geometry rotation trigonometry calculus integration

I think this problem is really interesting.

A conducting loop of area $ A$ and total resistance $ R$ is suspended by a torsional spring of constant $ k$ in a uniform magnetic field $ \text{B}\equal{}B\text{e}_{y}$.  The loop is in the $ yz$ plane at equilibrium and can rotate about the $ z$-axis with moment of inertia $ I$.  The loop is displaced by a small angle $ \theta$ from equilibrium and released.  Assume the torsion spring is non-conducting and neglect self-inductance of the loop.  Find the equation of the loop in terms of the given parameters.

could you please specify what you mean by "equation of the loop" here?

Oh sorry.  Equation of motion.  I guess knowing that would have been helpful. :oops:

I'm getting an equation of the form

$ I \ddot{\theta} \plus{} \frac {B^2A^2}{R} \cos^2 \theta \dot{\theta} \plus{} \kappa \theta \equal{} 0$.

If we make the approximation $ \cos \theta \approx 1$, this is solvable as a constant coefficient linear ordinary differential equation such as that of a damped spring. I'm looking at how to rigorously get the area term right now, after which I'll post my solution.

blargh I was not able to get the area as cleanly as I wanted to

BLARGH

anyway

We have a flux $ \Phi_B \equal{} BA \sin \theta$ and hence an induced electromotive potential $ \mathcal{E} \equal{} \minus{} \frac{d\Phi_B}{dt} \equal{} \minus{} BA \cos \theta \dot{\theta}$ for a current of $ I \equal{} \frac{\mathcal{E}}{R} \equal{} \minus{} \frac{BA}{R} \cos \theta \dot{\theta}$. The overall force $ F \equal{} \oint I dl \times B \equal{} \minus{}I B \times \oint dl \equal{} 0$ but we have a torque $ \tau \equal{} \oint r \times F \equal{} I \oint r \times (dl \times B)$. We note $ r \times (dl \times B) \equal{} dl (r \cdot B) \minus{} B (r \cdot dl)$. The latter term is $ \frac{1}{2} d(r \cdot r)$ which integrates to zero so the integral reduces to $ \oint dl (r \times B)$, which is in fact just the initial value of this integral at $ \theta \equal{} 0$ modulated by $ \cos \theta$. Now, assuming $ \theta \equal{} 0$ so as to work in the $ yz$-plane, we can pass $ B$ outside the integral so that it takes the form $ \langle 0, \oint y dy, \oint y dz \rangle$ where we are writing out the three components of the vector. The middle term may be rewritten as $ \frac{1}{2} \oint d(y^2)$ and hence integrates to zero around the closed loop while the last term evalutes to $ \int \int dy dz \equal{} A$ so that the overall torque is $ \tau \equal{} IBA \cos \theta \equal{} \minus{}\frac{B^2A^2}{R} \cos^2 \theta \dot{\theta}$ in the $ z$-direction. We also have the restoring torque $ \minus{} \kappa \theta$, and the total torque is $ I \alpha \equal{} I \ddot{\theta}$ to give us an equation of motion.

Your first post is getting there.  You'll eventually have to solve that equation, and your final equation will be pretty nasty.

what

I do not plan on actually solving that equation

well first of all can I assume $ \cos \theta \equal{} 1$

because then it is straightforward to proceed and apply initial conditions

if I cannot assume that then, well, shucks

I do not think I know how to solve the differential equation

If I remember correctly, the equation itself is not actually that difficult to sovle (second-order, constant-coefficient, homogeneous I think), it's just that the constants are ugly.

so by "constant coefficient" I assume you are making the leading-order approximation $ \cos \theta \equal{} 1$. In which case, yes, this is a straightforward algebra exercise.

Actually, if I remember correctly, I believe the approximation is $ \cos^{2}\theta\approx 1$.  But otherwise yes.

by yelling "to first order!" until you're exhausted (as is standard practice in these sorts of problems), one finds that both approximations are equivalent.

I know they're equivalent.  Just what it says in the solution.


geometry
incenter
cyclic quadrilateral
projective geometry
geometry unsolved

Let ABCD be an cyclic quadrilateral inscribed in a circle C. An other circle G and C internally tangent at T.DB and AC are tangent the circle G at E and F respectively. EF intersects AB at P. Prove that TP is the bisector of ATB

A well-known result: $ TP$ passes through the midpoint of arc $ AB$.
http://www.mathlinks.ro/viewtopic.php?t=303520

Another solution.

Let $ Q\equiv TP\cap (O),R\equiv TF\cap (O), I\equiv AR\cap DQ$. Consider the hexagon $ AQBRDT$ inscribed in circle $ (O)$, by Pascal's theorem we have $ P\equiv TQ\cap AB,I\equiv AR\cap DQ,F\equiv TR\cap BD$ are collinear. But from the fact that $ R\equiv TF\cap (O)$ is the midpoint of arc $ BD$ which does not contain $ A$ and by Sawayana's lemma, we conclude that $ I$ is the incenter of $ \triangle ABD$. This means that $ Q\equiv DI\cap (O)$ is the midpoint of arc $ AB$ and the result is followed. QED.

[quote]mestav, it is very easy and i solved it in 5 minutes[/quote]
please dont boast if you dont post your solution


Consider a triangle ABC with angles less than 90 and AB<AC. Let AD,BE,CZ its heights. The parallel from D to EZ meets AB,AC at K,L respectively. EZ meets CB at N. Prove that the circumcircle of the triaangle NKL bisects BC. :)  I hope you will find this exercise interesting.


If $3x-3y+z-w=7$ and $5x-y+3z+w=3$, what is the value of $w+x+y+z$?

[quote="The QuattoMaster 6000"][hide] Subtract the first equation from the second equation to get 2w+2x+2y+2z=2(w+x+y+z)=-4, so w+x+y+z=$-2$. [/hide][/quote]
I always miss easy things like that...


I might be redundant here, but, where are all you guys going for college?

Yeah im pretty sure im going to Stanford. Dont know tough i still have a bit of a soft spot for princeton.

[quote="manuel"]im going to MIT  :)[/quote]Congrats!! :)

Thanks, Valentin!

but i think i already told you, and you gave me congrats...  :?
but thanks again, i am still excited...


Let $a$, $b$, $c$ be three nonzero integers.
It is known that the sums $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$, and $\frac{a}{c}+\frac{c}{b}+\frac{b}{a}$, are integers. Find these sums.

[hide="Hint"]Let $A = \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $B = \frac{a}{c}+\frac{c}{b}+\frac{b}{a}.$

Then $A-B = \frac{(a-b)(b-c)(c-a)}{abc}.$[/hide]

Hmm.

Consider $f(x)=(x-\frac{a}{b})(x-\frac{b}{c})(x-\frac{c}{a})$. Then, using hydra's notation, $f(x) = x^{3}-Ax+Bx-1$. Suppose $A$ and $B$ are indeed integers; then by the RR theorem, all the rational roots (which all the roots are) must be either $\pm 1$. Then we get $a=\pm b$, $b=\pm c$, and $c=\pm a$. From here it is just casework to find the possible sums, which are either $3$ or $-1$.

The best solution I saw, was one similar to Elemennop's. Though I did not actually find this during the PAMO (it is the only question I did not solve). There is another solution using number theory as well, but it is long and much more complicated.

Hi Melissa,its Hendrie. Can you maybe e-mail me the PAMO questions or tell me where I might find them.How did you do?

BUT WE KNOW THAT  the sum is$>$than 6 :(

[quote="arkhammedos"]BUT WE KNOW THAT  the sum is$>$than 6 :([/quote]$\frac{a}{b}+\frac{b}{a}\ge2$ is not valid if $a,b$ can have different signs.

thanks for this help .because im not a good englesh speakr
than i had not understound whats integer :oops:

we know that    a/b + b/c + c/a >=3 and b/a +a/c+ c/b>=3 using the chebychevs ınequality for real nos....
   using the nth Chebychevs polynomial for the fact that: x+1/x=2cost for (t) a real no...
    therefore, substituting x=a/b, y=b/c, and z=c/a.   (x+1/x) + (y+1/y)+ (z+1/z)>=6 from the 1st arguement and
    LHS<=6 using the fact that (cost<=1) therfore.LHS=6 and it shows that the ineguality must be each 3....
                      Q.E.D


I got this from the Mathcounts forum but it was a little hard for there, so I wanted to see what you guys thought of it:

A $n$ by $n$ equilateral triangle is split up into $1$ by $1$ equilateral triangle.  How many triangles are there?

Hint: its not $n^2$

Picture:
[img]http://wpc.puzzles.com/history/tests/2000wpc/images/triangles-1.gif[/img]

Well, can the triangles overlap?? If so, there are more than $24$ I think.

There are 27, 16 1x1, 7 2x2, 3 3x3, 1 4x4.  You have to count the upside down ones and yes, they can overlap.  but I mean how many triangles would there be in an $n$ by $n$ triangle in terms of $n$, like a formula

Hmm, you're right, I missed 3 of the 2x2 triangles.

if we have any equilateral tri,and we split it by joining the midpoints of the sides ,4 equi triangles are formed.similarly if we split the small tris further in the same way,again 4 tiny equi tris are formed.soo...the tris are formed of the order of 4.does this help? :D  i guess not.

Hey... that was my problem!!!! :mad:

[I have edited my clues out of this post by special request of nat mc. :P ]

Following your definition of $n$, $a_n$ would represent the number of triangles formed from the subdivision of an equilateral triangle with length $n$.  So:

$a_n = n^2 - n + 2 + a_{n - 1}$, $n \ge 2$

$a_1 = 1$

I'm not sure where to go from there.   :?

[quote="shadysaysurspammed"]
if we have any equilateral tri,and we split it by joining the midpoints of the sides ,4 equi triangles are formed.similarly if we split the small tris further in the same way,again 4 tiny equi tris are formed.soo...the tris are formed of the order of 4.does this help? :D i guess not[/quote]
It doesn't always work that way.  When $n$ is odd you won't have the midpoint being used.  You split each side into
$n$ segments of length 1 and connect the points made by these to the other sides

[quote="hmoonster"]hmm... does the sides of the counted triangles have to be on the given lines? if you have an equilateral triangle by taking every other vertex of a hexagon, does it count?[/quote]
I don't know what u mean.  You can only count the triangles if they are made up by the lines drawn

[quote="mathnerd314"]Hey... that was my problem!!!! :mad:

Every other term satisfies a cubic with first coefficient 4. The number of right-side-up triangles has to do with the sum of certain triangular numbers, and the upside-down triangles follow a different pattern. That's all I've found out so far.[/quote]
sorry :oops:   I just wanted to see what they thought of it, seems like theyre not getting what I'm trying to say. I wanted to put it into a different forum cause I already solved it in the MathCounts forum, I just want to see some more solutions.
hey, don't give them clues  :lol:  :P


Jennifer received a 25 dollar gift certificate from the stereo store. she can use it to buy 2 albums and 5 singles or 1 album and 15 singles. Find theh cost fo an album and the cost of a single.


Express the sum of: (1/ 1*2) +  (1/2*3) + (1 / 3*4)..........( 1 / 9*10  as a common fraction.

  This is a mathcount's state sprint problem.  It took me about sixty seconds to solve.

[hide]It telescopes.  $\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)...\left(\frac{1}{9}-\frac{1}{10}\right)$.  Everything cancels out except the first and last terms, leaving $1-\frac{1}{10}=\frac{9}{10}$.[/hide]

[quote="G-UNIT"]Express the sum of: (1/ 1*2) +  (1/2*3) + (1 / 3*4)..........( 1 / 9*10  as a common fraction.[/quote]
Following the rules of PEMDAS, the way this problem is written, it means:
$\Big(\frac11\times 2\Big)+\Big(\frac12\times 3\Big)+\Big(\frac13\times 4\Big)+\cdots+\Big(\frac19\times 10\Big)$

[quote="236factorial"]1/1*2 +1/2*3 +.... + 1/n(n+1) = n/n+1[/quote]
And this means:
$\frac11\times 2+\frac12\times 3+\cdots+\frac1n(n+1)=\frac{n}{n}+1$

According to PEMDAS, multiplication and division get equal priority, and are performed from left-to-right.  Since I know the problem you refer to, I know this is not your intent.  However, if this were presented to you in a contest, how would you interpret the question?

The point is that it's very important to write mathematical expressions clearly and unambiguously.  In addition, I urge you to learn a little bit of LaTeX.  [It doesn't take very much to turn an ugly ambiguous expression into a terrific-looking typeset mathematical expression.]

I think this is probably what G-UNIT meant:
$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots+\frac{1}{9\times 10}$


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let $R$ a finite and comutative ring. Prove that $Spec(R)=Max(R)$. ;)

I guess what you're trying to say is that all prime ideals are also maximal in this particular case?

If $I$ is a prime ideal, then $R/I$ is a finite integral domain, and it's easy to prove that a commutative finite integral domain is actually a field (for every non-zero $a$, the map $x\mapsto ax$ is injective, so it's also surjective, meaning that it takes the value $1$ for some $x$, which will then be the inverse of $a$; I suppose the ring has a unit, right? :?), meaning that $I$ is actually maximal.

yes, you are right. I forgot to say that.  ;)


function
calculus
integration
trigonometry
real analysis
real analysis unsolved

A really funny problem: what are the Fourier coefficients of the function $f(x)=\sum_{n\geq 2}{\frac{cos(nx)}{nlnn}}$ ?

Those of the series. Note that the sum of the squares of the coefficients of the series is finite, which allows to conclude that the series converges in $L^2$, which is more than enough to justify the exchange of summation and integration in the formula for the Fourier coefficients. Seems like they give simple problems on oral exams sometimes :)

Yes, of course, sometimes problems are simple, sometimes they are hard... actually, this was the first question. The second one was: suppose that the coefficients of the series are decreasing (no more sum of squares that converges). Does the conclusion remain true?

This is a much more interesting question. I'm inclined to give other people a chance to figure it out. I also suspect that the third question was "Suppose that the coefficients decrease to $0$ and the sum of the series is an integrable function. Is the conclusion true then?" :P

Actually, as far as i know the condition that $na_n \to 0$ and $a_n$-monotone decreaing tending to $0$s sequence  is a necessary and sufficient conditon for the series $\sum_{n \geq 1} a_n cos(nx)$ to converge uniformly on $[-\pi,\pi]$.
For the second question: it is not sufficient condition
There is more general result : For the series $\sum_{n \geq 1}a_ncos(nx)$ with $a_n$-decreasing $\to 0$ to be a Fourier series there is  a [b]necessary[/b] condition $\lim(a_n-a_{n+1})lnn=0$

[quote="eugene"]Actually, as far as i know the condition that $na_n \to 0$ and $a_n$-monotone decreaing tending to $0$s sequence  is a necessary and sufficient conditon for the series $\sum_{n \geq 1} a_n cos(nx)$ to converge uniformly on $[-\pi,\pi]$.[/quote]
That is for $\sin$ series. The $\cos$ series diverges at $0$ unless $\sum_k a_k<+\infty$. Eugene is right about the second question though: there are $\cos$ series with decreasing to $0$ coefficients that are not Fourier series. But all of them converge pointwise to non-integrable functions, so the answer to question $3$ is "Yes". Can anybody prove it?


geometry
circumcircle
perpendicular bisector
geometry unsolved

In triangle $ABC$, $k$ is the circumcircle and $O$ is the centre of $k$. In triangle $BCO$, $k_1$ is the circumcircle and $D$ is the centre of $k_1$. It is known that $D$ lies on $k$.
Let $AD\cap BC=M$ and $OH\cap BC=P$. Find $\angle(AP,MH)$.

[quote="Beat"]In triangle $ABC$, $k$ is the circumcircle and $O$ is the centre of $k$. In triangle $BCO$, $k_1$ is the circumcircle and $D$ is the centre of $k_1$. It is known that $D$ lies on $k$.
Prove that if $I$ is the centre of the incircle and $H$ is the orthocentre of $\triangle ABC$ then both $H$ and $I$ lie on $k_1$, just like $O$ and prove that $OI=IH$.
  Let $AD\cap BC=M$ and $OH\cap BC=P$. Find $\angle(AP,MH)$.[/quote]
$B,C,O,H,I$lie on $k_1$
notice that
$\angle ABI=\angle CBI$
and
$\angle ABO=\angle CBH$
Therefore $\angle OBI=\angle IBH$

Yes, I see and how can i proove that (AP,MH)=90

i'll give just some data i've found...

since $D$ lies in the perpendicular bisector of $BC$ we must have $D$ to be equal to the midpoint of minor arc $BC$. Now, $\angle COD=\angle A$ and $\angle OCD=\angle OCB+\angle BCD=90-\angle A+\angle \frac{A}{2}=90-\angle \frac{A}{2}$, and since $DO=DC$ we must have $\angle COD=\angle OCD \Rightarrow \angle A=90-\angle \frac{A}{2}\Rightarrow \angle A=60$

and since $D$ is the midpoint of arc $BC$ we must have $AD$ to be bisector, hence $M$ is the foot of the $\angle A$ bisector.

..that's all i have by the moment


inequalities
function
integration
induction
calculus
calculus computations

Prove that $ e^x\geq 1\plus{}x$ for $ x\geq 0.$

At $ x \equal{} 0$, $ e^0 \equal{} 1 \plus{} 0$.
$ \frac {d}{dx} (e^x) > \frac {d}{dx} (1 \plus{} x)$, $ \forall x > 0$
$ e^x > 1$, $ \forall x > 0$
Thus, $ e^x\ge 1\plus{}x$ for $ x\ge 0$

( Similar to EastyMoryan's )

Consider the function

$ f(x) \equal{} e^x \minus{} x \\  \\ f '(x) \equal{} e^x \minus{} 1$

$ for \ x \geq 0, f '(x) \geq 1$

So, f(x) is an increasing function.

$ \Rightarrow \ f(x) \geq f(0) \ \forall \ x \geq 0 \\ \\
or \ e^x \minus{} x \geq 1$

proved.  :|

How about $ e^x\geq 1\plus{}x\plus{}\frac{x^2}{2}\plus{}\frac{x^3}{6}\ (x\geq 0)?$

Hmm, i think we can use :

$ e^x\equal{}\sum_{k\equal{}0}^{\infty} \frac{x^k}{k!}$, so the problem become clear..

Yes, mathmatically, that's doesn't have problem, generally speaking, Japanese high school students don't study Taylor expansion, Maclaurin Series.

Since $ e^x\geq1$ for $ x\geq0$,
$ \int_0^x e^t dt \geq \int_0^x 1 dt$
$ e^x\minus{}1\geq x$
$ e^x\geq 1\plus{}x$.
If we repeat this,
$ \int_0^x e^t dt \geq \int_0^x 1\plus{}t dt$
$ e^x \geq 1\plus{} x \plus{} x^2/2$

And using this method and induction, we have
$ e^x \geq \sum_{k\equal{}0}^n \frac{x^k}{k!}$

[quote="Ronald Widjojo"]Yes, but Indonesian high school students don't study it too..

I know this formula when i study combinatorics, this identity is well known in generating function..[/quote]

Yes, but how was in proven then?

There exists $ c$ with $ 0<c<x$  such that $ e^x \minus{} x \minus{} 1 \equal{} x(e^c \minus{} 1) \geq 0$, for all $ x >0$.

I just wanted to say hello to all of you friends. I am writing from the remotest of all villages on the eastern bank of the river nile in this part of the world - Southern Sudan.

It has seemed to be more harder solution out of my expectation.

[quote="An0maly"][quote="Nora.91"]

$ for \ x \geq 0, f '(x) \geq 1$
[/quote]

Shouldn't it be:

$ f '(x) \geq 0$

since $ f'(0) \equal{} e^0 \minus{} 1 \equal{} 1 \minus{} 1 \equal{} 0$[/quote]

Thanks for the correction !

[b]SILLY ME[/b]  :blush:


there is a box with sides a and b in a corridor of width c.find the conditions in which it can be pushed through

a door of width d.


modular arithmetic
factorial
number theory
AMC

[quote="azjps"][hide]Multiply out the $ 7!$ to get
\[ 5 \cdot 6! \equal{} (3 \cdot 4 \cdots 7)a_2 \plus{} (4 \cdots 7)a_3 \plus{} (5 \cdot 6 \cdot 7)a_4 \plus{} 42a_5 \plus{} 7a_6 \plus{} a_7 .
\]
By Wilson's Theorem, $ 5 \cdot 6! \equiv \minus{} 5 \equiv 2 \pmod{7}$, so $ a_7 \equal{} 2$. Then we move $ a_7$ to the left and divide through by $ 7$ to obtain
\[ \frac {5 \cdot 6! \minus{} 2}{7} \equal{} 514 \equal{} 360a_2 \plus{} 120a_3 \plus{} 30a_4 \plus{} 6a_5 \plus{} a_6.
\]
We then repeat $ \pmod{6}$, and so forth, to find the unique solution to be $ (a_2, a_3, a_4, a_5, a_6, a_7) \equal{} (1,1,1,0,4,2)$, and so the answer is $ 9 \Longrightarrow \mathrm{(B)}$.

In 1999, calculators were allowed, so the multiplication is pretty simple.[/hide][/quote]

If I have no mod/number theory background, do you think I should skip this question?

Nope, no modular arithmetic background is necessary.

Again, we multiply both sides out by $ 7!$: \[ 3600 \equal{} 2520a_2 \plus{} 840a_3 \plus{} 210a_4 \plus{} 42a_5 \plus{} 7a_6 \plus{} a_7 .\]

Note that when we multiplied out by $ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7$, the only fraction where the $ 7$ cancelled was with the $ a_7$. So all the other terms are divisible by $ 7$: \[ 3600 \minus{} a_7 \equal{} 7(360a_2 \plus{} 120a_3 \plus{} 30a_4 \plus{} 6a_5 \plus{} a_6)\]

We are given that $ 0 \le a_7 < 7$, and we require that $ 3600 \minus{} a_7$ is divisible by $ 7$ (in other words, $ a_7$ is the remainder upon division of $ 3600$ by $ 7$). Thus, we find that $ a_7 \equal{} 2$.

We are now left with  \[ \frac{3600 \minus{} 2}{7} \equal{} 514 \equal{} 360a_2 \plus{} 120a_3 \plus{} 30a_4 \plus{} 6a_5\] Now the same thing is true about $ 6$ as it had been with $ 7$ before! So we re-write this as \[ 514 \minus{} a_6 \equal{} 6(60a_2 \plus{} 20a_3 \plus{} 5a_4 \plus{} a_5)\] And then the remainder upon division of $ 514$ by $ 6$ is just $ 4$. We repeat with $ a_5, a_4, a_3, a_2$, and eventually come up with the solution set described above.

If you are familiar with computing the "decimal" expansion of fractions in bases other than $ 10$, this is what is called [url=http://planetmath.org/encyclopedia/Factoradic.html]factorial base[/url].  And although divisibility is a nice way to phrase the argument, it is actually completely unnecessary:  this is because the value of, say, $ a_2$ is determined uniquely by the fact that

$ 840a_3 \plus{} 210a_4 \plus{} 42a_5 \plus{} 7a_6 \plus{} a_7 < 2520$.

In other words, factorial-base representations are unique because the largest $ n$-digit number is smaller than the smallest $ n\plus{}1$-digit number.

[quote="azjps"]Nope, no modular arithmetic background is necessary.

Again, we multiply both sides out by $ 7!$:
\[ 3600 \equal{} 2520a_2 \plus{} 840a_3 \plus{} 210a_4 \plus{} 42a_5 \plus{} 7a_6 \plus{} a_7 .
\]
Note that when we multiplied out by $ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7$, the only fraction where the $ 7$ cancelled was with the $ a_7$. So all the other terms are divisible by $ 7$:
\[ 3600 \minus{} a_7 \equal{} 7(360a_2 \plus{} 120a_3 \plus{} 30a_4 \plus{} 6a_5 \plus{} a_6)
\]
We are given that $ 0 \le a_7 < 7$, and we require that $ 3600 \minus{} a_7$ is divisible by $ 7$ (in other words, $ a_7$ is the remainder upon division of $ 3600$ by $ 7$). Thus, we find that $ a_7 \equal{} 2$.

We are now left with
\[ \frac {3600 \minus{} 2}{7} \equal{} 514 \equal{} 360a_2 \plus{} 120a_3 \plus{} 30a_4 \plus{} 6a_5
\]
Now the same thing is true about $ 6$ as it had been with $ 7$ before! So we re-write this as
\[ 514 \minus{} a_6 \equal{} 6(60a_2 \plus{} 20a_3 \plus{} 5a_4 \plus{} a_5)
\]
And then the remainder upon division of $ 514$ by $ 6$ is just $ 4$. We repeat with $ a_5, a_4, a_3, a_2$, and eventually come up with the solution set described above.[/quote]

I get all of the steps, but not why they are done. How do we know:

So all the other terms are divisible by $ 7$:
\[ 3600 \minus{} a_7 \equal{} 7(360a_2 \plus{} 120a_3 \plus{} 30a_4 \plus{} 6a_5 \plus{} a_6)
\]

Thanks!


trigonometry
function
inequalities
inequalities unsolved

[quote="mecrazywong"]Let $x,y,z\le\sqrt{2},4+\sum x^2y^2=x^2y^2z^2+2\sum x^2$. Prove that \[x+y+z\le3\][/quote]
[b]case 1:[/b] One of the numbers is $\leq 0$.
\[\Rightarrow x+y+z\leq 0+\sqrt{2}+\sqrt{2}<3\]
[b]case 2:[/b] $0<x,y,z\leq 1$
\[\Rightarrow x+y+z\leq 1+1+1=3\]

That was easy, wasn't it?

[b]case 3:[/b] Exactly one of the numbers is $>1$, the other $\in ]0;1]$.
\[\Rightarrow 0\leq (x^2-1)(y^2-1)(z^2-1)=x^2y^2z^2-x^2y^2-y^2z^2-z^2x^2+x^2+y^2+z^2-1\\\quad =3-(x^2+y^2+z^2)
\\\Rightarrow \frac{x+y+z}{3}\leq\sqrt{\frac{x^2+y^2+z^2}{3}}\leq 1\]
[b]case 4:[/b] One of the numbers is equal to $\sqrt{2}$.
wlog $x=\sqrt{2}$.
\[\Rightarrow 4+2y^2+2z^2+y^2z^2=2y^2z^2+2(2+y^2+z^2)
\\\Rightarrow y^2z^2=0 \Rightarrow y=0\vee z=0
\\\Rightarrow x+y+z\leq 0+\sqrt{2}+\sqrt{2}<3\]
[b]case 5:[/b]All other cases.
This is equivalent to: $x,y,z>0$, at least two of the numbers are $>1$, none of the numbers is equal to $\sqrt{2}$.

wlog $x,y\in ]1;\sqrt{2}[$.

The given equation is equivalent to \[(x^2-1)(y^2-1)(z^2-1)=-(x^2-1)-(y^2-1)-(z^2-1).\]
It is well known, that if $abc=a+b+c$ then there are $\alpha,\beta,\gamma$ with $a=\tan{\alpha},b=\tan{\beta},c=\tan{\gamma},\alpha+\beta+\gamma=\pi$. But what about this case?
It turns out that there is something similar with $\tanh$:
As $x^2-1\in ]0;1[$ and $y^2-1\in ]0;1[$, there are $a,b\in\mathbb{R}^+$ such that $x^2-1=\tanh{a},y^2-1=\tanh{b}$.
Thus we get
\[z^2-1=-\frac{(x^2-1)+(y^2-1)}{(x^2-1)(y^2-1)+1}=-\frac{\tanh{a}+\tanh{b}}{\tanh{a}\tanh{b}+1}
\\=-\frac{\sinh{a}\cosh{b}+\sinh{b}\cosh{a}}{\sinh{a}\sinh{b}+\cosh{a}\cosh{b}}
\\=-\frac{(e^a-e^{-a})(e^b+e^{-b})+(e^b-e^{-b})(e^a+e^{-a})}{(e^a-e^{-a})(e^b-e^{-b})+(e^a+e^{-a})(e^b+e^{-b})}
\\=-\frac{e^{a+b}-e^{-(a+b)}}{e^{a+b}+e^{-(a+b)}}=-\tanh{(a+b)}=\tanh{c}\]
Where $c=-(a+b)$.

Let's consider the function \[f(x)=\sqrt{1+\tanh{x}}=\sqrt{1+\frac{e^x-e^{-x}}{e^x+e^{-x}}}=\sqrt{2}(1+e^{-2x})^{-1/2}\]
\[f'(x)=\sqrt{2}(-\frac{1}{2})(1+e^{-2x})^{-3/2}(-2e^{-2x})=\sqrt{2}(1+e^{-2x})^{-3/2}e^{-2x}
\\f''(x)=\sqrt{2}\left((-\frac{3}{2})(1+e^{-2x})^{-5/2}(-2e^{-2x})e^{-2x}+(1+e^{-2x})^{-3/2}(-2e^{-2x})\right)
\\=\sqrt{2}e^{-2x}(1+e^{-2x})^{-5/2}(3e^{-2x}-2(1+e^{-2x}))
\\=\sqrt{2}e^{-2x}(1+e^{-2x})^{-5/2}(e^{-2x}-2)\]
Thus, $f$ is convex in $\mathbb{R}^+$.
And now ... Jensen:[img]http://www.smiley-channel.de/grafiken/smiley/gewalt/smiley-channel.de_gewalt004.gif[/img]
\[f(a)+f(b)\leq 2f\left(\frac{a+b}{2}\right)=2f(-c/2)\]
For $x\leq 0$ we have \[\frac{\sqrt{2}(1+e^{-2x})^{-3/2}e^{-2x}}{\sqrt{2}(1+e^{x})^{-3/2}e^{x}}=\left(\frac{1+e^{2x}}{1+e^{x}}\right)^{-3/2}\geq 1
\\\Rightarrow [f(x)+2f(-x/2)]'=\sqrt{2}(1+e^{-2x})^{-3/2}e^{-2x}+2(-\frac{1}{2})\sqrt{2}(1+e^{x})^{-3/2}e^{x}\geq 0\]
Thus, $f(x)+2f(-x/2)$ is increasing for $x\leq 0$.
\[\Rightarrow f(c)+2f(-c/2)\leq f(0)+2f(-0/2)=3
\\\Rightarrow x+y+z=f(a)+f(b)+f(c)\leq 2f(-c/2)+f(c)\leq 3\]

So we have killed the ineq.[img]http://www.smiley-channel.de/grafiken/smiley/gewalt/smiley-channel.de_gewalt039.gif[/img]
It's a really ugly solution :( , but it works. :roll: At least I hope it works but maybe there are still some mistakes in my solution. If you find a mistake, you may keep it.


What have you come up with in your plan of a Mathlinkers' day in Hanoi, Valentin? I think we should put it down here for everyone's ideas.

We were planing to organize somewhat like that in Slovenia too but as far as I know we failed :( .


Please do not discuss any problems from the MATHCOUNTS chapter competition until after the last Saturday in February.  Chapter competitions happen on different dates around the country.  Students who discuss any information including scores and difficulty level of the test will be put in the AoPS penalty box and not be allowed to post for a considerable period of time.

[quote="frost13"]We might even want to wait until MC gives the okay. I think it was last year when a couple states had planned for late in February and got snowed out all the way into March.  It may actually have been states when that happened, but either way better safe than sorry.[/quote]

This is probably safest.  As time gets closer, we will update students on MATHCOUNTS schedules and requests about sharing information.

I would also request, during the competition windows, [b]please identify the source of any problem you post[/b].  Such as "2002 Chapter Target #4".

good point
also:
rather than posting chapter questions after 26th:
MATHCOUNTS posts the chapter contests on the web... as soon as it gets there, we can discuss.
normally it is just a week after the last chapter

Yes, please, no online discussion until EVERYONE is done testing. I will be watching, and I'm sure the administrators of the site and moderators of this board will be extra busy on all forums for a whole month.  :(

On a lighter note, good luck to everyone who is competing this year. It will be great to see other AoPSers at the meets.

One thing.

Whats the penalty box?

(But I'm BURSTING!!!)
(Cant...talk...about...chapter...)
(Don't think I'm the only one)

Be patient. I feel your pain, but I found out last year that it is actually much more satisfying to talk about competitions when everyone can talk about them legally, fairly, and together here on AoPS. So the waiting you are putting up with now will have a sweet reward at the end--much like preparing for math competitions in general.

I'm sure it's tough to schedule the AMC within a few hours as well.  But really, I'm not at all the source or expert on competition scheduling.

[quote="theaznkungfu"]yea... I REALLY WANT TO TALK ABOUT IT!!!
but i was wondering... i was at the san diego chapter competition, and the guy who made AoPS website was there to talk about it... is it u MCrawford? cuz i couldn't hear the guys name after our team was cheering so hard...(which is kind of sad when there are only 8 ppl :D )[/quote]

That was Richard Rusczyk, not me.  I might however be at the Southern California MATHCOUNTS competition.

[quote="theaznkungfu"]yea... I REALLY WANT TO TALK ABOUT IT!!!
but i was wondering... i was at the san diego chapter competition, and the guy who made AoPS website was there to talk about it... is it u MCrawford? cuz i couldn't hear the guys name after our team was cheering so hard...(which is kind of sad when there are only 8 ppl :D )[/quote]

That was Richard Rusczyk who was speaking. I was there too.

[quote="Scrambled"][quote="MCrawford"][quote="Scrambled"]hmmmm.. does anyone know why it happens on different dates? it would be so much easier if it happened on one[/quote]

Because scheduling several hundred competitions to be on the same date is a nightmare -- possibly impossible.[/quote]what about the amc?[/quote]

Well, the AMCs are often administrated by schools, so that may make a difference.

[quote]That was Richard Rusczyk who was speaking. I was there too.[/quote]
hey... what school are you from?

REMINDER

Until such time as MATHCOUNTS releases the Chapter Competitions or until Richard or Mathew tell me otherwise, I will relentlessly delete all comments containing your own score, your own rank, and [b]especially[/b] solicitations for others to discuss the Chapter Competition.

While discussion of rank and advancement to State are probably harmless, it is a slippery slope.  Solicitations to others to discuss their results is especially risky, since they may not know the rules of this forum.  When the administrators of this forum are ready to ask "Who advances to State?", they will ask the question.

[size=75][The only exception I know of in this forum is a legitimate discussion of a possible rules violation / dispute at a Chapter.  An administrator is involved in that discussion.][/size]

[quote="LuCy4EvA"]we haven't taken it yet, is all the chapters the same everywhere?[/quote]

No... It depends on when your Chapter Coordinator can find the time and location to hold the event... You may not have taken the chapters yet... but I have, for example...  It's usually based on Venue timing and when everyone can come on and pull the thing off... Schools host the AMCs, so it's much easier to coordiate the timing (especially because it's on a weekday) to respond to Scrambled's question...

--Adi

[quote="GoBraves"][quote="Hexahedron"]We can talk about the chapter exam tomorrow........
I can't wait.......
Must resist urge to talk........[/quote]

I dont think so.  We're not allowed to talk about it untill AFTER tomorrow[/quote]

Yeah, your right  :(

[quote="Hexahedron"]We can talk about the chapter exam tomorrow........
I can't wait.......
Must resist urge to talk........[/quote]

[b][size=150]Don't you DARE[/size][/b] talk about the Chapter Competitions until an Administrator or an adult Moderator has given you permission!

Such permission will not occur until MATHCOUNTS has posted the competition on their Web Site or until Chapter Coordinators have been notified by MATHCOUNTS that the embargo is lifted.

[b]Don't even think about it![/b]

i think yesterday was the last chapter competition date... but I agree with rcv.
once MATHCOUNTS puts the contest in the site, lets talk about it.
but how hard can it be to resist...

Have you ever heard of a school being closed unexpectedly?

Yesterday, February 26, was the last [b]scheduled[/b] day for Chapter Competitions.  In the past, there have been extenuating circumstances in which a Chapter was unable to hold their competition.  [Heavy snow would be such an example.  Problems (no heat, no water, no electricity, ...) at the scheduled facility might be another example.]

There are hundreds of Chapters across the nation.  The National MATHCOUNTS organization asks each Chapter to send them notification that their competition has been completed.  Only after the [b]very last[/b] Chapter sends their notification, does the National organization release the competition materials.

If National has not received all of the notifications by Monday, they will attempt to contact the remaining Chapters.  MATHCOUNTS knows that students and coaches and even their own Chapter Coordinators want to talk about the competition.  But it is most important to allow every Chapter to have a fair competition.

It's frustrating for me, too.  But, DO NOT TALK ABOUT CHAPTER COMPETITIONS UNTIL FURTHER NOTICE.

Can we talk about them now?  I'm dying to ask a question about a sprint round question!

Be patient. Learning patience is part of learning how to succeed in math at a higher level.

Thank you for your words of wisdom :D
I will try and learn patience :D

The test is up on mathcounts.org! We can start talking now!

Yes!  Feel free to discuss any aspect of the Chapter Competitions.

YAY!!!!! THE WAIT IS OVER!!!!! WOOT WOOT

Yeah, it's after the last saturday now. Yeah


Find all function $f: \mathbb{R}\to\mathbb{R}$ sastifying condition:
$2005f(x-1)+2004f(1-x)=x$  $\forall x\in\mathbb{R}$


ok there are two problems im having trouble figuring out

[b]1st one[/b]

f(x)=1/x

f(x+1)= ([u]1[/u])/([u]x+1[/u])
f(f(x))= (____)/([u]1[/u])

cant figure out the blank one, i guess on the denom.,

[b]2nd one[/b]

f(x) = 1/(x+1)

f(x+1) = ([u]1[/u])/([u](x+1)+1[/u])
f(f(x))= ([u](x+1)[/u])/(____)

again cant figure out the blank and i guess on the (x+1)[/u]

yeah i figured it out right after i posted it. I find a problem close to those that we did in class.


geometry
3D geometry
probability
symmetry
counting
distinguishability
rotation

You are given eight identical $ 1 \times 1 \times 1$ cubes. Each has a pair of opposite (i.e. parallel) faces colored red, a pair colored green and a pair colored blue. You put the cubes in a bag, shake them up and, blindfolded, assemble them into one $ 2 \times 2 \times 2$ cube. What is the probability that the big cube has each of its square faces colored uniformly (e.g. all red)?

Think of the cubes' finals positions as fixed, and their orientations as variable (by symmetry there is no reason to consider the cubes as distinct).  Each cube has eight possible orientations, so there are $ 8^8$ possible arrangements (some indistinguishable, of course).  Given any particular arrangement, it remains unchanged if any of the cubes is rotated by 180 degrees along any axis; there are four distinct such combinations of rotations, so there are $ 4^8$ indistinguishable copies of each arrangement.  There are three 'good' distinct arrangements (one for each color), so there are $ 3\cdot 4^8$ good arrangements in total, and hence the probability in question is $ \frac {3}{2^8}$.

Hmm, I think any two adjacent sides determines the orientation of the entire cube, so there should be 6 orientations per cube.

Shouldnt be $ 6 \cdot \frac{6^8}{8^8}$ ?

Generalization for an $ n$-dimentional space: $ 2\cdot n \frac{(2 \cdot n)^{2^n}}{(2^n)^{2^n}}$.


Sorry if this is a re-post but searched around and had no luck so far. Any way I am having a little trouble with my work as of lately, I missed a few days of class due to medical reasons and during the time that I was gone I have gone completely lost in math. I managed to get some notes and keep trying various ways to go about it, but still no luck. Any how the topic I am lost on is mentioned in the topic. I managed to get the basics of setting up the problem, but am lost when it comes to the actual solving of it. I wish I had time to post what I have so far, but am pressed for time as of now, so any insight or help would be aprechiated.

Thanks in advance
---------------------
Lemings

Do you mean questions like this:

solve for x and y:

x+y=4
3x+2y=9

Yes, as a matter of fact that is the very thing I am talking about.

Okay, so we have $x+y=4$ and $3x+2y=9$. We just want to find the intersection of those two lines. When do two lines intersect? When both their $y$ and $x$ values are equal.

So let's look at $x+y=4$. We can write that as $y=-x+4$. Now let's simply substitute that into $3x+2y=9$. We get:
\[ 3x+2(-x+4)=9 \]
\[ 3x-2x+8=9 \]
\[ x+8=9 \]
\[ x=1 \]

So now we know what the common $x$ value is! To find the y, we know that $x+y=4$, so if $x=1$, then $y=3$, right? So the answer is $(1,3)$. And if you were to sub that point in to both equations, you'd see it satisfied both. It is the only point to do that!

There are of course a million other techniques to do this, but this is the most straightforward.

Another basic way is elimination:

We are trying to get rid of a variable, so $x+y=4...(1)$ and $3x+2y=9...(2)$ (the number in the parentheses is just a way for me to refer to an equation)

Multiply $(1)$ by 3 and it will become $3x+3y=12$.

So now, we have:

$3x+3y=12...(3)$
$3x+2y=9...(2)$

After $(3)-(2)$, we get $y=3$.  After we plug $y=3$ into our first equation ($x+y=4$), we get $x=1$.

I am starting to understand it better now. Again, thanks guys for the help.


inequalities logarithms calculus derivative integration

I ran into this inequality while doing a contest problem:

\[ \ln \left( \frac{x\plus{}1}x \right) < \frac{x\plus{}\frac{1}2}{x^2\plus{}x}\]
for positive integers.

and now I am interested in it.  I am pretty certain that it is true, but haven't been able to prove it.  The way in which I did try to prove it is the following:

replace LHS w/A and RHS w/B:

A < B for x = 1, then if dA/dx < dB/dx (over positive integers) then that means B is growing faster than A.  Combined w/the fact that B was initially greater than A, then this shows that B > A for all positive integers.

However, it turned out that dA/dx > dB/dx (for all positive integers x).  I concluded that this does not mean that the inequality is false since it only means that A slowly approaches but B (B-A is decreasing and approaches 0).

Does anyone know of another way to show that the above inequality is true or if there is some way to alter my method to make it work?

Thank you!

Substitute $ y\equal{}\frac{1}{x}$, so it's equivalent to showing $ \ln (1\plus{}y)<\frac{y\plus{}\frac{1}{2}y^2}{1\plus{}y}$ when $ \frac{1}{y}$ is a positive integer.

Now $ \ln (1\plus{}y)$ has derivative $ \frac{1}{1\plus{}y}$ and $ \frac{y\plus{}\frac{1}{2}y^2}{y\plus{}1}$ has derivative
\[ \frac{1\plus{}y}{1\plus{}y}\minus{}\frac{y\plus{}\frac{1}{2}y^2}{(1\plus{}y)^2}\equal{}1\minus{}\frac{y\plus{}\frac{1}{2}y^2}{(1\plus{}y)^2}\]
So we obtain
\[ \frac{y\plus{}\frac{1}{2}y^2}{1\plus{}y}\minus{}\ln (1\plus{}y)\equal{}\int_0^y 1\minus{}\frac{t\plus{}\frac{1}{2}t^2}{(1\plus{}t)^2}\minus{}\frac{1}{1\plus{}t}\ dt\]
from the derivatives and the fact that the left hand side is 0 when $ y\equal{}0$. Now put the integrand over a common denominator:
\[ (1\plus{}t)^2\minus{}(1\plus{}t)\minus{}(t\plus{}\frac{1}{2}t^2)\equal{}(1\plus{}2t\plus{}t^2)\minus{}(1\plus{}t)\minus{}(t\plus{}\frac{1}{2}t^2)\equal{}\frac{1}{2}t^2\]
\[ \Rightarrow \frac{y\plus{}\frac{1}{2}y^2}{1\plus{}y}\minus{}\ln (1\plus{}y)\equal{}\frac{1}{2}\int_0^y \frac{t^2}{(1\plus{}t)^2}\ dt>0\]
for $ y>0$.

thanks scorpius thats an awesome solution.


function
real analysis
real analysis solved

Find all continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f\left(\mathbb{R}\right) \subset \mathbb{Q}$.

I hope this is not too well-known...

The real line is connected, so also its image under continuous function must be connected. But the connected components of the rationals are singletons. This means that $f\equiv q$, with $q$ fixed rational number.


calculus
derivative
real analysis
real analysis unsolved

Let $f: \mathbb{R} \to \mathbb{R}$ be three times differentiable. Suppose that $f(-1) = f(0) = f'(0) = 0$ and $f(1) = 1$.
Show that $f'''(c) \geq 3$ for some $c \in (-1,1)$.

By Taylor, we have there is a $\xi\in]-1,1[$ for which $f(x)=f(0)+f'(0)x+\frac{f''(0)}2 x^2+\frac{f'''(c)}6 x^3=0+0+\frac{f''(0)}2 x^2+\frac{f'''(\xi)}6 x^3$.

Thus $1=f(1)-f(-1)=\frac{f'''(\xi_1)+f'''(\xi_2)}{6}$, and since derivatives have darboux property, there is a $c\in]\xi_1,\xi_2[$ such that $f'''(c)=3$. Right?

[quote="Peter VDD"]By Taylor, we have there is an $c\in]-1,1[$ for which $f(x)=f(0)+f'(0)x+\frac{f''(0)}2 x^2+\frac{f'''(c)}6 x^3=0+0+\frac{f''(0)}2 x^2+\frac{f'''(c)}6 x^3$.

Thus $1=f(1)-f(-1)=\frac{f'''(c)}{3}$, which means $f'''(c)=3$. Right?[/quote]
It seems not unless you omitted one the parts of you proof. I think you put in the relation
$f(x)=f(0)+f'(0)x+\frac{f''(0)}2 x^2+\frac{f'''(c)}6 x^3=0+0+\frac{f''(0)}2 x^2+\frac{f'''(c)}6 x^3$ at first $x=1$ and then $x=-1$, but you may have got different $c$, so there could be done a slight modification of this:
$f(1)=\frac{f''(0)}2 +\frac{f'''(c_1)}6$  and $f(1)=\frac{f''(0)}2 -\frac{f'''(c_2)}6$ and substractng form the first relation the second relation you will get $f'''(c_1)+f'''(c_2)=6$, so one of the $f'''(c_i),i=1,2$ must be $\geq 3$ and you are done.

Yes, I just thought of that too. :)

It is easy to fix though. (and totally identical to last year's IMC question)

That's exactly how I did it, nice question.

I hadn't noticed the link with the IMC problem :blush:


trigonometry
geometry
trig identities
Law of Sines

We have right angle $ABD$ with right angle at $A$. Extend $BD$ past $D$ to an arbitrary point $C$ (on ray $BD$), and let $\angle{CAD}=a^{\circ}$. Find $\angle{ADB}$ in terms of $a$, if $BD=2AC$.

[hide="hint"]Draw the midpoint $M$ of $BD$, and then draw $AM$.[/hide]

bump?

The message is too small. Please make the message longer before submitting.

[hide="Solution"]
WLOG $AC=1$, so $BD=2$.

Draw the midpoint $M$ of hypotenuse $BD$. We have that $AM=1$, which can be seen by inscribing $\triangle BAD$ in a circle. Also, $MD=1$.

Label $\angle ADB = \beta$.

Now, just some angle-chasing in the isosceles triangles:
\[\angle MAD = \beta \]

\[\angle ADC = 180-\beta \implies \angle MCA = \beta-\alpha \implies \angle AMC = \beta-\alpha \]

\[\angle MAC+\angle ACM+\angle CMA = (\beta+\alpha)+(\beta-\alpha)+(\beta-\alpha) = 180 \]

\[\implies \boxed{ \beta = 60+\frac{\alpha}{3}}\]
[/hide]

Well I got about that far, but I tried to get the sides by bashing with the law of sines, most ended up being really complicated trig expressions and I wasn't sure what to do with them.


geometry
circumcircle
trigonometry
inequalities
inequalities unsolved

$a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}\le 3^{\frac{7}{4}}R^{\frac{3}{2}}$
$a,b,c$ are sides of triangle with circumradius $R$.

After applying the Sine rule, this becomes \[2^{\frac{3}{2}}((\sin A)^{\frac{3}{2}}+(\sin B)^{\frac{3}{2}}+(\sin C)^{\frac{3}{2}})\leq 3^{\frac{7}{4}}\] But by Jensen's on $f(t)=(\sin t)^{3/2}$, obviously concave in an acute triangle, $(\sin A)^{\frac{3}{2}}+(\sin B)^{\frac{3}{2}}+(\sin C)^{\frac{3}{2}}\leq 3(\sin\frac{A+B+C}{3})^{\frac{3}{2}}=3(\frac{\sqrt 3}{2})^{\frac{3}{2}}=\frac{3^{\frac{7}{4}}}{2^{\frac{3}{2}}}$ and the solution obviously follows.

The only remaining case is the obtuse triangle, but in this case, WLOG let A be the obtuse angle, we simply set $A'=\pi-A$, and apply the same logic, except now the RHS comes out of Jensens as sine of less than $\pi/3$ which is clearly smaller.

Hmmm... Sine is actually concave over all of $(0,\pi)$. Oh well, if it wasn't my proof would still work :D.


$PACIFIC \ + \ BALTIC \ + \ ARCTIC \ = \ CCCCCCC$

Each letter represents a distict decimal digit. Find the digits that correspond to each of the letters, show ALL your work, using as little words and paragraphs as possible but showing sufficient work to prove you didn't guess and check. Have fun.

[hide]Because Pacific, Baltic, and Arctic all end in $C$, and so does the answer:

$3*C = C \text{ (mod 10)}$

This is only true when $C=0$ or $C=5$, but $C$ cannot equal $0$ because the sum is $CCCCCCC$.

Now looking at $I$ as the second to last letter in all three words, with the knowledge $C=5$ (and remembering that a $1$ is carried over, as well):

$3*I+1 = 5 \text { (mod 10)}$

So $I=8$.  Now looking at the 3rd to last letter:

$2*T+F+2 = 5 \text { (mod 10)}$
$2*T+F = 3 \text { (mod 10)}$

So $F$ is odd.  The possible ordered pairs $(T, F)$ are: $(9,5), (8,7), (7,9), (6,1), (5,3), (4,5), (3,7), (2,9), (1,1), (0,3)$.  Now, disregarding the cases when either $T$ or $F$ equals $5$ or $8$ or $T=F$: $(7,9), (6,1), (3,7), (2,9), (0,3)$.

When $(T,F)=(7,9)$, then the 4th to last letters have the equation:

$8+L+5+2 = 5 \text { (mod 10)}$
$L = 0 \text { (mod 10)}$
$L = 0$

When $(T,F)$ is any of the remaining ordered pairs besides $(0,3)$, then the 4th to last letters have a different equation because $1$ is carried instead of $2$:

$8+L+5+1 = 5 \text { (mod 10)}$
$L = 1 \text { (mod 10)}$
$L = 1$

So now the ordered pair $(6,1)$ can be thrown out because $F$ cannot equal $L$.

And when $(T,F)=(0,3)$, then the 4th to last letters have the final equation:

$8+L+5 = 5 \text { (mod 10)}$
$L = 2 \text { (mod 10)}$
$L = 2$

So now the possible ordered triplets for $(T,F,L)$ are: $(7,9,0), (3,7,1), (2,9,1), (0,3,2)$.

Regardless of what $T$, $F$, and $L$ equal, $10<I+L+C<20$, so the 5th to last letters have the equation:

$5+A+R+1 = 5 \text { (mod 10)}$
$A+R = 9 \text { (mod 10)}$
$A+R = 9$

Regardless of what $A$ and $R$ equal, $10<C+A+R<20$, so the 6th to last letters have the equation:

$2*A+B+1 = 5 \text { (mod 10)}$
$2*A+B = 4 \text { (mod 10)}$

So the possible ordered triplets $(A,B,R)$ are: $(9,6,0), (8,8,1), (7,0,2), (6,2,3), (5,4,4), (4,6,5), (3,8,6), (2,0,7), (1,2,8), (0,4,5)$.  But none of them can equal $5$ or $8$, leaving: $(9,6,0), (7,0,2), (6,2,3), (2,0,7)$.  Remembering the $(T,F,L)$ ordered triplets, it can be shown that $(A,B,R)$ cannot be $(7,0,2)$ or $(2,0,7)$ because this would leave no possiblities for $(T,F,L)$, since all the digits must be distinct.  So $(A,B,R)$ can only be $(9,6,0)$ or $(6,2,3)$.

Now finding the possibilities for $P$ is very similar to finding that of $L$:

When $(A,B,R)=(9,6,0)$, then

$P+2 = 5 \text { (mod 10)}$
$P = 3 \text { (mod 10)}$
$P = 3$

But this cannot be because there would be no possibilities left for $(T,F,L)$.  So $(A,B,R)=(6,2,3)$.

And when $(A,B,R)=(6,2,3)$:

$P+1 = 5 \text { (mod 10)}$
$P = 4 \text { (mod 10)}$
$P = 4$

And the only possiblility left for $(T,F,L)$ is $(7,9,0)$.

So $\text{L=0, B=2, R=3, P=4, C=5, A=6, T=7, I=8,}$ and $F=9$.

[/hide]


trigonometry
vector
trig identities
Law of Cosines

Find the angles of a triangle with vertices at (0,0), (17,6), and (-12,3).

[hint]
Distance Formula + Law of Cosines

or

Vector Dot product
[/hint]

still need assistence. can someone walk me through this.

Let the points be $A(0,0), B(-12,3)$ and $C(17,6)$. Using the distance formula, we find that $AB=\sqrt{153}$, $BC=\sqrt{850}$, and $AC=\sqrt{325}$. Applying the law of cosines twice to angles A and B, we find that they are $146.52^{\circ}$ and $19.94^{\circ}$ respectively. We then find  $m\angle C$ to be $13.54^{\circ}$.

You can, of course, do it with vectors, which may arguably be easier.

Methinks this should go in HSB.


I don't know if this problem has been solved. Find all integers $a,b,m,n$ taht: \[a^n-b^m=1\]

You see some non-obvious solutions like $a=3,\ b=2,\ n=2,\ m=3$
Another question:
Are the solutions of $a^n-b^m=k$ finite?

The question as to whether 8 and 9 are the only consecutive powers is [url=http://dmoz.org/Science/Math/Number_Theory/Open_Problems/Catalan_Conjecture/]Catalan's Conjecture[/url] and it was recently proved by [url=http://www-math.uni-paderborn.de/~preda/]Preda Mihailescu[/url].  It is D9 in UPINT3.  The problem of powers with a fixed distance was resolved by Robert Tijdeman who showed that the set was finite by giving an effective bound for the solutions.  Erdos asks whether there are finitely or infinitely many numbers not the difference of two powers: for example, is 6 a difference?  See [url=http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A074981]A074981[/url].


LaTeX
inequalities
inequalities proposed

[quote="fermat3"]1_  prove that for a b c  positive numbers   abc=1     1\1+a+b   +    1\1+b+c  +1\1+c+a  >= 1\2+a  +1\2+b  +1\2+c                                                                                2_if a b c >2  prove that  (a² +bc)(b²+ac)(c²+ab) >= 125abc

               3-prove or desprove that   a²+bc\(a+b)(b+c)     +b²+ac\(b+a)(b+c)   +c²+ab\(c+a)(c+b)    >= 1.5 :ninja:[/quote]

Dear fermat3, I please you as honestly as I can, please use LaTeX. It very more readable on AoPS to work with expressions written in LaTeX. For some LaTeX help, please visit [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]this site[/url]. Nice day.

[quote="fermat3"]1_  prove that for a b c  positive numbers   abc=1     1\1+a+b   +    1\1+b+c  +1\1+c+a  >= 1\2+a  +1\2+b  +1\2+c                                                                                2_if a b c >2  prove that  (a² +bc)(b²+ac)(c²+ab) >= 125abc

               3-prove or desprove that   a²+bc\(a+b)(b+c)     +b²+ac\(b+a)(b+c)   +c²+ab\(c+a)(c+b)    >= 1.5 :ninja:[/quote]

If your first inequality is:
$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\geq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$ then it is wrong. Let $a=1, b=2, c=\frac{1}{2}$ then LHS is $\frac{1}{4}+\frac{2}{7}+\frac{2}{5}=\frac{131}{140}$ and RHS is $\frac{1}{3}+\frac{1}{4}+\frac{2}{3}=\frac{5}{4}$ which is larger.

Your second inequality is also wrong. Let $a=b=c=\frac{12}{5}>2$ then we have $\left(\frac{288}{25}\right)^3<\left(\frac{300}{25}\right)^3=12^3=125*\frac{12}{5}*\frac{12}{5}*\frac{12}{5}$

[quote="Tiks"][quote="eLeM"][quote="fermat3"]1_  prove that for a b c  positive numbers   abc=1     1\1+a+b   +    1\1+b+c  +1\1+c+a  >= 1\2+a  +1\2+b  +1\2+c                                                                                2_if a b c >2  prove that  (a² +bc)(b²+ac)(c²+ab) >= 125abc

               3-prove or desprove that   a²+bc\(a+b)(b+c)     +b²+ac\(b+a)(b+c)   +c²+ab\(c+a)(c+b)    >= 1.5 :ninja:[/quote]

If your first inequality is:
$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\geq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$ then it is wrong. Let $a=1, b=2, c=\frac{1}{2}$ then LHS is $\frac{1}{4}+\frac{2}{7}+\frac{2}{5}=\frac{131}{140}$ and RHS is $\frac{1}{3}+\frac{1}{4}+\frac{2}{3}=\frac{5}{4}$ which is larger.

[/quote]
Correct wersion is
$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$
Which is the well-know problem from Bulgarian NMO,and it is discussed and solved befor ;)[/quote]

I haven't seen it before  :blush: I will try to solve it for myself  :)


How many pairs of positive integers $ (a,b)$ are there such that $ \gcd(a,b) \equal{} 1$ and
\[ \frac {a}{b} \plus{} \frac {14b}{9a}
\]is an integer?

$ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ \text{infinitely many}$

[hide="For anyone who loves good old algebra and is in a rush"]
$\frac{9a^2+14b^2}{9ab}=k$

$9a^2+14b^2=(9ab)k$

$9a^2-9kba + 14b^2 = 0$

$a = \frac{9kb \pm \sqrt{81k^2b^2-504b^2}}{18}= \frac{3kb \pm b\sqrt{9k^2-56}}{6}$

For $a$ to be an integer,
$9k^2-56=m^2 \Leftrightarrow (3k+m)(3k-m) = 56$
Now $56 = 1\cdot56 = 2\cdot28 = 4\cdot14 = 7\cdot8$
But only $2\cdot28$ and $4\cdot14$ are possible as the factors have to be disivible  by $6$ because $3k+m + 3k-m = 6k$. Hence we get $m = 13,5$ and $k=5,3$ respectively.
Then,
$a = \frac{9\cdot5b \pm 3\cdot13b}{6},\frac{9\cdot3b \pm 3\cdot5b}{6}$

$a =\frac{14b}{3},\frac{b}{3},\frac{7b}{3},\frac{2b}{3}$
Now if we let $b = 3k$
$a =14k,k,7k,2k$. But unless $k=1$, $ \gcd(a,b) \neq 1 $
Hence our 4 solutions are $(1,3), (2,3), (7,3), (14,3)$[/hide]

Could someone please explain me why $b$ has to be $3$. It must be divisible by $3$, but why can't it be $27$, for example?

[quote="tjhance"]
Let $\frac{a}{b}+\frac{14b}{9a}=n$ for some integer $n$.
$9a^{2}+14b^{2}=9abn$

[b]$9$ has to divide $b^{2}$. Otherwise, since $9$ clearly divides $9a^{2}$ and $9abn$, the left hand side would not be divisible by $n$ and the right hand side would. For $9$ to divide $b^{2}$, either $3$ or $9$ divides $b$. If $9$ divides it, then the right hand side would be divisible by $81$ but the left hand side would still only be divisible by $9$. That means only $3$ can divide $b$. If any other number divides it, it would also divide the right hand side but not the left hand side, because $a$ and $b$ are relatively prime. Thus the only possible value of $b$ is $3$.[/b]
[/quote]

[quote="hsbhatt"]easier if you note that a|14 and b|9[/quote]

what does the bar between the a and 14 and b and 9 stand for?

thx

$a|b$ means that $b$ is divisible by $a$.

[quote=tjhance][hide]
Let $\frac{a}{b}+\frac{14b}{9a}=n$ for some integer $n$.
$9a^{2}+14b^{2}=9abn$

$9$ has to divide $b^{2}$. Otherwise, since $9$ clearly divides $9a^{2}$ and $9abn$, the left hand side would not be divisible by $n$ and the right hand side would. For $9$ to divide $b^{2}$, either $3$ or $9$ divides $b$. If $9$ divides it, then the right hand side would be divisible by $81$ but the left hand side would still only be divisible by $9$. That means only $3$ can divide $b$. If any other number divides it, it would also divide the right hand side but not the left hand side, because $a$ and $b$ are relatively prime. Thus the only possible value of $b$ is $3$.

If something divides $a$ except for $1,2,7,$ and $14$ then it would not divide the term $14b^{2}$, but it would divide the right hand side and the term $9a^{2}$ and so this is impossible. So the only possible values of $a$ are $1,2,7,14$

We then have four possibilities of $(a,b)$: $(1,3),(2,3),(7,3),$ and $(14,3)$. If we plug them in we see that they all work, so the answer is $\boxed{4}$
[/hide][/quote]

Hey Guys sorry for revival but I have a question.

In the third paragraph, it states that something divides a except for 1,2,7 and 14, it wouldn't divide 14b^2. Where is this all coming from? Thanks.

It states that if $I|a$ for some integer $I$, $I$ must be a factor of $14$, or else the value of the original expression isn't an integer. Combined with the previous proof that the only valid value for $b$ is $3$, we can find the four solutions $(1,3), (2,3), (7,3), (14,3)$.


If $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$, then
\[ \frac1{(a\plus{}b)^2\plus{}6}\plus{}\frac1{(b\plus{}c)^2\plus{}6}\plus{}\frac1{(c\plus{}a)^2\plus{}6}\ge\frac3{10}\]
When equality occous?

[quote="dduclam"]If $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then
\[ \frac1{(a \plus{} b)^2 \plus{} 6} \plus{} \frac1{(b \plus{} c)^2 \plus{} 6} \plus{} \frac1{(c \plus{} a)^2 \plus{} 6}\ge\frac3{10}
\]
When equality occous?[/quote]
$ \sum_{cyc}\frac1{(a \plus{} b)^2 \plus{} 6}\ge\frac3{10}\Leftrightarrow\sum_{cyc}\left(\frac {1}{(3 \minus{} a)^2 \plus{} 6} \minus{} \frac {1}{10} \minus{} \frac {1}{25}(a \minus{} 1)\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(a \minus{} 1)^2(5 \minus{} 2a)}{a^2 \minus{} 6a \plus{} 15}\geq0.$
Thus, our inequality is proven for $ \max\{a,b,c\}\leq2.5.$
Let $ 2.5 < a\leq3.$ Hence, $ 0\leq b \plus{} c < 0.5.$
But $ \left(\frac {1}{x^2 \minus{} 6x \plus{} 15}\right)'' \equal{} \frac {6(x^2 \minus{} 6x \plus{} 7)}{(x^2 \minus{} 6x \plus{} 15)^3} > 0$ for $ 0\leq x < 0.5.$
Hence, $ \sum_{cyc}\frac {1}{a^2 \minus{} 6a \plus{} 15}\geq\frac {2}{\left(\frac {b \plus{} c}{2}\right)^2 \minus{} 6\cdot\frac {b \plus{} c}{2} \plus{} 15} \plus{} \frac {1}{a^2 \minus{} 6a \plus{} 15} \equal{}$
$ \equal{} \frac {2}{\left(\frac {3 \minus{} a}{2}\right)^2 \minus{} 6\cdot\frac {3 \minus{} a}{2} \plus{} 15} \plus{} \frac {1}{a^2 \minus{} 6a \plus{} 15} \equal{} \frac {8}{a^2 \plus{} 6a \plus{} 33} \plus{} \frac {1}{a^2 \minus{} 6a \plus{} 15}.$
Id est, it remains to prove that
$ \frac {8}{a^2 \plus{} 6a \plus{} 33} \plus{} \frac {1}{a^2 \minus{} 6a \plus{} 15}\geq\frac {3}{10},$ which is equivalent to
$ (a \minus{} 1)^2(3 \minus{} a)(a \plus{}5)\geq0,$ which is true for $ 2.5 < a\leq3.$ Done! :)

[quote="arqady"][quote="dduclam"]If $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then
\[ \frac1{(a \plus{} b)^2 \plus{} 6} \plus{} \frac1{(b \plus{} c)^2 \plus{} 6} \plus{} \frac1{(c \plus{} a)^2 \plus{} 6}\ge\frac3{10}
\]
When equality occous?[/quote]
$ \sum_{cyc}\frac1{(a \plus{} b)^2 \plus{} 6}\ge\frac3{10}\Leftrightarrow\sum_{cyc}\left(\frac {1}{(3 \minus{} a)^2 \plus{} 6} \minus{} \frac {1}{10} \minus{} \frac {1}{25}(a \minus{} 1)\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(a \minus{} 1)^2(5 \minus{} 2a)}{a^2 \minus{} 6a \plus{} 15}\geq0.$
Thus, our inequality is proven for $ \max\{a,b,c\}\leq2.5.$
Let $ 2.5 < a\leq3.$ Hence, $ 0\leq b \plus{} c < 0.5.$
But $ \left(\frac {1}{x^2 \minus{} 6x \plus{} 15}\right)'' \equal{} \frac {6(x^2 \minus{} 6x \plus{} 7)}{(x^2 \minus{} 6x \plus{} 15)^3} > 0$ for $ 0\leq x < 0.5.$
Hence, $ \sum_{cyc}\frac {1}{a^2 \minus{} 6a \plus{} 15}\geq\frac {2}{\left(\frac {b \plus{} c}{2}\right)^2 \minus{} 6\cdot\frac {b \plus{} c}{2} \plus{} 15} \plus{} \frac {1}{a^2 \minus{} 6a \plus{} 15} \equal{}$
$ \equal{} \frac {2}{\left(\frac {3 \minus{} a}{2}\right)^2 \minus{} 6\cdot\frac {3 \minus{} a}{2} \plus{} 15} \plus{} \frac {1}{a^2 \minus{} 6a \plus{} 15} \equal{} \frac {8}{a^2 \plus{} 6a \plus{} 33} \plus{} \frac {1}{a^2 \minus{} 6a \plus{} 15}.$
Id est, it remains to prove that
$ \frac {8}{a^2 \plus{} 6a \plus{} 33} \plus{} \frac {1}{a^2 \minus{} 6a \plus{} 15}\geq\frac {3}{10},$ which is equivalent to
$ (a \minus{} 1)^2(3 \minus{} a)(a \minus{} 5)\geq0,$ which is true for $ 2.5 < a\leq3.$ Done! :)[/quote]
 :coolspeak:
Very beautiful solution arqady

Thank you, Dimitris X! I have fixed a little mistake.


geometry
analytic geometry
complex numbers
absolute value

analytic geometry! yea, think of $x$-axis as the real axis and $y$-axis  as the imaginary axis. so most of the things but weird things can happen in the complex plane...

I know a little bit about the complex plane. For example, complex numbers aren't "orderly" like normal numbers, AKA you can't say that, for example, $3i$ is larger than $2i$ or that $7+3i$ is larger or smaller than $8+7i$--because what is size when we're talking about a number that's imaginary?

I only know of one tie to geometry: finding the absolute value of a complex number, or the distance it is from the origin. If the real numbers run along one axis and the imaginary numbers run along the other, a complex number will be somewhere in the plane; then you can just draw a line from the point it lies on to the origin, make a right triangle with either axis, and use the pythagorean theorem. So if your complex number is in the form $a+bi$, its absolute value is $\sqrt{a^2 +b^2}$.

[quote="chess64"]Where can I learn how to use complex numbers in Geometry?[/quote]

The Art of Problem Solving's Intermediate Trigonometry/Complex Numbers class is great.  That's where I learned about complex numbers and how to use them in geometry.

Sounds interesting. I'll probably get to that point somewhere in the next school year.


For each positive integer $n$ , let $f(n)$  be the maximal natural number such that: $2^{f(n)}$ divides $[(1+\sqrt3)^n]$.find $n$ so that $f(n)=2005$


Joseph wrote on a big piece of paper all the numbers from 1 to 9999. How many zero digits did he have to write down?

[hide]The number of zeros that Joseph has to write from 1 to 999 is
189
The number of zeros that Joseph has to write from 1 to 9999 is
2889
The number of zeros that Joseph has to write from 1 to 99999 is
38889
The number of zeros that Joseph has to write from 1 to 999999 is
488889.

In our case, the answer is 2889.

***This is not my solution.  I don't know why the above trick works.  Could someone explain, please?***

Thanks.[/hide]

What it is trying to show is a pattern.

0s from 1 to 9
0

0s from 1 to 99
9

etc.

Is there a generalized formula/algorithm to remember for this type of problem?

Thanks.


[hide="Solution"] I got 3/5 for the answer; There are 30 total possibilities (6 on the first draw, 5 on the second) for your denominator. From here you have to take it situationally. If you pick a quarter, any of the five other coins will work (as you only have quarters, dimes, and pennies). With a dime, only quarters will work and the same goes for nickels. So, you now have $5+2+2=9$ working situations. However, you have two of each coin, so the fraction is $(2*9)/30=18/30=3/5$ giving you the answer.[/hide]

[quote="Treething"][quote="GoBraves"][hide]I got 5/8[/hide][/quote]
Uhm... How could you get 8 as the denominator when the answer's x/30? The denominator would have to be a factor of 30...[/quote]

Actually I was wondering the same thing... O well everyone's human...

[quote="Treething"][quote="GoBraves"][hide]I got 5/8[/hide][/quote]
Uhm... How could you get 8 as the denominator when the answer's x/30? The denominator would have to be a factor of 30...[/quote]

My mistake, i completly misunderstood the problem.

[quote="jli"][hide]he needs to choose a quarter which is 2/8

then he can choose a ...

quarter=1/7

dime=2/7

nickel=2/7

penny=2/7

2/8(1/7+2/7+2/7+2/7)=2/8=1/4  it think[/hide][/quote]

There were no pennies in the problem... i think that was the same mistake that GoBraves made... idk... o well i think you'll get the answer if you take out the pennies.

[quote="The problem"]A driver approaches a toll booth and randomly selects two coins from his pocket. If the pocket contains 2 quarters, 2 dimes, and 2 nickels, what is the probability that the value of the 2 coins he selects will be at least enough to pay the 30-cent toll? Express your answer as a common fraction.[/quote]

[hide]The way I see it, he has to pick a quarter and something else. He has 4*3 chances of not having enough and a 2*5*(2 different ways to arrange it) chances of having the right amount, for a $\frac{5}{8}=\frac{5}{11}$ probability.[/hide]

Billy

[quote="solafidefarms"][quote="The problem"]A driver approaches a toll booth and randomly selects two coins from his pocket. If the pocket contains 2 quarters, 2 dimes, and 2 nickels, what is the probability that the value of the 2 coins he selects will be at least enough to pay the 30-cent toll? Express your answer as a common fraction.[/quote]

[hide]The way I see it, he has to pick a quarter and something else. He has 4*3 chances of not having enough and a 2*5 chances of having the right amount, for a $\frac{10}{22}=\frac{5}{11}$ probability.[/hide]

Billy[/quote]

But aren't there 30 combinations? I see your logic, and you're right about the number of combinations that don't work, but you omitted (I think) combinations that have the quarter picked second.

[hide="Answer"] $3/5$, if i didn't leave out anything. My solution is above, on like the third message.  [/hide]

[hide]
Okay, in order to get 30 cents you must draw at least one quarter.  The next highest possibility, two dimes, isn't enough, while a quarter and nickel will.  So the chance of drawing a quarter is 1 minus the chance that neither coin is a quarter:
$\ \frac{4}{6}*\frac{3}{5}=\frac{2}{5}$

$\\1-\frac{2}{5}=\frac{3}{5}$.[/hide]

But each sum has different probabilities. For example, the 25/25 one has a $\frac{1}{30}$ prob, while the 10/5 has a $\frac{2}{15} $chance

[quote="cj"]What is wrong with this way:

There are a total 6 different values that can be obtained:

25+25 = 50
25+10 = 35
25+5 = 30
10+10 =20
10+5 = 15
5+5=10

Only 3 of them are enough to pay the 30 cents toll.

So  p= 3/6 = 1/2[/quote]

Adding to what solafidefarms said, there is more than one way to attain a ten and a five, which leads to different weightings. If you took into account the ways that you could get thirty cents and the order in which they would be attained, you would get $3/5$ (see my solution above)

[hide]I got 3/10 because:
As all agreed, there are 30 ways to pick two coins out of six coins. The combinations of desired results are as following:

Q D N
1 1 0   -Four ways to get a quarter and a dime
1 0 1   -Four ways to get a quarter and a nickel
2 0 0   -One way to get two quarters

So, the answer is 3/10.

Or am I wrong?[/hide]

My weak point is probability so I might not have gotten this right...
[hide] I first started by finding the number of possibilities that he could obtain with two coins and I got 9...so that will be the denominator.  The only possibilites that do not work are nickels and dimes, because there are two of them there are 4 possibilities there.  therfore your answer is 5/9

Wow. Looking at all these solutions is scary. Here's mine, and it achieves yet another different answer.
[hide]
The only possiblilities are QQ, QN, QD:

QQ: 2/6 x 1/5
QN: 2/6 x 2/5
QD: 2/6 x 2/5

Multiply and add, and you get 1/3. [/hide]

[quote="hockeyadi23"] There are 30 total possibilities (6 on the first draw, 5 on the second) for your denominator. [/quote]

I thought we need to draw two coins at the same time. So it should be C(6,2) = 15 possibilities, not 30.

[quote="hockeyadi23"]
CJ, do you have the answer to the problem, because there are tons of answers that hav popped up, which I have listed below:
-3/5 2x
-1/3
-5/9
-3/10
-1/2
-5/11
-1/4
-5/8
[/quote]

Yes the answer is 3/5.

My answer is 3/5.

For the first coin, let's say you got a quarter (a 1/3 chance)
Then, for the second coin, you can select any of the coins remaining. (1/1)
Getting a quarter and having the second coin to have the 2 values to add up to 30 cents is 1/3 times 1/1 or 1/3.

For the first coin, let's say you get a nickel (1/3).
Then, you have to get one of the two quarters out of the 5 remaining (2/5)
The chance of this happening is 1/3 times 2/5 or 2/15.

Finally, you can get the dime (1/3).
Then, you would also have to get one of the two quarters out of the 5 remaining (2/5)
The chance of this happening is 1/3 times 2/5 or 2/15 again.

Adding these, 1/3 + 2/15 + 2/15 = 9/15, simplifying to 3/5.

Or you could've made a list. That would've been shorter to type, but this is the way to compute the answer (at least how I think. I'm love math, but I just started doing mathcounts related questions).


Here is a nice question I need help on:

A real esate office handles 50 apartment units.  When the rent is $540 per month, all units are occupied.  However, for each $30 increase in rent, one unit becomes vacant.  Each occupied unit requires an average of $18 per month for service and repairs.  What rent should be charged to obtain the most profit.

If possible, equation too.

I got a different answer than Rep123max.  First off, I used x to be the number of rooms occupied instead of the number of rooms unoccupied.  Then I basically did everything the same as he did.  If you substitute k in for x in his equation, where k=50-x, you will get the equation I have.  The max for this equation is when x=33.7, however trying out x=33 and x=34 gives 34 to be slightly higher.  The rent corresponding to x=34 is 2040-30*34=$1020.

Rep's error may have been that he substituted 33 and 34 in for x, when he was using the equation that has x as the number of unused rooms, so what he should have checked was 50-33 and 50-34.


Let $ a$, $ b$ and $ k$ be three positive integers.
We define two sequences $ \left(  a_{n}\right)$ and $ \left(  b_{n}\right)$ by the starting values $ a_{1}\equal{}a$ and $ b_{1}\equal{}b$ and the recurrent equations $ a_{n\plus{}1}\equal{}ka_{n}\plus{}b_{n}$ and $ b_{n\plus{}1}\equal{}kb_{n}\plus{}a_{n}$ for each positive integer $ n$.
Prove that if $ a_{1}\perp b_{1}$, $ a_{2}\perp b_{2}$ and $ a_{3}\perp b_{3}$ hold, then $ a_{n}\perp b_{n}$ holds for every positive integer $ n$.

Here, the abbreviation $ x\perp y$ stands for "the numbers $ x$ and $ y$ are coprime".

[hide="Hint"]$ a_n\equal{}t_na\plus{}t_{n\minus{}1}b$ and $ b_n\equal{}t_nb\plus{}t_{n\minus{}1}a$ where $ t_0\equal{}0, t_1\equal{}1$, and $ t_{n\plus{}1}\equal{}kt_n\plus{}t_{n\minus{}1}$.[/hide]

I used a different method than Krishanu.  His way is probably faster, but here's my solution:

[hide="Full Solution"]Suppose for the sake of contradiction that $ a_{n\plus{}1}$ and $ b_{n\plus{}1}$ are not coprime for some $ n\ge 3$.  Then for some prime $ p$, we have $ p|a_{n\plus{}1}$ and $ p|b_{n\plus{}1}$.  Using the given recursions, $ p|ka_{n}\plus{}b_{n}$ and $ p|kb_{n}\plus{}a_{n}$.

Adding these gives $ p|(k\plus{}1)(a_{n}\plus{}b_{n})$, so $ p|k\plus{}1$ or $ p|a_{n}\plus{}b_{n}$.  But $ a_{n}\plus{}b_{n}\equal{}(k\plus{}1)(a_{n\minus{}1}\plus{}b_{n\minus{}1})$, so $ k\plus{}1|a_{n}\plus{}b_{n}$.  Therefore $ p|a_{n}\plus{}b_{n}$.  Then $ p|(ka_{n}\plus{}b_{n})\minus{}(a_{n}\plus{}b_{n})$, so $ p|(k\minus{}1)a_{n}$.  Analogously, $ p|(k\minus{}1)b_{n}$.

Case 1: $ p|k\minus{}1$.  We have $ k\equiv 1\pmod{p}$.  Then for all $ m\ge 1$, we have $ a_{m\plus{}1}\equiv a_{m}\plus{}b_{m}\equiv b_{m\plus{}1}\pmod{p}$.  Also, $ a_{m\plus{}2}\equiv a_{m\plus{}1}\plus{}b_{m\plus{}1}\equiv 2a_{m\plus{}1}\pmod{p}$.  Therefore, $ a_{n\plus{}1}\equiv 2^{n\minus{}1}a_{2}\pmod{p}$.  Since $ p|a_{n\plus{}1}$, either $ p\equal{}2$ or $ p|a_{2}$.  Both of these lead to contradictions:

If $ p\equal{}2$, then $ a_{3}\equiv 2a_{2}\equiv 0\pmod{2}$.  Using $ a_{m\plus{}1}\equiv b_{m\plus{}1}\pmod{p}$ for all $ m\ge 1$, we have $ 2|b_{3}$, a contradiction since $ a_{3}$ and $ b_{3}$ are coprime.

If $ p|a_{2}$, then using $ a_{m\plus{}1}\equiv b_{m\plus{}1}\pmod{p}$ for all $ m\ge 1$, we have $ p|b_{2}$, a contradiction since $ a_{2}$ and $ b_{2}$ are coprime.

Case 2: $ p$ does not divide $ k\minus{}1$.  By our reasoning above, if $ p|a_{m\plus{}1}$ and $ p|b_{m\plus{}1}$, then $ p|(k\minus{}1)a_{m}$ and $ p|(k\minus{}1)b_{m}$.  Therefore $ p|a_{m}$ and $ p|b_{m}$.  We apply this argument repeatedly until we get $ p|a_{3}$ and $ p|b_{3}$, a contradiction.

Therefore for all $ n$, $ a_{n}$ and $ b_{n}$ are coprime.[/hide]


probability
real analysis
real analysis solved

a ballot box contains 6 numbered pawns from 1 to 6 one draws the balls with one with handing-over an infinity of time. which is the probability so that a number fixed never appears

Sorry, but I don't understand what you ask for...  :?
And it doesn't look like a problem that belongs to Superior Algebra...  :huh:

If you're asking what I think you're asking, the probability is obviously zero.


geometry
incenter
circumcircle
inequalities
Euler
ratio
angle bisector

Let  $H$, $I$, $O$ be the orthocenter, incenter and circumcenter of a triangle. Show that $2 \cdot IO \geq IH$. When does the equality hold ?

[color=blue][b]Problem.[/b] Let H be the orthocenter, I the incenter and O the circumcenter of a triangle ABC. Prove that the inequality $2\cdot IO\geq IH$ holds, and that it becomes an equality if and only if the triangle ABC is equilateral.[/color]

[i]Solution.[/i] If the triangle ABC is equilateral, then H = I = O, so that IO = 0 and IH = 0, and thus the inequality $2\cdot IO\geq IH$ becomes a trivial equality. Remains to show that if the triangle ABC is not equilateral, then the inequality $2\cdot IO\geq IH$ holds and does not become an equality, i. e. we have $2\cdot IO>IH$.

So consider the case when the triangle ABC is not equilateral. Note that in the following, we will use [i]non-directed[/i] segments.

According to http://www.mathlinks.ro/Forum/viewtopic.php?t=16552 , the incenter I of triangle ABC lies inside the circle with diameter GH, where G is the centroid of triangle ABC. Thus, $\measuredangle GIH>90^{\circ}$. But on the other hand, by the famous Euler line theorem, the points H, G and O lie on one line, the so-called Euler line e of triangle ABC, and, more precisely, the point G lies on the segment HO and divides it in the ratio $\frac{HG}{GO}=2$.

Let the angle bisector of the angle OIH meet the Euler line e at a point W. Then, $\measuredangle WIH=\frac{\measuredangle OIH}{2}$; since $\measuredangle OIH<180^{\circ}$, we have $\measuredangle WIH=\frac{\measuredangle OIH}{2}<\frac{180^{\circ}}{2}=90^{\circ}$. Comparing this to $\measuredangle GIH>90^{\circ}$, we get $\measuredangle WIH<\measuredangle GIH$. Hence, the point W lies between the points H and G on the Euler line e. Therefore, HW < HG and WO > GO. Thus, $\frac{HW}{WO}<\frac{HG}{GO}$.

Now note that the point W is the point of intersection of the angle bisector of the angle OIH with the line e = OH. Since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we therefore have $\frac{HW}{WO}=\frac{IH}{IO}$. With the help of this equality and also of $\frac{HG}{GO}=2$, we can transform the inequality $\frac{HW}{WO}<\frac{HG}{GO}$ into $\frac{IH}{IO}<2$. Thus, $IH<2\cdot IO$, so that $2\cdot IO>IH$. This completes the proof.

[color=red][Edit: Problem also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=54012 .][/color]

  Darij


[quote="Extremal"]how can I take parts in international olympic in mathematics? some body can tell me or give me information about this? my cauntry is so poor I don't no what can I do  and haven't money. But in mathematics I think I am not bad I can give you different , nice and interesting sulution every problems wich was be in international mathematical olympic by myself[/quote]

I would think that most countries have an IMO team. What country are you from?

His/Her flag is Georgia's flag.

According to the [url=http://erdos.fciencias.unam.mx/results.htm]2005 IMO Results[/url], Georgia had their IMO team.
I don't know how the team selection works in Georgia, though...

You should try and find out what tests your country uses to determine its IMO team.  You might check the memberlist to see if there are other Georgian members of the forum, although you should watch out for people from the American state of Georgia, and you should maybe ask in the Russian subforum where I know residents of a number of former Soviet republics have posted.  I wasn't able to find anything about the Georgian IMO team with a Google search, but presumably you will have better luck if you search in the Georgian language.


Hi.

I've recently written a rather long report in LateX, with images included as .eps. When i "precompile" the text to DVI, the images show fine. When I compile into a PDF, they all disappear.

The error I seem to be getting is "cannot find file", where apparently the DVI could find the file just fine.

Anybody know how to solve this? I really need it in PDF format.

Regards,
Roger

You need to compile to PS first then to PDF. If you are using TexnicCenter then choose LaTeX => PS => PDF instead of LaTeX => PDF. That converts it to DVI then to PS then to PDF so you should be able to see your images.

Am indeed using TexNiCCenter but my version does not seem to have that option.

Anyway, I tried compiling directly from DVI using dvipdfm, with that solution the boundingbox shows but none of the actual pictures.

I'll try to compile to PS first and then do PS->PDF manually if I can, compiler executables gotta be on my harddrive somewhere ..

It may be missing that option if you don't have Ghostscript which does most of the work and is easy and free to download. If you want to add such an option and the options wizard is not helpful, then let me know and I'll send you screen shots of what is required.


If the warrior princess Mistar were to ascertain how many men and how many horses she has under her command by counting both the legs and heads, she would count 45 heads and 120 legs. How many horses are under Mistar's command ?

[i]from "Fantastic book of Math Puzzles" by Margaret C. Edmiston[/i]

[u]Please hide Your solution . Write inside Hide ... /Hide[/u]

[hide="Solution"]

$ m \plus{} h \equal{} 45$

$ 2m \plus{} 4h \equal{} 120$

$ m \plus{} 2h \equal{} 60$

$ h \equal{} 15$ and $ m\equal{}30$

[/hide]

Hide $ men\plus{}horse\equal{}45$so

$ 2(men)\plus{}4(horse)\equal{}120$

$ 4(men)\plus{}4(horse)\equal{}180$

$ 2(men)\equal{}60$

$ men\equal{}30 and horse\equal{}15$ \Hide

m-men h-horses

m+h=45
4h+2m=120
so 2h+m=60
subtract first equation from third
h=15
notice the problem doesnt ask for number of men.

[hide]
Supposing that every one of them was a man and disregarding the fact that there must be $ 120$ legs for now, there would be $ 45\times 2 \equal{} 90$ legs (that sounds rather suspicious).

If we go trade a man for a horse, the number of legs would increase by $ 2$.
We want to increase the number of legs by $ 120\minus{}90\equal{}30$, so we trade $ \boxed{15}$ men for horses.
[/hide]

Let the number of horses=x.
                                               (x+y=45).
Let the number of men=y.

A horse has 4 legs, and a man has 2 legs.     (no. of legs=120).

So,
    x+x+y+y+y=120

    x+y+x+y+y+y=120   (here x+y=45)

    45+45+y+y=120

    90+2y=120

    2y=30(120-90)

    y=30/2=15

    So,number of horses=15

m=man, h=horse

man=2 legs, horse=4 legs

m+h=45

2(m)+4(h)=120

4(m)+4(h)=180

2(m)=60

m=30 and [b]h=15[/b] NOTICE THAT THE QUESTION ASKED FOR HORSES ONLY!!!

I sincerely hope that each horse and man has one head.

[hide]If everyone was a man, there would be 45 heads and 90 legs.
For every horse that replaces a man, 2 extra legs are gained.
120-90 = 30, 30/2 = [b]15 horses[/b][/hide]

why can't someone just put a simple
15 horses
and it should be warrior princess Loor


dirooz hodoodaye 5 bad az zohr maloom shod.
محمدرضا ارمندپور
مهسا تازيكي
امير سپهري
محمدسعيدسرافراز
علي صادقيان
محمد جهانگشاهي
نيما حميدي
نادر عطرچين
محمدعلي كرمي
 اميرمسعود گيوه چي
محمودوحيدنيا
محمد مهدي يزدي
man be hamashoon tabrik migam.

manam be hameye bax makhsoooosan baxee alllame tabatabaeii ( ali sadeghian va milad bakhshizade) tabrik migam ke milad parsal tala shod dg fek konam emsal esmesh nemiad dgggg :wink:

salam
manam be hame makhsoosan tak dokhtareshun tabrik migam.
inja kasi az noghrehaye com khabar nadare??
mamnoon

salam
comia faghat boronzashoon maloom mishe

[quote="sahba"]salam
kasi midoone Msal max & min nomre baraye ghabooly marhale 2 olamp riazi ghad boode?
khahesh mikonam begin. :([/quote]
[quote="mehdi"]kaf 34 boode[/quote]

be hameye bache ha tabrik migam :D

movafagh bashid


inequalities
geometry
circumcircle
inradius
trigonometry
geometry solved

Let $R,r$ respectively the circumradius and the inradius of a triangle $\triangle ABC$ and $I$ be its incenter. Prove that
\[
\sum\limits_{cyclic} {\frac{1}{{IA \cdot IB}}}  \le \frac{{5R + 2r}}{{8Rr^2 }}
\]

It's not a hard problem .
 $ \sum\frac{1}{IB.IC} \leq \frac{5R+2r}{8Rr^2} $
$ \iff \sum{\sin\frac{B}{2}.\sin\frac{C}{2}} \leq \frac{5}{8} + \frac{1}{4}(\sum \cos A - 1) $
$ \iff (\sum \sin\frac{A}{2})^2 \leq \frac{9}{4} $ : OK


so how did your school do?

i dont know my schools exact score, because i left right after the test and didn't wait to see the results, but i answered around 70 myself,
but i think my school scored pretty high. :)

I wasnt there, and our team sucked. They only got 50.

[quote="Elemennop"]81 here (vestavia).

Bummer, we didn't do that well. we missed a whole lot in the 50-75 range.[/quote]

wow, that is still pretty good....   (probably first place) :)

we got 70.

weird,
they have not yet updated their website, about when the results will be out. ;)


trigonometry
Functional Analysis
real analysis
real analysis unsolved

Let $ A$, $ B$ bounded, convex, closed sets. The $ A\plus{}B$ is convex ,closed, bounded in $ \mathbb{R}^n$? And in a Banach-space?

In R(n), yes. In a Banach space, I think A + B might not be closed. I'll think about it tomorrow after some sleep..

Even in separable Hilbert spaces there are closed, disjoint subspaces whose sum is not closed

It's something like this (example obviously not due to me):

Let $ H \equal{} \ell^2$. $ M \equal{} \{ (x_n) \, : \, x_{2k} \equal{} 0\}$, $ N \equal{} \{ (x_n) \, : \, x_{2k} \equal{} y_k\cos k , x_{2k \minus{} 1} \equal{} y_k\sin k , (y_k) \in \ell^2 \}$

These are closed subspaces, and the orthogonal complement of $ M \plus{} N$ is empty, so it's dense, so all we have to do is exhibit $ (a_n) \in \ell^2$ not in $ M \plus{} N$.

Let $ (z_k) \in \ell^2$ and let $ a_{2k} \equal{} z_k\sin k$ and $ a_{2k \minus{} 1} \equal{} 0$. If this were of the form $ (m_k) \plus{} (n_k)$ we would have $ y_k\sin k \equal{} \minus{} m_{2k \minus{} 1}$ and $ z_k\sin k \equal{} y_k\cos k$, some $ (y_k) \in \ell^2$.

So $ m_{2k \minus{} 1} \equal{} \minus{} z_k\frac {\sin^2 k}{\cos k}$, but we can choose $ z_k$ in $ \ell^2$ so that the RHS is not in $ \ell^2$

I think it's very clear from what I've written.  :?:

Each $ (y_k) \in \ell^2$ defines an element in $ N$.  $ (a_k)$ is defined in the text, and $ (m_k) \in M$ ($ (n_k) \in N$) such that $ (m_k) \plus{} (n_k) \equal{} (a_k)$

Why is the set M convex,bounded? N is closed convex,bounded?


Trochę inny problem. Znacie jakieś dobre książki z zakresu Liceum? Powtarzam sobie do matury, ale jak tylko trafi się coś trudniejszego, to wszystko leży, a nawet jak zrobię zadanie to nie ma do tego zadania odpowiedzi. Najdoskonalsze by były na wzór tej którą mam do fizyki, pełne odpowiedzi do wszystkich zadań. Ok. Jak znacie to zapostujcie tutaj.


Solve the ecuation $ x!\equal{}x^3\minus{}5$ , where $ x!\equal{}1\cdot 2\cdot 3\cdot 4\cdot...\cdot x$

[hide="how do I make this solution concrete?"]

Factorial grows much faster than a cubic function.  For values greater than zero, both are concave up, so there is at most one solution.

For any $ x>5$, $ x!$ is divisible by $ 10$.  Thus, we only need to check multiples of $ 5$ and values less than it: $ 0 \rightarrow 4$.

$ 5$ works.[/hide]

use this:
$ \forall x \ge 6 \Longrightarrow x! > x^3$

[b]@azn:[/b] You need to change the angle of your approach. Consider $ 2^x$ and $ x$ -- both are increasing, the first is convex, and grows much faster than $ x$, but there are still two solutions.

[hide="Expanding on mathson"]Suppose $ n! > n^3$ for some $ n \in \mathbb{N}$. Then, $ (n \plus{} 1)! \equal{} (n \plus{} 1) n! > (n \plus{} 1) n^3 \equal{} n^4 \plus{} n^3 \geq 2 n^3$

Note that for $ n > 4$, we have $ 2n^3 > (n \plus{} 1)^3$. Thus, for $ n > 4$, we have $ n! > n^3 \implies (n \plus{} 1)! > (n \plus{} 1)^3$

Note that $ 6! > 6^3$, and $ 6 > 4$, so inductively, $ x! > x^3$ for all $ x \geq 6$.

It suffices to check $ x \in \{ 0 ..5 \}$, and the only solution is $ x \equal{} 5$.[/hide]


geometry
function
probability
logarithms
AMC
AMC 12
AMC 12 A

All the best for all of u guys writing the AMC's.

Please continue the discussion of the AMC problems in this forum.

Befuddlers

120: missed #6 and #12 :(, #24, #25, and one other. (but correctly guessed the radical on #23 :D)

I got 129 (23 answered - 2 mistakes). Personally I feel this AMC is much easier than the B test last year (in which I got 108). Can't discuss problems right now because I'm in school.

P.S.: #25 was really surprisingly easy, despite my division mistake.

Failed.  Got 111?
1-18 and left two blank.  Figured out how to do 19, 21, and 24 afterwards.  AAGH!

The lack of probability was annoying for me.

[quote="1076207"]Yay got 130.5 (22 answered - 1 mistake). I do think it was easier... but Walton except me failed the geometry portion. There was more geometry, wasn't there?

By the way, for #21, did you know that cubics have 4 roots? It's because a cubic function has 4 terms, so it obviously must have 4 roots.[/quote]

Yep. And why are negative roots considered real? The negative sign obviously indicates its nonrealness.

JK.  :P

I personally thought this test's geometry was easier because of decreased requirement of bruteforcing. (None of the last 5 were geometry, anyways)

Hehe, I, being the slacker I am, took the 10 and failed.  Let's not talk about my score just yet.   :(

[quote="1076207"]By the way, for #21, did you know that cubics have 4 roots? It's because a cubic function has 4 terms, so it obviously must have 4 roots.[/quote]

I don't think I understand... Could you enlighten me? :)

@Oliver: he's being sarcastic about his mistake, I suppose.

ohh... yeah I must've missed the sarcastic tone of voice  :lol:

[quote="1076207"]Yay got 130.5 (22 answered - 1 mistake). I do think it was easier... but Walton except me failed the geometry portion. There was more geometry, wasn't there?
[/quote]

So my reaction was "where's all the geometry??! :(" - I thought exactly the opposite, I thought it was ridiculously algebraic and not a bit geometric. :(

I was probably influenced by my fellow Walton teammates. Henry kept on complaining how there were 11 (OMG 11) geometry questions. But yea, most of the harder questions were algebraic.

By the way, how many people missed #18? It was the one with $ 10^{k\plus{}2} \plus{} 64$ and about factors of 2. I think that was probably one of the questions where the most points were lost. I did like the problem though; simple but tricky.

#21: Somehow added 2+2+4=10 and answered D) 10
#23: Got $ 150(1 \plus{} \sqrt {2})^2 \equal{} 150(3 \plus{} 2\sqrt {2}) \equal{} 750 \plus{} 500\sqrt {2}$, and didn't know what I did wrong, and left it blank
#24: Got that $ \log^{2011}_2(B) \equal{} 1 \plus{} \epsilon$, and thought that $ \log^{2012}_2(B) < 0$, and answered D) 2012.

So, I got a 133.5 on the AMC12A for the 3rd year in a row.

I'm going to guess that Miles and Eric got 135+. Then again, Eric could have gotten yet another 132.

Now for some positives & other observations:

I actually got 7 hours of sleep the night before.

#10: nice "duh" problem (Triangle inequality and quadrilateral) after 5 mins of overcomplicated it.

#14, #16, and #17 all used the sum of roots of a quadratic.

#18: my initial guess was 6, but after some investigation, I answered 7.

#19: Even though Miles showed me the derivation of $ \cos(\frac {2}{5}\pi) \equal{} \frac { \minus{} 1 \plus{} \sqrt {5}}{4}$ just two days earlier, I was smart enough to test a regular triangle, square, and regular hexagon first. Neat result that $ A \equal{} B \equal{} \pi$ for all number of sides. Edit: but not smart enough to use the Pythagorean theorem in the same way as finding the area of a doughnut given the length of the chord of the outside that is tangent to the hole.

#20: Spent like 10 mins on this problem until I realized that the answer was $ (\frac {9}{9 \plus{} 12})14 \equal{} 6$.

#21: I love how $ 2009 \plus{} 9002\pi i$ was used for intimidation.

#22: I never would have guessed, until the test, that the intersection was a regular hexagon. I used vectors after the test to figure it out.

#24: Again, nice intimidation type problem. Key points to know are that $ \log_{2}(T(n)) \equal{} T(n \minus{} 1)$ and $ T(n) > > T(n \minus{} 1)$ for large $ n$.

#25: Immediately let $ \theta_n \equal{} \arctan(a_n)$, started adding angles in degrees, and got $ \theta_{2009} \equal{} 0$.

[hide="Sketch_for_19"]Let the center of the polygon be O, the midpoint of a side be M, and the next vertex be V. Note that OM is the inradius and OV is the circumradius and that OMV is a right angle. Then we can use the Pythagorean Theorem to find that $ VM^2\minus{}OV^2\equal{}VM^2.$ We know that VM is 1, since we were given that the side length is 2. Since the area of the annulus is $ \pi(VM^2\minus{}OV^2),$ the answer is $ \pi.$[/hide]

By the way, I believe Eric did say he continued his streak.

For #22, you could also notice that it was a planar hexagon that was equilateral and equiangular, due to the symmetry.

I suppose y'all are talking about the 12, but here's what I thought bout the 10

1-20: pretty trivial?

21: trivial, however i forgot about rationalizing the denominator, guessed correctly :P

22: another trivial problem, however, i epic failed it because i thought it was the same as some old ahsme problem

23. didn't think to use a trapezoid, so random guessed correctly

24. used casework instead of complementary counting...ugh but still got right answer

25. blah, made the same mistake they wanted everyone to make...

138 total score...ideally i need 144 or 150 to be confident for aime

Yikes, you were lucky with odds of 24 to 1!  :o

You would've gotten 126 if you weren't that lucky.

For me:

1-24: Really trivial
25: Trivial

Dudes, let me introduce you to the forum name: Community Forum -> Georgia. And I don't think we made either of ya'll honorary Georgians. :P

Had I more time, I would have answered the last 3, but I left them all blank. But I am still happy with my 136.5, which I got assuming I didn't mis-bubble any of my answers.

I hope the AMC 12B will make up for the lack of geometry & counting/probability problems on the AMC 12A.

[hide="minor comment"]I'm sorry I'm such a grammar freak (on a mathematics forum), but I think it's punctuated as "y'all."[/hide]

Oh don't worry, people complain about ya'll all the time. It's an abbreviation of ya all, so technically it's redundant (ya is Minnesotan), but w/e. Congrats on getting the highest score we've heard so far from GA!

I thought the problem distribution was nice and even, typical for AMCs.

I did indeed get a 135 exactly. :D   Might have been a 141 if I'd gotten more sleep, but oh well, there are more important things in life.

I don't have a problem with non-Georgians posting here, but I do have a problem with abuse of the word "trivial".  This is a word with a specific meaning.  It does not mean "easy".  It means something like "degenerate".  So, for instance, we can call $ n\equal{}0$ cases "trivial" cases; we can call the group with one element the "trivial" group and the ring with one element the "trivial" ring.  I cannot think of an instance when it would be appropriate to use this word to describe a well-designed contest problem.  It is also rude and inconsiderate to people who have a hard time with the contest.

For problem 19, you might have known that the trigonometric functions of $ \pi/7$ are not nice numbers.  The cosine of $ 2 \pi / 7$ is a root of a pretty general third-degree polynomial.  In fact, the polynomial is
\[ 2(4x^3 \minus{} 3x) \plus{} 2(2x^2 \minus{} 1) \plus{} 2x \plus{} 1 \equal{} 0 , \]
and I think Santhosh can explain how you get this.  In particular, there could have been almost no simple geometric arrangement that would have converted it into something nice without converting about everything into something nice.  In other words, you might have guessed that there was nothing special about the number 7.

For problem 22, it can be helpful to coordinatize the octahedron.  You could also look at the graph (vertices, edges, faces) and realized that the plane had to pass through all the faces except the ones to which it was parallel.

I thought the test was perhaps easier or about the same as last year's, but I've also heard people saying the opposite.  Who am I to judge?

I agree with Boy Soprano II; also, there is no point to post in a regional forum not your own to say a test was easy without any breakdown at all. Please don't. On the other hand, if you have something worthwhile to say, please do.

By the way, is that polynomial derived from $ 2\cos\frac{6\pi}{7}\plus{}2\cos\frac{4\pi}{7}\plus{}2\cos\frac{2\pi}{7}\plus{}\cos0\equal{}0?$ Those formulas look very familiar... That's the cosine triple angle formula, right? I can never keep the sine and cosine triple angle formulas straight...

[quote="solafidefarms"]107...: Observe that cos(3x)=Re(e^3xi) which can easily be calculated whenever needed. :)[/quote]
Or faster, Chebyshev recursion.

Just curious, Miles, which problems did you do?

P.S.: Nice job, Ben.

EDIT: I just tried 23 and 24 and felt glad I didn't attempt them, especially 23 (it took me 15 min!). Very computation-heavy and error-prone.

Ah, thanks for reminding me. I would have done something more algebra-intensive... Although I have no idea what Chebyshev recursion is. I vaguely recognize the name and get the feeling I should learn it. Time to add it to my list of things to do...

Guess who wants to cry after finding how much he failed??? :D

MEEEEEEEEEEEEEEE

[hide="Problem 18"]
$ 2^n / 10^n$ and so we have a divisibility test for powers of $ 2$ that a number $ M$ is divisible by $ 2^k$ iff the number formed by the last $ k$ digits of $ M$ in the same order is divisible by $ 2^k$.

Let $ A\equal{}1000000..........64$.
By the above mentioned property we know that $ A$ is divisible by 64 if the number of zeroes is greater than or equal to $ 4$.
 However if the number of zeroes is greater than or equal to $ 5$ then $ A$ is not divisible by any power of 2 greater than $ 64$.
 So we have just one more number to check because we donot know if $ 1000064$ is divisible by $ 128$.

However upon checking we find that $ 128$ divides $ 1000064$ and $ 256$ does not divide $ 1000064$.
Since $ 2^7 \equal{} 128$ the answer is $ 7$

P.S. Is there a way to prove that the recursive sequence in #25 has a period of 24 without writing out the first 24 terms. If anyone knows a method please do post it.

[/hide]

@befuddlers: maybe you want to have a look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1219673118&t=213864]this discussion[/url].

The method given there, however, doesn't work for this particular problem, because the sequence never reaches $ 0 \pmod {24}$. But maybe you can do something with $ \pmod {12}$ instead?


Hi

Let $ a$, $ b$, and $ c$ be nonnegative real numbers such that $ ab\plus{}bc\plus{}ca\equal{}1$.  Prove that $ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a} \ge \frac{5}{2}$.

Thanks

Assume $ q \equal{} 1$.
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \ge \frac {5}{2} \iff 2p^2 \minus{} 5p \plus{} 5r \plus{} 2 \ge 0$.
By Schur's inequality we have : $ r \ge \frac {4p \minus{} p^3}{9}$
$ 2p^2 \minus{} 5p \plus{} 5r \plus{} 2 \ge \minus{} \frac {1}{9}(5 p^3 \minus{} 18 p^2 \plus{} 25 p \minus{} 18) \equal{} \minus{} \frac {1}{9} (p \minus{} 2) \left(5 p^2 \minus{} 8 p \plus{} 9\right)$
We consider 2 case.
[b]Case 1[/b]: When $ p\ge 2$. Sinse:
$ 2p^2 \minus{} 5p \plus{} 5r \plus{} 2 \ge 2p^2 \minus{} 5p \plus{} 2 \equal{} 2 \left(p \minus{} \frac {5}{4}\right)^2 \minus{} \frac {9}{8} \ge 0$

[b]Case 2[/b]: When $ p\le 2$. Sinse:
$ \minus{} \frac {1}{9} (p \minus{} 2) \left(5 p^2 \minus{} 8 p \plus{} 9\right) \equal{} \minus{} \frac {1}{9} (p \minus{} 2) \left(5 \left(p \minus{} \frac {4}{5}\right)^2 \plus{} \frac {29}{5}\right) \ge 0$

It is correct. we are done.

you can't assume $ p \equal{} 1$ , my friend because we have already a condition , besides the ineq is not homogenous

EDIT: sorry , I guess you mean $ q\equal{}1$, not $ p\equal{}1$

[quote="anas"]you can't assume $ p \equal{} 1$ , my friend because we have already a condition , besides the ineq is not homogenous

EDIT: sorry , I guess you mean $ q \equal{} 1$, not $ p \equal{} 1$[/quote]
You're right, I made a typing error. I meant "$ q\equal{}1$"

[quote="mathgeniuse^ln(x)"]Hi

Let $ a$, $ b$, and $ c$ be nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1$.  Prove that $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \ge \frac {5}{2}$.

Thanks[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101476

It's easy
[hide]square it and use IRAN's 1996 Inequality[/hide]