grobber"]This is indeed very famous :). You probably have a generalization up your sleeve :D.
[/quote]

 :D  :D  :D  Well, not really a generalization, but ...

[b]Problem. [/b] (maybe original, if not already discovered) Let $G$ be an infinite planar $3$-regular graph. Initially, $k$ vertices of $G$ are infected. The infection then spreads as follows: a vertex becomes infected iff at least two of its three neighbors are already infected (what a surprise  :lol: ). Determine, as a function of $k$, the maximum possible number of vertices of $G$ that eventually become infected.

--Vesselin

Is it $4k-1$?

In this case, the "perimeter" is also a monovariant, but, unlike the previous problem, in this one it strictly decreases. The perimeter of a configuration of infected vertices is the number of edges connecting an infected vertex with an uninfected vertex.

At the beginning, we have $k$ infected vertices, so the perimeter is at most $3k$. Each step either decreases the perimeter by $1$ or by $3$, so we can perform at most $3k$ steps, and, since in each step we infect a vertex, at the end we have at most $k+3k=4k$ infected vertices. Let's assume we can reach the value $4k$. This means that everything was sharp: in the beginning the perimeter was exactly $3k$, and it decreasesd by $1$ in each of the $3k$ steps. However, before we make the last step, the perimeter is $4k-1$, meaning that there is only one edge connecting a sick and a healthy vertex. We need two such edges in order to perform a step, so we have a contradiction: we cannot reach $4k$, so the answer is at most $4k-1$.

An example for $4k-1$ is easy to construct :).

[quote="grobber"]This is indeed very famous :). You probably have a generalization up your sleeve :D.

Anyway, we look at the perimeter of the infected zone. When a cell becomes infected, the perimeter can only decrease (it's a monovariant). Since in the end the perimeter is $ 4n$, it must have been at least $ 4n$ at the beginning, so that's that :).

As an aside, t's also clear that $ n$ infected cells suffice to infect all the others (take the ones on a diagonal, for instance).[/quote]

I was just wondering what do you/they mean about "perimeter" and the perimeter decreasing?

The perimeter is the set of edges that separate an infected square from a non-infected one (including edges that separate an infected square from the outside of the board). The length of this perimeter is not necessarily decreasing but it can never increase: a new infected square touches at least two old ones that are already infected, so it erases two edges or more from the perimeter, and it adds at most two new ones (if it adds three, it can only touch a single infected square which contradicts the rules).

[quote="grobber"]Is it $ 4k \minus{} 1$?

In this case, the "perimeter" is also a monovariant, but, unlike the previous problem, in this one it strictly decreases. The perimeter of a configuration of infected vertices is the number of edges connecting an infected vertex with an uninfected vertex.

At the beginning, we have $ k$ infected vertices, so the perimeter is at most $ 3k$. Each step either decreases the perimeter by $ 1$ or by $ 3$, so we can perform at most $ 3k$ steps, and, since in each step we infect a vertex, at the end we have at most $ k \plus{} 3k \equal{} 4k$ infected vertices. Let's assume we can reach the value $ 4k$. This means that everything was sharp: in the beginning the perimeter was exactly $ 3k$, and it decreasesd by $ 1$ in each of the $ 3k$ steps. However, before we make the last step, the perimeter is $ 4k \minus{} 1$, meaning that there is only one edge connecting a sick and a healthy vertex. We need two such edges in order to perform a step, so we have a contradiction: we cannot reach $ 4k$, so the answer is at most $ 4k \minus{} 1$.

An example for $ 4k \minus{} 1$ is easy to construct :).[/quote]
What exactly do they mean by infected/healthy vertices?  How do you define them?

Vertices = cells on the board. "Infected" is just the way the problem is phrased - it could be the cells where you have coins, with each additional coin being adjacent to at least two others. The perimeter is the edges around the coins / infected cells, there are at most $ 4n$ such edges to begin with if you start with $ n$ coins / infected cells.

[quote="grobber"]Is it $ 4k \minus{} 1$?

In this case, the "perimeter" is also a monovariant, but, unlike the previous problem, in this one it strictly decreases. The perimeter of a configuration of infected vertices is the number of edges connecting an infected vertex with an uninfected vertex.

At the beginning, we have $ k$ infected vertices, so the perimeter is at most $ 3k$. Each step either decreases the perimeter by $ 1$ or by $ 3$, so we can perform at most $ 3k$ steps, and, since in each step we infect a vertex, at the end we have at most $ k \plus{} 3k \equal{} 4k$ infected vertices. Let's assume we can reach the value $ 4k$. This means that everything was sharp: in the beginning the perimeter was exactly $ 3k$, and it decreasesd by $ 1$ in each of the $ 3k$ steps. However, before we make the last step, the perimeter is $ 4k \minus{} 1$, meaning that there is only one edge connecting a sick and a healthy vertex. We need two such edges in order to perform a step, so we have a contradiction: we cannot reach $ 4k$, so the answer is at most $ 4k \minus{} 1$.

An example for $ 4k \minus{} 1$ is easy to construct :).[/quote]

as you said wouldn't the perimeter be at most 4k instead of 3k?
Also if a vertice=cell, isn't it at least 2 of the 4 adjacent vertices that must be infected instead of out of 3?  I'm getting slightly lost here.

Different problem. Read vess's second message in this thread (the one with the infinite planar 3-regular graph).

haha this is similar to a problem that I had to solve for the nzimo
solution was similar as well.
In case you don't get it handrake this is what they mean(I think so as I solved the other problem this way)

[hide]Initially there are S number of cells
When each cells touch each other, the total perimeter decreases.
Thus when none of them are touching, the maximum perimeter obtained is 4S units.

Each move the perimeter can only decrease or stay the same.

finally when all the cells are filled, the whole perimeter is 4n.

Thus the smallest possible value is S=n
Then testing for S=n, fill in all the squares diagonally from top right to bottom left hand corner.
This works so S=n is minimum and it's true[/hide]


Let $p;q;r$ be real numbers satisfying $p^{2}-q=q^{2}-r=r^{2}-p$
Caculate the value of $A=(p+q+1)(p+r+1)(q+r+1)$

We have $A=(p+q)(p+r)(q+r)+(p+q)(p+r)+(p+q)(q+r)+(p+r)(q+r)+(p+q)+(p+r)+(q+r)+1$.
$p^{2}-q^{2}=q-r\Leftrightarrow (p-q)(p+q)=q-r \ \ (1)$
$p^{2}-r^{2}=q-p\Leftrightarrow (p-r)(p+r)=q-p \ \ (2)$
$q^{2}-r^{2}=r-p\Leftrightarrow (q-r)(q+r)=r-p \ \ (3)$
If $p=q\Rightarrow q=r\Rightarrow A=(2p+1)^{3},p\in\mathbf{R}$
Suppose that $p\neq q\neq r$. Then:
$(1)\cdot (2)\cdot (3)\Rightarrow (p+q)(p+r)(q+r)=1$
$ (1)\cdot (2)\Rightarrow (p+q)(p+r)=\frac{q-r}{r-p}$
$ (1)\cdot (3)\Rightarrow (p+q)(q+r)=\frac{r-p}{p-q}$
$ (2)\cdot (3)\Rightarrow (p+r)(q+r)=\frac{p-q}{q-r}$
$ (1)\Rightarrow p+q=\frac{q-r}{p-q}$
$ (2)\Rightarrow p+r=\frac{p-q}{r-p}$
$ (3)\Rightarrow q+r=\frac{r-p}{q-r}$
So $ A=1+\frac{q-r}{r-p}+\frac{r-p}{p-q}+\frac{p-q}{q-r}+\frac{q-r}{p-q}+\frac{p-q}{r-p}+\frac{r-p}{q-r}+1=1-1-1-1+1=-1$


How many ordered pairs of primes $ (a,b)$ satisfy $ a\plus{}b\equal{}20$?

Not to sure how to do this except for listing them out.
$ (3, 17)$
$ (7, 13)$
At this point, you don't have to do anymore listing, since you got to the "halfway" point.  Therefore, we just multiply how many we have by $ 2$ (since order pairs count as different sets), and get that we have a total of $ \boxed{4}$ ordered pairs.


Αποδείξτε ότι το σύστημα $ x^6 \plus{} x^3 \plus{} x^3y \plus{} y \equal{}$147^(157)

                                    $ x^3 \plus{} x^3y \plus{} y^2 \plus{} y \plus{} z^9 \equal{}$157^(147)    δεν έχει ακέραιες λύσεις $ x,y,z$. :)

Επιγραμματικά η λύση μου

Αν τισ προσθέσουμε κατά μέλη έχουμε $ (x^3\plus{}y\plus{}1)^2\plus{}z^9\equal{}147^{157}\plus{}157^{147}\plus{}1$  (1)

Θα δουλέψουμε mod 19 . Einai $ z^9\equiv \minus{}1,0,1 (\mod 19)$  kai $ s^2\equiv \minus{}8,\minus{}3,\minus{}2,0,1,4,5,6,7,9 (\mod 19)$  (2)

Από το Fermat $ 147^{157}\plus{}157^{147}\plus{}1\equiv 14(\mod 19)$  (3)

Από (1), (2) ,(3) άτοπο . :)

[quote="silouan"]
Θα δουλέψουμε mod 19 . Einai $ z^9\equiv \minus{} 1,0,1 (\mod 19)$  kai $ s^2\equiv \minus{} 8, \minus{} 3, \minus{} 2,0,1,4,5,6,7,9 (\mod 19)$  (2)[/quote]
Τι σημαινει ο συμβολισμος $ z^9\equiv \minus{} 1,0,1 (\mod 19)$ ?

Δεν εννοω το mod αλλα τα -1,0,1

απλώς έχουμε οτι $ z^9 \equal{} \minus{} 1(mod19)$   ή   $ z^9 \equal{} 0(mod19)$    ή    $ z^9 \equal{} 1(mod19)$ :)


probability
LaTeX
MATHCOUNTS
conditional probability

[hide]first 4/9 you pick a green then 3/8 so overall 12/72 simplifies into 1/6.[/hide]

[hide]
first pick:4/9
second pick: 3/8
overall: $\frac{1}{6}$[/hide]

[quote="236factorial"]There are 9 marbles in a bag, with 5 yellow and 4 green. You pick 2 marbles at random without replacement. what is the probability both of your marbles are green?[/quote]

[hide]
$P(G_1\cap G_2)=P(G_1)\cdot P(G_2|G_1)\\
\\
P(G_1)=\frac{4}{9}\\
\\
P(G_2|G_1)=\frac{3}{8}\\
\\
P(G_1\cap G_2)=\frac{4}{9}\cdot\frac{3}{8} =\boxed{\frac{1}{6}}$
[/hide]

I hope I get the correct answer...
[hide]The probability for first time you try to pick out a green: $\frac{4}{9}$
The probability for the second time you try to pick out a green: $\frac{3}{8}$.

Multiply them togather (I still don't know why...) and get $\frac{1}{6}$.[/hide]

[quote="math92"][quote="shobber"]
Multiply them togather (I still don't know why...)[/quote]
I don't know why either, and i've always wondered why...[/quote]

The reason you multiply them together is because the joint probability of two events $A \cap B$ is defined as: $P(A \cap B) = P(A)\cdot P(B|A)$, where $P(B|A)$ is the conditional probability of B given that A occurs.

In this case A is the event that you choose a green one first. $P(A) =\frac{4}{9}$

B is the event that you choose a green one second, however the conditional probability that you choose green second given that you chose green first, $P(B|A)=\frac{4-1}{9-1}=\frac{3}{8}$

Therefore the probability that you choose a green one first and another green one second is $P(A \cap B) = \frac{4}{9}\cdot\frac{3}{8} = \frac{1}{6}$

For my explanation, I suggest you start reading where the bold words start seeing as how the explanation before those bold words doesn't really help much.

Well it all goes back to multiplication.  If you have a pizza that has 10 equal slices and you want 7 of them, you want (7/10) * 1 = 7/10 of the pizza.

THEN, out of these 7 slices, you want to give 3 to your friend.  If you give 3 slices to your friend, he will have 3 slices and out of the 7 you originally had, you will now have 4.

Now, if we want to compare the amount of pizza your friend has to the whole pizza aka all 10 slices, we represent the whole pizza by 1/1 since it is whole and we represent your friend's slices by 3/10 since he has 3 slices out of the total 10.  So he has 1/1 * 3/10 = 3/10 of the total pizza.

Okay, I understand that wasn't the best explanation and it probably went nowhere, but here's another attempt to explain it.

Say you have a whole pizza again with 10 slices.  You take 7 of them or 7/10 of the pizza.  Think about this.  It's like multiplying independent events in probability.  We all know 7 slices out of a total 10 is 7/10.  But why?  If we think about the 7 slices as a PART of the whole 10, then we can find an answer.  Our total pizza is just 1/1, and OUT OF that 1/1 whole pizza, we want a PART of 7/10 slices.
Now imagine you have your 7 slices and your friend who happens to be very hungry and sly says you cant possibily eat all 7 slices and that you should give him 1/3 of your 7 slices.

So to find your friend's part, say Kyle's part, we do 1/3 * 7 = 7/3 slices.  Now that we know Kyle has 7/3  aka 2 1/3 slices of pizza, how could we find how much of the whole pizza aka all 10 pieces he really has?

We can think of this as a chain of events.  There are 10 slices out of 10 slices, aka 10/10 aka 1/1.  We took 7, so we have (1/1) * (7/1) = 7/1 = 7 slices.  Then Kyle, my friend, wanted 1/3 of my 7 slices aka my (1/1) * (7/1) = 7 slices of the pizza.  So Kyle gets (7/1) * (1/3) = 7/3 of my slices.  To find Kyle's amount of the total 10 slices, we would do (10/10) / (

[b]Okay forget all of that mumbo jumbo.[/b]  Here's my 3rd and I think my best attempt to explain why we multiply independent events instead of, say, adding them.

Say we have a pizza with 20 slices.  My friend and I are going to eat some of them, but because we both want to eat pizza, I'm not going to get to eat all 20 myself.  So we're going to have to split the pizza up.  We decide to split it up evenly.  From there, I decide I'm going to eat 6 slices.  So we can represent how much I'm eating altogether as a chain of fractions that are being multiplied together, aka:

(20/20) * (1/2) * (6/10) where the 20/20 is the original 20 sliced pizza, the 1/2 comes from my friend and I splitting the pizza, and the 6/10 comes from me wanting 6/10 of my new pizza which has 10 slices.

(20/20) * (1/2) *(6/10) = 1 * 1/2 * 3/5 = 3/10.  Thus I am eating 3/10 of the original 20 sliced pizza.

Notice why we chose to multiply these numbers.  Like I said, it's a chain reaction.  We had (20/20) * (1/2) * (6/10).  The (6/10) doesn't care about the (20/20) or the (1/2).  All it cares about is what (20/20) * (1/2) comes out to be.  We've all heard the terms "independent events" and "dependent events".  This case is an example of independent events since (1/2) doesn't care about (20/20), just (20/20) * (1/2).  Also, with independent events we multiply.  Why?  This is explained in the next paragraph.  So then when (20/20) * (1/2) is evaluated, (6/10) finally takes action or takes "part" in the fun and multiplies with (20/20) * (1/2) = (1/2) to get (3/10).

We multiply when all of our fractions are related to each other in a sense.  It's a chain reaction because in our example, 6/10 slices relates to 1/2 which relates to 20/20 slices.  Remember, chain reaction.  I'm not sure what else to say, so hopefully this helps!

Good, the problem is solved, and nice writing, NightFlarer

i guess it sort of helps after reading it 5 times.

And who helped you with the $\LaTeX$?
coughcoughcough ME coughcoughcough

[quote="gr8sk8r"]wow, i might be stupid, but i dont understand that explanation at all[/quote]

I'll try re-structuring the problem a bit to see if it helps. First, forget about the initial draw of the card for a moment. Let's go straight to the second draw. You have 8 cards total and 3 of them are green. Therefore, the probability that you would choose a green card is $\frac{3}{8}$. That is you find a green card in only $\frac{3}{8}$ of the space in which you could possibly find one. The question is how big is that space?

In the first draw, there are 9 cards and 4 are green. You choose a green card. Therefore to find a second green card, you looked for one in only $\frac{4}{9}$ of the total space where one could be found.

This means $\frac{3}{8}\cdot\frac{4}{9}=\frac{1}{6}$ of the total space available for two successive draws contains two green cards.

ooooooooooooohhhhhhhhhhhhhh, i get it! thanks a bunch :thumbup:

[quote="MCrawford"]Please do not post problems like this in Classroom Math.  This kind of problem is not typically taught in an ordinary middle school classroom except at very unusual schools.[/quote]
i guess my school is unusual then. :P

So must mine. My class just learned this topic today.

Nearly all AoPSers are from better than average schools.  One of the goals of the Classroom Math forum is to invite students in who have not had the advantage of as nice a background in the fundamentals of mathematics.

Very few middle schools teach what I'll call MATHCOUNTS level material in their regular classrooms.  And we don't mean for math team classes to be considered as classroom math.

But it's good to know there are some schools where this material is taught to at least some students.

My school is below the 70th percentile in the nation, though. Is that really that good? It is better than the average NY state school.

Two step probability is not even expected of high school students taking the SAT or ACT.

There is no more discussion.  Just use the Basics forum for these problems.  That makes the community work for more people.

I will no longer discuss.

I understanding your point, because probability can be very confusing to learn. Before I joined mathcounts, I didn't understand the combination concept.


Hi,
   When we say that the two computers have exponential time of failure, what it really means ???

    Why are we classifying this type of distribution as "exponential" ?? :blush:

[quote="latapaka"]Hi,
When we say that the two computers have exponential time of failure, what it really means ???

Why are we classifying this type of distribution as "exponential" ?? [/quote]

um i dont understand, could you explain more clearly

hmm... ok... when we say that some type is exponentially distributed...what does it really signify ??

I had already gone through it...but couldn't understand why the following are said to have exponential distribution .... :(  :(

    * the time until a radioactive particle decays, or the time between beeps of a geiger counter;
    * the time it takes before your next telephone call
    * the time until default (on payment to company debt holders) in reduced form credit risk modeling
    * the distance between mutations on a DNA strand;
    * the distance between roadkill on a given street;

[quote="hunter34"]In times like this, Google may help you more

http://en.wikipedia.org/wiki/Exponential_distribution[/quote]

Yeah i did as per your suggestions...but still cannot get much of it.. :(
[quote="hunter34"]Eh? Did you try to read the references? They usually go into things in more detail.[/quote]

try this website
[url]http://mathworld.wolfram.com/ExponentialDistribution.html[/url]

theres more info on wikipedia @ http://www.wikipedia.org

Well if the explanation on the websites are too complicated, just ask your math teacher or something...
To be honest I don't really get this either...


logarithms
calculus
calculus computations

A tank contains water which is heated by an electric water heater working under the action of a thermostat. The temperature of the water, $ \theta$ degrees C, may be modelled as follows. When the water heater is first switched on, $ \theta \equal{} 40$. The heater causes the temperature to increase at a rate $ k_{1}$ degrees C per second, wher $ k_{1}$ is a constant, until $ \theta \equal{} 60$. The heater then switches off.

(i) Write down, in terms of $ k_{1}$, how long it takes for the temperature to increase from $ 40$ to $ 60$.

The temperature of the water then immediately starts to decrease at a variable rate $ k_{2}(\theta \minus{} 20)$ degrees C per second, where $ k_{2}$ is a constant, until $ \theta \equal{} 40$.

(ii) Write down a differential equation to represent the situation as the temperature is decreasing.

(iii) Find the totla length of time for the temperature to increase from $ 40$ to $ 60$ and then decrease to $ 40$. Give your answer in terms of $ k_{1}$ and $ k_{2}$.

[I make the answers: (i) $ \frac{20}{k_{1}}$; (ii) $ \frac{d\theta}{dt} \equal{} \minus{}k_{2}(\theta \minus{} 20)$ and (iii) $ \frac{20}{k_1} \plus{} \frac{\ln 2}{k_{2}}$. Are these right?]

yes , i suppose , even i get the same answers .


Let $a \not = 0, b, c$ be real numbers. If $x_{1}$ is a root of the equation $ax^{2}+bx+c = 0$ and $x_{2}$ a root of $-ax^{2}+bx+c = 0$, show that there is a root $x_{3}$ of $\frac{a}{2}\cdot x^{2}+bx+c = 0$ between $x_{1}$ and $x_{2}$.

A direct use of the intermediate values theorem?  :wink:

Pierre.

; Let $f(x) =ax/2 + bx +c $ Then $f(x1)f(x2) = -3 a^2(x1)^2(x2)^2/4 < 0 $
 Hence the result follows by intermediate value theorem.

Do you know,how I get Ukrainian MO problems?


function
geometry
parallelogram
3D geometry
sphere
national olympiad

VietnamMO1993
Day1
1.Find the maximum and minimum values of the fuction
                    $f(x)=x(1993+\sqrt{1995-x^2})$ On its range.
2.Let$ABCD$be given a quadrilateral whose no two side are parallel.A parallelogram$MNPQ$varies so that$M,N,P,Q$are interior points od sides$AB,BC,CD,DA$respectively.Find the locus of the centroid of the parallelogram.
3.contruct a fuction$f: N\to N$such that$f(f(n))=1993n^{1945}$ for all $n$
Day2
1.Find all tetrahedra $ABCD$ inscribed in a given sphere for which $AB^2+AC^2+AD^2-BC^2-CD^2-BD^2$ attains its minimum 2.Each vertex of a convex polygon$A_{1}A_{2}...A_{1993}$ is assigned by + or - .In each step we reassigne all the vertices simultaneously,putting at $A_i$ + if the signs at $A_i$ and $A_{i+1}$ are the same ( where$A_{1994}=A_1$), and – ortherwise.Prove that after several steps the positionafter the first step will repeat.
3.Two sequenses ($a_n$) and ($b_n$) are defined by $a_0=2, b_0=1$ and $a_{n+1}=\frac{2a_n.b_n}{a_n+b_n}$
                                  $b_{n+1}=\sqrt{a_{n+1}.b_{n}}$
                   for all  nis positive integer
 Prove that two sequence have the same limit and evaluate this limit

\[\text{Day 1}\]
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=134114
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=758902#758902
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=758900#758900
\[\text{Day 2}\]
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=758904#758904
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=758906#758906
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=758911#758911


$ \begin{array}{ccccccc}
\mathrm{No.} & \mathrm{q} & \mathrm{w} & \mathrm{\Delta U} & \mathrm{\Delta H} & \mathrm{\Delta S} &{ \mathrm\Delta S_{\mathrm{univ}}} \\
a) & + & + & + & + & - & 0 \\
b) & + & + & + & + & - & 0 \\
c) & 0 & - & - & - & 0 & 0 \\
d) & + & - & 0 & 0 & + & -
\end{array}$
[code]
$
\begin{array}{ccccccc}
\mathrm{No.} & \mathrm{q} & \mathrm{w} & \mathrm{\Delta U} & \mathrm{\Delta H} & \mathrm{\Delta S} &{ \mathrm\Delta S_{\mathrm{univ}}} \\
a) & + & + & + & + & - & 0 \\
b) & + & + & + & + & - & 0 \\
c) & 0 & - & - & - & 0 & 0 \\
d) & + & - & 0 & 0 & + & -
\end{array}
$
[/code]

1. You are mixing maths and text mode. \Delta and the subscript _ command cannot go in text mode. You should only apply text where you want an upright symbol.
2. You need another column to cope with lines c & d
Consider using tabular rather than array for tables with lots of text in them.[code]\begin{tabular}{cccccccc} No. & q & w & $\Delta$U & $\Delta$H & $\Delta$ & $\Delta$S$_{\text{univ}}$ \\
a) & + & + & + & + & - & 0 \\
b) & + & + & + & + & - & 0 \\
c) & 0 & - & - & - & 0 & 0 & 0 \\
d) & + & - & 0 & 0 & + & - & 0
\end{tabular}[/code]

Certainly. Just count how many elements you have in each row. The number should be the same and it isn't. You have to put some empty elements if you want jagged tables. Stevem made this remark but forgot to incorporate it into the code ;)
[code]
\begin{tabular}{cccccccc} No. & q & w & $\Delta$U & $\Delta$H & $\Delta$ & $\Delta$S$_{\text{univ}}$ & \\
a) & + & + & + & + & - & 0 & \\
b) & + & + & + & + & - & 0 & \\
c) & 0 & - & - & - & 0 & 0 & 0 \\
d) & + & - & 0 & 0 & + & - & 0
\end{tabular}
[/code]

[quote="fedja"]Certainly. Just count how many elements you have in each row. The number should be the same and it isn't. You have to put some empty elements if you want jagged tables. Stevem made this remark but forgot to incorporate it into the code [/quote]In fact you don't need those extra &s in lines a & b so the code I gave is correct :) You are allowed to have too few & but not too many.

[quote="pardesi"]this form is not working stevem[/quote]What is wrong? Did you remember to use the amsmath package, which allows the use of \text? Try this complete document which works fine.
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\begin{tabular}{cccccccc} No. & q & w & $\Delta$U & $\Delta$H & $\Delta$ & $\Delta$S$_{\text{univ}}$ \\
a) & + & + & + & + & - & 0 \\
b) & + & + & + & + & - & 0 \\
c) & 0 & - & - & - & 0 & 0 & 0 \\
d) & + & - & 0 & 0 & + & - & 0
\end{tabular}
\end{document}[/code]

Yes, stevem, you are right!  :o It was a great surprise for me though... :)  Then I think pardesi referred to his attempt to test your construction on the forum (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1022252#1022252]here[/url]. I hope at least the explanation I gave there is correct :blush:

I agree with your explanation. It's even worse for this forum as it uses the dollar sign to signify the beginning and end of latex code so the first line of the tabular version would be seen as lots of little bits of code and thus gives a mess.
tabular can be made to work if all the dollar signs are removed and repaced by the alternative \(...\) which are not misinterpreted. Not much point doing this though except that the same code can then be used both on a forum and in a document.
[code]$\begin{tabular}{cccccccc} No. & q & w & \(\Delta\)U & \(\Delta\)H & \(\Delta\) & \(\Delta\)S\(_{\text{univ}}\) \\
a) & + & + & + & + & - & 0 \\
b) & + & + & + & + & - & 0 \\
c) & 0 & - & - & - & 0 & 0 & 0 \\
d) & + & - & 0 & 0 & + & - & 0
\end{tabular}$[/code]
$ \begin{tabular}{cccccccc} No. & q & w & \(\Delta\)U & \(\Delta\)H & \(\Delta\) & \(\Delta\)S\(_{\text{univ}}\) \\
a) & \plus{} & \plus{} & \plus{} & \plus{} & \minus{} & 0 \\
b) & \plus{} & \plus{} & \plus{} & \plus{} & \minus{} & 0 \\
c) & 0 & \minus{} & \minus{} & \minus{} & 0 & 0 & 0 \\
d) & \plus{} & \minus{} & 0 & 0 & \plus{} & \minus{} & 0
\end{tabular}$


...What?   :huh:

Do you mean $\left(-\frac{1}{n}\right)!$?

http://mathworld.wolfram.com/GammaFunction.html

[quote="john lai"]I meant$-(\frac{1}{(-n)!})$ for $n \in \mathbb{N}$[/quote]

that expression is undefined as it is, however, if you took the limit as n approaches an integer, you will get 0


What is the largest integer value of $ n$ for which $ 8^{n}$ evenly divides $ 100! ?$
I looked at the solution but didn't understand it.
Thanks for your help.

We will deal with this by prime factors. Since $ 8\equal{}2^3$, we will find the number of factors of $ 2$ in $ 100!$.

Every multiple of $ 2$ has at least one factor of two, and there are $ \frac{100}{2}\equal{}50$ multiples of two being multiplied in $ 100!$.

Every multiple of $ 4$ has at least two factors of two, but we already counted one of them in the above case. Thus, we have to count them once more, so we have $ \frac{100}{4}\equal{}25$ more factors of two.

Every multiple of $ 8$ has at least three factors of two, but we already counted two of these in the above cases. Thus, we have to count them once more, so we have $ \frac{100}{8}\equal{}12.5$, but we have to take the floor value of this because a half multiple of $ 8$ (which is 4) is already counted above. Thus, we have $ 12$ more factors.

We continue similarly for the rest of the powers of two, getting $ 6$ for sixteen, $ 3$ for thirty-two, and $ 1$ for sixty-four. Since the next power of two is greater than $ 100$, we can stop.

Now we add the number of factors of two: $ 50\plus{}25\plus{}12\plus{}6\plus{}3\plus{}1\equal{}97$. Thus, in the prime factorization of $ 100!$, we have a $ 2^{97}$. Now, think of the problem like this, where $ n$ and $ k$ are both integers:
\[ \frac{100!}{8^n}\equal{}\frac{2^{97} \cdot 3^a \cdots}{8^n}\equal{}k\]
Now, we cancel:
\[ \frac{2^{3 \cdot 32\plus{}1} \cdot 3^a \cdots}{8^n}\equal{} \frac{8^{32}\cdot 2^1 \cdot 3^a \cdots}{8^n}\equal{}\frac{8^{32}}{8^n} \cdot 2^1 \cdot 3^a \cdots \equal{}k\]
Since $ k$ is an integer, and the stuff not in the fraction, on the LHS, is also an integer, we must have the fraction be an integer. Thus, $ \frac{8^{32}}{8^n}\equal{}8^{32\minus{}n}$ is an integer as long as $ 32\minus{}n$ is not negative. The greatest value of $ n$ for which this is true is $ \boxed{32}$, so this is our answer.

Ho gogators. Just seen I've been beaten to it. My solution is much the same except that PowerOfPi has explained it in more detail and at the end I have  $ 2^{97} \equal{} (2^{3})^{32\frac{1}{3}}$, giving $ 8^{32}$ as the highest power of $ 8$.


geometry
parallelogram
power of a point
radical axis
geometry proposed

Let a trapez $ ABCD$ with $ AB //CD$. A circle through $ A,D$ is tangent to $ BC$ at point $ E$ ,while a circle through $ B,C$ is tangent to $ AD$ at point $ Z$ .Prove that the common chord of these two circles besects segment $ ZE$.(prove that the inverse also holds).

 [u]Babis[/u]

let $ MN$ be the common chord and $ w_{1},w_{2}$ be the circles which are tangent to $ BC,DA$ respectively,and also let $ T$ be the intersection of $ BC,AD$...
now we know that:
$ TZ^{2}\equal{} TB.TC (1)$
$ TE^{2}\equal{} TA.TD (2)$
now from $ AB\parallel CD$ we get that:
$ \frac{TB}{TC}\equal{}\frac{TA}{TD}\Rightarrow TA.TC \equal{} TB.TD (3)$
now from $ 1,2,3$ we get that
$ TE.TZ \equal{} TA.TC\Rightarrow AE\parallel CZ$
similary we can show that $ DE\parallel BZ$
now let $ P,Q$ be the intersection points of $ AE,BZ$ and $ DE,CZ$ repectively...
thus $ PEQZ$ is parallelogram...
now we have:
$ CZ\parallel EA\Rightarrow\angle ECZ \equal{}\angle BEZ \equal{}\angle EDA$
so we have $ \angle ECZ \equal{}\angle EDA$ so $ ECDZ$ is cyclic so $ PC.PZ \equal{} PE.PD$
but in the other hand we know that the power of point $ P$ to circle $ w_{1}$ is equal to $ PC.PZ$ and also the power of $ P$ to $ w_{2}$ is equal to $ PE.PD$ thus we get that the power of point $ P$ to circles $ w_{1},w_{2}$ are equal
so $ P$ lies on the radical axis of $ w_{1},w_{2}$ so $ P$ lies on $ MN$...
similary we can prove that $ Q$ lies on $ MN$...
so the problem is solved,because we proved that $ PEQZ$ is parallelogram so $ PQ$ bisects $ EZ$ but $ P,Q$ lie on $ MN$ so $ MN$ bisects $ EZ$ as wanted...

[b]SOME REMARKS[/b]

 I have improved my solution , after some nice results by BaBak

From the above solution we keep ready that $ AE//CZ$ ,  $ ZB//DE$ and that  $ DZEC,ABEZ$ are inscribed. Let line $ ZE$ meets the circles at points $ F,G$ in the order $ F\minus{}Z\minus{}E\minus{}G$ and $ KL$ is the common chord. If $ KL , ZE$ meet each other at $ M$  , then we have:

a)	$ MF.ME\equal{}MK.ML \equal{}MZ.MG$ or $ (MZ\plus{}ZF).ME\equal{}MZ.(ME\plus{}EG)$ , or $ ME.ZF\equal{}MZ.EG$

b)	We want to prove that $ MZ\equal{}ME$ , so it suffices to prove that $ ZF\equal{}EG$

c)	But $ ZF \equal{}\frac{ZA.ZD}{ZE}$ and $ EG \equal{}\frac{EB.EC}{EZ}$ ,

so we want to prove that

$ ZA.ZD \equal{} EB.EC$  (1)

d)	From $ AE//ZC$ we take $ \frac{TA}{AZ}\equal{}\frac{TE}{EC}$ and from $ ZB // DE$ we take

       $ \frac{TZ}{ZD}\equal{}\frac{TB}{BE}$

 We multiply these two relations and since $ TA.TZ \equal{} TE.TB$ ( from the inscribed ABEZ) we receive $ AZ.ZD\equal{} EC.EB$.
 Thus relation (1) is true and the solution is complete.

[u] Babis[/u]

[quote="Altheman"]I didn't read your solution. I will reformulate the problem as follows:

Problem: Let $ \triangle ABC$ have $ D\in AB$, $ E\in AC$, so $ DE//BC$. Let $ G\in AB$, $ H\in AC$ so $ \mathcal{C}(CEG)$, $ \mathcal{C}(BDH)$ touch $ AB$, $ AC$, respectively. Prove that the radical axis of $ \mathcal{C}(CEG)$, $ \mathcal{C}(BDH)$ bisects $ GH$.

Proof: Use power of a point [$ AG^{2}\equal{} AE*AC$, parallel lines] to prove that $ AD*AG \equal{} AE*AH$ and $ AB*AG \equal{} AC*AH$. This implies that $ DEHG$ and $ GHCB$ are cyclic. Let $ \mathcal{L}$ be the radical axis in the problem. It trivially follows the following triples of lines are concurrent: $ DH$, $ GE$, $ \mathcal{l}$ and $ GC$, $ BH$, $ \mathcal{L}$ by radical axis theory.

Let $ X\in DH\cap GE$; $ Y\in GC\cap BH$. This means that $ \mathcal{L}\equiv XY$. Again we can use power of a point and the parallel lines to prove that $ GE//BH$ and $ DH//GC$. It follows that $ XHYG$ is a parallelogram so $ XY$ bisects $ GH$, as desired.

Comment 1: $ BE\cap CD\in\mathcal{L}$ by pascal's on $ BGDEHC$.

Comment 2: If we take the points of tangency of the other tangents from $ A$ to $ \mathcal{C}(CEG)$, $ \mathcal{C}(BDH)$, then $ \mathcal{L}$ bisects this segment as well.[/quote]

what is the radical axis theory?

I was lazy and didn't feel like explaining it...

Here is the basic idea:

The power of a point, $ P$, wrt a circle center $ O$, radius $ R$ is $ OP^{2}\minus{}R^{2}$.

The locus of points with equal powers wrt two circles is called their radical axis. It is clear that if the radical axis of circle 1 and circle two intersects the radical axis of circle 2 and circle 3, then the radical axis of circle 1 and circle 3 passes through that intersection point.


If a, b, c, d are non-negative real numbers, prove:
$ \frac{a\minus{}b}{a\plus{}2b\plus{}c}\plus{}\frac{b\minus{}c}{b\plus{}2c\plus{}d}\plus{}\frac{c\minus{}d}{c\plus{}2d\plus{}a}\plus{}\frac{d\minus{}a}{d\plus{}2a\plus{}b} \geq 0$

nice one, this is my solution,
the inequality is equivalent as:
$ \sum {\frac{a-b}{a+2b+c}+\frac{1}{2} } \geq 2$
$ \Leftrightarrow \sum \frac{3a+c}{a+2b+c} \geq 4$
from Cauchy-Shwarz,
$ (\sum \frac{3a+c}{a+2b+c})(\sum (3a+c)(a+2b+c) ) \geq 16(a+b+c+d)^2$
and we have the identity $ \sum (3a+c)(a+2b+c)= 4(a+b+c+d)^2$ from where we deduct our inequality,

The following inequality is true too.
Let $ a,$ $ b,$ $ c$ and $ d$ are positive numbers. Prove that:
\[ \frac{a}{a\plus{}3b}\plus{}\frac{b}{b\plus{}3c}\plus{}\frac{c}{c\plus{}3d}\plus{}\frac{d}{d\plus{}3a}\geq1\]


calculus
integration
real analysis
real analysis unsolved

O.K. !
[quote="Virgil Nicula"]$ \boxed {\ \lim_{x\to \infty}\int_0^x\frac {1}{\left(x^2 \plus{} 1^2\right)\left(x^2 \plus{} 2^2\right)\ldots\left(x^2 \plus{} n^2\right)}\ \mathrm {dx} \equal{} \pi\cdot\sum_{k \equal{} 1}^n\frac {( \minus{} 1)^{k \minus{} 1}\cdot k}{(n \minus{} k)!\cdot(n \plus{} k)!}\ }$ .[/quote]
I hope to be no mistakes !

Let $ f(z) \equal{} 1 \left\slash \prod_{k \equal{} 1}^n \left( z^2 \plus{} k^2 \right) \right.$ and $ \gamma$ be the contour $ [ \minus{} R,R] \cup \mathcal C$, where $ \mathcal C$ is a semicircle of radius $ R$. As $ R \to \infty$, the integral over the semicircle goes to $ 0$, so $ \int_{ \minus{} \infty}^\infty f(x) \, dx \equal{} 2 \pi i \cdot \sum_{k \equal{} 1}^n \text{Res} \left( f, ki \right) \equal{} \pi \cdot \sum_{k \equal{} 1}^n \frac {( \minus{} 1)^{k \minus{} 1} \cdot 2 k^2}{(n \minus{} k)! \cdot (n \plus{} k)!}$. The desired integral is half of this, namely
\[ \pi \cdot \sum_{k \equal{} 1}^n \frac {( \minus{} 1)^{k \minus{} 1} \cdot k^2}{(n \minus{} k)! \cdot (n \plus{} k)!} .
\]
I'm not getting the same answer  :blush:

[quote="Virgil Nicula"]$ \boxed {I_n\equiv\int_0^{\infty}\frac {1}{\left(x^2 \plus{} 1^2\right)\left(x^2 \plus{} 2^2\right)\ldots\left(x^2 \plus{} n^2\right)}\ \mathrm {dx} \equal{} \pi\cdot\sum_{k \equal{} 1}^n\frac {( \minus{} 1)^{k \minus{} 1}\cdot k}{(n \minus{} k)!\cdot(n \plus{} k)!}\ }$ .[/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $ x^2 \equal{} y$ . Thus, $ \prod_{k \equal{} 1}^n\frac {1}{\left(x^2 \plus{} k^2\right)} \equal{} \prod_{k \equal{} 1}^n\frac {1}{\left(y \plus{} k^2\right)} \equal{}$ $ \sum_{k \equal{} 1}^n\frac {A_k}{y \plus{} k^2}$ $ \implies$

$ 1 \equal{} \sum_{k \equal{} 1}^n\left[A_k\cdot\prod_{j \equal{} 1\ ,\ j\ne k}^n(y \plus{} j^2)\right]$ for any $ y\in\mathbb R$ . Particularly, for $ y \equal{} \minus{} k^2$ obtain :

$ 1 \equal{} A_k\cdot\prod_{j \equal{} 1\ ,\ j\ne k}^n(j^2 \minus{} k^2) \equal{}$ $ A_k\cdot\prod_{j \equal{} 1}^{k \minus{} 1}(j \minus{} k)\cdot\prod_{j \equal{} k \plus{} 1}^n(j \minus{} k)\cdot\prod_{j \equal{} 1\ ,\ j\ne k}^n(j \plus{} k)$ $ \implies$

$ 1\equal{}A_k\cdot ( \minus{} 1)^{k \minus{} 1}(k \minus{} 1)!\cdot(n \minus{} k)!\cdot\frac {(n \plus{} k)!}{k!\cdot 2k}$ $ \implies$ $ A_k \equal{} \frac {( \minus{} 1)^k\cdot 2k^2}{(n \minus{} k)!(n \plus{} k)!}$ . Therefore,

$ I_n \equal{} \int_0^{\infty}\prod_{k \equal{} 1}^n\frac {1}{x^2 \plus{} k^2}\ \mathrm {dx} \equal{}$ $ \sum_{k \equal{} 1}^nA_k\int_0^{\infty}\frac {1}{x^2 \plus{} k^2}\ \mathrm {dx} \equal{}$ $ \frac {\pi}{2}\cdot\sum_{k \equal{} 1}^n\frac {A_k}{k}$ .

In conclusion, $ \boxed {I_n \equal{} \pi\cdot\sum_{k \equal{} 1}^n\frac {( \minus{} 1)^{k \minus{} 1}\cdot k}{(n \minus{} k)!(n \plus{} k)!}}$ .[/color]

[quote="milin"]perfect_radio, if I'm not mistaken, your integral runs from $ \minus{} \infty$ to $ \infty$ where Virgil Nicula's goes from $ 0$ to $ \infty$ - this may be the reason you are not getting the desired result.[/quote]
That was never the problem :)

I just misplaced a $ k$ when I computed the residues. Virgil Nicula's result is correct.


abstract algebra
college
invariant
group theory
superior algebra
superior algebra unsolved

Just for fun.  This appeared on the UC Berkeley prelim today:

Let $ F$ be a field, $ p$ a prime, and $ n$ a positive integer.  Show that the number of abelian groups of order $ p^n$, up to isomorphism, is equal to the number of $ n$x$ n$ matrices $ A$ over $ F$, up to similarity, such that $ A^n \equal{} 0$.

I'm sure that a few of you can solve this instantly, so try to save it for everyone else. :)  I don't know if it's a very deep fact, but there are some cute solutions...

are the matrices $ A$ of a fixed dimension? and i assume they are over the field $ F$, otherwise why mention the field at all...

that being said, the number of abelian groups of order $ p^n$ is in bijection with the partitions of $ n$ things (number of ways to factor $ p^n$, or maybe $ x^n$...).

if the dimension of $ A$ is $ n$, then $ A^n \equal{} 0$ means $ Char(A) \equal{} x^n$.

i assume the bijection you are implying is between the invariant factors of the finite abelian group $ G$ and the invariant factors of linear transformation associated with $ A$. both of these imply uniqueness, up to isomorphism in the former case, and up to similarity in the latter.

[quote="jmerry"]Those are $ n\times n$ nilpotent matrices, and the field doesn't matter at all. It should have been specified that way.[/quote]

It's called making mistakes. :)  Anyway, the problem is fixed, and of course shallow is correct.

structure theorem of abelian groups and jordan normal form links the two numbers.


Dear everyone.

studying middle school problems, I found thing odd :)

for a triangle ABE, let O be the midpoint of AB.
circle (O,OB) meets EA, EB at C, D respectively.
CA=CB and ED:DB=2:3
F $ \in$ EB and FD=DB. circle (O,OB) meets FA at G.
prove that GD, FO, CB are concurrent.

I guess this is true, wishing it is unknown.
May I ask you a favor? :)

Thanks for reading my shoddy post.

I think the key of the solution of this problem, is the relation $ \frac {ED}{DB} \equal{} \frac {2}{3}$ $ ,(1)$

$ \bullet$ Let be the points $ K\equiv AB\cap FC$ and $ L\equiv FB\cap GC.$

Applying the [b]Menelaos theorem[/b], we can prove that $ \frac {OK}{KA} \equal{} \frac {1}{2} \equal{} \frac {DL}{LF}$ $ ,(2)$ and so, we have that $ OD\parallel KL\parallel AF.$

Because of now, the collinearity of the points $ K\equiv BO\cap CF$ and $ L\equiv BD\cap CG$ and $ \infty \equiv OD\cap FG,$ based on the [b]Desarques theorem[/b], we have that the triangles $ \bigtriangleup BOD,\ \bigtriangleup CFG$ are perspective.

Hence, we conlude that $ BC\cap OF\cap DG\equiv N$ and the proof is completed.

Kostas Vittas.

PS. I will post here later more details about the relation $ (2)$ I mentioned ( with so much applying the [b]Menelaos theorem[/b]. :ewpu:)

There is an easy construction of the triangle $ \bigtriangleup EAB$ as the problem states, beginning from an arbitrary right isosceles triangle $ \bigtriangleup CAB$  with $ CA \equal{} CB.$

So, we define the point $ D'$ on the side-segment $ CB$ such that $ \frac {CD'}{D'B} \equal{} \frac {2}{3}$ and we denote the point $ D$, as the point of intersection of the circle $ (O)$ with diameter $ AB,$ from the line through $ D'$ and parallel to $ CA.$

We denote the point $ E\equiv CA\cap BD$ and it is easy to show that $ \frac {ED}{DB} \equal{} \frac {2}{3},$ by [b]Thales theorem[/b].

$ \bullet$ But, we will construct the triangle $ \bigtriangleup EAB$ by another way, useful for the solution of the proposed problem, as follows :

Let $ \bigtriangleup CAB$ be, the right isosceles triangle as before, with circumcircle $ (O)$.

We define the point $ M$ between $ A,\ B,$ such that $ AM \equal{} 2MB$ and let $ H,\ E$ be, the points of intersection of $ CB,\ CA$ respectively, from the line through the point $ M$ and perpendicular to $ AB.$

It is easy to show that he points $ B,\ D\equiv (O)\cap AH,\ E$ are collinear and we will prove that $ \frac {ED}{DB} \equal{} \frac {2}{3}.$

$ \bullet$ Applying the [b]Menelaos theorem[/b], in the triangle $ \bigtriangleup CAB$ with transversal $ MHE,$

we have that $ \frac {EC}{EA}\cdot \frac {MA}{MB}\cdot \frac {HB}{HC} \equal{} 1$ $ \Longrightarrow$ $ \frac {EC}{EA} \equal{} \frac {1}{4}$ $ ,(1)$

because of $ MA \equal{} 2MB$ and $ HB \equal{} 2HC.$

From $ (1)$ $ \Longrightarrow$ $ \frac {CE}{CA} \equal{} \frac {1}{3}$ $ ,(2)$

Applying again the [b]Menelaos theorem[/b], in the triangle $ \bigtriangleup BCE$ with transversal $ AHD,$

we have that $ \frac {DE}{DB}\cdot \frac {HB}{HC}\cdot \frac {AC}{AE} \equal{} 1$ $ \Longrightarrow$ $ \frac {DE}{DB} \equal{} \frac {2}{3}$ $ ,(3)$.

So, we have now, the configuration of the problem $ \bigtriangleup EAB$ as the problem states and because of $ (3)$, in the triangle $ \bigtriangleup ABD$ with transversal $ MHE$,

we have $ \frac {HD}{HA}\cdot \frac {MA}{MB}\cdot \frac {EB}{ED} \equal{} 1$ $ \Longrightarrow$ $ \frac {HD}{HA} \equal{} \frac {1}{5}$ $ ,(4)$

$ \bullet$ We draw the circle $ (D),$ centered at point $ D$ with radius $ DB$ and we denote the points $ F\equiv BE\cap (D)$ and $ G\equiv (O)\cap (D).$

Also, we denote the points $ K\equiv AB\cap FC$ and $ L\equiv FB\cap GC$ and we will prove that $ \frac {OK}{KA} \equal{} \frac {DL}{LF} \equal{} \frac {1}{2}$.

It is easy to show that the points $ F,\ G,\ A$ are collinear, because of $ BG\perp GF.$

Applying the [b]Menelaos theorem[/b] now, in the triangle $ \bigtriangleup EAB$ with transversal $ KCF,$

we have that $ \frac {KA}{KB}\cdot \frac {FB}{FE}\cdot \frac {CE}{CA} \equal{} 1$ $ \Longrightarrow$ $ \frac {KA}{KB} \equal{} \frac {1}{2}$ $ \Longrightarrow$ $ \frac {OK}{KA} \equal{} \frac {1}{2}$ $ ,(5)$ because of $ (2)$ and $ \frac {FB}{FE} \equal{} \frac {6}{1}$ ( easy to prove ).

$ \bullet$ We denote the point $ H'\equiv AD\cap BG,$ as the orthocenter of the triangle $ \bigtriangleup FAB$ and it is easy to show that $ HD \equal{} 2HH',$ from $ ED \equal{} 2EF$ and $ FH'\parallel EH.$

In the right triangle $ \bigtriangleup ADF$ with transversal $ GH'B$ we have that $ \frac {GF}{GA}\cdot \frac {H'A}{H'D}\cdot \frac {BD}{BF} \equal{} 1$ $ \Longrightarrow$ $ \frac {GF}{GA} \equal{} \frac {2}{3}$ $ ,(6)$

because of $ \frac {BD}{BF} \equal{} \frac {1}{2}$ and $ \frac {H'A}{H'D} \equal{} \frac {3}{1}$ ( $ H'$ is also the centroid of the isosceles triangle $ \bigtriangleup ABF$ with $ AB \equal{} AF$ ).

In the triangle now, $ \bigtriangleup AEF$ with transversal $ GCL,$ we have that $ \frac {LE}{LF}\cdot \frac {GF}{GA}\cdot \frac {CA}{CE} \equal{} 1$ $ \Longrightarrow$ $ \frac {LE}{LF} \equal{} \frac {1}{2}$ $ \Longrightarrow$ $ \frac {DL}{LF} \equal{} \frac {1}{2}$ $ ,(7)$

From $ (5),\ (7)$ $ \Longrightarrow$ $ OD\parallel KL\parallel AF$ and the proof is copleted by [b]Desarques theorem[/b], as presented in my previous post ( [url=http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=23559]schema [b]t=309289[/b][/url] ).

Kostas Vittas.


A question I can't solve:

Let A, B, C, D be four points in the plane.

Let P be a point on line segment AC and Q be a point on line segment BD such that AP = BQ.

Prove or disprove that |PQ| <= sqrt(2) max(|AB|, |CD|).

I can show that if sqrt(2) is replaced by any smaller constant, there is a counterexample to show that the inequality does not hold.  But I can't prove that this inequality holds for k=sqrt(2).

That's a cutie :). We can assume that $ |AC|\le |BD|$. By convexity, it is enough to consider the cases $ P \equal{} A, Q \equal{} B$ (trivial) and $ P \equal{} C, |BQ| \equal{} |AC|$. Now, if the inequality does not hold, we should have $ \angle CQB > 135^\circ$. Assuming that $ |AC| \equal{} 1$, $ |CQ| \equal{} u$, and using that $ |AB|\ge |BC| \minus{} |AC|$, we see that we need to prove the inequality $ \sqrt {1 \plus{} u^2 \plus{} \sqrt 2\,u} \minus{} 1\ge \frac u{\sqrt 2}$, which is fairly obvious.

The argument works in space as well.


Let $ a;b;c\geq 0 satisfying: \sum a^2\equal{}\sum a$.Prove that:
       $ \sum a^2b^2\leq\sum ab$

[quote="karis"]Let $ a;b;c\geq 0 satisfying: \sum a^2 \equal{} \sum a$.Prove that:
       $ \sum a^2b^2\leq\sum ab$[/quote]
 We must prove
$ (a^2b^2\plus{}b^2c^2\plus{}c^2a^2)(a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2bc\plus{}2ac)\le (ab\plus{}bc\plus{}ac)(a^4\plus{}b^4\plus{}c^4\plus{}2a^2b^2\plus{}2b^2c^2\plus{}2c^2a^2)$
$ (a^2b^2\plus{}b^2c^2\plus{}c^2a^2)(a^2\plus{}b^2\plus{}c^2)\le (a^4\plus{}b^4\plus{}c^4)(ab\plus{}bc\plus{}ac)$ thiss we could prove using Muirhead
$ 2F(4,2,0)\plus{}F(2,2,2)\le 2F(5,1,0)\plus{}F(4,1,1)$
$ F(4,1,1)\ge F(2,2,2)$
$ F(5,1,0)\ge F(4,2,0)$ and we are done :lol:


could it be something is missing in there? :D

sorry, solve for x:
$\frac{4}{x-3}-\frac{6}{x-2}=0$

[hide]

If you add $  \frac{6}{x-2} $ on each side, the equation becomes $  \frac{4}{x-3} = \frac{6}{x-2}  $
You multiply by $ x - 2 $ on each side.

$ \frac{4x-8}{x-3} = 6  $

$  x = 5  $[/hide]


probability
advanced fields
advanced fields theorems

Let X1 ~N(2, 3) and X2  ~N(2, 1) be two independent normal random
variables. Find the probability, P(2X1 + X2 > 8).

Hi, im really unsure how to do this question, but this is what ive tried.

im adding the two normal random var. using a formula i found here (http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables)

$\(X1+X2 \sim N ( 2+2, 3+1) \)$

$\(= X1+X2 \sim N ( 4, 4) \)$

but im stuck here, how do i times X1 by 2 to show that this is > 8, or is what im doing wrong?

thanks for any help :).

I assume that $N(2, 3)$ means the normal distribution with mean 2 and variance 3.

When you double a random variable, you double its mean and standard deviation, and hence quadruple its variance.  Because $X_{1}$ has the distribution $N(2, 3)$, its double $2X_{1}$ has distribution $N(4, 12)$.

Now you can apply your previous method.

Early on, you wrote $X1$ in two places where you meant $2X_{1}$.  Fortunately, that typo didn't cause you any errors.

Later, when you defined $Z$, you divided by $13$, the variance.  Aren't you supposed to divide by $\sqrt{13}$, the standard deviation?

At the end, when you wrote $P(X < 0.153)$, you need $Z$ instead of $X$ and, more importantly, greater-than instead of less-than.


geometry
incenter
vector
trigonometry
perimeter
angle bisector
geometry theorems

About angle bisectors & the incenter of a triangle :

-Is there a well - known formula for the angle bisector length ?
(Like for a median : $m_a=\frac12\sqrt{2b^2+2c^2-a^2}$)

-How do you express the incenter of a triangle with vectors ?

Is there a well - known formula for the angle bisector length ?

yes, you can use angle bisector theorem and stewart's theorem to derive the formula

[b]About the angle bisectors :[/b]

Let a, b, c be the lengths of the sides of triangle ABC.

the formula is:

$AD^2=\frac{4bc}{\left(b+c\right)^2}\cdot p\left(p-a\right)$ where AD is the bisector corresponding to angle A and p is the semiperimeter of ABC and it can be proved using Stewart and the bisector theorem!

[b]About the incenter :[/b]

we know that for any M in the same plane of triangle ABC (inside the triangle or outside) we have the following relation:

$a\cdot\overrightarrow{MA}+b\cdot\overrightarrow{MB}+c\cdot\overrightarrow{MC}=\left(a+b+c\right)\cdot\overrightarrow{MI}$ using vectors.

Hereby, I is the incenter of triangle ABC.

cheers@!

Here's another formula for the bisector: $l_a=\cos\frac{A}{2}\cdot\frac{2bc}{b+c}$, where $l_a$ is the bisector of the angle $A$ of a triangle $ABC$. This often comes in handy.

Indeed In both cases $l_a=\cos\frac{A}{2}\cdot\frac{2bc}{b+c}$ and $AD^2=\frac{4bc}{\left(b+c\right)^2}\cdot p\left(p-a\right)$ we have the interior bisectors! :D

yes there is the a problem that I have proposed for the Institute of Mathematics Test for High-School Students in 2002:

let p denote the semi-perimeter of a triangle ABC, and denote by L the sum of the lengths of all interior bisectors. Prove that
$p\sqrt 3\geq L$.

[color=red][Moderator edit: Just to note that this problem has recently been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=14431 .][/color]

I guess that I solved this problem somewhere on this forum! :D :D

$l_a\leq\sqrt{p\left(p-a\right)}$ and we apply Cauchy! :D

[quote="Simo_the_Wolf"]But why we can say that [tex]\displaystyle a\cdot\overrightarrow{MA}+b\cdot\overrightarrow{MB}+c\cdot\overrightarrow{MC}=\left(a+b+c\right)\cdot\overrightarrow{MI}[/tex] ? Can you show it please??[/quote]

 It follows imediately from bisector theorem applied in the triangles $ABC$ and $ABD$ where $AD$ is the bisector of
 the triangle $ABC$. ;)


earlier my teacher gave us this math problem as extra credit on a test but he could not solve it, i was wondering if there is a mathamatical way to solve this:

A man goes to a 711 store and he buys 4 items. He goes to the cashier and the cashier mulitplies the prices and asks for 7.11. The customer argues that the numbers should be added and not multiplied. The cashier then says that it does not matter; [b]the prices of the 4 items multiplied and the sum of the numbers equals 7.11[/b]. What are the prices of each of the items?

i cannot put the dollar signs because of the latex in but the prices are in dollars so like 1.23 dollars could be a price

Very difficult problem, but couldn't there be more than $1$ possibillity?
I don't see how you can do this without trial and error, which is very tedious...

I think:

Say A,B,C, and D are cents

$A+B+C+D=711$, and $ABCD=711000000$, and $711000000$, when prime factored is $3^2(79)(5^6)(2^6)$, and so one of the prices has to be a multiple of $79$, I think...and each of the others all have to be multiples of $5,3$ and $2$. From there, it's just trial and error...

Well, after that, you can say the following...
$A+B+C+D$ is odd, and therefore, one of them is odd, or three are.  Then, we see that the fourth root of 711000000 is around 163, while one fourth of 711 is around 175.  So, with the arithmetic mean pretty close to the geometric mean, we know that the numbers aren't too far out.
You need to be able to estimate to do the rest, i guess...
First, assume that one of the numbers has 2^6 as a factor, and the rest are odd.  Then, the numbers might look something like
$2^63=192$
$3*79=237$
$5^3=125$
$5^3=125$
Sum:679
We see that the sum of these numbers is not enough: they must be farther apart.  The sum must increase by 32.
Just keep using logic after that to get it, i suppose...

yeah, thanks that helped, basically what i did was that you can only produce a few odd numbers with 3's and 5's the units digit that can be produced

$1$ (79*9) then the other numbers have to add up to 0 in the units digits place but
   that does not work

$5$ 5x (where x is any of the odd numbers, since any odd number times 5 ends in 5)
   so if the units digit is 5 for the odd number, the other number has to end in 6
   the possibilities of the second number is 6,316,36,16,256,96

$7$ 79*3 = 237, then the other number has to end with 4, those options give 2^2=4, 2^6 = 64

$9$ 79 or 3*3 = 9 but i assumed that was too low, also the second number has to end in a 2, where the possibilities are just 2^1 and that is too low to be a factor
   to sum up to 1 in the units digit

this analysis gives you either to use:
5x and 316, 5x and 256 (since the other possibilities are too small)
or, 237 and 64 (but this does not work since the factors left and 3 and 5's cannot produce 711 - 237 - 64 = 410

so i tested 5x and 316
$WXYZ=711000000$
$WXY=2250000$
$W+X+Y+Z=711$
$W+X+Y=395$
and the remaining factors are $2^5*3*5^6$
so 316, then 5,25,125,15,75,375 are the possibilites for the price 5x
since the 125 seemed the most reasonable, since the other numbers were either too high or too low, i tried, (316,125,y,z)
$WXY=2250000$
$WX=18000$
$W+X+Y+395$
$W+X=270$
Then substitute those together,
$(270-x)x = 18000$
$x^2 -270x + 18000 = 0$
$(x-120)(x-150)=0$
$x = 120, 150$
so the solutions are $1.20 , 1.25, 1.50, 3.16$

in retrospect, i see that it was a large assumption that 2 of the factors created the whole last digit however, it was a reasonable assumption since there were 6 0's at the end of the product


linear algebra
matrix
linear algebra unsolved

Prove that : the determinant of anti-symmetric square matrix of even order is some perfect square.
The 'perfect square' above means ,for example n=2, $ \left |\begin{matrix} 0 & a\\ \minus{}a & 0 \end{matrix}\right|\equal{}a^2$, for n=4,
$ \left|\begin{matrix} 0 & a & b & c\\ \minus{}a & 0 & d & e\\ \minus{}b & \minus{}d & 0 & f\\ \minus{}c & \minus{}e & \minus{}f & 0\end{matrix}\right|\equal{}(af\minus{}be\plus{}dc)^2$


integration
trigonometry
calculus
calculus computations

Ascertain $ \int\frac {x\sin x\plus{}n\cos x}{x^{n\plus{}1}}\ \mathrm {dx}\ ,$ where $ n\in\mathcal N\ .$

$ \frac{\minus{}cos x}{x^n}$. Use integration by parts on the first part of integral.


Let $ a_{1},a_{2},...,a_{n}$ be non-negative real numbers. Prove that
\[ \sum_{i\equal{}1}^{n}a_{i}^{2}\plus{}(n\minus{}1)\sum_{i<j<k}a_{i}a_{j}a_{k}\plus{}1\ge 2\sum_{i<j}a_{i}a_{j}.\]
[Moderator edit: Corrected according to Vasc's post below. Thanks for the dedication, by the way. ;) - Darij]

For particular case $n=3$, we get the following one (of Darij)
\[a^{2}+b^{2}+c^{2}+2abc+1 \ge 2(ab+bc+ca). \]

This is a question Manlio raised here
http://www.mathlinks.ro/Forum/viewtopic.php?t=148721&sid=7551b896e15dcf9f90e6d9e6341852c8

What is CID theorem? Where is it?  :lol:

A theorem I posted here

http://www.mathlinks.ro/Forum/viewtopic.php?p=841912#841912

[quote="larry09"]hungkhtn, what does CID mean? Is this the most important and interesting result in your book?[/quote]

 :blush: My stupidity. It should be SID = Symmetric Inequality of Degree 3 theorem. I am mistaking between two theorems CID and PID. I am so sorry.

@Larry. It is not the most important and interesting result. It is only in 12 pages of my book (with about nearly 500 pages for both volumes, +150 page from the old Vietnamese book). I think you are Vietnamese? Am I right?

[quote="Vasc"][quote="hungkhtn"]Let $a_{1},a_{2},...,a_{n}$ be non-negative real numbers. Prove that
\[\sum_{i=1}^{n}a_{i}^{2}+2\sum_{i<j<k}a_{i}a_{j}a_{k}+1 \ge 2\sum_{i<j}a_{i}a_{j}. \]
[/quote]
The inequality doesn't hold for $n>3$. See the case $n=4$ and $x_{1}=x_{2}=x_{3}=x_{4}=\frac 4{5}$.
I found that
\[\frac 1{2}\sum_{i=1}^{n}a_{i}^{2}+\sum_{i<j<k}a_{i}a_{j}a_{k}+\frac{2n(n-2)}{3(n-1)^{2}}\ge \sum_{i<j}a_{i}a_{j}. \]
[/quote]

My head is really wrong these days. I made a lot of Computing mistake. The problem here is

\[\sum_{i=1}^{n}a_{i}^{2}+(n-1)\sum_{i=1}^{n}a_{i}a_{j}a_{k}+1\ge 2\sum_{i<j}a_{i}a_{j}.\]

It is Ok now, VASC.

[quote="hungkhtn"] The problem here is
\[\sum_{i=1}^{n}a_{i}^{2}+(n-1)\sum_{i=1}a_{i}a_{j}a_{k}+1\ge 2\sum_{i<j}a_{i}a_{j}. \]
It is Ok now, VASC.[/quote]
This inequality is strict and weak for $n>3$. For $n>3$ we may replace $n-1$ with $\frac 4{\sqrt 3}$.
The best inequality of type
\[\sum_{i=1}^{n}a_{i}^{2}+p\sum_{i<j<k}a_{i}a_{j}a_{k}+1\ge 2\sum_{i<j}a_{i}a_{j}\]
is for $p=\frac 4{n-1}\sqrt{\frac{n(n-2)}{3}}$.

Yes, I think that, too. Since natural numbers are better and alternatively, I proved my inequality basing on the fact the
\[p=\frac 4{n-1}\sqrt{\frac{n(n-2)}{3}}\le n-1\]
if $n\ge 3$. Just use SID theorem, then AM-GM inequality.


Euler
induction
Gauss
number theory
prime numbers
number theory unsolved

We define $\omega(n)$ the number of prime divisors of $n\in N$ and $\phi(n)$ the number of  prime is equal or smaller than $n\in N$.Find all $n\in N$ such that $\phi(n)-\omega(n)=\varphi(n)$ ($\varphi$ is the Euler function).

simple reasoning, Phi(n) has at least the number 1 and all primes <= n and coprime with n, Omega(n) has all the primes <=n that divide n. So these two together always have one more than all the primes <= n. So no solution!  :)

I suppose it's simple to proof, that the number of prime numbers $<n$ is less than the number of $k<n$ s.t. $(k;n)=1$, or by induction or using the relation $\sigma_0(n)=n-\varphi(n)$ or the Gauss formula $\pi\sim \frac{n}{ln n}$..what do you say?


Prove that for any $\alpha \in [0,\infty)$ there exists a sequence of positive integers $(k_{n})_{n\geq 1}$ such that: $\lim_{n\to\infty}\sqrt[n]{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k_{n}}}= \alpha$

Isn't this obvious? I mean, once you know that the $n$-th term of the harmonic series behaves as $log n+O(1)$, the construction is clear.

$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln (n)\to c$ was the whole thing here....

Uh, is there something else that's meant to be here? As written, it's obviously false; the limit is always equal to 1, since the $n$th term is between 1 and $n^{1/n}$.

jmerry, I think you had the same initial misreading that I did.  We're taking $\lim_{n \to \infty}\left(H_{k_{n}}\right)^{\frac{1}{n}}$.

OK, I see that. It's still false as stated; you can't get anything less than 1.
Also, you only need that the sum of the harmonic series diverges, while $\frac1n$ stay bounded.

The problem was stated for $\alpha \in [0,\infty)\cup \{ \infty\}$. I think you are right though in saying that $\alpha$ should belong to $[1,\infty]$.


On the plane s a triangle with red vertices and a triangle with blue vertices.O is a point inside both triangles such that the distance of O to any red vertex is less than distance of O to any blue vertex. can all three blue vertices and all three red vertices lie on same circle?

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assume they lie on a circle $ P$. $ A,B,C$ are the blue vertices, and $ D,E,F$ the red. WLOG, let $ D$ be on $ \stackrel{\frown}{BC}$ and $ E$  on $ \stackrel{\frown}{AC}$. Let $ k_1$,$ l_1$,$ k_2,l_2$ be the perpendicular bisectors of $ BD,DC,AE,CE$. Clearly, $ O$ lies inside the triangle formed by $ k_1,l_1,BC$, but it must also lie inside the triangle formed by $ k_2,l_2,AC$, which are nonoverlapping triangles. Contradiction.[/hide]