7.) There are 9 rectangles in a 2x2 lattice square. 4 of them are 1x1 squares, four of them are 2x1 rectangles, and 1 is a 2x2 square.  The sum of their areas is 16.

a) show that there are 36 rectangles in a 3x3 square.
b) show that the sum of these 36 rects is 100.
^^^ really easy
c) Prove that the number of rectangles in an NxN square is ((n+1)C2)^2

d) Prove that the sum of the areas of the ((N+1)C2)^2 rectangles in an NxN square is ((N+2) C 3 )^2

during the test i tried to do part c through induction, but i think now that its probably best not to use induction, it seems just like a summation inside another summation now. any ideas?

we can imagine the NxN lattice square as an NxN grid made up of N+1 horizontal lines and N+1 vertical lines

for each rectangle, two of the horizontal lines will be the top and bottom sides of the rectangle and two of the vertical lines will be the left and right sides of the rectangle

there are (N+1)C2 ways of choosing 2 horizontal lines and (N+1)C2 ways of choosing 2 vertical lines

multiplying the number of ways to choose horizontal lines and the number of ways to choose vertical lines we get ((N+1)C2)^2

we can also extend this method to find the number of rectangles in a NxM grid


Given $ n$ positive real numbers $ a_1,a_2,...,a_n$ such that

$ \sum_{i\equal{}1}^ka_i\leq \sum_{i\equal{}1}^ki(i\plus{}1)$ for $ k\equal{}1,2,3,...,n$.

Prove that $ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_{n}}\geq \frac{n}{n\plus{}1}$

I think when $ a_i\to 0^ + \Rightarrow\frac {1}{a_i)\to + \infty}$
$ a_i\to 0^+\Rightarrow\frac{1}{a_i}\to +\infty$


function
integration
real analysis
real analysis solved

Let $f:[a,b]\to\mathbb R$ be a continous function with the property that there exists a constant $\lambda\in\mathbb R$ so that for every $x\in[a,b]$ there exists a $y\in[a,b]-\{x\}$ s.t. $\int_x^yf(x)dx=\lambda$. Prove that the function $f$ has at least two zeros in $(a,b)$.
[i]Eugen Paltanea[/i]

Assume first that $f$ has no $0$'s in $(a,b)$. We may assume WLOG that $f>0$ on $(a,b)$. If $\lambda<0$, we can see that it's $\ge 0$ by pushing $x$ as close to $a$ as we want. On the other hand, if it's $>0$, we can see that it's $\le 0$, by pushing $x$ as close to $b$ as we want. We have thus got a contradiction, so $f$ has at least a root $x_0$ in $(a,b)$.

Assume now that $f>0$ on $(x_0,b)$ and $f<0$ on $(a,x_0)$. Then $\lambda>0$, since we can take $x=x_0$. Again, we push $x$ arbitrarily close to $a$ and $b$ to find $\int^b_{x_0}f(t)dt\ge\int^a_{x_0}f(t)dt$ and $\int^b_{x_0}f(t)dt\le\int^a_{x_0}f(t)dt$ respectively. This means that $\int^b_{x_0}f(t)dt=\int^a_{x_0}f(t)dt$. Making once more $x\to a$, $\sup_{y\in[a,b]}\lambda^y_x f(t)dt$ can be made smaller than any positive real, and, in particular, smaller than $\lambda$, which is a contradiction.

Sorry I didn't post the details of those claims, but they look pretty intuitive :).

I think it is sufficient.
this was my solution also. but I made some typos at the contest
the official solution: it can be demonstrated(with sup and inf)  that $\lambda=0$ after this the conclusion comes immediately.


geometry
circumcircle
trigonometry
similar triangles
Pythagorean Theorem

Here's an easy one I made: Let there be a right triangle with side lengths a, b, and c (with c being the hypotenuse). When you draw a polygon $ P_A$ with side length "a" , a polygon $ P_B$ with side length "b" , and a polygon $ P_C$ with side length "c", with $ P_A, P_B,$ and $ P_C$ being regular n-gons, prove that $ [P_A]\plus{}[P_B]\equal{}[P_C]$.

[hide="Solution"]We can find the area of a regular $ n$-gon of side length $ s$ like so:

$ \text{Circumradius}\equal{}\frac{s}{2\sin\frac{\pi}{2n}}$

$ K\equal{}\frac{\text{Circumradius}^2\cdot\sin\frac{\pi}{n}}{2}\equal{}\frac{s^2\sin\frac{\pi}{n}}{8\sin^2\frac{\pi}{2n}}$

Therefore, $ [P_A]\plus{}[P_B]\equal{}\left(\frac{a^2\sin\frac{\pi}{n}\plus{}b^2\sin\frac{\pi}{n}}{8\sin^2\frac{\pi}{2n}}\right)\stackrel{\text{Pythagoras}}{\equal{}}\left(\frac{c^2\sin\frac{\pi}{n}}{8\sin^2\frac{\pi}{2n}}\right)\equal{}[P_C]$[/hide]

:)

[b]Generalization:[/b] Given a right-angled triangle $\triangle ABC$ at $A,$ if we construct three similar polygons $P_a , P_b, P_c$ on the sides $BC, CA, AB,$ having $BC,CA,AB$ as homologous sides, then one of these polygons is equivalent to the sum of the other three.

The principle of equivalency states that if $a$ is a side of the polygon $P,$ then there exists a triangle $p$ with side $a$ equivalent  to the polygon $P.$ Using this principle, we have three similar triangles $P_a, P_b , P_c$ (with the same area of the referred polygons) constructed on the sides of $ \triangle ABC.$ By Pythagorean theorem $a^2 \equal{} b^2 \plus{} c^2,$  but since the triangles are similar we have:

$\frac {P_a}{P_b} \equal{} \frac {a^2}{b^2} \ , \ \frac {P_a}{P_c} \equal{} \frac {a^2}{b^2} \Longrightarrow \ P_b \plus{} P_c \equal{} P_a.$


integration
geometry
function
calculus
vector
parameterization
calculus computations

I was recently looking at a problem regarding how wet you will get if you run in the rain vs. walk in the rain.  The rate at which you are becoming wet is

\[\frac{dW}{dt}=-\int_{S}\sigma\nu\,dA,\]
where $\sigma$ is the atmospheric density of the water, $\nu$ is the velocity of the rain relative to the velocity of the person, and $S$ is the surface of the human body (obviously, impossible to [i]actually[/i] compute).  How do we solve this to get an expression explicitly in terms of the wetness, $W$?

how can you measure the atmosphere?

i think you said Air density or Atmospheric Pressure.

but for pratical purpose we can considere the air density constant the the velocity constant and use a formula for the Area of a surface of a person in function of his heigth

That would be one way but I am more interested in solving the general differential equation rather than making physical assumptions about the rain and velocity.  I haven't encountered a differential equation with such a surface integral and would like to know how to solve it, if a fairly simple solution exists (I don't know if more information is needed about $\nu$, $\sigma$, and $\Sigma$ first).

first of all

this equation is completely wrong for high velocitys

if v tend to infinity, i will absorve all water in the rain!?

the relation is not too linear like that.

a beter approach would be

$\frac{dW}{dt}=-\int_{S}\sigma f(\nu) \,dA$

did you understand? the denpendece on the variable of $v$ is more complex.

we can imagine a better equation if we think in vectors.

the vector field of the raion and the position of a person.

so we can measure the Flux in the person.

here we can think, the flux of rain just "enter" in the person but not leave, because the person have a physical body.

so the wetness is proportional to the Flux of Rain through the person

we can make a surface integral.

but this is a little bit impratical. because become more difficult.

did you understand?

dont make sense use your formule because it is not better approximation than make everything constant.

Let $R$ be the vector field of the rain and $S_{t}$ the function "area of person" in function of time (we are assumin that S change in the time, because the person is runin)

So we can define a parameter $W$ such as

${W=c_{1}\iiint \left \| \mathbf{R}\cdot \left(\frac{\partial \mathbf{S}_{t}}{\partial u}\times \frac{\partial \mathbf{S}_{t}}{\partial v}\right)\right\|dudvdt}$

is equal to How much the person get wet

We must not forget the saturation limit of body hair clothes etc. Then we can must have some $W_{\text{on person}}$ versus $W_{in}$ function.

If person is moving do we need to add his relative speed to $\mathbf{R}$?

not at all

$\mathbf{R}$ is dependent on $x,y,z,t$. I was assuming it varies in 4 dimensions.

the same case for $S_{t}$. since the closed curve of the person is changing in time, we don't need to know exactly the velocity of the person, but we could, if we want.

on my equation we are conting flux approximatelly twice, can you see that?

so, with the constant we can correct this and anothers things.

we can consider the constant a function of $x,y,z,t$ too and put it inside the integral. we are assumin the body (clotes) of the person have different permability and that it varies in time.

but, since to determine this function $c_{1}$ is too hard,i put it as a constant outside the integral :)


geometry
geometric transformation
homothety
similar triangles
geometry proposed

Let $ABC$ and $A_{1}B_{1}C_{1}$ two similar triangles such that the first one is inthe interior of the second one. Prove that there is a point that corresponds to itself.

If their sides are Parallel then , the center of Homothecy works.

If not , The center of their Spiral Homothecy works .

EDIT: Center of Spiral homothecy is not always INSIDE the bigger triangle ,So if you want a point Inside them , It doesnt work.


How many five-element subsets $S$ of set $A = \{0, 1, 2, ..., 9\}$ are there which satisfy
$\{r(x + y) | x,y \in S, x \neq y\} = A$,
where $r(n)$ denotes the remainder when $n$ is divided by $10$?

[hide]
Let $B=\{r(x + y) | x,y \in S, x \neq y\}$.
Suppose $S$ contains $0$. Then for $x\neq0$, $r(x+y)=r(x+0)=x$, so $x\in B$. For $x=0$, $r(0+y)|0$ no matter what $y$ we pick since anything divides $0$. So any subset containing $0$ works.
Now suppose $S$ does not contain $0$. Consider the case where $x=1$. Then $r(1+y)|1$, so $r(1+y)=1$ and $r(y)=0$. But, given that $y\le9$, this means $y=0$, a contradiction.
So, the number of 5-element subsets $S$ which work is the number of 5-element subsets containing $0$, of which there are ${9\choose4}$.
[/hide]


There are 2008 balls in the bag, numbered 1,2,3,...,2008. Three balls are randomly picked up from the bag and let $ T$ denote the largest number of these three balls. What's the expected value of $ T$?

[hide="not sure"]Let $ a \equal{} 3 \plus{} 4\times C^3_2 \plus{} 5\times C^4_2 \plus{} 6\times C^5_2 \plus{} ... \plus{} 2008\times C^{2007}_2$
$ 2a \equal{} 3\times2\times1 \plus{} 4\times3\times2 \plus{} 5\times4\times3 \plus{} 6\times5\times4 \plus{} ... \plus{} 2008\times2007\times2006$

$ 8a \equal{} ( \minus{} 0\times1\times2\times3 \plus{} 1\times2\times3\times4) \plus{} ( \minus{} 1\times2\times3\times4 \plus{} 2\times3\times4\times5) \plus{} ( \minus{} 2\times3\times4\times5 \plus{} 3\times4\times5\times6) \plus{} ... \plus{} ( \minus{} 2005\times2006\times2007\times2008 \plus{} 2006\times2007\times2008\times2009)$

$ a \equal{} \frac {2006\times2007\times2008\times2009}{8}$

So the expected value is $ \frac {2006\times2007\times2008\times2009}{8\times C^{2008}_3}\equal{}1506.75$[/hide]

[quote="murgi"]no. in fraction?[/quote]

If you mean expressing the answer as a fraction, I think you can do it yourself (multiply by 4) :wink:

Chrischris' answer is right. :)   But I've seen someone  doing this way:

$ \frac{2005}{4}\plus{}1\plus{}\frac{2005}{4}\plus{}1\plus{}\frac{2005}{4}\plus{}1\equal{}\frac{6027}{4}$

which is also correct. Can somebody explain this in detail?

[hide="Solution"]

$ ET \equal{} \frac{1}{\binom{2008}{3}}\left [ 2008 \binom{2007}{2} \plus{} 2007 \binom{2006}{2} \plus{} \cdots \plus{} 3 \binom{2}{2}\right]$

I will use the identity: $ \binom{n}{m} \equal{} \frac{n}{m}\binom{n\minus{}1}{m\minus{}1}$ on each of the terms,

$ ET \equal{} \frac{3}{\binom{2008}{3}}\left [ \frac{2008}{3} \binom{2007}{2} \plus{} \frac{2007}{3} \binom{2006}{2} \plus{} \cdots \plus{} \frac{3}{3} \binom{2}{2}\right]$

$ \equal{} \frac{3}{\binom{2008}{3}}\left [ \binom{2008}{3} \plus{} \binom{2007}{3} \plus{} \cdots \plus{} \binom{3}{3}\right]$

Now the sum is simply Hockeystick Theorem, giving, $ ET \equal{} \frac{3 \binom{2009}{4}}{\binom{2008}{3}} \equal{} \frac{6027}{4}$

 :)
[/hide]


[quote="SuperMath"]Let $\large x;y;z > 0$ and $\large x+y+z=1$. Prove that:
$\large \sum \frac{x}{ \sqrt{4y+(x-z)^2} \sqrt{4z+(x-y)^2}} \geq\ \frac{3}{4}$[/quote]

Let's prove such inequality:
$S=\sum_{cyclic}x\sqrt{4y+(x-y)^2}\sqrt{4z+(x-z)^2} \leq \frac{4}{3}$ :
$S \leq \sum_{cyclic}x\frac{4y+4z+(x-z)^2+(x-y)^2}{2} = 4(xy+xz+yz) +\sum_{cyclic}\frac{(x+y)(x-y)^2}{2}=$
$4(xy+xz+yz)+\sum_{cyclic}\frac{(1-z)(x-y)^2}{2}=$
$(x^2+y^2+z^2+3(xy+xz+yz))-\sum_{cyclic}\frac{z(x-y)^2}{2}=$
$1+(xy+xz+yz)-\sum_{cyclic}\frac{z(x-y)^2}{2} \leq 1+(xy+xz+yz) \leq$
$1+\frac{x^2+y^2+z^2+2(xy+xz+yz)}{3} =1+\frac{1}{3}=\frac{4}{3}$
So it's done

Then under the Caushy - Shvarts ineq we have
$\sum_{cyclic}\frac{x}{ \sqrt{4y+(x-z)^2} \sqrt{4z+(x-y)^2}}*\sum_{cyclic}x\sqrt{4y+(x-y)^2}\sqrt{4z+(x-z)^2} \geq$
$(x+y+z)^2=1 \Rightarrow$
$\sum_{cyclic}\frac{x}{ \sqrt{4y+(x-z)^2} \sqrt{4z+(x-y)^2}} \geq$
$\frac{1}{\sum_{cyclic}x\sqrt{4y+(x-y)^2}\sqrt{4z+(x-z)^2}} \geq \frac{3}{4}$
V:)


group theory
abstract algebra
vector
algebra
polynomial
linear algebra
matrix

Determine all finite groups $G$ that have an automorphim $f$ such that $H$ is not a subset of $f(H)$ for all proper subgroups $H$ of   $G$.

I think none, since $G \subset f(G) = G$ for all such groups  ;)
Do you mean proper subgroups¿

Given [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40132]this problem[/url], I think I have a solution for this one, but I'll use a rather strong theorem. A proof can be found in Rotman's "An Introduction to the Theory of Groups". The proof is neither nasty nor hard to understand, but I don't like to copy proofs, so I won't :).

The theorem states that a finite group without characteristic subgroups is a direct product of several copies of a certain simple group. Using this, plus the problem posted by sam-n to guide us :), we immediately find that our groups have to be direct products of several copies of $\mathbb Z_p$, or, in other words, finite-dimensional vector spacs over $\mathbb Z_p$.

On the other hand, it's not hard to show that all such groups satisfy the requirements: all we need is to show that an $n$-dimensional vector space over $\mathbb Z_p$ has an invertible linear transformation without fixed subspaces. This can be achieved by picking a monic  irreducible polynomial of degree $n$ in $\mathbb Z_p[X]$ and considering the linear transformation whose matrix (in the standard basis, for example) is the companion matrix of this polynomial.


Suppose that necklace $\, A \,$ has 14 beads and necklace $\, B \,$ has 19. Prove that for any odd integer $n \geq 1$, there is a way to number each of the 33 beads with an integer from the sequence
\[\{ n, n+1, n+2, \dots, n+32 \}\]
so that each integer is used once, and adjacent beads correspond to relatively prime integers. (Here a ``necklace'' is viewed as a circle in which each bead is adjacent to two other beads.)

[hide]
From the set of integers, choose one of the integers $ k$, with $ k\ge n\plus{}3$ and $ k$ divisible by $ 15$. Now of $ k\minus{}1$ and $ k\minus{}2$ (both of which are greater than or equal to $ n\plus{}1$), at least one must not one less than a multiple of $ 13$. Let this one be called $ m$. Then we have

$ m$ is relatively prime to $ 15$
$ m\plus{}1$ is relatively prime to $ 13$
$ m\ge n\plus{}1$

I claim that the chains

$ m\plus{}1,...,m\plus{}14$ and
$ n,n\plus{}1,...m,~~~~~m\plus{}15,m\plus{}16,...,n\plus{}32$ work.

To see this, we first note that consecutive integers are always relatively prime. Also,

$ \gcd(m\plus{}1,m\plus{}14)\equal{}\gcd(m\plus{}1,m\plus{}14\minus{}(m\plus{}1))\equal{}\gcd(m\plus{}1,13)\equal{}1$
$ \gcd(m,m\plus{}15)\equal{}\gcd(m,15)\equal{}1$
$ \gcd(n,n\plus{}32)\equal{}\gcd(n,32)\equal{}1$ (since $ n$ is given to be odd)

Therefore, the adjacent beads will always be relatively prime.
[/hide]


Evalute for a positive integer n \geq 2: \product^{n-1}_{k=1} (1 - cos(2*k*pi/n)).

Using that 1-cos(2x) = 2sin 2 (x), the given expression is equal to 2 n-1 A 2 , where A = prod{1 \leq k \leq n-1} sin(k \pi /n).
But it is known that A = n/2 n-1
(see : http://www.mathlinks.ro/viewtopic.php?t=4105&highlight=sin )
Thus, the given expression is n 2 /2 n-1 .

Pierre.

pbornsztein..it's  the second time in this night when i wanted to post a solution and when i wanted to write it i saw that u've just written it :(
:D:D:D

:D  That's why I'm gone to bed...

Pierre.


How many ordered pairs $(a, b)$ of positive integers satisfy the inequality,
\[ 3a + 4b < 24? \]

hint

a lot, but have limits as well...

 try to graph this inequality .
( you don't have to write in y-intercept form.. )

good luck :D

[hide="hint"]What's the highest a and b can be? [/hide]


There are r urns each with different color balls. There are $ x_1, x_2, \dots ,x_r$ balls in each urn.
Let $ \min(x_1,x_2...x_r) < n < \sum_{i\equal{}1}^r x_i$
In how many ways can we choose n balls from the groups?

(This is similar to a problem where $ n < \min(x_1,x_2...x_r)$. In this case the solution is $ \dbinom{n\plus{}r\minus{}1}{r\minus{}1}$)

Thanks.

Does a given urn consist of balls of the same color?  I think problems of this type can be handled by a generalization of Burnside's lemma called the [url=http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem#Informal_PET_statement]Polya enumeration theorem[/url], although that might be too strong.

Yes balls in the same urn are the same color (and balls in different urns are different colors). I'll take a look at the link. Thanks.


If $ p$ is prime, then is $ 2^p \minus{} 1$  prime?

I have have a firm belief that it is, but I never prove it. T.T

plz help me.

It is an open question .
A result :
Every prime divisor of this number has form $ 2kp\plus{}1$

// TTsphn

Thanks for your interest.

But, I don't understand your comment?

What does mean "Every prime divisor of this number has form $ 2kp\plus{}1$"?

[quote="rain_maker"]If $ p$ is prime, then is $ 2^p \minus{} 1$  prime?[/quote]
Try $ p \equal{} 11$.

Note: $ 2^{11}\minus{}1\equal{}23\cdot 89$ is not prime.

Theorem: if $ 2^n\minus{}1$ is prime, then $ n$ is prime. As the example above shows, the converse is false.

A prime of the form $ 2^p\minus{}1$ is called a [b]Mersenne prime[/b]. The largest currently known prime is the largest currently known Mersenne prime.

Some key words for further searches:

Mersenne prime.
GIMPS (Great Internet Mersenne Prime Search).
Perfect number.
Marin Mersenne (the person).


algebra
difference of squares
special factorizations

Simplify:

$ \frac{\sqrt{a\minus{}\sqrt{a}} \cdot \sqrt{a\plus{}\sqrt{a}}}{\sqrt{a\minus{}1}}$

[hide="Answer"]$ \sqrt{a}$[/hide]

Recall that $ \sqrt a \cdot \sqrt b \equal{} \sqrt {ab}$. So we have in the numerator: \[ \sqrt {a \minus{} \sqrt a} \cdot \sqrt {a \plus{} \sqrt a} \equal{} \sqrt {(a \minus{} \sqrt a)(a \plus{} \sqrt a)}\] This should scream difference of squares, therefore we have $ \sqrt {a^2 \minus{} a}\equal{}\sqrt {a(a\minus{}1)}$ in the numerator. We can split this up into $ \sqrt a \cdot \sqrt {a\minus{}1}$. On the other hand, we have $ \sqrt{a\minus{}1}$ in the denominator, so: \[ \frac {\sqrt a \cdot \sqrt{a\minus{}1}}{\sqrt {a\minus{}1}}\equal{}\sqrt a \cdot \frac {\sqrt {a\minus{}1}}{\sqrt {a\minus{}1}} \equal{} \sqrt a \cdot 1 \equal{} \sqrt a\]


Eight consecutive positive integers  are divided into 2 sets, such that the sum of the squares of the elements in the first set is equal to the sum of the squares of the elements in the second set. Prove that the sum of the lements in the first set is equal to the sum of the elements in the second one.

let the numbers be $8x, 8x+1, \ldots, 8x+7$, then the sum of the squares of all the numbers is $512x^{2}+448x+140$ so that if the sum of the squares of the elements of the two sets are equal, they must equal $256x^{2}+224x+70$.

Look at all the squares mod 8, they form the sequence: $0,1,4,1,0,1,4,1$.  From the above calculation, we know that the sum of the squares in each of the sets is congruent to $6\mod 8$, so that we cannot have $8x+2$ and $8x+6$ in the same set.

Now look at all the squares mod 16, they form the sequence:
$0,16x+1,4,16x+9,16,16x+25,4,16x+17$.  From the above comment we know that each of the two sets must have an element whose square is $4\mod 32$, so that the sum of the squares of the other elements must be $2\mod 32$.

Thus we have to be able to split $16x+1,16x+9,16,16x+25,16x+17$ into two sets whose sums are $2\mod 32$.  Note that we must have an even number of odd terms in each set, so that we necessarily will have two odd terms in each set.  An easy check shows that the only way we can do this is to split them up as $\{16x+9,16x+25\}$ and $\{16x+1,16,16x+17\}$

Therefore one of the two sets in the original problem contains the elements $8x+1,8x+4,8x+7$ and one of $8x+2,8x+6$.  Given that the sum of the squares needs to be $256x^{2}+224x+70$, we must have $8x+2$, and one can see that the separation:

$\{8x+1,8x+2,8x+4,8x+7\}$ and $\{8x,8x+3,8x+5,8x+6\}$

is the only such separation in which the sums of squares are equal.  The sums of the elements of each set are equal as well, so the problem is solved.


Hope someone could help me with this, thanks!

Show that if n is a positive integer, then

Summation of  1/(a_1*a_2*a_3......a_k) = n where {a_1,....a_k} is a subset of {1,2,....n}

[hide="One line"] $\left( 1+\frac{1}{1}\right) \left( 1+\frac{1}{2}\right) \left( 1+\frac{1}{3}\right) ... \left( 1+\frac{1}{n}\right)-1 = n$ [/hide]


This is good practice for determining unit digits of exponental numbers.

4everwise, you posted the answer twice two different ways

I found a similar problem to this one in Volume 1 of the Art of Problem Solving Books.

 ;)   :(


Hi is there any formula to find this sum $ 1*2\plus{}2*3\plus{}3*4\plus{}4*5\plus{}...(n\minus{}1)n$

Hi;

There are a couple of ways to do this sum.

1) fit for the coefficients and prove by induction.

2) The summation calculus, the discrete analog to integral calculus.

3) The Euler Mclaurin summation formula which establishes a relationship between integrals, derivatives and sums. This is what I used to get your answer.

$ \sum _{k \equal{} 2}^n k (k \minus{} 1) \equal{} \frac {1}{3} n \left(n^2 \minus{} 1\right)$

One way to calculate formula:
$ 1*2\plus{}2*3\plus{}3*4\plus{}4*5\plus{}...(n\minus{}1)n\equal{}( 1^2\plus{}2^2\plus{}...\plus{}n^2)\minus{}(1\plus{}2\plus{}...\plus{}n)\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}\minus{}\frac{n(n\plus{}1)}{2}\equal{}2n(n\plus{}1)(n\plus{}2)$

Hi hipernova;

Good use of telescoping but I think you made an algebraic error.

$ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \minus{} \frac {n(n \plus{} 1)}{2} \neq 2n(n \plus{} 1)(n \plus{} 2)$

Yes, bobbym is correct.

In the same way, we have

$ 1\cdot 2\cdot 3\plus{}2\cdot 3 \cdot 4 \equal{}3\cdot 4 \cdot 5 \plus{}...\plus{} n(n\plus{}1)(n\plus{}2)\equal{} \frac{n(n\plus{}1)(n\plus{}2)(n\plus{}3)}{4}$

and so on, for $ 1\cdot 2 \cdot 3 \cdot 4 \plus{}... \plus{} n(n\plus{}1)(n\plus{}2)(n\plus{}3) \equal{} \frac{n(n\plus{}1)(n\plus{}2)(n\plus{}3)(n\plus{}4)}{5}$, et cetera.


Let $ a,b,c$ be real numbers so that $ abc\equal{}1.$ Prove that at most two of the numbers $ 2a\minus{}\frac{1}{b},2b\minus{}\frac{1}{c},2c\minus{}\frac{1}{a}$  are greater than $ 1$.

$ abc \equal{} 1$  $ \implies$  $ c \equal{} \frac {1}{ab}$
$ 2a \minus{} \frac {1}{b} > 1$ ,$ 2b \minus{} \frac {1}{c} > 1$,$ 2c \minus{} \frac {1}{a} > 1$
$ i)a,b,c > 0$  $ \implies$  $ 2ab \minus{} 1 > b,2b \minus{} ab > 1,2 \minus{} b > ab$ $ \implies$  $ b \equal{} x,ab \equal{} y$  $ \implies$
$ 2y > x \plus{} 1,2x > y \plus{} 1,x \plus{} y < 2$  $ \implies$   $ 4y \minus{} 2 > 2x > y \plus{} 1$ $ \implies$  $ y > 1$
$ \implies$  $ x < 1$  but $ 2x > 2$  contradiction!
$ ii)a < 0,b < 0,c > 0$  $ \implies$ $ 2b \minus{} \frac {1}{c} < 0$ contradiction!   :wink:

[quote="moldovan"]Let $ a,b,c$ be real numbers so that $ abc \equal{} 1.$ Prove that at most two of the numbers $ 2a \minus{} \frac {1}{b},2b \minus{} \frac {1}{c},2c \minus{} \frac {1}{a}$  are greater than $ 1$.[/quote]$ i)a < 0,b < 0,c > 0\implies 2b \minus{} \frac {1}{c} < 0<1.$

$ ii)a,b,c > 0: \sum_{cyc}bc\left(2a \minus{} \frac {1}{b}\minus{}1\right)\equal{}6abc\minus{}\sum_{cyc}{c}\minus{}\sum_{cyc}{bc}$

$ \leq 6abc\minus{}3\sqrt[3]{abc}\minus{}3\sqrt[3]{a^2b^2c^2}\equal{}0$

$ \implies\min\left\{2a \minus{} \frac {1}{b}\minus{}1,2b \minus{} \frac {1}{c}\minus{}1,2c \minus{} \frac {1}{a}\minus{}1\right\}\leq 0$

$ \implies\min\left\{2a \minus{} \frac {1}{b},2b \minus{} \frac {1}{c},2c \minus{} \frac {1}{a}\right\}\leq 1.$

Assume the contrary. Since $ abc\equal{}1$, we make the substitution $ a \equal{} x/y, b \equal{} y/z, c \equal{} z/x$. Thus,

$ 2a \minus{} 1/b > 1 \implies 2x/y \minus{} z/y > 1 \implies 2x \minus{} z > y,$
and similarly,
$ 2b \minus{} 1/c > 1 \implies 2y \minus{} x > z$ and
$ 2c \minus{} 1/a > 1 \implies 2z \minus{} y > x$.

Adding these three inequalities gives a contradiction.


vector
analytic geometry
parameterization
function
calculus
topology
real analysis

We are given set $M=[0,\infty)\times \mathbb{R}^{n-1}$. Let $M$ be oriented positively in $\mathbb{R}^{n}$. Check if parametrisation $F(x_1,...,x_{n-1})=(0,x_1,...,x_{n-1})$ is positive or negative.

Thanks a lot for help

This doesn't belong in this forum. I would not have been able to interpret and answer this before a graduate course I took this spring.

In order to have this make sense, we need to interpret some unclear parts of the question:
First, $\mathbb{R}^{n-1}$ and $\mathbb{R}^n$ are given the standard orientations. The boundary $\partial M$ inherits an orientation from the orientation on $M$; in my reference, that orientation is given by the form $i_N\Omega$, where $N$ is an outward-pointing normal vector and $\Omega$ is an orientation form for $M$.
The standard orientation form for $M$ (coordinates $(x^0,x^1,\dots,x^{n-1})$) is $dx^0\wedge dx^1\wedge\cdots\wedge dx^{n-1}$, and an outward-pointing normal vector $N$ is $-\frac{\partial}{\partial x^0}$. The standard orientation form on $\mathbb{R}^{n-1}$ is $dx^1\wedge dx^2\wedge\cdots\wedge dx^{n-1}$.
$i_N(dx^0\wedge dx^1\wedge\cdots\wedge dx^{n-1})$
$=dx^0(-\frac{\partial}{\partial x^0})dx^1\wedge dx^2\wedge\cdots\wedge dx^{n-1}+ dx^1(-\frac{\partial}{\partial x^0})dx^0\wedge dx^2\wedge\cdots\wedge dx^{n-1}+\cdots$
$=(dx^0(-\frac{\partial}{\partial x^0}))dx^1\wedge dx^2\wedge\cdots\wedge dx^{n-1}=-dx^1\wedge dx^2\wedge\cdots\wedge dx^{n-1}$.
The parametrization reverses orientation.

Thanks very much for solution.

We had this exercise given on the lecture from real analysis so I posted it in that forum firstly.
I have a question - what did you denote by $i_N$?

Thanks a lot once again

A $k$-form $\omega$ is a multilinear function from $k$-tuples of vectors to $\mathbb{R}$. $i_V\omega$ is the $k-1$-form we obtain by fixing $V$ in as the first vector: $i_V\omega(v_2,v_3,\dot,v_k)=\omega(V,v_2,v_3,\dots,v_k)$ by definition.

This doesn't look like real analysis to me- I learned this stuff in a course on manifolds. It's relevant for Stokes' theorem and related topics- this could conceivably be done in a multivariable-focused analysis course. That's not a first analysis course, and it's a long way from the first-year calculus that goes in the Calc. Computations/Tutorials forum.

Ok,

As you noticed it is not real analysis. You're right - it's connected with Stokes theorem and manifolds with boundary. We study such things on real analysis IV (the fourth term of this subject) so after two years of education in analysis. Actually, we did the all the required things connected with manifolds in 2 months so it's almost impossible to understand it deeply regarding the fact that we have other subjects. I agree that it's quite difficult but fortunately the exam I'm  having tommorow is the last exam from real analysis.

Thanks for help :)


algebra
polynomial
inequalities proposed
inequalities

Is it true for positive integer $n$ and $a, b, c \ge 0$ that

$(a^{n}+b^{n}+c^{n})^{2}\ge (a+b+c)(a^{2n-2}b+b^{2n-2}c+c^{2n-2}a)$

 :?:

It is only obvious to me for $n=1, 2$, but I think it should always hold...   :|  :huh:

Oh!  :o  :huh:  :blush: Can someone prove it for $n=3$ then? I think it is not easy (if it is true :P).

[quote="Xevarion"] Can someone prove it for $n=3$ then? I think it is not easy (if it is true :P).[/quote]
I think it is true, but my proof is very ugly.


Does anyone know how to create subscripts on labels in KSEG when drawing a picture?

ehhh the help file is awfully helpful...

it says 'you can create super/subscripts if you can figure out how....'

have you tried, opening the Edit Label box, and then copy pasting in some text with subscript from lets say, Word?

[quote="pkerichang"]that's why I use GCLC...everything is easy this way (although it generates pictures from code...)[/quote]
GCLC ftw  :lol:

http://www.matf.bg.ac.yu/~janicic/gclc/

Looks interesting; it's not very popular, because gentoo doesn't have an ebuild for it.

Maybe wine would work well for that, since the linux version is command line only? :?:


calculus
integration
function
limit
trigonometry
parameterization
real analysis

Let $f: [0,\infty)\to (0,\infty)$ a continuous function sunch that $\int_0^\infty f(x)dx$ is convergent. Prove that there exists $\lim_{x\to \infty}\frac 1 {\sqrt{x}} \int_0^x \sqrt{f(t)}dt=0$.

I guess we're dealing with the properties of continuously decreasing functions, no?

It is not provided that $f$ is decreasing. If you meant that..

Your limit doesn't even make sense- you can't use $x$ both outside the integral and as a dummy variable. Did you mean $\lim_{x\to\infty}\frac1{\sqrt{x}}\int_0^x\sqrt{f(t)}\,dt$?

to Adidasty: $f(x)$ is not necessarily continuously decreasing, as the example $\int_0^{\infty}\cos x^2 dx$ shows.

Ok,I have thought this problem for a long time.
And now here is my proof.hope it is right:
Solution:because $\int_{0}^{\infty}f(x)dx$ is convergent.
So:for $\forall \epsilon>0$, there exist $A>0$ satisfy:
$\int_{A}^{\infty}f(t)dt<\epsilon$,
Thus for $x>A$:\[ \frac 1 {\sqrt{x}} \int_0^x \sqrt{f(t)}dt = \frac{\int_0^A \sqrt{f(t)}dt+\int_A^x \sqrt{f(t)}dt}{\sqrt x}  \]
\[ \le \frac{\int_0^A \sqrt{f(t)}dt+\sqrt{\int_A^xdt \int_A^x f(t)dt}}{\sqrt x}\le \frac{\int_0^A \sqrt{f(t)}dt}{\sqrt x}+\sqrt{\epsilon}  \]
But let $x \to \infty$ you will find the first $\to 0$
So we proved the problem. :)

The proof looks good. Two things about notation:
[b]Never[/b] use the same letter for a dummy variable as you do for a parameter outside the integral. All of those integrals (not just the first)should be $dt$.
The long expression splits onto two lines, and not in a pretty way. You should probably break it up yourself, at one of the $\le$ signs.


I am, in truth, a yellow fork
From tables in the sky
By inadventant fingers dropped
The awful cutlery.
Of mansions never quite disclosed
And never quite concealed,
The apparatus of the dark
To ignorance revealed.

Who or what corresponds to "I" in the problem?

[quote="mustafa"][hide]Seems like lightning to me[/hide][/quote]

sounds right to me  :lol:


Compute the following limit! :D

lim (n-->oo)  n^2(1-ch(1/n)*ch^2(2/n)*..*ch^k(k/n)), where k is natural and fixed and we have that ch(a)=(e^a+e^(-a))/2 and a is real.
(proposed for the local MO in Maramures, Romania in 2003).

cheers! :D :D

[quote="Lagrangia"]
lim (n-->oo)  n^2(1-ch(1/n)*ch^2(2/n)*..*ch^k(k/n)), where k is natural and fixed and we have that ch(a)=(e^a+e^(-a))/2 and a is real.
(proposed for the local MO in Maramures, Romania in 2003).
[/quote]
ch(u)=1+u^2/2+o(u^2) when u tends to 0
(1+t)^k = 1+kt+o(t) when t tends to 0
ch^k(k/n)= (1+1/2.k^2/n^2 +o(1/n^2))^k=1+1/2.k^3/n^2+o(1/n^2)
when n tends to oo
ch(1/n)*ch^2(2/n)*..*ch^k(k/n)=(1+1/n+o(1/n))(1+1/2.2^3/n^2+o(1/n^2))....(1+1/2.k^3/n^2+o(1/n^2))
=1+1/(2n^2)(1+2^3+...+k^3)+o(1/n^2) so the limit of the sequence
is -1/2(1+2^3+...+k^3)
Lagrangia for wich grade was this problem ?

Moubinool

It was proposed for the 11th grade! :D

cheers!

I am begining to like solutions with Landau, they are really cute! :D :D

Lets consider the following fact:

lim(n-->oo) (1-ch^k(k/n))/(1/n^2)=-k^3/2, k is natural.

now  considering our limit to be L_k=lim(n-->oo) ( (1-ch^k(k/n))/(1/n^2) +n^2(1-ch(1/n)*ch^2(2/n)*..*ch^k(k/n)) from which we get that L_k=-k^3/2+L_(k-1) now we give values and we get that:
the limit is -k^2(k+1)^2/8.

cheers! :D :D


trigonometry
function
inequalities
geometry proposed
geometry

Prove that

$ a\sin \frac{B}{2}\sin \frac{C}{2}\plus{} b\sin \frac{C}{2}\sin \frac{A}{2}\plus{} c\sin \frac{A}{2}\sin \frac{B}{2}\leq \frac{s}{2}$

Prove that $ \boxed {a\sin \frac {B}{2}\sin \frac {C}{2} \plus{} b\sin \frac {C}{2}\sin \frac {A}{2} \plus{} c\sin \frac {A}{2}\sin \frac {B}{2}\leq \frac {3r\sqrt 3}{2}}\le \frac {s}{2}$ .

[hide="Proof."]
$ \sum a\cdot\sin\frac B2\sin\frac C2 \equal{}$ $ \prod\sin \frac A2\cdot\sum \frac {a}{\sin \frac A2} \equal{}$ $ \frac {(s \minus{} a)(s \minus{} b)(s \minus{} c)}{abc}\cdot\sum 4R\cos \frac A2 \equal{} r\cdot\sum\cos\frac A2\le \frac {3r\sqrt 3}{2}$ because

$ \cos\ : \ \left(0,\frac {\pi}{2}\right)\rightarrow\mathrm R$ is a concave function $ \implies\ \frac 13\cdot \sum \cos\frac A2\le$ $ \cos\frac {\sum\frac A2}{3}$ ([u][b]Jensen[/b]'s inequality[/u]) $ \implies$ $ \sum\cos\frac A2\le \frac {3\sqrt 3}{2}$ .

[b]Well-known remarks.[/b]

$ \frac {(s \minus{} a) \plus{} (s \minus{} b) \plus{} (s \minus{} c)}{3}$ $ \ge \sqrt [3]{(s \minus{} a)(s \minus{} b)(s \minus{} c)}$ $ \implies s^3\ge 27sr^2\implies \boxed {\ 3r\sqrt 3\le s\ }$ .

$ 3\left(r_ar_b \plus{} r_br_c \plus{} r_cr_a\right)\le\left(r_a \plus{} r_b \plus{} r_c\right)^2\implies$ $ 3s^2\le (4R \plus{} r)^2\implies \boxed {\ s\sqrt 3\le 4R \plus{} r\ }$ .

$ \left\|\begin{array}{c} 3r\sqrt 3\le s \\
 \\
9s\le 3(4R \plus{} r)\sqrt 3\end{array}\right\|\bigoplus\implies 8s\le 12R\sqrt 3\implies \boxed {\ s\le \frac {3R\sqrt 3}{2}\ }$ . In conclusion, $ \boxed {\ 3r\sqrt 3\le s\le \frac {3R\sqrt 3}{2}\ }$

and $ 2r\le R$ . Directly $ a^2\ge 4(s \minus{} b)(s \minus{} c)$ a.s.o. $ \implies$ $ abc\ge 8(s \minus{} a)(s \minus{} b)(s \minus{} c)\implies$ $ 4Rsr\ge 8sr^2\implies \boxed {\ 2r\le R\ }$ .[/hide]


floor function
number theory proposed
number theory

[quote="sinajackson"]Let $ m,n$ be positive integers. Prove that:

$ \left\lfloor\frac{m^{2}}{n}\right\rfloor\plus{}\sum^{m\minus{}1}_{k \equal{} 0}\left(\left\lfloor\frac{k}{n}\right\rfloor\minus{}\left\lfloor\frac{m\plus{}k}{n}\right\rfloor\right)$

$ \equal{}\left\lfloor\frac{\min\left\{m\;\text{mod}\;n,(\minus{}m)\;\text{mod}\;n\right\}^{2}}{n}\right\rfloor$[/quote]
we suppose that $ m\equal{}r\plus{}hn$ where $ r\equal{}m\ (mod\ n)\in\{0,1,2,..,n\minus{}1\}$
$ X\equal{}\left\lfloor\frac{m^{2}}{n}\right\rfloor\plus{}\sum^{m\minus{}1}_{k \equal{} 0}\left(\left\lfloor\frac{k}{n}\right\rfloor\minus{}\left\lfloor\frac{m\plus{}k}{n}\right\rfloor\right)\equal{}2rh\plus{}h^{2}n\plus{}\left\lfloor\frac{r^{2}}{n}\right\rfloor\minus{}h(r\plus{}hn)\plus{}\sum^{r\plus{}hn\minus{}1}_{k \equal{} 0}\left(\left\lfloor\frac{k}{n}\right\rfloor\minus{}\left\lfloor\frac{r\plus{}k}{n}\right\rfloor\right)$
$ \equal{}\left\lfloor\frac{r^{2}}{n}\right\rfloor\plus{}\sum^{r\minus{}1}_{k \equal{} 0}\left\lfloor\frac{k}{n}\right\rfloor\minus{}\sum^{2r\minus{}1}_{k \equal{} r}\left\lfloor\frac{k}{n}\right\rfloor$

if $ 2r\minus{}1< n$; $ X\equal{}\left\lfloor\frac{r^{2}}{n}\right\rfloor$ and we have $ min(m\ (mod\ n),\minus{}m\ (mod\ n))\equal{}min(r,n\minus{}r)\equal{}r$

if $ 2r\minus{}1\ge n$ so $ 2r\minus{}1\equal{}n\plus{}r'$ where $ r'\in\{0,1,2,...,r\minus{}1\}$ then $ \left\lfloor\frac{r^{2}}{n}\right\rfloor\plus{}\sum^{r\minus{}1}_{k \equal{} 0}\left\lfloor\frac{k}{n}\right\rfloor\minus{}\sum^{2r\minus{}1}_{k \equal{} r}\left\lfloor\frac{k}{n}\right\rfloor\equal{}\left\lfloor\frac{r^{2}}{n}\right\rfloor\minus{}r'\plus{}1\equal{}\left\lfloor\frac{r^{2}}{n}\right\rfloor\plus{}n\minus{}2r\equal{}\left\lfloor\frac{(n\minus{}r)^{2}}{n}\right\rfloor$
and we have $ min(m\ (mod\ n),\minus{}m\ (mod\ n))\equal{}min(r,n\minus{}r)\equal{}n\minus{}r$

[quote="aviateurpilot"]
we suppose that $ m \equal{} r\plus{}hn$ where $ r \equal{} m\ (mod\ n)\in\{0,1,2,..,n\minus{}1\}$
[/quote]

thats right :) and thats the common and helpful method in solving problems including Floor Functions...
for example the Problem1 in this Topic>>>> http://www.mathlinks.ro/viewtopic.php?t=164410 is solved by this method :)


The sum of N real numbers
(not necessarily unique)
is 20.
The sum of the 3 smallest of these numbers is 5.
The sum of the 3 largest is 7.
What is N?

Among the 3 numbers that add up to 5, at least 1 is  positive. Thus, all the three largest numbers adding up to 7 must be positive. if N < 6, then the sum of all the numbers will be 5 + 7 - a = 12 - a, where a is the sum of numbers which belong to both the sets of 3 smallest and 3 largest numbers. But the set of 3 largest numbers contains all positive numbers, so
a > 0. Hence the total sum is < 12, which is not true.
If N = 6,the total sum is 5 + 7 = 12, which doesn't work.
So, N > 6. The total sum is 12 + b, where b is the sum of the numbers  that don't belong to the sets of 3 smallest and 3 largest numbers. But for every element  k of those "excluded" numbers, k =<  7/3 (the maximum value of the smallest number among the 3 largest numbers)  and
k >= 5/3 (the minimum value of the largest number among the 3 smallest numbers).
Now, b = 20 - 12 = 8. If the number of "excluded" numbers is 4, everything works (have each excluded number = 2, 5/3<2<7/3). If it's less than 4, then
b =< 3k =< 21/3 < 8 -- impossible. If it's larger than 4, then b>= 5k>= 25/3>8 -- also impossible.
Hence the number of excluded numbers is 4.
N = 3 + 3 + 4 = 10.
Nice puzzle!   :D

[b]The sum of N real numbers (not necessarily unique) is 20[/b].
Let the N real numbers in increasing order be x1, x2, x3, x4, ...., xK, ...., X(N-2), X(N-1), and x(N).

[b]The sum of the 3 smallest of these numbers is 5.[/b]
x1 + x2 + x3 = 5. This implies that the largest of the 3 smallest numbers i.e. x3 >= 5/3.

[b]The sum of the 3 largest is 7[/b].
X(N-2) + X(N-1) + x(N) = 7. This implies that the smallest of the 3 largest numbers i.e. x3 <= 7/3.

[b]What is N[/b]?
In accordance with the above deductions, we can say the following.

(5/3) <= {x4, ...., xK, ...., X(N-3)} <= (7/3) ....(1)
The boundary conditions are inclusive as it has been specified that the N real numbers may not be unique.
Also, {x4 + ... +  xK +  ... + X(N-3)} = 20 - (5 + 7) = 8  ....(2)

Say, there are 'm' other real numbers in between the 3 smallest and the 3 largest numbers. N = m + 3 + 3 = m + 6.

From (1) and (2), we have the following:
8/(7/3) <= m <= 8/(5/3)
=>  (24/7) <= m <= (24/5)
=>  3.428571429  <= m <= 4.8
As m is an integer it can me nothing but 4.

Therefore, N = 4 + 6 = 10.


consider $a_1,...,a_{2k+1}$ integers and an arbitrary permutation of
these numbers we denote by $b_1,...,b_{2k+1}$
is $P=\prod_{i=1} ^{2k+1} (a_i-b_i)$
even or odd ?

even

Suppose it is odd, then all $(a_i-b_i)$ are odd. So all $a_i \neq b_i (mod 2)$. Yet there are odd number of terms, so this is impossible.

i feel also that even is the answer i tried to give a proof starting with
$p+q=2k+1$ where p number of even terms and q number of odd terms.
now without changing the problem suppose $p<q$:
we can couple the p evens in $a_i$ with o of the q odd in $b_i$ and so on couple them...
until remains a pair of even-even or odd-odd wich will make the product even.
can you give a more rigorous proof lightrhee ?

Call $c_i = a_i - b_i$.  $\sum_{i=1}^{2k+1} c_i = 0$, so at least one even term must be present.  :)

Assume that all summands are odd, then you get an odd number as sum of an odd number of odd numbers. But $0$ is not odd.


Can some one please help me with recursive sequences. i can not figure out how to make the equation   .PLEASE SOMEONE HELP ME.

It might be a little more helpful if you showed us something specific that you had a problem with, or if you posted a sample problem for someone to solve.  Just having us explain a full topic of mathmatics doesn't work to well :)


Prove that in a non-isosceles triangle we have:

$ \minus{}\frac 18<\frac{(h_a\minus{}h_b)(h_b\minus{}h_c)(h_c\minus{}h_a)}{(r_a\minus{}r_b)(r_b\minus{}r_c)(r_c\minus{}r_a)}<0$, where $ h_a,\ h_b,\ h_c$ are the altitudes of $ \triangle ABC$ and $ r_a\ r_b,\ r_c$ are the radii of the excircles of the triangle.

hello, we need the following formulas
$ r_a \equal{} \frac {A}{s \minus{} a}$, $ r_b \equal{} \frac {A}{s \minus{} b}$, $ r_c \equal{} \frac {A}{s \minus{} c}$
and
$ h_a \equal{} \frac {2A}{a}$, $ h_b \equal{} \frac {2A}{b}$, $ h_c \equal{} \frac {2A}{c}$ inserting this formulas in our proposition we get after some algebra
$ \minus{} \frac {1}{8} < \frac { \minus{} 8(s \minus{} a)^2(s \minus{} b)^2(s \minus{} c)^2}{a^2b^2c^2} < 0$
The right-hand side is clear, the left-hand side is equivalent with
$ abc > ( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)$ and this is also true.
Sonnhard.


geometry
circumcircle
trigonometry
ratio
geometric transformation
reflection
power of a point

Let $ABC$ be a triangle, let $B_{1}$ be the midpoint of the side $AB$ and $C_{1}$ the midpoint of the side $AC$. Let $P$ be the point of intersection, other than $A$, of the circumscribed circles around the triangles $ABC_{1}$ and $AB_{1}C$. Let $P_{1}$ be the point of intersection, other than $A$, of the line $AP$ with the circumscribed circle around the triangle $AB_{1}C_{1}$. Prove that $2AP=3AP_{1}$.

Let (O) be circumcircle of the $\triangle ABC.$ Let the circumcircles (U), (V) of the $\triangle ABC_{1},\ \triangle AB_{1}C$ meet BC again at F, E. Let PE, PF meet (U), (V) again at X, Y. Let AP meet BC at K and (O) at Q. $EB_{1},\ FC_{1}$ are antiparallel to AB, AC with respect to the $\angle B,\ \angle C,$ respectively, so that $EB_{1}\perp OB,\ FC_{1}\perp OC.$ From (U), $\angle BXE \equiv \angle BXP = \angle BAP \equiv \angle B_{1}AP$ and from (V), $\angle B_{1}AP = 180^\circ-\angle B_{1}EP = \angle B_{1}EX.$ It follows that $BX \parallel EB_{1}$ and similarly, $CY \parallel FC_{1}.$ Consequently, $BX \perp OB,\ CY \perp OC$ are tangents of (O) at B, C. From (U), (V), $\angle XC_{1}A = \angle XPA \equiv \angle EPA = \angle ECA,$ hence $XC_{1}\parallel BC$ and similarly, $YB_{1}\parallel BC.$ Thus $X,\ B_{1},\ C_{1},\ Y$ are collinear and $XY \parallel BC.$ Consequently, BX = CY. Let the tangents BX, CY of (O) meet at L. Then powers of L to (U), (V) are equal, $LB \cdot LX = LC \cdot LY,$ so that L is on the radical axis AP of (U), (V) and $AP \equiv AL \equiv AK$ is the A-symmedian of the $\triangle ABC.$ This can also be proved by a direct calculation of powers of the A-symmedian foot to (U), (V).

From (O), (V), $\angle QBC = \angle QAC \equiv \angle PAC = \angle PEC \equiv \angle PEF$ and similarly from (O), (U) $\angle QCB = \angle PFE.$ Thus $\triangle PEF \sim \triangle QBC$ are similar with coefficient $\frac{EF}{a}.$ Powers of B, C to (V), (U) are $BE \cdot a = \frac{c^{2}}{2},\ CF \cdot a = \frac{b^{2}}{2},$ yielding

$\frac{EF}{a}= 1-\frac{BE+CF}{a}= 1-\frac{b^{2}+c^{2}}{2a^{2}}$

By cosine law for the $\triangle ABK$ or $\triangle ACK,$ the A-symmedian length is

$AK = \frac{bc}{b^{2}+c^{2}}\ \sqrt{2(b^{2}+c^{2})-a^{2}}$

K divides BC in the ratio $\frac{KB}{KC}= \frac{c^{2}}{b^{2}},$ yielding KB, KC. Power of K to (O) is $KB \cdot KC = AK \cdot KQ,$ yielding

$\frac{AK}{KQ}= \frac{AK^{2}}{KB \cdot KC}= 2 \frac{b^{2}+c^{2}}{a^{2}}-1$

As a result,

$\frac{AP}{PQ}= \frac{AK+KP}{KQ-KP}= \frac{AK+\frac{KQ \cdot EF}{a}}{KQ-\frac{KQ \cdot EF}{a}}=$

$= \frac{\frac{AK}{KQ}+\frac{ER}{a}}{1-\frac{EF}{a}}= \frac{2 \frac{b^{2}+c^{2}}{a^{2}}-1+1-\frac{b^{2}+c^{2}}{2a^{2}}}{1-1+\frac{b^{2}+c^{2}}{2a^{2}}}= 3$

$AQ = 2AP_{1},$ because the triangles $\triangle ABC \sim \triangle AB_{1}C_{1}$ are similar with center A and coefficient 2, so that

$\frac{AP}{AP_{1}}= \frac{2AP}{AQ}= \frac{2AP}{AP+PQ}= \frac{6PQ}{4PQ}= \frac{3}{2}$

Consider inversion with center $A$ and coeficient $\sqrt{\frac{AB\cdot AC}{2}}$ combinated with reflection through the axis of angle $\angle BAC$. Points $B$ and $C_1$ and poitns $C$ and $B_1$ will change places. $P'$ will be point, where $BC_1$ and $CB_1$ meet, therefore it'll be centroid of $\triangle ABC$. $Q'$ will be the point, where $AP'$ and $BC$ meet, therefore it'll be midpoint of $BC$. Therefore $\frac{AQ}{AP}=\frac{AP'}{AQ'}=\frac{2}{3}$, which gives what we wanted.

Q.E.D.


Let $ a,b,c>0$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that:

\[ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\ge \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\]

[quote="Pain rinnegan"]Let $ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that:
\[ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\]
[/quote]
Because $ \left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right)(a^2\plus{}b^2\plus{}c^2)\geq\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right)^2$ and $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq3.$

$ LHS = \frac 1 3 (\sum{a^2})\left(\sum{\frac 1 {a^2}}\right)$

$ \ge \frac 1 3 (\sum{ab})\left(\sum{\frac 1 {a^2}}\right)$

After multiplying out, we are left to prove $ \sum{\left( \frac b a + \frac {bc}{a^2}\right) \ge 2\sum{\frac a b}}$ which is just AM-GM since $ \frac {bc}{a^2} + \frac c b \ge 2 \frac c a$

Our desired inequality can be written in the following way:

    $ \frac {1 \minus{} ca}{a^2} \plus{} \frac {1 \minus{} ab}{b^2} \plus{} \frac {1 \minus{} bc}{c^2} \geq 0$.

Now by adding and substracting the terms $ (ca)^2,(ab)^2,(bc)^2$ into the fractions above respectively,we get:

     $ \frac {(ca)^2 \minus{} ca \plus{} 1}{a^2} \minus{} c^2 \plus{} \frac {(ab)^2 \minus{} ab \plus{} 1}{b^2} \minus{} a^2 \plus{} \frac {(bc)^2 \minus{} bc \plus{} 1}{c^2} \minus{} b^2 \geq 0$ or equivalently:

     $ \frac {(ca)^2 \minus{} ca \plus{} 1}{a^2} \plus{} \frac {(ab)^2 \minus{} ab \plus{} 1}{b^2} \plus{} \frac {(bc)^2 \minus{} bc \plus{} 1}{c^2} \geq 3$

Using $ AM \minus{} GM$ on the last inequality it suffices to prove that:

     $ ((ab)^2 \minus{} ab \plus{} 1)((bc)^2 \minus{} bc \plus{} 1)((ca)^2 \minus{} ca \plus{} 1) \geq a^2b^2c^2$ or equivalently

     $ (\frac {(ab^2) \minus{} ab \plus{} 1}{ab}) (\frac {(bc^2) \minus{} bc \plus{} 1}{bc}) (\frac {(ca^2) \minus{} ca \plus{} 1}{ca}) \geq 1$

But we have that : $ \frac {(ab^2) \minus{} ab \plus{} 1}{ab} \equal{} ab \minus{} 1 \plus{} \frac {1}{ab} \geq 2 \minus{} 1 \equal{} 1$.


inequalities
inequalities solved
3-variable inequality
cyclic inequality
JBMO
cauchy schwarz

We will use $a^2+1 \geq 2a$ and the analogous ones. We have to prove that:
\[\sum \frac{1+a ^2}{1+c^2 +\frac12(1+b ^2)} \geq 2\]
Denote $x=1+a ^2, y=1+b ^2 , z=1+c ^2$ and we have to prove that
\[\sum\frac{x}{y+2z} \geq 1\] But by Cauchy Schwarz
\[\sum \frac{x}{y+2z}= \sum \frac{x ^2}{xy+2zx} \geq \frac{(x+y+z) ^2}{3(xy+yz+zx)} \geq 1\]


inequalities
trigonometry
inequalities proposed

Let $ ABC$ be an arbitrary triangle. Prove that
\[ \prod{\frac{\sqrt{1\plus{}8\cos^{2}{A}}}{\sin{A}}} \geq 8.\]

Any simple proof?

Equivalently we have to prove that \[ \prod \sqrt{1\plus{}8\cos^{2}{A} }\geq 64 \prod \sin ^{2}A\] or \[ \prod \sqrt{1\plus{}8\cos^{2}{A} }\geq 64 \prod (1\minus{} \cos ^{2}A)\] or \[ 8\prod \sqrt{1\plus{}8\cos^{2}{A} }\geq \prod (8\minus{}8 \cos ^{2}A).\] Now, set $ x\equal{}8 \cos ^{2}A$  etc. Then, we have to prove that \[ 8(1\plus{}x)(1\plus{}y)(1\plus{}z) \geq (8\minus{}x)(8\minus{}y)(8\minus{}z)\] or \[ 8(x\plus{}y\plus{}z\minus{}8)\plus{}xyz \plus{}8\geq 0.\] Is it a known fact that in every triangle \[ \sum \cos ^{2}A \plus{}2 \prod \cos A \equal{}1,\] hence

\[ (\sum \cos ^{2}A \minus{}1)^2\equal{}4 \prod \cos ^{2}A,\] which becomes \[ (\frac{x\plus{}y\plus{}z}{8}\minus{}1)^{2}\equal{}\frac{xyz}{2}.\] Substituting this into the desired inequality we arrive at $ (x\plus{}y\plus{}z\minus{}6)^2 \geq 0,$ which is true. It is easily seen that we have equality iff  the triangle is equilateral.


can anyone suggest me a nice book full of conceptual mcq's and illustrative examples for physics.
PS: particularly for iit-jee :roll:

Please check here- http://www.mathlinks.ro/viewtopic.php?t=256304&sid=ee3a170f2b20616f57067948e71543cc
 :wink:

hi, I'm SHIBIN GEORGE FROM C.G.
i think 'university physics' will be nice if u r really interested to go inyo the depths of the subject...i myself have tried it...others r 'halliday &resnick',i.e. irodov',etc...
please suggest some maths bools for........

for conceptual mcq's try IIT Physics by Deb Mukherjee(Bharti bhavan).
and u'll see,its bloody small and thin :D

[quote="shibin.george"]please suggest some maths [b]bools[/b] for........[/quote]
whats that??? :maybe:  :rotfl:

blindsight by robin cook and the prisoner of....by jeffrey archer
excellent books

these are physics books :maybe:  :maybe:

they will certainly improve ur  physics :D

i checked amazon and cudn't understand the physics :rotfl:  :rotfl:

hey, even arihant has plenty of errors in answers :D

[quote="nuclear_alchemist"]Nootan ISC Physics-
Best book if you want to test your conceptual error spotting abilities  :rotfl:[/quote]

Please Don`t mind that is one of the worst book..........

[quote="murgi"]Please Don`t mind that is one of the worst book..........[/quote]
Maybe thats why I said 'error spotting' abilities in the text.  :|


limit
trigonometry
calculus
calculus computations

I'm supposed to calculate the limit if it exists. Can you can firm that this one doesn't exist:

limit as x approaches 0 of [1 - cos(4x)] / [9x^2]

I know that

lim as x approaches 0 of [1 - cos ax] / [ax] = 0, however, I don't think you can manipulate the above to help you.

$ \lim_{x \to 0} \frac {1\minus{}\cos(4x)}{9x^2} \equal{} \lim_{x \to 0} \frac {2\sin^2(2x)}{9x^2} \equal{} \lim_{x \to 0} \frac {8} {9} \left (\frac {\sin(2x)}{2x}\right )^2 \equal{} \frac {8} {9}$.

Alternatively, using the Taylor series expansion $ \cos(4x) \equal{} 1 \minus{} 8x^2 \plus{} O(x^3)$, we get the same result.

Remark \[ \boxed{\boxed{(1\minus{}\cos \theta)(1\plus{}\cos \theta)\equal{}\sin ^ 2 \theta}}\]


Prove that if A is a diagonalizable matrix, then the rank of A is the number of nonzero eigenvalues of A.

This is relatively well known from linear algebra theory.  If $ A$ is diagonalizable, an inversible matrix $ P$ exists such that $ P^{\minus{}1}AP\equal{}D$, where $ D$ is the diagonal matrix whose main diagonal is formed by the eigenvalues of $ A$.  Since multiplying by $ P^{\minus{}1}$ to the left and $ P$ to the right is equivalent to performing line and column transformations that do not affect rank, then $ D$ and $ A$ have the same rank.  The conclusion follows.  Kent's remark is important, you need to consider all eigenvalues with their multiplicity.

OK, but how do we prove that multiplying $ P$ and $ P^{\minus{}1}$ does not affect the rank?

well, $ P$ and $ P^{\minus{}1}$ are invertible matrices, so you can represent them as a product of elementary matrices which don't change the rank

Now we have to show that elementary operations do not change the rank of a matrix. So, let us have a m x n matrix A in which we have made a finite number of elementary row operations, obtaining in that process the matrix B. Due to the nature of the elementary operations, each row of B is a linear combination of rows of A, and so we must have $ \dim \mathcal{L} (B) \leq \dim \mathcal{L} (A)$*. But elementary operations are reversible, in the sense that if B can be obtained from A by a finite sequence of elementary operations, then also A can be obtained from B by a finite number of elementary operations, and so $ \dim \mathcal{L} (A) \leq \dim \mathcal{L} (B)$. These two inequalities imply $ \dim \mathcal{L} (B) \equal{} \dim \mathcal{L} (A)$. Therefore, A and B have the same rank, which means that elementary row operations do not change the rank. For elementary column operations, we just have to observe that, performing a finite sequence of such operations on the columns of a matrix A is the same as applying the same operations to the rows of $ A^T$, and a given matrix and its transpose have the same rank. The previous argument now allow us to conclude that elementaty row or column operations do not change the rank of a matrix.

* Here, $ \mathcal{L} (P)$ denotes the row space of matrix P.

[quote="mathwizarddude"]OK, but how do we prove that multiplying $ P$ and $ P^{ \minus{} 1}$ does not affect the rank?[/quote]
This should be intuitive if you think of rank as the dimension of the image of a linear transformation, which is a basis-independent notion.


is there a formula to do the problems that ask

"Find the number of distinct paths from point A to point B traveling only to the right or slanted up and right"

( i don't know how to add pictures once i have scanned them, so i couldn't actually show you guys a problem. If someone tells me how to attach a scanned pic, then i will, or you can just supply one  :lol: )[/img]

The easy way, if you're restricted to moving in two ways like that, is just to write on each lattice point the number of paths from point $ A$ to that point, and then add along the paths.  For example, if you do this on the first quadrant with steps going to the right or up, you get Pascal's triangle.  If the number of points in between isn't too big, you can even use this technique on those weird questions where they don't let you move to a certain point.


integration
calculus
calculus computations

[quote="bos1234"]
$-\frac{x}{4}sin2x+\boxed{\frac{x^{2}}{2}}-\frac{1}{8}cos2x-\boxed{\frac{x^{2}}{4}}$

$-\frac{x}{4}sin2x-\frac{1}{8}cos2x+\boxed{\frac{x^{2}}{4}}$
[/quote]

clearly, you need to get some sleep, i think... LOL ! :P

lol yeah i think i should.. i was stuck on that qn for a looooong time :roll:

thanks for the help


function
floor function
calculus
calculus computations

Construct a monotone function $ f: [0,1]\rightarrow\mathbb{R}$ that is discontinuous at [i]exactly[/i] $ \mathbb{Q}\cap[0,1]$.

Construct as many classes of functions as you can.

Let $ q_n$ be an enumeration of $ \mathbb{Q} \cap [0, 1]$ and let $ f(x) \equal{} \sum_{q_n \in [0, x]} g(n)$ where $ g(n)$ is any positive function with the property that $ \sum_{n\equal{}1}^{\infty} g(n)$ converges.  We can take, for example, $ g(n) \equal{} 2^{\minus{}n}$.

This is the same contruction that I had.  Can you contruct any other types of functions? (not necessarily addressed to you, of course)

Clearly the sum of $ f$ and a continuous monotonic function has the same property.  But I strongly suspect that all such $ f$ arise in this way; at every discontinuity $ q$ let $ g(q)$ be the difference between the right limit and the left limit, and after subtracting off all $ g$s what remains is continuous and monotonic.  Moreover, the choice of enumeration is irrelevant since $ \sum g$ converges absolutely.

Or let $ f(x) \equal{} \begin{cases} 0 \ \ \ \text{if} \ x \in \mathbb{R} \minus{} \mathbb{Q} \\
1/q \ \text{if} \ \ \ x \in \mathbb{Q} \end{cases}$

where $ x \equal{} p/q$ is in lowest terms.

If you're going to use the cases environment, why not use its full power (in particular, the alignment character &)?

$ f(x) \equal{} \begin{cases} 0 & \text{if } x \in \mathbb{R} \minus{} \mathbb{Q} \\
1/q & \text{if } x \equal{} \frac{p}{q} \in \mathbb{Q}, \ (p, q) \equal{} 1. \end{cases}$

So, this function is clearly not monotonic, as t0r noted, and we can't turn it into a monotonic function just by, say, adding a linear function to it.  I'm curious, though -- is there any way to rescue a construction of this form?  Or will doing so just boil down to a construction like the one t0r gave?

I think the following works:
\[ f(x) \equal{} \sum_{k \equal{} 1}^\infty \frac {\lfloor kx \rfloor}{k^3}.
\]
(I might need to redefine $ f(0)$ as $ \minus{}1$.)
Although we can view it as another presentation of tor's function.

[quote="The Captain"]Theorem $ 6.16$ of Rudin's Principles of Mathematical Analysis might help too, if you have not already seen it.[/quote]
How does this matter as far as this topic is concerned?

I found the following, although it may just be the same thing as t0r's, just with different notation:

Let $ D\equal{}\{x_{1},x_{2},\ldots\}$ and let $ (\varepsilon_{n})$ be a sequence of positive numbers with $ \sum_{n\equal{}1}^{\infty}\varepsilon_{n}<\infty$.  Define $ f(x)\equal{}\sum_{x_{n}\leq x}\varepsilon$, where the sum is taken over the set $ \{n: x_{n}\leq x\}$ and where $ f(x)\equal{}0$ if this set is empty.  If $ x<y$, then $ f(y)\equal{}\sum_{x_{n}\leq y}\varepsilon_{n}\equal{}\sum_{x_{n}\leq x}\varepsilon_{n}\plus{}\sum_{x<x_{n}\leq y}\varepsilon_{n}\equal{}f(x)\plus{}\sum_{x<x_{n}\leq y}\varepsilon_{n}\geq f(x)$.

In particular, this function is discontinuous at the point of $ D$, whatever $ D$ happens to be.

Yes, that's the same.  I'm reasonably certain that this is the only possible construction modulo continuous functions.


calculus
integration
number theory
greatest common divisor
induction

Theorem:

if a_1, a_2, .... a_n are integers and gcd(a_1,a_2, ... , a_n) = d, there exist integers x_1, x_2....x_n such that

a_1x_1 + a_2x_2 + .... + a_nx_n = d.

Have fun proving it; and try _not_ to use prime factorizations, since this theorem is "integral" :) in proving the fundamental theorem of arithmetic.

A try :

[hide]

. First notice that it's enough to prove the theorem for d=1. Otherwise divide all a_i by d.    Then branch and repeat (I did some assembler programming recently ) ...

. gcd (a,b,c)=gcd(gcd(a,b),c) and by induction gcd (a1,..,an)=gcd (gcd(a1,...an-1),an))

. I assume we can use Bachet-Bezout theorem.

Induction :

. The theorem is true for n=2 : it's Bachet-Bezout !

. If the theorem is true for n-1.

Since gcd (a1,..,an)=gcd (gcd(a1,...an-1),an)=1, there exist u,v such as :   u*gcd (a1,...an-1) + v*an = 1.

But gcd (a1,...an-1) = sum w_i a_i (i=1..n-1) for some  w_i by induction hypothesis (dividing by gcd (a1,...an-1) to be in the strict induction hypothesis).

Then sum u*w_i*a_i + v*a_n = 1.[/hide]

using Bachet-Bezout  is a little cheap... i just consider this the generalized form of it, since this and that have almost the exact same proof.


inequalities algebra polynomial calculus integration

If the arithmetic mean of some 2 numbers is greater than the geometric mean, is it necessary that the numbers are positive reals? :idea:

The answer is no. In the case of two numbers, you can have one be zero:

$ \frac{0\plus{}1}{2}>\sqrt{0\cdot 1}$

However neither can be negative: if exactly one is negative then $ GM$ is not defined. If both are negative, then $ AM<0<GM$.

But you can have negative reals with more than two numbers:

$ \frac{(\minus{}1)\plus{}(\minus{}1)\plus{}1\plus{}9}{4}>\sqrt[4]{(\minus{}1)(\minus{}1)(1)(9)}$

So I've got a question:  if the [url=http://en.wikipedia.org/wiki/Maclaurin%27s_inequality]Maclaurin inequalities[/url] all hold for a set of real numbers, must they all be non-negative?

I have a pre-requisite question for answering t0r's question: is there a (multi-)set $ \{a_1, \ldots, a_n\}$ of real numbers including some negative values such that the elementary symmetric polynomials on those number are all positive?

(The elementary symmetric polynomials on $ a_1, \ldots, a_n$ are $ e_1 \equal{} a_1 \plus{} \ldots \plus{} a_n$, $ e_2 \equal{} a_1a_2 \plus{} \ldots \plus{} a_1a_n \plus{} a_2a_{3} \plus{} \ldots \plus{} a_2a_n \plus{} \ldots \plus{} a_{n \minus{} 1}a_n$, ..., $ e_n \equal{} a_1a_2\cdots a_n$.)

... and I answer my own question:  this is impossible by Descartes' law of signs.  But for Maclaurin's inequalities to hold it is possible that $ n$ is odd and $ S_n$ is negative.

Okay, good -- and the set $ \{2, 2, \minus{} 1\}$ is a counter-example to your question as $ \frac {2 \plus{} 2 \minus{} 1}{3} > \sqrt {\frac {4 \minus{} 2 \minus{} 2}{3}} > \sqrt [3]{ \minus{} 4}$.

What if we replace Maclaurin with (integer, or nonnegative integer) power-mean inequalities?  (I've given this exactly 0 thought, so it may be obvious.)

And the answer to my question is that this is also insufficient: for example, the nonnegative integer power means of $ \{ \minus{} 1, 1/2, 2\}$ are in increasing order, and all the integral power means of $ \{ \minus{} 1, \minus{} 1, 1/4, 1/4, 4, 4\}$ are in increasing order.  (Well, I haven't proven this, just numerically verified it, but it's easy to see that it should be true if it's true for "small" power means.)

So here's the meta-question, then: is there [i]any[/i] condition of this sort that holds for all sets of positive real numbers but is violated for any set that contains negative real numbers?  One property such a chain of inequalities must have is that whenever some element of the set is 0, some of the inequalities must be equalities.  I suppose $ a_1 \geq 0$, $ a_2 \geq 0$, ..., $ a_n \geq 0$ has the desired properties, but is there something more ... satisfying?

It looks like the relevant idea is the notion of a [url=http://en.wikipedia.org/wiki/Semialgebraic_set]semialgebraic set[/url]; then we're talking about some sort of appropriate generalization of an ideal and asking what kind of generators it can be given.  Here, the set of real polynomials that are non-negative whenever all of their variables are non-negative is closed under multiplication and non-negative linear combinations and contains $ a_1 \ge 0, a_2 \ge 0, ...$ as well as the elementary symmetric polynomials.

Here's an idea for Joel's question that may or may not work.  Using Sturm sequences, we can test whether a polynomial has any negative roots.  The test should boil down to checking the signs of certain combinations of the elementary symmetric polynomials.

Apparently [url=http://en.wikipedia.org/wiki/Stengle%27s_Positivstellensatz]there is a theorem[/url] about questions of this nature.


Hey, I am currently in College taking Calculus III.  Though I am in College, I've never studied Probability, Statistics, and Number Theory.  I'd like to know if it would serve me to even study the AoPS texts, or perhaps I should stick to Undergraduate level texts.  I am pretty sure that the AoPS book would still prove to be somewhat a challenge, but I just want an honest opinion so I can save myself some money.

For Number Theory:

Elementary Number Theory by Gareth Jones
http://www.amazon.com/Elementary-Number-Theory-Gareth-Jones/dp/3540761977/ref=pd_sim_b_title_4

Introduction to Analytic Number Theory by Tom Apostol
http://www.amazon.com/Introduction-Analytic-Number-Undergraduate-Mathematics/dp/0387901639/ref=pd_sim_b_title_6

For Probability and Statistics:

Probability and Statistics by Morris DeGroot
http://www.amazon.com/Probability-Statistics-3rd-Morris-DeGroot/dp/0201524880/ref=pd_bbs_1?ie=UTF8&s=books&qid=1203386122&sr=8-1

bump!

Any advice is appreciated, would like some feedback before I spend money.

i have not read any of those books but are you looking to buy those books just to learn for fun or do you need probability and stats in calculus III.


probability
combinatorics theorems
combinatorics

If in a bantumi mach, you begin the first and you are only informed of the rules of bantumi completely, and your opponent is super power master in bantumi, will you have any hope to win, or in mathematical language, is there any probablity for you to win, ha?
Thanks.

Hey, doesn't exist any one understanding of GAME THEORY? It's not only related with maths, but also depends on with biology. And please excuse me for me to apply a double standard. Not really. I am a democratic. And I'm from Europe!

Why donna write down? Feelin' annoyed of me? So I'll be appropriate about all of you which not thinking to post a reply to me.

Your general posting habits are extremely annoying, and this thread is no exception.  You've posted a question without providing sufficient definitions for anyone to make sense of your ramblings, and then you complain that no one responds as if we are obligated to solve your problems for you.  Grow up.


geometry
circumcircle
trigonometry
ratio
parallelogram
geometric transformation
homothety

We have an acute-angled triangle $ABC$, and $AA',BB'$ are its altitudes. A point $D$ is chosen on the arc $ACB$ of the circumcircle of $ABC$. If $P=AA'\cap BD,Q=BB'\cap AD$, show that the midpoint of $PQ$ lies on $A'B'$.

From Menelaos it's enough to have:
$\frac{B'Q}{B'H}=\frac{PA'}{A'H}$.

But $\frac{B'H}{HA'}=\frac{\cos{A}}{\cos{B}}$.
Now we use tha fact that $HQDP$ is cyclic to obtain that $\frac{AB'}{B'Q}=\frac{A'B}{A'P}$ and since
$\frac{AB'}{A'B}=\frac{\cos{A}}{\cos{B}}$ we are done!

The problem is more or less trivial using the following fact:

[color=blue][b]Fundamental theorem on directly similar polygons:[/b] If $P_1P_2...P_n$ and $Q_1Q_2...Q_n$ are two directly similar polygons, and $R_1$, $R_2$, ..., $R_n$ are points on the lines $P_1Q_1$, $P_2Q_2$, ..., $P_nQ_n$ dividing the segments $P_1Q_1$, $P_2Q_2$, ..., $P_nQ_n$ in the same ratio $\frac{P_1R_1}{R_1Q_1}=\frac{P_2R_2}{R_2Q_2}=...=\frac{P_nR_n}{R_nQ_n}$, then the polygon $R_1R_2...R_n$ is directly similar to the polygons $P_1P_2...P_n$ and $Q_1Q_2...Q_n$.[/color]

How does this solve the problem? Well, first some trivialities: Let M, T and R be the midpoints of the segments AB, A'B' and PQ. We have to show that the point R lies on the line A'B'.

We will use directed angles modulo 180°. Since < AA'B = 90° and < AB'B = 90°, the points A' and B' lie on the circle with diameter AB; the center of this circle is the midpoint M of the segment AB. Thus, MA' = MB', so the point M lies on the perpendicular bisector of the segment A'B'. Hence, $MT\perp A^{\perp}B^{\perp}$ (as T is the midpoint of the segment A'B').

Now, since the point D lies on the circumcircle of triangle ABC, we have < DBC = < DAC; in other words, < PBA' = < QAB'. Also, < PA'B = < QB'A (since < PA'B = 90° and < QB'A = 90°). Thus, the triangles PBA' and QAB' are directly similar. Now, the points R, M, T lie on the lines PQ, BA, A'B' and divide the segments PQ, BA, A'B' in the same ratio $\frac{PR}{RQ}=\frac{BM}{MA}=\frac{A^{\prime}T}{TB^{\prime}}=1$ (since they are the midpoints of these segments); thus, according to the fundamental theorem on directly similar polygons, the triangle RMT is directly similar to the triangles PBA' and QAB'. Hence, < RTM = < PA'B. Since < PA'B = 90°, we get < RTM = 90°, so that $RT\perp MT$. Together with $MT\perp A^{\perp}B^{\perp}$, this yields RT || A'B'; but since the point T lies on the line A'B', this simply yields that the point R lies on the line A'B', and we are done.

  Darij

Great!!!!!! :D

Is there any easy proof I can look at for that "Fundamental theorem on directly similar polygons"?

I was asked to supply a proof for the Fundamental theorem on directly similar polygons I quoted above. I will begin with some prerequisites, because everyone seems to define the notion of similitude differently, and this matters when proving the theorem.

[color=red]Note: The following is unproofread text. Please notify me of errors you find.[/color]

In the following, we use directed angles modulo 180°, as defined in [url=http://www.cip.ifi.lmu.de/~grinberg/NeubergMineur.pdf]my paper [i]The Neuberg-Mineur circle[/i][/url] (and lots of other sources).

Let $ X_1$, $ X_2$, ..., $ X_n$, $ Y_1$, $ Y_2$, ..., $ Y_n$ be $ 2n$ points in plane. The two polygons $ X_1X_2...X_n$ and $ Y_1Y_2...Y_n$ (which may be degenerate) are called [i]directly similar[/i] if and only if for every triple of elements $ i$, $ j$, $ k$ of the set $ \left\{1,2,...,k\right\}$ with $ i\neq j$ and $ j\neq k$, the following two conditions hold:

[i]The angle condition:[/i] We have $ \measuredangle X_iX_jX_k  =  \measuredangle Y_iY_jY_k$.
[i]The degeneracy condition:[/i] If the points $ X_i$, $ X_j$, $ X_k$ are collinear, then the points $ Y_i$, $ Y_j$, $ Y_k$ are also collinear and satisfy $ \frac{X_iX_j}{X_jX_k}  =  \frac{Y_iY_j}{Y_jY_k}$ with directed segments.

Note that the second condition is there to cover degenerate cases, when some vertices of the polygons are collinear.

Instead of saying "the polygons $ X_1X_2...X_n$ and $ Y_1Y_2...Y_n$ are directly similar", we will simply write $ X_1X_2...X_n\approx Y_1Y_2...Y_n$.

Five very easy facts on similitude:

[b]1.[/b] Let $ X_1$, $ X_2$, ..., $ X_n$, $ Y_1$, $ Y_2$, ..., $ Y_n$ be $ 2n$ points. Then, $ X_1X_2...X_n\approx Y_1Y_2...Y_n$ if and only if for any three pairwise distinct integers $ i$, $ j$, $ k$ from the set $ \left\{1,2,...,n\right\}$, we have $ X_iX_jX_k\approx Y_iY_jY_k$.

[b]2.[/b] Let $ X_1Y_1Z_1$ and $ X_2Y_2Z_2$ be two triangles. Then, $ X_1Y_1Z_1\approx X_2Y_2Z_2$ if and only if
- either $ \measuredangle X_1Y_1Z_1  =  \measuredangle X_2Y_2Z_2\neq 0^{\circ}$ and $ \measuredangle X_1Z_1Y_1  =  \measuredangle X_2Z_2Y_2\neq 0^{\circ}$
- or the points $ X_1$, $ Y_1$, $ Z_1$ are collinear, the points $ X_2$, $ Y_2$, $ Z_2$ are also collinear, and we have $ \frac{X_1X_2}{X_1X_3}  =  \frac{Y_1Y_2}{Y_1Y_3}$ with directed segments.

Mind the $ \neq 0^{\circ}$ part in the first condition - in fact, the equalities $ \measuredangle X_1Y_1Z_1  =  \measuredangle X_2Y_2Z_2$ and $ \measuredangle X_1Z_1Y_1  =  \measuredangle X_2Z_2Y_2$ alone are not enough to make two triangles similar, because degenerate triangles always have the same angles ($ 0^{\circ}$, $ 0^{\circ}$, $ 0^{\circ}$) but not always are similar to each other.

[b]3.[/b] The relation $ \approx$ is an equivalence relation. This means:
Let $ X_1$, $ X_2$, ..., $ X_n$ be $ n$ points. Then, $ X_1X_2...X_n\approx X_1X_2...X_n$.
Let $ X_1$, $ X_2$, ..., $ X_n$, $ Y_1$, $ Y_2$, ..., $ Y_n$ be $ 2n$ points. If $ X_1X_2...X_n\approx Y_1Y_2...Y_n$, then $ Y_1Y_2...Y_n\approx X_1X_2...X_n$.
Let $ X_1$, $ X_2$, ..., $ X_n$, $ Y_1$, $ Y_2$, ..., $ Y_n$, $ Z_1$, $ Z_2$, ..., $ Z_n$ be $ 3n$ points. If $ X_1X_2...X_n\approx Y_1Y_2...Y_n$ and $ Y_1Y_2...Y_n\approx Z_1Z_2...Z_n$, then $ X_1X_2...X_n\approx Z_1Z_2...Z_n$.

[b]4.[/b] If $ A_1B_1C_1$ is a triangle with $ A_1\neq B_1$, and if $ A_2$ and $ B_2$ are two points, then there exists one and only one point $ C_2$ in the plane such that $ A_1B_1C_1\approx A_2B_2C_2$.

[b]5.[/b] If $ A$, $ B$, $ C$, $ D$ are four points with $ A\neq B$ and $ ACB\approx ADB$, then $ C  =  D$.

A less known but even more important fact on directly similar triangles:

[color=blue][b]Lemma 1.[/b] Let $ A$, $ B$, $ C$, $ D$ be four points such that $ ABDC$ is not a parallelogram. Then, there exists one and only one point $ P$ in the plane such that $ PAB\approx PCD$.[/color]

This point $ P$ will be called the [i]Miquel point[/i] of $ \left(A,B;C,D\right)$.

[i]Proof of Lemma 1.[/i] We assume that $ AC\not\parallel BD$ (leaving the case $ AC\parallel BD$ to the reader). Let $ Q$ be the point of intersection of the lines $ AC$ and $ BD$. Let $ R$ be the point of intersection of the circumcircles of triangles $ QAB$ and $ QCD$ apart from $ Q$ (if these two circumcircles are tangent to each other, we set $ R  =  Q$). Then, $ \measuredangle RAB  =  \measuredangle RQB$ (by chordal angles in the circle $ RABQ$) and $ \measuredangle RCD  =  \measuredangle RQD$ (by chordal angles in the circle $ RCDQ$). Hence, $ \measuredangle RAB  =  \measuredangle RQB  =  \measuredangle RQD  =  \measuredangle RCD$. Similarly, $ \measuredangle RBA  =  \measuredangle RCD$. Hence, $ RAB\approx RCD$.

Hence, there exists a point $ P$ in the plane such that $ PAB\approx PCD$, namely $ P  =  R$. In order to prove Lemma 1, it thus remains to show that this point is determined uniquely, i. e. to show that if $ S$ is a point in the plane such that $ SAB\approx SCD$, then $ S  =  R$.

In fact, let the circumcircles of triangles $ SAB$ and $ SCD$ intersect at a point $ V$ (apart from $ S$). Then, $ \measuredangle SVA  =  \measuredangle SBA$ (by chordal angles in the circle $ SVAB$) and $ \measuredangle SVC  =  \measuredangle SDC$ (by chordal angles in the circle $ SVCD$). But $ \measuredangle SBA  =  \measuredangle SDC$ because of $ SBA\approx SDC$. Hence, $ \measuredangle SVA  =  \measuredangle SVC$, and thus the point $ V$ lies on the line $ AC$. Similarly, $ V$ lies on the line $ BD$. Hence, $ V$ is the point of intersection of the lines $ AC$ and $ BD$. But the point of intersection of the lines $ AC$ and $ BD$ is $ Q$. Thus, $ V  =  Q$. Since the point $ V$ lies on the circumcircle of triangle $ SAB$, we thus conclude that the point $ Q$ lies on the circumcircle of triangle $ SAB$. In other words, the point $ S$ lies on the circumcircle of triangle $ QAB$. Similarly, $ S$ lies on the circumcircle of triangle $ QCD$. Thus, the point $ S$ is a point of intersection of the circumcircles of triangles $ QAB$ and $ QCD$; but these two circumcircles intersect at the points $ Q$ and $ R$. Consequently, $ S  =  Q$ or $ S  =  R$. But $ S  =  Q$ would yield $ QAB\approx QCD$ (since $ SAB\approx SCD$), so that $ \measuredangle QAB  =  \measuredangle QCD$ and thus $ AB\parallel CD$, what would imply that the circumcircles of triangles $ QAB$ and $ QCD$ are tangent to each other at $ Q$, and thus $ Q  =  R$ and hence $ S  =  R$. So we have $ S  =  Q$ or $ S  =  R$, but if $ S  =  Q$ then $ S  =  R$. Therefore, we must have $ S  =  R$. This completes the proof of Lemma 1.

Something else we can easily show is:

[color=blue][b]Lemma 2.[/b] Let $ S$, $ A$, $ B$, $ C$, $ D$ be five points. Then, $ SAB\approx SCD$ if and only if $ SAC\approx SBD$.[/color]

[i]Proof of Lemma 2.[/i] Let the circumcircles of triangles $ SAB$ and $ SCD$ intersect at a point $ V$ (apart from $ S$). The same way as in the proof of Lemma 1, show that $ V$ is the point of intersection of the lines $ AC$ and $ BD$. Conclude that $ \measuredangle SAC  =  \measuredangle SAV$ and $ \measuredangle SBD  =  \measuredangle SBV$. But $ \measuredangle SAV  =  \measuredangle SBV$ by chordal angles in the circle $ SABV$. Thus, $ \measuredangle SAC  =  \measuredangle SBD$. Similarly, $ \measuredangle SCA  =  \measuredangle SDB$. Hence, $ SAC\approx SBD$, and Lemma 2 is proven.

[color=blue][b]Lemma 3.[/b] Let $ A_1$, $ B_1$, $ C_1$, $ D_1$, $ A_2$, $ B_2$, $ C_2$, $ D_2$ be eight points. Then, $ A_1B_1C_1D_1\approx A_2B_2C_2D_2$ if and only if $ A_1B_1C_1\approx A_2B_2C_2$ and $ C_1D_1A_1\approx C_2D_2A_2$.[/color]

[i]Proof of Lemma 3.[/i] One direction of Lemma 3 is fairly obvious: if $ A_1B_1C_1D_1\approx A_2B_2C_2D_2$, then it is clear that $ A_1B_1C_1\approx A_2B_2C_2$ and $ C_1D_1A_1\approx C_2D_2A_2$.

It thus remains to prove the other direction only - i. e. it remains to show that if $ A_1B_1C_1\approx A_2B_2C_2$ and $ C_1D_1A_1\approx C_2D_2A_2$, then $ A_1B_1C_1D_1\approx A_2B_2C_2D_2$.

In order to prove this, let $ S$ be the Miquel point of $ \left(A_1,C_1;A_2,C_2\right)$ (this Miquel point exists only if $ A_1C_1C_2A_2$ is not a parallelogram, but in the case when $ A_1C_1C_2A_2$ is a parallelogram, you can easily find that the quadrilaterals $ A_1B_1C_1D_1$ and $ A_2B_2C_2D_2$ are homothetic and therefore $ A_1B_1C_1D_1\approx A_2B_2C_2D_2$ is trivial). Then, $ SA_1C_1\approx SA_2C_2$, so that Lemma 2 yields $ SA_1A_2\approx SC_1C_2$.

Let $ Y_2$ be the point in the plane such that $ SA_1A_2\approx SB_1Y_2$ (this point $ Y_2$ exists because of fact [b]4.[/b] above). Together with $ SA_1A_2\approx SC_1C_2$, this yields $ SA_1A_2\approx SB_1Y_2\approx SC_1C_2$.

Now, by Lemma 4 [b](b)[/b] below (we may use this lemma because its proof does not rely on Lemma 3), this yields $ A_1B_1C_1\approx A_2Y_2C_2$. On the other hand, $ A_1B_1C_1\approx A_2B_2C_2$. Thus, $ A_2Y_2C_2\approx A_2B_2C_2$. By fact [b]5.[/b] above, this yields $ Y_2  =  B_2$. Hence, $ SA_1A_2\approx SB_1Y_2\approx SC_1C_2$ rewrites as $ SA_1A_2\approx SB_1B_2\approx SC_1C_2$. Similarly, $ SC_1C_2\approx SD_1D_2\approx SA_1A_2$. Hence, $ SA_1A_2\approx SB_1B_2\approx SC_1C_2\approx SD_1D_2$. Again using Lemma 4 [b](b)[/b] below, we conclude that $ A_1B_1C_1D_1\approx A_2B_2C_2D_2$, and Lemma 3 is proven.

[color=blue][b]Lemma 4.[/b] Let $ X_1$, $ X_2$, ..., $ X_n$, $ Y_1$, $ Y_2$, ..., $ Y_n$ be $ 2n$ points.

[b](a)[/b] If $ X_1X_2...X_n\approx Y_1Y_2...Y_n$ holds, but $ \overrightarrow{X_1Y_1}  =  \overrightarrow{X_2Y_2}  =  ...  =  \overrightarrow{X_nY_n}$ does [b]not[/b] hold, then there exists a point $ S$ in the plane such that $ SX_1Y_1\approx SX_2Y_2\approx ...\approx SX_nY_n$.

[b](b)[/b] If there exists a point $ S$ in the plane such that $ SX_1Y_1\approx SX_2Y_2\approx ...\approx SX_nY_n$, then $ X_1X_2...X_n\approx Y_1Y_2...Y_n$.[/color]

[i]Proof of Lemma 4.[/i] [b](b)[/b] From $ SX_1Y_1\approx SX_2Y_2\approx ...\approx SX_nY_n$, it follows that $ SX_iY_i\approx SX_jY_j$ for any two distinct integers $ i$ and $ j$ from the set $ \left\{1,2,...,n\right\}$, so that $ SX_iX_j\approx SY_iY_j$ (by Lemma 2), and thus $ \measuredangle SX_iX_j  =  \measuredangle SY_iY_j$; thus, for any three integers $ i$, $ j$ and $ k$ from the set $ \left\{1,2,...,n\right\}$, we have $ \measuredangle SX_jX_i  =  \measuredangle SY_jY_i$ and $ \measuredangle SX_jX_k  =  \measuredangle SY_jY_k$, and thus

$ \measuredangle X_iX_jX_k  =  \measuredangle SX_jX_k  -  \measuredangle SX_jX_i  =  \measuredangle SY_jY_k  -  \measuredangle SY_jY_i  =  \measuredangle Y_iY_jY_k$.

This "almost" yields that $ X_1X_2...X_n\approx Y_1Y_2...Y_n$ (I say "almost" because we have proven the angle condition, but not the degeneracy condition - but I am leaving the proof of the degeneracy condition to the reader). This proves Lemma 4 [b](b)[/b].

[b](a)[/b] Let $ S$ be the Miquel point of $ \left(X_1,X_2;Y_1,Y_2\right)$ (this Miquel point exists if $ X_1X_2Y_2Y_1$ is not a parallelogram; I let the reader show that the case of $ X_1X_2Y_2Y_1$ being a parallelogram is exactly the case when $ \overrightarrow{X_1Y_1}  =  \overrightarrow{X_2Y_2}  =  ...  =  \overrightarrow{X_nY_n}$, which we excluded in the statement of Lemma 3). Then, $ X_1SX_2\approx Y_1SY_2$. On the other hand, $ X_2X_kX_1\approx Y_2Y_kY_1$ for every $ k\in\left\{1,2,...,n\right\}$ with $ k\neq 1$ and $ k\neq 2$ (because $ X_1X_2...X_n\approx Y_1Y_2...Y_n$). Hence, Lemma 3 yields $ X_1SX_2X_k\approx Y_1SY_2Y_k$, so that $ SX_1X_k\approx SY_1Y_k$. Lemma 2 yields $ SX_1Y_1\approx SX_kY_k$. So we have shown that $ SX_1Y_1\approx SX_kY_k$ for every $ k\in\left\{1,2,...,n\right\}$ with $ k\neq 1$ and $ k\neq 2$. Also, $ SX_1Y_1\approx SY_1Y_2$ (this is just a rewriting of $ X_1SX_2\approx Y_1SY_2$). Thus, the point $ S$ satisfies $ SX_1Y_1\approx SX_2Y_2\approx ...\approx SX_nY_n$. This proves Lemma 4 [b](a)[/b].

Hence, both parts of Lemma 4 are now proven.

[color=blue][b]Fundamental theorem on directly similar polygons:[/b] If $ P_1$, $ P_2$, ..., $ P_n$, $ Q_1$, $ Q_2$, ..., $ Q_n$ are $ 2n$ points such that $ P_1P_2...P_n\approx Q_1Q_2...Q_n$, and if $ R_1$, $ R_2$, ..., $ R_n$ are points on the lines $ P_1Q_1$, $ P_2Q_2$, ..., $ P_nQ_n$ dividing the segments $ P_1Q_1$, $ P_2Q_2$, ..., $ P_nQ_n$ in the same ratio $ \frac{P_1R_1}{R_1Q_1}  =  \frac{P_2R_2}{R_2Q_2}  =  ...  =  \frac{P_nR_n}{R_nQ_n}$, then $ R_1R_2...R_n\approx P_1P_2...P_n$ and $ R_1R_2...R_n\approx Q_1Q_2...Q_n$.[/color]

[i]Proof.[/i] If $ \overrightarrow{P_1Q_1}  =  \overrightarrow{P_2Q_2}  =  ...  =  \overrightarrow{P_nQ_n}$, then the proof is easy. Hence, in the following we will only consider the case when the relation $ \overrightarrow{X_1Y_1}  =  \overrightarrow{X_2Y_2}  =  ...  =  \overrightarrow{X_nY_n}$ does [b]not[/b] hold. Then, using Lemma 4 [b](a)[/b], we can conclude from $ P_1P_2...P_n\approx Q_1Q_2...Q_n$ that there exists a point $ S$ in the plane such that $ SP_1Q_1\approx SP_2Q_2\approx ...\approx SP_nQ_n$. Hence, for any two distinct integers $ i$ and $ j$ from the set $ \left\{1,2,...,n\right\}$, we have $ SP_iQ_i\approx SP_jQ_j$. In other words, $ P_iSQ_i\approx P_jSQ_j$. But on the other hand, $ Q_iR_iP_i\approx Q_jR_jP_j$ because the points $ Q_i$, $ R_i$, $ P_i$ are collinear, the points $ Q_j$, $ R_j$, $ P_j$ are collinear and we have $ \frac{P_iR_i}{R_iQ_i}  =  \frac{P_jR_j}{R_jQ_j}$. Thus, Lemma 3 yields $ P_iSQ_iR_i\approx P_jSQ_jR_j$, so that $ SR_iP_i\approx SR_jP_j$.

So we have $ SR_iP_i\approx SR_jP_j$ for any two distinct integers $ i$ and $ j$ from the set $ \left\{1,2,...,n\right\}$. Thus, $ SR_1P_1\approx SR_2P_2\approx ...\approx SR_nP_n$. Hence, Lemma 4 [b](b)[/b] yields $ R_1R_2...R_n\approx P_1P_2...P_n$. Hence, $ R_1R_2...R_n\approx Q_1Q_2...Q_n$ (because $ P_1P_2...P_n\approx Q_1Q_2...Q_n$). Thus, the Fundamental theorem on directly similar polygons is proven.

Sorry for the length, the typos, the obscure notations and the use of results proven later, but I hope it was halfways understandable. It is not quite appropriate to speak of an "elementary" proof of the theorem without first specifying what one actually means by "similar". For instance, if one allows the use of similitude transformations, then the theorem becomes very easy, but then again, working with (direct) similitude transformations is the same as working with complex numbers, and you don't want this. The proof above more or less rewrites the basics of complex numbers algebra in elementary geometric terms.

  darij

[quote="Sailor"]From Menelaos it's enough to have:
$ \frac {B'Q}{B'H} \equal{} \frac {PA'}{A'H}$.

But $ \frac {B'H}{HA'} \equal{} \frac {\cos{A}}{\cos{B}}$.
Now we use tha fact that $ HQDP$ is cyclic to obtain that $ \frac {AB'}{B'Q} \equal{} \frac {A'B}{A'P}$ and since
$ \frac {AB'}{A'B} \equal{} \frac {\cos{A}}{\cos{B}}$ we are done![/quote]

denote H = PA ∩BQ

Is the Fundamental Theorem on directly similar polygons a well known one? Like can a person make use of it in a contest? (TO do so, what is the name of the theorem now?).
Here in this case, since it is the midpoints, can do we some homothety argument?

[i][color=#0000FF]Sketch of a solution using bashing[/color] [/i]---
WLOG , the circumcircle be the unit circle centred at origin.suppose $AA'$ cut's the circumcircle again at $A''$
then $a''=-bc/a$ . similarly , $b''=-ac/b$ . then
$p=\frac{a+c+bc-bc/a}{c+1}$ and $q=\frac{b+c+ac-ac/b}{c+1}$. note that , $\mu A'B'=-c^2$
now , note that $t=2a'-p-q=a+b+c-bc/a-\frac{ac+bc+a+b+2c-ac/b-bc/a}{c+1}$ .
now , it would be enough to show that , $\frac{t}{\overline{t}}=-c^2$ , which is a piece of not so lengthy calculations. :D  :blush:

$\angle B'AQ=\angle A'BP$ $\angle BA'H=\angle AB'H=\angle AB'Q=90$ and $\angle HAB'=\angle HBA'$ where $H=A'A\cap B'B$. so $B'QA\sim A'PB$ and $BA'H\sim B'AH$ so $B'Q/A'P=B'A/BA'=B'H/HA'$ now if $PQ\cap A'B'=T$ by Menelaus theorem on $HPQ$ $PT=TQ$

We may try to prove that $S_{A'B'P}=S_{A'B'Q}$, $S_{XYZ}$ meaning area of $XYZ$.
Let $X, Y$ be the projections of $A', B'$ onto $BB', AA'$ respectively, and we need to prove $QB'\cdot A'X=PA'\cdot B'Y\ (\ *\ )$.

From $\triangle AQB'\sim\triangle BPA'$ we get $\frac{QB'}{A'P}=\frac{AB'}{BA'}\ (\ 1\ )$, and from $\triangle AB'Y\sim\triangle BA'X$ we obtain $\frac{A'X}{B'Y}=\frac{A'B}{AB'}\ (\ 2\ )$; multiply $(1)\wedge (2)$ side by side and we get $(*)$, done.

Best regards,
sunken rock

[b]Notation:[/b] For two variable points $V,W$ moving on a pair (not necessarily distinct) of conics, we say that the map $V \rightarrow W$ is projective if and only if the cross-ratio of any four positions of $V$ is the same as that of the corresponding four positions of $W$.

[quote=grobber] Lines $AA',BB'$ are the altitudes in an acute angled triangle $ABC$. Point $D$ is chosen on the arc $ACB$ of the circumcircle of triangle $ABC$. If $P=AA'\cap BD,Q=BB'\cap AD$, show that the midpoint of $PQ$ lies on $A'B'$.[/quote]

[i]Proof.[/i] By perspectivity, the maps $P \rightarrow D$ and $D \rightarrow Q$ are projective. Thus, the composition, $P \rightarrow Q$ is also a projective map. Let $R$ be the reflection of $P$ in point $B'$ and $Q'$ be the point on line $AA'$ such that $RQ' \parallel A'B'$. Observe that the midpoint of $PQ'$ lies on line $A'B'$. Note that the maps $P \rightarrow R$ and $R \rightarrow Q'$ are projective, hence the map $P \rightarrow Q'$ is projective.

Let lines $AA'$ and $BB'$ meet the circumcircle of triangle $ABC$ again at $A_1$ and $B_1$ respectively. Lastly, observe that for $D=A_1,$ $D=B_1,$ and $D=C,$ we have $Q \equiv Q'$. As $Q \rightarrow Q'$ is projective, we conclude that $Q \equiv Q'$ for all positions of $D$ on the arc $ACB$ of the circumcircle. $\square$

[b]Remark:[/b] Observe how nicely the idea of a projectivity between two lines being uniquely determined by the images of three points on it made this problem into a trivial case-check.

In this kind of problem that we have 90
We should use similitude,cyclic.
And Menelaus.


Let $ a,b$ be positive real numbers.
Prove that
$ \frac {1}{1 + a^2} + \frac {1}{1 + b^2}\geq\frac {2}{1 + ab}$
if either
(1)
${ 0 < a,b\leq1}$
or
(2)
$ ab\geq3$

(1) try $ a \equal{} 0,001$ and $ b \equal{} 1,1$
we have $ LSH\minus{}RSH$=$ \minus{} 0,545...$
(2) We can prove the Inequality when $ ab \ge 1$. The condition $ ab \ge 3$ doesn't need


floor function
induction
graph theory
combinatorics proposed
combinatorics

For $n \in N$, call $K_{n}$ is complete graph of $n$ vetices. Find the largest integer $n$ which satisfies the following conditions.

1. We color each edge of $K_{n}$ by one of three colors: Blue, Red, Yellow

2. There is no monochromatic cycle of even length.

This is an interesting problem. I am pretty sure this has a very nice solution. (VMEO = Vietnamese Maths Olympiad, right?)

Haven't been able to solve it fully, but here is a partial solution, which shows that $7 \le n \le 9$

We can show that any simple graph on $n > 2$ vertices with more than $\lfloor \frac{3(n-1)}{2}\rfloor$ edges has a cycle of even length: If the graph is 2-connected, it has an ear decomposition comprised of at least one ear, giving an even cycle. If the graph is not 2-connected, then we can find a subgraph on $2 < m < n$ vertices which has more than $\lfloor \frac{3(m-1)}{2}\rfloor$. By induction, this would have an even cycle.

Now given any 3-colouring of $K_{10}$, at least one mononchromatic subgraph must contain at least 15 edges, and hence by above statement it must have an even cycle.

Thus $n \le 9$

Now we show that there is a three colouring of $K_{7}$ with no even cycles.

It is well known that we can decompose $K_{2r+1}$ into $r$ hamiltonian cycles. For $r=3$, we see that $K_{7}$ can be decomposed into 3 cycles of length 7. Colour each cycle with a different colour.

Thus $n \ge 7$

In fact the required answer is $n=9$.

Attached is a 3 colouring of $K_{9}$ which has no even cycles.

The colouring is 3 isomorphic graphs, which sort of looks like the nuclear radiation symbol.


geometry
incenter
trigonometry
inradius
geometry unsolved

Let $ AA_{1},BB_{1},CC_{1}$ are 3 altitudes of triangle $ ABC$ and $ O_{A},O_{B},O_{C}$ respectively be the incenter of triangle $ AB_{1}C_{1},BC_{1}A_{1},CA_{1}B_{1}$. Call $ T_{A},T_{B},T_{C}$ respectively be the point of tangency of incircle of triangle $ ABC$ with 3 sides $ BC,CA,AB$. Prove that all side of hexagon $ T_{A}O_{C}T_{B}O_{A}T_{C}O_{B}$ are equal

the quad $ CBC_{1}B_{1}$ is cyclic so $ \angle AC_{1}B_{1}\equal{} 90\minus{}(90\minus{}\gamma) \equal{}\gamma$ etc.
for the similitude with $ \triangle AB_{1}C_{1}$ and $ \triangle ABC$ we have that $ AO_{A}\equal{} AI\cdot\cos{\alpha}$ and so

$ IO_{A}\equal{} AI\cdot (1\minus{}\cos{\alpha})$ and $ IT_{C}\equal{} AI\sin{\frac{\alpha}{2}}$ so $ \triangle IO_{A}T_{C}$ is isosceles ($ T_{C}I \equal{} T_{C}O_{A}$) couse $ OT_{C}\cdot 2\sin{\frac{\alpha}{2}}\equal{} AI\cdot 2\sin^{2}{\frac{\alpha}{2}}\equal{} AI\cdot (1\minus{}\cos{\alpha}) \equal{} IO_{A}$

equally for the others triangles we prove that the side are equal  :wink:

Let $ X$ be the point of tangency of triangle $ AB_{1}C_{1}$ with side $ AC_{1}$. Since $ \triangle AB_{1}C_{1}$and$ \triangle ABC$are inversely similar and$ X,T_{B}$are

corresponding point,$ \frac{XC_{1}}{AX}=\frac{T_{B}C}{AT_{B}}$. Therefore,$ XT_{B}$ and $ C_{1}C$are parallel, But $ XO_{A},C_{1}C$ are parallel. So$ X,O_{A},T_{B}$are collinear. Let$ I$be the

incenter of$ \triangle ABC$which lie on$ AO_{A}$and line that pass$ T_{B}$and perpendicular to$ CA$. Since$ IA$bisect$ \angle A$, $ -\angle AO_{A}X =\angle AIT_{B}$.

 But$ -\angle AO_{A}X =\angle T_{B}O_{A}I$. Therefore, $ T_{B}O_{A}= T_{B}I = r$,where r is inradius of $ \triangle ABC$. Similarly, all side of $ T_{A}O_{C}T_{B}O_{A}T_{C}O_{B}$ are r. Therefore, proven


I heard from some of my relative a couple months ago that India was going to pass a law or something that allowed dual citizenship.  Does anyone know what's going on with that right now?

Well if I  find some time, I may comment here but there is lot of discussion in various india-us web forums.
Here is the link which  gives the detaisl:
[url=http://www.indiacgny.org/php/showContent.php?linkid=355]Link[/url]

i dont think they have put it in to action.  i heard about that to.  its already in some countries, like Austrailia, I think


trigonometry
function
algebra
domain
geometry
circumcircle
floor function

how would you go about finding the number of solutions to $x=y\sin{x}$ for $y>0$ and $x\ge0$?

There's infinite solutions because you can always add a multiple of $2\pi$ to $x$, and it will still satisfy the given equation. :lol: What's the interval of $x$??

no
i know thats not right. mayby if i state it this way
$f(x)=x$
$g(x)=asinx$
find the number of times that $f(x)=g(x)$ in terms of $a$
as you notice $f(x)$ is always increasing and $g(x)$ is between $-a$ and $a$. so we know there is a set amount of solutions.

hope that clerified thins.

[quote="maokid7"]$f(x)=x$
$g(x)=a\sin x$
Find the number of times that $f(x)=g(x)$ in terms of $a$.[/quote]

You can do it given some $a$, but it's difficult to generalize, I think.

graph it...there is no other good way to do it, also you know that if x is a solution, then -x is a solution because the function is odd
but you might not want to have the domain of x being x>=0 because if x=0, then we have that

0=ysin{0}
0=0
so there are infinite solutions since any y value can accompany x=0

yeah i know.
here let me make it more specific.

For how many real values does $x=2310sinx$
i would think about graphing but it would be relly tedious.

[quote="chess64"][hide]We need to find the number of solutions to \[ x=2005\sin x.  \]

[hide="possible approach maybe?"]
We can always take a geometric approach and divide by $\sin x$. We get \[ \frac{x}{\sin x} = 2005  \] and by extended Law of Sines, we have \[ R=1002.5  \] where $R$ is the circumradius of a triangle with one side $x$. How many ways can the circumradius of a triangle be equal to $1002.5$? $\infty$, maybe? I don't know... :(
[/hide]

Anyway, if we graph $y=x$ and $y=2005\sin x$, we get something like

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=367[/img]

It's pretty clear from this that there are infinite solutions...[/hide][/quote]

The law of sines is $2R=\frac{a}{\sin A}$, not $2R=\frac{a}{\sin a}$. You find the sine of the angle, not of the side length.

The graph $y=2005\sin x$ is bounded by the lines $y=\pm 2005$. Therefore, all solutions are in the interval $x\in [-2005,2005]$.

I think that you should approach this in a different way.
we have:
$x = y \sin x$
$1 = \frac{y \sin x}{x}$
$\frac 1y = \frac{\sin x}{x}$
$y = \frac{x}{\sin x}$
$y = x\csc x$

now i think we can see that there are an infinite number of solutions. This is because this is a function. Now, if $y$ is a constant:

$y = x \csc x$

now, because $y > 0$ and $x > 0$, we only think about positive solutions.
Now, $\csc x$ by itself has a minimum of $1$ and a maximum of $\infty$ but because it's being multiplied by $x$, we know that eventually the minimun of one "bump" of the graph will exceed $y$. Therefore, we can generalize a formula for the number of solutions, but it will not be $\infty$ if $y$ is a constant.

(I think it's something like $\lfloor \frac y\pi \rfloor$ but I'm just guessing. I know that you'll have to round and $\pi$ will be in it somewhere but I don't know where).


Hi everyone,

We'd like to wish a Happy New Year for our Asian Members (even if it's a little bit late :blush:)  and wish you the best in the upcoming Rooster Year :)
We also wanted to give greetings for those of you celebrating the Luna Holiday (We are sorry if that's just another name for the New Year's celebration and we hope you'll straighten us out).

Greetings,
The AoPS-MathLinks Team

[quote="Valentin Vornicu"][quote="LuCy4EvA"][size=150]hApPy nEw YeAr!!!![/size]

why do u say it's late, it's today in US[/quote]It's not "today" anymore in China :( ... well better a little late then never :)[/quote] ooo, gotta get my dates straight...

[size=200][color=red]新[/color][color=yellow]年[/color][color=red]快[/color][color=yellow]?[/color]??? ?大家一年比一年?明???[/size]

Happy New Years. It's still the 9th in the US so it's technically not late for us.

I'm a korean but koreans celebrate Chinese new year

I just ate the celebrations meal and talked with my grandma and my grandpa

HAPPY NEW YEAR TO EVERYONE!!!!

by the way, Lucy, how much money did you get for chinese new year?
my friend got 500 dollars or something..

Happy new year ! :D .
Money in red bags :D ,thank "The AoPS-MathLinks Team "very much!  :)

[quote="liyi"][quote="LuCy4EvA"]wow, 500 is alot... especially dollars, but when I was in China, I used to get like 1000 yuan from like my parents, and others give like 100 yuan each.... miss the good old days... but I got like 100 something dollars this year.... which is pretty good seeing that I didn't get anything teh last two years...[/quote]

I don't have it for many years...[/quote]

Don't worry liyi, you're not alone :)

[quote="Arwen"]I think when  you cannot get money from the adults mean that you are already a adult.it's a wonderful thing, isn't it? I always wanna be a grown-up. but families still give money to me, maybe they forget I am a senior high school student. too bad:?  :([/quote]

are you conplain because you get money??? wow, I would never complain

[quote="tokenadult"]Happy Chinese New Year.

There's a [url=http://www.taipeitimes.com/News/taiwan/archives/2005/02/13/2003222920]news story[/url] today about Chinese New Year in foreign countries with a rather odd joke at the end.[/quote]
That is weird  :?

[size=75][color=orange]:rotfl:

well actually....... i'm not quite sure what to make of this story >_<[/color][/size]

HAPPY CHINESE NEW YEAR EVERYONE....

i don't get money anymore... really sad about that...
before i came to the US...
people gave me money every place i go...for the chinese new year..
200 from this person... 500 from another... but the best i got was 8000....
(all in chinese of course..)
I never got to keep the money though... i handed them all to my mom..
i regret it now.. but in *those days* money meant nothing to me... after all.. i did get everything i wanted....

Gosh i miss it... i wish my parents never decided to stay here... but o well.. can't do anythin now...

Happy Chinese New Year!!  Well I never gotten any money from my parents in my life

Ya I wish I can go back to China some time...

Happy New Year, though it is too late to say it.
Anyway, anyone here has red pocket to give us :D

Now, the 2006 Lunar New Year is coming, so fast....
well, the memory is so great, this should me great time for us to sum up our experiences in the past year...
Personally, I moved from Shanghai to Vancouver, a distance of more than 10 thousand miles, a good 10-hour flight with the fastest civil plane, and also experienced a lot...
And the age of MathLinks is approaching 4, i guess, am i right?
Happy new (lunar) year to everybody...

大家新年好！万事如意！心想事成！

狗年快乐，想想大家相逢在mathlinks也是一种缘分 :)

[quote="Kalimdor"]And the age of MathLinks is approaching 4, i guess, am i right?
Happy new (lunar) year to everybody...[/quote]Actually we are going to make 3 years in February right now :) Happy Lunar Year! :P

PS Since this is an old thread, a new thread has been created: http://www.mathlinks.ro/Forum/viewtopic.php?t=72407 Please give your cheers there! :P


Let $ ABCDEFGHILMN$ be a regular dodecagon, let $ P$ be the intersection point of the diagonals $ AF$ and $ DH$. Let $ S$ be the circle which passes through $ A$ and $ H$, and which has the same radius of the circumcircle of the dodecagon, but is different from the circumcircle of the dodecagon. Prove that:
1. $ P$ lies on $ S$
2. the center of $ S$ lies on the diagonal $ HN$
3. the length of $ PE$ equals the length of the side of the dodecagon

An ugly solution  :) .
Easy to prove that :
$ \angle{ADH}\equal{}\frac{5.\pi }{12}$
$ \angle{DAP}\equal{}\frac{\pi}{6}$
Therefore
\[ \angle{ADP}\equal{}\frac{5\pi}{12}
\]
It gives $ AD\equal{}AP\equal{}AL$
Similar we can show that  : $ HP\equal{}HL$
$ O_1,L$ is symmetric of $ O$ and $ P$ through line $ AH$
Therefore $ O_1P\equal{}OL\equal{}R$ , means that $ P$ lies on circle S.
b) It is quite easy from $ \angle{AHO_1}\equal{}\frac{\pi}{24}\equal{}\angle{AHN}$ so $ P$ lies on HN.

c) Easy to check that $ (AH,LE)\equal{}\frac{\pi}{2}$ . So $ AH\perp LE$
Therefore $ L,P,E$ are collinear .
$ \angle {DPE}\equal{}\angle{PDE}\equal{}\frac{\pi}{4}$
So $ PE\equal{}DE$
Our problem claim .

$P$ lies on $S$ means that $\angle AKH=\angle APH$ because $K$ lies on minor arc $AH$.
By symmetry, $AK||DH$ and $KH||AF$. So $\angle AKH=\angle APH$.

Let's call the center of $S$ $J$.
$J$ is the reflection of $P$ across $AH$, so $APHJ$ is a rhombus. $B$, $P$, and $H$ are on a line parallel to $AI$ by symmetry. $PH||AJ$ means that $J$ lies on diagonal $AI$.

PEF is equilateral
https://en.wikipedia.org/wiki/Rhombitrihexagonal_tiling