probability function algebra polynomial calculus integration

In how many ways can a 100-dollar bill be changed into 1-, 2-, 5-, 10-, 20-, and 50-dollar bills?

It is a good candidate for computer-oriented solution, though we can easily consider all cases manually ;)

And it is really popular problem on programming contests.

there is a nice manual solution to this problem

Interestingly, they went over this in the Intermediate Counting and Probablity class.

It is the coefficient of $z^{100}$ in power serie expansion of
$f(z)=\frac{1}{(1-z)(1-z^2)(1-z^5)(1-z^{10})(1-z^{50})}$

[quote="DPopov"]Has that started!?! :?[/quote]

I took the one which recently ended, and they talked about generating functions in it.

We can make 10 dollars 10 different ways with the 1s 2s and 5s.

Similarly we can make 100 dollars 10 ways with the10s 20s and 50s.

Now, every time we have a ten dollar bill we can replace it in 10 ways.  We might have 1, 2, 3, 4, 5, 6, 8, or 10 ten dollar bills to replace.

Hence,
10 + 10*(1+2+3+4+5+6+8+10) = 400 ways to make 100.

[quote="DPopov"]hint: the answer is >4000[/quote]

4001?   :lol:

Edit: What?!  I don't want a santa smiley!  I'm Jewish.  And offended.  Okay, I'm not really offended.  You've caught me.  But still...

hehe...ur closer than all the rest have been...well actually i think there was a correct solution but no answer :P ...keep going...haha...answer>4001

I took the sum
$\displaystyle\sum_{m=0}^{10}(5m^2+4m+1)f(10- m)$
where $f(n)$ is the number of ways to get $n$ with 1s 2s and 5s.  That polynomial is how many ways there are to get $10m$ with 1s 2s and 5s.
This is still a rather tedious operation.
I am very curious about your "nice" solution.

Well...heh..."nice" is not the exact word :blush:  :lol: ....hehe....I basically did what you did and it's not that hard but tedious yes

Solution

[hide]
Since only an integral multiple of 10 dollars can be made up of out of 10, 20, and 50 dollar bills, the 1, 2, 5 dollar bills must also add up to a multiple of 10 dollars. Suppose, that the 1's, 2's, and 5's add up to $10m$ dollars, and the 10's, 20's and 50's to $100-10m$ dollars. Here $m$ can range from 0 to 10. We will thus determine the number of solutions for each fixed value of $m$, and then add these together to get the answer to the problem.
     The number of ways of breaking the integer $10m$ into a sum of 1's, 2's, and 5's is $\frac{N(10m+4)^2}{20}$. On the other hand the problem of breaking $100-10m=10(10-m)$ into 10's, 20's and 50's is equivalent to that of writing $10-m$ as a sum of 1's, 2's, and 5's because everything has only been multiplied by a factor of 10. Therefore the number of solutions is \[N\left(\frac{(10-m+4)^2}{20}\right)=N\left(\frac{(14-m)^2}{20}\right)\]

Each of these solutions can be combined with each of the representations of $10m$ as a sum of 1's, 2's, and 5's; thus, with this value of $m$ there are
\[N\left(\frac{(10m+4)^2}{20}\right) \cdot N\left(\frac{(14-m)^2}{20}\right)\]

ways of changing the 100 dollars. Summing over the vlues of $m$ from 0 to 10 we get a total of

\[1 \cdot 10 + 10 \cdot 8 + 29 \cdot 7 + 58 \cdot 6 + 97 \cdot 5 + 146 \cdot 4 + 205 \cdot 3 + 274 \cdot 2 + 353 \cdot 2 + 442 \cdot 1 + 541 \cdot 1\]
\[=4562\]

[/hide]

[quote="darij grinberg"][quote="Myth"]And it is really popular problem on programming contests.[/quote]

Why programming? One doesn't need to write a new program for expanding the product of six polynomials.

  Darij[/quote]

Its a simple dynamic programming question, and questions of this type show up a lot (very rarely something this simple, but it is a classic example of a dynamic programming question). These kinds of questions show up on the USACO almost every contest.

$N(x)$ denotes the integer closest to $x$.


inequalities
logarithms
inequalities open

Here is my proof,
because the inequality is equal to:
\[ x^{(x-y)(x-z)}y^{(y-z)(y-x)}z^{(z-x)(z-y)} \ge 1  \]
we may assume:$z \ge y \ge x > 1$
then we have:
$x^{(x-y)(x-z)} \ge 1$
and
\[ y^{(y-z)(y-x)}z^{(z-x)(z-y)} \ge y^{(y-z)(y-x)}y ^{(z-x)(z-y)}=y^{(y-z)(y-x)+(z-x)(z-y)} \ge 1  \]
So we proved. :)

Another solution is to write the ineq equivalent in this form

\[ x^{(x-y)(x-z)}y^{(y-z)(y-x)}z^{(z-x)(z-y)} \ge 1  \]

and then take the logarithm and use Vornicu-Schur ineq.


geometry
circumcircle
trapezoid
perpendicular bisector
geometry proposed

two equileteral triangle ABC and A'B'C' lie on the same circle. AA', BB', CC' intersect at points X,Y,Z. Prove that the circumcenter of triangles ABC and A'B'C' is also a circumcenter of triangle XYZ

Let $C''$ be the intersection of $BB'$ and $AA'$

See that $AB=A'B'$ then $ABA'B'$ is an isoceles trapezoid (or whatever it's called) then $BA'$ is parallel to $AB'$
Also triangles $BA'C''$, $AB'C''$, $C''B''A''$ are equilateral so $B''A''$ is parallel to $AB'$ and the perpendicular bisector of $AB'$ is the same line as the perpendicular bisector of $B''A''$ (because of Thales)

With that we know that the perpendicular bisector of each side is equal to the perpendicular bisector of three chords of the circle, so they concur at the center.


group theory
abstract algebra
search
superior algebra
superior algebra unsolved

If $H$ is a maximal subgroup of a finite solvable group $G$,prove that the index $[G: H]$ is prime power.

Thank you for your nice solution in the link,and sorry for my repeated post.


In any $\bigtriangleup ABC$, $E$ and $D$ are interior points of $\overline{AC}$ and $\overline{BC}$, respectively. $\overline{AF}$ bisects $\angle CAD$, and $\overline{BF}$ bisects $\angle CBE$. Prove that $m\angle AEB + m\angle ADB = 2m \angle AFB$.

I'm going to leave out the $m$ when I say something like $m\angle AFB$

[hide]Proof: Let $\angle CAF=\angle FAD=\alpha$ and $\angle CBF=\angle FBE=\beta$. Looking at $\triangle AEB$ and $\triangle BDA$ we have $\angle AEB=180-(\angle DAB +2\alpha +\angle EBA)$ and $\angle ADB=180-(\angle EBA +2\beta +\angle DAB)$. Adding these two equations yields $\angle AEB + \angle ADB=360-2\alpha-2\beta-2\angle DAB-2\angle EBA=2(180-(\alpha+\beta+\angle DAB+\angle EBA))=2(180-((\alpha+\angle DAB)+(\beta+\angle EBA)))=2(180-(\angle FAB +\angle FBA))=2\angle AFB$. QED[/hide]

Correct  :) Though you might want to hide your solution in case anyone else wants to look at the problem.


inequalities conics parabola algebra function domain ellipse

[quote="10000th User"][quote="shobber"]Real numbers $x$ and $y$ satisfy $2x^2+y^2=6x$. Find the max of $x^2+y^2+2$.[/quote]

$2x^2+y^2=6x$

$x^2+y^2=6x-x^2$

Find max of $x^2+y^2+2=6x-x^2+2$, this is a parabola and max is at vertex of parabola :)[/quote]

Note the domain of $x$. ;)

[quote="shobber"]Real numbers $x$ and $y$ satisfy $2x^2+y^2=6x$. Find the max of $x^2+y^2+2$.[/quote]

[hide="solution"]clearly, it is sufficient to maximize $\sqrt{x^2+y^2}=r$, consider the locus of points such that $k$ is constant, it is a circle at the origin, if we want to maximize r, then we make the circle have as large a radius as it can while still touching the ellipse, consider when we have the intersection of the circle $x^2+y^2=r^2$, we want to maximize $r$

$x^2-6x+r^2=0$, since the intersection must be real, using the discriminant, $36-4(r^2) \ge0 \implies r\le 3$, and this max occurs at the point $(3,0)$, so then the max is $\boxed{11}$[/hide]

hmm nice solution, the circle / ellipse thing is unnecessary i think but its a nice touch  :lol:

[quote="10000th User"][quote="shobber"]Real numbers $x$ and $y$ satisfy $2x^2+y^2=6x$. Find the max of $x^2+y^2+2$.[/quote]

$2x^2+y^2=6x$

$x^2+y^2=6x-x^2$

Find max of $x^2+y^2+2=6x-x^2+2$, this is a parabola and max is at vertex of parabola :)[/quote]
are u unable to finish that thought or just to lazy?

its simple \[ -x^2+6x+2 = -(x-3)^2 + 11 \le 11 \]
and also for $x=3$ it is $11$ so Altheman's solution is very right ;)

We should refer to the domain of $x,$ since $y$ is real numbers, so $y^2\geq 0\Longleftrightarrow 0\leq x\leq 3.$ ;)

[quote="peeta"]are u unable to finish that thought or just to lazy?[/quote] Yea I'm lazy :P
I'll try to finish it next time  :blush:


Asymptote
analytic geometry
function
algorithm
trigonometry
parameterization

fedja: Thanks for the work you've been doing with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149650]cse5[/url] which can be used in the forum. Can the package also be downloaded to use on a PC, or have I missed a download link?

I'm wondering. What did you do to make finding intersections not crash when there are none?

Also, where can I view the code for this package? I'm pretty sure the code you were using before would have caused the crash still.

@ Hamster1800 Returned nullpair. And, please, could the mods split this topic to put Hamster1800's question and this answer of mine to a separate thread. I will answer the questions, but, for God's sake, let us keep the description separately and questions separately in two different threads :roll:

[quote="stevem"]Can the package also be downloaded to use on a PC, or have I missed a download link?[/quote]I can answer that: yes! :) The link is http://www.mathlinks.ro/Forum/latexrender/cse5.asy

More details on Hamster1800 question about why the code doesn't crash. The answer is that IP() looks at the return of intersectionpoints() and, if it is empty, returns a special pair nullpair=(1.2345,6.7890). All other commands in cse5 recognize this exceptional case and act accordingly. Of course, the commands outside the package will treat it just as an ordinary point but still you can see what happened sometimes. So, using IP() will never crash the code, just give some fancy result if the paths do not intersect. Actually, it would be nice if other commands would also recognize nullpoint. It is not too hard to do, just redefine all operators once, but I was too lazy to turn $\mathbb R^{2}$ into a projective plane myself.

Yes, I looked at the code and figured out what it was doing. I was asking because I wanted to create a similar package which I am currently calling csel for Compass-Straight-Edge-construction-with-Long-names.

What it features that yours doesn't: It will find the intersection point of two lines or a line and a circle without requiring the line segment to be intersecting it. It does this with a lot of coordinate computation (manually done). And of course, since it's long names, it uses more intuitive functions like draw(), line(), circ(), intersect(), toPath(), etc. I've used the same nullpair as you so that both packages can be used together without problems. I'm currently hosting it [url=http://bhamrick.homelinux.com:33333/csel.asy]here.[/url] Feel free to comment on it. Please note that the webserver is on my laptop, so the site will go down often.

EDIT: Unfortunately, the package is over 2000 characters, so I can't demonstrate on this forum, but the code to bisect a line segment would be this:

[code]
import csel;
size(400);
pair a, b;
circ c, d;
draw(line(draw((a=(0,0))),draw((b=(2.5,0)))));
draw(extend(line(draw(intersect(draw((c=circ(a,2.0)),blue),draw((d=circ(b,2.0)),blue))),draw(intersect(c,d,1)))),red);
[/code]

Of course, this is the super-packed-with-syntax-abuse format. A more logical format would be

[code]
import csel;
size(400);
pair a=draw((0,0)), b=draw((2.5,0)), x, y;
circ c = circ(a,2.0), d=circ(b,2.0);
draw(c,blue);
draw(d,blue);
x = intersect(c,d);
y = intersect(c,d,1);
draw(x);
draw(y);
draw(extend(line(x,y)),red);
[/code]

It may be beneficial to combine two packages in one. Long names are extremely easy to add to cse5: just add
[b]
path line(pair A,pair B,real a=0.6, real b=a)=L;
[/b]
and similar things to the code.
It is a good idea to have both a long and a short name for any routine connected by some easy to understand converting algorithm. I suggest the long names like MarkPoint synonymous to MP with the conversion just leaving the capital letters. Everybody who knows the long name will immediately know the short one as well and when he is tired of typing the long ones, or when he needs to use a construct in the style of cse5, he will just switch to short ones. Actually, everybody can elaborate on the current version of cse5 by adding as much subroutines as he wishes. If he comes with a nice routine, we can just add it to the site version. I have no intention to keep the code unchanged or to claim any sort of copyright on it. Want long names available in cse5? OK, will be done as soon as somebody else seconds the idea and I and Valentin find time to update the code :). With your add-on on intersection of lines, the story is more complicated:  I wanted the lines to be regular paths (so they could be treated by other asymptote commands and you could intersect a line and a path, etc. by using standard routines). Your approach requires to keeping them as structures of two pairs and two reals with special versions of D, IP, etc. This may be a good idea and this could be incorporated into the code but you'll need to add a special type line and to add quite a few definitions. I just wanted the code to be as simple and clean as possible. If you ever need the intersection of the extended lines L1 and L2 in cse5, and you know two points A and B on the first line and two points C,D on the second just write IP(L(A,B,1000),L(C,D,1000)). It won't draw anything, but will create your point. :). Since the lines in cse5 and anywhere else in Asymptote are designed from pairs of pairs, you always know two points on each available line. Just take care of giving them some names when they arise before they are called by the L routine and disappear. ;) Anyway, cse5 is just for drawing nice pictures, not for scientific investigations of Compass and Straight Edge constructions, which seems to be what you are after :lol:

I second the idea of long and nice names for cse5 and actually renaming it to csed (compass and straight edge drawings). The thing of having less than 2000 lines of code is a Microsoft thing, and I'd rather not get into details :)

Long names are OK with me then. I'll add the lines to the code, test its workability, and send it to Valentin as soon as I find time for it. So, both long and short names for each procedure will be available with the conversion algorithm that just leaves the capital letters with one exception. The suggested names are

distance for d
Draw for D
IntersectionPoint for IP
OtherPoint for OP
CirclebyRadius for CR
CirclebyPoint for CP

(alas, I cannot make both of them into Circle because real is too easily cast into pair by Asymptote. Otherwise the short name would be C for both routines :()

Line for L
MarkPoint for MP

At any rate, that is how I referred to them myself and how I chose the letters. Can you offer improvements? By now the short names seem to catch on and I would like to keep them as they are. Microsoft or Valentin, but the limit is there anyway :P.

Renaming can be done, of course, but, again, it would be nice to keep both names available. I would hate to go and change all the names in my posts now and who knows how many other posts have import cse5; by this moment. So, we'd better see first if somebody can come with a really good name (I personally do not see much improvement in csed compared to cse5 and to create more than one alias would be stupid) So, what would be a really good name for the package?

[quote="fedja"]So, what would be a really good name for the package?[/quote][i]cosec[/i] from [b]Co[/b]mpass-[b]S[/b]traight-[b]E[/b]dge-[b]c[/b]onstruction is a good mathematical name though it might give the impression that it's a trig package.
Similar ideas would be:
[i]cosed[/i] [b]Co[/b]mpass-[b]S[/b]traight-[b]Ed[/b]ge-construction
[i]cosedge[/i] [b]Co[/b]mpass-[b]S[/b]traight-[b]Edge[/b]-construction
[i]comsec[/i] [b]Com[/b]pass-[b]S[/b]traight-[b]E[/b]dge-[b]c[/b]onstruction
[i]compact[/i] [b]Comp[/b]ass-Str[b]a[/b]ight-Edge-[b]c[/b]ons[b]t[/b]ruction (another nice mathematical name)
or even just [i]compass[/i]?

There is a problem with establishing an alias Draw for D (apparently, Asymptote has a subroutine Draw in it somewhere, so the call to Draw becomes ambiguous; Hamster1800 should have the same problem with his redefinition of draw. The only longname I can invent is Drawp (with p standing for both point or path). I'm not completely happy with this name, so I ask whether anybody can offer a better one. The only restriction is that it should start with D. The rest names worked just fine, so, as soon as we decide what to do about the long name for D, the package will be updated to make long names available as well.

You could always just name it constructions...(or construct if you prefer a shorter name).

Anyway, I haven't had any problems with my overloaded draw command because I'm not using normal paths. draw(pair, pen) doesn't exist by default, rather you use dot(pair, pen), so I just included draw to make it easier. Also, as line and circ hadn't been defined yet, of course draw(line, pen) and draw(circ, pen) didn't exist.

Yep, pair draw(pair,pen) is easy but if you want to use the standard version of draw for paths, you have to adjust the default pen (unless you are happy with black lines, of course) and forfeit addition of extra features like my disabling of D by adding negative parameter in the end. Anyway, should I go ahead with Drawp?

Out of the names suggested by stevem, I like compact most for several reasons. So, unless Valentin strongly objects this new name and unless somebody comes up with even a better one in a couple of days, I suggest that this alias be created in the middle of the next week. I'll still refer to it as cse5 in the manual meanwhile.

Thanks for your interest, guys! :lol:

Below are the changes I'd like to introduce to cse5/compact code. The new code ran smoothly in all situations I could think of (except that with 2 paths intersecting at infinitely many points, of course). You may especially like what happens if you put
[b]fullstop=true;[/b]
after the import command and try some code (any Asymptote code, not necessarily that with only cse5/compact commands) in which some paths fail to intersect ;) Of course, to see the nice picture and not a mess, you have to use D instead of draw and dot and MP instead of label everywhere and to intersect your paths with IP. As to long names, since no better suggestions have been made so far, I finally chose [b]Drawing[/b] for D. The rest of them are as above. So, what do you think of this new version? If there are no objections, bug reports or suggestions for improvements, I'd like Valentin to implement it when he finds time for that. :)
[code]
//Version of 05/20/2007

//Default size

size(400);

//Adjustable variables

pair nullpair=(1.2345,6.7890);
pair messageplace=(0,5);
pen pointpen=red,pathpen=blue,errorpen=black+linewidth(1.5);
bool fullstop=false;

//Auxiliary variables

bool stopflag=false;

//Basic routines

real d(pair A,pair B)
{
if ((A==nullpair)||(B==nullpair)) return -1;
return sqrt((A.x-B.x)^2+(A.y-B.y)^2);
}

path D(path p, pen q=pathpen, real m=1)
{
if (fullstop&&stopflag) return p;
if(m>0) draw(p,q);
return p;
}

pair D(pair p, pen q=pointpen, real m=1)
{
if (fullstop&&stopflag) return p;
if ((p!=nullpair)&&(m>0)) dot(p,q);
return p;
}

pair MP(string s, pair A, int f=11,pair B=plain.S)
{
if (fullstop&&stopflag) return A;
if (A!=nullpair) label("$"+s+"$",A,B,fontsize(f));
return A;
}

path CR(pair A, real R)
{
if((A==nullpair)||(R<=0)) return nullpath;
return circle(A,R);
}

path CP(pair A, pair B)
{
if (B==nullpair) return nullpath;
return CR(A,d(A,B));
}

path L(pair A, pair B, real a=0.6, real b=a)
{
if((A==nullpair)||(B==nullpair)) return nullpath;
return (b*(B-A)+B)--(a*(A-B)+A);
}

pair IP(path A, path B, int m=0)
{
pair[] intpoints=intersectionpoints(A,B);
int len=intpoints.length;
if ((m>=len)||(m<0))
 {
 if (fullstop)
  {
  D(A,errorpen);
  D(B,errorpen);
  MP("\mbox{Paths have no common point with index }"+string(m),messageplace,9);
  stopflag=true;
  }
 return nullpair;
 }
return intpoints[m];
}

pair OP(path A, path B)
{
return IP(A,B,1);
}

//Basic routine long names

real distance(pair A,pair B)=d;

path Drawing(path p, pen q=pathpen, real m=1)=D;

pair Drawing(pair p, pen q=pointpen, real m=1)=D;

pair MarkPoint(string s, pair A, int f=11,pair B=plain.S)=MP;

path CirclebyRadius(pair A, real R)=CR;

path CirclebyPoint(pair A, pair B)=CP;

path Line(pair A, pair B, real a=0.6, real b=a)=L;

pair IntersectionPoint(path A, path B, int m=0)=IP;

pair OtherPoint(path A, path B)=OP;

[/code]


analytic geometry
trigonometry
integration
geometry
rectangle
function
calculus

I have a question that I'm unable to figure out.

To derive a formula for arc length, I know that you start with $x=f(\theta)\cos \theta$, $y=f(\theta)\cos \theta$, differentiate with the product rule, and eventually enf up with $L=\int^{b}_{a}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta$

I was thinking of another way to derive the formula and decided to use the fact that a sector of a circle with central angle $\theta$ has outside length $r\theta$.

To find the length of a curve in polar coordinates, you would use an infinite number of these segments, much like you use an infinite number of rectangles to find the area under a curve. However, the result is simply $\int^b_a r~d\theta$, which is slightly off from the initial formula.

Do you guys know what I did wrong?  :huh:

What you did was to replace the small pieces of the curve with the small arcs of the same angular size. But it is a really bad approximation for the pieces that go almost radially. It is like saying that the length of the graph of the function $y=f(x)$ is $\int dx$ (here each small piece of the graph is replaced by the horizontal interval with the same projection to the $x$-axis, which is a really bad approximation for the oblique pieces of the graph).

Be aware that the reasonings with "$dr$" as pieces of length and what not are really just informal motivations. A real proof of the formula is quite lengthy and nothing fun.

I see.

I'm still kind of new to calc, I guess.  :D


I have isolated the cause!

[img]http://2.bp.blogspot.com/_tQ74IfNZmZQ/R9C-HytY1UI/AAAAAAAAEYY/kwiwKrecdFQ/s400/kiss-pig.jpg[/img]

Nobel Prize here I come!

i literally started cracking up when i saw that picture.

[quote="Mewto55555"]Oh dear! I forgot all about her brother!

[img]http://www.worldofstock.com/slides/PCH9631.jpg[/img][/quote]

no, no, no, mew.
why do you fail to see that the pig is fake in this picture?

Dang it! I knew i shouldn't have done that!  Now i'm sick D=

congrats mew to win the Noble Peace Prize

[quote="AIME15"]Blah lol

[img]http://images2.wikia.nocookie.net/wikiality/images/0/03/Loli%2BDonald.jpg[/img][/quote]

The bill should be pointed the other way right? I hope it should.

What?? Swine flu isn't related to pigs???
http://www.thefreedictionary.com/swine

Swine is apparently meaning hog or pig. And it originated from a disease that pigs get. And then some people decided to eat some sausage or hot dogs which the pig that they ate I think was sick.

No, they're renaming the flu to North American Influenza. So it is no longer swine flu.

Wasn't that because of pork companies lobbying the WHO to change the name because it hurt business?

PS jonathan, a piece of advice, don't assume that because you can utilize dictionary.com you know everything.

It's actually not called Swine Flu, its true name is H1N1.

Yes, this current outbreak is the H1N1 virus. I believe the cause of the spanish flu was the R2D2 virus, and the black plague was the C3PO virus.

[quote="Mewto55555"]... was the R2D2 virus, ...[/quote]

Isn't R2D2 that round robot in Star Wars?

@buzzer: HEY!! WAS THAT SARCASM???  :huh:  :huh:  :wink:

[quote="Yongyi781"]C3PO virus........... Can it speak in 20 million languages?[/quote]

With gold plating!

[quote="buzzer11"]and six ears![/quote]

And a really long [size=9]nose[/size]

I caught it. All I have to do is wait until I get better and I am IMMUNE!!!

Dude really? Get off our forum, there are few cases in MO.

And the Missouri Forum created S1N1 -- Spam Influenza

[quote="1=2"]*humph*

Now the word "Bronchitis" is being spread around. :(

/me gives MO Bronchitis[/quote]

Dudes we should like photoshop a picture of Mabbitt kissing a pig or something.


Hi.

Can anyone prove this:
ABC is a triangle. Let T(a), T(b) and T(c) be the circles, with centres A, B and C respectively and are mutually tangent on the sides of ABC. Let T be the circle externally tangent to T(a), T(b) and T(c). Let S be the centre of T. Prove the triangles ABS, CAS and BCS have the same perimeter.

Well, this is very easy: Let r(a), r(b), r(c) and r be the radii of the circles T(a), T(b), T(c) and T, respectively. The perimeter of triangle BCS is

  BS + CS + BC = (r - r(b)) + (r - r(c)) + (r(b) + r(c)) = 2r,

and similarly, the perimeters of triangles CAS and ABS are 2r, too.

  Darij

I'm sorry, once again I've inadvertently deleted a posting.

You asked how to prove the converse of theorem.

The converse is a bit trickier:

If the perimeters of triangles BCS, CAS and ABS are equal - say, they are all equal to 2r -, then we have to show that the circle with center S and radius r touches all three circles T(a), T(b) and T(c).

In fact, we have

  2 AS = (AS + BS) + (CS + AS) - (BS + CS)
          = (AS + BS + AB) + (CS + AS + CA) - (BS + CS + BC) + BC - AB - CA
          = 2r + 2r - 2r + (r(b) + r(c)) - (r(a) + r(b)) - (r(c) + r(a))
          = 2r - 2r(a),

so that AS = r - r(a), and the circle with center S and radius r touches the circle T(a). Similarly, it touches the circles T(b) and T(c), too.

  Darij


linear algebra
matrix
function
vector
group theory
abstract algebra

Let's $M \in S_3^{++}(\mathbb{Z})$ a positive definite symetric matrix. Prove that there is a matrix $U \in GL_3(\mathbb{Z})$ such that $M=U^tU$.

[quote="Xevarion"]I have a suggestion, I am not sure if it will work though:

How about you diagonalize M as $A = SMS^t$, and then since it is positive-definite, you can write $A = B^2$ where $B$ is also diagonal. Then $M = (S^tB)(BS)$.

Very sketchy, I guess. Is it okay?[/quote]

Xevarion, i have a question: surely it would be done if the basic field was $\mathbb{C}$, but where is the quarantee that $BS$ is in $M_n(\mathbb{Z})$ ?  We cannot take a square root being always in $M_n(\mathbb{Z})$ as well as diagonalize.

of course it does not work liek this because we want here to have matrices with integer coefficients. Or at least, this proof does not work. (and you don't use the fact that $n=3$ ..)

I proved it also for $n=4$ using Minkowski's theorem. It can be also proved for $n=5$ using more accurate estimates. I wonder what happens for 6 and 7 (for 8 it does not work).

great ! and which Minkowski theorem are you talking about ? there are several such theorems  :D
(PS: from this result one can deduce the well known theorem about sum of $3$ squares)

The one about the existence of non-trivial lattice points in convex bodies of volume greater than $2^n$.

Harazi, could you please give a sketch of you proof for $n=4$.It seems very interesting.

I wrote a small article for Gazeta Matematica, but unfortunately it is written in romanian, so I cannot post it here. I will just give a few steps (I've just realized that my assertion holds for matrices of determinant 1, at least this is what I proved: I don't know a proof for other values of the determinant)
  1) The main idea is the following lemma: for $n\leq 4$ and $A$ symmetric and positively defined of determinant 1, if $g$ is the billinear form associated to $A$, one can find $x\in Z^4$ such that $g(x,x)=1$;
    In order to prove this lemma, one considers the ellipsoide defined by $g(x,x)<2$ and computes its volume: small computation gives $\frac{(\sqrt{(2\cdot\pi)})^n}{\gamma(1+n/2)}$ (the gamma function) and one can easily check this is greater than $2^n$. Minkowski's theorem gives the result.
  2) One proves that one can find a basis of $Z^n$ called $v_1,...,v_n$ (where $v_1$ is the one found on lemma) such that $g(v_1, v_i)=0$ for $i\geq 2$. In order to prove this, observe that the set of vectors $x\in Z^n$ orthogonal (with respect to $g$) to $v_1$ is a subgroup of $Z^n$, thus of the form $Zv_2+...+Zv_r$. One can easily prove that $v_1,...,v_r$ is then a basis of $Z^n$ and has the required properties.
  3) A small induction allows then to conclude.
   By the way, what is the proof for arbitrary value of the determinant?

:blush:  you're right, I forgot to say say that $\det A = 1$ ..

But my question was: is still right even without this assumption?

$det(M)=det(U)^2$ ? so the determinant must be a square . But it is not sufficient. For example in $M_{3}(\mathbb{Z})$, the matrix $A=diag(1,7,7)$ is not of the form $U^tU$ because otherwise $7=A_{3,3}=U_{1,3}^2+U_{2,3}^2+U_{3,3}^2$ would be a sum of $3$ square.


Can anyone explain post office box? I dint understand anything of it.

A post office box is... something that you can put mail in.  lol.  Is that really a physics question?

http://en.wikipedia.org/wiki/Post_office_box

Little bit of googling and I found out what the post box is:

http://curiousminds.co.uk/product_info.php/products_id/669

http://www.zodiacindia.com/pobox.htm

But it seems to me like a simple variable conductor (?).


Kvant M1250.
I can write the solution here, if you want...

Myth, would you be so kind to write the solution here or could you give a link ?

Oh, my god, it is so nice and so difficult to find. I tried to solve it and failed. How many people managed to solve this problem during the contest?  :(

In the same way we can prove that $LHS> \left(\frac{2a}{\sqrt{a+1}}-a\right)n$.

myth, how could u find the solution? It looks ad hoc to me..

It is not mine solution  ;) It was published 14 years ago in Kvant.  :P  :P  :P

oh ic... the solution is definitely incredible and amazing to me.. :what?:

In fact, the stronger inequality holds: $\prod_{i=1}^{n}{\frac{x_i+x_{i+1}+x_{i+2}}{x_{i+1}+x_{i+2}}>\sqrt{2^n}}$.  ;)

harazi, nobody solved this one.
What about proof of this stronger equality?

Use the fact that $ (x_i+x_{i+1}+x_{i+2})^2\geq (2x_i+x_{i+1})(2x_{i+2}+x_{i+1})$, make the product of these and then use the fact that $ (2a+b)(2b+a)>2(a+b)^2$. Ok, I thought that at most two or three people could solve that in a contest, but it seems it was even worse. Hard inequality.

A generalized version of this inequality is to find the best constant $ c$ for which the $ RHS\equal{}cn$..So I posted this as a problem here [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1752842#1752842[/url]...

For the interested reader, the proof given by Myth may be found at page 101 of Andreescu-Dospinescu' "Problem from the book"


1.Let M be a subst of {1,2,...,2006} with the following property: For any three elements x,y and z (x<y<z) of M, x+y does not divide z. Determine the largest possible size of M. Justify your claim.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=693038[/url]

2. For a positive integer k, let $f_{1}(k)$ be the square of the sum of the digits of k. (For example $f_{1}(123)=(1+2+3)^{2}=36$.) Let $f_{n+1}(k)=f_{1}(f_{n}(k))$. Determine the value of the $f_{2007}(2^{2006})$. Justify your claim.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=693035[/url]

3. A convex quadrilateral $ABCD$ with $AC \neq BD$ is inscribed in a circle with center $O$. Let $E$ be the intersection of diagonals $AC$ and $BD$. If $P$ is a point inside $ABCD$ such that $\angle PAB+\angle PCB=\angle PBC+\angle PDC=90^\circ$, prove that $O$, $P$ and $E$ are collinear.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=693044[/url]

4. Let a_1, a_2, a_3,... be a sequence of positive numbers. If there exists a positive number M such that for n = 1,2,3,...,

$a^{2}_{1}+a^{2}_{2}+...+a^{2}_{n}< Ma^{2}_{n+1}$

then prove that there exist a positive number M' such that for every n = 1,2,3,...,

$a_{1}+a_{2}+...+a_{n}< M'a_{n+1}$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=693041[/url]

[quote="radio"]What about TSTs 1,2?[/quote]

TST 1 :

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=107256[/url]

No one post the problems of TST 2 since the question paper was collected after the TST.......


geometry
circumcircle
projective geometry
cyclic quadrilateral
geometry proposed

Let $ H_{1},H_{2},H_{3}$ be the feet of the altitudes of triangle $ ABC$ from vertices $ A,B,C.$ Let $ M_{1},M_{2},M_{3}$ be midpoints of its sides $ BC,CA,AB.$ Lines $ H_{i}H_{i \plus{} 1}$ intersect $ M_{i}M_{i \plus{} 1}$ at $ P_{i \plus{} 2}$. Show $ AP_{1}$∥$ BP_{2}$∥$ CP_{3}$.

Thank you dear [b]Fang-jh[/b] for your nice problem.

[b]Solution.[/b] Applying [b]Pascal's theorem[/b] to the cyclic hexagon $ M_2H_3M_1H_2M_3H_1$, we have   $ X \equal{} M_2H_3\cap M_3H_2$, $ Y \equal{} M_3H_1\cap M_1H_3$ and $ Z \equal{} M_1H_2\cap M_2H_1$ are collinear.

Now, denote by $ E_1$, $ E_2$, $ E_3$ be the midpoints of the segments $ AH_1$, $ BH_2$, $ CH_3$, respectively. And $ H$, $ E$ be the orthocenter and the center of the nine-point circle of triangle $ ABC$.  Applying the [b]Pascal's theorem[/b] to the cyclic hexagon $ E_1H_1M_3E_3H_3M_1$, we have $ H \equal{} E_1H_1\cap E_3H_3$, $ Y \equal{} H_1M_3\cap H_3M_1$ and $ E \equal{} M_3E_3\cap M_1E_1$ are collinear. Similarly, $ H$, $ E$ and $ Z$ are collinear. Therefore we have $ X$, $ Y$, $ Z$ and $ E$ are collinear.

Now, let's consider the cyclic quadrilateral $ M_2H_2H_3M_3$, we have $ A$, $ P_1$ are the intersections of its opposite sides, and $ X$ is the intersection of its diagonal. It follows that $ AP_1$ is the polar of $ X$ relative to the nine-point circle of triangle $ ABC$, which implies that $ EX\perp AP_1$. Similarly, $ EY\perp BP_2$ and $ EZ\perp CP_3$. On the other hand, we have $ E$, $ X$, $ Y$ and $ Z$ are collinear, hence $ AP_1\parallel BP_2\parallel CP_3$, as desired.

[b]Remark.[/b] We have six points $ X$, $ Y$, $ Z$, $ H$, $ E$ and the circumcenter of triangle $ ABC$ are collinear.

More generally: http://www.mathlinks.ro/Forum/viewtopic.php?t=16317 .

  darij


How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not
12?

Why can't you just find the # of multiples for 3 and 4, and then just subtract out the multiples of 12?

You have to subtract twice the multiples of $ 12$ to get what you are looking for. Note that when you count the multiples of $ 3$ and the multiples of $ 4$, you count the mutliples of $ 12$ twice. You don't want the multiples of $ 12$ so you subtract $ 2$ times the number of multiples of $ 12$.

If you subtract the multiples of $ 12$ once, then you will get all the multiples of $ 3$ or $ 4$ including the multiples of $ 12$.

To make sense of how this works, consider a smaller set of numbers, such as the numbers from $ 1 \minus{} 24$.

A good way to visualize these things is to draw a Venn diagram.  One circle is the multiples of $ 3$ and one is the multiples of $ 4$.  You want to count the parts of the Venn diagram that don't include the overlap in the middle, but if you just add the areas of the circles you get the part in the middle twice.


This is a very annoying problem. If anybody could help me, I would be extremely grateful.

Whenever I log onto any website I will randomly get logged off a few minutes later while browsing the site as normal. This happens on at least 4 websites, AoPS included. In some cases, I will become logged off as soon as I go to any other page from the "You have successfully logged in" page. I can't find any browser setting that would explain this problem -- can anybody help? I am completely unable to use certain websites because of this.

Edit >> Preferences >> Privacy >> Cookies: "Accept cookies from sites" should be selected. Also, if you wish to use the "keep me logged in" feature, either make sure "Always clear private data when I close Firefox" is not selected, or remove cookies from the list of data cleared each time (Private Data >> Settings).

All of these setting are already set as you ask.

I'm experiencing this same problem on my old computer. Firefox and XP too.

However, this problem only occurs when I hit the "Forum" button (near the Math Jams link) - in which it'll redirect me to the log in screen. If I simply hit "Forum Index" it won't log me out.

Ok well I always hit Forum Index. Anyway, the problem with AoPS is mainly that it will log me out when i close my browser (though it shouldn't?). For some sites, thought, it is constantly logging out.

Unlcess you hit remember it'll log you out on exit

I tried clearing cookies. I'm not sure yet if it fixed the problem or not.

There still seems to be a problem. There is one site which still fails on my computer, both with Firefox and with IE, but it works fine on my other computer. I think the problem is that my cookies are failing even though they are set on.

EDIT- problem fixed. stupid new security software...


How  to  draw  the  following   picture? ( the  picture  is  not  mine)
Thanks!

The incircle is easy (and it is one of the custom tools on AoPS): Just create 3 points, draw 3 lines joining them, choose the incircle tool, click on the 3 vertices of the triangle and then obtain all other points as intersections of a couple of lines (tangency points can be obtained as intersections of the circle and the side segments). To change the line style, click on the line and then choose "Properties", etc.: [geogebra]a33060c849331b932b152f0d5d4adbd79cd5b335[/geogebra]
Now, as to the inscribed ellipse, you should tell which one you are looking for (they are infinitely many and the picture you posted gives no hint whatsoever as to how the ellipse should depend on the triangle). :?


group theory
abstract algebra
superior algebra
superior algebra solved

G is a group with 120 elements and H is a subgroup of G with 24 elements. Show that if one of the congruence classes to the left (with respect to H), different from H, is equal to one of the congruence classes to the right, then H is normal in G (in other words: if there is a in G but not in H and b in G s.t. aH=Hb then blabla..).

A standard way to show that a subgroup is normal is to consider its normalizer.  Lets look at how this works in this case.  I read the problem as saying there exists an a \not  \in H s.t. aH = Ha, but lets consider Grobber's version of aH = Hb.  Then aHa^{-1} = H(ba^{-1}).  The group on the left contains the identity, e, so we must have  ba^{-1} \in H.
Therefore aHa^{-1} = a and hence a \in N_G(H).  Also we always have that H \in N_G(H)  \leq  G.  So |N_G(H)| \geq 25.  But by Lagrange's Theorem we must have |N_G(H)| dividing 120, so |N_G(H)| = 120
which implies H \normal \in G.

I think there's a step missing in that proof. Yes, it's a triviality, but I think I should add it for the sake of completeness:

[quote="nikheel"]But by Lagrange's Theorem we must have |N_G(H)| dividing 120[/quote]

Then what prevents $|N_G(H)|$ from being something like... let's say, 60? The answer is: again by Lagrange's theorem, since $H\leq N_G(H)$, then 24 divides $|N_G(H)|$. But 120/24 = 5, which is a prime. Therefore, the only number between 24 and 120 (in divisibility, not in magnitude), which is greater than 24, is 120. [i]Ergo[/i], $|N_G(H)|=120$, q.e.d.


calculus AMC AIME Princeton college algorithm

um...does anyone know how exactly who gets the AP Scholar Awards is determined?  I am particularly interested in the National Scholar one.  Generally, you need 8 grades that are 4 or above to get this.  Let's say last year I got a 4 on BC.  This year I retook it and got a 5.  Does that count as one or two exams - I'm guessing it's one, but does anyone know for sure?

It's one

You need to acheive a grade of 4 or higher on 8 subjects....

I don't know if AB/BC counts as one....

are you sure though, or are you just guessing?  I mean, someone told me it was two.  Where do they say that it must be eight different subjects...maybe that's what everyone just assumes...though that's probably my wishful thinking.

It wouldn't make sense if you could retake... then theoretically you could take 2 AP tests each year you are in HS and get the award? I think not.

I am not sure, but I think that there is also: 'AP Scholar with Distinction' for those that have taken 12 APs.

Edit: Never mind; it sounded ridiculous anyway

[quote="MithsApprentice"]Why can't you retake? I thought you could retake but they just erase your old score or at least just have it not count towards the awards.[/quote]
I'm pretty sure he means for it to count towards the award.  Of course you can retake though, it says so somewhere on one of the 21423 sheets of information they give you.

[quote="MithsApprentice"]Why can't you retake? I thought you could retake but they just erase your old score or at least just have it not count towards the awards.[/quote]

Yeah, sorry, I was being unclear. I meant retake for it to count as an additional test for the award.

so has any AoPSer gotten this AP award??? ( I have no chacnce, becuase my school offers IB, so I will probably only take 2 or 3 more APs.)

i got AP Scholar last year for 3 APs- can't get any higher with school's many requirements :(

[quote="probability1.01"]Undoubtedly.  Many of us are taking APs as early as 7th or 8th grade.[/quote]

Many? Are you sure... are there really that many middle schools that allow students to take APs? I know of one eighth grader who is taking Calculus AB right now, but that's all... and my district pushes AP really hard. I'm not sure if that many people do (though, from what you say, you probably are).

I guess I'll be trying to get the National AP Scholar Award; shouldn't be too difficult, if I do well enough I can probably have it by the end of next year - they should have more, for like 10 or 12 tests... that would be quite something to get.

One of my friends who is in 8th grade took the BC Calc exam. Then again, she got around 144 on her AMC 10B and 8 or 9 on the AIME (not sure).

[quote="aznphatso"]They see your composite after taking into account how many 5s you got.[/quote]

In that case, just get as many 5's as you can!!!  :lol:

I'm not sure where to find the exact scoring algorithm... If anybody knows where it is, I'd greatly appreciate a link.

i wonder if i have a chance at getting state ap scholar. i live in south carolina so the level shouldn't be that high. i will have 7 aps at the end of my senior year. (i would assume most will be 5's). but there's always this one kid who will take 10 ap tests by self-studying in his sophmore year.  :wallbash_red:

Can anyone clarify that we must take all the APs by the end of junior year? If it is true than I am pretty much screwed. I am only taking one AP as a sophomore this year. Stupid school...

[quote="MithsApprentice"]DEFINITELY not all the APs ever. If you take 5 by junior year, it's a pretty big deal. 8 is a huge deal. More than that is extremely difficult. It may not seem that way on this forum but that's because we don't exactly have a random sample of high school kids.[/quote]

Yeah... I'll have 9 by the end of junior year and 14 or 15 by senior year... I'm going to get that AP Award :D

I may have confused things by throwing the Siemen's award into a thread about the College Board AP Scholar awards. The rules are different.

For Siemen's, only tests taken thru the end of junior year count, because the award is given the following school year, and you have to still be in high school to receive it. For the College Board awards, the senior year AP scores do count, although by the time you get the award you are already in college and don't care.

For Siemen's, female regional winners may have as few as 4 exams, and males may have as few as 5. But it only takes 2 people the same gender in your region who have really gone after APs to move the mark up. Last year, for example both female winners from the region including Texas had all 7, but that was unusual. Male (and probably female) national winners always have all 7. This year there were at least 3 males with all 7 exams in the running for the Nat'l award, and that's just kids I know of.

Also for Siemens - if student A has 5s on five exams with perfect scores, (gets every possible point), and student B has 5s on six exams but barely squeaked over the cut-off (missed a third of the points), student B wins. And they do not distinguish among the 7 exams - Env Sci counts just as much as Calc BC.

[quote="paladin8"]
Yeah... I'll have 9 by the end of junior year and 14 or 15 by senior year... I'm going to get that AP Award :D[/quote]

That's crazy...
for the Siemen's thing, I don't think the environmental science should count...many schools don't offer it and it's not something most people think about self-studying; same goes for psych.

[quote="polymorphic"]
for the Siemen's thing, I don't think the environmental science should count...many schools don't offer it and it's not something most people think about self-studying[/quote]

I don't think Env. Sci. should count because it is a stupid exam. But, hey, it's not my award.  :roll: People who are going after the Siemen's award figure out what exams they need specifically for the award, and self study if there is no course at their school. (that's why I'm telling you guys about it) If you just cruise along taking the APs you think you need for college, it's unlikely anyone would get all the Siemens ones by the end of junior year.

a few of my friends asked this genius who is in middle college (having a college schedule in junior/senior year) and he told them to take enviornmental science AP because it was easy. that class isn't offered at my school, but a lot of people juz take AP tests by buying princeton review and barron's prep books and using them as texts. i did that myself with world history, and i died, but that was mostly because i spent only one month studying, and i never took a practice test. i guess i was just trying to see what was required to break the college board. competetive public schools like mine believe that the college board is naturally evil

but if youre in a math forum like here, you should be at the very least aiming for AP calc (preferably BC) in your high school career. you should also be aiming for AP physics and AP stats. AP bio is kind of optional, but its such an easy course (compared to the others) that you should take it anyway

[quote="Hessm"]yah.... i nearly killed myself this year, lol

sophomore year:

comp sci A - 4 (grrrrrr... very angry)
statistics - 5

junior year: (year i just had)

calc bc-
physics b-
government-
macroecon-
chem-
comp sci AB-

(still waiting on these scores... i think i got 5's in all EXCEPT for comp sci AB... hoping for a 3 or 4)

so that makes 8 so far

and my senior year, i've signed up for these AP's so far:

-euro history
-us history
-physics C: E&M
-physics C: Kinematics
-Bio

so that makes 13 total (14 if you count the sub-score of calc bc).... assuming i pass comp-sci and next years

my school offers almost every AP... i believe in the high 20's...? not exactly sure... we take them as classes, and then take the AP in may

NOTE: did anyone take the compsci ab exam this year? i crammed the whole course in one night  :(  ... oh man was i tired...[/quote]

Just wondering, which school?

[quote="yif man12"]

Just wondering, which school?[/quote]

Orange High School, in Pepper Pike, Ohio

my friends though are beasts... they will (at least 3 of them) have 20+ credits

bleh... so much for "state scholar", lol

Hmm.... I was wondering if this is ok:...

I took Computer Science AB test this year, and got a 3 (because the teacher didn't taught us the ab materials and was absent 2 days prior to the test, so we got no review), and I'm not quite satisfied with the score.

So I think I will take the Computer Science A (not AB) test next year. I am pretty confident I will get a 5, because they're soo easy. Will that erase my AB score? If you were me would you do it? Is it even possible?

I am not sure if it will erase your score, but I would study again for AB. The A test doesn't have much value; it pretty much only tests syntax stuff. I doubt that colleges will be interested in someone who only knows how to write Java syntax and not know any algorithms or data structures. It might be tough and time consuming, though, so keep that in mind (definitely time consuming, actually).

After searching the College Board Web site for a while, I couldn't find an official statement about taking the Computer Science A after taking the AB test, but I agree with Paladin's advice: just do over the test you took the first time, and thoroughly learn the material. That seems more strategic, and more educationally sound, than going "backward."


Math-M-Addicts Season 3 is starting in mid-October; to apply, please download the application, fill it out, and Email it to breydo@gmail.com.

To obtain more information about the program, please go to [url]http://www.pushkinacademy.com/en/pamm/[/url], although the website is still being modified from last year.

Ruvim

Seems that people got their act together a bit earlier this year -- and so we are ready to start one week earlier (October 13th rather than October 20th) as well.  The group of "6th - 9th graders without significant math team experience" is formed; the group of "9th - 12th graders with math team experience" is still being formed.  For those interested in either, it is still not too late to apply -- please Email breydo@gmail.com for details.

Ruvim

I thought I would give a bump to this thread about the Math-M-Addicts program on Saturday mornings from 10 am to noon on East 81st Street in Manhattan.  My daughter and her friends have been participating in this program, and have enjoyed it a lot.

The second semester is about to start, so now would be a good time to join.  Feel free to contact Ruvim at the address above to apply.


function
topology
calculus
calculus computations

[quote="JRav"]Suppose that we are given a point $ x$ and a sequence $ (x_{n})$ in a metric space $ M$ and suppose that $ f(x_{n})\to f(x)$ for every continuous real-valued function $ f$ on $ M$.  Prove that $ x_{n}\to x$ in $ M$.[/quote]
What about to take $ f$ as
$ f(y)\equal{}d(y,x),\forall y\in M$
where $ d$ is the metric.

Since it holds for every continuous function and we know that the metric function is continuous, $ f(x_{n})\equal{}d(x_{n},x)\to f(x)\equal{}d(x,x)\equal{}0$, we have that $ d(x_{n},x)\to 0$ so $ x_{n}\to x$, right?

The only problem is, if we can do it with that function, why not the identity map?

I see no problem with the identity map except that it is not real-valued in abstract $ M$  :)


$ 1\blacktriangleright$ Let $ ABC$ be a triangle. For a point  $ M\in (BC)$ denote the points

$ \{\begin{array}{c} F\in (AB)\ ,\ \widehat {FMA}\equiv\widehat {FMB} \\
 \\
E\in (AC)\ ,\ \widehat {EMA}\equiv\widehat {EMC}\end{array}$ . Show that the lines $ BE$ , $ CF$ , $ AM$ are concurrently.

$ 2\blacktriangleright$ Let $ ABC$ be a triangle. Denote the midpoint $ M$ of the side $ [BC]$ . For two mobile points $ X$ , $ Y$

so that $ A\in XY$ and $ XY\parallel BC$ consider the intersections $ E\in AB\cap XM$ and $ F\in AC\cap YM$ .

Ascertain the geometrical locus of the intersection $ L\in XF\cap YE$ .

$ 3\blacktriangleright$ Let $ ABC$ be an acute triangle. For a point $ M\in (BC)$ denote $ \{\begin{array}{c}
x=m(\widehat {MAB})\\\
y=m(\widehat {MAC})\end{array}$

and consider the points $ \{\begin{array}{c} F\in AB\ ,\ MF\perp AB \\
 \\
E\in AC\ ,\ ME\perp AC\end{array}$ . Prove that

the lines $ BE$ , $ CF$ , $ AM$ are concurrently $ \Longleftrightarrow\ \tan x-\tan y=\tan B-\tan C$  .

[hide]Since ray $ MF$ bisects angle $ \angle{AMB}$, it is known that $ \frac{BF}{FA}\equal{}\frac{BM}{MA}$ by the Angle Bisector Theorem.  Similarly, $ \frac{AE}{EC}\equal{}\frac{AM}{MC}$ because ray $ ME$ bisects angle $ \angle{AMC}$.

Therefore $ \frac{BF}{FA}*\frac{AE}{EC}*\frac{CM}{MB}\equal{}\frac{BM}{AM}*\frac{MA}{MC}*\frac{CM}{MB}\equal{}1$.
By Ceva's Theorem, the three cevians are concurrent.[/hide]

I added two problems more in my upper message !


trigonometry
algebra
polynomial
inequalities
linear algebra
matrix

Is there a more trivial way using basic properties to solve the determinant:

$ \left|\begin{array}{ccc}\cos^2a & 2\sin^3a & 1 \\
\cos^2b & 2\sin^3b & 1 \\
\cos^2c & 2\sin^3c & 1\end{array}\right|$

other than expanding? The [hide="answer"]0[/hide], if it helps.

That answer can't be right, unless there's some relation between the angles you haven't mentioned.

There is indeed a typo, but it shouldn't affect the answer. There is no relation between the angles in the problem :(

Fixed the typo anyways

That changes nothing. It's still wrong.

In fact, this is quite similar to a Vandermonde determinant: the value is $ \minus{} 2(x \minus{} y)(y \minus{} z)(z \minus{} x)(xy \plus{} yz \plus{} zx)$, where $ x \equal{} \sin a,y \equal{} \sin b,z \equal{} \sin c.$

Added in response to t0rajir0u: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=241442]Reference[/url]

One can partially solve this using the same "factoring" argument as one method for computing the Vandermonde determinant.  If any of $ a, b, c$ are equal to each other, the determinant evaluates to zero.  Furthermore, the determinant must be a polynomial in the variables jmerry wrote down; that tells you the first three factors jmerry wrote down.  The remaining factor is probably best computed straight.

There should be an interesting relationship between determinants of this type and the determinants used to define [url=http://en.wikipedia.org/wiki/Schur_polynomial]Schur polynomials[/url].

$ \left| \begin{matrix}
   \cos ^{2}a & 2\sin ^{3}a & 1  \\
   \cos ^{2}b & 2\sin ^{3}b & 1  \\
   \cos ^{2}c & 2\sin ^{3}c & 1
\end{matrix} \right|\equal{}\left| \begin{matrix}
   \cos ^{2}a & 2\sin ^{3}a & 1  \\
   \cos ^{2}b\minus{}\cos ^{2}a & 2\sin ^{3}b\minus{}2\sin ^{3}a & 0  \\
   \cos ^{2}c\minus{}\cos ^{2}b & 2\sin ^{3}c\minus{}2\sin ^{3}b & 0
\end{matrix} \right|.$

Thus we need to compute $ \left| \begin{matrix}
   \cos ^{2}b\minus{}\cos ^{2}a & 2\sin ^{3}b\minus{}2\sin ^{3}a  \\
   \cos ^{2}c\minus{}\cos ^{2}b & 2\sin ^{3}c\minus{}2\sin ^{3}b
\end{matrix} \right|.$


[b]PRO[/b]: find all $f(x):\mathbb {N^*}\to{\mathbb{N^*}}$ such that:
$f(f(m)+f(n)+f(p))=m+n+p$ for all possitive integer $m,n,p$

Then $f$ is clearly injective.
Since $f(f(m)+f(n-1)+f(p+1)) = m+n+p = f(f(m)+f(n)+f(p))$, it follows that $f(n-1)+f(p+1) = f(n)+f(p)$, so that the function $g(a) = f(a+1)-f(a)$ is constant.
Thus $f(n) = an+b$ for some constant $a,b$. Back to the initial equation, we easily see that $a=1,b=0$ so that $f(n)=n$.

Pierre.

okie--thanks a lot----[b]pbornsztein[/b], your solution very nice
now, solve new pro:
[b]PRO[/b]: find all $f:\mathbb{N^*}\to{\mathbb{N^*}}$ such that: $f((f(m))^2+(f(n))^2)=m^2+n^2$ for every possitive integer $m,n$

Pretty! At first I was going to ask why injectiveness is obvious, but I found this: $f(x) = f(y) = z$, then $f(3z) = 3x = 3y$, so $x = y$. So similarly the new function is injective?


Hi

Given that vectors $ x$ and $ y$ satisfy

$ x\plus{}y(x\cdot y)\equal{}a$

for a fixed vector $ a$, show that

$ (x\cdot y)^2\equal{}\frac{|a|^2\minus{}|x|^2}{2\plus{}|y|^2}$

Deduce using the Schwarz inequality (or otherwise) that

$ |x|(1\plus{}|y|^2)\geq |a| \geq |x|$

:)

1. power the data equality to the square.
2. Multiply the data equality by y.[/code]

hi
i'm not sure i follow what you're saying. i don't think i can square the equation because it is a vector equation.
thanks

I think he means that you use $ \| x\plus{}\langle x,y\rangle y \|^2 \equal{} \| a \|^2$.


search
function
videos
SFFT
geometry
rectangle
special factorizations

I'm not sure what forum to put this, so moderators, please move this to the appropriate forum.

I was wondering - what's Simon's Favorite Factoring Trick? I know it was named after an AoPS User, ComplexZeta, who's real name is Simon, but after launching a search on Wikipedia, Google, and the Search Function on the Forum, I still couldn't figure out exactly what it is. Can someone please explain it to me?

Thanks in advance!  :)

Are you sure you checked google?  It got me to page I wanted after 1 search.

http://www.artofproblemsolving.com/Wiki/index.php/Simon's_Favorite_Factoring_Trick

Watch the 2 videos at the bottom to see two examples worked out.

Here is an example of SFFT:
Find all positive integer solutions to $ xy \minus{} 3x \minus{} 5y \equal{} 1$
The key in this case is to add 15 so both sides factor
$ xy \minus{} 3x \minus{} 5y \plus{} 15 \equal{} 16$
$ (x \minus{} 5)(y \minus{} 3) \equal{} 16$
So now we can just sift through every possible set of factors of 16
16,1 $ \implies$ $ x \equal{} 21, y \equal{} 4$ or $ x \equal{} 6, y \equal{} 19$
8,2 $ \implies$ $ x \equal{} 13, y \equal{} 5$ or $ x \equal{} 7, y \equal{} 11$
4,4 $ \implies$ $ x \equal{} 9, y \equal{} 7$
If you check negative factors, you find there are no solutions in that case.
So SFFT is just adding the number to factor it, then possibly sifting through the factors to find integer values.

I was teaching and using this idea before I'd ever heard of Simon; my name for it was always "completing the rectangle" as an analogue to completing the square.

Here's another problem, a little one: find all rectangles with integer sides which have area numerically equal to perimeter.

Okay.
$ x$ and $ y$ are the sides
$ xy \equal{} 2x \plus{} 2y$
$ xy \minus{} 2x \minus{} 2y \equal{} 0$
$ xy \minus{} 2x \minus{} 2y \plus{} 4 \equal{} 4$
$ (x \minus{} 2)(y \minus{} 2) \equal{} 4$

$ (3,6) (6,3) (4,4)$


Find all primes $p$ such that $\frac{2^{p-1}-1}{p}$ is a square.

Obviously $p=2$ doesn't work, so assume $p>2$, then $p=2k+1$. We have $2^{2k}-1=n^2p=(2^k+1)(2^k-1)$. Then one of the factors is a perfect square, and the other $p$ times a perfect square, because they're coprime. If $2^k-1$ is a perfect square we have $k=1$, as for bigger $k$ it's a contradiction mod 4. Also $k=1, p=3$ works. If $2^k+1$ is a perfect square, $2^k+1=a^2 \Rightarrow 2^k=(a+1)(a-1)$, from which $a-1=2, a+1=4, 2^k=8, k=3, p=7$ as $a-1, a+1$ are both powers of $2$. $p=7$ is the only other solution.


algebra
polynomial
binomial theorem
algebra unsolved

Prove that:$ \sum_{k\equal{}0}^{n} \binom{n}{k}(k\plus{}1)^{k\minus{}1}(n\minus{}k\minus{}1)^{n\minus{}k}\equal{}n^n$

[quote="ywl"]I am not absolutely sure (I have not checked)
but I think it is a consequence of Abel's Binomial theorem

[url]http://en.wikipedia.org/wiki/Abel%27s_binomial_theorem[/url][/quote]

Thank you very much!Can you help me to find a proof for it?
My classmates used two different methods to slove an interesting combinatorics problem,one directly get $ n^n$ the other get $ LHS$
but none of us can prove the indentity :D
Isn't it interesting?

see it as a identity about w
and then it is easy

Nice :lol:
substitute $ k$ with $ n \minus{} k$ and let $ w \equal{} 1,z \equal{} \minus{} 1$ which immediately get our desired identity :)

P.S.Can you post a full proof for [b]Abel's Binomial theorem[/b] ?

In fact,we can do derivation to w(=-z-m) i times(i=0,1,2,...,m-1),then we wil find that the form is a difference of a polynomial,and it equals to 0.So it equals to (z+w+m)^m.
Sorry for that my English is poor so that I can't post the full proof.[/url]


I was wondering if anyone here had any recommendations for a good pair of binoculars for astronomy (around or less than 400 dollars). I know there's lots of places online that sell them, but I was wondering what works for people from personal experiences.

Thanks!

I am not that familiar with specific brands and such.  However, you should get a pair with coated lens to ensure for a clearer, crisper image.  The other specifications depends on what you plan to see with them.  :wink:

$ \$ 400$ touches the "premium" binocular market (I got mine for $ \$ 80$). So if you are sure you want to stick to binoculars, you have plenty of choices.

You can either go for 'large' ($ 20\times 100$ and above) sized pairs from main-line brands (Orion, Meade, etc.) or smaller ones ($ 10\times 50$ etc.) from 'elite' brands (Takahashi, Russian manufacturers etc.) The large pairs are essentially two refractors put together, and require all sorts of mounts etc. I haven't personally used the 'elite' ones, but it is said that they are well worth the cost (typical specifications include flourite lenses, extra-sturdy design etc.). Image-stabilising binoculars (from Canon) also fall in this range, but I don't know much about them.

Definitely look for "fully coated/multicoated" lenses rather than just "coated".
Sorry for the long post, hope it was helpful :)

I've never gone shopping for binoculars, so I don't know which type you should buy, but I've been very happy with my 7x50's over the years.  Are you familiar with the book [u]Touring the Universe Through Binoculars[/u]?  It contains a good discussion of the relative merits of different types of binoculars for astronomy, I think.

I would recommend that you put some of your binocular money towards a good tripod to mount the binoculars.  Even relatively small binoculars, like my 7x50's, get tiring to hold over the course of an evening.  If you want to look at one object for a long period of time, the tripod keeps the image steady.  A tripod also allows you to show the view to other people.

Thanks so much for your input! :D

Any other suggestions?

More about the tripod.  You might consider a parallelogram-style mount for the binoculars, instead of a traditional tripod mount.

Here's another idea.  The book [u]Unusual Telescopes[/u] features a pair of 17-inch "binoculars," really two 17-inch Newtonian telescopes mounted side by side.  Maybe you could do something along those lines?    :P

Just kidding.  Do let me know what you decide and how it works-- you make me want to go out and get a nice pair of binoculars!

(http://www.amazon.com/Unusual-Telescopes-Peter-L-Manly/dp/052148393X)


modular arithmetic
number theory proposed
number theory

Show that the product of any set of $ n$ consecutive integers is divisible by $ n$.

We can do much better:  the product of $ n$ consecutive integers is divisible by $ n!$.

[hide="Proof"] $ {k \plus{} n \choose k}$ is an integer.  :) [/hide]
[hide="The argument originally intended"] $ n$ consecutive integers leave the remainders $ 0, 1, 2, 3, ... n \minus{} 1 \bmod n$, so their product is congruent to $ 0 \bmod n$.  In fact, if $ n$ isn't prime then their product is congruent to $ 0 \bmod n^2$.

But look here:  $ n$ consecutive integers include $ n \minus{} 1$ consecutive integers which leave remainders $ 0, 1, 2, ... n \minus{} 2 \bmod n \minus{} 1$, so we can repeat this argument for any integer from $ 1$ to $ n$.  And then the fun part is to show that we can combine these results (which isn't obvious). [/hide]

[quote="t0rajir0u"][hide="Proof"] $ {k \plus{} n \choose k}$ is an integer.  :) [/hide][/quote]
I was in the midst of replying with a question as I was unclear on how this works, but it popped into my head as I was typing. I really like that answer; it's quite elegant.

My solution is a little less exciting.[hide="Solution:"]Lemma 1: Any set of $ n$ consecutive integers forms a complete set of residues modulo $ n$.
[i]Proof:[/i] Let $ S$ be a set of $ n$ consecutive integers. Then $ S \equal{} \{c,c\plus{}1,\ldots,c\plus{}(n\minus{}1)\} \equal{} \{c \plus{} i \mid i \in [0,n) \subset \mathbb{N}_0\}$ for some $ c \in \mathbb{Z}$. Assume, contrary to the hypothesis, that $ S$ does not form a complete set of residues modulo $ n$. Then there exist distinct positive integers $ a$ and $ b$ with (without loss of generality) $ a < b < n$ such that $ c \plus{} a \equiv c \plus{} b \pmod{n}$. It follows that $ a \equiv b \pmod{n}$, which is a contradiction since $ 0 < b\minus{}a < n$. QED.
Let $ S$ be a set of $ n$ consecutive integers. Then $ S \equal{} \{c \plus{} i \mid i \in [0,n) \subset \mathbb{N}_0\}$ for some $ c \in \mathbb{Z}$. Let $ N$ be the product of the elements of $ S$; then $ N \equal{} \prod_{i\equal{}0}^{n\minus{}1} (c \plus{} i)$. By Lemma 1, there exists $ k \in [0,n) \subset \mathbb{N}_0$ such that $ c \plus{} k \equiv 0 \pmod{n}$, which means that $ N \equiv 0 \pmod{n}$ since $ c \plus{} k \in S$. QED.[/hide]


geometry
geometric transformation
pigeonhole principle

Given 50 segments on a line show that you can always find either 8 segments which are disjoint or 8 segments with a common point.

:D  Pigeon-hole principle strikes again....

Pierre.

My solution doesn't use Pigeon-hole principle.
Is Pigeon-hole principle the same as Dirichlet's principle in Russia?

I do not know about Russia, but in France it is.

Pierre.

the pigeon-hole principle is for sure the "principe des tiroirs" for th french, isn'it ?

arthur

yes, that's three different names for the same principle.

Pierre.

it's called "box principle" in romania :) (direct translation), and sometimes "Dirichlet Principle" ...

Well here it's simply "Dirchlet's PigeonHole Principle" :)

In Chinese, it is called 'drawer principle' (direct translation)...  :) (exactly same to French)
of course, pigeon-hole principle is known too.
I've never seen 'Dirichlet's Princple'....


calculus
derivative
function
real analysis
real analysis unsolved

We know geometric mean of the first and second derivative of a function. Are there any geometric explanation about  3. and 4. derivative of a function?

One can say: yes, the third derivative is related to the rate of change of curvature, etc. But I prefer the answer: no, our eyes cannot see the third derivative. Use your favorite program to accurately plot the graphs of two functions: $f(x)=\begin{cases}x^{3},\quad x>0; \\ 0, \quad x\le 0\end{cases}$ and $g(x)=\begin{cases}x^{4},\quad x>0; \\ 0, \quad x\le 0\end{cases}$. Only one of two functions has the third derivative at $0$, but there is no way to tell which one just by looking at the graphs.


vector
geometry
geometric transformation
parallelogram
trapezoid

Let $\Gamma$ and $\Gamma'$ be two congruent circles centered at $O$ and $O'$, respectively, and let $A$ be one of their two points of intersection. $B$ is a point on $\Gamma$, $C$ is the second point of intersection of $AB$ and $\Gamma'$, and $D$ is a point on $\Gamma'$ such that $OBDO'$ is a parallelogram. Show that the length of $CD$ does not depend on the position of $B$.

Let $K$ be second intersection point of $BD$ with Г'. then $\cup AD= \cup AK$  so $\measuredangle{ABD}=\measuredangle{ACD}$ and $CD=BD=OO'$.

Sorry, but I don't see how exactly it was concluded that $\angle DBA = \angle DCA$.. Could someone explain this?

This is clearly a problem on translation... Let $OO' = d$.

Translate the circle $\Gamma'$ through the vector $O'O$. This is just moving the circle $\Gamma'$ in the direction of $O'O$ and a distance $d$ such that it coincides with circle $\Gamma$. Suppose the translation sends $C$ to $C'$ and $A$ to $A'$. Clearly, because of the way it was constructed, $D$ is sent to $B$.

Clearly, we have $DC= BC'$. Also $AA' = CC' = d$.

Since $ACC'A'$ is a paralelogram, we have $A'C' \parallel AC$. Hence $A'C' \parallel AB$. Therefore, $BAC'A'$ is an isosceles trapezium, which means $BC' = AA'$.

In conclusion, $CD = BC' = AA' = d$, which is a constant.

Let $E$ the second point of intersection of $\Gamma$ and $\Gamma'$. If we consider the homothecy that sends $\Gamma$ to $\Gamma'$, it is clear that sends $B$ to $D$ and $A$ to $E$, now notice that this implies that sends $BA \cap \Gamma' = C$ to a point in $\Gamma$ that sattisfies $DE \cap \Gamma = F$, then it's clear that $BA\parallel DE$ hence $AC \parallel DE$ which directly implies that $CD=AE$ which is clearly constant

Define: $OO' \cap \Gamma '=H $, Let $BO \cap \Gamma =G$ and Let $BD \cap \Gamma =E$, then we have,
$$\angle EBD=\angle ABP-\angle EBO=\frac{1}{2} \angle AOG-\angle EBO=\frac{1}{2} \angle AOO'-\frac{1}{2} \angle EBO=\frac{1}{2} \angle AO'O-\angle DCH=\angle ACH-\angle DCH=\angle ACD$$
Hence, $CD=BD=OO' \longrightarrow$ which is obviously fixed no matter where $B$ lies!


Let $M,N,P$ is on the side $BC,CA,AB$ of triangle $ABC$ such that:
    $r(NAP) = r(PBM) = r(MCN)$ = $\frac{1}{2}$ $r(ABC)$

Prove that $M,N,P$ is the midpoint of $BC,CA,AB$

Denote $r(XYZ)$ is the radius of incircle of triangle $XYZ$

No one interested in , please help me this .


trigonometry
inequalities proposed
inequalities

$\sqrt{\sin{x}}+\sqrt{\cos{x}}>\sin^{2}{x}+\cos^{2}{x}=1$, and we are done  :D


let P be an interior point of triangle ABC,and $r_{1},r_{2},r_{3}$be the inradii of triangle BPC,CPA,APB,r be the inradii of triangle ABC,prove
$\frac{r}{r_{1}}=1+\frac{1}{sin A}$,respectively

$r_{1}$- constant $\implies$ absurdity !


Find all functions f:R-R satisfying f(x)+f(y) diferent from 0 and
(f(x)-f(x-y))/(f(x)+f(x+y))+(f(x)-f(x+y))/(f(x)+f(x-y))=0 for all x,y real numbers

The only functions are f(x)=c (c<>0). This is how we prove it:

If we calculate the common denominator for the expression which is equal to 0 we get 2*(f(x))^2=(f(x+y))^2+(f(x-y))^2 for any real x and y. Let g(x)=(f(x))^2. Then g(a)+g(b)=2*g((a+b)/2) and g(x)>=0 for any real x. It?s easy to see that g must be constant to satisfy both these conditions, so g(x)=c^2 for any x. We found that f(x)=either c or ?c. If there exist a, b s.t. f(a)=c, f(b)=-c then f(a)+f(b)=0, which is false, so f is constant. Because 2*f(x)<>0 we get that f(x)=c<>0. If we check we find that all the functions of this form are good solutions. It?s a cute one. Did you compose it?

No I didn't compose this problem. It was given at the BMC contest. If you want I'll post more functional equations cause you said you like them.

Have a nice day! :)

This problem was proposed by Gabriel Caroll, a perfect scorer at IMO2001

I knew he was a perfect scorer but I didn't know he proposed this problem. Thanks for the info.

Cheers :)

hey grobber,

how do you prove that g(x)=(f(x))^2 must be constant. It's easy
to prove it if g is continue but how do you do without this condition ??
thanxx


I didn't search through forum but has anyone in this site did some sort of mock Mathcounts? Like mock AMC in AMC competition forum? I'm over with mathcounts but I might need to get some of the questions to do something so...

If there is, please tell...

And if I get time... I might make one but o well..

Thanks.. I see now..

In case anyone wants to do this, I'll save your time:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16398

Well, I did a mock MathCounts at school for the school round (it's a past states competition) and I got a 45 :D ! A 29 and an 8 :) .


geometry
circumcircle
trigonometry
power of a point
radical axis
geometry unsolved

Let $ X$, $ Y$ be the second intersections of $ AB$ with the circumcircles $ \rho_{1}$, $ \rho_{2}$ of triangles $ ACF$ and $ BCE$. Let also $ Z$ be the intersection of $ CD$ with $ AB$. Since $ Z$ lies on the radical axis of $ \rho_{1}$ and $ \rho_{2}$,
\[ \frac {ZA}{ZB} \equal{} \frac {ZY}{ZX} \equal{} \frac {AY \minus{} ZA}{BY \minus{} ZX} \equal{} \frac {AY}{BX}.
\]
On the other hand, since $ BF \cdot BC \equal{} BX \cdot BA$, $ AY \cdot AB \equal{} AE \cdot AC$ and $ AE \equal{} BF$, it follows that $ AY/BX \equal{} AC/BC$. Thus, $ ZA/ZB \equal{} AC/BC$, which proves that $ CD$ bisects $ \angle{ACB}$.

[quote="orl"]In triangle $ ABC$ points $ E$ and $ F$ lie on sides $ AC$ and $ BC$ such that segments $ AE$ and $ BF$ have equal length, and circles formed by $ A,C,F$ and by $ B,C,E,$ respectively, intersect at point $ C$ and another point $ D.$ Prove that that the line $ CD$ bisects $ \angle ACB.$[/quote]
[hide="Solution"]
Notice that $ BDEC$ is cyclic, so $ \angle AED\equal{}\angle DBF$.  Also, notice that $ DACF$ is cyclic, so $ \angle DFB\equal{}\angle DAE$.  These, combined with the fact that $ AE\equal{}FB$ implies that $ \triangle DAE\cong \triangle DFB$.  Thus, $ DF\equal{}DA$.  Since $ DA\equal{}DF$ and $ DACF$ is cyclic, we see that $ D$ is the midpoint of minor arc $ AF$, so $ DC$ bisects $ \angle ACB$. [/hide]

[quote="orl"] [color=darkred]Consider in $ \triangle ABC$ the points $ E\in (AC)$ , $ F\in (BC)$ so that $ AE \equal{} BF$ .

The circumcircles of $ \triangle ACF$ , $ \triangle BCE$  intersect at $ C$ , $ D$ . Prove that that $ \widehat {DCA}\equiv\widehat {DCB}$ .[/color] [/quote]

[color=darkred][b][u]An equivalent enunciation (an old, well-known and very nice problem !).[/u][/b]

Let $ w_1 \equal{} C(O_1,R_1)$ , $ w_2 \equal{} C(O_2,R_2)$ be two circles which are secant at $ C$ , $ D$ . Let $ d_1$ , $ d_2$ be two lines through $ C$ for which denote

$ \left\|\begin{array}{ccc} B\in d_1\cap w_1 & , & F\in d_1\cap w_2 \\
\ E\in d_2\cap w_1 & , & A\in d_2\cap w_2\end{array}\right\|$ . Prove that $ AE \equal{} FB\ \Longleftrightarrow\ \widehat {DCB}\equiv\widehat {DCA}$ and in this case $ \boxed {\ (*)\ \sin\alpha \equal{} \frac {l}{2d}\ }\le 1$ ,

where $ \left\|\begin{array}{c} \alpha \equal{} m(\widehat {DCA}) \equal{} m(\widehat {DCB}) \\
 \\
l \equal{} AE \equal{} BF\ \ ;\ \ d \equal{} O_1O_2\end{array}\right\|$ . The relation $ (*)$ allows to construct the points $ A$ , $ B$ for which $ AE \equal{} BF$ !

[b]Remark.[/b] Here appears four cases :

$ 1.\blacktriangleright$ the point $ C$ doesn't separate the points $ A$ , $ E$ and doesn't separate the points $ B$ , $ F$ .

$ 2.\blacktriangleright$ the point $ C$ doesn't separate the points $ A$ , $ E$ and separates the points $ B$ , $ F$ .

$ 3.\blacktriangleright$ the point $ C$ separates the points $ A$ , $ E$ and doesn't separate the points $ B$ , $ F$ .
\sin y}=
$ 4.\blacktriangleright$ the point $ C$ separates the points $ A$ , $ E$ and separates the points $ B$ , $ F$ .[/color]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \left\|\begin{array}{c}
m(\widehat {DCB})\equal{}\alpha\\\\
m(\widehat {DCA}0\equal{}\beta\end{array}\right\|$ and $ \left\|\begin{array}{c}
m(\widehat {CBD})\equal{}x\\\\
m(\widehat {CAD})\equal{}y\end{array}\right\|$ . Observe that $ \widehat {BDE}\equiv\widehat {ADF}$ $ \implies$ $ m(\widehat {BDF})\equal{}m(\widehat {ADE})\equiv\phi$ . Show easily that

$ \left\|\begin{array}{c}
DF\equal{}2R_2\sin \alpha\\\
DE\equal{}2R_1\sin \beta\\\
R_1\sin x\equal{}R_2\sin y\equal{}\frac d2\end{array}\right\|$ and $ \left\|\begin{array}{c}
\frac {DE}{\sin y}\equal{}\frac {AE}{\sin\phi}\\\\
\frac {DF}{\sin x}\equal{}\frac {BF}{\sin\phi}\\\\
(\ AE\ \equal{}\ BF\ )\end{array}\right\|$ . Therefore,  $ \frac {\sin x}{\sin y}\equal{}\frac {DF}{DE}\equal{}\frac {R_2}{R_1}\cdot \frac {\sin\alpha}{\sin \beta}\equal{}\frac {\sin x}{\sin y}\cdot \frac {\sin\alpha}{\sin \beta}$ , i.e. $ \sin\alpha \equal{}\sin\beta$ $ \implies$ $ \alpha \equal{}\beta$ .[/color]


If $ 10^{2y} \equal{} 25$, then $ 10^{ \minus{} y}$ equals:

$ \textbf{(A)}\ \minus{} \frac {1}{5} \qquad\textbf{(B)}\ \frac {1}{625} \qquad\textbf{(C)}\ \frac {1}{50} \qquad\textbf{(D)}\ \frac {1}{25} \qquad\textbf{(E)}\ \frac {1}{5}$

[hide="Another possible solution"]We take the square root of both side and get $ 10^y \equal{} 5$. So $ 10^{ \minus{} y} \equal{} \frac {1}{5}$.  So answer is E.[/hide]

edit: added latex.


LaTeX
inequalities proposed
inequalities

[quote="bodom"]0=a0<a1<a2<...<an  and 1>=a1-a0, 1>=a2-a1 ,...., 1>=an-a(n-1)
show that (a1+a2+...+an)^2>=a1^3+a2^3+...+an^3
i'm sorry if this was posted before.if so tell me and i will delete this post.i would enjoy somebody posting a solution because mine is kind  of ugly.[/quote]

Could you use LaTeX please?!

He meant:

Given that
$0=a_0<a_1<a_2<...<a_n$ and
$1\ge a_i-a_{i-1}$ for all $i=1,2,\ldots,n$

Show that $(a_1+a_2+...+a_n)^2\ge a_1^3+a_2^3+...+a_n^3$.

conceive thanks for puting that in LaTeX.i did not do it myself because i do not know how to use it yet. :blush:


inequalities
geometry
trigonometry
function
geometry solved

Do you know an algebrical inequality which is solving with geometry? Or do you know a famous inequality's  geometric solution or interpretation? For example A.M-G.M, cauchy,jensen,hölder...? I prefer more original one.

[quote="Arne"]I know plenty of them. For example, if a, b, c are positive reals, prove that \[\sqrt{a^2 - ab + b^2} + \sqrt{b^2 - bc + c^2} \geq \sqrt{a^2 + ac + c^2}.\][/quote]

I know it - I solved some time ago - it's funny use of cosines law  :D

[quote="Arne"]Hey, that's a cute one Armo, it is just Ptolemy :)[/quote]
I know and I like it very much. :)

[quote="Armo"][quote="Arne"]I know plenty of them. For example, if a, b, c are positive reals, prove that \[\sqrt{a^2 - ab + b^2} + \sqrt{b^2 - bc + c^2} \geq \sqrt{a^2 + ac + c^2}.\][/quote]
This is very well-known and here is another one: $a,b,c$ are positive reals , prove that:
$c\sqrt{a^{2}-ab+b^{2}}+a\sqrt{b^{2}-bc+c^{2}}\geq{b\sqrt{a^{2}+ac+c^{2}}}.$[/quote]

As I was asked, I am writing my solution.

Consider the segments $OA, OB, OC$ such that $OA=a, OB=b, OC=c$ and $\measuredangle{AOB}=\measuredangle{BOC}=60^{\circ}$. Now by cosines theorem we have:$AB=\sqrt{a^{2}-ab+b^{2}}, BC=\sqrt{b^{2}-bc+c^{2}}, AC=\sqrt{a^{2}+ac+c^{2}}$. So our inequality becomes : $AO\cdot{BC}+OC\cdot{AB}\geq{AC\cdot{OB}}$, and this is obviously Ptolemy.

To eren1989: I don't think that AM-GM can be proven by geometry.

[quote="zhaoli"]Another example, which also very famous (used in russia olympiad):

If 0 < x, y, z, <1, then x(1 - y) + y(1 - z) + z(1 - x) < 1.[/quote]

Let $ABC$ is equilateral triangle with sidelength 1. Let $X\in{AB}, Y\in{BC}, Z\in{CA}$  such that $BX=x, CY=z, AZ=y$. Now $S_{AXZ}+S_{BXZ}+S_{CYZ}<S_{ABC}\Rightarrow{x(1 - y) + y(1 - z) + z(1 - x) < 1}$.
Hereby $S$ is the area.

[quote="Armo"] I don't think that AM-GM can be proven by geometry.[/quote]

Not only AM-GM can be proved by geometry, but even HM<GM<AM<QM . If you need a hint: consider a right triangle, take point P on the side opposite to the right angle and take function S(P) which describes distance of P from the other 2 sides of the triangle. There exist such 4 points that S(P) equals HM, GM, AM, QM - you get some obvious inequality and you're done.   :)  Though it only proves HM<GM<AM<QM for 2 numbers. I'll try to find proof for n numbers - I think it won't be so hard, we just have to go to n-dimensional space and generalize the method above  :D If anybody find it quicker than me, please post it on forum because I'm curious as well.  :)

[quote="Megus"]Not only AM-GM can be proved by geometry, but even HM<GM<AM<QM . If you need a hint: consider a right triangle, take point P on the side opposite to the right angle and take function S(P) which describes distance of P from the other 2 sides of the triangle. There exist such 4 points that S(P) equals HM, GM, AM, QM - you get some obvious inequality and you're done.   :)[/quote]

Can you elaborate, Megus?  The AM is obtained from the mid-piont of the hypotenuse, and the HM is obtained from the inscribed square, but I don't see anything nice for GM or QM.

Because it is not said that these points must be nice  :D It is only said that such points exist [ie where S(P) equals GM, QM]. Maybe there are in some way nice but I just didn't check it  :P

Given $AC = a$, $CD = \frac{b-a}{2}$, $DB = \frac{b+a}{2}$, we can calculate the lengths to be $CE = \frac{2ab}{a+b}$, $CF = \sqrt{ab}$, $DF = DG = \frac{a+b}{2}$, $CG = \sqrt{\frac{a^2 + b^2}{2}}$.

I am still thinking that AM-GM for general case can't be proved by geometry . :?


calculus
function
logarithms
geometry
limit
integration
calculus computations

Let f be the function defined by $ f(x)\equal{}\minus{}\ln{x}$ for $ 0<x\leq{1}$ and let R be the region between the graph of f and the x-axis.
 :huh:
a. determine whether region R has finite area. [b]justify your answer.[/b] (please?)
b. determine whether the solid generated by revolving region R about the y-axis has finite volume. [b]justify your answer[/b] (pretty please?)

Thanks.

$ a.)$

The area, if it is finite, is given by $ \lim_{a \to 0^\plus{}} \int_a^1 \minus{}\ln(x)dx$.  If the area is finite, this integral exists!

$ b.)$

Use the shell method to find the volume, if it exists.

The volume, if it is finite, is given by $ \lim_{a \to 0^\plus{}} \int_a^1 \minus{}2\pi x \ln(x) dx$.


I think we should have a search button for blogs, since many blogs have lots of information and it's really annoying to go through pages of blog posts to get to a particular blog entry.

I agree. You could also try searching for it through Google, including the blog title.


Euler
geometry
incenter
projective geometry
geometric transformation
homothety
circumcircle

Let $ABC$ be an acute triangle and let $A_{1}, B_{1}, C_{1}$ be the feet of its altitudes. The incircle of the triangle $A_{1}B_{1}C_{1}$ touches its sides at the points $A_{2}, B_{2}, C_{2}$. Prove that the Euler lines of triangles $ABC$ and $A_{2}B_{2}C_{2}$ coincide.

i) first, with easy angle chasing, we find that $H$ is the incenter of $A_{1}B_{1}C_{1}$, so, is the circuncenter of $A_{2}B_{2}C_{2}$!

now, see that $AA_{2}$, $BB_{2}$ and $CC_{2}$ are concurrent (by Desargues Theorem), because $ABC$ and $A_{2}B_{2}C_{2}$ have parallel sides (we see this in the angle chasing)! Let X the concurrence point.
do one homotety with center X and maping $A_{2}B_{2}C_{2}$ in $ABC$. the circuncenter of $A_{2}B_{2}C_{2}$ will be map in the circuncenter of $ABC$, so, X, H and O are colinear!
the ortocenter $Y$ of $A_{2}B_{2}C_{2}$ will be map in the ortocenter of $ABC$, so, X, H and Y are collinear!

this implies that Y, H and O are collinear, killing the problem!

i think the simplest solution for this problem is the following: we reformulate the problem, taking $I_{a}I_{b}I_{c}$ the excenters and the problem now is asking to prove that the triangle $DEF$ and $I_{a}I_{b}I_{c}$ have the same euler line, where $DEF$ is the contact triangle of $\triangle{ABC}$.

but from [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=6228[/url] the euler line of $DEF$ is $OI$, and now we just simply observe that $O$ in $I_{a}I_{b}I_{c}$ is the euler point, and $I$ is the ortocenter, thus the euler lines coincide.

Or we known Euler line of contact triangle of $ABC$ is connecting circumcenter and incenter of $ABC$ see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=824323#824323]the post of me[/url]
Now can see $H$ is inceter of $A_{1}B_{1}C_{1}$ and circumcenter of $A_{1}B_{1}C_{1}$ is $E$(center of nine point circle) thus Euler line of $A_{1}B_{1}C_{1}$ is connecting $H,E$ which is coincide Euler line of $ABC$.

[hide="Solution"]Let $n$ be the nine-point circle of $\triangle ABC$, which at the same time is the circumcircle of $\triangle A_1 B_1 C_1$. Let $r$ be the radius of the inscribed circle of $\triangle A_1 B_1 C_1$, then an inversion $\varphi$ with center $H$ and power $r^2$ maps $n$ into $n'$, where $n'$ is the nine-point circle of $\triangle A_2 B_2 C_2$. That is $\varphi =I(H,r^2): n \mapsto n'$. Let $N$ be the center of $n$ and $N'$ the center of $n'$. Then since $\varphi :n \mapsto n'$ we have that $H,N,N'$ are collinear. The Euler line of $\triangle ABC$ passes through $H$ and $N$, and the Euler line of $\triangle A_2 B_2 C_2$ passes through $H$ and $N'$. Since $H, N, N'$ are collinear, the Euler lines coincide. (Q.E.D.)[/hide]

Triangles A2B2C2 and ABC are homothetic . Let R be the center of this homothety and S be the orthocenter of A2B2C2. R-H-O are collinear and R-S-H are collinear. Hence the Euler lines of both the triangles coincide.

Let $O$ and $H$ be the circumcenter and orthocenter of $ABC$ respectively.

By duality, we know $ABC$ is the excentral triangle of $A_1B_1C_1$. Thus, [url]https://artofproblemsolving.com/community/q2h42412p22304668[/url] yields $A_2B_2C_2$ (the contact triangle of $A_1B_1C_1$) is homothetic to $ABC$.

Let $AA_2$, $BB_2$, and $CC_2$ concur at the homothetic center $P$. It's well-known that $H$ is the incenter of $A_1B_1C_1$, so $H$ is the circumcenter of $A_2B_2C_2$. Thus, the homothety mapping $A_2B_2C_2$ to $ABC$ maps $H$ to $O$, implying $P, H, O$ are collinear. Now, observe $OH$ is precisely the Euler Line of $ABC$.

Let $H_1$  be the orthocenter of $A_2B_2C_2$. Obviously, the homothety from $A_2B_2C_2$ to $ABC$ maps $H_1$ to $H$, so $P, H_1, H$ are collinear. Since $H_1H$ is the Euler Line of $A_2B_2C_2$ and $O \in PH$, the result easily follows. $\blacksquare$


calculus algorithm MIT college derivative search

Where could a middle schooler find information about physics. Is there a book for the physics olympiad similar to the AoPS? I have experience with the subject but it facinates me.

unfortunately, there is nowhere near as much hype to the physics olympiad like there is to the math olympiad.  nevertheless, there is still tons of physics out there and there are some problem solving-ish books.  if you're a middle schooler, i'm assuming you have no physics background.  the best way to begin is to learn basic physics without calculus (as you probably have not had calculus anyway).  a course is a great way to do this (i recommend EPGY Introductory Physics, an online course found at http://epgy.stanford.edu).  but, if you prefer to just work from a book or don't wish to spend the money on a course, you could simply buy a good basic physics textbook and work your way through it.  Giancoli is probably the best basic physics textbook that does not use calculus (this is the textbook that EPGY uses).  in basic physics without calculus, you learn basic newtonian mechanics, electricity and magnetism, thermodynamics, very basic fluid mechanics, waves and optics, and are introduced to some modern physics such as special relativity.  although it seems like an abundance of topics, they are all only covered on the surface.  a more fun-ish book to read which explains only qualitatively about special relativity, general relativity, quantum mechanics, and the ideas of string theory, is The Elegant Universe by Brian Greene.  if you do have a calculus background, post again and I will give you a completely different slew of suggestions.

I have *this* much calculus backround. look at my name.

Also, wher ecan I find these sort of books for chem and informatics?
I wish to be well rounded. Not world class, but well rounded.
 I have 'Introduction to algorithms" by MIT press for informatics, but I need more info on that.

well i don't know what *this* much background in calculus is, but i'm assuming you know derivatives and integrals.  in that case, you might choose to go into the "calculus route" of physics, meaning that you'd go into more depth in Newtonian Mechanics and Electricity and Magnetism, but not cover so much of the other topics.  there are some good books for these that are more abstract textbooks, in that their problems require a great deal of problem solving.  calculus is used freely throughout the books, and there is much more proving and deriving done.  Introduction to Mechanics by Kleppner and Kolenkow is the book to start with and Electricity and Magnetism by Purcell would be the next book to go into.  These would certainly be tough although, with little or no physics background.  A more general textbook which uses calculus and covers a wider range of topics but does not deal with much problem solving, is Physics for Scientists and Engineers by Tipler and Mosca.  In terms of chem and informatics, those questions would be better placed in their respective forums =).

[quote="chess64"]I'm in the same situation as PenguinIntegral. Thanks for the suggestions! :)[/quote]
Yes, thanks bunches.

I don't know much about how physics courses work where you live, but my experience (1 year in US HS) says that you cannot learn too much just studying from a textbook on your own, and a bad teacher doesn't help much. I got most of my physics knowledge on physics competitions, correspondence seminars and their physics camps. Now I see that I wouldn't learn anything in school. OK, maybe I'd know all the equations and know the laws word by word, but that's of no use. Concerning textbooks, I don't remember exactly, but I think what I went through was a Mc Graw Hill textbook, I didn't like it much. It pretended to explain a lot of topics, but I'd rather it explained maybe half of them, but thoroughly. Not just like a fairy-tale. I think a kind of basic college textbook will help you much more, there is one that is co-authored by Jearl Walker and it looked quite good - about the IPhO level, and most of the concepts are quite well-explained.

I know this my "advice" doesn't help much, but that's my impression of how one can learn physics ... just find some good external resources, i.e. friends who study physics, correspondence seminars like we have in Slovakia (I haven't found something like that in the US), and try hard. And everytime you learn somehing more you find out that you know much less than you thought you do;)

[quote="Spoon"]Purcell is much too advanced for any middler schooler at any level. If you truly could understand it, then there is no reason for you to be in middle school, you are merely wasting your time. The fact that you *know* calculus is just dumbfounding. Might I ask why you are in middle school? Surely your time would be better spent elsewhere.

I think a good move from non calculus to calculus based physics is [i]Physics for Scientists and Engineers[/i] by Giancoli. It goes all the way up to modern physics, and is not quite at the Purcell level, yet still uses *some* calculus to explain the theories. Also, [i]Physics: Calculus[/i] by Hecht is good as well, and only uses *some* calculus to explain stuff. It's a nice place to start.[/quote]

Yes, I am still in middle school. Going into seventh grade.

The American education system sucks. I already wrote that sentence a few times in my blog. I know kids, even non gifted ones, who routinely get A's and B's just because you really have to do no thinking(and I help them out on tests). All my education consists of is memorizing facts about the topic. Even then it is just surface treatment. A 50 page book would cover the amout of material.

The thing is, I live in a rural community, and my parents work at the local college. I have no money, I have to literally buy my education from amazon, and even then I have to literally beg for anything over $15.

Heck, getting the second Aops book ordered last night filled my book quota for a while.
The only reason I was able to learn calculus was I read an old book I found in the college library. Something by Morris Kline. Its called 'Calculus: an intuitive and physical approach'

I feel opressed.
 :lol:

well in terms of spending money, on amazon you can buy used books from sellers (you probably already know this).  if you want to get a textbook, this is probably the best way to go because a new textbook such as Tipler and Mosca's Physics for Scientists and Engineers is quite expensive.  Agreeing with Spoon, you should probably begin with a simpler textbook that does use calculus.  Physics for Scientists and Engineers by Giancoli should be ok, but I prefer and recommend Physics for Scientists and Engineers by Tipler.

I understand your situation.

... but, you seem to be online pretty often, so that means you have access to the Internet. Which is a very valuable resource. Although you usually have to look really hard to find something, and even more often it somehow just falls into your lap than you intentionally search for it and ultimately find it. I know that;)

But guess what, I found quite a good college book on physics in the Long Creek school library. That's where I stayed in the US, and Long Creek would definitely be something I call a rural community;) Maybe that was pure luck, but since your parents work at local college, you should be able to find something there ...

I'll repeat myself, I have never seen anything like correspondence seminars in the US, which is too bad ... I have learned almost all my math and physics and informatics from those. They are run by college students who do it for free. Usually mainly because THEY were raised on those seminars and they know that they owe them a lot. But I guess that sooner or later they'll become available in English also.

I know that all this looks quite pessimistic and it's not really valuable advice, but that's how it is;) I'm spending a lot of my time and effort working for the correspondence seminars (MS and HS), because I know that also here in Slovakia, if you're not lucky with a teacher who's actually good and WANTS to teach well (there are not many of those, the good people usually work in better-paid fields), you have almost no chance to learn anything well. Especially the higher-level math, physics, and so on...

And my personal opinion is, that however good textbook you can buy, even better for you is to practice, take part in competitions you can find in your community or online or wherever, make friends with the same interest and learn on your own mistakes. Me myself, if I read a textbook, I'd forget it all in a week ... ;-) (I'm not trying to tell you not to buy them!!)

Good luck!

Thanks for the help! I pretty much figured out school was never going to teach me anything, so I guess books and maybe getting a tutor is the way to go.

you just said you have no money, how are you going to afford a tutor? In any event, you're going into 7th grade and alreay know calculus. I was learning how to solve 2(x+2) = 4x when I was in 7th grade!!!!!!! Crazy.

I wish you the best of luck.

Some nice college student probably will.

I have connections and frequent their picnics and such.


geometry
circumcircle
parallelogram
trapezoid
geometry unsolved

$ (C_1)$, $ (C_2)$ two circle intersect in $ A$ and $ B$.
the tangent of $ (C_1)$ in $ A$ intersect  $ (C_2)$ in $ D$ and the tangent of $ (C_2)$ in $ A$ intersect $ (C_1)$ in $ C$.
the semi right passer by $ A$ and it is in the angle $ CAD$ intersect $ (C_1)$  in $ M$ and $ (C_2)$ in $ N$ and the inscriptible circle of $ ACD$ in $ P$.
prove that $ AM\equal{}NP$

[quote="greatestmaths"]
the semi right passer by $ A$ and it is in the angle $ CAD$ [/quote]
Sorry, I don't understand this.  What does it mean?

Two circles $ (C_1)$, $ (C_2)$ intersect in $ A$ and $ B$. The tangent of $ (C_1)$ at $ A$ intersects $ (C_2)$ in $ D$ and the tangent of $ (C_2)$ at $ A$ intersects $ (C_1)$ in $ C$. Arbitrary line through $ A$ in the angle $ \angle CAD$ intersects $ (C_1)$ in $ M$ and $ (C_2)$ in $ N$ and the circumcircle of $ \triangle ACD$ in $ P$. Prove that $ AM \equal{} NP$.  :?:  :?:  :?:

Inversion with center $ A$ and arbitrary power $ r^2$ takes the lines $ AC, AD, AM \equiv AN \equiv AP$ passing through the inversion center $ A$ into themselves. Let $ B', C', M', N', P'$ be the inversion images of the points $ B, C, D, M, N, P.$ The circles $ (C_1) \equiv (ACM), (C_2) \equiv (ADN), (O) \equiv (ACPD)$ passing through the inversion center $ A$ go into the lines $ c_1 \equiv C'M' \parallel AD', c_2 \equiv D'N' \parallel AC', o \equiv C'P'D'.$ The images $ C'M', D'N'$ of $ (C_1), (C_2)$ intersect at $ B',$ $ AC'B'D'$ is a parallelogram, the image $ C'D'$ of $ (O)$ is one of its diagonals. The arbitrary line through $ A$ cuts $ C'D', C'B', D'B'$ in $ P', M', N',$ respectively.

Let $ D'M'$ intersect $ AC'$ at $ Q'.$ From the trapezoid $ AC'M'D'$ with $ C'M' \parallel AD',$ we have

$ \frac {AM'}{AP'} \minus{} 1 \equal{} \frac {AM' \minus{} AP'}{AP'} \equal{} \frac {P'M'}{AP'} \equal{} \frac {C'Q'}{AC'} \equal{} \frac {C'M'}{AD'} \equal{} \frac {AM'}{AN'},$

Reducing by $ AM'$ and multiplying by the inversion power $ r^2,$

$ \frac {r^2}{AP'} \minus{} \frac {r^2}{AN'} \equal{} \frac {r^2}{AM'}\ \Longleftrightarrow\ NP \equal{} AP \minus{} AN \equal{} AM$

We restate the given circles as $ (O_{1}),$ $ (O_{2}),$ instead of $ (C_{1}),$ $ (C_{2})$ respectively and let $ (O)$ be, the circumcircle of the triangle $ \bigtriangleup ACD.$

We draw the circle with diameter the segment $ AO,$ where $ O$ is the center of $ (O)$ and it is easy to show that it passes through the points $ E\equiv AC\cap OO_{1}$ and $ F\equiv AD\cap OO_{2},$ as the midpoints of the segments $ AC,$ $ AD,$ respectively.

Because of $ OO_{1}\perp AC$ and $ O_{2}A\perp AC,$ we have $ OO_{1}\parallel O_{2}A$ and similarly $ OO_{2}\parallel O_{1}A.$

So, from the parallelogram $ AO_{1}OO_{2},$ we conclude that its diagonals $ AO,$ $ O_{1}O_{2},$ bisect each other at the point so be it $ L,$ as the center of the circle $ (O).$

Let be the point $ K\equiv (O)\cap AP$ and then, from $ OK\perp AP,$ we have that the point $ K$ is the midpoint of the segment $ AP.$

Let $ O'_{1},$ $ L',$ $ O'_{2}$ be, the orthogonal projections of the points $ O_{1},$ $ L,$ $ O_{2}$ respectively , on the line segment $ AP.$

Because of $ LO_{1} \equal{} LO_{2}$ $ ($ from the parallelogram $ AO_{1}OO_{2}$ $ ),$ we conclude that $ L'O'_{1} \equal{} L'O'_{2}$ $ ,(1)$

From $ (1)$ and because of the points $ O'_{1},$ $ L',$ $ O'_{2},$ are the midpoints of the segments $ AM,$ $ AK,$ $ AN$ respectively, we conclude that $ KM \equal{} KN$ $ ,(2)$

From $ (2)$ and because of $ AK \equal{} KP,$ we conclude that $ AM \equal{} NP$ $ ($ also $ AN \equal{} MP$ $ )$ and the proof is completed.

Kostas Vittas.

[quote="greatestmaths"][color=darkred]$ (C_1)$, $ (C_2)$ two circle intersect in $ A$ and $ B$. The tangent of $ (C_1)$ in $ A$ intersect  $ (C_2)$ in $ D$ and the tangent of $ (C_2)$ in $ A$ intersect $ (C_1)$ in $ C$. A ray by $ A$ inside of the angle $ CAD$ intersect $ (C_1)$  in $ M$ and $ (C_2)$ in $ N$ and the inscriptible circle of $ ACD$ in $ P$. Prove that $ AM = NP$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $ R\in AP\cap CD$ . Observe easily that $ \|\begin{array}{c} \widehat {ACM}\equiv\widehat {PCR} \\
 \\
\widehat {ADN}\equiv\widehat {PDR}\end{array}$ .

Using the [b]Steiner's theorem[/b] in the triangles $ ACP$ and $ ADP$ obtain :

$ \|\begin{array}{c} \frac {MA}{MP}\cdot\frac {RA}{RP} = (\frac {CA}{CP})^2 \\
 \\
\frac {NA}{NP}\cdot \frac {RA}{RP} = (\frac {DA}{DP})^2\end{array}$ $ \implies$ $ \|\begin{array}{c} \frac {MA}{MP} = \frac {RP}{RA}\cdot(\frac {CA}{CP})^2 \\
 \\
\frac {NA}{NP} = \frac {RP}{RA}\cdot (\frac {DA}{DP})^2\end{array}$ . Therefore,

$ AM = NP$ $ \Longleftrightarrow$ $ \frac {MA}{MP} = \frac {NP}{NA}$ $ \Longleftrightarrow$ $ \frac {RP}{RA}\cdot(\frac {CA}{CP})^2 = \frac {RA}{RP}\cdot (\frac {DP}{DA})^2$ $ \Longleftrightarrow$

$ \frac {RA}{RP} = \frac {CA\cdot DA}{CP\cdot DP}$ , what is truly. See an [b]equivalent enunciation[/b] of this

easy and nice problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=188310 [/color]

Dear Mathlinkers,
you can see

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=199642

Sincerely
Jean-Louis