geometry
trapezoid
parallelogram
circumcircle
geometry proposed

[color=darkred]Let $ABCD\ (AD\parallel BC)$ be an isosceles trapezoid inscribed in the circle $w(O)$. Consider a fixed point $K\in w$ so that the line $BC$ separates the points $A$ and $K$. Consider a variable parallelogram $BCEF$ so that $BF\parallel CE\parallel AK$. Define the second intersections $B'$, $C'$ of the lines $AE$, $AF$ respectively with the circle $w$. Prove that the geometrical locus of the intersection $L\in BB'\cap CC'$ is the line $KD$.

[b][u]Particular case[/u][/b] :[/color] $AK\perp BC$, i.e. $O\in KD$ $\Longrightarrow$

http://www.mathlinks.ro/Forum/viewtopic.php?t=122340

[hide="Proof."]
[color=darkred][b][u]Lemm[/u][/b] (prove easily). Let $ABCD$ be a cyclical convex quadrilateral for which denote

$\{\begin{array}{c}AB=a\ ,\ BC=b\ ,\ CD=c\ ,\ DA=d\\\ AC=e\ ,\ BD=f\ ,\ I\in AC\cap BD\end{array}$. Then

$\frac{IA}{da}=\frac{IB}{ab}=\frac{IC}{bc}=\frac{ID}{cd}=\frac{e}{ad+bc}=\frac{f}{ab+cd}=\sqrt{\frac{-p(I)}{abcd}}$, where

$p(I)$ is the power of the point $I$ w.r.t. the circumcircle of the given quadrilateral.[/color]

[color=darkblue][b][u]Proof of the proposed problem.[/u][/b] From $BF\parallel AK\parallel CE$ obtain :

$\begin{array}{ccc}\{\begin{array}{c}\widehat{BAF}\equiv\widehat{BAC'}\equiv\widehat{C'KB}\\\ \widehat{BFA}\equiv\widehat{KAC'}\equiv\widehat{C'BK}\end{array}\| & \Longleftrightarrow & \triangle ABF\sim\triangle KC'B\\\\ \{\begin{array}{ccc}\widehat{CAE}\equiv\widehat{CAB'}\equiv\widehat{B'KC}\\\ \widehat{CEA}\equiv\widehat{KAB'}\equiv\widehat{B'CK}\end{array}\| & \Longleftrightarrow & \triangle ACE\sim\triangle KB'C\end{array}$.

Thus, $\{\begin{array}{ccc}AD\parallel BC & \Longrightarrow & DB=AC\ ,\ DC=AB\\\\ \triangle ABF\sim \triangle KC'B & \Longrightarrow & \frac{C'B}{C'K}=\frac{BF}{BA}\ \ (1)\\\\ \triangle ACE\sim\triangle KB'C & \Longrightarrow & \frac{B'K}{B'C}=\frac{CA}{CE}\ \ (2)\end{array}$.

Therefore, $BB'\cap CC'\cap KD\ne\emptyset$ $\Longleftrightarrow$ $\frac{C'B}{C'K}\cdot\frac{B'K}{B'C}\cdot\frac{DC}{DB}=1$,

what is truly because $(1)$, $(2)$, $BF=CE$ and $\frac{DC}{DB}=\frac{AB}{AC}$.

[b][u]Remark.[/u][/b] Denote $\{\begin{array}{c}S\in BC\cap AK\ ,\ T\in BC\cap DK\\\ S'\in (BC)\ ,\ S'B=SC\end{array}$. From the above lemma,

$\{\begin{array}{ccc}(ABKC) & \Longrightarrow & \frac{SB}{SC}=\frac{BA}{CA}\cdot\frac{BK}{CK}\\\\ (DBKC) & \Longrightarrow & \frac{TB}{TC}=\frac{BD}{CD}\cdot \frac{BK}{CK}\end{array}\|$ $\Longrightarrow$ $\boxed{\ \frac{TB}{TC}=(\frac{AC}{AB})^{2}\cdot\frac{SB}{SC}\ }$.

Thus, $\frac{TB}{TC}\cdot\frac{S'B}{S'C}=(\frac{AC}{AB})^{2}\cdot \frac{SB}{SC}\cdot \frac{SC}{SB}=(\frac{DB}{DC})^{2}$, i.e.

[b][u]the rays[/u][/b] $[\overline{DTK}$ and $[DS'$ [b][u]are isogonally in the angle[/u][/b] $\widehat{BDC}$ [b][u]![/u][/b]

[/color][/hide]

Thank you dear Virgil, for your beautiful generalization of the problem I presented three months ago
and also for the dedication (specially honorific to me).

We can also use the same methods about the solution ( as in the particular case of $AD\perp BC,$
http://www.mathlinks.ro/Forum/viewtopic.php?t=122340 )
and here your method with respect to inscribed hexagon, works again better, as the simplest of all.

Best regards, Kostas Vittas.

[color=darkblue]Thank you for the your drawing ! I liked very much the your problem. The my extension is easily. I observed after all that $AD\parallel BF$ is essentially and not $AD\perp BC$. [b]Thank you ![/b][/color]

[quote="Virgil Nicula"][color=darkblue]Let $ABCD\ (AD\parallel BC)$ be an isosceles trapezoid inscribed in the circle $w(O)$. Consider a fixed point $K\in w$ so that the line $BC$ separates the points $A$ and $K$. Consider a variable parallelogram $BCEF$ so that $BF\parallel CE\parallel AK$. Define the second intersections $B'$, $C'$ of the lines $AE$, $AF$ respectively with the circle $w$. Prove that the geometrical locus of the intersection $L\in BB'\cap CC'$ is the line $KD$.[/color][/quote]

[size=117][color=darkred][b]A "strong" dedication ( to [u]Vittasko[/u] ) :[/b][/color][/size]

$\boxed{\left\{\begin{array}{c}M\in AC\cap BD\\\ I\in BB'\cap CC'\\\ N\in AB'\cap DC'\end{array}\right\|\Longrightarrow I\in MN\ .}$

[color=darkred][b]Remark.[/b] $P\in AK\cap BC'\Longrightarrow P\in \overline{IMN}\ .$[/color]

Thank you dear Virgil, for a nice new result (collinearity), in this problem.

I see now, that we must change the method about the solution. We can show the collinearity of $M\equiv AC\cap BD,$ $N\equiv AB'\cap DC'$ and $L\equiv BB'\cap CC',$ by Pascal’s theorem, from the non-convex inscribed hexagon $ACC'DBB'.$

Also the collinearity of $M,$ $N,$ $P\equiv AK\cap BC',$ by Desarques’s theorem, from the perspective triangles $\bigtriangleup DBC',$ $\bigtriangleup ACQ,$ where $Q,$ is the intersection point of $AE,$ from the line through the point $C'$ and parallel to $BC$ $($ It is easy to show that the point $Q,$ lies on $CP$ $).$

I will post here later (after the work day) more details, about all the above.

Best regards, Kostas Vittas.

[color=darkblue]Yes, it is and my proof. I wanted to define more accurately the position of the point $I$ w.r.t. the given informations from  the hypothesis. Have a good day ![/color]

[quote="vittasko"]
Also the collinearity of $M,$ $N,$ $P\equiv AK\cap BC',$ by Desarques’s theorem, from the perspective triangles $\bigtriangleup DBC',$ $\bigtriangleup ACQ,$ where $Q,$ is the intersection point of $AE,$ from the line through the point $C'$ and parallel to $BC$ $($ It is easy to show that the point $Q,$ lies on $CP$ $).$[/quote]

From $C'Q\parallel BC$ $\Longrightarrow$ $\frac{QA}{QE}= \frac{C'A}{C'F}= \frac{AP}{BF}=\frac{AP}{CE}$ $\Longrightarrow$ $P,$ $Q,$ $C,$ collinear points.

So, because of the triangles $\bigtriangleup ACQ,$ $\bigtriangleup DBC',$ are perspective $($ from $AD\parallel CB\parallel QC'$ $),$ we have that the points $M\equiv AC\cap DB,$ $N\equiv AQ\cap DC'$ and $P\equiv CQ\cap BC',$ are collinear.

Kostas Vittas.


Problems from Tournament of The Towns 2004 Senior A Level:

#3

The results of operations

,     ,       ,

are written on four cards that are placed on a table in random order.  Prove that one can restore both x and y given that x and y are positive numbers.

#4

A circle with the center I is entirely inside of a circle with center O.  Consider all possible chords AB of the larger circle which are tangent to the smaller one.  Find the locus of the centers of the circles circumscribed about the triangle AIB.

#5

There are two rectangles A and B.  It is known that one can tile a rectangle similar to B using copies of A.  Prove that one can tile a rectangle similar to A using copies of B.

My proof:   Can anyone check it?

[hide]Let the sides of A be a and b, the sides of B be x and y.

By the problem,

         , where m, n are positive integers and k is a positive real.

Now:

we can rewrite it as:

           , where

This implies using copies of B one can tile a rectangle similar to A since n and m are integers.[/hide]

#6

Let n be an integer divisible by neither 2 or 3.  Let us call a triangle admissible if all its angles are in the form  where m is an integer.  Triangles which are not similar we call essentially different.

In the beginning there is one admissible triangle.  The following procedure is applied:  we chose a triangle the set obtained on the previous stage and cut it into two admissible triangles so that all the triangles in the new set are essentially different.

This procedure repeats itself until it is possible.  Prove that in the end we get all possible admissible triangles.

Pretty sure about 5, not sure about 6.  Both are the ones I did on the contest, but 6 I did in the last 30 mins so I dont know.

#5.
We need A and B as 1,p and 1,q for convenience.

A, B has dimensions a,b ; x,y.  If a,b tiles kx,ky for some k, then surely 1/q, b/aq tiles kx/aq, ky/aq.  So wlog A has dimensions 1,p.
If 1,p tiles (1,q) then it also tiles kx,ky.  So wlog B has dimensions 1,q.

Now take p/q rational.  Then we can clearly exhibit a tiling of (n,nq) with (1,p) tiles if we can find k such that kp = nq, for integers k, n.  But p/q is rational.  So we can exhibit such a tiling and similarily for a tiling of (m,mp) by (1,q).

Take p/q irrational.  Then a tiling of (n,nq) by (1,p) is composed of either 1 components or p components.  But a multiple of 1 doesnt go into a multiple of p.  Thus all tiles must be oriented in the same way.  Therefore, we cannot have a tiling (because using the reasoning above for when p/q was rational, we had to find kq=nq with integer soln, but we know that is impossible when p/q is irrational.)

But p/q is either rational or irrational.  So we are done.

Too lazy to type up number 6 today.

6.

k = 180/n, if I write (a,b,c) I actually mean the tuple ka,kb,kc.  Note a+b+c = n.

Let (a,b,c) be angles of an admissible triangle ABC.  If we cut it at vertice A, with an angle x, we get two new triangles, simply (x, b, a+c-x), and (a-x,c,b+x).  Call these subtraingles of ABC tri-1 and tri-2 (wrt to ABC).

So take the triangle ABC and observe tri-1 to get (x,b, n-x-b).  With this triangle, you can repeat the cut to get (y,b,n-y-b) for a new variable y.  So it is clear that given a triangle (a,b,c) we can get (x,b,n-b-x) for all x (call this *).

Take the triangle ABC and observe tri-2 to get some new triangle, a-x, c, b+x.  If we had cut at some other vertice, we essentially change the tuple (from now on) to (b+x,a-x,c).  If we observe the tri-2 of this triangle, we get (b+x-y, c, a-x+y).  Shifting again, we get (a-x+y, b+x-y, c).  Now it is possible given Q, to generate a tuple (P, Q, R).  But using * once we have generated one tuple with a Q, we have generated all.

Now we are done. (??)

ps. this is what i wrote with 30 mins left on the contest so its gonna be probably bad.  please check it


1.
If $ f(x)\equal{}\sqrt[3]{\frac{9}{\log_2 (3\minus{}2x)}\minus{}1}$, determine the value of $ a$ that satisfies $ f^{\minus{}1}(2a\minus{}4)\equal{}\frac{1}{2}$.

2.
Given $ f(x)\equal{}2x\minus{}1$ dan $ h(x)\equal{}1\minus{}3x$. If $ (f \text{ o } g \text{ o } h)^{\minus{}1}(x)\equal{}\frac{46x}{15x\plus{}3}\minus{}3$, determine the formula of $ g^{\minus{}1}(x)$.

[hide="Hint"]If $ f(x)\equal{}a$ then $ f^{\minus{}1}(a)\equal{}x$

$ (fog)^{\minus{}1}(a)\equal{}(g^{\minus{}1}of^{\minus{}1})(a)\equal{}x$[/hide]

[quote="silvergrasshopper"]1.
If $ f(x) \equal{} \sqrt [3]{\frac {9}{\log_2 (3 \minus{} 2x)} \minus{} 1}$, determine the value of $ a$ that satisfies $ f^{ \minus{} 1}(2a \minus{} 4) \equal{} \frac {1}{2}$.

2.
Given $ f(x) \equal{} 2x \minus{} 1$ dan $ h(x) \equal{} 1 \minus{} 3x$. If $ (f \text{ o } g \text{ o } h)^{ \minus{} 1}(x) \equal{} \frac {46x}{15x \plus{} 3} \minus{} 3$, determine the formula of $ g^{ \minus{} 1}(x)$.[/quote]
[hide="1"]Since $ f^{\minus{}1}(2a\minus{}4)\equal{}\frac{1}{2}$ this means that $ f(\frac{1}{2})\equal{}2a\minus{}4$. So substituting in $ \frac{1}{2}$ for x we find that $ f(\frac{1}{2})\equal{}2$. Thus $ 2a\minus{}4\equal{}2$ so $ a\equal{}3$[/hide]

For f o g (x), is it f(g(x)) or g(f(x))
and its inverse isnt that clear to me as well.
Help appreciated. :)


trigonometry
trig identities
complex numbers

Hello,

I want to know if there a specific part to calculate such stuff like this $ \tan 45\plus{}\cos 180\equal{}\sin 180$ I worked this by you know using a calculator. I hope you get my idea  :maybe:

[quote="hasan4444"]Hello,

I want to know if there a specific part to calculate such stuff like this $ \tan 45 \plus{} \cos 180 \equal{} \sin 180$ I worked this by you know using a calculator. I hope you get my idea  :maybe:[/quote]
Those are very well known trig identities, which you are supposed to have memorized  :wink:
$ \tan 45 \equal{} 1 , \cos180 \equal{} 0, \sin180 \equal{} 1$. You can learn all this by using 30-60-90 and 45-45-90 triangles and an unit circle

Yes I know that they are famous but I mean like other unclear things like $ \tan 35 \left(\sin 10\plus{}\cos 10\right)\equal{}\cos 10 \minus{} \sin 10$

[quote="hasan4444"]Yes I know that they are famous but I mean like other unclear things like $ \tan 35 \left(\sin 10 \plus{} \cos 10\right) \equal{} \cos 10 \minus{} \sin 10$[/quote]
unless you can somehow manipulate them, Ya use a calculator.. But those questions aren't asked too often, because all they do is test if you know how to use the trig buttons on a calc iunno

[quote="hasan4444"]I mean if you are in a contest without a claculator and they say prove it  :lol:[/quote]
Then it can be proven using trig identities  :roll: and a bit of insight if it is harder competitions. But the best way to deal with trig is complex numbers.

[quote="hasan4444"]OK can you do an example for me?
Thanks[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=392631#392631
First post has a trig bash, last has a complex number interpretation.. As you can see, the complex numbers are easier  :) This may not be exactly what you asked for, but it will be similar.

[quote="Complex_Ninja"]$ \cos180 \equal{} 0, \sin180 \equal{} 1$[/quote]

$ \cos 180\equal{}\cos (90\plus{}90)\equal{}\cos 90 \cos 90\minus{}\sin 90 \sin 90\equal{}\minus{}1$

$ \sin 180 \equal{}\sin (90\plus{}90)\equal{}\sin 90 \cos 90 \plus{}\sin 90 \cos 90\equal{}0$

That's cool can it work with this "$ \tan 35\left(\sin 10\plus{}\cos 10\right)\equal{}\cos 10\minus{}\sin 10$"  :?:

[quote="vhagiro tiko"]I think : tan 35 \left(\sin 10+\cos 10\right)=\cos 10 - \sin 10

[url=http://productreviewsby.me/cellulite-removal/]Cellulite[/url][/quote]
Spam right?
I prefere no one enter his site


Find all natural numbers $ n$,such that there exist $ x_1,x_2,\dots,x_{n\plus{}1}\in\mathbb{N}$,such that
$ \frac{1}{x_1^2}\plus{}\frac{1}{x_2^2}\plus{}\dots\plus{}\frac{1}{x_n^2}\equal{}\frac{n\plus{}1}{x_{n\plus{}1}^2}$.

Easily we can prove that if for $ n$ there exist these numbers then there exist for $ n \plus{} 1$ too :
$ \frac {1}{x_{1}^2} \plus{} \cdots \plus{} \frac {1}{x_{n}^2} \equal{} \frac {n \plus{} 1}{x_{n \plus{} 1}^2}$
then  $ \frac {1}{x_{1}^2} \plus{} \cdots \plus{} \frac {1}{x_{n}^2} \plus{} \frac {1}{x_{n \plus{} 1}^2} \equal{} \frac {n \plus{} 2}{x_{n \plus{} 1}^2}$ .
now for the cases $ n \equal{} 1,2$ we haven't any solution .
but for $ n \equal{} 3$ we can choose $ x_{1} \equal{} 3$ , $ x_{2} \equal{} 3$ , $ x_{3} \equal{} 6$ and $ x_{4} \equal{} 4$
so for all $ n\geq 3$ this equation has a solution and the proof is finished .


A set of 100 cubes, each 1 cubic unit, are stacked to form a rectangular prism. What is the number of square units in the least possible surface area of the prism?

The cube root of $ 100$ lies between $ 4$ and $ 5$. Thus, you look for a possibility of getting $ 100$ multiplying three of only $ 4$s and $ 5$s. This can be done, $ 4\times5\times5$. Thus the surface area is $ 4*5*2\plus{}4*5*2\plus{}5^2*2\equal{}5(2(8)\plus{}10)\equal{}\fbox{130}$.


topology
complex numbers
real analysis
real analysis unsolved

Is there a homeomorphism: $ [0,\plus{}\infty)\times [0,\plus{}\infty)\to  R\times [0,\plus{}\infty)$?

I'm not entirely certain which topology is being imposed, but I think that you can remove the point $ (0,0)$ from the first one and it remains connected, yet there is no point from the second that you can remove in which the space stays connected. So they are not homeomorphic.

Um, what?  How exactly do you propose to disconnect $ {\bf R}\times [0,\infty)$ by removing a point?

How about $ (a, b)\mapsto (a^{2}\minus{}b^{2}, 2ab)$?  (That is, make things into complex numbers, square them, and then make them back into real numbers again.)

Along these lines: the Cartesian product of manifolds-with-boundary is a manifold-with-boundary.

Wait, just so I know I'm not going crazy. If you were using the dictionary order you would always disconnect that second space by removing a point, right?

"Disconnect" in the sense of order, yes. But the product topology has nothing to do with the dictionary order.


it seems INMO 2005 problem 1 is from olympiad of spain 2004? dont indians make their own geo. problems?

May be they felt a pinch of pity for the candidates :rotfl:

It is not a pity,rather it is injustice  :(
Though i don't think many know the source before exam.
In my opinion,it applies to many problems in INMO(one example is prev year(2008) combinatorocs prob,abt colouring)
but they have some beautiful indigeneous problems also,and some of indian proposals have alos been in IMO SL and IMO itself.  :rotfl:

I meant if someone had,by chance,practised Spanish problems
He/she definetely had the upper hand that year :rotfl:

also INMO 2003 problem 1 seems trivial with harmonic conjuation. but i see a long proof with trig.


function
topology
complex analysis
complex analysis unsolved

Let $ A\subseteq\mathbb{C}$ be an open set and $ (g_{n}) ,n = 1,2,3..$ a [b]normal[/b] family of analytic functions $ g_{n}: A\rightarrow\mathbb{C}$ and ${ g}: A\rightarrow\mathbb{C}$ an analytic function such that
$ Reg_n(z)\rightarrow Reg(z)$ for all $ z\in A$.

Prove that $ g'_{n}\rightarrow g'$ uniformly on the compact subsets of $ A$

Two questions:
- Do you mean by "normal sequence" a normally convergent sequence or a normal family of analytic functions?
- Do you assume that the analytic function $ g$ already exists? Or is it meant that the sequence $ (Re(g_n))$ converges pointwise to some function $ v: A\rightarrow \mathbb R$?

Sorry for the notation  :blush:  I edited my post
I mean a normal family and g already exists and is analytic

[hide="Hint"]Wlog we can assume that $ Re(g)\equiv 0$ - if not then consider the normal family $ \{g_n \minus{} g\}$ instead.

[b]1. Step:[/b] Assume that $ g_n$ converges normally on $ A$ to $ g$. Then $ g_n^\prime$ converges normally to $ g'$. By the open mapping theorem, $ Re(g) \equal{} 0$ implies that $ g$ is locally constant, such that $ g'\equiv 0$. Hence $ g_n^\prime$ converges normally to 0.

[b]2. Step:[/b] For the general case use Step 1 and the normality of $ \{g_n^\prime\}$.[/hide]


probability
expected value
advanced fields
advanced fields unsolved

Hello,

I have trouble solving these problems.

Problem 1:

A producer of fluorescent tubes has learned from experience that for tubes of a certain type, A, the number of hours this tube can burn before it should be renewed is a normal distributed random variable with a mean, μ, of 9,600 hours and a standard deviation, σ, of 1,200 hours.

A chain of shops has recently opened a new supermarket, in which 40 fluorescent tubes of type A have been installed.

Question1: What is the probability that a tube is to be renewed before it has burnt for 7,500 hours?

Question2: What is the probability that none of the 40 tubes are to be renewed before having burnt for 7,500 hours?

Question3: What is the probability that the average time of burning of the 40 tubes exceeds 10,000 hours?

Problem 2:

A company has moved to new premises and they would like to know if this has resulted in an increased number of daily internal telephone calls.

At the old premisses the daily number of calls could be described by a normal distribution with a mean of 110. For 5% of the days, the number of calls was either less than 80 or abvove 140.

Question 1: Calculate the standard deviation.

On the basis of 16 daily registrations of the internal calls at the new premisses, a daily average of 120 calls was found with the same standard deviation as before the removal.

Question 2: Can an increase in the number of calls be concluded, when the probability of postulating an increase must not exceed 5%, if in fact there is not any?

Question 3: Construct a 90% confidence interval for the expected number of calls.

Question 4: With reference for the requirement in question 2, what is the probability of concluding an increase, if the number of calls has in fact increased by 10% after the removal?

Thanks

P.S.: Is this problem really even to hard for guys  in this forum. Man I must have got a real difficult one then, thanks anyway.

Please help with this one. A brother cant figure it out.

Stank you


geometry
circumcircle
calculus
trigonometry
integration
symmetry
real analysis

Suppose there are n (n>2) people and each one of them stands on an vertex of a regular n-gon (with circumradius 1). Then all of them simultaneously starts to walk towards the person who stands on the adjacent vertex on the right with same speed. How long does each person have to walk so that everyone meets one another? (Note they are allowed to go to the interior of the polygon)

My calculus suggests $ \frac{1}{\cos(\pi/n)}$, but my computer program suggests $ \frac{1}{\sin(\pi/n)}$

Solution with no integral signs and no differential equations:

At every moment, the configuration of the $ n$ people remains a regular $ n$-gon. A such, the angle between paths remains the same. Person 1 walks toward person 2 at a speed of $ 1.$ Person 2 moves obliquely away from person 1; the component of person 2's motion in the direction of person 1's travel is $ \cos\left(\frac{2\pi}{n}\right).$ Hence the rate at which the distance between these two people is decreasing is a constant $ 1-\cos\left(\frac{2\pi}{n}\right).$

The original distance between two adjacent people is the side length of the $ n$-gon, which is $ 2\sin\left(\frac{\pi}{n}\right).$

Elapsed time $ =\frac{2\sin\left(\frac{\pi}{n}\right)}{1-\cos\left(\frac{2\pi}{n}\right)}=\frac{2\sin\left(\frac{\pi}{n}\right)}{2\sin^{2}\left(\frac{\pi}{n}\right)}=\frac{1}{\sin\left(\frac{\pi}{n}\right)}.$

The elapsed time is also the the distance each one travels. By symmetry, the point at which they meet is the center.

Alternatively: we know that the point of intersection will be the center. Compute the radial component of the velocity. Draw a right triangle with angle ${ \frac{\pi}{n}}.$ The person's velocity is directed along the leg opposite that angle. The radial component of the velocity is that velocity projected onto the hypotenuse. That component (for a velocity of $ 1$) is $ \sin\left(\frac{\pi}{n}\right).$ The rest follows.


limit
logarithms
induction
real analysis
real analysis unsolved

Let $(a_n)$ : $a_n >0 \forall n$ and $a_{m+n}\leq a_m.a_n$
 Prove that : there exists $\lim_{n\to \infty}\sqrt[n]a_n$

Let $b_n$ such that $b_{n+1} \leq b_n + b_1$ . It's easy to prove that $b_n \leq nb_1$, so  $\frac{b_n}n$ is decreasing , so $\displaystyle e^{\frac{b_n}n}$ is convergent
If you put ${b_n = \frac{\ln a_n}n}$ you obtain that $\sqrt[n]a_n$ is convergent
 :cool:

:cool:  Sorry a little mistake ! Take :  $b_n = \ln a_n$:cool:

I'm not sure how you get from $b_n \leq n b_1$ to $\frac{b_n}{n}$ is decreasing.

Letting Setting $b_1=b_3=2, b_2=b_4=b_5=b_6=...=1$ gives a sequence which is subadditive but has $\frac{b_n}{n}$ increase between $n=2$ to $n=3$

[quote="kevinatcausa"]I'm not sure how you get from $b_n \leq n b_1$ to $\frac{b_n}{n}$ is decreasing.[/quote]
You are right kevinatcausa, and I was wrong  : $\frac{b_n}{n}$  is not necessarily decreasing !!  :blush:

I hope that the following proof is correct ..${a_{m+n}\leq a_ma_n}$..   By induction we have : $a_r \leq a_1^r ,\ a_{kn} \leq a_n^k,\ \ a_{kn+r}\leq a_n^ka_1^r$

The sequence  $(a_n)^{\frac 1n}$ is manifestly bounded and $L$= limit points set of $(a_n)^{\frac 1n}$ is non empty and bounded, where limit point = limit of a sub-sequence.

Let $S$ be the superior limit : $S = \sup (L)$. So there is a sub-sequence $\displaystyle (a_{u_n})^{\frac 1{u_n}}$ convergent to $S$

$u_n$ is an increasing sequence of natural numbers so I can find  $p_n$,  a sub-sequence of $u_n$,  such that $p_n > n^2$

I can write $p_n = k_nn+r_n$, with $1 \leq r_n \leq n ,\ n \leq k_n$ , so : $\lim_{n\rightarrow\infty}\frac{r_n}{p_n}=0 ,\ \lim_{n\rightarrow\infty}\frac{k_nn}{p_n}=1$, so :

${\displaystyle a_{p_n}=a_{k_nn+r_n} \leq a_n^{k_n}a_1^{r_n} \Longrightarrow (a_{p_n})^{\frac 1{p_n}}\leq ((a_n)^\frac 1n})^{\frac{k_nn}{p_n}}(a_1)^{\frac{r_n}{p_n}}$

Now, if ${n\rightarrow\infty}$, we obtain that for all limit point $\lambda\in L,\ \sup (L)= S \leq \lambda$ , so $L=\{S\}$, and $(a_n)^{\frac 1n}$ converges to $S$ .. (q.e.d)..

 :cool:


1.Find all $ \ f: R\to R^\plus{}$ satisfy
$ \ f(x)^2 \equal{}1\plus{}xf(x\plus{}1)$ for all $ \  x \epsilon R$

2.Find all $ \ f: R\to R$ satisfy
a.$ \ f(x)\plus{}f(y) \neq 0$
b.$ \ \frac{f(x)\minus{}f(x\minus{}y)}{f(x)\plus{}f(x\plus{}y)}\plus{}\frac{f(x)\minus{}f(x\plus{}y)}{f(x)\plus{}f(x\minus{}y)}\equal{}0$

3.Find all $ \ f: R\to R$ satisfy:
a.If $ \ x\neq y$ then $ \ f(x)\neq f(y)$;
b.$ \ f(xf(y)\minus{}y)\equal{}f(yf(x)\minus{}x)\plus{}f(x\minus{}y)$.

[quote="duongptnkvn"]1.Find all $ \ f: R\to R^ \plus{}$ satisfy
$ \ f(x)^2 \equal{} 1 \plus{} xf(x \plus{} 1)$ for all $ \ x \epsilon R$

2.Find all $ \ f: R\to R$ satisfy
a.$ \ f(x) \plus{} f(y) \neq 0$
b.$ \ \frac {f(x) \minus{} f(x \minus{} y)}{f(x) \plus{} f(x \plus{} y)} \plus{} \frac {f(x) \minus{} f(x \plus{} y)}{f(x) \plus{} f(x \minus{} y)} \equal{} 0$

3.Find all $ \ f: R\to R$ satisfy:
a.If $ \ x\neq y$ then $ \ f(x)\neq f(y)$;
b.$ \ f(xf(y) \minus{} y) \equal{} f(yf(x) \minus{} x) \plus{} f(x \minus{} y)$.[/quote]
The first is very easy because It is not original problem ,The author of that book may have some mistake (I think so )
With $ x\equal{}0$ we get $ f(0)\equal{}1$
And $ x\equal{}\minus{}1$ we have $ f(\minus{}1)$=0 It is contracdition
2and 3 were posted on tis forum,Search tool will help you


Does anyone know of a way to test whether a given number is a palindrome?  Obviously you can tell, at least with smaller numbers, by looking at it, but with numbers in the range of 10^100, this would be very hard to see.

I don't think there is a way to test for it.  And why would you need to?

-interesting_move

Whether or not an integer is a palindrome depends on what base you write the integer in.  If you write the natural number n in base n-1 (assuming there is one), it will be the palindrome 11_(n-1).  All natural numbers are palindromes in some bases.


Amy has three Powerful Pink geraniums and three Roaring Red geraniums. If she plants two flowers in each of the three pots, how many distinct 3-pot arrangements could she make? (Consider the pots distinct.)

There are $ 3$ ways to pick two pots from the three. There are $ 4$ ways to pick two flowers for the two pots. $ 4\times3\equal{}\fbox{12}$ ways.

that's the wrong answer the answer is 7...

There are two cases.
Either there are one of each type in each of the boxes, which has $ 1$ way, because the types are indistinct,
or there are two Pink, two Red, and one of each type, which has $ 3! \equal{} 6$ ways, because the types are distinct.
The final answer is $ 1 \plus{} 6 \equal{} \fbox{7}$.


Can you help me prove that there exists a bijection between A and A*N where A is an infinite set??? (if A is countable then it is trivial, but what happens if A is uncountable?) (N are the naturals)

The idea is to split $ A$ into disjoint countable sets  $ A \equal{} \bigcup_\alpha A_\alpha$. Then use a bijection between $ A_\alpha$ and $ A_\alpha\times \mathbb N$.

But how to split $ A$ into countable sets? Hint: $ A$ can be well-ordered.

Can you be more precise please? Thanks. :)

Ok. Suppose $ A$ is well-ordered. Define
\[ d(x,y) \equal{} \sup\{n: \exists a_1,\dots,a_n \text{ s.t. }x < a_1 < \dots < a_n < y\}\]
Let $ x\sim y$ iff $ d(x,y) < \infty$.

This (together with some minor adjustments) gives the required split.


function
induction
algebra solved
algebra

Find all real functions such that for all reals x, y one have:

f(f(x)+y)=f(x 2 -y)+4f(x)y

We prove that the solutions are the functions $f=0$ and $f(x) = x^2$.
Note that these functions are solutions. Now let $f$ be a solution which is not identically $0$.
- For $y=0$, it commes $f(f(x))=f(x^2)$.
- For $y=-f(x)$, we obtain $f(0)=f(x^2 + f(x))-4f^2(0).$ (1)
Thus for $x=0$ it leads to $f(0)=0$.

It follows easily that $f$ is even.

- For $y=x^2$, we get $f(f(x)+x^2) = f(0)+4f(x)x^2 = 4f(x)x^2$.
But, from (1), we have $f(x^2 + f(x))=4f(x).$ Then $f(x)(f(x)-x^2)=0.$
It follows that for each $x$ we have $f(x)=0$ or $f(x)=x^2$. (2)

Note that $f(x) \geq 0$ for each $x$ (3)

Now, from our hypothesis, there exists $a$ such that $f(a) =/= 0$. Since $f$ is even and $f(0)=0$, we may assume that $a>0$.
From (2), we then have $f(a)=a^2$.

Let $y>0$.
Then $f(a^2+y)=f(f(a)+y)=f(a^2-y) + 4a^2y > f(a^2 - y) \geq 0$ (from (3)).
Using (2), it follows that $f(a^2+y) = (a^2 + y)^2.$Then $f(x) = x^2$ for all $x > a^2$. (4)

Supopose, for a contradiction, that $f(x)=0$ for all $x \in [0,1[$.
Thus, since $f$ is even, $f(x)=0$ for $x \in ]-1,1[$.
Moreover $f(1) = f(f( \frac 1 2 ) + 1) = f( \frac 1 4 - 1) + 4f( \frac 1 2) = 0$. Thus $f \equiv 0$ on $[-1,1]$.
It follows that $f(2) = f(f(1)+2) = f(1-2) + 8f(1) = 0$. An easy induction on $n$ leads to $f(2^{2^n}) = 0$ for all $n \geq 0$.
Choose $n$ such that $2^{2^n} > a^2$ and use (4) to get the desired contradiction.
Thus, we may assume that $a \in ]0,1[$.

Let $(U_n)$ be the sequence defined by $U_0 = a$ and $U_{n+1} = \frac {U_n + U^2_n} {2}$.
An easy induction leads to $U_n \in ]0,1[$ and $U^2_n < U_{n+1} < U_n$ for all $n \geq 0$.
Moreover, if $f(U_n) =/= 0$ for some $n$ then, as above, we deduce that $f(x) =/= 0$ for all $x > U^2_n$. Thus another induction leads to $f(U_n) =/= 0$ for all $n \geq 0$.
Now, since the sequence $(U_n)$ is positive an decreasing, this sequence converges to some real number $L \in [0,1[$. From $L = \frac{L + L^2}{2}$, we deduce that $L=0$.

Let $x>0$. There exists $n \geq 0$ such that $0 < U^2_n < U_n < x$. Then, from above $f(x) > 0$ which leads to $f(x) = x^2$.
Since $f$ is even, it follows that $f(x) = x^2$ for all $x$, and we are done.

Pierre.

Oh pbornsztein! I have shorter one,
put instead of y ,$f(x)$ then we have $f(2(f(x))=f(x^2-f(x))-4(f(x))^2$
now substitude $x^2-2f(x)$ in y we conclude
$f(x^2-f(x))=f(2f(x))+4f(x)f(x^2-xf(x))$ so we have
$f(2f(x))-f(x^2-f(x))=4(f(x))^2$
$f(2f(x))-f(x^2-f(x))=(2f(x)-x^2)(4f(x))$ so
$f(x)=0 $or$ f(x)=x^2$

Well, it's shorter, but it's not complete. You have only proved that  the value of f in any point x is 0 or $ x^2$. But this doesn't show you the function.

:blush: yes you are ight Harazi but now I try to complete it
WLOG: assume that there are $x,y$ such that $f(x)=0$,$f(y)=y^2$ without lose the generality assume $y<0$ now according to the problem we have
$f(f(x)+y)=f(x^2-y)+4yf(x)$ so
$y^2=f(x^2-y)$
since $f(x^2-y)  <> 0$so we have $y^2=f(x^2-y)=(x^2-y)^2$ so $x^4-2x^2y=0$ this is impossible so we are done
is it complete now Harazi :maybe:

Yes, now it's complete. I understood that you have already proved that f is even.

In fact the result holds for any commutative ring without zero divisors ..
:)


induction
combinatorics unsolved
combinatorics

Find all integers $m$ and $n$ such that an $m*n$ board can be tiled with corners (2*2 squares without a corner)

(Edit: I wanted to post it in proposed and own problems because i have a solution, but i made a mistake)

Yes! I got it...

We will prove that the $(m,n)$ board whcih can be tiled are those of the form (at a permutation of $m,n$) :
$(3p,q)$ with $p,q \geq 2$ or $(3,2q)$ with $q \geq 1$.

First, it is clear that if $(m,n)$ can be tiled then $m,n \geq 2$ and $mn = 0 \mod [3]$.
Moreover $(3,2)$ can be tiled so that $(3p,2q)$ can be tiled for all $p,q \geq 1$.

It is easy to verify that $(6,2),(6,3)$ can be tiled.
Since $(6,2)$ can be tiled, it follows that if $(6,n)$ can be tiled then so is $(6,n+2)$. Then, using induction, $(6,q)$ can be tiled for all $q \geq 2$, and then $(6p,q)$ can be tiled for all $p,q \geq 1$.

Now, we may assume that $m,n$ are odd. As above, we assume that $m=3p$.

By induction we prove that $(3,2q+1)$ cannot be tiled :
Considering all the possiible dispositions for the tile which covers a $1 \times 1$ square at a corner of the board, we easily see that if $(3,2q+1)$ with $q \geq 2$ can be tiled then so is $(3,2q-1)$. But it is easy to verify that $(3,3)$ cannot be tiled by direct checking.

The key step (which took me a couple of hours to see that! I was convinced that there was an extra parity condition to be fulfilled...) is that $(9,5)$ can be tiled! But I do not know how to draw it here.... :?
Anyway, since from above $(9,2)$ can be tiled, we deduce that $(9,2q+1)$ can be tiled for all $q \geq 2$ and then $(9k,2q+1)$ can be tiled for $k \geq 1,q \geq 2$

Since $3(3k+1) = 9(k-1) + 12$ and both $(9(k-1),2q+1), (12,2q+1)$ can be tiled, we deduce that $(3(3k+1),2q+1)$ can be tiled for all $k\geq 1, q \geq 2$.

Since $3(3k+2)=9k+6$ and both $(9k,2q+1),(6,2q+1)$ can be tiled we deduce that $(3(3k+2), 2q+1)$ can be tiled for all $k \geq 0, q \geq 2$.

If I have omitted no case, the initial claim follows.

erre.

Yes, I heard that this problem was considered really tough because of this (I didn't actually take those TST's).


logarithms
probability
number theory proposed
number theory

Show the existence of a set of integers $M$ such that every positive integer
$n$ can be represented in at most $(c_{1}+o(1))\log n$ ways as a sum of two elements of $M$ and in at most $(c_{2}+o(1))\log n$ such ways. Here $c_{1},c_{2}$ are two positive absolute constants.

Here's the rough idea.  We'll choose $M$ probabilistically.  Given a positive integer $k$, we put it in $M$ with probability $10 \sqrt{\frac{\log k}{k}}$.  Then we show that $M$ will satisfy the conditions with high probability.


yeah,I have added you.
but I would like you send me a pm to tell your qq :)


a and b are positive reals and AB a line segment in a plane. For how many distinct points C in the plane will it happen that for triangle ABC, the median and the altitude through C have lengths a and b respectively

note:how can we write AB as line segment in latex

problem 2

4 if $b<a$, 2 if $b=a$, 0 if $b>a$.  (Think of the intersection of (i) a circle centred at the midpoint of AB, with radius $a$, and (ii) the two lines parallel to, and separated by distance $b$ from, AB.)


Prove that if $ f$ is defined for $ |x| < r$ and there exists a constant $ B$ such that $ |f^{(n)} (x) | \le B$ for all $ |x| < r$ and $ n \in N$, then the Taylor series expansion $ \displaystyle\sum_{i\equal{}1}^{\infty}\frac {f^{(n)} (0) } {n!} x^n$ converges to $ f(x)$ for all $ |x| < r$.

My work:
Let the remainder term be $ R_n$.
$ |R_n| \le \frac {B |x|^n} {n!}$.
So taking limit for both sides,we get $ lim (R_n(x)) \equal{} 0$,so we get the Taylor expansion.

Am I right?


geometry
analytic geometry
Pythagorean Theorem

Square $ ABCD$ is positioned on the axes as shown.  Compute the area of the square [NOTE: The coordinate of $ C$ (not indicated in the diagram) is (13,8)]

[hide]You can circumscribe a square with side length 13 around the current square. You'll see that the circumscribed square consists of the smaller square, and four congruent right triangles, which each have legs measuring 5 and 8. Using the Pythagorean Theorem, the side length of the smaller square is $ \sqrt{5^2\plus{}8^2}$, and so the area of the square is $ 5^2\plus{}8^2\equal{}25\plus{}64\equal{}\boxed{89}$.[/hide]


arithmetic sequence
algebra proposed
algebra

Let $ S_{n}=x_{0}+x_{1}+\cdots+x_{n}$, $ nx_{n}=-2007S_{n}\Longleftrightarrow n(S_{n}-S_{n-1})=-2007S_{n}\Longleftrightarrow \frac{S_{n}}{S_{n-1}}=\frac{n}{n+2007}$, yielding $ S_{n}=\frac{2007\cdot 2007!n!}{(n+2007)!}\ (n\geq 1)$.
$ \therefore x_{n}=S_{n}-S_{n-1}=-\frac{(2007)^{2}\cdot 2007!(n-1)!}{(n+2007)!}\ (n\geq 1)$.

[color=red]Comment $ nx_{n}=-2007S_{n}\Longleftrightarrow n(S_{n}-S_{n-1})=-2007S_{n}$ is wrong.[/color]

Take a look the following [b]aviateurpilot[/b]'s post.

kunny

[quote="aviateurpilot"]

$ kunny$ you made a small fault, but what you have fact gives like same an idea on the solution
for $ n<2008$
$ S_{n}=x_{0}+x_{1}+..+x_{n}$
$ \frac{S_{n+1}}{S_{n}}=\frac{n-2006}{n+1}$[/quote]

Oops! It seems to plead that we are taught the initial term of a sequence is not $ a_{0}$ but $ a_{1}$ in Japan.

I was surprised at that the general term of an arithmetic sequence is written as $ a_{n}=a_{0}+dn\ (n=0,\ 1,\ 2,\ \cdots)$ in French mathematics book of Dunod.
In this conection, in Japan, Natural number is taught as $ 1,\ 2,\ 3,\ \cdots$ in elementary, junior, high school.
Anyway, your solution is nice.


algebra
polynomial
limit
calculus
calculus computations

Let $P_n (x)$ be a polynomial such that $x^{2n}=P_n(x)\left(x^2-x+\frac{n-1}{n^2}\right)+a_n x+b_n\ (n=2,3,\cdots)$

Evaluate $\lim_{n\to\infty} a_n,\lim_{n\to\infty} b_n$

We substitute in $x = \frac{1}{n}$ to find $\frac{1}{n^{2n}} = \frac{a_n}{n} + b_n$ and substitute in $x = \frac{n-2}{n}$ to see $\frac{(n-2)^{2n}}{n^{2n}} = \frac{(n-2)a_n}{n} + b_n$.  So, this is now a linear system that we can solve.  Subtracting equations, ${\frac{(n-3)a_n}{n}} = \frac{(n-2)^{2n}-1}{n^{2n}}$ or $a_n = \frac{(n-2)^{2n}}{n^{2n-1}(n-3)} - \frac{1}{{n^{2n-1}(n-3)}}$.  Thus, we see we must calculate $\lim_{n \to \infty} a_n = \lim_{n \to \infty}\frac{(n-2)^{2n}}{n^{2n-1}(n-3)}$.  So, I don't have a rigorous explanation for this, but it's clear that n - 3 is effectively n here, so this just becomes $e^{-4}$.  This gives us with epsilonically more work that $\lim_{n\to\infty}b_n = 0$.

[quote="JBL"]...substitute in $x = \frac{n-2}{n}$... [/quote]

Just a typo the other root is $\frac{n-1}{n}$

Yes, I didn't divide by 2 properly, and that propagates all the way through.  Ah, well.


$ A, B, C, D$ are points in the plane, no three collinear. The lines $ AB$ and $ CD$ meet at $ E$, and
the lines $ BC$ and $ DA$ meet at $ F$. Prove that the three circles with diameters $ AC$, $ BD$ and $ EF$
either have a common point or are pairwise disjoint.

Do not double post: http://www.mathlinks.ro/viewtopic.php?p=1386762#1386762


Let ABC be a triangle, the middlepoints M, N, P of the sides BC, CA, AB respectively, the projections D, E, F of the orthocentre H on the sides BC, CA, AB respectively and the intersections X, Y, Z between the pairs of lines (BC, EF), (CA, DF), (AB, DE) respectively. I know and can prove that Z belongs to XY and that the lines AM, BN, CP are perpendiculary on the lines HX, HY, HZ respectively. [b]Is it then true that the line GH[/b][b] is perpendiculary on the line [u]XYZ[/u] ?.[/b][/quote]

You are correct.  The line $XYZ$ is called the orthic axis of triangle $ABC$.  You can find a proof here: [url=http://de.geocities.com/darij_grinberg/Schroeder/Schroeder.html]de.geocities.com/darij_grinberg/Schroeder/Schroeder.html[/url]

$GH$ is $OH$, the line connecting $H$ to the circumcircle of $ABC$, while $XYZ$ is the polar of $H$ wrt the circumcircle, so yes, they are perpendicular.


search
combinatorics proposed
combinatorics

Police want to arrest on the famous criminals of the country whose name is Kaiser. Kaiser is in one of the streets of a square shaped city with $ n$ vertical streets and $ n$ horizontal streets. In the following cases how many police officers are needed to arrest Kaiser?
[img]http://i38.tinypic.com/2i1icec_th.png[/img]          [img]http://i34.tinypic.com/28rk4s3_th.png[/img]
a) Each police officer has the same speed as Kaiser and every police officer knows the location of Kaiser anytime.
b) Kaiser has an infinite speed (finite but with no bound) and police officers can only know where he is only when one of them see Kaiser.
Everybody in this problem (including police officers and Kaiser) move continuously and can stop or change his path.

It was already posted (but not answered) with an additional proposed problem that wasn't on the exam: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1468877292&t=224004

You must specify that Kaiser may not cross any policeman. This problem can be extended on any board and more generaly on graphs.
To get pointers  google something like "tree width" "grid width" and cops Seymour .
Try the following simpler ( yet more general) problem : If you play the game on a tree ( any finite tree ) what is the minimum number of cops you need.
Then of course you may try rather but narrow grid ( 1xN) obvious , 2xn , 3xn ..
Hope it helps.

Sorry to revive a 10 year old topic, But this is very interesting...
any solution ? :maybe:

wow the last post in 2008 i was born around that time -.-

also im pretty sure you just post them along the boundaries or something


modular arithmetic
Divisibility
number theory
prime numbers
composite numbers
IMO Shortlist

A natural number $n$ is said to have the property $P,$ if,  for all $a, n^2$ divides $a^n - 1$ whenever $n$ divides $a^n - 1.$

a.) Show that every prime number $n$ has property $P.$

b.) Show that there are infinitely many composite numbers $n$ that possess property $P.$

Non-$\LaTeX$ed problems/solutions: http://www.mathlinks.ro/Forum/viewtopic.php?t=266

[hide="Solution by grobber"]

Solution by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=26]grobber[/url]:

Numbers of the form $n=pq$ with $p,q$ distinct primes should work: if $p|a^{pq}-1$, then $p|a^q-1$ because $a^q\equiv (a^q)^p\pmod p$, which means that $a^{tq}\equiv 1\pmod p,\ \forall t\ge 0$, from which we get $p|a^{q(p-1)}+\ldots+a^q+1=\frac{a^{pq}-1}{a^q-1}$. This is just another way of saying that $p^2|a^{pq}-1$, and we prove that $q^2|a^{pq}-1$ in the exact same way.

[/hide]

[hide="Solution"]

[hide="(a)"]

If $n=p$ then we arrive at \[a^p-1\equiv a-1\equiv 0\pmod{p}\]
Next, notice that when $p\neq 2$ we arrive at \[v_p(a^p-1)=v_p(a-1)+v_p(p)\ge 2\]
When $p=2$ we arrive at \[a\equiv 1\pmod{2}\implies a^2-1\equiv 0\pmod{4}\] [/hide]

[hide="(b)"]
I claim that all $n=pq$ for $p,q\neq 2 \text{  and  } p\neq q$ satisfy this condition.  First off notice that from the given condition we have:
\begin{eqnarray*} a^{pq}-1\equiv a^q-1\equiv 0\pmod{p} \\ \text{ord}_{p}(a)\mid q \end{eqnarray*}
Therefore $\text{ord}_{p}(a)\in \{1, q\}$.
$\begin{cases} \text{ord}_{p}(a)=1 & \implies v_p(a^{pq}-1)=v_p(a-1)+v_p(pq)\ge 2  \\ \text{ord}_{p}(a)=q &\implies v_p[(a^q)^p-1]=v_p(a^q-1)+v_p(p)\ge 2\end{cases}$
We use a similar argument to prove $q^2\mid a^{pq}-1$ and we are done $\Box$. [/hide] [/hide]

Observe that $2$ has the property $P$, since whenever a square is odd, it is also congruent to $1$ modulo $4$.  Henceforth assume that $n$ is odd.

Let $n$ be any odd positive integer such that $\gcd(n,\varphi(n)) = 1$. [color=red](EDIT 1/29/2023: This implies $\color{red}n$ is squarefree.)[/color] The condition that $n$ divides $a^n - 1$ implies that $\gcd(a,n) = 1$, so $a^{\varphi(n)} \equiv 1\pmod n$.  This yields $a\equiv a^{\gcd(n,\varphi(n))} \equiv 1\pmod n$.  Now for any (odd) prime $p$ dividing $n$, we see that
\[
\nu_p(a^n - 1) = \nu_p(a-1) + \nu_p(n) \geq 1 + 1 = 2,
\]
and so $n^2$ divides $a^n - 1$.

This makes both parts easy.  For part (a), observe that any prime $p$ satisfies $\gcd(p,\varphi(p)) = \gcd(p,p-1) = 1$.  For part (b), recall that the are infinitely many primes $p$ which are $2$ modulo $3$.  Now let $n = 3p$; then
\[
\gcd(n,\varphi(n)) = \gcd(3p,2p-2) = \gcd(p+2,2p-2) = \gcd(p+2,6) = 1,
\]
and so $n$ has property $P$.

a) The problem is trivial if $n=2$. Otherwise, we have $n|a^n-1\implies a^n\equiv 1\text{(mod n)}$. By FLT, this implies $a\equiv 1\text{(mod n)}\implies v_n(a^n-1)=v_n(a-1)+v_n(n)\geq 2$, as desired.

b) Let $n=p_1p_2$ for odd primes $p_1,p_2$ such that $(n,\phi (n))=1$. Then $a^{p_1p_2}\equiv 1\text{(mod }p_1p_2\text{)}$. Since $(n,\phi (n))=1$, we must have $a\equiv 1\text{(mod }p_1p_2\text{)}$. Then, $v_{p_1}(a^n-1)=v_{p_1}(a-1)+v_{p_1}(n)\geq 2$, and similarly $v_{p_2}(a^n-1)\geq 2$, so we're done.

We know that there are infinitely many twin primes $(p,p+2)$ and then showed that for all such twin primes $n=p(p+2)$ works....It's okay right? This does prove the infinititude....

(a) Let $p$ be any prime. By Fermat's Little Theorem, $a \equiv a^p \equiv 1 \pmod p$. In particular, $a = kp + 1$ for some integer $k$. Then, \[a^p - 1 \equiv (kp + 1)^p - 1 \equiv 0 \pmod{p^2}\] by binomial expansion, as required.

(b) We claim that all $n$ of the form $2p$ satisfy $P$, where $p$ is any odd prime. By Fermat's Little Theorem, \[a^2 \equiv (a^2)^p \equiv a^{2p} \equiv 1 \pmod{2p},\] so $a^2 = kp + 1$ for some integer $k$. Then, \[a^{2p} - 1 \equiv (kp + 1)^p - 1 \equiv 0 \pmod{p^2}\] by binomial expansion. All perfect squares leave a residue of $1$ modulo $4$, so \[a^{2p} - 1 \equiv (a^2)^p - 1 \equiv 1^p - 1 \equiv 0 \pmod 4.\] Since $p$ is odd, $4p^2 \mid a^{2p} - 1$, as required.

[hide=Solution]
a) For the sake of clarity, let $n = p$, where $p$ is prime.
If $p = 2$, then $2|a^2 - 1$ implies that $a$ is odd. Therefore, we need that $4|(a+1)(a-1)$. This is evident as $2|a+1$ and $2|a-1$.
Now assume that $p$ is odd. By Fermat's Little Theorem, $a^{p} \equiv a \pmod{p}$. However, $n|a^p - 1$ implies that $a^p \equiv 1 \pmod{p}$; therefore, $a \equiv 1 \pmod{p}$, or $p|a-1$.
Therefore, $v_p(a-1) > 0$. By LTE, $v_p(a^n - 1) = v_p(a-1) + v_p(p)$. Note that $v_p(a-1) \geq 1$ since $p|a-1$, and clearly $v_p(p) = 1$. Therefore, $v_p(a^p - 1) = v_p(a-1) + v_p(p) \geq 2$, which implies that $p^2 | a^p - 1$, as desired. $\square$
b) We claim that all $n = 3p$ work, where $p$ is a prime such that $p > 3$ and $p \equiv 2 \pmod{3}$. There are infinitely many such primes by Dirchlet.
First, we consider divisibility by $3$.
Note that $3|a^{3p} - 1 \implies a^{3p} \equiv 1 \pmod{3}$. Since $3-1=2$, and the order of a mod $3$ must divide this, the order is either $1$ or $2$. If the order is $2$ then $2|3p$, which is a contradiction since $p$ is odd. Therefore, the order of a mod 3 is $1$. Therefore, $a \equiv 1 \pmod{3}$. From here we can proceed in a similar way as a) to prove the given.
For divisibility by $p$, we can note that the order must divide $\gcd(3p, p-1) = \gcd(p-1, 3)$. This is $1$ since $p \equiv 2 \pmod{3}$, then we can proceed in a similar way as a), and we are done. $\square$
[/hide]


[quote="biffanddoc"]A farmer has 60 meters of fence with which to build a rectangular animal run for her cows, horses and pigs. She wants each type of animal to have the same area. What is the largest number of square meters which can be enclosed? Express your answer as a decimal[/quote]

Uhhhhhhhhhhhhhh...what the
Doesnt' each animal get 20 meters of fencing...which makes the side length 5 and the area of each pen 25...and the total is 75...? I know i'm wrong and I think i'm interpreting the question wrong...

[hide]Oh.
So I labeled the lengths of the total pen x, so the area is $x^2$. there are six x's and each x=10...so the area is 100....what the heck? Can someone tell me what I'm doing wrong?[/hide]

I'm getting [hide]108, from 6*18[/hide] but I know that's wrong because it says decimal.

How did you get six $x$s?

heres how I did it.

Draw a square with side $x$

You have to divide it into 3 equal parts.

Draw a line near the top of the square, and call the side $y$. The area is thus $xy$

Divide the rest not in the rectangle we just divided in half down the middle.

We know that one of the dimensions is $x/2$. The area must be $xy$, so we have to multiply by $2y$.

Thus the other dimension is $2y$.

We can now see that $3y=x$ and $x+x+x+x+x+2y=5x+2y=60$.

Solve to get $(180/17,60/17)$.

Thus the total area is $\frac{180}{17}^2$

[quote="Ubemaya"]Why is everyone making squares? What about a circle?
Draw a circle with radius r. The total fencing is $3r + 2\pi r$. He has 60 meters of fencing, so set this equal to 60. Then $r(3 + 2\pi) = 60$, and $r = \frac{60}{3 + 2\pi}$. Solving with a calculator gives the total area to be 131.24.
(Biffandoc's way gives $112.11$)[/quote]

It says "rectangular animal run."  :)

also I think you have to use our 60 meters of fencing to divide up the animals too, so not all of the fencing could go on the outside edge of the circle, you'd have to have some to divide up the interior.

The original problem includes an illustration and the words "...build a rectangular animal run (as shown)..."

The illustration is a long (horizontal) rectangle, divided into three equal parts.  Each part  is a slightly tall (vertical) rectangle.

That method would give $108$
But mine is more.
And when they say "as shown" I don't think that the way they showed was the only way the pens could be made

The top illustration (without labels) is what the National Competition gave "as shown".

The lower illustration is rather similar to the clever arrangement biffanddoc described.

An optimal arrangement in one configuration can be transformed to an optimal arrangement in the other configuration.

We're still awaiting a correct solution, since the answer key gives an answer greater than 112.11.

Problems requiring an area to be maximized can often be solved by graphing the area versus some parameter.  For area problems, the graph is frequently a parabola.  If you can find the vertex of the parabola, you can find the maximum area.

Attached is a graph for this problem, showing several choices for $x$, ranging from zero to 10 meters.  At $x=0,$ all of the fencing is used for the $y$ sides of the pens $(y=15),$, and the area is zero.  At $x=10,$ all of the fencing is used for the $x$ sides of the pens.  At $x=10,\thinspace y=0,$ so the area is zero.

Somewhere between the extremes, the area is maximized.

As far as an algebraic solution, we use the equations for the total length of the fencing, $6x+4y=60,$ and total area, $A=3xy.$  We can get an equation for the area in terms of $x$:

$6x+4y=60\quad\Longrightarrow\quad y=\frac{30-3x}{2}=\frac{3}{2}(10-x)$
$A=3xy$
$A=3x\times \frac{3}{2}(10-x)$
$A=-\frac{9}{2}(x^2-10x)$

If you put this parabola in standard form, the correct answer should pop out.

I'm sorry to revive an old topic, but no one correctly answered the problem.

So the farmer has 60 meters to build a rectangular animal farm. one side of the animal farm is x, and the other is y. From one of the y sides you can draw two lines that are perpendicular to it to divide the rectangle into 3 equal sections. Each of these lines is equal to x. So 4x+2y=60.

You want 4x=2y to get xy to be the greatest possible, so solving, you get x=7.5 and y=15, so 7.5*15=112.5


geometry
circumcircle
incenter
angle bisector
geometry unsolved

In an acute triangle [b]ABC[/b] the angle at [b]C[/b] is equal to [b]45[/b] , [b]O[/b]- is circumcenter and [b]H[/b] orthocenter.  the line through [b]O[/b] perpendicular to [b]CO[/b]  meets the lines [b]AC[/b] and [b]BC[/b] at [b]K[/b] and [b]L[/b] respectively. prove that [b]OK+KH=OL+LH[/b]

Fortunately I find out the way to the roof.
Let M is the intersection of  BH and KL.
Let N is the intersection of AH and KL.
We will prove that M and N belong circle center O.
We have $ \hat {ONE}$ = $ \hat {OCA}$ = $ \hat {OAC}$ = $ \hat {OBE}$
  -> ON = OB -> N belong circle center O.
With similar way we also have that M belong circle center O.

With all that, we find the proof easily. Goodluck.
My english is very bad. So if I have some mistakes, Please show it for me!

In fact it is OK,  since MN is a diameter perpendicular to ray OR, the angle <NAB = <A-45, and AN contains H, and as well known, N is the symmetrical of H w.r.t. BC and LN=LH.
Same way KM=KH and we are done, the sum being the radius of the circle (ABC).

Note:
We can see that HC is angle bisector of <KHL and the circle (C,CO) is the H excircle for the triangle KHL, right angled at H.

Best regards,
sunken rock

[quote="kimnimalar"][color=darkred] Let $ ABC$ be an acute triangle with $ C = 45^{\circ}$ . Denote the its circumcircle $ w = C(O,R)$ and the its orthocenter $ H$ .

Define the points $ \{\begin{array}{c} K\in CA \\
\ L\in CB\end{array}$ so that $ O\in KL$ and $ \overline {KOL}\perp OC$ . Prove that $ OK + KH = OL + LH$ .[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \{\begin{array}{c} D\in AH\cap CB \\
\ E\in BH\cap CA\end{array}$ and $ \{\begin{array}{c} M\in w\cap KL\ ,\ L\in (OM) \\
\ N\in w\cap KL\ ,\ K\in (ON)\end{array}$ . Thus, $ C = 45^{\circ}$ $ \implies$ $ OA\perp OB$ $ \implies$

$ ABDOE$ is cyclically $ \implies$ $ \{\begin{array}{c} m(\widehat {ODC}) = m(\widehat {OAB}) = 45^{\circ} \\
\ m(\widehat {OEC}) = m(\widehat {OBA}) = 45^{\circ}\end{array}\|$ $ \implies$ $ \{\begin{array}{c} \widehat {ODC}\equiv\widehat {OMC} \\
\ \widehat {OEC}\equiv\widehat {ONC}\end{array}\|$ $ \implies$

$ ODMC$ and $ OENC$ are cyclically $ \implies$ $ \{\begin{array}{c} DM\perp CB \\
\ EN\perp CA\end{array}\|$ $ \implies$ $ \{\begin{array}{c} M\in AH \\
\ N\in BH\end{array}\|$ $ \implies$ $ \{\begin{array}{c} LH = LM \\
\ KH = KN\end{array}\|$ $ \implies$

$ \{\begin{array}{c} OL + LH = OL + LM = R \\
\ OK + KH = OK + KN = R\end{array}\|$ $ \implies$ $ \boxed {\ OL + LH = OK + KH = R\ }$ .

[b][u]Remarks.[/u][/b]

$ 1\blacktriangleright$ Prove easily that $ HMCK$ , $ HNCL$ are cyclically. Thus,  $ \{\begin{array}{c} m(\widehat {CHK}) = m(\widehat {CMK}) = 45^{\circ} \\
\ m(\widehat {CHL}) = m(\widehat {CNL}) = 45^{\circ}\end{array}\|$ $ \implies$

$ HK\perp HL$ and the ray $ [HC$ is the bisector of the angle $ \widehat {KHL}$ .

$ 2\blacktriangleright\ \ ABC\sim LKC\sim HKA\sim LHB\sim NBC\sim CMA$ .

$ 3\blacktriangleright\ \{\begin{array}{c} m(\widehat {BLH}) = m(\widehat {CLO}) = A \\
\ m(\widehat {AKH}) = m(\widehat {CKO}) = B\end{array}\|$ $ \implies$ the incenter of the triangle $ KHL$ belongs to the line $ CH$

and the vertex $ C$ is the $ H$ - exincenter of the triangle $ KHL$ .[/color]

[color=darkred][b]P.S. [u]Kimnimalar[/u][/b], why do you wish a proof with Sinus' theorem for this nice problem ?![/color]

[quote="kimnimalar"]Because many problems we can easily solve with sinus theorem. Usually if we use all terms, the problem becomes only with sinuses.  I'm tryed to solve this with sinus theorem but I can't solve.[/quote]
Hence isn't easy to solve only with the Sinus' theorem this problem !
I"ll try for to see it is possibly. But I don't think that I could do it.

[quote="mstoenescu"][quote="caothujjj"]
We have $ \hat {ONE}$ = ...[/quote]

Who's E ?  :maybe:[/quote]

Sorry. E = BH X AC
          D = AH X BC

I think my proof is the shortest

Very nice proof M. Virgil Nicula ! I had a lot of pleasure to read it !

I add a figure to thank you !


function
floor function
parameterization
algebra

Find the least positive integer $k$ for which the equation $\lfloor \frac{2002}{n}\rfloor = k$ has no integer solutions for $n.$ (The notation $\lfloor x \rfloor$ means the greatest integer less than or equal to $x.$)

[hide="I think this is it"]

let $floor(n)$ denote the floor function as stated in the problem

I have a hunch that the least possible $k$ will have something to do with to the least $n$ that satisfies
$\frac{2002}{n}-\frac{2002}{n+1}>2$
because then we have an open integer between $floor(\frac{2002}{n})$ and $floor(\frac{2002}{n+1})$
32 is the least possible $n$ that satisfies the equation, but sure enough
$floor(\frac{2002}{32})=62$, $floor(\frac{2002}{33})=60$, and just to check, $floor(\frac{2002}{34})=58$, so we need to keep going until we dont skip any integers. $floor(\frac{2002}{35})=57$, so we know every $k$ less than or equal to 57 has an integer $n$ so that $floor(\frac{2002}{n})=k$.
Our answer is 058.
[/hide]

if i'm not mistaken....isnt it 059 since
$\lfloor \frac{2002}{34}\rfloor = 58$
So according to your solution $k \le 58$ has a solution therefore $\boxed{059}$

yea..i think thats the lowets how did you come with that..

Note that the value of $n$ is at most the solution to $\frac{2002}{n}-\frac{2002}{n+1}=1 \implies n \leq 44$.

If it were any more, then two values of $n$ would produce values of $2002/n$ that differed by less than 1, which would mean that we have 2 or values of $n$ producing $k$... far from what we want.

Also notice that we [b]must[/b] be skipping values if a difference of $2$ or more is created in $2002/n$ when we change $n$ by 1, so the solution to $\frac{2002}{n}-\frac{2002}{n+1}=2 \implies n \geq 32$.

However, the latter is a very threshold parameter, so it makes more sense to start trying values of $n$ starting from $44$. We soon reach $n=41 \implies k=48$ and $n=40 \implies k=50$, so we have found our first skipped value of $k$, $\boxed{k=49}$

Not that elegant a solution, I'm sure there's something better out there.

(I know it's a long bump, but I found it through contest collections so it's alright.)

[hide=Solution]There is some value of $n$ called $a$ such that $\frac{2002}{a}<k$ but $\frac{2002}{a-1}\geq k+1$. Multiplying through and combining, we have $(k+1)(a-1)\leq2002< ak$. We want the minimum value of $k$ such that there exists $a$ that satisfies the inequality. Note that $a<k+1$ by the first and last parts of the chain, and $ak>2002$, so we start testing at $46$, which is the smallest $k$ such that there is an $a$ that satisfies both inequalities ($44$). However, it doesn't satisfy the LHS of the original inequality, so we keep going up and testing until we find that $k=\boxed{49}$ because $49\cdot41>2002$ but $50\cdot40<2002$.[/hide]

@below 2002 AIME II #8

Which contest and problem is this from? Pretty sure it's AIME but I'm just wondering which problem.

@above okay thanks

p2: We are looking for numbers $k$ such that there is no integer between $\frac{2002}{k}$ and $\frac{2002}{k+1}$, so there is no integer $p$ such that $\frac{2002}{k+1} < p < \frac{2002}{k}$. If there was, we'd have $2002<p(k+1)$ and $2002>p(k)$, or $pk < 2002 < p(k+1)$, so we want this to have no solution for $p$. We start testing at the ceiling at $\sqrt{2002} = 46$, and checking we see that $k =\boxed{49}$ is the smallest value.

Basically we are looking for $k$ such that $\left\lfloor\frac{2002}{k}\right\rfloor=\left\lfloor\frac{2002}{k+1}\right\rfloor.$

[pywindow]
for k in range(1,1000):
    r_1=2002%k
    r_2=2002%(k+1)
    b=(2002-r_1)/k
    c=(2002-r_2)/(k+1)
    if b==c:
        print(k)
        break
    continue
[/pywindow]


8. How many ions are there in 30.0 grams of sodium fluoride?
(Relative atomic masses: Na, 23.0; F, 19.0; Avogadro's constant = $6 \times 10^{23}$ per mole)

A. $\frac{30 \times 6 \times 10^{23}}{42 \times 2}$
B. $\frac{30 \times 6 \times 10^{23}}{42}$
C. $\frac{30 \times 2 \times 6 \times 10^{23}}{42}$
D. $30 \times 6 \times 10^{23}$

P.S. I'm leaving out all the questions with diagrams because I don't know how to scan and upload them.  :)

$n_{NaF}= \frac{30}{42}$ mol

Since there are 2 moles of ions per mol of $NaF$.
There are $2 \times \frac{30}{42}$ moles of ions.

$\frac{6 \times 10^{23}ions}{1mol}= \frac{x}{2 \times \frac{30}{42}mol}$

$x = \frac{30 \times 2 \times 6 \times 10^{23}}{42}ions$

$\boxed{C}$ is the correct answer.


probability
geometry
integration
algebra unsolved
algebra

Two real numbers x and y are chosen at random in the
interval (0,1) with respect to the uniform distribution.
What is the probability that the closest integer to x/y is
even? Express the answer in the form  $ r\plus{}s\Pi$, where r
and s are rational numbers.

[hide="Answer"]$ \dfrac{5}{4} \minus{} \dfrac{1}{4}\pi$[/hide]
[hide="Solution"]
Consider the unit square $ (x,y) \in (0,1)^2$
Note that $ \frac {x}{y} \equal{} \cot(\theta)$ where $ \theta$ is polar argument of the point.
We can neglect the probability that $ \cot(\theta) \equal{} \frac {n}{2}$ for some $ n \in \mathbb{N}_0$.

Let $ P(n)$ be the probability that $ \cot(\theta)$ is closest to $ n$, i.e. $ \cot(\theta) \in \left(n \minus{} \frac {1}{2} , n \plus{} \frac {1}{2} \right)$

Thus for all $ n \in \mathbb{N}$, we have $ P(n)$ is the area between the lines $ \cot(\theta) \equal{} n \minus{} \frac {1}{2}$ and $ \cot(\theta) \equal{} n \plus{} \frac {1}{2}$, that is inside the unit square.

Note that the line $ \cot(\theta) \equal{} \minus{} \frac {1}{2}$ does not pass through the unit square. So, $ P(0)$ is the area inside the unit circle between the lines $ \cot(\theta) \equal{} 0$ and $ \cot(\theta) \equal{} \frac {1}{2}$.

The area between $ \cot(\theta) \equal{} 0$ and $ \cot(\theta) \equal{} \frac {1}{2}$ inside the unit square is a triangle with vertices $ (0,0)$; $ (0,1)$; and $ \left(\frac {1}{2},1\right)$.
Thus, $ P(0) \equal{} \frac {1}{4}$

For all $ n \ge 2$:
The area between $ \cot(\theta) \equal{} n \minus{} \frac {1}{2}$ and $ \cot(\theta) \equal{} n \plus{} \frac {1}{2}$ inside the unit square forms a triangle with vertices $ (0,0)$; $ \left(1,\frac {2}{2n \minus{} 1}\right)$; and $ \left(1,\frac {2}{2n \plus{} 1}\right)$.
Thus, $ P(n) \equal{} \frac {1}{2n \minus{} 1} \minus{} \frac {1}{2n \plus{} 1}$ for all $ n \ge 2$.

The probability that $ \cot(\theta)$ is closest to an even integer is:
$ \displaystyle \sum_{j \equal{} 0}^{\infty}P(2j) \equal{} P(0) \plus{} \sum_{j \equal{} 1}^{\infty}P(2j)$ $ \equal{} \frac {1}{4} \plus{} \sum_{j \equal{} 1}^{\infty}\left( \frac {1}{4j \minus{} 1} \minus{} \frac {1}{4j \plus{} 1} \right)$ $ \equal{} \frac {1}{4} \plus{} \left( \frac {1}{3} \minus{} \frac {1}{5} \plus{} \frac {1}{7} \minus{} \frac {1}{9} \plus{} \cdots \right)$

Note that:
$ \displaystyle \int^1_0 \dfrac{1}{1\plus{}x^2} \,dx \equal{}  \int^1_0 1 \minus{} x^2 \plus{} x^4 \minus{} x^6 \plus{} x^8 \minus{} \cdots \,dx$

$ \left[ \arctan(x) \right]^1_0 \equal{}  \left[x \minus{} \frac{x^3}{3} \plus{} \frac{x^5}{5} \minus{} \frac{x^7}{7} \plus{} \frac{x^9}{9} \minus{} \cdots \right]^1_0$

$ \frac {\pi}{4} \equal{} 1 \minus{} \frac {1}{3} \plus{} \frac {1}{5} \minus{} \frac {1}{7} \plus{} \frac {1}{9} \minus{} \cdots$

Therefore, $ \frac {1}{3} \minus{} \frac {1}{5} \plus{} \frac {1}{7} \minus{} \frac {1}{9} \plus{} \cdots \equal{} 1 \minus{} \frac {\pi}{4}$

And finally, $ \displaystyle \sum_{j \equal{} 0}^{\infty}P(2j) \equal{} \boxed{\dfrac{5}{4} \minus{} \dfrac{1}{4}\pi}$
[/hide]


I have just been sent a tex file that a colleague who uses MicTeX sent me (don't know which interface though). I use MicTeX together with TeXnicCenter (the setup proscibed by AoPS). It gives him no errors and the PDF he's produced is fine. However I need to add material to the document, but my setup gives me 75 errors when I try to build it and it shows a PDF output this is vastly different from the one he made with the same code.

Errors include things like:

"Paragraph ended before \\title was complete."

[code]\title[Periodic unique beta-expansions]

{Periodic unique beta-expansions: \\

the Sharkovski ordering}[/code]

(apparently this was legal to his setup but not mine). And

"missing \begin{document}."

He put this after the title.

All sorts of othe problems occur such as the text not being justified, the references looking grotesque and footnotes not displayed.

Any suggestions?
[/code]

If you haven't made any changes to the document then it sounds like you don't have a package that he is using. The log file will tell you which one it is.

I've worked it out now. He was using a blank line between every line of his code. My compiler kept thinking this was a new paragraph but his must have allowed it for some reason.

It turns out that we both use MicTeX but I use TeXnicCenter as an interface whereas he uses winedt. Is'nt it strange that the "blank line = new paragraph" rule which I've come to know and love seems not apply in winedt. I thought the language was universal and didn't depend on shell. Any thoughts on this?

Also, out of interest, which is generally the most popular?

[quote="Degenerate Polygon"]Is'nt it strange that the "blank line = new paragraph" rule which I've come to know and love seems not apply in winedt.[/quote]That is what LaTeX always does unless you tell it to do otherwise eg using parskip. There may be some setting that removes hard carriage returns before compiling, but that is a property of the editor and nothing to do with LaTeX.


[quote="heaven_us"]A number is said to be a good number if it is a composite number and it isn't a multiple of either 2,3,5 . How many good number which are less than 1000?[/quote]
$ N\equal{}500\plus{}333\plus{}200\minus{}166\minus{}66\minus{}100\plus{}33\equal{}734$


At my school, I'm currently taking Chemistry H. We don't have a AP Chemistry course available. However, the instructor says the course is quite rigorous, and if we're interested in the AP exam, we should probably get a review book and follow along during the year.

Also, I'm taking AP Phys C. Does the B test contain that much other material compared to C?

I'm planning to self study statistics, as many many people have told me that it is quite an easy class and test.

In summary:
Which AP review books/online courses/other resources would you guys recommend for the following:
Chemistry
Physics C/B
Statistics

Thanks!

Physics B/C - Princeton Review
Statistics - Barrons
Chemistry - Princeton Review or Barrons

Statistics is incredibly easy. I did not study at all (i.e., I did not know any of the tests that you do on sets of data), and I got a 4.

If you want to use APs for college credit and you are interested in the subject (and therefore would continue in teh subject), make sure you actually learn the material, not just focus on passing AP. If you only train yourself to pass the test, you will have difficulty applying your knowledge in other situations, which will be a problem if you take more advanced courses in the subject.

Zumdahl is a very good textbook for Chemistry.

Zumdahl is good stuff, it got me a 5 and I barely payed attention in class

for physics, use the halliday/resnick book.

Could you please specify which Zumdahl book is best for preparing for AP/SAT tests? I mainly need something to reinforce what I'm learning in my chem class, and then maybe some light reading to cover what I need to learn on my own time.

thank you

[quote="Xevarion"][url=http://www.amazon.com/gp/product/0618035915/sr=8-2/qid=1156372055/ref=pd_bbs_2/103-7985618-3328648?ie=UTF8]This[/url] is the one I used. There might be a newer edition out.[/quote]

yea, we used the same one this year too. It's pretty good and covers everything on the AP exam.

Yea, I used that same text, got me an ez five.

alright, i'm going to get the zumdahl book

another question, since i have a textbook from class and will get work from there too, what would be a good way of studying from zumdahl, as opposed to just reading it through 100%(which probably wouldn't be a bad idea, cuz it probably covers all of my classes curriculum...)

zumdahl book is good for text, the study guide that comes with it is much better for hte problems. I never even looked at the problems in the back of the zumdahl text.

the study guide has the answers too, anyways.

Regarding AP Physics B vs. C, there's definitely a big difference in material. B covers quite a few more topics (fluid mechanics, thermodynamics, modern physics, etc.). I wouldn't recommend taking the B test knowing only the C material (or vice versa).

On chemistry, get the latest edition of the Zumdahl text, and you will have an outstanding resource for both the AP test and for the SAT II test in chemistry.

On physics, the physics B syllabus does indeed cover topics (at algebra-level mathematical sophistication) that aren't covered at all in the physics C syllabus. A good physics B text will also be good preparation for the physics SAT II. Nobel laureate Leon Lederman said in a public speech here last year about education reform that he wishes any of his graduate students would have had a "conceptual physics" course before they started college. Paul Hewitt is the outstanding author of such physics books. You can read through a book like that in a reasonable amount of time before any other physics course to get familiarity with the subject.


inequalities
triangle inequality
geometry theorems
geometry

Very intersting observation

[color=Blue]Fact[/color].[hide="Ptolemy Inequality"]http://mathworld.wolfram.com/PtolemyInequality.html[/hide] is equivalent to[hide="triangle inequality"]http://mathworld.wolfram.com/TriangleInequality.html[/hide]

I used to belive that Ptolemy inequality is Very stronger, But it is not .

oh I don't understand, when we apply them in difference case we can't rate the streng of two inequality in this case!
here we cann't say Ptolemy inequality is as strong as triangle inequality.

I don't see how one can prove the Ptolemy's inequality without using the triangle inequality.

because, actually $aMA,bMB,cMC$ are three side of a triangle! but in this case we cann't say about the streng of them.

By equivalent i mean you can,By a Nice trick ,change the Triangle inequality to Ptolemy and , Vice versa.

[quote="gemath"]because, actually $aMA,bMB,cMC$ are three side of a triangle! but in this case we can say about the streng of them.[/quote]

They are , But how can you show it  :D

A Hint : Look at the Equality cases , LINE for trinagle and CIRCLE for ptolemy.

ok if you think that ,this is a way,we can see it as triangle inequality by complex number
$(m-a)(b-c)+(m-b)(c-a)+(m-c)(b-a)=0$
but maybe you don't understand me, we can use triangle inequality to prove Ptoleme's ineq but we can't say about the streng of them because they are two different case.

[quote="lomos_lupin"]Ok , the key is inversion.

Assume an inscribed quidrilateral , Write the Ptolemy equation for it.

With a SUITABEL invertion , Ptolemy will transform to triangle inequality .

The reverse process is similar.[/quote]

mr.vakili(teacher) have been said this realtionship,

i wondered,why u didnt remember!


What is the least value of $ x$ for which $ |x\plus{}1|\leq5$?

We either have $ x \plus{} 1\le5$ or $ x \plus{} 1\ge \minus{} 5$. Clearly, $ x \plus{} 1\ge \minus{} 5$ will produce a smaller minimum value. Thus,
$ x \plus{} 1\ge \minus{} 5 \implies x\ge \minus{}6$. So the minimum is $ \boxed{ \minus{} 6}$.

hello, we have
$ |x\plus{}1|\le5$ is equivalent to $ \minus{}5\le x \plus{}1\le 5$ therefore we get $ \minus{}6\le x \le 4$ so the  answer is $ \minus{}6$
Sonnhard.


let a>0 and  (xn)n \in N be a sequence such that x0>0, xn+1=xn+ax_n for every n>0.
a)compute (when n->oo):
lim xn
lim xn/n
lim xn/lnn

b)Study the convergence of the sequence:
(-x1+x2/2+...+(-1)nxn/n)

[i]Nicolae Paun 2003 11th grade[/i]

I only thought about a (during the contest, even if I'm in the 12'th grade  :D ). A general notation: a_n~b_n if a_n/b_n -> 1.

x_n is increasing, so it has limit l, so l=l+a l so l is 00.

For x_n/n we use Cesaro-Stolz:
x_n/n ~x_(n+1)-x_n=a x_n , so the limit is 00 if a>1, 1 if a=1 and 0 if 0<a<1.

Let a_n=a x_n  We have a_(n+1)=a_n*a a_n
x_n/lnn ~ (x_(n+1)-x_n)/ln(1+1/n) ~ n(x_(n+1)-x_n) = na_n = n/(1/a_n)
If a>=1 the limit is 00. If 0<a<1 we can use Cesaro-Stolz because 1/a_n increases to 00, so na_n ~ a_n*a_(n+1)/(a_n-a_(n+1)). Do the calculations and you'll see the answer is -lna (if I remember correctly).

For 2) with the result of 1)

If a \geq 1,  the general does not tend to 0 so the serie \sum_{n>0}(-1)^nx_n/n does not converge.
If 0<a<1 , x_n ~-lna.lnn the serie is the same nature
\sum_{n>0}(-1)^n.lnn/n wich is convergent.

if x_n~-lna.lnn we can consider instead of the original serie \sum {n>0}-lna.(-1)nlnn/n
??
i mean... i didn't know that we can do that :(
:D :D
thx for second point moubi :D

i think that i found other sol

using Leibniz critery for alternant series
"s_n=u_1-u_2+u_3-u_4+...+(-1)^nu_n.
if u_n->0 ,u_n>0 and u_n is strictly decreasing then (s_n) is convergent".

and we have that x_n/n ->0 and is decreasign for a<1 so
(x_1-x_2/2+...(-1)^nx_n/n) is convergent for a<1.


In training, Petra runs $ \frac 16$ of a mile in 60 seconds. At this rate, how many seconds does it take Petra to complete a run of $ \frac 14$ of a mile?

If Petra runs 1/6 of a mile in 60 seconds, she runs a mile in 360 seconds.
Thus, 1/4 of a mile will take 360/4 or 90 seconds.

You could also set up a proportion.

$ \frac{\frac{1}{6}}{60} \equal{} \frac{\frac{1}{4}}{x}$

$ \frac{1}{6}x \equal{} 15$

$ x \equal{} \boxed{90}$


I am not a Math student whatsoever :S this is in my Geography course and I'm very confused can anyone help me out? I'm a History Major *shy face*

Imagine that a woodlot 2x3km in size has been marked on two separate maps. Map A has a scale (in mixed units) of 1cm: 80m and Map B has a scale of 5mm:110m.

a. What are the Representative Fractions of Map A and Map B
b. Which map has the larger scale? (Where 1/10 is a larger scale than 1/20)
c. Which ma shows the most detail of woodlot
d. What are the dimensions of the woodlot (in mm) on each of the two maps...

Any help and guidance would help. I want to learn this, I need to learn it...

Well to compare these maps - we first have to get in the same terms.  Map A is given as having a scale such that every 1 centimeter represents 80 meters in real life. Map B says that 5 millimeters represents 110 meters in real life. To get them in the same terms - we know from metrics that 5 millimeters is one half of a centimeter (a centimeter is 10 millimeters). We want to change the 5 millimeters to 1 centmeter - so we multiply it by two. When we multiply it by two, we also have to multiply the corresponding distance it represents (in this case 110m) by two. So we rearrange the ratios to this: Map A is 1cm:80m and Map B is 1cm:220m. I suppose when you say "Representative Fractions" you mean these ratios in the form of fractions. Then I suppose they would be for Map A: $\frac{1}{80}$ and Map B: $\frac{1}{220}$. For question b, we have to look at which at the fractions. A smaller denominator indicates a larger fraction, so $\frac{1}{80}$ would be considered larger than $\frac{1}{220}$ because 80 is smaller that 220. For c, we think of detail as how big the woodlot is in each map. Think of it this way: Map A crams 80 meters in 1 centimeter and Map B crams 220 meters in 1 centimeter. Therefore the woodlot would be bigger in Map A, because less is being crammed in. Since it appears bigger in Map A, it follows that Map A would show more detail. Finally, to get these dimensions, we have to go back to our fractions. Once again, we need to get everything in the same terms to work with them, so lets change 2x3 km to 2000mx3000m. Then we can divide each by 80 (remember the scale!) to get 25cmx 37.5cm. In millimeters, this is 250mmx375mm as the dimension of the woodlot on Map A. For Map B, we do the same thing - except we use the scale for Map B. Thus we divide 2000mx3000m by 220 (remember the scales!) to get approximately 9.0909cmx13.6363cm. In millimeters this is approximately 90.9mmx136.3mm. And now - we are done.


Well, I like the way that simple problems can be very difficult, and i also appreciate the elegance in clever solutions. What about you?

Math is just so...elegant at times!  And it takes creativity too.

I also like how its not all memorization, you can derive one thing from another, its logically correct.

I like math just because I do.  Honestly, I think I like it so much because I understand it.  Those who don't like math do not fully understand it or just fail to see it's beauty.  I also think that inspiration early in my life helped me love math.  When I was in about 2nd or third grade I envied the math skills my sister possessed.  Her ability to multiply and divide mystified me.  At first I thought it would be impossible for me to do math like she did.  But once I learned it I was amazed at how simple such a seemingly complicated process turned out to be.  But, to tell the truth, I didn't have much more critical inspiration like that until middle and mostly until high school.

[quote="joml88"]I like math just because I do.  Honestly, I think I like it so much because I understand it.  Those who don't like math do not fully understand it or just fail to see it's beauty.[/quote] Yep. Me, too.
For me, understanding has always been the source of inspiration for all subjects (well, maybe except for art, music and stuff like that)
That explains why I hate literature and history, which I don't understand at all :(

well, at the beginning, i thought math was what we learned in the cirriculum.  This is why i believe most people dont like math. Fortunately for me, my elementary school had a teacher that would actually prove things (like area of a circle)
i just like the cleverness and elegance in which solutions are found. There is only reason i am slightly frustrated in mathematics (and this goes for most professional mathematicians too), that some things arent even provable using our modern techniques (like the Continuum hypothesis), and that some things that seem true really arent (like the overestimated prime conjecture) which is false when the value goes above 10^10^100000000

damn, youre lucky. In third grade, I knew the multiplication tables already but that was the standard math curriculum, so the teacher gave me fifth grade packets of math to do. Once I finished their curriculum, she had nothing else to give me, so I just sat in the back of the room and calculated 2^50...........................by hand.

i dislike the ambiguity associated with literature and history... with math, most things are very objective.

who says I like math?...Just kidding :D

because it is fun and entertainig andyou need it for practiclaly anything

[quote="mcalderbank"]-, youre lucky. In third grade, I knew the multiplication tables already but that was the standard math curriculum, so the teacher gave me fifth grade packets of math to do. Once I finished their curriculum, she had nothing else to give me, so I just sat in the back of the room and calculated 2^50...........................by hand.[/quote]

Ouch, That sounds painful, but easy if you do it in base two, I mean
$50 = 32 + 16 + 2$  $2^{50}=2^{32}\cdot2^{16}\cdot2^2 = (2^16)^2\cdot(2^8)^2\cdot2^2$ $ 4294967296\cdot65536\cdot4=1125899906842624$ and you reduce the multiplication to 7 instead of 50  :D , I don't know if you though about it in third grade, but I didn't.

Well, why do I like math, Because ...
1 - ... it is challenging.
2 - ... it is the mother of all sciences.
3 - ... it is, in my point of view, beautifull.
4 - ... shows you that there is always more than a way to do anything.
5 - ... it applies to basically anything.
6 - ... mathematicians are always considered wierd!  :D
7 - ... it is amusing.
8 - ... it is logical.
9 - ... it is universal.
A - ... it is most of the times not easy.
B - ... most of the people don't like it, what makes it even more interesting. :ninja:
C - ... developes in you schemes of thought that you can't develope elsewhere.
D - ... It is very extence.
E - ... there is math for most of the tastes.
F - ... in society, you're considered by definition smart if you like math (at least your IQ should be high!)
10 - ... Did I mention that it is beautiful?
11 - ... Did you notice that this list is in hexadecimal basis?

Best,


Let $ a,b,c$ be non-negative real numbers.Prove that

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


calculus
real analysis
topology
geometry
vector
Gauss
rudin

Hi all!

sorry if it isn't the right place to ask. :oops:

so here's the deal. I'd been studying Spivak's Calculus and I finished it. where to go now? what book to study next? should I tackle baby rudin? or maybe other books, advanced calculus?

pls help!

 :lol: thanks so much,
Gal.

[quote="ragingbullrocky"]yuo've done all of the problems? how long did that take?[/quote]

a very long year... :wink:

[quote="karthik_k"]If you've done only the book titled "Calculus", you could try this one -- "Calculus on Manifolds: A Modern Approach to Classical Theorems of Advanced Calculus." It's not really all that interesting, but all the same, makes a fairly challenging read.[/quote]

 :) thanks!! I will check it out. any more suggestion?

If you absolutely loved the book, then you might want to move on to a real analysis text.  I don't recommend baby Rudin, because it is notoriously difficult to read on your own, but you're probably well-prepared, having read Spivak on your own.  (Also, the last couple chapters are not so well-regarded.)

On the other hand, you might want to check out other interesting branches of mathematics.  In particular, every math person should learn abstract algebra at some point, and most math people enjoy it.  I think that Gallian has a pretty readable book on that subject.  I should also mention that if you haven't learned linear algebra yet, then you should do that.

Being in high school also gives you the luxury of reading fun math books outside the standard sequence.  For example, you could read up on number theory (Davenport) or knot theory (Adams) or totally random stuff ("Mathematics and Logic," by Kac and Ulam).

Also, I object to karthik_k's characterization of calculus on manifolds as non-interesting.  However, given the choice, I'd read about differential geometry (curves and surfaces) over calculus on manifolds.  Do Carmo's book is one of the standard texts.

Hi yenlee!

Thanks so much for your answer! I read Lang's “Introduction to Linear Algebra”, I think the next natural step would be to read Lang's “Linear Algebra”. (I also read Halmos's "I want to be a mathematician - an automathography" and would like to check out his "Finite Dimensional Vector Spaces" have you heard about it?)

I heard a lot about the huge Dummit & Foote "Abstract Algebra" how is it comparing to Gallian?

In Davenport do you mean "the higher arithmetic...”? (Since all the other books on number theory are Graduate Text)? Kac and Ulam are famous, and Knot theory sounds interesting. I gather they are all popular-mathematics (as in popular-science) books? I've been reading popular math for years and I'm always happy to learn about new books (or ones I haven't heard of before) :)

I think I'm going to read the Spivak’s book anyway (I like the guy). Talking about differential geometry, doesn't he have like 4 volumes in the subject? How are they comparing to Carmo?

Sorry, for all the questions. I heard alot of staff from a lot of different people and I'm trying to integrate it all into a solid opinion. Your answer helped me alot, hope you'll find the time to answer me again.

Thanks!
Gal.

I think you should also look at mathematical statistics/ combinatorics/discrete math and algorithms/differential equations (i.e. more applied stuff). Also check out Apostol Calculus vol.2, or Courant vol2. Intro to Calc and Analysis.

I reccommend abstract algebra.  Its a completely different type of math compared to calculus or linear algebra, and a lot more interesting in my opinion.

The question has might started as a question about calculus and analysis. Howevet,  the answers weren't about calculus but about learning mathematics form books in general. Answers about abstract algebra and popular-math have been suggested and received gratefully.

Please consider moving the post back into the round table or alternatively more suitable forum, since Calculus isn't the one.

Thank you for your time and kind management,
Gal.

I have no personal knowledge of Halmos's linear algebra book, but I imagine that it's good since Halmos is a good writer.  I grew up on Hoffman & Kunze, which is pretty decent.

Dummit & Foote is written on a higher level than Gallian; it can be used for a graduate course.  However, I like Dummit & Foote and find it to be very readable compared to grad texts such as Lang.  Artin is also very good, but also probably harder than Gallian.  I mentioned Gallian because I've heard that it's very friendly, which is important for self-study.

Yes, Davenport's "Higher Arithmetic" is quite readable (and doesn't even require a background in calculus, so youngsters can give it a try).  He has a tendency to use a lot of words when equations would do, but that's part of the book's style.

Spivak's "Comprehensive Intro to Diff Geom" series is probably more advanced than what you're looking for, but they're pretty good.  Weird but good.  Differential geometry comes in two main flavors: curves and surfaces in R^3, and general abstract differentiable manifolds.  The former (e.g. do Carmo) is accessible with a good background in calculus, while the latter (e.g Spivak's series) shouldn't be attempted without some mathematical maturity.

[quote="yenlee"]
Spivak's "Comprehensive Intro to Diff Geom" series is probably more advanced than what you're looking for, but they're pretty good.  Weird but good.  Differential geometry comes in two main flavors: curves and surfaces in R^3, and general abstract differentiable manifolds.  The former (e.g. do Carmo) is accessible with a good background in calculus, while the latter (e.g Spivak's series) shouldn't be attempted without some mathematical maturity.[/quote]
I have been thinking about something related to this recently. Before I write my thesis paper next spring (it is not nearly an as serious thesis as a PhD thesis, but you devote a whole semester to writing it) I feel like I should study differentiable manifolds. Is it necessary to have studied gaussian geometry in $\mathbb{R}^{3}$ before tackling differentiable manifolds? The official course requirements at my university say this, but they are not always right.

Definitely not necessary.  Is it helpful to understand submanifolds of R^n before tackling abstract differentiable manifolds?  Yes.  Is it helpful to know about Gauss curvature and mean curvature of surfaces in R^3 before tackling Riemannian geometry?  Yes.  But in neither case is it actually necessary.  (Personally, I learned these things in the "wrong" order.  It was difficult, but I persevered.)  I would analogize the difficulties to those of learning general topology before understanding the topology of R^n.

That said, I should mention that if your university has a curves and surfaces course as a required prerequisite for your differentiable manifolds course, then that could cause some problems because your prof might assume that you know various things.

Btw, how on earth did this thread end up in "Calculus Computations and Tutorials?"


geometry
geometric transformation
rotation
calculus
integration
3D geometry
sphere

A square of side lenght $s$ lies in a plane perpendicular to a line $L$. One vertex of the square lies on $L$. As this square moves a distance $h$ along $L$ the square turns one revolution about $L$ to generate a corkscrew like column with square cross sections.

a) Find the volume of the column
b)What will the volume be if the square turns twice instead of once? Explain.

The volume can be computed by integrating the area of cross-sections perpendicular to $L$. Each such cross-section is a square of sidelength $s$. Rotation makes no difference at all.

Can you  explain a little more?? What is the integral you need to set up in order to get the volume?

What he's quoting is called "Cavalieri's principle". Two solids which have the same cross sections everywhere (with the cross sections taken pependicular to a given axis) have the same volume. (And you can relax that "same cross sections" to "cross sections of the same area.") This figure has the same cross-sections as a rectangular box, and has the same volume as that box.

[Note: Cavalieri, an early 17th-century follower of Galileo, used this idea with great cleverness to compute some areas and volumes - calculations that we would now call calculus, although this is a generation or two before Newton and Leibniz. But the same principle occurs in a 5th century A.D. Chinese computation of the volume of a sphere, and it is implicit it the Archimedes work called the [i]Method[/i], discovered in the 20th century on the famous Archmedes palimpset.]

If you insist on writing it as an integral, you can say

$\int_{a}^{a+h}s^{2}\,dx=s^{2}h.$

Personally, I don't see that writing it as an integral adds much to the understanding.


There are n buckets, each initially containing one ball. Balls are thrown, and must land in one of the buckets. The probability that a ball will land in a particular bucket is the ratio of the number of balls in that bucket to the total number of balls. How many balls must be thrown so that there is greater than 50% chance that the "fullest" bucket will have at least k balls?

This is a Polya's urn process. There are a lot of references available online. It is not hard to get an estimate for the number of balls, but I'm not sure that there exists an exact simple formula for that.


inequalities
complex numbers
linear algebra
linear algebra theorems

I recently decided to go back and really take a look at the introductory sections in my textbook on inner product spaces, and I realized that the proof of the Cauchy-Schwarz inequality (the $ | < x,y > | \le \parallel{}x\parallel{}\parallel{}y\parallel{}$ version) that they give feels largely unmotivated (consider $ \parallel{}x \minus{} cy\parallel{}^2 \ge 0$, do a little algebra, choose a nice $ c$, and it pops out!) and unlike the dot product version in $ R^n$, I can't find a nice motivating factor or way to think of this inequality in a deeper way at this level of generality. I'd be grateful if anybody could provide any insights?

Thanks!

It's natural to try to prove the inequality by expanding $ \langle x \minus{} y,x \minus{} y\rangle\ge 0$. At first this does not work: we only get that the real part of $ \langle x,y\rangle$ does not exceed $ \frac {\|x\|^2 \plus{} \|y\|^2}{2}$.

Notice, however, that all is fine if $ \langle x,y\rangle$ happens to be a nonnegative real number and $ \|x\|$ and $ \|y\|$ happen to be equal. So we have proved a special case of the C-S inequality.

Finally, we reduce the general case of C-S to the special case using the homogeneity of C-S: if $ x$ or $ y$ is multiplied by a scalar $ \lambda$, both sides get multiplied by $ |\lambda|$. (By the way, if an inequality of this kind does not exhibit some degree of homogeneity, it's unlikely to be true.)


[quote="ashegh"]let (C) be a circle and M be a point in the interior of it.and let the mid point of a variable chord,wich goes through M,

be N.  find the locus of N.[/quote]

Locus is the circle with M and center of the given circle as extremities of a diameter

i will be glad if u post u solution ;) .i have two solutions for it,and i will post them ,if u like.

[quote="ashegh"]let (C) be a circle and M be a point in the interior of it.and let the mid point of a variable chord,wich goes through M,

be N.  find the locus of N.[/quote]

Let the equation to the circle is C is $x^2+y^2=r^2$.  Let M is the fixed point $(h, k)$.
Let $N(x_1, y_1)$ is the mid point of the chord.  Then its equation is given by

$xx_1+yy_1=x_1^2+y_1^2$.  If it passes through $M(h, k)$, then we have

$hx_1+ky_1=x_1^2+y_1^2$.  Hence the locus of N is given by

$x^2+y^2=hx+ky$ which is the circle with M and centre of the given circle as extremities of diameter.

And it's even easier in polar coordinates...

of course it is...

when i started to learn iversion and pole and polar saw this problem...

it has very nice solutions by inversion and pole and polar...

Oh, yeah... right angles....

Let O the center, then all triangles formed by MNO, have a right angle N. Then, all N are on a circle, with MN as a diameter.

Angel "Java" Lopez


[quote="Rose_joker"]$ a,b,c\in R \plus{}$ show that
$ \sum_{cyc} \frac {a^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\leq 2$
I was wondering whether there is a solution without brute force for this problem[/quote]
Actually, the following more general inequality holds:

Let $ a,$ $ b,$ $ c$ be positive real numbers. Prove that, if $ k \ge 1,$ then
$ \frac{a^2}{a^2\plus{}kab\plus{}b^2}\plus{}\frac{b^2}{b^2\plus{}kbc\plus{}c^2}\plus{}\frac{c^2}{c^2\plus{}kca\plus{}a^2}\plus{}\frac{2k\plus{}1}{k\plus{}2}\cdot \frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \le 2.$

This inequality is [b]Problem 60[/b], page 71 in book (where it is presented with a simple solution by Cauchy-Schwarz):

Võ Quốc Bá Cẩn, Trần Quốc Anh, [url=http://canhang2007.wordpress.com/2009/10/28/new-book-b%E1%BA%A5t-d%E1%BA%B3ng-th%E1%BB%A9c-va-nh%E1%BB%AFng-l%E1%BB%9Di-gi%E1%BA%A3i-hay/][b][i]Bất đẳng thức và những lời giải hay[/i][/b][/url], Nhà xuất bản Hà Nội, 2009 (Vietnamese).

[quote="can_hang2007"][quote="Rose_joker"]$ a,b,c\in R \plus{}$ show that
$ \sum_{cyc} \frac {a^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\leq 2$
I was wondering whether there is a solution without brute force for this problem[/quote]
Actually, the following more general inequality holds:

Let $ a,$ $ b,$ $ c$ be positive real numbers. Prove that, if $ k \ge 1,$ then
$ \frac {a^2}{a^2 \plus{} kab \plus{} b^2} \plus{} \frac {b^2}{b^2 \plus{} kbc \plus{} c^2} \plus{} \frac {c^2}{c^2 \plus{} kca \plus{} a^2} \plus{} \frac {2k \plus{} 1}{k \plus{} 2}\cdot \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \le 2.$

This inequality is [b]Problem 60[/b], page 71 in book (where it is presented with a simple solution by Cauchy-Schwarz):

Võ Quốc Bá Cẩn, Trần Quốc Anh, [url=http://canhang2007.wordpress.com/2009/10/28/new-book-b%E1%BA%A5t-d%E1%BA%B3ng-th%E1%BB%A9c-va-nh%E1%BB%AFng-l%E1%BB%9Di-gi%E1%BA%A3i-hay/][b][i]Bất đẳng thức và những lời giải hay[/i][/b][/url], Nhà xuất bản Hà Nội, 2009 (Vietnamese).[/quote]
Dear Can_hang2007,
I was wondering whether you can provide your beautiful solution for this problem...
After seeing the solution appeared in the book (which is exactly the same as Tourish's one), I with My friends who used to be Thailand representatives for IMO were inspired to solve it without using that nice identity and brute force by trying to solve it by elementary method but the result is a failure...
Sorry for the inconvenient if this bothers you too much...
Best Regards,
Rose_joker

[quote="can_hang2007"]
Let $ a,$ $ b,$ $ c$ be positive real numbers. Prove that, if $ k \ge 1,$ then
$ \frac{a^2}{a^2\plus{}kab\plus{}b^2}\plus{}\frac{b^2}{b^2\plus{}kbc\plus{}c^2}\plus{}\frac{c^2}{c^2\plus{}kca\plus{}a^2}\plus{}\frac{2k\plus{}1}{k\plus{}2}\cdot \frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \le 2.$[/quote]
Here is my proof(I haven't got your nice book yet):
The inequality is equavilent to
\[\sum{\left(1-\frac{kab+b^2}{a^2+kab+b^2}\right)}+\frac{(2k+1)(ab+bc+ca)}{(k+2)(a^2+b^2+c^2)}\leq2\]
\[\Longleftrightarrow \sum{\frac{kab+b^2}{a^2+kab+b^2}}\geq 1+\frac{(2k+1)(ab+bc+ca)}{(k+2)(a^2+b^2+c^2)}\]
By using Cauchy-Schwarz inequality we have
\[\sum{\frac{kab+b^2}{a^2+kab+b^2}}=\sum{\frac{b^2}{\frac{b(a^2+kab+b^2)}{ka+b}}}\geq \frac{(a+b+c)^2}{\sum{\frac{b(a^2+kab+b^2)}{ka+b}}}\]
Then we are required to prove that
\[\frac{(a+b+c)^2}{\frac{a^2b}{ka+b}+\frac{b^2c}{kb+c}+\frac{c^2a}{kc+a}+a^2+b^2+c^2}\geq 1+\frac{(2k+1)(ab+bc+ca)}{(k+2)(a^2+b^2+c^2)}\]
\[\Longleftrightarrow \frac{(k+2)(a+b+c)^2\sum{a^2}}{(k+2)\sum{a^2}+(2k+1)\sum{ab}}-\sum{a^2}\geq \sum{\frac{a^2b}{ka+b}}\]
\[\Longleftrightarrow \left(\sum{a^2}\right)\cdot\frac{3\sum{ab}}{(k+2)\sum{a^2}+(2k+1)\sum{ab}}\geq \sum{\frac{a^2b}{ka+b}}\]
\[\Longleftrightarrow \left(\sum{a^2}\right)\left(1-\frac{3\sum{ab}}{(k+2)\sum{a^2}+(2k+1)\sum{ab}}\right)\leq \sum{\left(a^2-\frac{a^2b}{ka+b}\right)}\]
Using Cauchy-Schwarz again
\[\sum{\left(a^2-\frac{a^2b}{ka+b}\right)}=\sum{\frac{ka^3}{ka+b}}\geq \frac{k(a^2+b^2+c^2)^2}{k(a^2+b^2+c^2)+ab+bc+ca}\]
It suffices to show that
\[\frac{k\sum{a^2}}{k\sum{a^2}+\sum{ab}}\geq 1-\frac{3\sum{ab}}{(k+2)\sum{a^2}+(2k+1)\sum{ab}}\]
\[\Longleftrightarrow \frac{3\sum{ab}}{(k+2)\sum{a^2}+(2k+1)\sum{ab}}\geq \frac{\sum{ab}}{k\sum{a^2}+\sum{ab}}\]
which is equavilent to
\[(2k-2)\sum{a^2}\geq (2k-2)\sum{ab}\]
which is obvious true since $k\geq 1$. So the inequality is proved.


geometry
angle bisector
geometry unsolved

Three mutually and externally disjoint circles centred at X1, X2 and X3 respectively on the same plane. The two internal common tangents of Circle 2 and 3, (circle 3 and 1, circle 1 and 2) meet at
P, (Q, R respectively).
Prove that X1P;X2Q and X3R are concurrent.

name the circles $ w_1,w_2,w_3$ with centers $ X_1,X_2,X_3$ respectively,now for $ \triangle PQR$,circle $ X_1$ plays the roll of the excircle wrt $ P$,thus $ PX_1$ is the angle bisector of $ \angle QPR$ and similary $ QX_2,RX_3$ are the angle bisectors of $ \angle PQR,\angle QRP$ respectively,so the problem asks us to show that the angle bisectors of triangle $ \triangle PQR$ are concurrent.

QED