function
real analysis
calculus
integration
derivative
inequalities
floor function

Prove that a monotone function defined on $\mathbb{R}$ is differentiable almost everywhere.

i dont think it's true , but my be you can show that there is at least one point that f is differentible in  :!:

Actually I have no ideea.  :D That's why I posted it. When I read about functions with bounded variation this fact was simply invoked. It was invoked another fact: any function with bounded variation can be written as difference of two monotone functions. I have no ideea for a proof for this fact too. I will be glad if someone can show it.

yassinus: this theorem is true. I'd look for it in analsysis textbooks that do Lebesgue measure and integration, possibly in a chapter labelled "differentiation."

Bear in mind that the function doesn't have to be the integral of its derivative. A famous example is the Cantor "Devil's Staircase" function which is a non-decreasing continuous function mapping $[0,1]\mapsto[0,1].$ It is differentiable at every point in the complement of the Cantor set (which is "almost everywhere" because the Cantor set has measure zero), but everywhere that the derivative exists, it is zero.

O-o...
[quote="xirti"] any function with [u]continuous[/u] variation[/quote].
I read it as bounded variation...
Or you mean exactly bounded variation? If no, then please write what is you mean "continuous variation"?

xirti meant bounded variation; he's just not familiar with the term yet.

the problem is that i dont know what does mean the measure  ( i will study it the next year )   :?

[quote="anik_andrew"]on the $[a,b]$
$f(x) = g(x) -h(x)$ where $g(a)=0$ and $g(x)={\bf V} (f,[a,x]),\ x\in [a,b]$
and $h(x)= g(x)-f(x)$. One can show that $h$ is monotonous.[/quote]

I only can prove that $g$ is an increasing function. We have that, for any $x<y$, each variation of $f$ with respect to a division of the interval $[a,x]$: $\sum_{i=1}^n |f(x_{i+1})-f(x_i)|<\sum_{i=1}^{n+1} |f(x_{i+1})-f(x_i)|$ ,where $x_{n+1}=y$, which is the variation of $f$ with respect to a division af the interval $[a,y]$.
From this fact \[ g(x)=\sup_{a=x_1<x_2<..<x_n=x}\sum_{i=1}^n |f(x_{i+1})-f(x_i)|< \]
\[ \sup_{a=x_1<x_2<..<x_n<x_{n+1}=y}\sum_{i=1}^{n+1} |f(x_{i+1})-f(x_i)|=g(y) \]

To prove that $h$ is monotonus we need two statement
1) $|f(a)-f(b)|<{\bf V}(f,[a,b])$
2) if $f\in {\bf BV}([a,b])$ then $\forall c\in [a,b]$ $f\in {\bf BV}([a,c])$ and $f\in {\bf BV}([c,b])$, and
\[ {\bf V}(f,[a,b])={\bf V}(f,[a,c])+{\bf V}(f,[c,b])  \].
The first statement is follow from division interval $[a,b]$ by points $\{a,b\}$. And the second one i will prove later (if you need :) ) (it's not  small...)
So now we can write $h(y)-h(x)=g(y)-f(y)-g(x)+f(x)={\bf V}(f,[x,y])-(f(y)-f(x))\ge 0.$
Thats all :)

[quote="anik_andrew"]
2) if $f\in {\bf BV}([a,b])$ then $\forall c\in [a,b]$ $f\in {\bf BV}([a,c])$ and $f\in {\bf BV}([c,b])$, and
\[ {\bf V}(f,[a,b])={\bf V}(f,[a,c]+{\bf V}(f,[c,b]))  \]
[/quote]

What do you understand by $\bf{BV([a,b])}$?(f has bounded variation on $[a,b]$?) And by $\bf{V}([a,b]+V...)$?

I think that you want to prove that:

\[ {\bf V}(f,[a,b])={\bf V}(f,[a,c])+{\bf V}(f,[c,b])\Leftrightarrow \]
\[ \sup_{a=x_1<x_2<...<x_n=b}\sum_{i=1}^n|f(x_{i+1})-f(x_i)|= \]
\[ \sup_{a=y_1<y_2<...<y_m=c}\sum_{i=1}^m|f(y_{i+1})-f(y_i)|+\sup_{c=z_1<z_2<...<z_p=b}\sum_{i=1}^p|f(z_{i+1})-f(z_i)|. \]
It is obvious that $LHS\ge RHS$ because for every pair of divisions of $[a,c]$ and $[c,b]$ their reunion is a division of $[a,b]$.
The converse inequality, $LHS\le RHS$, follows from the fact that $|f(y_{m-1})-f(c)|+|f(c)-f(z_1)|\ge | f(y_{m-1})-f(z_1)|$.
From these two follows that
Thank you for hints. But can someone prove the fist posted problem?

Yes xirti, you are right. Thanks, i'm edited

So... the first posted problem is't solved yet? Does anyone have an ideea?

So.... Have you come up with any other ideas?

[quote="blahblahblah"]Except that it isn't true.[/quote]
Why??  :(

[quote="perfect_radio"]Take the $\lfloor x \rfloor$ function or something similar if you want strictly monotonic.[/quote]
Yes, I forgot, I want say strictly monotonic.
Thanks perfect_radio.

Why are you thanking me? It's still not true. I think you should search for the right statement.

[quote="leibniz"]This is an other one like xirti's problem:
[code]Proove that if a real function is strictly monotonic, so it's discontinuous in a finit number of points[/code]
I think that is exciting too![/quote]

It should be that the function is discontinous in a set of numbers which is numerable, not necesarily finite.

Indeed.  Monotonic functions only have jump discontinuities, and we can associate a rational to each of these, so there can be only countably many for a given function.


Let $p(n)$ be the number of ways of expressing the natural number n as the sum of three positive integers.
Show that
(a) $p(n)=\lfloor{\frac{n^2+3}{12}}\rfloor$
(b) The number of ways of expressing n as no more that three positive integers is given by   $\lfloor{\frac{(n+3)^2+3}{12}}\rfloor$

I am assuming by the number of different ways you mean $(1,2,3) \not = (1,3,2)$.

Given $n$ start with $\forall \ (a,b,c) \ a = 1$. Thus we are left with the number of ways to partition $n-1$ into two postive integers. Thus $\forall \ b \leq \ n-2, c \ \text{exists}$. However half of these are repeats (technically there are $\lfloor \frac{n-1}{2} \rfloor$. So if $a=1$ there are $\lfloor \frac{n-1}{2} \rfloor$ solutions. Likewise when $a = 2$ there are $\lfloor \frac{n-2}{2} \rfloor$ solutions, and so on.

Let me clarify for what I mean by "different".
e.g. Expressing 6 as a sum of three positive integers.
Then
6 = 1 + 1 + 4 = 1 + 2 + 3 = 2 + 2 + 2.
There are only 3 possible ways.
1 + 1 + 4 and 4 + 1 + 1 are considered as the same pattern.


Prove that $\left| sin\ a\cdot cos\ b\cdot sin\ c\right|+\left| cos\ a\cdot sin\ b\cdot cos\ c\right|\leq 1$

[hide] LS is  than    which is  than the maximum of   which is  [/hide]


limit
trigonometry
function
algebra
polynomial
algebra proposed

But why would you not want to use H'opital's rule or Taylor polynomials?  These are extremely useful tools in mathematics.  That's like manipulating triangle formulas without trigonometric functions!


The base of the decimal number system is ten, meaning, for example, that $ 123 \equal{} 1\cdot 10^2 \plus{} 2\cdot 10 \plus{} 3$. In the binary system, which has base two, the first five positive integers are $ 1,\,10,\,11,\,100,\,101$. The numeral $ 10011$ in the binary system would then be written in the decimal system as:

$ \textbf{(A)}\ 19 \qquad
\textbf{(B)}\ 40\qquad
\textbf{(C)}\ 10011\qquad
\textbf{(D)}\ 11\qquad
\textbf{(E)}\ 7$

[hide="Solution"]
$ 10011_2 \equal{} x_{10}\implies x \equal{} 1\cdot 2^4 \plus{} 0\cdot 2^3 \plus{} 0\cdot 2^2 \plus{} 1\cdot 2^1 \plus{} 1\cdot 2^0 \equal{} \boxed{19}$

So the answer is $ \boxed{\textbf{(A)}}$
[/hide]


inequalities
algebra
polynomial
absolute value

Problem 1

I just need little help on finishing this proof.

Problem says;

If $a+b+c = 0$, then prove that $a^3 + b^3 + c^3 = 3abc$.

My answer:

From AM-GM Inequality on three variables, we have $\frac{a^3 + b^3 + c^3}{3} \geq abc$

$\Rightarrow a^3 + b^3 + c^3 \geq 3abc$.

All I need to do now is since I have right approach (I think I do.. Don't I?), I just have to do "=" part.

I tried to do expansion of $(a+b+c)^3$ and I got:

$a^3 + b^3 + 6abc + 3(ab^2 + ac^2 + a^2b + a^2c + b^2c + bc^2) = 0$

Now the problem is I don't know whether this fact is useful or not and if it does, how do I apply to my original inequality that I found to finish the proof?

Thank you a lot!

Well, AM-GM works only for positive (non-negative?) numbers anyway, so a^3+b^3+c^3 is minimized when a, b, c are each 0.

You can't finish the proof with only AM-GM because it doesn't apply to negative numbers.

I'm not quite sure how to proceed from there, but...

[hide="hint"]
Look for a factoring of the polynomial $a^3+b^3+c^3-3abc$. This is a very common and useful factoring.[/hide]

Thanks to TMildorf for helping me on this factoring.. :D

After I saw that, this problem is so easy... ;)

[hide="final proof"]

$a^3+b^3+c^3-3abc = (a+b+c)(a^2 - ab + b^2 - bc + c^2 - ac)$

Since $a+b+c=0$, we get $a^3 + b^3+c^3 -3abc = 0$ and moving $-3abc$ to right, we get:

$a^3+b^3+c^3 = 3abc$.QED.[/hide]

Thanks for your help guys!

Question: So.. AM-GM only appears to natural numbers?

No, it applies only to non-negative reals.  (If you wish to extend it to include the harmonic mean, you have to decide how to define HM when some of the numbers are 0.)  You can make it apply to all reals by throwing in a lot of absolute value signs.  The reasons for this should be clear from the proof, which requires that you be able to take square roots.

Source: Romanian NMO 1970 (9th grade)

Since $x+y+z=0$ we have $x+y=-z$, $y+z=-x$, $z+x=-y$

Knowing that $(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)$ $\Rightarrow x^3+y^3+z^3=(x+y+z)^3-3(-z)(-y)(-x)=3xyz$
Not hard  ;)


Suppose that for any point P chosen in the first quadrant of the xy plane, and infinitely many points $a_1,a_2,a_3...a_n$ chosen on the x-axis. Show that there exist infinitely many points such that the distance between P and each infinite $a_k$ is not rational.

If I understand correctly, there are infinitely many points $a_1, a_2, \ldots$, and we must show the distances to an infinite subset of these points are irrational, which is false.

Take P as (m,n).  Then choose rationals $r$ and $s$ such that $2rs=n^2$.  Then take the x-coordinate of $a_i$ to be $m+r^2-s^2$.  The distance is then $r^2+s^2$, which is rational, and there are infintely many such $r$ and $s$, so just let these $r$ and $s$ determine the x-coordinates of the points for a counterexample.

EDIT:  Corrected problem: the x-coordinates of the $a_i$ are positive integers, and we must show an infinite number of distances non-integer.

[quote="inteluser"]Then choose rationals $r$ and $s$ such that $2rs=n^2$.[/quote]

What if $n^2$ isn't rational?   :)

But the question asked for an arbitrary P, so I chose a P that led to a counterexample, though I clearly should have specified $n^2$ and $m$ rational.

No, no, I think the problem asks to prove that an infinite number of points $P$ exist that do satisfy the problem, so counterexamples don't affect the validity of the question.


function
algebra
floor function
number theory unsolved
number theory

Let $ g(x) \equal{} [\frac {x}{2}]$denote the floor function of $ \frac {x}{2}$, eg $ g(5) \equal{} 2, g(6) \equal{} 3$, etc.
Find an explicit formula for
$ A\equal{}2g(x) \minus{} 4g^{2}(x) \plus{} 8g^{3}(x) \minus{} 16g^{4}(x) \plus{} \ldots \equal{} \minus{} \sum_{k \equal{} 1}^{\infty} ( \minus{} 2)^{k}.g^{k}(x)$
where $ g^{k}(x) \equal{} g(g^{k \minus{} 1}(x))$.

Hello, can anyone help please?
I really need the solution or maybe an explanation whether the problem is doable...
Thanx.. :)

Let $ x\equal{}\sum_{i\equal{}0}^{\infty}{a_{i}2^{i}}$ where $ a_{i} \in \{ 0,1 \}$ for every $ i \ge 0$.

Then we can see

$ A\equal{}2g(x) \minus{} 4g^{2}(x) \plus{} 8g^{3}(x) \minus{} 16g^{4}(x) \plus{} \ldots \equal{} \minus{} \sum_{k \equal{} 1}^{\infty} ( \minus{} 2)^{k}.g^{k}(x)$

$ \equal{}\sum_{i\equal{}0}^{\infty}{a_{2i\plus{}1}2^{2i\plus{}1}}$.

But I have no idea to say this expression .


Solve

$ \begin{cases}|x^2\minus{}y|\plus{}|2x\plus{}1|\equal{}4\\|2x\plus{}y|\equal{}4\end{cases}$

the second one gives 2x+y=4 or 2x+y=-4 so y=4-2x or y=-4-2x so we set these values seperately into the first equation take cases for the polynomials inside the absolute values and after solving for each case seperately we get the solutions (x,y):
(-2+11^.5,8-2*11^.5)
(1,2)
(-2+3^.5,-2*3^.5)
(-1,-2)


geometry
circumcircle
incenter
geometry open

In a triagle ABC,BC is the smallest edge.Choose M and N on AB and Ac respectively such that BM=BC and CN=CB.Prove that $ \frac{MN}{BC}\equal{}\sqrt{3\minus{}2(cosA\plus{}cosB\plus{}cosC)}$

[b]Proposition.[/b] Let $\triangle ABC$ be a scalene triangle and $BC$ is its shortest side. $M,N$ are two points on $AC,AB,$ such that $ BN \equal{} CM \equal{} BC.$ Then $MN \perp OI,$ where $O,I$ are the circumcenter and incenter of $\triangle ABC$ and $\frac {_{MN}}{^{BC}} \equal{} \frac {_{OI}}{^{R}}.$

[b]Proof.[/b] In the isosceles triangles $\triangle OAC$ and $\triangle OAB,$ we have

$R^2 \minus{} OM^2 \equal{} AM \cdot MC , \ R^2 \minus{} ON^2 \equal{} AN \cdot BN$

$\Longrightarrow ON^2 \minus{} OM^2 \equal{} AM \cdot MC \minus{} AN \cdot BN$

$\Longrightarrow ON^2 \minus{} OM^2\equal{} (CA \minus{} BC)BC \minus{} (AB \minus{} BC)BC \equal{} BC(CA \minus{} AB)$

Since $\triangle MIC$ and $\triangle MIB$ are isosceles, we have $IM \equal{} IB,$ $IN \equal{} IC.$ Then

$IN^2 \minus{} IM^2 \equal{} IC^2 \minus{} IB^2 \equal{} BC(CA \minus{} AB)$

$\Longrightarrow ON^2 \minus{} OM^2 \equal{} IN^2 \minus{} IM^2 \Longrightarrow \ IO \perp MN.$

On the other hand, $\angle NBM \equal{} \angle OCI \equal{} 90^{\circ} \minus{} \frac {_1}{^2}\angle C \minus{} \angle A.$ If  $L \equiv{} CI \cap BM,$ then it follows that  $\angle NMB$ and $\angle OIC$ are supplementary, which implies that $\triangle IOC$ and  $\triangle MNB$ are pseudo-similar, hence

$\frac {MN}{OI} \equal{} \frac {BN}{OC}  \Longrightarrow \frac {MN}{BC} \equal{} \frac {OI}{R}.$


AMC
AIME
geometry
analytic geometry
quadratics
calculus
inequalities

you could just use hero's formula and maximize it, btw what is the circle of apollonius?
$9,40x,41x$ $s=\frac{9 + 81x}{2}$
$A = \sqrt{ (\frac{9 + 81x}{2})(\frac{9 + 81x}{2}-9)(\frac{9 + 81x}{2}-40x)\left(\frac{9 + 81x}{2}-41x)}$
$A = \sqrt{ (\frac{9 + 81x}{2})(\frac{-9 + 81x}{2})(\frac{9+x}{2})(\frac{9-x}{2}) }$
$A = \frac{9}{4} \sqrt{ (81x^2-1)(81-x^2)}$
maximizing this is a pain to say the least, but consider the am - gm
$\frac{ (81x^2-1) + (81-x^2)}{2} \ge \sqrt{ (81x^2-1)(81-x^2)}$
we want to maximize the right so we make that an equals sign, i am not sure that this works, but it seems that it could work, i dont want to go through all of the algebra

[quote="white_horse_king88"]You get 820 using the circle of Apollonius. (I got this problem wrong the first time I did a while back...)[/quote]

What is the circle of Apollonius?

http://planetmath.org/encyclopedia/ApolloniusCircle.html

Doesn't say much about maximizing though.

Think about which triangle would have the greatest area when the base is fixed.  However, I think there should be a good way without the circle of appolonius or however you spell it.  Actually, if you assign coordinates and use the pythagorean theorem, the problem becomes a quadratic (although with somewhat large numbers).  Still a bit harder but requires less knowledge (and can be used to derive the circle of appolonius).

as for maximizing that, the only way that i see doing that would be with simple calculus (all it really involves is finding the inflection point, and using the chain rule to differentiate), but of course that is not aime level math, but can anybody offer a good explanation of that circle of Apollonius (is seems like such a random theorem, where did you learn it)

Altheman: you have to consider than x cannot be greater equal to 9, as if it is, the 3 sides of the triangle won't satisfy the triangle inequality

9+360=369

hmm, i see, and i thought of another way to do this problem,
consider the triangle in a cooridinate system with (0,0) (9,0) and (a,b)
then $\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}$
and we obviously want to maximize b as that is the height and 9 is the base
simplifying gives $-a^2 -\frac{3200}{9}a +1600 = b^2$
if you remember, the max/min value of a quadratic is $\frac{4ac-b^2}{4a}$
so $\frac{4(-1)(1600) - \left(\frac{3200}{9}\right)^2 }{(-1)(4)} = b^2$
$1600+\left( \frac{1600}{9} \right)^2 = b^2$
$1600\left(1+\frac{1600}{81} \right) = b^2$
$b = \frac{40*41}{9}$
$9*\frac{1}{2} * \frac{40*41}{9} = \boxed{820}$

oh, and krustytheclown, that is built into the equation as is, if $x \ge 9$, then then it becomes imaginary or 0 think of the part that is $81-x^2$, i love how math works...i did not think of this, but for it to be a correct expression, then everything has to be true about it

Well, coincidentally  :P  (OR NOT!), if you use coordinates, the point C(x,y) such that AC:BC = 41:40 which is the maximum area of this triangle is on the circle.

Given A(0,0), and B(c, 0),

If the origin of circle O is on AB, then you can use coordinate geometry to find that the locus of all points C(x,y), that satisfy the original problem, lies on the circle (called circle of Apollonius). The maximum is therefore, when height is maximized, which is radius of circle. Using coordinate geometry you can find the radius to be (we want it in terms of a,b, and c) $\frac{abc}{b^2-a^2}$.

yes, i actually remember doing that locus in my geometry class just we did not discuss the significance of the problem, and we never did any generalization in my geo class
if we have points $A,B,C$ that are $(0,0),(a,0),(x,y)$ respectively (a is a constant, x and y are arbitrary), find the locus of points that satisfy $\frac{AC}{BC} = k$
using the distance formula, $\frac{ \sqrt{x^2+y^2} }{ \sqrt{ (x-a)^2 + y^2 } } = k$
and after some algebra, $\left(x + \frac{ak^2}{1-k^2}\right)^2 + y^2 = \left(\frac{ak}{1-k^2}\right)^2$
i wonder if this is something worth memorizing, or maybe just the idea
the only thing that bothers me is if we do $\lim_{k\rightarrow1}$(i.e. if they are equal), then we get a problem

[quote="qweretyq"]kalva didnt have a solution

13.  The triangle ABC has AB = 9 and BC/CA = 40/41. What is the largest possible area for ABC?

What do you get as answer?[/quote]

I don't know about my method, but using triangle inequality, we have $40x+9>41x$, so $x<9$, and so using heron's formula:

$\sqrt{(\frac{81x+9}{2})(\frac{81x-9}{2})(\frac{x+9}{2})(\frac{9-x}{2})}$, but I don't know how to maximize this, at first, I thought I could plug in $9$ and then find the greatest integer less than the answer, but obviously that wouldn't work :?  :(

i think the easiest way to do this is use coordinate geometry, with the locus of points that are the fixed ratio from 0,0 and 0,9, then maximize the height
the only thing that i wonder: is there a use for the fact that 9,40,41 is a pythagorean triple, because that jumped out a me, but it seems to have no use

I guess you could search for a better (elegant) solution using 9,40,41 as a pyth. triplet by finding the side lengths of the triangle going backwards knowing that the area is 820.

[quote="Altheman"]
the only thing that bothers me is if we do $\lim_{k\rightarrow1}$(i.e. if they are equal), then we get a problem[/quote]

Yeah, it's then an isoceles triangle, right?  Which makes sense.

[quote="Iversonfan2005"]$\sqrt{(\frac{81x+9}{2})(\frac{81x-9}{2})(\frac{x+9}{2})(\frac{9-x}{2})}$, but I don't know how to maximize this[/quote]

Not difficult. For maximization you can forget about the square root. Then multiply the first two and the last two factors:

$\frac{1}{16}(6561x^2-81)(81-x^2)$

Now forget about $\frac{1}{16}$ too and put $y=x^2$:

$(6561y-81)(81-y)=-6561y^2+531522y-6561$

And now it's the maximization of quadratic function: The maximum is reached in $y=\frac{531522}{2\cdot 6561}=\frac{3281}{81}$ etc.


I am using LaTeX to produce quiz papers. I would like to have a few version of this. One straightforward way of doing is to have individual .tex file for each version. Does anyone know if there's a way that I can produce different version from just one .tex file?

You can't produce more than one pdf file from each compilation.

1. You could make a single tex file which has different versions combined one after each other. Thus you would have
preamble
document 1
new page and restart page numbering
document 2
new page and restart page numbering
document 3
etc

The parts of each document that are common could be put in an external file so as to save having to write them out multiple times. The external file is then included using \include.

2. Another way is to put your questions in a database (eg csv format) and then use [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=datatool]datatool[/url] to include the questions you select according to some cirteria. I have used the previous version called [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=csvtools]csvtools[/url] (which may be included in your LaTeX distribution) successfully for mail merge.


probability counting distinguishability AMC

A parking lot has $ 16$ spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires $ 2$ adjacent spaces. What is the probability that she is able to park?

$ \textbf{(A)} \ \frac {11}{20} \qquad \textbf{(B)} \ \frac {4}{7} \qquad \textbf{(C)} \ \frac {81}{140} \qquad \textbf{(D)} \ \frac {3}{5} \qquad \textbf{(E)} \ \frac {17}{28}$

[hide="Solution"] The probability that she is not able to park is the probability that the $ 4$ empty parking spaces arrange themselves so that at least one car is between each.  The total number of ways the $ 4$ empty spots can arrange themselves is $ {16 \choose 4}$.  (The key here is that the cars are indistinguishable.)

[b]Case:[/b]  The leftmost spot is empty.  The other $ 3$ empty spots must each have a car immediately to its left, so group each empty spot with the car to its left.  The number of ways the other $ 3$ empty spots can be distributed is then $ {12 \choose 3}$ (because we have $ 3$ empty-spot-cars and $ 9$ cars).

[b]Case:[/b]  The leftmost spot has a car.  The $ 4$ empty spots must each have a car immediately to its left, so group each empty spot with the car to its left.  The number of ways the $ 4$ empty spots can be distributed is then $ {12 \choose 4}$ (because we have $ 4$ empty-spot-cars and $ 8$ cars).

The complement of our desired probability is then

$ \frac{ {12 \choose 3} \plus{} {12 \choose 4} }{ {16 \choose 4} } \equal{} \frac{11}{28}$

so the answer is $ \boxed{E}$. [/hide]

Actually, by your method it should be $ \frac{ {13 \choose 4} }{ {16 \choose 4} }$.  Incidentally, this gives a proof of Pascal's rule.

yeah, that's what I meant...I typed it up without reviewing my notes...my bad.

[hide="basically the solution above"]Consider the parking lot as twelve cars parked, and we insert 4 empty spaces between them (including at the ends).

If there are not to be two empty spaces next to each other (and Auntie Em will not be able to park), we simply choose 4 of the 13 possible places to put an empty space: $ \binom{13}{4}$.

For the total number of ways to put four empty spaces in 13 places, use the balls-and-urns argument* (4 balls, 13 urns) which gives us $ \binom{16}{4}$

The probability that she will NOT be able to park is therefore $ \frac{\binom{13}{4}}{\binom{16}{4}}\equal{}\frac{11}{28}$ and the probability she will be able to park is the complement, $ \boxed{\frac{17}{28}}$. (E)

*Balls and urns: If we have $ n$ identical balls to be placed in $ m$ distinct urns, we can biject any placing of the balls to a word made up of $ n$ "o"s (for balls) and $ m\minus{}1$ "|"s (dividers)- the "|"s divide the o's into the ways they will be distributed into urns. (ex.: oo||ooo means 2 balls in the first urn, 0 balls in the second, and 3 in the third). The number of possible words is $ \binom{n\plus{}m\minus{}1}{n}$, and therefore that is the number of ways to put balls into urns. [/hide]

[quote="t0rajir0u"][hide="Solution"] The probability that she is not able to park is the probability that the $ 4$ empty parking spaces arrange themselves so that at least one car is between each.  The total number of ways the $ 4$ empty spots can arrange themselves is $ {16 \choose 4}$.  (The key here is that the cars are indistinguishable.)

[b]Case:[/b]  The leftmost spot is empty.  The other $ 3$ empty spots must each have a car immediately to its left, so group each empty spot with the car to its left.  The number of ways the other $ 3$ empty spots can be distributed is then $ {12 \choose 3}$ (because we have $ 3$ empty-spot-cars and $ 9$ cars).

[b]Case:[/b]  The leftmost spot has a car.  The $ 4$ empty spots must each have a car immediately to its left, so group each empty spot with the car to its left.  The number of ways the $ 4$ empty spots can be distributed is then $ {12 \choose 4}$ (because we have $ 4$ empty-spot-cars and $ 8$ cars).

The complement of our desired probability is then

$ \frac { {12 \choose 3} \plus{} {12 \choose 4} }{ {16 \choose 4} } \equal{} \frac {11}{28}$

so the answer is $ \boxed{E}$. [/hide][/quote]
"The key here is that the cars are indistinguishable."

But how do we know if the objects are indistinguishable in a problem?  I think the cars are distinguishable because they are different.

Treating them as indistinguishable doesn't affect the computation; any parked car is an obstruction, regardless of what car it is.  In other words, treating the cars as distinguishable just multiplies the numerator and denominator of the computation by $ 12!$.

Yep, I know they usually don't matter if probability is required, but what if they are asking you about the number of combinations in the numerator?

Then they specify whether the cars are distinguishable or indistinguishable :wink:

I think for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83102]this problem[/url], (the babies) being whether distinguishable makes a difference. Why?

For computing probabilities the answer is not the balls-and-urns answer because the babies are not being assigned to distinct parking spaces (if I might mix problems here) and the probability of a given balls-and-urn case is different depending on how many cars are born on a given day (if I might mix problems again).

Edit:  I should be more specific.  In this problem, each arrangement of cars occurs with equal probability, and the computation is essentially the same regardless of whether the cars are treated as distinct or indistinct because all we have to do is divide by $ 12!$.  On the other hand, in the babies problem we have to divide by things like $ \frac {6!}{1!2!2!1!}$ and the factor changes depending on which arrangement of babies we're looking at, and it's the distinguishable computation that gives the actual [b]probabilities.[/b]

Yeah I sorta see your argument, but since two babies can be born on different times during a day (in different order), shouldn't we count their number of permutations in factors like $ \frac {6!}{1!1!2!2!}$?

We are.  Let me explain.  Instead of numbering by day, number by baby:  call the babies $ 1, 2, 3, 4, 5, 6$.  Then a birth order is an assignment of different days to each baby.  A particular assignment of days corresponds to a string

$ WMWTMM$

of letters from the alphabet $ \{M, T, W, R, F \}$.  In this example, the number of permutations of the above assignment of days to babies is $ \frac{6!}{2!3!1!}$.

No. That's not what I meant.  For example, if we label $ M_1$ as the time for the first baby born on $ M$, and similarly for $ M_2$ and $ M_3$, then each of the $ M$'s would be distinct, and we shouldn't divide by $ 3!$.  Similar argument for $ M$.

Exactly!  That's the reason your original argument is wrong.  The balls-and-urns argument corresponds to not taking those factors into account.  (My apologies; I still wasn't being particularly clear.)

But neither the methods treat the same day as distinct, like dividing the 3 M's into distinct $ M_1,M_2,M_3$ and the 2 W's into $ W_1,W_2$.  So in your previous example with baby 1 in the first position and so on for baby 6 there should be $ 6!$ permutations instead of just $ 6!/(3!2!1!)$.

Hmm.  Okay, ignore that argument completely; it gives a misleading view of the probabilities involved even if it can be corrected.  So let me go back to your original question:  [b]computing actual probabilities essentially requires that you treat every object involved as distinct.[/b]

What's wrong with the following solution?

The total number of configurations of the 12 cars with 4 parking spaces is $16!/4!$ if you consider the spaces to be indistinguishable. Now arrange 13 cars with 2 spaces (count the configurations in which Auntie Em has already parked) which yields $15!/2!$. But if you divide the two quantities, it does not result in an answer choice.

[quote="Zellex"][hide="faster?"]

Okay, consider the cases where she CANNOT park her car.  There are $ 4$ empty parking spots.  In order for her to not be able to park, just place a car between each of the parking spots.

So now, we have this | o | o | o |, where | = parking spots, and  o = cars.

We have 9 cars left, and we want to fit them in wherever.  That's $ 13C3$ because we have $ 9$ cars and $ 4$ dividers.  There are $ 16C4$ maximum ways to choose those 4 spots out of the 16.

So, the probability that she can't park there is $ \frac {13C3}{16C4}$.

Take one subtract that probability to get the answer.[/hide][/quote]

Why do we have 4 dividers? Can someone help me if I'm interpreting this solution wrong: so, since we are trying to prevent her parking we place a car between the open parking spaces. ( we placed 4 cars) Thus, we have 9 cars left.... So it's simply a matter of how many ways we can arrange these remaining cars? but they are not distinguishable and so.... this is where I am confused at...

^^^^^^^^^bump

[color=#FF0000][you just posted yesterday. There is no need for a bump] -David[/color]

[quote="AlanLiangTheAsian"]Why do we have 4 dividers?[/quote] We're counting sequences of the form [code]o|ooo|oo|ooo|ooo[/code] where an o represents a car and a | represents a parking space, such that there are 4 spaces, 12 cars, and no two parking spaces are adjacent.

To do this, we start with [code]|o|o|o|[/code] and then add in the remaining nine cars to the five "slots".  By Balls and Urns the number of ways to do this is $\textstyle\binom{9+4}{4}$.

This is what I hate about dealing with probabilities of indistinguishable objects. The problem is that when dealing with these objects, calculating probabilities by taking the number of correct outcomes over the number of possible outcomes often does not generate true probabilities of events. Some cases which are each considered as one event may not have equal probabilities. For example, suppose a problem asks the probability of drawing a red ball out of a box which contains 1 white ball and 100 indistinguishable red balls. The answer would be 1/2, even though this obviously does not correspond to the true probability. Luckily, for this reason these types of problems are rarely given.


geometry
conics
ellipse
geometric transformation
reflection

Farmer John is inside of an ellipse with reflective sides, given by the equation x^2/a^2 + y^2/b^2 = 1, with a > b > 0. He is standing at the point (3, 0), and he shines a laser pointer in the y-direction. The light reflects off the ellipse and proceeds directly toward Farmer Brown, traveling a distance of 10 before reaching him. Farmer John then spins around in a circle; wherever he points the laser, the light reflects off the wall and hits Farmer Brown. What is the ordered pair (a, b)?

[quote]Farmer John then spins around in a circle; wherever he points the laser, the light reflects off the wall and hits Farmer Brown.[/quote]
From this, we know that the two farmers are at the foci of the ellipse. Because the sum of the distances from the foci to the ellipse is 10, the major axis is 10. $ a\equal{}\frac12(10)\equal{}5$. There is no $ h$ or $ k$, so the center is at the origin. With one of the foci at $ (3, 0)$, $ c$ must be 3. $ b\equal{}\sqrt{a^2\minus{}c^2}\equal{}\sqrt{5^2\minus{}3^2}\equal{}4$. $ (a, b)$ is $ (4, 3)$.

[quote="ThinkFlow"]How do you prove that the laser reflected off of the ellipse must pass through the other foci? :maybe:[/quote]

Because the normal line to the ellipse at P is bisector of the angle with vertex P that passes through the focus. then according to the rule of reflected rays , the reflected ray from one foci passes through the another foci. thet is the reason

[quote="ThinkFlow"]How do you prove that the laser reflected off of the ellipse must pass through the other foci? :maybe:[/quote]

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=267665[/url] for the proof.


hi im Ze-S from germany
i just wanna tell abt a person Hazrat Ali he was the czn of The Holy Prophet Mohammad (p.b.u.h) Ali was a Gr8 mathematican, once to him came some ppl n asked a question, they said one man had 3 sons & 17 camels , now he doesnt live anymore his & his will is that the eldest one must get 1/2 of those camels, the second one 1/3 n the 3rd one 1/9 , no body in the whole city could solve it so they came to Ali , & he solved it in minutes. ok i dont wanna tell the answer now if anybody can solve it , so plz do it.

Is this a kind of a mathematical riddle?  :D

Given the combination of poor english and chatspeak, I can't make heads or tails of it.

Problem:

a guy had 17 camels and 3 sons.  when he dies, he wanted 1/2, 1/3, 1/9 of the camles be given to sons I, II, III respectively.  how is this possible without spliting camels?

[hide="solution"]
add one camel from yourself, now you have total 18 camles.
so the first son gets 9
the second son gets 6 a
and third son gets 2
but notice 9+6+2=17. so you take your own camle back

btw, i don't think that this problem was originated by "Hazrat Ali"...[/hide]

yeah, 1/2+1/3+1/9=17/18.  If you put your own camel in the 17 to get 18, the three sons will get 9,6, and 2 camels respectively.  Since they only took 17 camels (17/18 of the 18) the one camel remaining, which was yours, you can take back.
btw, i think the fact that 1/2+1/3+1/9=17/18 and $18\nmid17$ is the reason it's a [i]problem[/i] ;)

Well this is a famous riddle I've heard before...

But technically, it's implied that the father wants the remainined 1/18 to be let free.  Thus, his will cannot be fulfilled exactly.

The answer is very simple, once you know it.

Add one camel to make it 18.
the son who gets 1/2 gets 9 camels
the son who gets 1/3 gets 6 camels
the son who gets 1/9 gets 2 camels

totaling to 17 camels and return the borrowed camel to its owner

It's a BS problem because the character is adding something of his own
to achieve something, then taking it back.

That's a major example of fudging.

first add 1 camel now it is 18 camels
1/2 of 18 is 9 and it is the elders share
1/3 of 18 is 6 and it is the second child share
the last child takes 2 camels which is 1/9 of 18
then you can take the camel you add first

I think it is a great idea and the best you can do it. If you think of it like you can only have whole camels, then you are rounding to "the nearest camel". But it is not exact. If you take each fraction and multiply it by 17, here is what you get:
1/2*17=8.50
1/3*17=5.66
1/9*17=1.89
Since the fractions don't add up to a whole, you could take the leftover 1/18 and distribute it among the three sons to round off. I think the answer is the best because this way the most benefit is received and I think the man would also be happy with it. Also, the will was not exact so no exact answer can be found.

this is simple . and very common puzzle indeed.

If it's so simple and so common, what's the answer?   :mad:
To Ze-S, why does your signature translate into the words of a Hindi love song?
"I will remember you / my heart does not forget / I will not forget my love"  :o


inequalities
real analysis
real analysis solved

I'd like to post a few problems from the book "Selected problems from real analysis" just to see if you like them. I apologize in advance if some of them have already appeared on this forum...

So here comes the first one. Find all sequences $x_n$ ($n=1,2,\dots$) such that $|x_{m+n}-x_m-x_n|\leq \frac 1{m+n}$ for all $m,n\geq 1$.

I have already solved this one. It was posted on the inequalities forum. By the way, really nice problem. Please, post more problems.

We get $x_m=mx_1$, of course :).

First we get, from $|x_{n+1}-x_n-x_1|\le\frac 1{n+1}$, that $x_{n+1}-x_n\to x_1$ as $n\to\infty$. Call this $(*)$. We can now fix $m$, and write the inequality as $|(x_{n+m}-x_{n+m-1})+\ldots+(x_{n+1}-x_n)-x_m|\le\frac 1{m+n}$. When $n\to\infty$, we get $(x_{n+m}-x_{n+m-1})+\ldots+(x_{n+1}-x_n)\to x_m$, and also to $mx_1$, according to (*).

Actually it is a very nice problem!! I may want to use it, what is the SPRA, I mean, to know the source and make the quotation.

Best,

I can assure you that if SPRA is not something very old, then it is not the source of the problem. I've seen it in an Leningrad olympiad long time ago and I guess it is not the first time it was used.

SPRA is my abbreviation for the book "Selected Problems from Real Analysis"  :D
Harazi is right, the problem is much older than the book and was known in Leningrad (where the book was written) for a long time. To find the true origin of the problem seems quite a hopeless task now...

Take natural numbers $ a,b,n $. Define $ P:= x_{n+a} - x_{n} - x_{a} $ ,
$ Q:= x_{n+a+b} - x_{n+a} - x_{b} $ , $ R:= x_{n+a+b} - x_{n} - x_{a+b} $.
Then $ |P| \leqslant \frac{1}{n+a} \leqslant \frac{1}{n} $,
$ |Q| \leqslant \frac{1}{n+a+b} \leqslant \frac{1}{n} $,
$ |R| \leqslant \frac{1}{n+a+b} \leqslant \frac{1}{n} $,
so $ |x_{a+b} - x_{a} - x_{b}| = | P + Q - R | \leqslant |P| + |Q| + |R| \leqslant
\frac{3}{n} $ so we get that for every natural $ a,b $ for every natural $ n $,
that $ |x_{a+b} - x_{a} - x_{b}|  \leqslant \frac{3}{n} $. Therefore
$ x_{a+b} = x_{a} + x_{b} $, so $ x_{n} = n x_{1} $.


[quote="Phelpedo"]If $a+b+c=1$, find (with proof) the minimum and maximum possible value of $ab+bc+ca$.[/quote]
I think a minimum requires positive reals. :huh:
[hide="Because"] We can have the numbers be $-n$, $0$, $n+1$, $n\to\infty$. Thus, $ab+bc+ca=-n^{2}-n$
$\lim_{n\to\infty}-n^{2}-n=-\infty$ [/hide]

[quote="lotrgreengrapes7926"]
I think a minimum requires positive reals. :huh:
[hide="Because"] We can have the numbers be $-n$, $0$, $n+1$, $n\to\infty$. Thus, $ab+bc+ca=-n^{2}-n$
$\lim_{n\to\infty}-n^{2}-n=-\infty$ [/hide][/quote]

Yeah, in the positive reals case we'll get that maximum is $\frac{1}{3}$ and minimum approaches $0$.

[hide="Maximum"]
From Cauchy, we have $(a^{2}+b^{2}+c^{2})(1^{2}+1^{2}+1^{2}) \geq (a+b+c)^{2}\implies a^{2}+b^{2}+c^{2}\geq \frac{1}{3}$.

Then, $(a+b+c)^{2}= a^{2}+b^{2}+c^{2}+2(ab+bc+ac)$
$\implies \frac{2}{3}\geq 1-(a^{2}+b^{2}+c^{2}) = 2(ab+bc+ac) \implies ab+bc+ac \leq \frac{1}{3}$.[/hide]

[hide="Minimum"]
(Going with the positive reals assumption):
$\lim_{a \rightarrow 0, \ b \rightarrow 0}(ab+b(1-a-b)+a(1-a-b)) =0$.[/hide]


geometry
circumcircle
incenter
geometry solved

In triangle $ABC$, $\angle C = 30^{\circ}$, $O$ and $I$ are the circumcenter and incenter respectively, Points $D \in AC$ and $E \in BC$, such that $AD = BE = AB$. Prove that $OI = DE$ and $OI \bot DE$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=14655 .

  darij


Putnam
analytic geometry
geometry
circumcircle
graphing lines
slope
geometric transformation

What is the maximum number of rational points that can lie on a circle in $ \mathbb{R}^2$ whose center is not a rational point? (A [i]rational point[/i] is a point both of whose coordinates are rational numbers.)

Just an illustrative example: The circle with center at $ (\sqrt{2},2\minus{}\sqrt{2})$ that passes through the origin also passes through $ (2,2).$

[hide]

The answer is $ \boxed{2}$. The circle with center (0, $ \frac{\sqrt{3}}{2}$) passes throught (0,-1/2) and (0,1/2). 3 is impossible because the circumcenter of a triangle lies at the intersection of the perpendicular bisectors of the sides, which is a rational point if all the vertices are.

[/hide]

edit: Realized I can't add.

Meh, spent way too much time on that problem...

It doesn't pass through $ (1/2, \pm 1/2)$. Note the circumcircle of that (right) triangle is centered at $ (1/2,0)$, the midpoint of the hypotenuse.

All of my work on this started from saying WLOG the circle passes through the origin. After that, the issue was, for a circle centered at $ (a,b),$ whether $ a$ and $ b$ were linearly dependent over the rationals.

Algebraically, a generic circle has equation $ x^2 \plus{} y^2 \equal{} 2ax \plus{} 2by \plus{} c$.  Substituting in the values of three rational points and solving the resulting system of linear equations (which has all rational coefficients) results in rational values of $ a, b, c$ - in particular, the center $ (a, b)$ has rational coefficients.

[quote="Kent Merryfield"]Just an illustrative example: The circle with center at $ (\sqrt {2},2 \minus{} \sqrt {2})$ that passes through the origin also passes through $ (2,2).$[/quote]

I came up with a similar example, with 2 being the maximum number of rational points.

How many points do you think I'm likely to get if I couldn't find an example of a circle with 2 rational points, but provided a proof that 3 rational points was not possible?

Since it is a B1 the graders might be harsh.

So, like an idiot I didn't think to try to find a contradiction if there are 3 rational points. But I think what I did is going to end up being effectively what t0rajir0u said. If you have a rational point, then you can write the radius as a sum of rational multiples of the center coordinates (plus a rational number minus the distance of the center from the origin). But then this representation is unique since all of those components are going to be independent over the rationals, so you can easily show that you can have at most two rational points as you can have at most 2 free variables in that representation.

I feel like my argument is essentially correct but my explanation of it on the exam was even worse than what I just wrote here, so I'll be grateful for whatever points I receive on this problem.

[quote="FibonacciFan"][hide]

the circumcenter of a triangle lies at the intersection of the perpendicular bisectors of the sides, which is a rational point if all the vertices are.

[/hide][/quote]

Why is this true? I had the same idea but couldn't find a proof and eventually resorted to messy coordinates instead.

Basically, the perpendicular bisectors have rational equations. Each midpoint is a rational point, and the perpendicular to something with rational slope has rational slope (or is vertical).

Here's the solution from the notes I made while proctoring. It's fairly inefficient and certainly not the best solution, but I thought I'd share it.

Circles with no rational points do not concern us. Suppose that the circle in question has at least one rational point. Since translation by a rational vector does not change which points are rational and which not, we may assume WLOG that the circle passes through $ (0,0)$ and has $ (a,b),$ where at least one of $ a,b$ is irrational as its center. The circle is

$ x^2\plus{}y^2\minus{}2ax\minus{}2by\equal{}0.$

Now suppose there is a rational point $ (x,y)\ne (0,0)$ on this circle. Then we would have $ ax\plus{}by\in\mathbb{Q}.$ Suppose there is a second such point $ (x',y')$ such that $ ax'\plus{}by'\in\mathbb{Q}.$ That is, we have the system

$ \begin{cases}ax\plus{}by\equal{}u&\\ax'\plus{}by'\equal{}v&\end{cases}$ , where $ u,v\in\mathbb{Q}.$

Consider this a linear system to be solved for $ a$ and $ b.$ If the determinant $ xy'\minus{}x'y\ne0,$ then we can write $ (a,b)$ and the product of a rational matrix with $ (u,v),$ which would make $ (a,b)$ a rational point - which it's not. We conclude that $ xy'\minus{}x'y\equal{}0$ which means that any two rational points (other than the origin) on the circle must lie on one particular line through the origin. But every line intersects the circle in at most two places, which is at most one place other than the origin. Hence the number of rational points on the circle is at most 2.

I thought I'd share my solution.

[hide]
Assume a circle has three rational points, call them $ (x_1,y_1), (x_2,y_2), (x_3,y_3)$. Then, there exist lines with rational coefficients that pass pair of these points. In other words, there exist

$ y\equal{}m_{12}x\plus{}b_{12}$
$ y\equal{}m_{23}x\plus{}b_{23}$
$ y\equal{}m_{31}x\plus{}b_{31}$

where $ m_{ij},b_{ij}\in\mathbb{Q}$ and it's clear which line correspond to which points.

Let the center of the circle be $ (\alpha,\beta)$. Then,

$ (x_1\minus{}\alpha)^2\plus{}(y_1\minus{}\beta)^2\equal{}(x_2\minus{}\alpha)^2\plus{}(y_2\minus{}\beta)^2$
$ \frac{1}{2}(x_1^2\minus{}x_2^2\plus{}y_1^2\minus{}y_2^2)\equal{}\alpha(x_1\minus{}x_2)\plus{}\beta(y_1\minus{}y_2)$

Because $ LHS\in\mathbb{Q}$,
$ \alpha(x_1\minus{}x_2)\plus{}\beta(y_1\minus{}y_2)\equal{}\alpha(x_1\minus{}x_2)\plus{}m_{12}\beta(x_1\minus{}x_2)\equal{}(x_1\minus{}x_2)(\alpha\plus{}m_{12}\beta)\in\mathbb{Q}$.

At most two of $ x_1,x_2,x_3$ are equal; so WLOG, let $ x_2$ be the unique one.
Because $ 0\neq x_1\minus{}x_2\in\mathbb{Q}$, $ \alpha\plus{}m_{12}\beta\in\mathbb{Q}$.

Similarly, $ \alpha\plus{}m_{23}\beta\in\mathbb{Q}$. We know the $ m_{12}\neq m_{23}$ because the points form a triangle and a triangle cannot have parallel sides.

So from then on, you just take linear combinations to show that $ \alpha,\beta\in\mathbb{Q}$. Of course, there's the case when one of the lines are vertical, but then the $ x$s would be equal, and that has already been accounted for.
[/hide]

This question seems mysterious.  Take,

$$(x - \sqrt{2})^2 + (y - \sqrt{2})^2 = 1$$

This doesnt pass through any rational points does it?

What does your question have to do with the problem?

@JBL, I was wondering if the radius of the circle would make a difference. thats all

I do not understand the connection between your first and second comments.  Perhaps you could write a few sentences to clarify exactly the nature of your question.


Let $x_1,x_2,....,x_n \in\mathbb R$ satisfying the condition:
\[\sum_{i=1}^n x_i=2n+1\]
Prove that $\exists k\in \{1,2,...,n\}$ such that:
\[\sum_{i=k}^{k+r} x_i\leq 2r+3,\forall r=0,1,2,...,n-2,(x_{n+i}=x_i)\]

Then we just need to sum up over all $k=1,2,...,n$ to obtain result. Right?

???I don't know your idea (because my English is so poor),hix! :blush:  :(

Don't mind my post. I didn't see that you need inequality for ALL $r$ simultaneously.

Let $y_i=x_i-2$. Then $\sum_{i=1}^n y_i=1$ and we need to prove that thre exists $k$ s.t. $\sum_{i=k+1}^{k+r}y_i\leq 1$ for all $r=1,2,...,n-1$.
Let $u$ and $v$ ($u\leq v$) be such indexes that $S=\sum_{i=u}^{v}y_i$ is minimal and $v-u$ is maximal possible ($u$ and $v$ exist!). Then $S_r=\sum_{i=v+1}^{v+r}\leq 1$ for all $r=1,2,...,n-1$. Indeed, consider complimentary sum $S'_r=\sum_{i=v+r+1}^{u}$. We know $S_r+S'_r=1$, but also $S'_r\geq 1$, otherwise $S\geq S'_r+S=\sum_{i=v+r+1}^{v}y_i$ -- contradiction, since we have managed to increase number of summands.
Thus $S'_r\geq 1$ and $S_r\leq 0$ (???).


OK, I don't know how hard this problem is... the way I solved it, it isn't that easy, but of course somebody else may come up with an easier solution :)

If $a,b,c$ are positive real numbers and $a+b+c=1$, prove that
$1-2ab-c^2\ge 6{\sqrt{\frac{a^2{b^2}+a^2{c^2}+b^2{c^2}}{3}}}$ and determine where equality occurs.

[quote="Melissa"]If $a,b,c$ are positive real numbers and $a+b+c=1$, prove that
$1-2ab-c^2\ge 6{\sqrt{\frac{a^2{b^2}+a^2{c^2}+b^2{c^2}}{3}}}$ and determine where equality occurs.[/quote]I'm afraid you applied AM-QM the reverse way, Melissa... right? QM>=AM :(

So, it is very nice, but not correct. Take $a=\frac14,b=\frac14,c=\frac12$. Then there stands $\frac58 \ge \frac{3\sqrt{3}}8$, or $\frac{25}{64}\ge\frac{27}{64}$

:blush:  :blush:  :blush:
Ummm... well somewhere in the middle I messed up some + and - signs, that's why I got this weird result. And because of that it looks as though I used QM-AM the wrong way round. I should have known it could not be right... I really thought it was weird, but then I tested it for a few values and it seemed OK.
Sorry about that, I'll think up a new one when I've got time.

I think this one is true.
Since: $(1-c)=a+b$, so if we put it in and expand, we will have:
\[ 4ab+4c \sqrt{ab} \leq a+b+c(a+b) \]
Since: $\sqrt{ab} \leq \frac{a+b}{2}$, so:
\[ 4ab+4c \sqrt{ab} \leq 4ab + 2c(a+b) \].
Now it is easy to compare. We will get $(a-b)^2 \geq 0$ in the end.

[quote="Melissa"]OK, this one is totally different of the first one (and probably a lot easier)... hope it's right this time!
If $a,b,c$ are positive real numbers and $a+b+c=1$, prove that
$ab+c{\sqrt{ab}}\le \frac{1-c^2}{4}$ :)[/quote]
[hide="my solution"]
Using AM-GM, we have

$ab + c\sqrt{ab} \leq \left(\frac{a+b}{2}\right)^2 + c \left(\frac{a+b}{2}\right)$

Now, substituting $a + b = 1 - c$, we have

$ab + c\sqrt{ab} \leq \frac{(1 - c)^2}{4} + \frac{c - c^2}{2} = \frac{1 - c^2}{4}$

as desired.
[/hide]


Provez que nous avons l'inegalite (1/1+abc)+(1/1+bcd)+(1+cda)+(1+dab)>3.Nous savons que abcd=1, d>1 et
a+b+c+d=2,a,b,c,d>0.

answer = réponse.

Ensuite, est-ce que ton inégalité est bien :
[quote="stancioiu sorin"]$\frac 1{1+abc}+\frac 1{1+bcd}+1+cda+1+dab>3$.[/quote]?

Ok, alors, cela se réécrit :
$\frac a{a+1} + \frac{b+1}b + \frac{c+1}c + \frac d{d+1} > 3$.

Maintenant, peux-tu clarifier les conditions?

Nous avons abcd=1, a+b+c+d=2, ab,c,d>0 et d>1. :)

Je crois que j'ai trouve la solution:
C'est evident que \[(1+cda)+(1+dab)>2.(rel 1)\] Je prouverai que l'inegalite suivant est vrai : \[ \frac{1}{1+abc}+\frac{1}{1+bcd}>1,(rel 2)\] donc \[ \frac{d}{1+d}+\frac{a}{1+a}>1,\]  \[ad+d+da+a>ad+d+a+1,\]  \[da> 1, abcd>bc ,1>bc.\] Vrai, parce que si \[d>1,a,b,c>0\] et \[a+b+c+d=2\] nous avons \[a+b+c<1\] et \[a,b,c<1\] et \[bc<1.\] Si nous faisons la somme de (rel1) et (rel2), nous obtenirons l'inegalite demandee.


What biology textbooks are you all using?

I'm using Starr/Taggart's [u]Biology: the Unity and Diversity of Life[/u]

 :D  :D

Hey that's exactly what I'm using! What grade are you in? :)

In AP Biology we use Campbell's Biology, 6th ed.

used campbell's 6th ed. as well in ap bio.

Grade 9, taking AP bio.

I also have campbell's, but only third edition. our ap bio teacher draws half the material from the campbell book, so i don't get how we are supposed to know it...
 :|

AP Bio? We only have GT Bio for 9th grade :(

Campbell's Biology, 6th ed. here as well too.

whats GT bio? do you have to take GT before you can take AP?

Yeah, it's "Gifted and Talented". In 9th grade, you take either Standard, Honors, or GT Bio.


vector
abstract algebra
superior algebra
superior algebra unsolved

Let V1, V2, U1, and U2 be vector spaces over a field F. Let f1 be in Hom(V1,U1) and f2 in Hom(V2,U2). The tensor product: f1 (X) f2 is a linear transformation from V1 (X) V2 to U1 (X) U2.
Note: (X) means tensor product.

Prove: If f1 and f2 are surjective, then f1 (X) f2 is surjective.
           If f1 and f2 are injective, then f1 (X) f2 is injective.
And, compute the kernel of f1 (X) f2 if neither f1 or f2 is injective.


[hide]1/2, because it doesn't matter about the previous throws and the coin is fair so we get 1/2.[/hide]

[hide][quote="236factorial"]A fair coin is thrown 6 times and lands heads everytime. What is the probability that the next coin throw will also be heads?[/quote]

The previous six aren't important in this problem.

The probability is 50/100, or 1/2. [/hide]

[hide]1/2. If you asked what was the possibility of getting 6 heads in a row, then it would be 2^-6. or (1/2)^6.[/hide]

[quote="Sinistra"][hide]

Ok. I think this question is just common sense.

A coin only has two major sides which it can land on so it must be a 1 in 2 chance.

[/hide]

[By the way, has anyone seen a coin land on it's really thin side?][/quote]

No, I have not. However, it is possible and I've seen people make the coin go that way...or else, no.

[hide]very simple, 1/2 :D  hope nobody missed that[/hide]

it be sad if anybody missed that.... anyway, my answer
[hide]1/2[/hide]

I think I've seen one of these problems before, unless this is the topic i saw from a long time ago... but when I first saw it, I got kinda confused, and started finding out $2^6$... and then i realized my mistake after seeing everybody's answer, and Kristy it's not nice to say that... people (me) might feel bad because they  didn't get it at first. but here's the answer

[hide]it would be $\frac{1}{2}$ because it doesn't really matter what the previous $6$ was, and the possibility of getting a head on a fair coin is $\frac{1}{2}$ [/hide]


I have a double subscripts in my equation. My code is:
M_{R_{b}}{}_R_{b}...The result in the attachment is exactly  what I want. BUT Latex showed two errors w.r.t.double subscripts.
What is the right code for double subscripts in my case?  BTW, \documentclass[a4paper,12pt]{article}
\usepackage{amsmath}
\usepackage[pdftex]{graphicx}
\usepackage{setspace}
\title{Second Chapter}
\begin{document}
\maketitle
\doublespacing

This is what I set for this paper.

The error is because in {}_R_{b}... {}_R_{b} means R is the only subscripted letter, so _{b} makes no sense and LaTeX ignores the second _ giving Rb instead of R_b (your pictures show that you [i]don't[/i] get what you want). Thus it should be {}_{R_{b}} as the whole of the subscript must be put in {...}.
But that is far too complicated; you should put the 2 R_b's together as M_{R_{b}R_{b}}... which gives $ M_{R_{b}R_{b}}...$. Thus again the main subscript is all together in one {...}.


The length of the diagonal of a square is $ 2\sqrt{6}$ cm. What is the number of square centimeters in the area of the square?

hello, let $ d$ be the length of the diagonal of the given square then we have after the Pythagorean theorem $ 2a^2 \equal{} d^2$ or $ 2a^2 \equal{} 24$ where $ a$ is the side length, from here we get $ a \equal{} 2\sqrt {3} cm$ and the area is $ 12cm^2$.
Sonnhard.

Since a square is a rhombus, we can do $ \frac{d_1 \cdot d_2}{2}\equal{}\frac{2\sqrt{6}^2}{2}\equal{}\frac{24}{2}\equal{}12$.

Or we can do it the way mentioned above. $ 2\sqrt{3}^2\equal{}12$. Don't forget to square the side length.

Or since you know that $\sqrt\frac{24}{2}$ is $\sqrt{12}$, $\sqrt{12}$ is the side length, and the formula for the area of the square is $\text A = bh$ where $\text A = Area$, you get

$\text A = \sqrt{12}^2$, or

$\text A = \sqrt{144}$.

That gives us $\boxed{A = 12}$


Diane is writing a book with ten chapters, which have 17, 23, 14, 26, 21, 32, 36, 19, 24 and 30 pages, respectively. The first chapter begins on page 1, and each subsequent chapter must also begin on an odd numbered page, with a blank page
between chapters if needed. What is the page number of the last page of the last chapter?

Okay, we have to picture this.

First chapter will end on page 17. We need on page in between, so the second chapter will start on page 19 and end on page 42. Third chapter begins on page 43 and ends on page 57. We need a page in between, so fourth chapter will begin on page 59 and end on page 85. Blank page in between, fifth chapter begins on page 87 and ends on page 108. Sixth chapter begins on page 109 and ends on page 141. Seventh chapter begins on page 143 and ends on page 179. Eighth chapter begins on page 181 and ends on page 200. Ninth chapter begins on page 201 and ends on page 225. The last chapter begins on page 227 and ends on page 246. Thus, our answer is $ \boxed{246}$.

If there are an odd number of pages in each chapter, we must add an extra page at the end. There are 4 odd-number-of-pages-chapters. Normally, without blank pages the last page would be $ 17\plus{}23\plus{}14\plus{}26\plus{}21\plus{}32\plus{}36\plus{}19\plus{}24\plus{}30\equal{}232$, but since there are 4 extra pages, the last page without table of contents, about the author, publications, more blank pages at the very end, title pages, and other things would be page 236.

Actually, that sum totals to $ 242$, so the answer is $ \boxed{246}$, like I said.


AMC
USA(J)MO
USAMO
number theory
prime numbers
number theory unsolved

[quote="orl"]Is it possible to arrange on a circle all composite positive integers not exceeding $ 10^6$, so that no two neighbouring numbers are coprime?

[i]Author: L. Emelyanov[/i]

[hide="Tuymaada 2008, Junior League, First Day, Problem 2."]Prove that all composite positive integers not exceeding $ 10^6$
may be arranged on a circle so that no two neighbouring numbers are coprime. [/hide][/quote]
This reminds me of [url=http://www.artofproblemsolving.com/Wiki/index.php/2005_USAMO_Problems/Problem_1]USAMO 2005, Problem 1[/url].
So you could arrange many numbers with a factor of 2, ending with 6, then arrange many numbers with a factor of 3, ending with 15, and so on... right?

$ ....x\cdot y,1,2\cdot a_1,2\cdot a_2,...,2\cdot3,3\cdot b_1,...,3\cdot5,5\cdot c_1,......$

where $ a_1$ is the biggest prime such that $ 2\cdot a_1\leq10^6$, and so on, until we arrive at $ x\cdot y$.

So in this way we have arranged all the composite numbers not exceeding $ 10^6$ that are a product of exactly two prime numbers.

So now if we take a remaining number $ \alpha$, then $ \alpha$ is a product of at least three prime numbers:

$ (\beta,\gamma,\delta)$ (not necessarily distinct).

Wlog $ \beta\leq\gamma\leq\delta$, so we can put $ \alpha$ beside $ \beta\cdot\gamma$.


analytic geometry
algorithm
LaTeX
number theory
relatively prime

Let $p$ and $q$ be relatively prime positive integers. A subset $S\subseteq \mathbb{N}_0$ is called ideal if $0 \in S$ and, for each element $n \in S$, the integers $n+p$ and $n+q$ belong to $S$. Determine the number of ideal subsets of $\mathbb{N}_0$.

First off, since $ p$ and $ q$ are relatively prime, any integer $ x$ can be represented in the form $ x \equal{} ap \plus{} bq$ where $ a,b$ are integer coefficients.
Second, it is clear that every ideal subset contains a subset $ T \equal{} \{ up \plus{} vq : u,v\in\mathbb{N}_0 \}$, implying that ideal subsets can differ only by elements $ ap \plus{} bq\not\in T$ (hence, it is necessary that either $ a$ or $ b$ is negative).

Notice that if $ x \equal{} ap \plus{} bq$ then $ x \equal{} (a\pm q)p \plus{} (b\mp p)q$, implying that every positive integer $ x$ that is not in $ T$ admit a reduced form:
$ x \equal{} ap \plus{} bq$ with $ \minus{} q < a < 0$, $ 0 < b < p$, and $ \frac {b}{ \minus{} a} > \frac {p}{q}$

It is clear that if an ideal subset contains $ x \equal{} ap \plus{} bq$ then it also contains $ up \plus{} vq$ for all $ u\geq a$ and $ v\geq b$.

Therefore, every ideal subset is uniquely defined by a pair of sequences:
$ q > \minus{} a_1 > \minus{} a_2 > \dots > \minus{} a_m > 0$ and $ p > b_1 > b_2 > \dots > b_m > 0$ with $ \frac {b_j}{ \minus{} a_j} > \frac {p}{q}$ for all $ j\equal{}1,\dots,m$.

It is easy to see that every pair of such sequences define a path from $ (0,0)$ to $ (q,p)$ in the integer grid $ [0,q]\times[0,p]$ where every step increments both coordinates and does not go below the line connecting the points $ (0,0)$ and $ (q,p)$.

I think your answer is incorrect, unless I'm misunderstanding the question.

Consider $ (p,q)\equal{}(3,5)$. Clearly the set contains at least $ \mathbb{N}_0/\{1,2,4,7\}$
Depending on our choices of which numbers to include in the set, we can also achieve each of:
$ \mathbb{N}_0/\{1,2,4\}$ (include only $ 7$ extraneously)
$ \mathbb{N}_0/\{1,2\}$ (include $ 4$ extraneously, which auto-adds $ 7$)
$ \mathbb{N}_0/\{1,4\}$ (include $ 2$)
$ \mathbb{N}_0/\{2\}$ (include $ 1$)
$ \mathbb{N}_0/$ (include $ 1$ and $ 2$)

This yields 6 different sets, which is NOT $ {6 \choose 2}$ as your solution states.

[b]me@home[/b], thanks for the (counter)example.
I've corrected the solution and got the answer as the number of paths in the integer grid but no closed formula yet.

In particular, the six ideal sets in your example with $ p\equal{}3$ and $ q\equal{}5$ correspond to the following paths:
(0,0) -> (5,3)
(0,0) -> (1,1) -> (2,3) -> (5,3)
(0,0) -> (1,1) -> (3,3) -> (5,3)
(0,0) -> (1,3) -> (5,3)
(0,0) -> (2,3) -> (5,3)
(0,0) -> (3,3) -> (5,3)

Previously, I found only 6 different sets, while there are actually seven - I missed $ \mathbb{N}_0/\{1\}$
I've thought about this a little more, and its actually a rather deep problem. For instance, I've solved the trivial case when $ (p,q)\equal{}(n,n\plus{}1)$. Then, the numbers can be arranged into a "triangle of implication" and the number of sets is equal to $ \frac{{ 2n \choose n}}{n\plus{}1}$, which is a Catalan number.

Extending the idea of a triangle of implication, I've come up with some ways to efficiently compute answers (in, say, a computer algorithm) but no nice closed form answers. I'm sort of confident about this because lots of particular answers are rather large primes... all in all i'm confused about how this could have been a proposed contest problem.

At kalva there is a page proposing similar ideas for this problem, and a closed formula: [url]http://web.archive.org/web/20060925154521/www.kalva.demon.co.uk/short/soln/sh00c6.html[/url].

However, it's very too long to read, which really decourages people. A well-structured proof (probably with lemmata and clear drawings) in LaTeX on this site would still be highly appreciated!

Another solution is given at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1218952


Let $ x\plus{}y\plus{}z\equal{}0$. Prove that $ 27^x\plus{}27^y\plus{}27^z\ge 3^x\plus{}3^y\plus{}3^z$.

Thanks for any help.

[quote="ppt"]Let $ x \plus{} y \plus{} z \equal{} 0$. Prove that $ 27^x \plus{} 27^y \plus{} 27^z\ge 3^x \plus{} 3^y \plus{} 3^z$.

Thanks for any help.[/quote]
$ a \equal{} 3^x, b \equal{} 3^y,c \equal{} 3^z$ so $ abc \equal{} 1$ and we must prove $ a^3 \plus{} b^3 \plus{} c^3\ge a \plus{} b \plus{} c$ and its obviously true by Muirhead

[quote="dima ukraine"][quote="ppt"]Let $ x \plus{} y \plus{} z \equal{} 0$. Prove that $ 27^x \plus{} 27^y \plus{} 27^z\ge 3^x \plus{} 3^y \plus{} 3^z$.

Thanks for any help.[/quote]
$ a \equal{} log_3(x), b \equal{} log_3(y),c \equal{} log_3(z)$ so $ xyz \equal{} 1$ and we must prove $ x^3 \plus{} y^3 \plus{} z^3\ge x \plus{} y \plus{} z$ and its obviously true by Muirhead[/quote]
Maybe $ a\equal{}3^x$ $ ?$ :wink:

[quote="arqady"][quote="dima ukraine"][quote="ppt"]Let $ x \plus{} y \plus{} z \equal{} 0$. Prove that $ 27^x \plus{} 27^y \plus{} 27^z\ge 3^x \plus{} 3^y \plus{} 3^z$.

Thanks for any help.[/quote]
$ a \equal{} log_3(x), b \equal{} log_3(y),c \equal{} log_3(z)$ so $ xyz \equal{} 1$ and we must prove $ x^3 \plus{} y^3 \plus{} z^3\ge x \plus{} y \plus{} z$ and its obviously true by Muirhead[/quote]
Maybe $ a \equal{} 3^x$ $ ?$ :wink:[/quote]
Yeah sorry it was a  :oops: typo


A:curved digits; B: straight digits; C: others?

do you know for sure if those are the desriptions for the sequences (given in the problem)?

well... going from those descriptions, 15has both a straight and a curved number, so it belongs in C, as does the number 16
17 contains only straight digits, so it b elongs in B

That's what I thought too, but I don't know the "official" answer.


Evaluate:

\[ \lim_{x \to \minus{}1} \frac{\sqrt{\pi}\minus{}\sqrt{\arccos{x}}}{\sqrt{1\plus{}x}}.\]

Put $ t \equal{} 1 \plus{} x$. Easy to show that $ \arccos\left(t \minus{} 1\right) \equal{} \pi \minus{} \sqrt {2t} \plus{} O(t^{3/2})$.

Finally: $ \lim_{t\to 0}\frac {\sqrt {\pi} \minus{} \sqrt {\arccos\left(t \minus{} 1\right)}}{\sqrt {t}} \equal{} \lim_{t\to 0}\frac {\pi \minus{} \arccos\left(t \minus{} 1\right)}{\sqrt {t}\cdot \left(\sqrt {\pi} \plus{} \sqrt {\arccos\left(t \minus{} 1\right)}\right)} \equal{} \frac {1}{\sqrt {2\pi}}.$

[hide="Another approach..."]$ L \equal{} \lim_{x \to \minus{} 1}{\frac {\pi \minus{} \arccos(x)}{\sqrt {1 \plus{} x} (\sqrt {\pi} \plus{} \sqrt {arccos(x)})}} \equal{} \lim_{x \to \minus{} 1} \frac {\arccos( \minus{} x)}{\sqrt {1 \plus{} x} ( \sqrt {\pi} \plus{} \sqrt {\arccos(x)})}$

$ L \equal{} \lim_{x \to \minus{} 1} \frac {\arcsin{\sqrt {1 \minus{} x^2}} }{\sqrt {1 \minus{} x^2}} \cdot \frac {\sqrt {1 \minus{} x}}{\sqrt {\pi} \plus{} \sqrt {\arccos(x)}}$

$ L\equal{} \lim_{x \to \minus{} 1} \frac {\sqrt {1 \minus{} x}}{\sqrt {\pi} \plus{} \sqrt {\arccos(x)}} \equal{} \frac {1}{\sqrt {2 \pi}}$.[/hide]


Let $n$ be the largest integer that is the product of exactly 3 distinct prime numbers, $d,$ $e,$ and $10d+e$, where $d$ and $e$ are digits. What is the sum of the digits of $n$?

$\left(A\right) 12$
$\left(B\right) 15$
$\left(C\right) 18$
$\left(D\right) 21$
$\left(E\right) 24$

well, your problem is that when you use $5$, $10d+e$ is not prime, so it doesnt work


Αφού έβαλα μια γεωμετρία για τους μικρούς , ας βάλω και μία για τους μεγάλους. Εδώ να υπενθυμίσω και τούτο: Ακόμα και οι σχετικά εύκολες ασκήσεις έχουν μια γοητεία και πολλές φορές κρύβουν συμπεράσματα που χρειάζονται για τη λύση των δύσκολων ασκήσεων. Ας βρούμε λοιπόν σε αυτό το πρόβλημα όσες πιο πολλές λύσεις γίνεται. Αλλά να μην αρχίσετε με ... πολικές :wink: (αν και μου αρέσουν !).

[b] ΕΚΦΩΣΗΣΗ[/b]

 Από σημείο $ A$ φέρουμε τις επαπτομένες $ AM,AN$ προς έναν κύκλο καθώς και μια τέμνουσα $ ABC$ του κύκλου. Αν $ D$ είναι το μέσο του $ MN$ , να αποδείξετε ότι η $ MN$ διχοτομεί τη γωνία $ BDC$.

[quote="silouan"]Γράφω τη λύση που μου ήρθε πρώτη στο μυαλό . :)
Έστω ότι η BC τέμνει την MN sto S . Τότε τα σημεία τα σημεία (C,B,S,A) σχηματίζουν αρμονική τετράδα (πάρτε τα όμοια ΑΒΝ και ACN για να το διαπιστώσετε (αν πάλι υπάρχει πρόβλημα πρίτε μου να γράψω τις σχέσεις))
Επιπλέον $ AD\bot DS$ άρα η DS είναι η διχοτόιμος της BDC .[/quote]

Σιλουανέ ,
 αυτή ακριβώς τη λύση έκανα και γω , μόλις την πάλαιψα για λίγο και ήμουν σίγουρος ότι εκεί θα πήγαινε και το δικό σου μυαλό. Για το λόγο αυτό έγραψα  '' μην μου αρχίσετε με πολικές ...''. Αν μπορέσουμε να κάνουμε τη λύση πιο στοιχειώδη , πχ με θεώρημα διχοτόμων (αντίστροφο),  θα ήταν χρήσιμο. Για συνέχισε λίγο το ψάξιμο με την ομοιότητα και τα ξαναλέμε.

Πολύ ωραία τελείως στοιχειώδης απόδειξη Μιχάλη :)

[quote="savvas_michael"]Βάζω ακόμα μια λύση.

Λοιπόν έχουμε $ AM^{2}\equal{} AD*AO \equal{} AB*AC$ ($ O$ το κέντρο του κύκλου), λόγω των ορθογωνίων τριγώνων $ AMO, AMD$ και της δύναμης σημείου.
Άρα το τετράπλευρο $ BDOC$ είναι εγγράψιμο, οπότε θα έχουμε (γωνίες)
$ ADB \equal{} BCO \equal{} CBO \equal{} CDO$, άρα και $ BDM \equal{} 90\minus{}ADB \equal{} 90\minus{}CDO \equal{} CDM$.[/quote]

 :lol:  :lol: . Πώς δε με πρόλαβε ο Vittαsko για να σου σφίξει το χέρι ; Σε αυτόν αρέσουν πολύ αυτά τα απλά , έξυπνα και σύντομα ! Αλλά και σε όλους !

[u] Μπάμπης[/u]

[quote="vittasko"][quote="stergiu"].....

 :lol:  :lol: . Πώς δε με πρόλαβε ο Vittαsko για να σου σφίξει το χέρι ; Σε αυτόν αρέσουν πολύ αυτά τα απλά , έξυπνα και σύντομα ! Αλλά και σε όλους !

[u] Μπάμπης[/u][/quote]

Με πέντε λεπτά μόνο καθυστέρηση, μας πρόλαβε ο [b][size=100]bilot4[/size][/b].

Όταν έχεις και τη ζαλάδα του [b][size=100]Omid Hatami[/size][/b] (που το βρήκαν οι άνθρωποι αυτό το πρόβλημα να το βάλουν στην Ολυμπιάδα τους http://www.mathlinks.ro/Forum/viewtopic.php?t=164548 ...), τι να πρωτοπρολάβεις...

Τη χάρηκα πάντως την απόδειξη αυτή και μπράβο στον Μιχάλη που τη σκέφτηκε.

Κώστας Βήττας.[/quote]

 Κώστα ,
 μόλις έριξα μια ματιά στο πρόβλημα από το Ιραν. Σίγουρα σε παίδεψε και σου πήρε πολύ χρόνο.
[b] Με την ευκαιρία ![/b]
 Θέλω κάποια στιγμή που θα είμαστε μαζί να βρούμε όλα σου τα προβλήματα που έστειλες στο φόρουμ , ελληνικό ή αγγλικό και να τα βάλω σε μια σελίδα. ΑΠό κει θα τα παρουμε έτοιμα για το βιβλίο , αφού γίνουν τυχόν αναγκαίες συμπληρώσεις. Αυτός ο κόπος με τις τόσες ωραίες ιδέες πρέπει να είναι στη διάθεση όλου του Ελληνικού κοινού !
[u] Μπάμπης[/u]


geometry
incenter
ratio
trigonometry
circumcircle
quadratics
geometric transformation

let ABC be a triangle such that AB<BC , AC not equal to BC and K be its circumcircle. The tangent line to K at the point A intersects the line BC in the point D. Let k be the circle tangent to K and to the segment AD and BD. We denote by M,N the points where k touches BD and AD respectively. Let J be the center of the exscribed circle which is tangent to the side AB. Prove that J,M,N are collinear.

Stating the problem in a more general way:

Let $\triangle ABC$ be a triangle with the sides $a = BC, b = CA, c = AB$, such that $b < c$, and let $(O)$ be its circumcircle. A tangent line to $(O)$ at the point $A$ intersects the line $BC$ at a point $D$ and the points $B, C, D$ follow on the line $BC$ in the this order. Let $(P_1), (P_2), (P_3)$ be circles tangent to $(O)$ and to the lines $AD, BD$, such that $DP_1 < DP_2 < DP_3$. Denote $M_1, M_2, M_3$ the tangency points of $(P_1), (P_2), (P_3)$ with the line $AD$ and $N_1, N_2, N_3$ the tangency points of these circles with the line $BD$. Let $I$ be the incenter of the $\triangle ABC$ and $E_1, E_2, E_3$ the excenters opposite to the vertices $B, A, C$. The points $(E_1, M_1, N_1), (M_2, I, N_2, E_2), (M_3, N_3, E_3)$, respectively, are collinear.

The case $(M_2, I, N_2, E_2)$ is trivial, because the point $M_2$ is necessarily identical with the vertex $A$. Since $\measuredangle ACD = \measuredangle B + \measuredangle A$, $\measuredangle DAC = \measuredangle B$, it follows that $\measuredangle ADC = 180^o - (\measuredangle A + 2 \measuredangle B)$, $\measuredangle ADP_2 = \measuredangle ADC/2 = 90^o - (\measuredangle A/2 + \measuredangle B)$. In addition, $\measuredangle DAI = \measuredangle B + \measuredangle A/2$. Consequently, the bisectors of the angles $\measuredangle ADC, \measuredangle A$ are perpendicular, which implies that the point $N_2$ lies on the internal bisector $AI$ of the angle $\measuredangle A$ and the points $(M_2 \equiv A, I, N_2, E_2)$ are all collinear.

Let $T_1, T_3$ be the tangency points of the excircles $(E_1), (E_3)$ with the line BC. Since $BT_1 = CT_3 = s(\triangle ABC)$, the semiperimeter of the $\triangle ABC$, the ratio of radii of the excircles $(E_1), (E_3)$ is

$\frac{r(E_1)}{r(E_3)} = \frac{s \tan{\frac{\widehat B}{2}}}{s \tan{\frac{\widehat C}{2}}} = \frac{\tan{\frac{\widehat B}{2}}}{\tan{\frac{\widehat C}{2}}}$

Denote $R$ the circumradius of the triangle $\triangle ABC$, $r(P_1), r(P_2), r(P_3)$ radii of the circles $(P_1), (P_2), (P_3)$, and $x$ the ratio $\frac{DM_1}{AD}$. Since the circles $(P_1), (O)$ touch, their common external tangent length is $2 \sqrt{Rr(P_1)}$. Hence,

$1 = \frac{AM_1}{AD} + \frac{M_1D}{AD} = \frac{2 \sqrt{Rr(P_1)}}{AD} + x$

$x^2 - 2x + 1 = \frac{4Rr(P_1)}{AD^2}$

Using the sine theorem for the triangles $\triangle ABC, \triangle ADC$, we have

$\frac b c = \frac{\sin{\widehat B}}{\sin{\widehat C}} = \frac{\sin{\widehat B}}{\sin{(\widehat A + \widehat B)}}$

$\frac{CD}{AD} = \frac{\sin{\widehat B}}{\sin{(\widehat A + \widehat B)}} = \frac b c$

Since $AD = N_2D$ and the point $N_2$ lies of the bisector of the angle $\measuredangle A$,

$AD = N_2C + CD = \frac{ab}{b + c} + AD \cdot \frac b c$
$AD = \frac{abc}{c^2 - b^2}$

The angle $\measuredangle ADP_2 = 90^o - \left(\frac{\measuredangle A}{2} + \measuredangle B\right) = \measuredangle C - \frac{\measuredangle B}{2}$. Substituting

$R = \frac{a}{2 \sin{\widehat A}} = \frac{a}{2 \sin{(\widehat B + \widehat C)}}$

$r(P_1) = AM_1 \tan{\frac{\widehat C - \widehat B}{2}}$

into the last expression for $AD$, we get the following quadratic equation for $x$:

$x^2 - 2x + 1 = \frac{2xa(c^2 - b^2) \tan{\frac{\widehat C - \widehat B}{2}}}{abc \sin{(\widehat C + \widehat B)}} = 2x\left(\frac c b - \frac b c\right) \frac{\tan{\frac{\widehat C - \widehat B}{2}}}{\sin{(\widehat C + \widehat B)}} =$

$= 2x\left(\frac{\sin{\widehat C}}{\sin{\widehat B}} - \frac{\sin{\widehat B}}{\sin{\widehat C}}\right) \frac{\tan{\frac{\widehat C - \widehat B}{2}}}{\sin{(\widehat C + \widehat B)}}$

Product of the two roots of this quadratic equation is 1. The coefficient $-q$ of the linear term is equal to the sum of these roots. Using some trigonometric formulas,

$-q = 2 + 2\left(\frac{\sin{\widehat C}}{\sin{\widehat B}} - \frac{\sin{\widehat B}}{\sin{\widehat C}}\right) \frac{\tan{\frac{\widehat C - \widehat B}{2}}}{\sin{(\widehat C + \widehat B)}} =$

$= 2 + 2\ \frac{\sin^2{\widehat C} - \sin^2{\widehat B}}{\sin{\widehat B} \sin{\widehat C} \sin{(\widehat C + \widehat B)}} \tan{\frac{\widehat C - \widehat B}{2}} =$

$= 2 + 2\ \frac{\sin{(\widehat C + \widehat B)} \sin{(\widehat C - \widehat B)}}{\sin{\widehat B} \sin{\widehat C} \sin{(\widehat C + \widehat B)}} \tan{\frac{\widehat C - \widehat B}{2}} =$

$= 2 + 4\ \frac{\sin^2{\frac{\widehat C - \widehat B}{2}}}{\sin{\widehat B} \sin{\widehat C}} =\frac{\tan{\frac{\widehat C}{2}}}{\tan{\frac{\widehat B}{2}}} + \frac{\tan{\frac{\widehat B}{2}}}{\tan{\frac{\widehat C}{2}}}$

The two the roots of the above quadratic equation are therefore $\frac{\tan{\frac{\widehat C}{2}}}{\tan{\frac{\widehat B}{2}}}$ and $\frac{\tan{\frac{\widehat B}{2}}}{\tan{\frac{\widehat C}{2}}}$. Since we could get the same quadratic equation for the ratio $\frac{M_3D}{AD}$, it follows that

$\frac{r(P_3)}{r(P_2)} = \frac{M_3D}{AD} = \frac{\tan{\frac{\widehat C}{2}}}{\tan{\frac{\widehat B}{2}}}$

$\frac{r(P_2)}{r(P_1)} = \frac{M_1D}{AD} = \frac{\tan{\frac{\widehat C}{2}}}{\tan{\frac{\widehat B}{2}}}$

$\frac{r(P_3)}{r(P_2)} = \frac{r(P_2)}{r(P_1)} = \frac{r(E_3)}{r(E_1)}$.

The circle pairs $(P_1), (P_2)$ and $(P_2), (P_3)$ are centrally similar with the same homothety coefficient (but different homothety center) as the excircle pair $(E_1), (E_3)$. The point $D$ on the line $BC$ is the external homothety center of the circles $(P_1), (P_2), (P_3)$. Let $F$ be the external homothety center of the excircles $(E_1), (E_3)$, the intersection of their common external tangent $BC$ and their center line $E_1E_3$. Since the external bisector $AE_1 \equiv AE_3$ of the $\measuredangle A$ is perpendicular to the internal bisector $AI \equiv AN_2$ of this angle and $AN_2 \perp DP_1$, the lines $FE_1 \equiv FE_3$ and $DP_1 \equiv DP_2 \equiv DP_3$ are parallel. The internal and external bisectors of the $\measuredangle A$ cut the opposite side $a = BC$ internally and externally in the ratio $\frac c b$:

$BN_2 = \frac{ac}{c + b}$, $CN_2 = \frac{ab}{c + b}$

$BF = \frac{ac}{c - b}$, $CF = \frac{ab}{c - b}$

The distance between the intersections $N_2, F$ of the internal and external bisectors of the angle $\measuredangle A$ with the line $BC$ is then

$N_2F = BF - BN_2 = CF + CN_2 = \frac{ab}{c - b} + \frac{ab}{c + b} = \frac{2 abc}{c^2 - b^2}$

Consequently, $AD = N_2D = \frac{N_2F}{2}$ and

$N_1F = N_2F - N_2N_1 = 2 AD - (AD - AN_1) = AD + AN_1 = AD + AM_1$

$N_3F = N_2F + N_3N_2 = 2 AD + (AN_3 - AD) = AD + AN_3 = AD + AM_3$

$N_1F = \frac{abc}{c^2 - b^2} \left(1 + \frac{\tan{\frac{\widehat B}{2}}}{\tan{\frac{\widehat C}{2}}}\right)$, $N_3F = \frac{abc}{c^2 - b^2} \left(1 + \frac{\tan{\frac{\widehat C}{2}}}{\tan{\frac{\widehat B}{2}}}\right)$, $\frac{N_3F}{N_1F} = \frac{\tan{\frac{\widehat C}{2}}}{\tan{\frac{\widehat B}{2}}}$

Let $N_1', N_3'$ be the intersections of the line $BC$ with normals to the bisector $DP_1 \equiv DP_2 \equiv DP_3$ of the angle $\measuredangle ADC$ through the excenters $E_1, E_3$. Obviously,

$T_1F = E_1F \cos{\frac{\widehat C - \widehat B}{2}} = N_1'F  \cos^2{\frac{\widehat C - \widehat B}{2}}$

Since the distance of the tangency point $T_1$ of the excircle $(E_1)$ with the line $BC$ from the opposite vertex $B$ is equal to the semiperimeter $s(\triangle ABC)$,

$T_1F = BF - BT_1 = \frac{ac}{c - b} - s =$

$= \frac{abc}{c^2 - b^2} \left(\frac{c + b}{b} - \frac{(a + b + c)(c^2 - b^2)}{2 abc}\right) = AD\ \frac{(c + b)[a(c + b) - (c^2 - b^2)]}{2 abc}$

Substituting $a = 2R \sin{\widehat A} = 2R \sin{(\widehat C + \widehat  B)}$, $b = 2R \sin{\widehat B}$, $c = 2R \sin{\widehat C}$ and using the following easy to verify trigonometric identities:

$\sin{\widehat C} + \sin{\widehat B} = 2 \sin{\frac{\widehat C + \widehat B}{2}} \cos{\frac{\widehat C - \widehat B}{2}}$

$\sin^2{\widehat C} - \sin^2{\widehat B} = \sin{(\widehat C + \widehat B)} \sin{(\widehat C - \widehat B)}$

$\frac{\sin{\widehat C} + \sin{\widehat B} - \sin{(\widehat C - \widehat B)}}{\sin{\widehat C}\sin{\widehat B}} = \frac{\sin{\frac{\widehat B}{2}}}{\cos{\frac{\widehat B}{2}}} + \frac{\cos{\frac{\widehat C}{2}}}{\sin{\frac{\widehat C}{2}}} = \frac{\cos{\frac{\widehat C - \widehat B}{2}}}{\sin{\frac{\widehat C}{2}} \cos{\frac{\widehat B}{2}}}$

this becomes:

$T_1F = AD\ \frac{(\sin{\widehat C} + \sin{\widehat B})[\sin{\widehat C} + \sin{\widehat B} - \sin{(\widehat C - \widehat B)}]}{2 \sin{\widehat C}\sin{\widehat B}} =$

$= AD\ \frac{\sin{\frac{\widehat C}{2}} \cos{\frac{\widehat B}{2}} + \cos{\frac{\widehat C}{2}} \sin{\frac{\widehat B}{2}}}{\sin{\frac{\widehat C}{2}} \cos{\frac{\widehat B}{2}}} \cos^2{\frac{\widehat C - \widehat B}{2}} = $

$= AD \left(1 + \frac{\tan{\frac{\widehat B}{2}}}{\tan{\frac{\widehat C}{2}}}\right) \cos^2{\frac{\widehat C - \widehat B}{2}}$

$N_1'F = \frac{T_1F}{\cos^2{\frac{\widehat C - \widehat B}{2}}} = AD \left(1 + \frac{\tan{\frac{\widehat B}{2}}}{\tan{\frac{\widehat C}{2}}}\right)$

But this is exactly the equation we previously obtained for $N_1F$. It follows that $FN_1' = FN_1$ and the points $N_1' \equiv N_1$ are identical. Since both $E_1N_1'$ and $M_1N_1$ are both perpendicular to the bisector $DP_1 \equiv DP_2$ of the $\measuredangle ADC$, the points $E_1, M_1, N_1$ are collinear. Similarly, the points $E_3, M_3, N_3$ are also collinear.
 :P

[b]Related problem:[/b]

Let $S_1, S_3$ be the tangency points of the circles $(P_1), (P_3)$ with the circumcircle $(O)$. The internal bisector $AI \equiv AN_2$ obviously cuts the arc $BC$ on the opposite side of the triangle vertex $A$ at its midpoint $K$. Prove that the lines $E_1S_1, E_3S_3$ also concur at the point $K$.

[quote="orl"][quote="Valiowk"]:D It seems that Darij has a competitor!  As far as I remember, Darij's longest post was 7 screens (!) long (on 1024 x 768), and this one is 6 but without LaTeX formatting... :?[/quote]

Did you expand all hide-scripts ?  ;)[/quote] :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

From $Caysey$'s theorem applied to points B,A, circle k, and point C we have the relations:
${b}{BM}={a}{AN}+{c}{CM}$.(1)
Let $MN $ intersect $AB$ at $X$. Then from Menelaus’s theorem applied to triangle BAD we have:
$\frac{MD}{BM}\frac{BX}{XA}\frac{AN}{ND}={1}$,
but $MD=ND$ hence,
$\frac{AN}{BM}=\frac{XA}{BX}$  and comparing it with (1) we obtain:
${b}={a}{\frac{ XA}{BX}}+{c}{\frac{CM}{BM}}$ and
from the converse of $Transversal$'s theorem we get that $J$ is on $MX$, hence on $MN$.

Hi, it's Darij speaking. ML is failing to load on my computer right now, so I have asked a friend of mine to submit this message (thanks!).
____________________________________________________________________

This thread is old, but maybe someone is still interested in a synthetic and elementary solution I found three days ago on the Rostock seminar:

[i]Preliminaries:[/i] [b]1.[/b] We will use directed angles modulo 180° throughout the following text.
[b]2.[/b] The abbreviation "circle $P_1P_2P_3$" always means the circle through three given points $P_1$, $P_2$ and $P_3$.

I reduced the problem to the following theorem:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle. Let D, E, F be the midpoints of its sides BC, CA, AB, and let A', B', C' be the feet of the altitudes to these sides. Also, let H be the orthocenter of triangle ABC, and let D', E', F' be the midpoints of the segments AH, BH, CH. It is pretty well-known that the points D, E, F, A', B', C', D', E', F' lie on the nine-point circle of triangle ABC.

Now, let the line AE' intersect this nine-point circle at a point E" (apart from the point E'). Let the tangent to the nine-point circle of triangle ABC at the point B' meet the line C'A' at a point S. Finally, let the perpendicular to the line CA at the point A meet the line B'S at a point V and the line C'A' at a point W.

Then,

[b](a)[/b] the circle VWE" touches the line B'S at the point V, the line C'A' at the point W, and the nine-point circle of triangle ABC at the point E";

[b](b)[/b] the points W and E both lie on the perpendicular to the line AE' at the point E".[/color]

[i]Proof.[/i] Since < CC'A = 90° and < CA'A = 90°, the points C' and A' lie on the circle with diameter CA. Thus, < AC'A' = < ACA'. On the other hand, since the point H is the orthocenter of triangle ABC, the lines AH and BH are altitudes of triangle ABC, and thus they are perpendicular to its sides BC and CA, respectively; thus, < (AH; BC) = 90° and < (CA; BH) = 90°. Finally, since the points E' and F are the midpoints of the sides BH and AB of triangle ABH, the line E'F is a midparallel in this triangle, and thus it is parallel to the side AH, so that AH || E'F, and thus < (BH; AH) = < (BH; E'F). Combining these results, we find that

  < AC'W = < AC'A' = < ACA' = < (CA; BC) = < (CA; BH) + < (BH; AH) + < (AH; BC)
           = 90° + < (BH; AH) + 90° = 180° + < (BH; AH) = < (BH; AH)
           = < (BH; E'F) = < B'E'F.

Since < AB'B = 90°, the point B' lies on the circle with diameter AB. Of course, the points A and B also lie on this circle with diameter AB. The center of this circle is clearly the midpoint F of the segment AB. Thus, B'F = AF. Hence, the triangle AFB' is isosceles, and thus, < B'AF = < FB'A. In other words, < (CA; AB) = < (FB'; CA). Now, we know that the line BB' is perpendicular to the line CA (as an altitude in triangle ABC); thus, < (CA; BB') = 90°. On the other hand, since the point W lies on the perpendicular to the line CA at the point A, we have $AW\perp CA$, so that < (AW; CA) = 90°. Thus,

  < WAC' = < (AW; AB) = < (AW; CA) + < (CA; AB) = 90° + < (CA; AB)
                = 90° + < (FB'; CA) = < (CA; BB') + < (FB'; CA) = < (FB'; BB') = < FB'E'.

Thus, we have found that < AC'W = < B'E'F and that < WAC' = < FB'E'. Consequently, the triangles AC'W and B'E'F are directly similar. Thus, $\frac{AW}{AC^{\prime}}=\frac{B^{\prime}F}{B^{\prime}E^{\prime}}$. In other words, $AW\cdot B^{\prime}E^{\prime}=B^{\prime}F\cdot AC^{\prime}$.

We know that the points A, V, W lie on one line - namely, on the perpendicular to the line CA at the point A. Let W' be the orthogonal projection of the point E' on this line. Then, < E'W'A = 90°. On the other hand, < W'AB' = 90°, since the point W' lies on the perpendicular to the line CA at the point A. Finally, < AB'E' = 90°, since the point E' lies on the altitude BB' of triangle ABC. Hence, the quadrilateral W'AB'E' has three right angles; therefore, it is a rectangle. Consequently, B'E' = AW'. We also know that B'F = AF. Thus, the equation $AW\cdot B^{\prime}E^{\prime}=B^{\prime}F\cdot AC^{\prime}$ we obtained before becomes $AW\cdot AW^{\prime}=AF\cdot AC^{\prime}$. But since the points F, C', E' and E" lie on one circle (namely, on the nine-point circle of triangle ABC), the intersecting chords theorem yields $AF\cdot AC^{\prime}=AE^{\prime}\cdot AE^{\prime\prime}$. Thus, we get $AW\cdot AW^{\prime}=AE^{\prime}\cdot AE^{\prime\prime}$. By the converse of the intersecting chords theorem, this implies that the points W, W', E' and E" lie on one circle. Since the point W' is the orthogonal projection of the point E' on the line through the points A, V, W, we have < E'W'W = 90°, and thus, the segment E'W is a diameter of the circle through the points W, W', E' and E". Consequently, < E'E"W = 90°. Thus, the point W lies on the perpendicular to the line AE' at the point E".

On the other hand, since the line BB' is an altitude in triangle ABC, we have < (CA; BB') = 90°, or, in other words, < EB'E' = 90°. Remembering that the points E, B' and E' lie on the nine-point circle of triangle ABC, it becomes clear that the equation < EB'E' = 90° yields that the segment EE' is a diameter of the nine-point circle. Hence, < EE"E' = 90° (as the point E" also lies on the nine-point circle). In other words, the point E lies on the perpendicular to the line AE' at the point E".

Thus, altogether, we have shown that the points W and E both lie on the perpendicular to the line AE' at the point E". This proves Theorem 1 [b](b)[/b]. Now, we will show Theorem 1 [b](a)[/b]:

Let the circle VWE" meet the line AE' at a point R (different from the point E"). Then, since the point W lies on the perpendicular to the line AE' at the point E", we have < WE"R = 90°, and this shows that the segment WR is a diameter of the circle VWE". In other words, the circle VWE" is the circle with diameter WR. On the other hand, since the segment EE' is a diameter of the nine-point circle of triangle ABC, the nine-point circle of triangle ABC can be seen as the circle with diameter EE'.

Since the line B'S is the tangent to the nine-point circle of triangle ABC at the point B', by the tangent-chordal angle theorem, we have < (B'S; B'E") = < B'E'E". On the other hand, the lines BB' and AV, both being perpendicular to the line CA, are parallel to each other, so that < (BB'; AE') = < (AV; AE'). Thus,

  < VB'E" = < (B'S; B'E") = < B'E'E" = < (BB'; AE') = < (AV; AE') = < VAE".

Thus, the points V, E", B' and A lie on one circle. Hence, < E"VA = < E"B'A. In other words, < E"VW = < E"B'E. But since the point R lies on the circle VWE", we have < E"VW = < E"RW, and since the points E", E, B' and E' lie on one circle (namely, on the nine-point circle of triangle ABC), we have < E"B'E = < E"E'E. Thus, the equation < E"VW = < E"B'E becomes < E"RW = < E"E'E. In other words, < (AE'; WR) = < (AE'; EE'). Hence, WR || EE'.

Now, let's look once more at our configuration: The points W, E and E" lie on one line, the points R, E' and E" lie on one line, and we have WR || EE'. Thus, there exists a homothety with center E" which maps the segment WR to the segment EE'. This homothety then, of course, must map the midpoint of the segment WR to the midpoint of the segment EE'. Since the point E" is the center of this homothety, and since the center of a homothety always lies on the line joining a point with its image under that homothety, it thus follows that the point E" lies on the line joining the midpoints of the segments WR and EE'. Since the midpoint of a segment is the center of the circle which has this segment as diameter, we can restate this result as follows: The point E" lies on the line joining the centers of the circles with diameters WR and EE'.

Remembering that the circle with diameter WR is the circle VWE" and that the circle with diameter EE' is the nine-point circle of triangle ABC, we thus can state: The point E" lies on the line joining the centers of the circle VWE" and of the nine-point circle of triangle ABC. But on the other hand, we know that the point E" lies on both of these circles, i. e. the point E" is a common point of these two circles. It is well-known that, if a common point of two circles lies on the line joining the centers of these circles, then these two circles touch each other at that point. Hence, we conclude that the circle VWE" and the nine-point circle of triangle ABC touch each other at the point E". In other words, we have shown that the circle VWE" touches the nine-point circle of triangle ABC at the point E".

In order to complete the proof of Theorem 1 [b](a)[/b], it remains to show that the circle VWE" touches the line B'S at the point V, and the line C'A' at the point W.

It is well-known that the perpendicular bisector of a chord in a circle always passes through the center of the circle. Thus, since the points C' and A' lie on the circle with diameter CA, the perpendicular bisector of the segment C'A' passes through the center of the circle with diameter CA, i. e. through the midpoint E of the segment CA. Similarly, since the points C' and A' lie on the circle with diameter BH (this is because < BC'H = 90° and < BA'H = 90°), the perpendicular bisector of the segment C'A' passes through the midpoint E' of the segment BH. Thus, the perpendicular bisector of the segment C'A' passes through both points E and E'. In other words, the perpendicular bisector of the segment C'A' is the line EE'. Hence, the line EE' is perpendicular to the line C'A'. Since WR || EE', the segment WR must also be perpendicular to the line C'A'. In other words, the line C'A' is the perpendicular to the segment WR at its endpoint W. But the segment WR is a diameter of the circle VWE", and the perpendicular to a diameter of a circle at one of its endpoints is the tangent to the circle at this endpoint; thus, the line C'A' is the tangent to the circle VWE" at the point W. In other words, the circle VWE" touches the line C'A' at the point W.

Remains to show now that the circle VWE" touches the line B'S at the point V.

Since the points V, E", B' and A lie on one circle, we have < E"VB' = < E"AB'. Since the point E lies on the perpendicular to the line AE' at the point E", we have $EE^{\prime\prime}\perp AE^{\prime}$, so that < (AE'; EE'') = 90°, and since the point W lies on the perpendicular to the line CA at the point A, we have $AW\perp CA$, so that < (AW; CA) = 90°. Thus,

  < (E"V; B'S) = < E"VB' = < E"AB' = < (AE'; CA)
                = < (AE'; EE") + < (EE"; AW) + < (AW; CA) = 90° + < (EE"; AW) + 90°
                = 180° + < (EE"; AW) = < (EE"; AW) = < E"WV.

Thus, by the converse of the tangent-chordal angle theorem, the line B'S is the tangent to the circle VWE" at the point V. In other words, the circle VWE" touches the line B'S at the point V. This completes the proof of Theorem 1.

Now, let's come to the solution of the problem; we will use the notations from Yetti's post #2. So, we have a triangle ABC and its excenters $E_2$, $E_1$, $E_3$ (opposite to the vertices A, B, C, respectively). We will apply Theorem 1 to the triangle $E_1E_2E_3$; but before we do this, we recall some simple facts about this triangle:

- The points B, A, C are the feet of the altitudes of the triangle $E_1E_2E_3$. (This is well-known - and pretty clear: Since the internal and the external angle bisector of an angle are always perpendicular to each other, we have $BE_1\perp E_2E_3$, $AE_2\perp E_3E_1$ and $CE_3\perp E_1E_2$, so that the points B, A, C are the feet of the altitudes of the triangle $E_1E_2E_3$.)
- The circumcircle of triangle ABC is the nine-point circle of the triangle $E_1E_2E_3$. (This is clear, since it passes through the feet B, A, C of the altitudes of triangle $E_1E_2E_3$.)
- The midpoints $Q_1$, $Q_2$, $Q_3$ of the sides $E_2E_3$, $E_3E_1$, $E_1E_2$ of the triangle $E_1E_2E_3$ are the midpoints of the arcs ABC, CAB, BCA on the circumcircle of triangle ABC. (Prove this for yourself - it's pretty easy.)
- The orthocenter of triangle $E_1E_2E_3$ is the incenter I of triangle ABC, and the midpoints $R_1$, $R_2$, $R_3$ of the segments $E_1I$, $E_2I$, $E_3I$ are the midpoints of the arcs CA, BC, AB on the circumcircle of triangle ABC. (Dito.)

Using these facts, after applying Theorem 1 to the triangle $E_1E_2E_3$, we get the following results:

[b]Application of Theorem 1 to the triangle $E_1E_2E_3$:[/b]

Let the line $E_1R_2$ intersect the circumcircle of triangle ABC at a point $L_2$ (apart from the point $R_2$). Let the tangent to the circumcircle of triangle ABC at the point A meet the line BC at a point D. Finally, let the perpendicular to the line $E_3E_1$ at the point $E_1$ meet the line AD at a point $V_2$ and the line BC at a point $W_2$.

Then,

[b](a)[/b] the circle $V_2W_2L_2$ touches the line AD at the point $V_2$, the line BC at the point $W_2$, and the circumcircle of triangle ABC at the point $L_2$;

[b](b)[/b] the points $W_2$ and $Q_2$ both lie on the perpendicular to the line $E_1R_2$ at the point $L_2$.

Well, what follows from these results? Very much. In fact, the result [b](a)[/b] shows that the circle $V_2W_2L_2$ is tangent to the circumcircle of triangle ABC and to the lines AD and BD. But there exist exactly three circles tangent to the circumcircle of triangle ABC and to the lines AD and BD - namely, the circles $\left(P_1\right)$, $\left(P_2\right)$ and $\left(P_3\right)$. Hence, the circle $V_2W_2L_2$ must coincide with one of these three circles $\left(P_1\right)$, $\left(P_2\right)$ and $\left(P_3\right)$. Actually, it is clear that the circle $V_2W_2L_2$ cannot coincide with the circle $\left(P_2\right)$ (since the circle $\left(P_2\right)$ touches the line AD at the point A, while the circle $V_2W_2L_2$ touches the line AD at the point $V_2$ which is clearly different from A). Hence, the circle $V_2W_2L_2$ coincides with one of the two circles $\left(P_1\right)$ and $\left(P_3\right)$. Looking at the arrangement of points, one concludes that the circle $V_2W_2L_2$ coincides with the circle $\left(P_1\right)$. Hence,

- the point $V_2$, at which the circle $V_2W_2L_2$ touches the line AD, coincides with the point $M_1$, at which the circle $\left(P_1\right)$ touches the line AD;
- the point $W_2$, at which the circle $V_2W_2L_2$ touches the line BD, coincides with the point $N_1$, at which the circle $\left(P_1\right)$ touches the line BD;
- the point $L_2$, at which the circle $V_2W_2L_2$ touches the circumcircle of triangle ABC, coincides with the point $S_1$, at which the circle $\left(P_1\right)$ touches the circumcircle of triangle ABC.

From here we can draw a lot of conclusions:

- We know that the points $V_2$ and $W_2$ lie on the perpendicular to the line $E_3E_1$ at the point $E_1$. Substituting $V_2=M_1$ and $W_2=N_1$, we can conclude that the points $M_1$ and $N_1$ lie on the perpendicular to the line $E_3E_1$ at the point $E_1$. In particular, this yields that the points $M_1$ and $N_1$ lie on one line with the point $E_1$. This solves the initial problem.
- The point $L_2$ was initially defined as the point where the line $E_1R_2$ intersects the circumcircle of triangle ABC. Hence, the point $L_2$ lies on the line $E_1R_2$. Keeping in mind that $L_2=S_1$, we conclude that the point $S_1$ lies on the line $E_1R_2$. In other words, the line $E_1S_1$ passes through the point $R_2$. This solves the "Related problem" proposed by Yetti (note that he uses the notation K for the point which I call $R_2$).
- Part [b](b)[/b] of our result above states that the points $W_2$ and $Q_2$ both lie on the perpendicular to the line $E_1R_2$ at the point $L_2$. Since $W_2=N_1$ and $L_2=S_1$, this can be rewritten as follows: The points $N_1$ and $Q_2$ both lie on the perpendicular to the line $E_1R_2$ at the point $S_1$. This fact doesn't seem to have been mentioned in this thread until now.

So much for the problem. Two remarks:

[b]1.[/b] Thanks, Yetti, for your "Related problem" - it did help me in finding the above solution of the initial one. This gives another nice example of how proving a stronger assertion is easier than proving a weak one.
[b]2.[/b] No guys, my longest post on ML was much more than 7 screens long (it takes 12 sides on paper) and didn't contain any hide-scripts. Guess which post I mean. (No, not this one.)

  darij

So much for hoping to see a shorter solution to the problem... :( :( :( ;)

Hello Pestich,

[quote="pestich"]  Your 'favorite' extraversion will be of help here.  You certainly
  remember that mixtilinear circle touches the sides of triangle at the
  points were perpendicular to angle bisector thru incenter intersects
  the sides.[/quote]

Though the analogy with the mixtilinear incircle configuration is striking, I fear that it cannot help to find a new solution of this problem. I don't see how extraversion could make the line AD (the tangent to the circumcircle of triangle ABC at the point A) out of a sideline of triangle ABC. Moreover, the incenter of triangle ABC is the midpoint between the points of tangency of a mixtilinear incircle with two sides of triangle ABC, while in our problem, the excenter is only collinear with the points of tangency, but is far from being their midpoint.

  Darij

[quote="Darij"][b]1.[/b] Thanks, Yetti, for your "Related problem" - it did help me in finding the above solution of the initial one. This gives another nice example of how proving a stronger assertion is easier than proving a weak one.
[b]2.[/b] No guys, my longest post on ML was much more than 7 screens long (it takes 12 sides on paper) and didn't contain any hide-scripts. Guess which post I mean. (No, not this one.)[/quote]

[b]1.[/b] WOW !!! :first: I did not know the solution of the related problem, it just showed on the Sketchpad and it was plausible - what is true for an incircle and one excircle should be true for the remaining two excircles.
[b]2.[/b] The guys were laughing at me. It was my 1st post at this forum, I did not know any LATEX at the time, the post was horrible (with [color=blue][b]a lot[/b][/color] of hide scripts) and only latexed at a later time.

[quote="pestich"]Will 2 tangent points (M,N) and the ex-center (J) still be collinear if tangent to the
 circumcircle at vertex A is replaced by any line thru vertex A intersecting opposite side
 BC?[/quote]

Wow, Pestich, a very good - and correct - observation! And it can be proven exactly in the same way as Sailor solved the original problem in post #8. But it can also be proven similarly to Lemma 1 in [1], §2 (actually, it's an extraversion of that lemma). So you were right that extraversion could help!

[b]References[/b]

[b][1][/b] [url=http://forumgeom.fau.edu/FG2003volume3/FG200325index.html]Jean-Louis Ayme, [i]Sawayama and Thébault's Theorem[/i], Forum Geometricorum 3 (2003) pp. 225-229[/url].

  darij

Dear Mathlinkers,
working about mixtilinear incircles, I found this interesting problem which I have proved last year on my site
see : http://perso.orange.fr/jl.ayme     vol. 2  Un remarquable probleme de Vladimir Protassov p.6  "Un alignement".
Sincerely
jean-Louis