Euler
geometry
circumcircle
incenter
geometry unsolved

Let $ ABC$ be an acute-angled triangle, $ O$ its circumcentre, $ H$ its orthocentre, and $ AH_a$, $ BH_b$, $ CH_c$ its altitudes. Denote by $ A_1$, $ B_1$, $ C_1$ the circumcircles of the triangles $ BOC$, $ COA$ and $ AOB$, respectively. Prove that lines $ A_1H_a$, $ B_1H_b$, $ C_1H_c$ intersect at a point lying on the Euler line of the triangle $ ABC$.

$ \bullet$ Thank you dear [b]Babis Stergiou[/b] and [b]Theodor Bolis[/b] for your translation.

Sides of the $ \triangle A_1B_1C_1$ are perpendicular bisectors of the segments OB, OC, OA. Sides of the orthic $ \triangle H_aH_bH_c$ are also perpendicular to OA, OB, OC, for example, $ \triangle AH_bH_c$ is oppositely similar to the $ \triangle ABC,$ AO is the A-altitude of the former one. Thus the $ \triangle A_1B_1C_1 \sim \triangle H_aH_bH_c$ are centrally similar, having parallel sides, and $ A_1H_a,\ B_1H_b,\ C_1H_c$ are concurrent in their similarity center S. O is the similarity center, i.e., the common incenter (or one of the common corresponding excenters) of the $ \triangle A_1B_1C_1$ and the tangential triangle $ \triangle A_2B_2C_2$ of the $ \triangle ABC,$ H is the incenter (or one of the corresponding excenters) of the $ \triangle H_aH_bH_c,$ line OH connecting incenters (or corresponding excenters) of the two centrally similar $ \triangle A_1B_1C_1 \sim \triangle H_aH_bH_c$ passes through their similarity center S.

Because of $ A_{1}B_{1}\parallel H_{a}H_{b}$ and $ A_{1}C_{1}\parallel H_{a}H_{c}$ and $ B_{1}C_{1}\parallel H_{b}H_{c},$ we can say also, by [b][size=100]Desarques’s theorem[/size][/b], that the triangles $ \bigtriangleup A_{1}B_{1}C_{1},$ $ \bigtriangleup H_{a}H_{b}H_{c},$ are perspective and then, we have the concurrency point $ S\equiv A_{1}H_{a}\cap B_{1}H_{b}\cap C_{1}H_{c}.$

Because of now, $ B_{1}C_{1}\parallel H_{b}H_{c}$ and $ OB_{1}\parallel HH_{b}$ and $ OC_{1}\parallel HH_{c},$ we conclude that $ \bigtriangleup OB_{1}C_{1},$ $ \bigtriangleup HH_{b}H_{c},$ are also perspective.

Hence, the line segment $ OH,$ as the [b][size=100]Euler’s line[/size][/b] of $ \bigtriangleup ABC,$ passes through the point $ S\equiv B_{1}H_{b}\cap C_{1}H_{c}$ and the proof is completed.

Kostas Vittas.


[hide="hint"]
I'm pretty sure you can find the set of general solutions $(x,y)\in \mathbb{Z}^{2}$ to $ax+by=c$ by reducing it in either mod $a$ or mod $b$, as long as $a$ and $b$ are relatively prime.
[/hide]

[quote="Franz Joseph"]Find all whole number solutions to:

$2893719372x+199999998y=1111934$.[/quote]

Are $x,y$ digits of the number or the numbers are $2893719372*x$ and $199999998*y$?

What do you mean by "whole number"?  If you're talking about integers in general, there are an infinite number of solutions if $(2893719372, 199999998)|1111934$, and 0 otherwise.  If you're talking about natural numbers, then there are no solutions, because $2893719372, 199999998>1111934$.

@ José: It's a product for sure, because otherwise LHS would have more digits than RHS.

@ K81o7: a whole number is an integer.

@ The problem: Modulo 3 kills it. ($0x+0y \equiv 2\mod{3}$)

[quote="Kurt Gödel"]@ The problem: Modulo 3 kills it. ($0x+0y \equiv 2\mod{3}$)[/quote]

Pah.  Then the problem is easy.  Here is an alternate which has solutions, from my calculations:

$2893719372x+199999998y = 1111938$

:)

Use number theory and think of a linear diophantine equation and how you can solve it.


algebra
polynomial
complex numbers
algebra proposed

Let $f$ be a complex polynomial of degree at least 2 which has at least a non-real coefficient. Then the equation $ P(z)P(-z)=P(z)$ has roots in both half-spaces $ Rez>0, Imz<0$.

hmm,I cant understand the problem.
Did you mean $f(z)f(-z)=f(z)$? :?
And what do you mean by halfspace?
You mean seperating complex numbers accoriding to the sign of their Image or the sign of their Real.

Yes, it is $f$ instead of $P$. And the two halfspaces in theis problem are the set of numbers $z$ with $Rez>0$ and the one with $Imz<0$.


integration
function
real analysis
real analysis unsolved

Find all continuous functions $ f: [0,1]\to\mathbb{R}$ such that
\[ \int_{0}^{1}g(x)dx\equal{}\int_{0}^{1}f(g(x))dx,\]
for any continuous function $ g: [0,1]\to [0,1]$.

Suppose that for some $ a,$ we have that $ f(a)\ne a.$

Then we can always take the constant function $ g(x)\equal{}a.$

$ \int_{0}^{1}g(x)\,dx\equal{}a$ but $ \int_{0}^{1}f(g(x))\,dx\equal{}f(a)\ne a.$

Thus if it the statement "$ f(x)\equal{}x$ for every $ x$" is false, then the conclusion fails. Hence the only function that stands a chance is the identity function $ f(x)\equal{}x.$ Since the identity function does work, it is the unique answer to the question.

for given $ f$ then the right hand determine a Radon measure on $ C[0;1]$ and this equiv to $ Id$ in mean of distribution. So from Riesz representation Theorem we could deduce that $ f\equiv Id$ almost every where in $ [0;1]$, and because $ f$ is continous so $ f\equiv Id$.


Robert has three fair coins and Dunkirke has two. Each round, they both flip all of their coins and give the coins showing heads to the other player to flip on the next turn. If either player gets all five coins, he wins. What are the odds in favor of Robert's eventual victory?

[hide="Solution"]
If Robert must win a given round, all of his coins must be tails and all of Dunkirke's coins must be heads. If Dunkirke must win a given round, all of his coins must be tails and all of Robert's coins must be heads.

Thus, the probability of any one player winning a round is $ (\frac {1}{2})^{5} \equal{} \frac {1}{32}$, as the winning sequence can only occur in one way. The probability that neither will win the round is
$ 1 \minus{} \frac {1}{32} \minus{} \frac {1}{32} \equal{} \frac {30}{32} \equal{} \frac {15}{16}$.

So, the probability Robert wins eventually is:
$ (\frac {1}{32})(1 \plus{} \frac {15}{16} \plus{} (\frac {15}{16})^{2} \plus{} (\frac {15}{16})^{3} \plus{} ...) \equal{} \frac {1}{2}$
(The probability that Robert wins the first round, or that nobody wins the first round and Robert wins the second round, or that nobody wins the first two rounds and Robert wins the third round...)

Not surprisingly, this works out to $ \frac {1}{2}$, which makes sense, as the probability of any single player winning a given round is equal to the probability of the other player winning the round.

[/hide]


My mom just told me this... can any of you answer it?
There's a king that has three horses, each of them having different rankings. One was fast, one was medium, and one was slow.
On the other hand, there was a warrior whom had three horses, which also had different rankings: fast, medium, and slow.
One day the king wanted to have a race with the horses against the warrior. The problem was that each of the warrior's horses weren't as good as the king's. For example, the warrior's fast one wasn't as fast as the king's, his medium horse wasn't as good as the king's medium, and his slow one also wasn't as good as the king's. It's sortta confusing, but---

[hide]If the warrior was wanting to win, he would send his slow horse against the king's fast one first, forcing himself to lose. Then the warrior would send out his medium one against the king's slow one, making him win. Next, the warrior would send out his fastest one against the king's medium one, winning 2/3.  :rotfl: [/hide]

[quote="jhredsox"]a this is spam
b this should be in riddles forum
c how do you know the warrior's med horse is better than the king's slow (same with fast to med)???
d this is to easy for mathcounts geniuses

jorian[/quote]
a) This is not spam.
b) Mathcounts often has logic problems that would be worded differently from this.
c) That's a good point.
d) Geniuses isn't a word, it's proper to use genii. Also, you should use the word too instead of to. Finally, mathcounts has questions that are easy, not all the questions are meant for the top students.

[quote="toadoncart"][quote="uldivad9"]Does the king know what horse the warrior is racing?
If not, then does the king randomly select horces to race?
If not, then what order does he race them in?[/quote]

The king lets the warrior choose his horses against his own. I forgot to tell you that.  :D[/quote]

THATS WEIRD, I JUST WAS ASKED THIS QUESTION (ITS LIKE AN OLD CHINESE LOGIC QUESTION)

YOU JUST DO 1 BETTER THEN THEM  AND RACE YOUR WORST AGAINST THEIR HARDEST.

HOWEVER, THIS QUESTION DOES NOT BELONG HERE (TOO EASY AND ITS RIDDLE)

AND DONT REVIVE OLD TOPICS.

THIS MATHCOUNTS FORUM IS TURING INTO THE KINDERGARDEN QUESTION FORUM

Not really... some mathcounts problems are like this (aka. sprint)...

Yeah, this is an old chinese logic question, and, I know this doesn't belong here, but back then, I was an new member and felt like posting whatever I wanted anywhere I wanted.... :D

[quote="kyyuanmathcount"]If one has a sticky caps lock key, your best bet would be to unplug the keyboard and wash it out with water - make sure you let it dry completely before using again.

It's ok to revive topic if the question was unanswered or you are expanding on the question.[/quote]

really?  I once revived an old topic that was like a couple of geometry questions (there were 8 of them or so but like 2 or 3 werent answered yet and I answered one) and some mod locked the topic..

Do you still have the thread?  I'll unlock it for you if that's comforting...

Uh, so does anybody want to discuss the riddle itself rather than debating over the possibility of it being in MATHCOUNTS?

...
i heard this before, and i'm korean..
i think my best friend (who's chinese) told me this..
but yeah..
u don't know if the medium is faster than the king's slowest...
 :read:

So then the ratings of the horse have to be:
(from fastest to slowest):

1. king's fastest
2. warrior's fastest
3. king's medium
4. warrior's medium
5. king's slowest
6. warrior's slowest

There! Case closed!  :D


Let $r,s$ be the roots of\[ x^2-(a+d)x+(ad-bc)=0 \]Prove that $r^3$ and $s^3$ are the roots of\[ y^2-(a^3+d^3+3abc+3bcd)y+(ad-bc)^3=0 \]

[hide]
since $r,s$ are roots of the first equation,
$rs=ad-bc$-----(1)
$r+s=a+d$-----(2)

Now we have only to prove that $r^3s^3=(ad-bc)^3$, that follows directly from (1) and that $r^3+s^3=a^3+d^3+3abc+3bcd$.

From (2) we have that

$(r+s)^3=(a+d)^3 \implies r^3+s^3=a^3+d^3+3a^2d+3ad^2-3rs(r+s)$ Substituting $rs$ from (1) and $r+s$ from (2) we have the proof.
[/hide]

isn't it a little bit simple for intermediate?


calculus
vector
function
ratio
calculus computations

I'm having a problem with this ODE.  The question asks to use the power series method to solve the given DE about the ordinary point x=0.

(1+x^2)y"+6xy'+6y=0

When I come up with a solution I get y=ao+0+0... because i'm getting a0=a0, a1=0 and a2=0.  When I plug these in to find a3, a4, etc. they all equal zero.  Any suggestions?  Thanx.

You have obviously made some mistake. I get $(6+6n+n(n-1))a_n+(n+2)(n+1)a_{n+2}=0$, which simplifies to $(n+3)a_n+(n+1)a_{n+2}=0$ or $\frac{a_n}{n+1}=-\frac{a_{n+2}}{n+3}$. This can be easily solved to give $a_{2k}=\frac{(-1)^kb}{2k+1}$ and $a_{2k-1}=\frac{(-1)^kc}{2k}$.

The solutions to this differential equation are rational functions.

oops i guess the code didn't work.  what i meant was i got up to what you have in the last part of the second line.

To get the code to work, put dollar signs around the items you want to have interpreted as $\text{\LaTeX}.$ I've gone ahead and edited your post to put them in. I'm also moving this to "Calculus Computations and Tutorials," which is were we prefer to deal with questions like this.

[quote="Zdog3333"]I got up to $\frac{a_n}{n+1}=-\frac{a_{n+2}}{n+3}.$  Do I solve for $a_n$ and then put in values for $n$ to get $a_0, a_1, a_2,$ etc...?[/quote]
No. This is a recursion. You have infinitely many equations for infinitely many unknowns. Think of these equations as being arranged in a latter.

You're not going to solve any of these equations for $a_0$ or $a_1.$ Those will be determined by your initial conditions.

Now solve the equation you do have for the coefficient of the highest index in terms of coefficients of lower index:

$a_{n+2}=-\frac{n+3}{n+1}a_n.$

So $a_2=-\frac31a_0=-3a_0.$

$a_4=-\frac53a_2=5a_0.$

$a_6=-\frac75a_4=-7a_0.$

Starting over, $a_3=-\frac42a_1=-2a_1.$

$a_5=-\frac64a_3=3a_1.$

$a_7=-\frac86a_5=-4a_1.$

You see the pattern here (jmerry had the fractions flipped).

so I could write my solution as:
y=$a_0+a_1x-3a_0x^2-2a_1x^3+5a_0x^4+3a_1x^5+...)$

OR

y=$a_0(1-3x^2+5x^4-...) + a_1(x-2x^3+3x^5-...)$

Thats how our prof. wants us to write the solution.  Would that be correct?

Yes this is correct, although you could spend a little more effort putting it into summation notation. (I wouldn't care, but I'm not your teacher.) The second form you list would be far more useful, for two reasons.

1. This is a second order linear homogeneous differential equation. Hence, its solution space os a two-dimensional vector subspace of the vector space of (smooth enough) functions. Remember from earlier in the course that you were writing the solution to $y''+3y'+2y=0$ as $y=C_1e^{-x}+C_2e^{-2x}$ ? What you were doing there was finding a basis for this solution space. The second form of your equation is clearly $y=a_0y_1(x)+a_1y_2(x).$ That's expressing the solutions as a linear combination of a two-function basis.

2. And it's a very convenient basis, if you have initial conditions given at zero. $a_0=y(0)$ and $a_1=y'(0).$

By the way, can you tell me what the radius of convergence of each of these power series is?

[hide="Stray comment on that."]The radius of convergence was predictable from the original differential equation.[/hide]

I remember doing power series in Calc 2 where we had to find if the power series converged or diverged and the Radius of convergence.  That was about 2 semesters ago for me and I forgot how to find the radius of convergence.  I'm guessing its 0 maybe?

Remember, we're looking for solutions to a differential equation. A power series with a radius of convergence of zero doesn't solve anything. It would be a function whose graph is a single dot, and that wouldn't do us any good. No, this had better have a positive radius of convergence.

Hint: the ratio test is very specifically well-suited to problems of this type.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Show that a group of order $ 12$ has a normal subgroup of order $ 4$ or $ 3$

This type of problem is usually solved with the Sylow theorems:

By the Sylow theorems, $ n_3 | 12$ and $ n_3 \equiv 1\ (mod\ 3)$, where $ n_3$ is the number of Sylow 3-subgroups of $ G$.  If $ n_3 \equal{} 1$, the Sylow 3-subgroup is normal and we are done.  So $ n_3 \equal{} 4$ is our only choice.  Similarly, assume $ n_2 \geq 3$.

The number of elements of order 3 must be $ 4*2 \equal{} 8$, and there are two distinct Sylow 2-subgroups, so their union is at least 5 elements.  We have shown that $ |G| \geq 13$, but $ |G| \equal{} 12$, contradiction.  Hence one of the Sylow subgroups must be normal.


Solve the equation:

sin(sin(sin(sin(x)))) = cos(cos(cos(cos(x))))

Can this problem be generalized to an arbitrary number of sin's and cos's?  And what about the more general problem

f(f(f(...f(x)))...) = g(g(g(...g(x)))...)  ?

If I remember correctly, you can sort of bound the regions where they could intersect and show that they don't. I know that it has no solution.

Anyone come up with a complete solution for this one?

what would happen when f(x) = sin x is the x of g(x) = sin (sin x))?
is there a certain pattern?

Here's the sketch of a proof that there are no solutions.  Any one else should feel free to make it complete.

Spoiler: [hide]It's not hard to prove that  is at most .  It's also not hard to prove that  is at least .  Unfortunately, those two bounds by themselves don't prove that there are no solutions, because the two intervals intersect.

Let c be the (unique) fixed point of cosine.  We will split the proof into two cases, depending on how |cos x| compares with c.

Suppose that .  Then it's not hard to prove that .  Because , there are no solutions in this case.

Suppose that .  Thus .  Then it's not hard to prove that .  Because , there are no solutions in this case either.[/hide]

Edited to convert no-longer-supported tags to LaTeX.

[quote="Michael V."]Can this problem be generalized to an arbitrary number of sin's and cos's?[/quote]

The upper row of unit circles shows successive operations of the sin function on real values of x.  The lower row of unit circles shows successive operations of the cos function on real values of x.

The blue arcs shows the domain of the function around the unit circle.  The red lines show the range of the function along the x- or y-axis.

The possible range of one operation of the function is shown as the possible domain of the next successive operation of the function.

The possible range of the nested sin function is symmetric, and decreases towards zero.

The possible range of the nested cos function decreases with each operation, and approaches approximately 0.739085133.

After five operations of the sin function, the maximum possible result is approximately 0.62757.  The maximum value can never exceed 0.62757, regardless of the number of additional operations of sin are performed.

After five operations of the cos function, the minimum possible result is approximately 0.65429.  The minimum possible value can never dip below 0.65429, regardless of the number of additional operations of cos are performed.

Thus, there can be no solutions when the functions are nested five or more deep.

Must I prove the ranges get smaller with each nesting?  Not formal, but...  For the sin function, each operation produces a range on the y-axis.  In the next operation, that range is wrapped around the unit circle, centered at the midpoint.  Obviously, the projection of the arc onto the y-axis must be shorter than the length of the arc, itself.

For the cos function, each operation produces a range on the x-axis.  In the next operation, that range is wrapped around the unit circle.  Alternate endpoints are anchored at one end of the previous range.  Once again, the projection of the arc onto the x-axis must be shorter than the length of the arc, itself.


geometry
incenter
Asymptote
function
parameterization
graph theory

In absense of a separate Geogebra Forum, I'm starting this thread here. It is not a secret that the default Geogebra tool list can hardly be called "full", but the good part is that we can add as many tools as we wish and change the default toolbar makng it more suitable for AoPS needs. The first step is to figure out which additional tools people who intend to use Geogebra for posting the pictures on AoPS may be interested in. So, please put your suggestions here :)

I can currently think of

[b]incenter(point,point,point)[/b] that takes 3 points and returns the incenter of the triangle
[b]incircle(point,point,point)[/b]
[b]circumcenter(point,point,point)[/b]
[b]centroid(point,point,point)[/b]
[b]foot(point,line)[/b] that returns the projection of the point to the line
[b]waypoint(point,point,number)[/b] that returns the point between two given points whose position depends on number (so $ 1/2$ will give you the midpoint, etc.)

(Needless to say, I just borrowed those from the Olympiad.asy :P).

Does anybody have other suggestions? Geogebra is not as powerful as Asymptote, but quite a lot is possible and we can do it if there is sufficient interest.

[b]fill(polygon,color)[/b] would help [custom color, color = FFFFFF for black and the like]

probably be just aesthetic purposes...

[b]fix(point,xfix,yfix)[/b]

point: point you want to fix
xfix: 1 = fixed for x [cannot move point in x direction]
yfix: 1 = fixed for y [cannot move point in y direction]

[b]func(function)[/b]

Graphs function y=function.

showaxes parameter [show axes on startup]
showgrid parameter [show grid on startup]

[b]pfunc(function,param)[/b]

Graphs function y=function plus extra parameter that can be changed with a slider:

e.g. pgraph(2x^2+ax-4,a)

[b]kgraph(number)[/b]

Draws complete graph of order n.

[b]bpgraph(number1,number2)[/b]

Draws complete bipartite graph of order n1,n2

more to come

Well, you should understand that Geogebra is not Asymptote: it is a simple point-n-click graphing tool, not a full-fledged programming language, so some things you want are just impossible to define as "user tools". On the other hand, some operations are easier than on Asymptote. There is no need to have any special command for graphing a function; just defining it produces a graph that you can either show or hide. The same with parameters: just say a=2, etc. in the command line, choose to show it to get the slider, and define your function using the values of the parameter(s). The picture below graphs $ x^3 \plus{} ax^2 \plus{} bx \plus{} c$ with adjustible $ a,b,c$:
 [geogebra]337c616b8bc8b1b5420f0398338386670ca1687f[/geogebra]
On the other hand, if you know how to design your tool and think that my "impossible" verdict is wrong, you can describe the way and we'll follow it. With all that in mind, let's go over your suggestions:

[b]fill(polygon, color)[/b] Just click on properties of the polygon and choose your color

[b]fix(point,xfix,yfix)[/b] Impossible.

[b]func(function)[/b] Not needed

[b]pfunc(function,parameter)[/b] Impossible but the standard way to do it (input the parameter variable first and define the function afterwards is not too complicated).

[b]kgraph()[/b] and [b]bpgraph()[/b] Possible but it would be nice to know what exactly you want here. What seems to make most sense to me is a command that takes a list of points and returns the list of all segments between them, so you'll have to write [b]kgraph({A,B,C,D,E})[/b] to get a complete graph with 5 vertices but then you'll be able to move your vertices. It is feasible to create a standard size full graph (see the aplet below) but it will be stationary then and always look like a regular polygon with all diagonals:
[geogebra]fda8d096bad636ec08bf9c17478b82d8978c0c92[/geogebra]
Same about [b]bpgraph()[/b].

Forgive me if this has already been requested, but adding this as a tool would save a lot of time for certain problems:

[b]commontangents(circle,circle)[/b] draws four (or however many are possible given circumstances) common tangent lines to two given circles.

All right. I'll make one for inner tangents and one for outer tangents (most of the time you do not need both). Anything else?

Thank you!
Another function that would be useful:
[b]tritangent(line, line, line)[/b] draws the four (or however many are possible given circumstances) circles tangent to all three lines.  This would be helpful for drawing excircles and such.

[quote="fedja"]

[b]incenter(point,point,point)[/b] that takes 3 points and returns the incenter of the triangle
[b]incircle(point,point,point)[/b]
[b]circumcenter(point,point,point)[/b]
[b]centroid(point,point,point)[/b]
[b]foot(point,line)[/b] that returns the projection of the point to the line
[b]waypoint(point,point,number)[/b] that returns the point between two given points whose position depends on number (so $ 1/2$ will give you the midpoint, etc.)

.[/quote]

What are your tools?
How to use  your  tools  in  my     GeoGebra ?

This is a rather old revive post, but I wanted to point out that the newest version of Geogebra contains a number of these commands. In particular, [b]SetColor[(Object),(Color)][/b] and [b]SetFixed[(Object),true][/b] are available commands.

How do I input these commands into the newest version of Geogebra?

[quote]
[b]commontangents(circle,circle)[/b] draws four (or however many are possible given circumstances) common tangent lines to two given circles.
[/quote]
This has been available in GeoGebra since release 4.0 by clicking the existing "Tangents" tool and then selecting two circles.

It looks like Geogebra got updated. Is there a way to add those in again?

Ask [b]levans[/b]. I can write them again, no problem, but I cannot implement them from the user end...


[quote="Chen241290"][quote="José"]Consider 3 real numbers $a,b,c$ such that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$. Find $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$.[/quote]
$a+b+c=(a+b+c)(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})=\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c$
Hence:$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0$[/quote]

Nice! Thanks!!


:?: Arne, For your last post about Cauchy's function That use for real function .
 I am confused. In my country Cauchy's function can use for positive integers function
 integers function and real function(but for real function It's must have other conditions.Every things I said I have the clear solution in my book. What about
  Cauchy's function in your country Arne.

When it's directly to Arne, a PM would be better (and specify what you are talking about) ;)
http://www.mathlinks.ro/Forum/privmsg.php?folder=inbox brings you to your inbox.

PS: I think it's about http://www.mathlinks.ro/Forum/viewtopic.php?t=14908


function
trigonometry
inequalities
real analysis
real analysis unsolved

We say that the functions$ f: R\to{R}$ have the property $ (P)$ if are derivable and $ f(x)\ge{f(x\plus{}sinx)}$ for every $ x\in{R}$.
$ a)$ Prove that exist functions with the property $ (P)$ which aren't constant
$ b)$Prove that if $ f$ have the property $ (P)$ then the equation $ f^{'}(x)\equal{}0$ has an infty of solutions

b) follows from the observation that $ f$ has a local maximum when $ \sin x\equal{}0$

a) Any [b]differentiable[/b] function that is decreasing when $ \sin x> 0$ and increasing when $ \sin x<0$ should work. Like ... $ \cos x$.

How can I prove that $ cosx\ge{cos(x\plus{}sinx)}$ for every $ x\in{\mathbb{R}}$

[quote="tarta"]How can I prove that $ cosx\ge{cos(x \plus{} sinx)}$ for every $ x\in{\mathbb{R}}$[/quote]

Maybe the derivates are usefull here   :lol:

Can you explain me how to use derivates? :)

Write it like this :

$ \cos x-\cos(x+\sin x)=2\sin \left(x+\frac{\sin x}{2}\right) \sin \frac{\sin x}{2} \geq 0$
It's enough to prove this inequality for $ x\in [0,2\pi]$

Let $ f(x)=x+\frac{\sin x}{2}$ and $ g(x)=\frac{\sin x}{2}$.
Since $ f'(x)>0$ it follows that $ f$ is increasing .From here we deduce that $ x\in [0,\pi] \Rightarrow f(x) \in [0,\pi] \Rightarrow \sin f(x) \geq 0$
Moreover , $ x \in [0,\pi] \Rightarrow g(x) \in [0,\pi] \Rightarrow \sin g(x) \geq 0$.
So $ \sin f(x) \sin g(x) \geq 0$ for $ x \in [0,\pi]$

In the same way $ x\in [\pi ,\2pi] \Rightarrow f(x) \in [\pi ,2\pi] \Rightarrow \sin f(x) \leq 0$
and $ x \in [\pi ,2\pi] \Rightarrow g(x) \in [-\pi,0] \Rightarrow \sin g(x) \leq 0$

So $ \sin f(x) \sin g(x) \geq 0$ for $ x \in [\pi ,2\pi]$


geometry
3D geometry
analytic geometry
rectangle

In a 120 × 150 × 180 cuboid (made out of unit cubes joined along their faces), how many unit cubes does its diagonal pass through?

(Just ”touching” at one point does not qualify as passing through).

This particular case can be reduced to the 4 x 5 x 6 case, but more generally,

[hide="hint"]Each time the diagonal passes through a edge, it passes through a face, and each time is passes through a corner, it passes through an edge.  Count the number of corners it goes through, the number of edges it goes through, and the number of faces it goes through.  Do some addition/subtraction, ie inclusion-exclusion.[/hide]

[hide]
$ 4\times 5\times 6$ case:
Starting from the diagonal corner box, you change x coordinate 3 times, y coordinate 4 times, z coordinate 5 times. So we go through 3 + 4 + 5 = 12 blocks, plus the original corner box, so 13 blocks. But we overcounted since, if you look from a birds view (at the 4 x 6 rectangle), that our diagonal passes through an edge ($ \gcd(4,6) = 2 \neq 1$, so we subtract 1, giving 12 boxes.

$ 120 \times 150 \times 180$ case:
It's just 30 times the previous case, so 360 boxes.

In general:
\[ 120+150+180-\gcd(120,150)-\gcd(120,180)-\gcd(150,180)+\gcd(120,150,180) = 360\]
[/hide]


Let AD || CB we know AC=a; BD=b the point where AC intersects BD is O, let OX be the height in AOB and OX=1
Knowing that  angle DAB=90 and BAC=90

Find AB!

This problem may be wrong,however i really need your help on this problem (important).Thank you!

Error fixed thanks to Quevvy, now can anyone solve?

[quote="alex__ib"]Let AD || CB we know AC=a; BD=b the point where AC intersects BD is O, let OX be the height in AOB and OX=1
Knowing that  angle ABC=90 and BAC=90

Find AB!

This problem may be wrong,however i really need your help on this problem (important).Thank you![/quote]
Knowing that $\angle{ABC}=90, \angle{BAC}=90$ you cannot draw $\triangle{ABC}$

But you don't specify nothing about a quadrilateral, triangle, etc. Or is there to find out?

If 3 points make a triangle (drawing auxilary lines), how can you draw $\triangle{ABC}$ if $m\angle{A}=m\angle{B}=90^{\circ}$?
Then $m\angle{C}=0^{\circ}$ and it is not a triangle...

Exactly, but just wondering  :whistling:

[quote="lotrgreengrapes7926"]Now C, D, and A are collinear.  :maybe:[/quote]
But since AD || BC, then that cannot be otherwise they would be the same...

[quote="alex__ib"]It looks like
 D............................................................

................................ ............................C
..................................... O......................

A....................................X........................B

Ox is height in $\triangle{AOB}$ and O is where ac intesrsects bd, basically abcd is a  trapezoid with a and b angle of 90 degrees, i am so sorry if i did not make myself clearly understood  :blush: and i don't have a program for geometry,....i guess now you can solve it.[/quote]
You editted your original post so now I understand.
I made a picture on GeoGebra and I have attached it. You can just go online for a free mathematics program.

Even now, you can't find out because segment BC can be stretched down when everything else stays the same...

Where did you get this problem? :?:

[quote]You editted your original post so now I understand.
I made a picture on GeoGebra and I have attached it. You can just go online for a free mathematics program.

Even now, you can't find out because segment BC can be stretched down when everything else stays the same...

Where did you get this problem? :?:[/quote]

My father asked me to do it, howwever i was supposed to find AB not BC. in terms of a and b   :?:  (a and b are the lengths).
I just want to be sure that i have done everything possible to solve, i tried similar triangles and didn't get far, but you may try too  :lol: .


Hey is the colume 2 book and solutions worth the money for one who is looking to score well on the amc 12?

Yes, it was probably the best money ever spent.  Definitely get the solutions too, I couldn't really inspire myself to do problems if I didn't know there was an answer I could find.  It is covers the most topics for the AMC 12, its written by people who have taken these exams, and is largely designed for them.

Notice which topics are important for AMC 12.  (use the search function on the forum)

Good luck to you, I strongly suggest you get the books!


geometry
combinatorics unsolved
combinatorics

For every lattice point $ (x,y)$ with $ x,y$ nonnegative integers, a square of side $ \frac{0.9}{2^x 5^y}$ with center at the point $ (x,y)$ is constructed. Compute the area of the union of all these triangles.

If I am not wrong, simple computations shows that no square intersects with the square centered at $ (0,0)$. This assures that none of the squares intersect. Hence, we have

\[ \sum \sum \frac{9}{2^{x\plus{}1}5^{y\plus{}1}}\equal{}\frac{5}{4}\]


[quote="anorithdialga"]
\[ \begin{array}{l} find\_largest\_y\_such\_that \\
\frac {1}{{1 \plus{} x^2 }} \ge \frac {{y \minus{} x}}{{y \plus{} x}} \\
\end{array}
\]

\[ \forall x > 0
\]
[/quote]
[hide="solution"]
$ y \equal{} 0$ satisfies the condition,so let $ y > 0$,in this case,the ineq is equivalent to

$ y\le x \plus{} \frac {2}{x}$　,

by AM-GM ineq we have $ x \plus{} \frac {2}{x}\ge 2\sqrt {x\cdot \frac {2}{x}} \equal{} 2\sqrt 2$ (when $ x \equal{} \sqrt 2$ eq holds.),

therefore we have the largest $ y \equal{} \boxed{2\sqrt 2}$.

[/hide]


integration
trigonometry
inequalities
induction
real analysis
real analysis theorems

Let $n\geq 0$ be integers. Answer the following questions $(1)$ through $(7).$

(1) Prove the following equality.
\[ \int_0^{\frac{\pi}{2}} \sin ^ n x\ dx=\int_0^{\frac{\pi}{2}} \cos ^ n x\ dx.  \]

(2) Let $I_n=\int_0^{\frac{\pi}{2}} \sin ^ n x\ dx=\int_0^{\frac{\pi}{2}} \cos ^ n x\ dx.$

Prove the following equality.
\[ I_n=\frac{n-1}{n}I_{n-2}\ (n\geq 2).  \]

(3) Prove the following equality.

\[ I_n=\begin{cases} \frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \cdot \cdot \frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi}{2}\ \ (n: even) \\ \frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \cdot \cdot \frac{4}{5}\cdot \frac{2}{3}\ \ \ \ \ \ \ (n: odd) \end{cases}  \]

(4) Let $S_n=\int_0^{\frac{\pi}{2}} x^2\cos ^ {2n} x\ dx.$ Prove the following equality.

\[ S_{n-1}-\frac{2n}{2n-1}S_n=\frac{2}{2n(2n-1)}I_{2n}\ \ (n\geq 1).  \]

(5) Let $N\geq 1$ be integers. Prove the following equality.

\[ S_N=\frac{(2N-1)(2N-3)\cdots 5\cdot 3\cdot 1}{2N(2N-2)\cdots 4\cdot 2} \cdot \frac{\pi}{4}\left(\frac{\pi ^2}{6}-\sum_{n=1}^N \frac{1}{n^2}\right).  \]

(6) Let $N\geq 1$ be integers. Prove the following inequality.

\[ S_N\leq \frac{1}{2N+2}\cdot \frac{(2N-1)(2N-3)\cdots 5\cdot 3\cdot 1}{2N(2N-2)\cdots 4\cdot 2} \cdot \left(\frac{\pi}{2}\right)^3.  \] You may use the inequality $x<\frac{\pi}{2} \sin x$ on $0<x<\frac{\pi}{2}.$

(7) Prove the following equality.

\[ \sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi ^2}{6}.  \]

It's a good exercise for people learning how to integrate by parts.

[hide="1."] We have that
\begin{eqnarray*}\int_{0}^\frac{\pi}2 \sin^{n}x \, dx &=& \int_{0}^\frac{\pi}2 \cos^{n}\left( \frac{\pi}2-x \right) \, dx \\ \ &=&-\int_\frac{\pi}2^{0}\cos^{n}x \, dx \\ \ &=& \int_{0}^\frac{\pi}2 \cos^{n}x \, dx .\end{eqnarray*}
[/hide]
[hide="2."] We integrate by parts:
\begin{eqnarray*}I_{n}&=& \int_{0}^\frac{\pi}2 \sin^{n}x \, dx \\ \ &=& \int_{0}^\frac{\pi}2 \sin^{n-1}x \, d \left(-\cos x \right) \\ \ &=&-\sin^{n-1}x \cos x \Big|_{0}^\frac{\pi}2+\int_{0}^\frac{\pi}2 (n-1) \sin^{n-2}x \cos^{2}x \, dx \\ \ &=& \left(-0+0 \right)+\int_{0}^\frac{\pi}2 (n-1) \sin^{n-2}x \left( 1-\sin^{2}x \right) \, dx \\ \ &=& (n-1) \int_{0}^\frac{\pi}2 \sin^{n-2}x \, dx-(n-1) \int_{0}^\frac{\pi}2 \sin^{n}x \, dx \\ \ &=& (n-1) I_{n-2}-(n-1) I_{n}.\end{eqnarray*}
Thus, $ I_{n}+(n-1) I_{n}= (n-1) I_{n-2}$, so $ I_{n}= \frac{n-1}{n}I_{n-2}$.
[/hide]
[hide="3."] We'll prove the statements by induction over $ k$.
It's clear that $ I_{0}= \frac{\pi}2$ and that $ I_{1}=-\cos x \Big|_{0}^\frac{\pi}2 =-0+1 = 1$.
Now let's suppose that we have proved the statement for $ k-1 \geq 0$, i.e. \[ \begin{cases}I_{2k-2}= \frac{2k-3}{2k-2}\cdot \frac{2k-5}{2k-4}\cdot \ldots \cdot \frac34 \cdot \frac12 \cdot \frac{\pi}2 ; \\ I_{2k-1}= \frac{2k-2}{2k-1}\cdot \frac{2k-4}{2k-3}\cdot \ldots \cdot \frac45 \cdot \frac23 . \end{cases}\] Then, we have that \[ \begin{cases}I_{2k}= \frac{2k-1}{2k}\cdot I_{2k-2}= \frac{2k-1}{2k}\cdot \frac{2k-3}{2k-2}\cdot \frac{2k-5}{2k-4}\cdot \ldots \cdot \frac34 \cdot \frac12 \cdot \frac{\pi}2 ; \\ I_{2k+1}= \frac{2k}{2k+1}\cdot I_{2k-1}= \frac{2k}{2k+1}\cdot \frac{2k-2}{2k-1}\cdot \frac{2k-4}{2k-3}\cdot \ldots \cdot \frac45 \cdot \frac23 . \end{cases}\] This completes the induction step.
[/hide]
[hide="4."] We have that
\begin{eqnarray*}S_{n}&=& \int_{0}^\frac{\pi}2 x^{2}\cos^{2n}x \, dx \\ \ &=& \int_{0}^\frac{\pi}2 x^{2}\cos^{2n-1}x \, d \left( \sin x \right) \\ \ &=& x^{2}\cos^{2n-1}x \sin x \Big|_{0}^\frac{\pi}2-\int_{0}^\frac{\pi}2 \sin x \left[ 2x \cos^{2n-1}x-(2n-1) x^{2}\cos^{2n-2}x \sin x \right] \, dx \\ \ &=& \left( 0-0 \right)-2 \int_{0}^\frac{\pi}2 x \cos^{2n-1}x \sin x \, dx+(2n-1) \int_{0}^\frac{\pi}2 x^{2}\cos^{2n-2}x \sin^{2}x \, dx \\ \ &=& \frac1{n}\int_{0}^\frac{\pi}2 x \, d \left( \cos^{2n}x \right)+(2n-1) \int_{0}^\frac{\pi}2 x^{2}\cos^{2n-2}x \left( 1-\cos^{2}x \right) \, dx \\ \ &=& \frac1{n}\left( x \cos^{2n}x \Big|_{0}^\frac{\pi}2-\int_{0}^\frac{\pi}2 \cos^{2n}x \, dx \right)+(2n-1) S_{n-1}-(2n-1) S_{n}\\ \ &=& \frac1{n}\left( \left( 0-0 \right)-I_{2n}\right)+(2n-1) S_{n-1}-(2n-1) S_{n}\\ \ &=&-\frac1{n}I_{2n}+(2n-1) S_{n-1}-(2n-1) S_{n},\end{eqnarray*}
so \[ \frac1{n}I_{2n}= (2n-1) S_{n-1}-2n S_{n}, \, \forall n \geq 1 .\] [/hide]
[hide="5."] The statement is equivalent to $ S_{N}= \frac{I_{2N}}2 \cdot \left( \frac{\pi^{2}}6-\sum_{n=1}^{N}\frac1{n^{2}}\right)$.
We'll prove this equivalent form by induction over $ N$.
The base case is clear: $ S_{0}= \int_{0}^\frac{\pi}2 x^{2}\, dx = \left. \frac13 x^{3}\right|_{0}^\frac{\pi}2 = \frac{\pi^{3}}{24}$.
Now let's suppose that we have proved it for $ N-1 \geq 0$, i.e. we know that \[ S_{N-1}= \frac{I_{2N-2}}2 \cdot \left( \frac{\pi^{2}}6-\sum_{n=1}^{N-1}\frac1{n^{2}}\right) .\] Then,
\begin{eqnarray*}S_{N}&=& \frac{2N-1}{2N}\cdot S_{N-1}-\frac{I_{2N}}{2 N^{2}}\\ \ &=& \frac{2N-1}{2N}\cdot \frac{I_{2N-2}}2 \cdot \left( \frac{\pi^{2}}6-\sum_{n=1}^{N-1}\frac1{n^{2}}\right)-\frac{I_{2N}}{2 N^{2}}\\ \ &=& \frac{I_{2N}}2 \cdot \left( \frac{\pi^{2}}6-\sum_{n=1}^{N-1}\frac1{n^{2}}\right)-\frac{I_{2N}}{2 N^{2}}\\ \ &=& \frac{I_{2N}}2 \cdot \left( \frac{\pi^{2}}6-\sum_{n=1}^{N}\frac1{n^{2}}\right) ,\end{eqnarray*}
which completes the induction step.
[/hide]
[hide="6."] Let's prove $ x < \frac{\pi}2 \sin x, \, \forall x \in \left( 0, \frac{\pi}2 \right)$. To this end, let $ f : \left[ 0, \frac{\pi}2 \right] \to \mathbb R$, \[ f(x) = \frac{\pi}2 \sin x-x .\] We have that $ f^{\prime \prime}(x) =-\frac{\pi}2 \sin x \leq 0, \, \forall x \in \left[ 0, \frac{\pi}2 \right]$ (with equality iff $ x = 0$), so $ f$ is strictly concave.
Thus, it reaches its minimum in one of its endpoints. But $ f(0) = f \left( \frac{\pi}2 \right) = 0$, so \[ f(x) > 0 \, \, \Longleftrightarrow \, \, \frac{\pi}2 \sin x > x , \, \forall x \in \left( 0, \frac{\pi}2 \right) .\] Hence,
\begin{eqnarray*}S_{N}&=& \int_{0}^\frac{\pi}2 x^{2}\cos^{2N}x \, dx \\ \ &<& \frac{\pi^{2}}4 \int_{0}^\frac{\pi}2 \sin^{2}x \cos^{2N}x \, dx \\ \ &=& \frac{\pi^{2}}4 \int_{0}^\frac{\pi}2 \left( 1-\cos^{2}x \right) \cos^{2N}x \, dx \\ \ &=& \frac{\pi^{2}}4 \left( I_{2N}-I_{2N+2}\right) \\ \ &=& \frac{\pi^{2}}4 \left( I_{2N}-\frac{2N+1}{2N+2}\cdot I_{2N}\right) \\ \ &=& \frac{\pi^{2}}4 \cdot \frac{I_{2N}}{2N+2}.\end{eqnarray*}
[/hide]
[hide="7."] Combining 5 and 6 we get that \[ \frac{I_{2N}}2 \cdot \left( \frac{\pi^{2}}6-\sum_{n=1}^{N}\frac1{n^{2}}\right) < \frac{\pi^{2}}4 \cdot \frac{I_{2N}}{2N+2},\] which implies \[ \frac{\pi^{2}}6-\sum_{n=1}^{N}\frac1{n^{2}}< \frac{\pi^{2}}{2(2N+2)}, \, \forall N \geq 0 .\] We have that $ S_{N}> 0$, so $ \frac{\pi^{2}}6-\sum_{n=1}^{N}\frac1{n^{2}}> 0, \forall N \geq 0$.
By the Pinching Theorem we get that $ \sum_{n=1}^\infty \frac1{n^{2}}= \frac{\pi^{2}}6$.
[/hide]

[quote="vijaymenon"][b]3)[/b] but how do we do this without induction.please help[/quote]
That's a fairly pointless request. Look at the thing to be proved there. Do you see the $ \cdots$ in the middle? For anything stated that way, with the dots in the middle, the very statement itself cannot be understood without induction or recursion. An induction will be present in any proof of this, either explicitly or implicitly; your only choice is how formal or informal your language regarding that induction will be. With respect to that induction, perfect_radio actually chose a middle path somewhere between strict formality and being so informal as to hide the method.

He splits off $ x^2 \cos^{2n} x \, dx \equal{} x^2 \cos^{2n\minus{}1} x  \cdot \cos x \, dx$ and then notes that $ d(\sin x) \equal{} \frac{d}{dx} \sin x \,dx \equal{} \cos x \, dx$


Φοβάμαι ότι χρειάζομαι τα φώτα σας γιατί είμαι τρελά σκουριασμένος στις διαφορικές μου...

Έστω $ w,\sigma \in \mathbb{R}$ σταθερές και θεωρήστε τη μερική διαφορική εξίσωση:
\[ \frac {\partial u}{\partial t} \equal{} x(1 \minus{} x)(w \plus{} \frac {1}{2}\sigma^2(1 \minus{} 2x))\frac {\partial u}{\partial x} \plus{} \frac {1}{2}\sigma^2 x^2 (1 \minus{} x)^2 \frac {\partial^2 u}{\partial x^2}
\]
με $ x\in[0,1]$, $ t\geq 0$ και τις συνοριακές συνθήκες:
1. $ u(x,0) \equal{} x$
2. $ u(0,t) \equal{} 0$
3. $ u(1,t) \equal{} 1$

Έχει μοναδική λύση η παραπάνω ΜΔΕ?

Cheerio,

Durandal 1707

Παναγιώτη
Αν θέσεις $ f\equal{}x(1\minus{}x) ,\frac{\sigma^2}{2}\equal{}A$ τότε έχεις $ \frac{u_t}{f}\equal{}u\plus{}A(f.u_x)_x$ και νομίζω ότι δουλεύει ο χωρισμός των μεταβλητών $ u(x,t)\equal{}X(x)T(t)$ αλλιώς ο όρος $ u.u_x$ μου θυμίζει εξίσωση Burgers αλλά μάλλον θα καταλήξεις σε κάποιο πρόβλημα ιδιοτιμών

[quote="r_boris"]Παναγιώτη
Αν θέσεις $ f \equal{} x(1 \minus{} x) ,\frac {\sigma^2}{2} \equal{} A$ τότε έχεις $ \frac {u_t}{f} \equal{} u \plus{} A(f.u_x)_x$ και νομίζω ότι δουλεύει ο χωρισμός των μεταβλητών $ u(x,t) \equal{} X(x)T(t)$ αλλιώς ο όρος $ u.u_x$ μου θυμίζει εξίσωση Burgers αλλά μάλλον θα καταλήξεις σε κάποιο πρόβλημα ιδιοτιμών[/quote]

Με φοβίζουν οι συντελεστές $ x(1 \minus{} x)$ που είναι singular στα όρια του διαστήματος $ [0,1]$ και δεν ξέρω κατά πόσο συγκλίνουν τα πράγματα που προκύπτουν κλπ.

Δηλαδή, αυτό που με ανησυχεί είναι ότι ο συντελεστής της πιο υψηλόβαθμης παραγώγου είναι singular και, άρα, το να διαιρούμε με αυτόν είναι κάπως επικίνδυνο... σωστά με φοβίζει αυτό ή είμαι paranoid?

Thanks for the input! :) Cheerio,

Durandal 1707

Edit: χμμ... ίσως δε βγαίνει νόημα από την προηγούμενή μου απάντηση. Αυτό που εννοώ είναι ότι μετά το χωρισμό μεταβλητών, καταλήγουμε για το $ x$ σε ένα πρόβλημα Sturm-Liouville το οποίο όμως θα είναι ιδιάζον (λόγω των συντελεστών του). Πόσο μας ενοχλεί αυτό για την ύπαρξη και τη μοναδικότητα της λύσης?

Edit 2: και όλα να είναι καλά με το Sturm-Liouville, έχουμε και τις μη ομογενείς οριακές συνθήκες που, αν πάμε να τις ξεφορτωθούμε με το συνηθισμένο τρόπο ($ u(x) \equal{} v(x) \plus{} x$ για να γίνουν ΟΟΣ ως προς την $ v$), θα έχουμε άλλον έναν όρο στη διαφορική εξαιτίας της πρώτης παραγώγου $ \frac {\partial u}{\partial x}$ και άρα θα πρέπει να πάμε στην αντίστοιχη μή ομογενή ΜΔΕ.

Χρρρρμμμμφφφφφφφ... :(

Έχεις δίκιο
Επειδή και μένα με παίδεψε αρκετά αν βρεις κάτι στείλτο να το δούμε


Suppose you have polynomials $ P(x),Q(x)$ (over any field $ K$ of characteristic 0) such that there exists a polynomial $ R(x,y)$ with $ R(P(x),Q(x))=x$. Prove that $ \deg P$ divides $ \deg Q$ or $ \deg Q$ divides $ \deg P$.

Note: This is really beautiful, though extremely hard(it is known as the Abhyankar-Moh theorem; it has e.g. implications on the structure of the group of polynomial automorphisms of the affine plane)! You should give it a try :-) .

All I can say right now is that it's really easy if $ \deg R \equal{} 1$ :)


Prove that the straight lines $ (a+b)x + (a-b)y=2ab, (a-b)x + (a+b)y=2ab$ and $ x+y=0$ form an isosceles triangle whose vertical angle is $ 2.tan\inv{-1}(\frac{a}{b})$. Please help.Thanks a lot for your help.

[color=blue][b]the result of the vertical angle   with me is
tan^-1 (2ab / b^2 - a^2 ) !!!!!!![/b][/color] :maybe:


trigonometry
inequalities proposed
inequalities

The answer is $A = 120 , B=C=30$ but I don't know how to prove it

[quote="jarod"]Let triangle $ABC$ satisfying :
 $2\cos A\sin B \sin C + \sqrt{3}(\sin A + \cos B + \cos C )= \frac{17}{4}$
  Format triangle $ABC$ ?[/quote]
$\frac{17}{4}=2\cos A\sin B \sin C + \sqrt{3}(\sin A + \cos B + \cos C )\leq 2\cos A (\sin {\frac{B+C}{2})^2+\sqrt{3}(\sinh+\frac32-\cos A)}$
I think you can solve A by this way,but have not enough time to ensure it.

I don't think your way is ok , HT .
  It's not true if $\cos A <0$ . Am i right ? [/quote]


counting
distinguishability
combinatorics unsolved
combinatorics

An ordinary cubical dice has $ 6$ faces with numbers $ 1$ to $ 6$ written on it. If there are $ n$ such distinct dice (say of different colours) then total number of outcomes is $ 6^n$.

But if the $ n$ dices are identical(same colour,same weight,...etc.) then what is the total number of possible outcomes?

For e.g. when $ 2$ identical dice are rolled number of outcomes is $ 21$(not $ 36$).

This is the number of ways to place $ n$ indistinguishable balls into $ 6$ distinguishable urns, or $ {n \plus{} 5 \choose 5}$ (if I'm not off by one).  There are four classical "balls in urns" problems that can be stated by changing which of the balls or the urns is distinguishable and more generally something called the [url=http://en.wikipedia.org/wiki/Twelvefold_way]twelvefold way[/url].

Thanks a lot.Such a simple explanation.WOW.


After spending one-half of her money and losing one-third of the remainder, Sally was left with just enough money to buy a magazine and a candy bar. The price of the magazine is 5 times the price of the candy bar, which is 75 cents. How many dollars did Sally have initially? Express your answer as a decimal.

Sally is left with $ 4.50$, so $ \frac{1}{3}x\equal{}4.5 \implies x\equal{}13.5$.


Let $ ABCD$ be a convex circumscribed quadrilateral, let $ I$ be its incenter, let $ P$ be the intersection of its diagonals $ AC,BD$. Denote by $ I_{1},I_{2},I_{3},I_{4}$ be the the incenters of triangles $ PAB,PBC,PCD,PDA$ respectively. Prove that
(I): $ I_{1}I_{2}I_{3}I_{4}$ is a cyclic quadrilateral. denote by $ M$ its circumcenter.
(II):$ I,M,P$ are collinear if and only if $ ABCD$ is a kite-shaped or cyclic quadrilateral.
(where $ ABCD$ is called a kite-shaped quadrilateral if a pair of its opposite vertices is symmetric with respect to its corresponding diagonal.)

(I) is very easy, can anyone give a proof of problem (II)?


Hi guys I hate logs and I really need help on some of these problems

1. 256^5x = 64^(x+3)

2. 2^(9x-5) = 13

for number 2 solve the equation and give an exact solution and also an appropriate solution to four decimal places

PLEASE HELP!!! Any advice would be awesome, and please let me know the answer

[hide="‘Hint’"]For the first problem, note that both $ 256$ and $ 64$ are powers of 4. To solve, change the bases to 4, and use the exponent rules (e.g., $ (a^{x})^{y} \equal{} a^{xy}$) to end with an equality in the exponents. Finally, solve for x.[/hide]

Other way, use $ logarit$:
$ 256^{5x}\equal{}64^{x\plus{}3}$
$ \Longleftrightarrow 5xlog_{2}256\equal{}(x\plus{}3)log_{2} 64$
$ \Longleftrightarrow 5xlog_{2}2^8\equal{}(x\plus{}3)log_{2} 2^6$
$ \Longleftrightarrow 5x.8\equal{}(x\plus{}3).6$
$ \Longleftrightarrow 20x\equal{}3(x\plus{}3)$
$ \Longleftrightarrow 17x\equal{}9$ $ \Longleftrightarrow \boxed {x\equal{}\frac{9}{17}}$

Use $ logarit$:

$ 2^{9x\minus{}5}\equal{}13$

$ \Longleftrightarrow 9x\minus{}5\equal{}log_{2} 13$

$ \Longleftrightarrow 9x\equal{}5\plus{}log_{2} 13$

$ \Longleftrightarrow \boxed{x\equal{}\frac {5\plus{}log_{2} 13}{9}}$

Yes, that works, but he said that he hates logarithms, so I tried to give him as little logarithmic expressions as possible. The second problem, of course, is easily solved if you use logarithms.


Can someone remind me how to change fractions to decimals ?


One way is induction.
Another way is using trig formulae's.
And one more way is this one (which is actually induction):
$a_{0}=\sqrt{2},a_{1}=\sqrt{2+\sqrt{2}},a_{n+1}=\sqrt{2+a_{n}}$. With these notations, it must be proven that:
$\frac{\sqrt{2}}{a_{1}\cdot a_{2}\cdot ... a_{n}}=\sqrt{2-a_{n-1}}\leftrightarrow \frac{\sqrt{2}}{a_{1}\cdot a_{2}\cdot ... \cdot a_{n-1}}=\sqrt{(2-a_{n-1})\cdot(2+a_{n-1})}\leftrightarrow \frac{\sqrt{2}}{a_{1}\cdot a_{2}\cdot ... \cdot a_{n-1}}=\sqrt{4-a_{n-1}^{2}}=\sqrt{2-a_{n-2}}\leftrightarrow .....\leftrightarrow \frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}}=\sqrt{2-\sqrt{2}}\leftrightarrow \sqrt{2}=\sqrt{4-2}$
It's a backwards induction.


Find the value of each of the following:
(1) $ 2^{32}\mod 7$
(2) $ 10^{35}\mod 7$
(3) $ 3^{35}\mod 7$

I know that $ a\equiv r$ (mod m) where $ 0\leq r < m\implies a\mod r$

So I'm thinking that for (1)
$ 2^{32}\equiv r$ (mod 7)

But I don't know how to get r.

i'll do the first one as an example
(1)Firstly, notice that $ 2^{3}\equal{}8\equiv 1(mod 7)$
so that $ 2^{3k}\equiv 1 (mod 7)$ for every natural k
so $ 2^{32}\equal{}2^{3.10}.2^{2}\equiv 2^{2}\equal{}4(mod7)$

[quote="Identity"]I don't know much about Fermat's Last Theorem. I've heard the exponents have to be prime but they aren't in this case. Can you do a question as an example? Thanks heaps.

EDIT: Woops I mean Little Theorem lololol![/quote]

It's the modulus that has to be prime.  Fermat's Little Theorem states that for any prime $ p$ we have

$ a^{p}\equiv a\bmod p$.

When $ p\not | a$ we can further conclude that

$ a^{p\minus{}1}\equiv 1\bmod p$.

In this case, $ a^{6}\equiv 1\bmod 7$.


calculus
integration
function
calculus computations

Is it $\frac{d^{2}y}{dx^{2}}$ at $2$ or is that it for all $x$?  Are you asked to find $\frac{dy}{dx}$?

My directions are to solve the differential equation. lol. Sorry i forgot to include y(1)= 2. I don't know if that helps, maybe my teacher wrote the problem wrong?

Well, I'm assuming that $\frac{d^{2}y}{dx}$ is $32$ for all real $x$.

Integrate once to get that $\frac{dy}{dx}= 32x+c$. Using the condition $y'(2)=3$, we see that the constant $c$ of integration is $-61$.

Integrating again, we see that $y = 16x^{2}-61x+c$. Applying the intial condition easily yields that $c = 47$.

Thus, $y = 16x^{2}-61x+47$.

Looks like a projectile problem in feet to me.


Do AMC A and B have corresponding questions with same concept?

I know they basically are same, including the same concepts, but I wonder if the question was corresponded.

 :)

Not from my knowledge. I like the concepts more on 10A last year despite the 10B test being easier overall (for the general population, I mean). That would kind of give away what to study, too, wouldn't it?

No.  That would really give too much information--for example, you'd know that there won't be much (any?) trig on the B test, whereas there could easily be.

I thought there are no trigs at all.

Hmm..

I thought the concepts were to be same.. So the test was fair either way... :)

Hmm..

I might use detective skill on past years but...

On the AMC10, there is no trig because the "average" 10th grader has not taken trig.

As for the topics being the same, they are in general, but sometimes one topic may be emphasized more than another (such as geometry over algegra). But there is definitely no question correspondence between the A and B tests.


inequalities
algebra
polynomial
IMO Shortlist
algebra solved

Let a,b,c,d positive reals. Prove that:

[(abc+abd+acd+bcd)/4]^{1/3} \leq [(ab+ac+ad+bc++bd+cd)/6]^0.5

possible approaches:

(1) consider the properties of symmetric functions' roots
(2) apply 'mean-inequalities'

See also 'Inequalities' by Hardy, Littlewood and Polya.

Here's what I did:

Let P be the polynomial (x-a)(x-b)(x-c)(x-d)=x^4-S1x^3+S2x^2-S3x+S4 where S1, S2, S3, S4 are the symmetric sums of a, b, c, d. P has all roots positive so P' must also have all roots real and positive. P'(x)=4x^3-3S1x^2+2S2x-S3. Its roots are x1, x2, x3 all positive. What we must prove is (S2/6)^3 >= (S3/4)^2, which here is equivalent to ((x1x2+x2x3+x3x1)/3)^3>=(x1x2x3)^2, which is exactly AM-GM for the 3 numbers x1, x2, x3.

The problem was also used as A. 270 in the Hungarian Kmal journal:

http://komal.elte.hu/verseny/2001-09/A.e.shtml

comment:

We also have to consider the fact that one has to think of a limited area. One can only use problems which can be solved within the field of elementary mathematics. That's why many problems are reproposed at different math journals and various competitions.
Sometimes the problems were slightly modified. Many problems are also rediscovered without knowing that it was already published anywhere else. And that occurs a lot of times ! I feel it's rather impossible to have a complete overview of the all problems ever published in any source, e.g. Moubi mentioned that the problem 'sequence in binary expansion' is already known from a French competition though I found it in a German math journal.
As you probably know several IMO shortlist problems are rejected because they have been already used in a different  competition. Very often the problem was known from Crux Mathematoricum and Russian competitions including Lenigrad, Moscow, Russian, USSR and their IMO selection tests.
There is the same issue with the proposed IMO problems (longlist). Neither the longlist is published nor will it be given to the IMO team leaders by the IMO Problem Selection Committee, so that problems which do not make the shortlist remain confidential and can be used for other competitions. The decision was established due to the lack of good competition problems.


As this is my first time posting something on this site I don't know if this question is going to be hard or easy.

I got this question from a island wide math competition here in Puerto Rico so here it goes.(These are not the exact words to the problem)

Question:

Bob has 6 friends, too each friend he gives half of the tokens he has and one more.

Example: I have 10 tokens. To my first friend I give him half(5) and one more in total I give him 6 and im left with 4, And then continue after with 4 using the same proccess(I don't know if that is spelled right).(this is not the answer)

After talking with all of his friends he has no tokens left. How many tokens did he have to begin with?

Note: He can only have interger numbers of tokens.

Welcome to AoPS! :welcomeani:

[hide="Solution"]Work backwards:

Each time, if he has $ x$ tokens, he gives $ x\minus{}\left(\frac{x}{2}\plus{}1\right)$, or $ \frac{x}{2}\minus{}1$.

He ended up with $ 0$ tickets, so solve for $ x$ if $ \frac{x}{2}\minus{}1\equal{}0$. We get $ x\equal{}2$, so he had $ 2$ tokens before he gave some to his 6th friend.

Repeating this procedure, we can determine that he originally had $ \boxed{126}$ tokens.[/hide]

repeated fraction method can solve this.


modular arithmetic
induction
combinatorics proposed
combinatorics

A convex $n$-gon $A_1A_2\dots A_n$ $(n>3)$ is divided into triangles by non-intersecting diagonals.
For every vertex the number of sides issuing from it is even, except for the vertices
$A_{i_1},A_{i_2},\dots,A_{i_k}$, where $1\leq i_1<\dots<i_k\leq n$. Prove that $k$ is even and
\[n\equiv i_1-i_2+\dots+i_{k-1}-i_k\pmod3\]
if $k>0$ and
\[n\equiv0\pmod3\mbox{ for }k=0.\]

Note that this leads to generalization of one recent Tournament of towns problem about triangulating of square.

I think the case $k=0$ has been posted before: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=cycle&t=29422[/url].

For the general case, we could do something like this (pretty similar to what was done in those other threads; this is just a sketch):

First of all, color the triangles chessboard-style (this means that we color the triangles with two colors so that no two triangles sharing a side are colored the same). We can immediately show this is possible by induction on the number of sides of the polygon. We can now orient each black triangle, since the sides of the black triangles do not interfere with one another. After we've done that, we can complete the orientation of the diagonals so that all triangles (both black and white) are oriented. We now observe that the sides between $A_{i_1}$ and $A_{i_2}$ (going in the direction in which the vertices of the polygon are numbered) all have the same color, which is the same as the color of those between $A_{i_3}$ and $A_{i_4}$, and so on. The other sides have the other color.

Because each triangle is oriented, each cycle has border $0$ (the border is defined as the number of sides you traverse in the right direction as you go around the cycle $-$ the number of sides you traverse in the wrong direction $\pmod 3$). However, the cycle going around the sides of the polygon has border $0$ (like all cycles), but also (modulo $3$) $2(i_2-i_1+i_4-i_3+\ldots+i_k-i_{k-1})-n$, and the conclusion follows.


Let S_n be sum obtained by adding largest odd divisors of each integer x with 1=<x=<2^n where n is a non-negative nteger. Find a formula for S_n.

Bomb

We have $S_n = 1+3+5+ \cdots (2^n-1) + S_{n-1}$ because the largest odd divisor of $2p$ is the same as the one of $p$.
Thus $S_n = S_{n-1} + 2^{2(n-1)} = S_{n-1} + 4^{n-1}.$
Since $S_0=1$ it easily follows that $S_n = \frac {4^n + 2} 3.$

Pierre.


Dear all
I need a collection of problems and solutions for The Vietnam Mathematics Olympiad in years 1990 to 2008. Could you help me? Thank you very  much!

[quote="kmath93"]You could find the book named "Các bài thi Olympic Toán Trung học Phổ thông Việt Nam 1990-2006), NXBGD" in "Tủ sách Toán học & Tuổi trẻ".[/quote]
Where do you live ??


summer program
Mathcamp
probability
LaTeX

Edit'd

[hide]
Ah, yes, the original problem doesn't make much sense, does it?

The only circumstances in which there would not be two culprits is if the same person did it both times.

The probability of Delong doing it both times is $\frac{1}{3}^{2}=\frac{1}{9}$.

The probability of Avive doing it both times is $\frac{1}{4}^{2}=\frac{1}{16}$.

The probability of Yichen doing it both times is $\frac{5}{12}^{2}=\frac{25}{144}$.

The probability that two of them conspired together, therefore, is $1-(\frac{1}{9}+\frac{1}{16}+\frac{25}{144})=\frac{47}{72}$.

Sorry, I'm not real proficient in LaTeX.
[/hide]

[size=75][color=darkred]Changed it to LaTeX
-nebula42[/color][/size]

The problem is incorrect. Since both Eric and I slept through the alarm, neither of us could have opened the emergency exit. The probability that Delong (nealth) opening the exit is $\frac13$ and the probability that a hyper Avive opening it is $\frac{1}{4}$. The probability of Yichen opening it is now $\frac{5}{12}$. Now what's the answer?

WTF??? I was downstairs and slept through it too, so it must have been the person who ran down stairs: Delong.

I was sleeping too. After investigation, a cheerleader opened the door and put everyone into a panic so that she had the chance of abducting Macky.

The last three posts were spam, please restrict such comments to private messages and the sort, or at least add some real content/do the problem in the same post.

Ooo, Mathcamp is going on right now?

[quote="nealth"]Some people at Mathcamp think it's extremely funny to open an emergency exit and make the alarm go off  :mad: . This happened last night and this afternoon. The probability of Eric playing this serious joke is $\frac{1}{3}$, the probabiliy of Brian is $\frac{1}{5}$ and the probabilty of Yichen is $\frac{1}{6}$. What is the probability that two of them conspired together?[/quote]
[hide]The probabilities are not adding up to 1 in the original question. So I am giving the answer using the probabilities given by Waxing Rhapsodic. Solution is using a different approach than Waxing R. The complementary approach used by Waxing R. is obviously simpler.

There are 6 possible pairs:

(1) Eric opens first and Brian opens second. This gives us, $p_{1}= \frac{1}{3}\times\frac{1}{4}= \frac{1}{12}$

(2) Brian opens first and Eric opens second. This gives us, $p_{2}= \frac{1}{4}\times\frac{1}{3}= \frac{1}{12}$

(3) Eric opens first and Yichen opens second. This gives us, $p_{3}= \frac{1}{3}\times\frac{5}{12}= \frac{5}{36}$

(4) Yichen opens first and Eric opens second. This gives us, $p_{4}= \frac{5}{12}\times\frac{1}{3}= \frac{5}{36}$

(5) Brian opens first and Yichen opens second. This gives us, $p_{5}= \frac{1}{4}\times\frac{5}{12}= \frac{5}{48}$

(6) Yichen opens first and Brian opens second. This gives us, $p_{6}= \frac{5}{12}\times\frac{1}{4}= \frac{5}{48}$

So the answer is $p_{1}+p_{2}+p_{3}+p_{4}+p_{5}+p_{6}= \frac{47}{72}$.[/hide]


1) The set of all solutions to the system
$ \begin{flushleft}x\geq0\\ y\geq0\\ 2x+5y\geq10\\ x+3y\leq9\end{flushleft}$
is a quadrilateral region.  How many units, to the nearest whole number, are in its perimeter?

ANSWER: [hide]20[/hide]

2) How many three-digit numbers are there such that no two digits differ by more than 3?

ANSWER: [hide]270[/hide]

3) How many even five-digit numbers have the property that the sum of the first four digits is the units digit?

ANSWER: [hide]200[/hide]

[hide="2"]i think, if no 2 digits differ by >3, then they must all or 2 of the same digits, be consecutive one digit #s or consecutive and one greater.(not including 0 yet) there are 7 triplets:123, 234, 345, 456, 567, 678, and 789 there are 3! ways each can be arranged, making 42. then, you have the 0s: 012. it can only be arranged in 4 ways, since 0 cant be a first #. that makes 46 total

now, the all the same ones: 111, 222, 333, 444, 555, 666, 777, 888, 999

then the one with range=3

124=6 ways
134=6 ways
235=6 ways
245=6 ways
346=6 ways
356=6 ways
457=6 ways
467=6 ways
568=6 ways
578=6 ways
679=6 ways
689=6 ways
013=4 ways
023=4 ways

so far, we have 135 total.

next, the 2 digit same

11...(0-4 but not 1)10
22...(0-5 but not 2)13
33...(0-6 but not 3)16
44...(1-7 but not 4)18
55...(2-8 but not 5)18
66...(3-9 but not 6)18
77...(4-9 but not 7)15
88...(5-9 but not 8)12
99...(6-8)9
00...(1-3)3

that makes 267. [/hide]


Find the smallest positive whole number that does not contain the digit 9, but it is divisible by 999.

the was just on the basics.
[hide]I believe 111888.[/hide]

Although that may be the answer, there is a logical solution to the problem.

The logical solution is to check the first multiple of 999, which is obviously 999, but that has 9's in it. $2(999)=1998,$ which got rid of the 9 in the units digit, but you have to do that for the tens and the hundreds still. Every proceeding multiple of 999 will make the units digit one lower, so $12(999)=10988.$ To get rid of the 9 in the hundreds place, it needs to be $112(999)=111888.$


inequalities
integration
induction
inequalities proposed

Given: $ 1\plus{}\int_0^xf(t)dt\le f(x).$

Let $ f_n(x): \equal{}\sum_{i\equal{}0}^n\frac{x^i}{i!}$. Then $ f_n(x)\equal{}1\plus{}\int_0^xf_{n\minus{}1}(t)dt$, thus by induction trivially $ f_n(x)\le f(x)$ for all $ n\in\mathbb N$. :)


Prove that \[ \frac{(y\plus{}z\minus{}x)^{3}}{x}\plus{}\frac{(z\plus{}x\minus{}y)^{3}}{y}\plus{}\frac{(x\plus{}y\minus{}z)^{3}}{z}\geq x^{2}\plus{}y^{2}\plus{}z^{2}\] for all $ x,y,z>0.$

[quote="bilarev"]Prove that
\[ \frac{(y\plus{}z\minus{}x)^{3}}{x}\plus{}\frac{(z\plus{}x\minus{}y)^{3}}{y}\plus{}\frac{(x\plus{}y\minus{}z)^{3}}{z}\geq x^{2}\plus{}y^{2}\plus{}z^{2}\]
for all $ x,y,z > 0.$[/quote]
$ \sum_{cyc}\frac{(y\plus{}z\minus{}x)^{3}}{x}\geq x^{2}\plus{}y^{2}\plus{}z^{2}\Leftrightarrow\sum_{cyc}(x^{4}y\plus{}x^{4}z\plus{}3x^{3}y^{2}\plus{}3x^{3}z^{2}\minus{}6x^{3}yz)\geq0,$
which true by Schur.

There is a little error, there is a $ 6$ that should be an $ 8$, but well, it is the same, it is true by Schur or by Muirhead
$ \sum_{cyc}\frac{(y\plus{}z\minus{}x)^{3}}{x}\geq x^{2}\plus{}y^{2}\plus{}z^{2}\Leftrightarrow\sum_{cyc}(x^{4}y\plus{}x^{4}z\plus{}3x^{3}y^{2}\plus{}3x^{3}z^{2}\minus{}8x^{3}yz)\geq 0$.

by chebyshev
$ LHS\geq\sum (x\plus{}y\minus{}z)^{2}\geq\sum x^{2}$  :)

because if $ x\geq y\geq z$ then $ (x\plus{}y\minus{}z)^{2}\geq (x\plus{}z\minus{}y)^{2}\geq (z\plus{}y\minus{}x)^{2}$ and $ \frac{x\plus{}y\minus{}z}{z}\geq\frac{x\plus{}z\minus{}y}{y}\geq\frac{y\plus{}z\minus{}x}{x}$


Let $ 0\leq a,b,c\leq3$,and $ a\leq b \leq c$.Prove that:
$ (a\minus{}b)(a^2\minus{}9)\plus{}(b\minus{}c)(c^2\minus{}9)\plus{}(a\minus{}c)(b^2\minus{}9)\leq36$

[quote="Inequalities Master"]Let $ 0\leq a,b,c\leq3$,and $ a\leq b \leq c$.Prove that:
$ (a \minus{} b)(a^2 \minus{} 9) \plus{} (b \minus{} c)(c^2 \minus{} 9) \plus{} (a \minus{} c)(b^2 \minus{} 9)\leq36$[/quote]
My solution :)

Very nice and easy proof, can_hang2007. :D

It seems that for you almost every inequality is just a warm-up exercise .... :)


The Lily family plans to build an outdoor garden with a
powerjet pump that will pump 375 gallons of water per hour.
How many quarts of water will be pumped per minute?

375 gallons = 375*4 quarts=1500
Divide by 60 to get minutes, 1500/60=25.

The rate is $ \frac {375 \text{ gallons}}{1 \text{ hour}}$
The rate in quarts/minutes is:
$ \frac {375\times 4 \text{ quarts}}{1\times 60 \text{ minutes}} \equal{} \frac {375}{15} \equal{} 25$ quarts/minute.

EDIT: Beaten to it.
EDIT: EDIT: Thanks AIME.

[quote="dragon96"]The rate is $ \frac {375 \text{ gallons}}{1 \text{ hour}}$
The rate in quarts/minutes is:
$ \frac {375\times 4 \text{ quarts}}{1\times 60 \text{ minutes}} \equal{} \frac {375}{15} \equal{} 25$ quarts/minute.

Beaten to it.[/quote]

Fixed your unparseable or "dangerous" latex.


inequalities
limit
real analysis
real analysis solved

The sequence $\{a_n\}$ satisfies the system of inequality as follows.

\[2004a_{n+1}-2003a_n<2005\]
\[2002a_{n+1}-2001a_n>2005\]

Evaluate $\lim_{n\to\infty} a_n$.

Let $U_n = a_{n+1} - a_n$.
Then $a_{n+1} + 2003U_n < 2005$ and $a_{n+1} + 2001U_n > 2005$, which is $-a_{n+1} - 2001U_n < -2005$.
Thus $2U_n < 0$ which means that the sequence $(a_n)$ is decreasing.

In another hand $a_{n+1} > 2005 - 2001U_n > 2005$, thus $(a_n)$ is bounded from below. Therefore $(a_n)$ is convergent to some real number $L$. It follows that $(U_n)$ has limit $0$.
Thus, from the initial inequalities, taking limits leads to $L \leq 2005$ and $L \geq 2005$, so that $(a_n)$ convreges to $L=2005$.

Pierre.

For a second argument, let $b_n=a_n-2005.$ The inequalities become:

$\frac{2001}{2002}b_n<b_{n+1}<\frac{2003}{2004}b_n.$

This bounds $b_n$ between two geometric sequences, both of which tend to zero.

Incidentally, this shows that initial value, $b_1$, must be positive (that is, $a_1>2005$) or the inequalities cannot be met.

Thank you!Farewell 2004!

Have a nice year,everyone!

A Happy new Year in 2005!


A ball or mass $ m$ is released from above a vertical spring. It attains a velocity $ v$ when it reaches the spring and compresses it by a distance of $ y$ before the spring sends the ball back up. What is the average force the spring exerts on the ball during the compression? Do we have to include the potential energy of $ mgy$ lost by the ball during the compression?

Of course you have to include the potential energy $ mgy$ of the ball in your calculation :)


Where can I find a *qualified* person to help me with some first year calculus problems?

I don't know if he teaches calculus but the admin [url=http://s215234424.onlinehome.us/index.php]here[/url] is a tutor and you can ask him about it.


Find all integers $a$ for which $x^3 -x+a$ has three integer roots.

Call $ x_1,x_2,x_3$ is three integer root of the polynomial.
Apply viete law we has:
$ x_1\plus{}x_2\plus{}x_3\equal{}0$
$ x_1x_2\plus{}x_2x_3\plus{}x_3x_1\equal{}\minus{}1$
Imply that :
$ x_1x_2\minus{}(x_1\plus{}x_2)^2\equal{}\minus{}1$
$ \Leftrightarrow x_1^2\plus{}x_1x_2\plus{}x_1^2\equal{}1$
$ \Rightarrow x_1x_2x_3\equal{}0$
$ \Rightarrow a\equal{}0$
With $ a\equal{}0$ then the equation has solution $ x\in\{\minus{}1,1,0\}$


geometry
rectangle
inequalities
geometry unsolved

It is given rectangle ABCD with sides 1, 2 and a point M inside it. What is minimal value of the expression AM+BM+CM+DM and when it is reached?

[quote="borislav_mirchev"][color=darkred]Prove that for any point $ X$ which belongs to the interior of the given rectangle $ ABCD$ exists

the inequality $ AX + BX + CX + DX\ge 2\cdot AC$ with equality iff $ X\in AC\cap BD$ .[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Define the projections $ M$ , $ N$ , $ P$ , $ Q$ of $ X$ on $ AB$ , $ BC$ , $ CD$ , $ DA$ and  denote

$ \|\begin{array}{c} AB = CD = a\ ,\ AD = BC = b \\
 \\
AM = DP = QX = x\ ,\ MB = PC = XN = y \\
 \\
AQ = BN = MX = u\ ,\ QD = NC = XP = v \\
 \\
(\ x + y = a\ ,\ u + v = b\ )\end{array}$ . From the [b]Minkowki's inequality[/b]

$ \boxed {n\in\mathcal N^*\ ,\ (\forall )\ k\in\overline {1,n}\ ,\ x_k\ge 0\ ,\ y_k\ge 0\implies\sqrt {\sum_{k = 1}^na_k^2} + \sqrt {\sum_{k = 1}^nb_k^2}\ge \sqrt {\sum_{k = 1}^n(a_k + b_k)^2}}$

obtain : $ XA + XB + XC + XD =$ $ \sqrt {x^2 + u^2} + \sqrt {y^2 + u^2} + \sqrt {y^2 + v^2} + \sqrt {x^2 + v^2} =$

$ (\sqrt {x^2 + u^2} + \sqrt {y^2 + v^2}) + (\sqrt {x^2 + v^2} + \sqrt {y^2 + u^2})\ge$ $ 2\sqrt {(x + y)^2 + (u + v)^2}$ ,

i.e. $ \sum XA\ge 2\cdot AC$ . We have equality iff $ X\in AC\cap BD$ , i.e. $ x = y = \frac a2$ and $ u = v = \frac b2$ .

[b][u]Remark.[/u][/b] Can extend you the proposed inequality to a given [b]parallelepiped[/b] ?![/color]

:)  Very good Virgin Nicula! ... sometimes I don't think at all ... I'm sorry. I want to show my proof at this fact for any convex quadrilateral. It is well known that the shortest distance between two points is a stright line (let our points are A, C) so the  sum AM+CM reaches its minimum when M belongs to AC. On the other side BM+DM belongs its minimum when M belongs to BD. AM+CM and BM+DM achieve its minimum simultaneously when M belongs to AC and BD simultaneously but only such a point is intersection point of AC and BD.

Is my proof correct?
I think Virgil Nicula's solution is very beautiful, but a little complicated.
Yes this statement may be extended for a parallelepiped.

[color=darkblue]Indeed [b]SpongeBob[/b] and [b]Borislav_Mirchev[/b],[/color] [color=darkred]HAHAHAHAHAHHA !!![/color]

[color=darkblue]I presented with the best of intentions this curious proof for to show, even with the risk to die with laughter, what has happened sometimes when someone knows too much ! [/color]

[color=darkred]See an curious proof of the little Fermat's theorem at[/color]

http://arxiv.org/PS_cache/arxiv/pdf/0801/0801.0805v1.pdf

I respect you Virgil Nicula - you are really a man with lots of knowledge. It is normal for a man who is dealing with very difficult problems to think in such a manner. Your way of thinking is more "universal".
When I posted this topic I had no idea how to solve the problem ... the solution came in my mind a little more time after you sent the solution. Problems I'm posting recently are easier than ususal - the reason is that I've no time to solve difficult problems and maybe no one will solve very difficult problems ... for example - see my last post with systems - no one is still solved. I'm surprised - these systems doesn't seems to be so difficult or boring.

@Virgil I am sorry if my "hahahaha" sounded like dissrespect, I perfectly understand you; The same thing happens to all of us; If you treat some problem like a hard one, you get long solution. It's enough that someone tells you that problem is easy, and you'll solve it in two lines :)
Btw, I know you from this forum and I am aware of your grate problemsolving skills :)

Bye


Let $ AS$ be a $ A$-symmedian in a triangle $ ABC$ , where $ S\in BC$ . Denote the harmonical conjugate $ D$ of $ B$

w.r.t. the pair $ \{S,C\}$ . Consider a point $ P\in (AD)$ . Prove that $ \angle BPD\equiv\angle BAC\ \Longleftrightarrow\ \angle ABP\equiv\angle CAD$ .

Let $ (O)$ be the circumcircle of $ \Delta ABC$.Let $ T$ be the intersection of the two tangents at points $ B,C$ to $ (O)$.obviously $ A,S,T$ are collinear and $ BOCT$ is concyclic.Let $ W$ be the circle passing through $ B,O,C,T$ and let $ F$ be the intersections of $ W$ and $ AS$.we can easily see that $ \angle BFA \equal{} 180^\circ \minus{} \angle BAC$ and also we have $ \angle ABF \equal{} \angle FAC$.
at first let $ P$ be a point on $ AD$ s.t.
$ \angle BAC \equal{} \angle BPD \implies \angle BPA \equal{} 180^\circ \minus{} \angle BAC \implies \angle BPA \equal{} \angle BFA$.
so $ BAPF$ is concyclic and we have $ \angle ABP \equal{} \angle ABF \minus{} \angle PBF \equal{} \angle FAC \minus{} \angle FAP \equal{} \angle PAC$.
so we are done.
the reverse is done exactly in the same way.
(if i'm not wrong,we don't need $ D$ to be the harmonical conjugate.just a point on $ BC$ for which $ AD$ intersect $ (AFB)$.
maybe i'm missing sth obvious...)

I'm sorry. The enunciation of my proposed problem is wrong. It is correctly thus :
[quote]Let $ AS$ be a $ A$-symmedian in a triangle $ ABC$ , where $ S\in BC$ . Denote the harmonical conjugate $ D$ of $ B$

w.r.t. the pair $ \{S,C\}$ . Consider a point $ P\in (AD)$ . Prove that $ \angle BPD\equiv\angle BAC\ \Longleftrightarrow\ \angle CAS\equiv\angle CPD$ .[/quote]


A few days ago, I tried doing a problem in MOSP 2002 and I found a beautiful synthetic proof. Here is my generalization :
     Let $k$ be a real number and $ABC$ be a triangle. Call $H$ and $O$ respectively the orthocenter and circumcenter of $ABC$. Let $K$ be a point lying on OH such that $\frac{KO}{KH} = -\frac{1}{2k}$. Let $A'$, $B'$, $C'$ be points lying on $OA$, $OB$, $OC$ such that $\frac{OA'}{OA} = \frac{OB'}{OB}= \frac{OC'}{OC} = k$. Let $A_1$ be the intersection of $BC$ and the line which is perpendicular to $OA$ at $A'$. Similarly, we obtain $B_1$, $C_1$.
     Prove that $A_1$, $B_1$, $C_1$ lye on a line which is perpendicular to $OK'$, where $K'$ is the isogonal conjugate of $K$ wrt $ABC$.
   1) The problem in MOSP is a special case when $k=\frac{1}{2}$.
   2) Another special case when $k = 1$ is well known. $K$ at that time is the centroid and $K'$ is the Lemoine point of $ABC$. Hence $OK$ is the Brocard axis of $ABC$. $A_1$ at that time is the intersection of $BC$ and the tangent of $(O)$ at$ A$ and $BC$.
   I 'll post my solution later for you to think about it. Anyway, it 's worth doing. :D

Oh, that means "with respect to"
 A point P' is called an isogonal conjugate point of P wrt ABC if $(P'A, AB) = -(P'A,AC),(P'B, CB) = -(P'B,AB)$ , where (a,b) is the direct angle of the two lines a and b (modulo $\pi$)

American Mathematical Monthly Advanced Problem 3926, by V. Thbault, Solution in AMM 1941 pages 281-283 could help.

See the attached PDF file.

No time to say more.

  Darij


Find all values of n or prove that there are an infinite amount of n values such that there is an integer x value greater than 1 that makes

$ \sqrt {\displaystyle\sum_{i \equal{} 0}^n x^i}$

an integer.

EDIT: oops forgot to specify the domain of x haha

well obviously there are infinitely many values of $ n$ that work: take $ n$ to be one less than a square and $ x\equal{}1$  :P

Haha, I forgot to specify what x could be.

Sorry about that. Fixed.

well now it's a hard problem, and it's definitely not Pre-Olympiad  :lol:

Oh, sorry about that. I had no idea how hard the problem was so I kinda just posted it somewhere. :blush:

Even so, I don't even know where to begin with this problem haha.

post it in the Number Theory section of the Olympiad forum if you haven't already.


[quote="cooley"]Let G be a group and let G' denote the subgroup of G generated by all commutators. Let N be a normal subgroup of G. Prove that G/N is abelian iff G' is a subset of N.

We know that G' is normal in G and G/G' is abelian. Is the solution a special consequence of these two facts?[/quote]

I let $C$ denote the commutator subgroup. Recall, that a "commutator" is an element of the form $aba^{-1}b^{-1}$. We want to show the set [b]generated[/b] by all commutators forms a group. (The actual set of commutators generally doesnt). And where $a,b \in G$

1)I think is is easy to see how that the set of all elements generated by the commutators is a subgroup. (To find inverse just reverse the elements and negate exponents).

2)We now show that $C$ is a normal subgroup of $G$. There are several equivalent statements to what it means "normal" one, which is best here, states $g^{-1}xg \in C$ for all $x\in C$ and $g\in G$.
Say $x=aba^{-1}b^{-1}$ then,
$g^{-1}(aba^{-1}b^{-1})g=(g^{-1}aba^{-1})(e)(b^{-1}g)$
$=(g^{-1}aba^{-1})(gb^{-1}bg^{-1})(b^{-1}g)$
$=[(g^{-1}a)b(g^{-1}a)^{-1}b^{-1}][bg^{-1}b^{-1}g] \in C$
Because, $(g^{-1}a)b(g^{-1}a)^{-1}b^{-1}\in C$ and $[bg^{-1}b^{-1}g]\in C$ so certainly their product.

This shows that if $x$ is a single commutator then the statement $g^{-1}xg \in C$ is true. But what if $x\in C$ is a product of more than one commutator, that is,
$x=c_{1}c_{2}...c_{n}$
Then,
$g^{-1}xg=g^{-1}(c_{1}c_{2}...c_{n})g = g^{-1}c_{1}(e)c_{2}(e)...(e)c_{n}g^{-1}$
$=(g^{-1}c_{1}g)(g^{-1}c_{2}g)...(g^{-1}c_{n}g)$
But we already shown above that,
$g^{-1}c_{i}g\in C$ for all $1\leq i\leq n$.
So then the product of all of them is in $C$ as well.
Thus,
$C$ is a normal subgroup.

The other part of the theorem $G/H$ is abelian if and only if $C\leq C$ is quite obvious.


Let $ a, b, c$ positive real numbers such that $ a\plus{}b\plus{}c \equal{} 1$. Prove or disprove that

$ \sum\frac{3a^5 \plus{} 2a^4 \plus{} 3b^5}{a^3 \plus{} b^3}\ge 2$ .

 :(

[quote="Marius Mainea"]Is true!

 ${ \sum\frac {3(a^5 + b^5)}{a^3 + b^3}}\ge\frac {3}{2}\sum{\frac {(a^3 + b^3)(a^2 + b^2)}{a^3 + b^3}} = 3\sum a^2\ge(a + b + c)^2 = 1$

 and

  $ \sum{\frac {a^4}{a^3 + b^3}\ge\frac {1}{2}}$ (is a old and hard problem)[/quote]

-----------
Dear [b]Marius[/b] you can render explicit the demonstration of the inequality:
$ \sum{\frac {a^4}{a^3 + b^3}\ge\frac {1}{2}}$
thanks


[quote="math10"]Let $ f$ be a function on $ [0;1]$ such that:
 $ f(0) \equal{} f(1) \equal{} 1$ and $ |f(a) \minus{} f(b)| < |a \minus{} b|$ for all $ a \not \equal{} b$ in the intevarl $ [0;1]$.
Prove that: $ |f(a) \minus{} f(b)| < \frac1{2}$[/quote]

Let max. value of $ f(x)$ attains when $ x\equal{}a$ and min. value attains when $ x\equal{}b$.

So now the statement $ \Leftrightarrow f(a)\minus{}f(b)<\frac{1}{2}$.And WLOG, $ a \ge b$.

Assume that the statement is not true, then $ a\minus{}b>f(a)\minus{}f(b) \ge \frac{1}{2}$
$ 1\equal{}(1\minus{}a)\plus{}(a\minus{}b)\plus{}(b\minus{}0)$
$ \Rightarrow \frac{1}{2}>|1\minus{}a|\plus{}|b\minus{}0|>[f(a)\minus{}f(1)]\plus{}[f(0)\minus{}f(b)]\equal{}f(a)\minus{}f(b) \ge \frac{1}{2}$, contradiction, done.

[quote="stephencheng"] $ a\in [0,1]$, $ |f(a) \minus{} f(0)| < |a \minus{} 0| \le 1$, so $ 0 < f(a) < 2$ for all $ a \in [0,1]$ . So there are max.  and min. values.[/quote]

 :blush: You're right. There are bounds, but you cant say that these bounds may be reached.

[quote="stephencheng"][quote="pco"][quote="stephencheng"] ...Let max. value of $ f(x)$ attains when $ x \equal{} a$ and min. value attains when $ x \equal{} b$....[/quote]

Since $ f(x)$ is not said to be continuous, we dont know if there is a maximum and a minimum values.[/quote]

$ a\in [0,1]$, $ |f(a) \minus{} f(0)| < |a \minus{} 0| \le 1$, so $ 0 < f(a) < 2$ for all $ a \in [0,1]$ . So there are max.  and min. values.[/quote]
I think can not prove the existence maximum and a minimum values so.
conditions $ |f(a)\minus{}f(b)|<|a\minus{}b|$ is to prove continuous of function.So solution of stephencheng is right.

[quote="math10"] ... conditions $ |f(a) \minus{} f(b)| < |a \minus{} b|$ prove continuous of function...[/quote]

 :blush:  You're right (and stephencheng too).
Sorry.


So first of all, about what score would i want to get on the AIME if i get a 132 on the AMC 10?

Second, how do u get those little things on the side of ur posts with like Riemann or Poincare???

The second question:More posting increases your rank.
The first:I don't know much, but I think you should aim for like 8 on the AIME to qualify by floor. A 10 will definitely qualify.

Actually, the AIME score you should aim for is different depending on your grade.  10th and 9th grade can qualify by the floor, which hopefully isn't higher than 8 this year(last year was 6, I think?) but I'm quite sure that there are stricter standards for juniors or seniors.

Well, if he's 11th or 12th grade, he can't take the 10, so...

Depends on difficulty of the AIME questions.. 8 would almost definitely make you into USAMO.

Unless they decide to restrict access this year like in 2005, which hopefully they won't.  In 2005 the cutoff was 9.

:o They better not.. after allowing the AoPS to sponsor 200 additional spots.

Could you explain what you mean by AoPS sponsoring 200 spots? :huh:

AoPS sponsored the expansion of the number of USAMO takers (thats what spots referred to) from approximately $ 250$ in 2005 to approximately $ 500$ in 2007. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1990663024&t=70669]thread[/url]

If the cutoff is 9 again (which it will most likely not), that would probably mean that the AIME would be easier than normal, so it might not actually be that bad.

[i]Edit[/i]: mountain.dew, it does? Where?  :P

Hmm.. [url]http://www.artofproblemsolving.com/Wiki/index.php/Current_events[/url] says it's 500 while the Wiki [url]http://www.artofproblemsolving.com/Wiki/index.php/USAMO[/url] says it's 400.


$(0<x,y<1)$

prove $x^y+y^x>1$

plz help cudnt get that

It is somewhere on this website. Please search.

Just use Bernouly's Inequality to get the result ...

Davron


The first and thirteenth terms of an arithmetic sequence are 5 and 29, respectively. What is the fifteth term?

The question is asking for the fiftieth (I think they typoed there but the answer on FTW is the fiftieth term) term.  Since $d=2$, then $a_{50}=5+2(49)=5+98=\boxed{103}$.

[hide=My Strategy]1st term=5
13th term=29
I use this strategy: $5-5$ and $29-5$
1st term (-5) is 0
13th term (-5) is 24
So terms go up by $\boxed{2}$
So 50th term is 98 and add the 5 we had subtracted.
Ans: $\boxed{103}$[/hide]


Three people A, B and C play a tournament of table-tennis. After each match, the winner plays the person who didn't play the last time. At the end of the tournament, A has played $34$ matches, B has played $47$ and C has played $57$.

(a) Determine, with proof, who lost the fourth match.

(b) Show that we do not have enough information to determine who won the fourth match.

[hide]There were $\frac{34+47+57}{2}=69$ total matches.  Since C played in $57$ of them, there were $69-57=12$ matches between A and B.  Therefore, A played $34-12=22$ matches against C, and B played $47-12=35$ matches against C.  Since B and C played 35 matches out of 69, and no consecutive games can have the same 2 players, all of the odd-numbered games were between B and C.  This means that A participated in all of the even-numbered games.  Furthermore, since A didn't play in the next match after each of these (because it was an odd-numbered match and was therefore between B and C), A lost every even-numbered match (including match 4).  Since A played games against both B and C, there is no way to know which of them was playing against A in that game.[/hide]


calculus
derivative
analytic geometry
conics
ellipse
graphing lines
slope

Let (a,b) be the coordinates of a point on the ellipse
x^2/36 + y^2/16 = 1
with a,b ≠ 0 and let L be the line tangent to the ellipse at (a,b) and let the slope of L be m.

a. In terms of a and b, what is the slope of L?
b. Which point on the ellipse has a tangent line parallel to L?
c. Which points on the ellipse have a tangent line perpendicular to L?
d. Which points on the ellipse have a tangent line with slope −m?
e. Which points on the ellipse have a tangent line with slope 1m?

this is what I have done so far.
a.  x/18 + yy'/8 = 0 ,  so
     y ' = -8x/18y  or   m = -8a/18b
     then L = -8a/18b(x - a) + b

I am not sure how to do the rest, all the different variables are confusing.  I was given 5 of these type problems to do and told that I might see one on the next test.  If you could, please show me how to do this problem so I can use it as a guide to solve the other 4 problems.

Thankyou
Gary

$b$ and $c$ are easiest to think of by graphing an ellipse.

$d$ and $e$ just require you to solve the system

$m =-\frac{8a}{18b}$
$\frac{a^{2}}{36}+\frac{b^{2}}{16}= 1$

(Remember, $(a, b)$ is on the ellipse.)

part e should read. Which points on the ellipse have a tangent line with slope 1/m?


geometry
incenter
trigonometry
circumcircle
parallelogram
geometry proposed

just a few computations show that:
$|MI|=|MT|\leftrightarrow |MI|^{2}=|MT|^{2}\leftrightarrow r^{2}+\left( \frac{b-c}{2}\right)^{2}=R^{2}(1-\cos{A})^{2}\leftrightarrow (\cos{A}+\cos{B}+\cos{C}-1)^{2}=(1-\cos{A})^{2}+(\sin{B}-\sin{C})^{2}\leftrightarrow$
$(1-\cos{A})(\cos{B}+\cos{C}-1)=0$ :wink:

[color=darkred][b]Lemma[/b] (well-known or prove easily). Let $ABC$ be a triangle for which denote the incircle $C(I,r)$, the orthocenter $H$,
the middlepoint $M$ of the side $[BC]$ and the intersection $S\in AH\cap MI$. Then there is the relation $AS=r\ .$[/color]

[color=darkblue][b]Proof of the proposed problem (in the case of an acute triangle).[/b] Are well-known the relations $OM=R\cos A$ and $\cos A+\cos B+\cos C=1+\frac{r}{R}\ \ \ (*)\ .$
Denote the circumcircle $C(O,R)$. Therefore, $MI=MT$  $\Longleftrightarrow$ $\widehat{SAO}\equiv\widehat{SMO}$, i.e. the quadrilateral $AOMS$ is parallelogram $\Longleftrightarrow$ $\boxed{\ OM=AS\ }$ $\Longleftrightarrow$
(using the first relation $*$ and the above lemma) $R\cos A=r$ $\Longleftrightarrow$ $\boxed{\ \cos A=\frac{r}{R}\ }$ $\Longleftrightarrow$ (using the second relation $*$) $\boxed{\ \cos B+\cos C=1\ }\ .$

[b]Remark.[/b] $MI=MT\Longleftrightarrow OA\parallel IM\Longleftrightarrow \cos A=\frac{r}{R}$ $\Longleftrightarrow$ $\cos B+\cos C=1$ $\Longleftrightarrow$ ${\frac{a}{b+c}+\frac{r}{R}}=1\ .$[/color]

[quote="Omid Hatami"]$M$ is midpoint of side $BC$ of triangle $ABC$, and $I$ is incenter of triangle $ABC$, and $T$ is midpoint of arc $BC$, that does not contain $A$. Prove that \[\cos B+\cos C=1\Longleftrightarrow MI=MT\] [/quote]

Sorry I have a solution at

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=640832#640832[/url]

But it's in Spanish  :oops:

And as $IO\parallel BC$ the conclusion follows

This is the solution of a my student:
We have $ \cos B \plus{} \cos C \equal{} 1$
$ \Longleftrightarrow \cos\angle BTY \plus{} \cos\angle CTY \equal{} 1$
$ \Longleftrightarrow \frac {TB^2 \plus{} TI^2 \minus{} IB^2}{2TB.TI} \plus{} \frac {TC^2 \plus{} TY^2 \minus{} IC^2}{2TI.TC}\equal{}1$
$ \Longleftrightarrow IB^2 \plus{} IC^2 \equal{} TB^2 \plus{} TC^2$
$ \Longleftrightarrow IM^2 \equal{} TM^2.\Box$
Note that $ TB \equal{} TI \equal{} TC$.

Here's a more intuitive approach:

Let the projections of $B,C,M$ onto $IT$ be $B',C',M'$. So obviously $M'$ is the midpoint of $B'C'$. By well-known relation $TI=TB=TC$, $\angle BTI = C$, $\angle CTI = B$, we have
$\cos B+\cos C=1 \Leftrightarrow B'T=C'I \Leftrightarrow M'$ midpoint of $B'C'$ and hence midpoint of $IT \Leftrightarrow MT=MI$


What are the last 3 digits of $ 111^{89}$?

hello, we have $ 111^{89}\equiv 391 \mod 1000$
Sonnhard.

Hi Sonnhard. I agree but my method is probably not as neat as yours:

[hide=" click to reveal hidden content"]

I kept multiplying to get powers of $ 11$ modulo $ 1000$ and quickly got $ 11^{10} \equal{} 601 (mod 1000)$ which looked promising.

I then wrote $ 111^{89}$ as $ (100\plus{}11)^{89}$.

Expanding by the binomial theorem, we see that all terms apart from the final two are divbisible by $ 1000$, so we only need to find $ 89\cdot 100\cdot 11^{88} \plus{} 11^{89} \equiv 900\cdot 11^{88} \plus{} 11^{89}\equiv 911\cdot11^{88} \equiv \minus{}89\cdot 11^{88}$.

Now $ 11^{2} \equiv 201, 11^{30} \equiv 801, 11^{5} \equiv 001$.

Therefore $ 11^{80} \equiv 801, 11^{8} \equiv 881 \implies 11^{88} \equiv 681$

$ \implies \minus{}89\cdot 11^{88} \equiv \minus{}609 \equiv 391$.

Therefore the last three digits of $ 111^{89}$ are $ 391$.

[/hide]

[hide="Alternative solution"]Notice that $ 111\cdot (\minus{}9)\equal{}\minus{}999\equiv 1\bmod 1000$, so we can write $ 111^{89}\equiv 111^{100}\cdot (\minus{}9)^{11}\bmod 1000$.

Then $ \lambda(1000)\equal{}100$, so $ 111^{100}\equiv 1\bmod 1000$, and we have $ (\minus{}9)^{11}\bmod 1000$

This isn't hard to compute. Just bash it out and drop everything but the last three digits (and subtract from $ 1000$ where appropriate). It reduces to $ \minus{}609\equiv\boxed{391}\bmod 1000$.[/hide]


linear algebra
matrix
geometry
geometric transformation
rotation
vector
analytic geometry

Page 124 (Might depend on edition)

4.2.9 a)

It easy to "simplify"
\[ \frac {1}{a \plus{} bi} \equal{} \frac {a \minus{} bi}{a^2 \plus{} b^2}
\]
by multiplying numerator and denominator by $ a \minus{} bi$. But one can also verify this without calculation. How?

Can someone please check my solution

[hide]
Let $ z \equal{} a \plus{} bi$. Therefore the RHS = $ \frac {\bar{z}}{|z|^2} \equal{} \frac {\bar{z}}{z\bar{z}} \equal{} \frac {1}{z}$

Is that what they want, or should I be searching for something geometric?
[/hide]
Thanks

That's valid  :)

You can also represent a complex number $ a \plus{} bi$ with the matrix $ \binom{a \; \; \; \minus{} b}{b \; \; \; \; a}$.
(Can you see why? Hint: this represents the complex number in terms of a scaling and a rotation in plane)

Then you basically have the statement $ \binom{a \; \; \minus{} b}{b \; \; \; \; a} ^{ \minus{} 1} \equal{} \frac {1}{\text{det} \binom{a \; \; \minus{} b}{b \; \; \; \; a}} \cdot \binom{a \; \; \; \; b}{ \minus{} b \; \; a}$

I can't see why at all (mainly since I have never done anything with matrices at all (Its not in the British A level syllabus))

I'll have a look at WP and see whether I can make sense of it :)

We can represent complex numbers with points on a plane, which can be represented by column vectors (which are just like points, except with the coordinates forming a column instead of a row).

We can write the point $ \binom{a}{b}$ as $ \binom{r \cos \theta}{r \sin \theta}$ for some $ r$, $ \theta$.

Now, the matrix $ \left( \begin{array}{cc} a & \minus{} b \\
b & a \end{array} \right)$ can be rewritten as $ \left( \begin{array}{cc} r \cos \theta & \minus{} r \sin \theta \\
r \sin \theta & r \cos \theta \end{array} \right) \equal{} r \left( \begin{array}{cc} \cos \theta & \minus{} \sin \theta \\
\sin \theta & \cos \theta \end{array} \right)$

Note that the scalar factor of $ r$ is a compression/dilation factor, akin to the radius of a complex number in polar form, and the remaining matrix is the rotation matrix for an angle of $ \theta$.

So, if you have a column vector $ \binom{x}{y}$, then the value  $ \left( \begin{array}{cc} a & \minus{} b \\
b & a \end{array} \right) \binom{x}{y} \equal{} r \left( \begin{array}{cc} \cos \theta & \minus{} \sin \theta \\
\sin \theta & \cos \theta \end{array} \right) \binom{x}{y}$, which is a rotation  of the original point by an angle of $ \theta$ followed by a dilation by a factor of $ r$ . This is exactly what multiplication by $ a \plus{} bi$ does in the complex plane.

Because matrices of this form are only combinations of rotations and dilations, they are also commutative. Their addition is fully analogous to addition of complex numbers, and their multiplication/exponentiation is analogous to that of complex numbers.


I know I have already posted this in another section but I am desperate.  I wont be able to continue without an answer.  And I figured the more sections it is in the more chances it has of being answered

If {A,B,N,f1-fn,g1-gn} are real functions. And an equality is in the form
A(f1)+B(f2)+...+N(fn)=A(g1)+B(g2)+...+N(gn)

does that mean that
f1=g1,f2=g2, and so on and so on

Any help on this would be greatly apreciated

Are there any more constraints? As it is odd to have so many equalities as the result of one equation.


Let $ a>0$, $ b>0$ and $ c>0$ such that $ a\plus{}b\plus{}c\equal{}2$. Prove that $ \frac{1}{a^3\plus{}a}\plus{}\frac{1}{b^3\plus{}b}\plus{}\frac{1}{c^3\plus{}c} \leq \frac{1}{abc}$.

[quote="Pilam"]Let $ a > 0$, $ b > 0$ and $ c > 0$ such that $ a \plus{} b \plus{} c \equal{} 2$. Prove that $ \frac {1}{a^3 \plus{} a} \plus{} \frac {1}{b^3 \plus{} b} \plus{} \frac {1}{c^3 \plus{} c} \leq \frac {1}{abc}$.[/quote]
The inequality is equivalent to
\[ \sum \frac{bc}{a^2\plus{}1} \le 1\]
See my solution here http://www.mathlinks.ro/viewtopic.php?t=57558&start=20 :)


What is the smallest positive integer that can be expressed as a sum of 9 consecutive integers, and as a sum of 10 consecutive integers, and as a sum of 11 consecutive integers?

[hide="Generalization"] The sum of $ 2k+1$ consecutive integers with middle term $ a$ is $ a(2k+1)$.  The sum of $ 2k$ consecutive integers with middle terms $ a, a+1$ is $ k(2a+1)$. [/hide]

[quote="sludgethrower"]Ok...there is probably a much easier way but here is my solution...

Answer

Let the integer be $ m$

$ m$ can be expressed as a sum of 9 consecutive integers, let the smallest of these be $ a$
$ m$ can be expressed as a sum of 10 consecutive integers, let the smallest of these be $ b$
$ m$ can be expressed as a sum of 11 consecutive integers, let the smallest of these be $ c$

The formula for the sum of the first $ n$ terms in an arithmetic sequence is:

$ S_{n}= \frac{1}{2}n(2\alpha+(n-1)d)$ where $ \alpha$ is the first term and $ d$ is the common difference (in this case $ d = 1$).

Using the formula: $ m = 9(a+4)$, $ m = 5(2b+9)$, $ m = 11(c+10)$

Since $ gcd(9,\,5,\,11) = 1$ then $ m$ has the form $ 495k, k \in Z$ (since $ 9\cdot5\cdot11 = 495$)

We have to be careful about our choice of $ k$: because $ m = 5(2b+9)$, then $ m$ is odd $ \Rightarrow$ $ k$ is odd.

All in all, this is not a worry, since the smallest postitive $ k$ is 1, which is odd. hence $ m = 495$[/quote]

yep, that's right. Good job.


[quote="FieryHydra"]Show that $2^{\frac1n}$ is irrational for all $n\geq2, n\in N$.[/quote]

$2^{\frac{1}{n}}=k$, for $k$ a positive integrer
Then, $2=k^{n}$. From here, we see that $2 \geq k$, from were $k=1$ or $k=2$. If $k=1$, there are no solutions. But, if $k=2$, then $n=1$. Contradiction.

$k$ need not be a integer, only rational.

[quote="José"]

$2^{\frac{1}{n}}=k$, for $k$ a positive integrer
[/quote]
Are you sure that's correct?

Also, $2^{\frac{1}{n}}=\sqrt [n]{2}$.

Then, you see that $1 < \sqrt [n]{2}\leq{2}$

You see that, from a bigger $n$, the result tends to $1$, without being $1$.

If $\frac{a^{n}}{b^{n}}$ is an integrer, in this case $2$, then $b^{n}\mid a^{n}$, from where $b\mid a$, so $b=1$...then, $k=1$ or $k=2$ is this correct? surely not

Hint:
[hide]Fermat's Last Theorem.[/hide]

This is a simple number theory problem, following directly from Euclids proof of $\sqrt{2}$ being irrational.

Let $\sqrt[n]{2}$ be rational by assumption (in order to reach a contradiction). Then $\sqrt[n]{2}=\frac{a}{b}$ for relatively prime integers a and b. So $2=\frac{a^{n}}{b^{n}}\implies 2b^{2}=a^{n}$. We see that 2 divides the left, so 2 must divide the right. Since 2 is prime, $2\mid a^{n}\implies 2\mid a$. Thus $a=2c$ for an integer c. Then we have $2b^{n}=2^{n}c^{n}\implies b^{n}=2^{n-1}c^{n}$ so (since $n \ge 2$) we have $2\mid b$ from the same argument as before. Since $2\mid a$ and $2\mid b$ then $(a,b) \neq 1$, contradiction.

(blahblahblah's post says it much more succinctly)

[quote="ZetaX"]Or use Eisenstein's criterion... or the rational-root-theorem...

But the real reason why I post here:
Fermat's Last Theorem ¿  :rotfl: ¿ What's going on¿[/quote]
 :) I know what you mean. Just look at the proof below.

We know $\sqrt2$ is irrational. I won't go into the standard proof here.
Now, for all $n\geq3$, let's assume that $2^{\frac1n}$ is rational. So, we have $2^{\frac1n}=\frac{p}{q}$, where $p\in N, q\in N \Rightarrow p^{n}= q^{n}+q^{n}.$ But this directly contradicts Fermat's Last Theorem, which we know is true. Hence, $2^{\frac1n}$ is irrational for all $n\geq2, n\in N$.

I am sorry if the problem was too trivial. I just wanted to see if anyone would post the above solution.

Cheers! :)

I like your proof fieryhydra. Too bad an elementary number theory student can't really back up that Fermat's last is actually true without referring to a book.  :P
Creative method!


number theory
relatively prime
number theory proposed

[quote="huang"]Many of you have probably seen this problem (since it is fairly easy) but I still found it fun to prove: Prove there is an infinite number of primes $ p$ such that $ p\equiv 3(mod 4)$[/quote]
Notice the sequence: $ a_1 \equal{} 3$, $ a_{n\plus{}1} \equal{} 4a_1a_2...a_n\minus{}1$. This is an infinite sequence, all with a prime divisor $ p$ such that $ 4 \mid p\plus{}1$ and all numbers in the sequence are relatively prime, so there exists infitily many primes $ p$ such that $ p \equiv 3 \bmod 4$.


algorithm
number theory
Euclidean algorithm

Could someone please explain to me, in the AoPS Introduction to Number Theory book, Chapter 3, 3.6 Excersize 3.6.1

It states:

Note that 7 divides 42 and 7 divides 77, meaning 42 and 77 are multiples of 7. Through addition of multiples of the integers, we can identify other multiples of 7. Fill in the blanks with whole numbers to create true statements of arithmetic in which all of the sums are multiples of 7.

__ x 42 + __ x 77 = 350 = 50 x 7
__ x 42 + __ x 77 = 357 = 51 x 7
__ x 42 + __ x 77 = 364 = 52 x 7
__ x 42 + __ x 77 = 371 = 53 x 7
__ x 42 + __ x 77 = 378 = 54 x 7
__ x 42 + __ x 77 = 385 = 55 x 7
__ x 42 + __ x 77 = 392 = 56 x 7

What i dont understand is how can you figure out the numbers its ment to be without using guess and check?

The [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]extended Euclidean algorithm[/url].  One way to test small cases is to pick consecutive integer values for the first blank and to see if you can fill in the second blank, which requires that, for example, $ 350 \minus{} 42x$ is divisible by $ 77$.  This is equivalent to modular division, and that's what the Euclidean algorithm is for.


trigonometry
analytic geometry
geometry unsolved
geometry

Point M is located inside triangle ABC. $ A_1$,$ B_1$,$ C_$ are foots of perpendiculars from M to BC,AC,AB respectively.Prove that:
$ \cot \widehat{{\rm{AA}}_1 B} + \cot \widehat{BB_1 C} + \cot \widehat{CC_1 A} = 0$

M can be any point on the plane ABC if we consider oriented angles. Let $(u:v:w)$ be the barycentric coordinates of $M$ with respect to $\triangle ABC.$ Then the coordinates of $A_1,B_1,C_1$ (in Conway's notation) are: $A_1(0:a^2v+uS_C:a^2w+uS_B),$  $B_1(b^2u+vS_C:0:b^2w+vS_A)$  and $C_1(c^2u+wS_B:c^2v+wS_A:0).$

Let $\alpha,\beta,\gamma$ denote the oriented angles between the cevians $AA_1,BB_1,CC_1$ and $BC,CA,AB,$ respectively

$\cot \alpha = \frac{-(uS_C+a^2v)S_B+(uS_B+a^2w)S_C}{-a^2S(u+v+w)}=\frac{vS_B-wS_C}{S(u+v+w)}$

By cyclic permutation of elements we get $\cot \beta$ and $\cot \gamma$ as

$\cot \beta =\frac{wS_C-uS_A}{S(u+v+w)} \ , \ \cot \gamma =\frac{uS_A-vS_B}{S(u+v+w)}$

$\Longrightarrow  \cot \alpha+ \cot \beta+ \cot \gamma=\frac{vS_B-wS_C+wS_C-uS_A+uS_A-vS_B}{S(u+v+w)}=0.$


logarithms
function
calculus
derivative
calculus computations

How do I go about trying to estimate the larger root of Exp(X)=X^1000 ?

First, I would rewrite the equation as $ x\equal{}1000\ln x.$ That helps keep the numbers under control, and will keep us from overflow errors in our calculator or spreadsheet. Then I would notice that $ 1000<1000\ln 1000$ so the answer we seek is greater than $ 1000.$ Finally, in this form, we can do a simple function iteration. Make an initial guess, $ x_0$ (and make $ x_0>1000$), then let $ x_{n\plus{}1}\equal{}1000\ln x_n.$ Since the derivative of $ f(x)\equal{}1000\ln x$ will be less than $ 1$ at the point at which the curves intersect, this iteration should converge to the root you want. Here's a run on a spreadsheet program (it's also easy to do this on a calculator):

$ 2000$
$ 7600.90246$
$ 8936.02226$
$ 9097.84583$
$ 9115.79294$
$ 9117.76368$
$ 9117.97984$
$ 9118.00355$
$ 9118.00615$
$ 9118.00644$
$ 9118.00647$
$ 9118.00647$

Take that last entry as your estimate.