I was curious about how to solve a quadratic conguence in the form $rx^2+sx+t\equiv u(mod m)$

The AoPS book says that it's too difficult to deal with it and excludes it.  :(

If this topic is too advanced for this forum, feel free to move it.

It's only been 3 hours since you asked, so I wouldn't expect a reply that quickly! :roll:

i think that there is something on this on mildorf's website, but i think that you just use the quadratic formula to factor it

Some of the ideas are not hard,  (specially if  m is samll ,  one can  try out  tral and error too)

What can be done:

Transfer the equation  to the form  $y^2 \equiv b \bmod m$

For example, in the case given,  multiply  whole thing  by $4r$ compelete the square so you would have some thing lilke:

$(2rx+s)^2 \equiv (some_simple_number_which_You_know_if_you_know_your_mod)$ etc

andnow  you can  use $y=(2rx+s)$ (point is:  if you can get  solve for y, you can solve for x)

Now comes the fun part.

First  factor $m$ (if it is not prime ) to bite-size (samller - prime) factors, find solution for each and apply CRT.

(Assume CRT - Chinese Remainder Theorem is something not considered too advanced)

( Now if you are lucky you are done here as trial and error would suffice  :)  (   Problem setters would generally use small easy numbers here :)
But if you are not  you have to do the hard part ...

So crux  of the matter is to  know how to solve $x^2 \equiv a \bmod p$ for prime  p.

And here is more fun.
Only for half the values of $a$ (quadratic residue) the solution exists,  For other   half - No luck.  and for each value, of couse,  there are two possible roots of $x$.

Now the part which is advanced , (and of practical value to solve such equations) - depends on Gauss's Reciprocavity  theorem  (Look in a good  - advanced - NT book ) It is easy to know and use (Certainly could be used by any high-schooler) but not easy to prove...This can help to know if a solution exist. To find a solution there are several methods but  none I know,  I can put it here without the moderator  coming out and reprimanding me (or worse banning me)   for putiting  too advanced material...:)

This might be more of an olympiad level topic but anyways, I am posting what I know.

First , it it easy to transform the any quadratic congrunece form $ax^2 + bx + c \equiv 0 (\bmod p )$ to $y^2 \equiv d (\bmod p )$

This is done as follows
$ax^2 + bx +c \equiv 0$ (all are $\bmod {p}$)
$4a(ax^2 +bx +c) = (2ax+b)^2 - (b^2 -4ac)$
Thus,$(2ax+b)^2 \equiv (b^2 - 4ac)$
Putting $(2ax+b) = y$ and $(b^2 -4ac = d$, we get the simple congruence.

Thus the problem of quadratic conruences is reduced to solving $x^2 \equiv a {\bmod {p} )}$

Solving this requires a no of properties and theorems including the wonderful Gauss's Quadratic Reciprocity Law.

If the congruence $x^2 \equiv a (\bmod{p})$ has a solution, then $a$ is the quadratic residur of $p$.

1) [b]Euler's Criterion[/b] -  Let $p$ be an odd prime and $gcd(a,p) = 1$. Then $a$ is a quadratic residue of $p$ iff $a^{ \dfrac{p-1}{2} } \equiv 1 (\bmod{p} )$

2) [b]Legendre's symbol[/b] $(a/p)$ = 1 if $a$ is the quadratic residue of $p$.

= -1 if $a$ is not the quadratic residue of $p$

3) [b]Quadratic reciprocity[/b] If $p$ and $q$ are distinct odd primes . then $(p/q) (q/p) = (-1)^{ \dfrac{p-1}{2} \dfrac{q-1}{2}}$.

Thus, there are no general methods of solving a general quadratuc reciprocity. You can use the above theroems and many others to simplify your work or guess etc.

Hope this helps!!!


If ((x+4)/(x-3))+((x+3)/(x-4)) is written as a mixed number, what is the numerator?
A)49
B)31
C)25
D)24
E)7

$ \frac{x\plus{}4}{x\minus{}3} \plus{} \frac{x\plus{}3}{x\minus{}4}$.
First find the common denominator:
$ \frac{(x\plus{}4)(x\minus{}4)}{(x\minus{}3)(x\minus{}4)} \plus{} \frac{(x\plus{}3)(x\minus{}3)}{(x\minus{}3)(x\minus{}4)}$
Expanding and adding we get:
$ \frac{2x^2\minus{}25}{x^2\minus{}7x\plus{}12}$. This simplifies to $ 2\frac{14x\minus{}29}{x^2\minus{}7x\plus{}12}$, so there is no integer answer. It all depends on what x value you give.

[quote="DiscreetFourierTransform"]The last simplification is simply untrue.
$ 2(14x \minus{} 29) \ne 2x^2 \minus{} 25$[/quote]
That's not what the answer is saying.
The problem asks to find the solution to the sum as a [b]mixed number[/b] (a number written in the form $ a\frac {b}{c}$, with $ b < c$, $ a\in \mathbb{N}$ where $ a\frac {b}{c} \equal{} \frac {ac \plus{} b}{c}$). Generally this is widely used with $ a, b, c$ as whole positive numbers.

What the statement $ \frac {2x^2 \minus{} 25}{x^2 \minus{} 7x \plus{} 12} \equal{} 2\frac {14x \minus{} 29}{x^2 \minus{} 7x \plus{} 12}$ is actually saying is

$ \frac {2(x^2 \minus{} 7x \plus{} 12) \plus{} 14x \minus{} 29}{x^2 \minus{} 7x \plus{} 12} \equal{} \frac {2x^2 \minus{} 14x \plus{} 24 \plus{} 14x \minus{} 29}{x^2 \minus{} 7x \plus{} 12} \equal{} \frac {2x^2 \minus{} 25}{x^2 \minus{} 7x \plus{} 12}$.

Better yet, I think the problem asks for $ 2\frac {14x \minus{} \color{red}49\color{black}}{x^2 \minus{} 7x \plus{} 12}$ (which can be verified with long division and polynomials).

I'm not wholeheartedly sure since it's late now, but I feel this statement is awkward, considering certain values of $ x$ like $ x \equal{} 3, 4$, $ x\le \frac {7}{2}$, and more (consider $ \exists x > \frac {7}{2}: 7(2x \minus{} 7) > (x \minus{} 3)(x \minus{} 4)$) . It isn't common for a mixed number to be written as $ 2\frac { \minus{} 1}{2}$, $ 2\frac {c}{0}$ or $ 2\frac {5}{2}$.


[color=darkblue] Find all pairs of natural numbers $ (m,n)$, for which numbers
$ A \equal{} 3m^2 \plus{} 2mn \plus{} n^2 \plus{} 3n$
$ B \equal{} m^2 \plus{} 3mn \plus{} 2n^2$
$ C \equal{} 2m^2 \plus{} mn \plus{} 3n^2$
are consecutive in some order.
[/color]

Let $ X,Y,Z$ be permutation of $ \{A,B,C\}$

Consider in three case

Case I : $ Y \equal{} A$, $ m^2 \plus{} 1 \equal{} (n \minus{} 1)^2$ so $ m \equal{} 0,n \equal{} 0,2$ but is contradiction

Case II : $ Y \equal{} B$, we obtain no solution

Case III : $ Y \equal{} C, n(n \minus{} m \minus{} 1) \equal{} 0$. we get $ (m,n) \equal{} (1,0),(m,m \plus{} 1)$

But m and n are natural so only (m,m+1) is the solution.


$ a,b,c>0$
$ a\plus{}b\plus{}c\equal{}3$ prove that
$ \sum_{cyc} \sqrt{9a\plus{}7(b\minus{}1)^2}\geq 9$

[quote="Rose_joker"]$ a,b,c > 0$
$ a \plus{} b \plus{} c \equal{} 3$ prove that
$ \sum_{cyc} \sqrt {9a \plus{} 7(b \minus{} 1)^2}\geq 9$[/quote]
Yes, it is very tough .
Can_hang posted it here :
http://www.mathlinks.ro/viewtopic.php?t=194632
But It was unsolved. :(


For any natural number n>=2, find out which of the real positive numbers a and b is greater, knowing that a^n=a+1 and b^(2n)=b+3a.

Solution:

I am going to prove that a>b, for any n>=2. we can see that a<>1 so (a+1)^2>4a. From where we get that (b/a)^(2n)=(b+3a)/(a+1)^2<(b+3a)/4a from which we have a>b because, from b>=a we get that (b+3a)/4a<=b/a<=(b/a)^(2n) which is false!

cheers!:D

This problem is simmilar to a problem that was given for the county MO and Bucharest in 2002 (Etapa judeteana si a municipiului Bucuresti):

Let a and be two real numbers that satisfy the relations: 3^a+13^b=17^a and 5^a+7^b=11^b. Prove that a<b.

This problem was proposed by Cristinel Mortici.

Cheers :D

The solution for the other problem is simmilar to the other!

cheers!


inequalities
trigonometry
inequalities unsolved

$ a,b,c>1$ are real numbers and $ a\plus{}b\plus{}c\equal{}abc$.prove that $ (a^2\minus{}1)(b^2\minus{}1)(c^2\minus{}1)<8$

I didn't solve it but i think AM-GM inequality can help

hi!!

there is my solution:

Let: $ a \equal{} \tan(\alpha)$ ; $ b \equal{} \tan(\beta)$ and $ c \equal{} \tan(\gamma)$

with : $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$ and $ \frac {\pi}{4} < \alpha,\beta,\gamma < \frac {\pi}{2}$

so :

$ (a^{2} \minus{} 1)(b^{2} \minus{} 1)(c^{2} \minus{} 1) < 8$ $ \Longleftrightarrow$ $ \prod((\tan(\alpha))^{2} \minus{} 1) < 8$

$ \Longleftrightarrow$  $ \prod(\cos(2\alpha)).\prod(\frac {1}{(\cos(\alpha))^{2})}) > \minus{} 8$

since $ \cos(\alpha) < \frac {\sqrt {2}}{2}$ so $ \prod(\cos(\alpha))^{2}) < \frac {1}{8}$

$ \Longleftrightarrow$ $ \prod(\frac {1}{(\cos(\alpha))^{2})}) > 8$

$ \Longleftrightarrow$ $ \prod(\cos(2\alpha)).\prod(\frac {1}{(\cos(\alpha))^{2})}) > 8.\prod(\cos(2\alpha))$

we have: $ 8.\prod(\cos(2\alpha)) \plus{} 8 \equal{} 8(\prod(\cos(2\alpha)) \plus{} 1) > 0$ $ (\prod(\cos(2\alpha)) < 1)$

so :

$ \prod(\cos(2\alpha)).\prod(\frac {1}{(\cos(\alpha))^{2})}) > 8.\prod(\cos(2\alpha)) > \minus{} 8$.

$ \Longleftrightarrow$ $ (a^{2} \minus{} 1)(b^{2} \minus{} 1)(c^{2} \minus{} 1) < 8$

thank you.but i think this problem has nice solution with use AM-GM.

[quote="MJ GEO"]$ a,b,c > 1$ are real numbers and $ a \plus{} b \plus{} c \equal{} abc$.prove that $ (a^2 \minus{} 1)(b^2 \minus{} 1)(c^2 \minus{} 1) < 8$[/quote]
It's wrong. Try $ a \equal{} b \equal{} c \equal{} \sqrt3.$
By the way, the following inequality is true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a\plus{}b\plus{}c\equal{}abc.$ Prove that
\[ (a^2 \minus{} 1)(b^2 \minus{} 1)(c^2 \minus{} 1) \leq 8\]

[quote="arqady"][quote="MJ GEO"]$ a,b,c > 1$ are real numbers and $ a \plus{} b \plus{} c \equal{} abc$.prove that $ (a^2 \minus{} 1)(b^2 \minus{} 1)(c^2 \minus{} 1) < 8$[/quote]
It's wrong. Try $ a \equal{} b \equal{} c \equal{} \sqrt3.$
By the way, the following inequality is true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} abc.$ Prove that
\[ (a^2 \minus{} 1)(b^2 \minus{} 1)(c^2 \minus{} 1) \leq 8\]
[/quote]

the inequality is true,my solution too,we have just to add $ \le$.

$ a,b,c >1$

$ (c^{2}\minus{}1)(b^{2}\minus{}1)(a^{2}\minus{}1)\le8$

[quote="arqady"][quote="reason"]

the inequality is true,my solution too,we have just to add $ \le$.

$ a,b,c > 1$

$ (c^{2} \minus{} 1)(b^{2} \minus{} 1)(a^{2} \minus{} 1)\le8$[/quote]
Now it's indeed true. By the way, how about the following constraints:
 $ a\geq0,$ $ b\geq0,$ $ c\geq0$ and $ a \plus{} b \plus{} c \equal{} abc$ $ ?$ :wink:[/quote]
The condition $ a,b,c > 1$ is not necessary, of course.


MATHCOUNTS
ratio
number theory
least common multiple
Pythagorean Theorem
geometry
similar triangles

For those who didn't see this problem on the AMC 10B, #30 had a nicer solution than the one that Mathcounts gave, in my opinion; the provided one had ridiculously complicated algebra compared to this solution, the Math Jam solution and the one I used on the test.

[hide]Starting from the top left vertex and going counterclockwise, label the vertices of the square $P,Q,R,S$. By some routine angle-chasing (let one of the acute angles be $x$, use the right angles to fill everything else in), we find that, by AA similarity,

$ABC\sim APS\sim QBP\sim RSC$.

Since $AB: AC: BC=3: 4: 5$ (by the Pythagorean Theorem), all the other triangles have that same ratio. For aviodance of two many fractions, let $AP=3x,AS=4x,SP=5x$. Using the square, we clearly have, from this, $PQ=QR=RS=SP=5x$. Then, we can write some ratios, and easily find that $BQ=\frac{5}{4}x$ and $RC=\frac{20}{3}x$. Hence, we have

$BC=\frac{15}{4}x+5x+\frac{20}{3}x=10$
$\frac{185}{12}x=10$
$x=\frac{24}{37}$.

The sidelength of the square is $5x$, yielding a final answer of $5\cdot\frac{24}{37}=\frac{120}{37}$.[/hide]

[hide]My way is to assume that the square has side length 120 which is the LCM of (6,8,10), and APS, BPQ, and RCS is similar for a 3,4,5 triangle. Triangle BQP has sides 90,120,150 since 120/8=15, which is what you multiply for sides 6 and 10 for it to be a 3,4,5 triangle. Triangle APS has sides 72,96,120 for the same reason, except that its 120/10=12. Side BA is 150+72=222 if the square has sides 120. 222/37=6, the original length for sides BA. Divide the square side by 37 for it to be proportionate, which is 120/37.[/hide]
Your way is probably simpler though.

oh, i remember this one
i thought it was pretty easy, but i got it wrong :maybe:

That was exactly my method. I can't remmeber if i got it right, 'cause i did the test backwards.

I've actually found that it works better on easier tests. This one wasn't that bad.

have you ever gotten in trouble w/ timing doing that?

jorian

Well it isn't TOO difficult to take a decent stab at or solve 30 problems in 40 minutes on a Sprint Round, if you're well-prepared. I did that once or twice on practice chapters, but I'vre never done it on a competition.

Yeah, time has never been a problem. When doing them 1-30, I've found that I often spent too much time on easy problems, and don't have enough for later ones. This way, I take my time on harder ones, not risking being forced to rush on them, as it's easier to avoid mistakes going quickly on easier ones. I only do this on easy tests, as harder ones require more time for earlier problems. It worked very well at chapters, and not too terribly at states.

[quote="Walk Around The River"]would anyone else agree that mathcounts has quite a few problems where they use a longer complicated method than what should have been used?[/quote]
Like that one target where Ptolomey's and Bhramagupta make up a  one-liner?

[hide="really fast"]In triangle ABC, let AB=6, BC=8 and AC=10. EFGH is a square where E is on AB and F is on BC.

We have a lot of similar triangles! CBA~FBE~CGF. Call the side length of the square s Let EF=10/10s so BF=8/10s. Let FC=6/6s so FC=10/6s. So now looking at BC we have 8/10s+10/6s=8, or s=120/37[/hide]


algebra
function
domain
complex numbers
complex analysis

-  Let $G$ be a simply connected plane domain with smooth boundary which contains $z=0$.  Let $h=\frac{x}{r^{2}}+h_{1}$ where $h_{1}$ is a regular harmonic function in $G$.  Let $f$ be the analytic function whose real part is $h$; what is the form of $f$ near $z=0$?

-  Now suppose $h$ and $f$ are as above and $h$ is constant on the boundary of $G$.  Describe the image of $G$ under the map $f$.

(I believe I can correctly solve the first one, but not the second)

The answer to the first question is clearly $ f(z)=1/z+g(z)$, where $ g(z)$ is regular in $ G$.  Than this form of $ f$ implies that the image domain contains infinity inside.  On the other hand if $ h=Re(f)$ is constant on the boundary than the image of the boundary must lie on a line parallel to imaginary axis. This seems to mean that the image domain is the whole complex plane with a slit along a line parallel to imaginary axis.  DOes it make sense to you?


group theory
abstract algebra
superior algebra
superior algebra unsolved

is subgroup notation $ \le$ an order?  Does bijection preserve this order?  I know subgroup notation $ \le$ is transitive, so ...


superior algebra
superior algebra unsolved

$ n\equal{}pqr$ is a Carmichael number. If $ q\minus{}1|pr\minus{}1$ and $ r\minus{}1|pq\minus{}1$ then show that q-1 is a divisor in
$ (d\plus{}p)(p\minus{}1).$

ok here it is the clear version of it.

n=pqr is carmicahelnumber. if we have that r-1|pq-1 then I have shown that
pq-1=d(r-1) where d is [2;p-1]. Moreover I have shown that q-1|d(r-1)-p+1.
Now I want to show that if  q-1|pr-1 is also fulfilled then q-1 is divisor in (d+p)(p-1). Do you understand it?


rotation
inequalities
complex numbers
absolute value
calculus
calculus computations

Let R be the real numbers, and let C be the complex numbers.

(i) Does there exist a power series Sum c(n)*z^n with coefficients c(n) ∈ R which diverges when z = (3/5) + (4/5)i, but converges for all other values of z ∈ C on the unit circle |z| = 1?
(ii) Does there exist a power series Sum c(n) * z^n with coefficients c(n) ∈ C, which diverges when z = (3/5) + (4/5)i, but converges for all other values of z ∈ C on the unit circle |z| = 1?
(iii) Does there exist a power series Sum c(n) * z^n with c(n) ∈ C which converges for all values of z ∈ C on the unit circle except for the values 1, i, −1 (at which the power series diverges)?

Answer all questions with an example or a proof of the impossibility.

I have a gut feeling that parts a and b have to do with the theorem that Sum (z^n)/n converges at all z with |z| = 1 except z = 1. We have to "rotate" the solution somehow. But I do not know how to actually do it. Can someone help? Also, I'm clueless about (c).

Thanks all.

Your intuition is absolutely correct; if you want to "rotate" your solution, the series you want is $ \sum \frac{(z/\alpha)^n}{n}$ where $ |\alpha| \equal{} 1$, which converges everywhere on the unit circle except $ z \equal{} \alpha$.  This is enough to do all three examples; for the first, take the real part, and for the third, add three series of this form.

Thank you! That should give me the second and third solutions. But how do I get the first? Which real part did you just refer to?

Sorry for being such a nuisance.

Also, one last thing. I have three power series, Sum a(n)*z^n, Sum b(n)*z^n, and Sum c(n)*z^n, which converge to 1, -1 and i respectively. I claim that the power series Sum (a(n) + b(n) + c(n))*z^n converges to all points on the unit circle except 1, -1 and i. What justifies this claim?

What is the radius of convergence of such a series? Consider the Cauchy-Hadamard formula and the way superior limits add.

All right, I basically proved b and c with your help -- thanks guys!

For part a, my proof assumes that such real coefficients DO exist. I'm trying to show that one of the following two series diverges, if all other conditions hold:

1. c(0) + c(1) + c(2) + c(3) + ...
2. c(0) - c(1) + c(2) - c(3) + ...

If I can show that, I'm through. Could someone help me out, please?

Thanks

dxdt

[quote="walter.rudin"]Which real part did you just refer to?[/quote]
The entire series.  If $ \sum \frac{(z / \alpha)^n}{n}$ converges everywhere except $ z \equal{} \alpha$ with $ |\alpha| \equal{} 1$ a complex number, the same should be true of the [i]real[/i] series

$ \sum \frac{\Re\left( (z / \alpha)^n \right)}{n}$

(where $ z$ is considered to be real.)  Actually, to be safe you should do something like add the absolute value of the real and imaginary parts.

In part (i), if the series diverges at $ \frac{3}{5} \plus{} \frac{4}{5} i$, doesn't it also diverge at the conjugate $ \frac{3}{5} \minus{} \frac{4}{5} i$?

Consider what it means for a series to diverge.  Then, examine what happens if you take the conjugate.

Completely unrelatedly, Ravi, your (or your and Joel Spencer's) proof of Janson's inequality is nice.  (Jeff Kahn just presented it in a class he's teaching on the probabilistic method.)

<off-topic>
Thanks!  Jeff was my colleague at Rutgers before I joined NYU -- he's a nice guy and an excellent combinatorialist.  Your post brings back good memories.  Maybe someday I will go back to research.
</off-topic>


pigeonhole principle
geometry unsolved
geometry

show that when we put ten difference points inside a circle with diameter 5, we will get two of that ten points have the distance less than 2.
thanks 1st for helping me...

Observe the picture below and apply Dirichlet box principle.

Lets divide circle equal parts so that a measure of central angle will be $40^{o}$.According to the Dirichlet principle, there are two points in the same part.Lets  be these two points $A_{1},A_{2}$.Lets take $A_{1}$ at the centre.For at least one of $i,(i=2,3,...,9)$, $m(A_{i}A_{1}A_{i+1})\leq 40^{0}$.We assume that $m(A_{2}A_{1}A_{3})\leq 40^{o}$.If  the point $A_{3}$ is at the yellow part, there is nothing to prove.If the point $A_{3}$ is at the green part, we need to prove $max|A_{2}A_{3}|< 2$.In the figure: $|A_{2}A_{3}|=\sqrt{2^{2}+5^{2}-2.5.2.cos40^{o}}<2$.

Lokman GÖKÇE


vector
linear algebra
linear algebra theorems

Prove that a transformation T defined by T([b]x[/b]) = [b]Ax[/b] maps $ \mathbb{R}^n$ onto $ \mathbb{R}^m$ if and only if the columns of [b]A[/b] span $ \mathbb{R}^m$.

EDIT:
Damned, why doesn't the mathbb code work???  :huh:

/mathbb {R}^n stands for [size=125]Rⁿ[/size] by the way...and how do you type Rⁿ using those special thingie in this forum???

EDIT: Yes!!!  :lol:

Separate $ x$ into its components; vectors of the form $ \mathbf{Ax}$ are linear combinations of the columns of $ \mathbf{A}$ (in other words, the range of $ T(\mathbf{x})$ is the column space of $ \mathbf{A}$).  The conclusion follows.

[quote="mathwizarddude"]
/mathbb {R}^n stands for [size=125]Rⁿ[/size] by the way...and how do you type Rⁿ using those special thingie in this forum???[/quote]

Put \mathbb{R}^n between $ \$$ signs


inequalities
algebra
system of equations
algebra solved

Find all positive solutions to the system of equations :
           a+b+c = (3 \sqrt 3)/2
           a 2 +b 2 +c 2  = 2xyz + 2
           a 2 +b 2 +c 2 +x 2 +y 2 +z 2  = 3

oh,easy system of equations! :)
From the second and third equation we have :
 x^2+y^2+z^2 +2xyz=1
Hence x=CosA ,y=cosB,z=CosC, where A,B,C are three angle of triangle ABC
thus x^2+y^2+z^2 >=3/4
otherwise : a^2+b^2+c^2>=(a+b+c)^2/3=9/4 => x^2+y^2+z^2 <=3/4
So we have : cos^A+cosB^2+cosC^2=3/4
             a=b=c
             a+b+c=3 \sqrt 3/2
this equivallent to :  A=B=C=pi/3 =>x=y=z=1/2
                       a=b=c=sqrt(3)/2

I guess from where Tunglam created this system. But this system can be solved algebraically. Two lines. Try ppl!

Let S=\sum a2.
We have S=3-\sum x2 and S=2xyz+2. [b](1)[/b]

We also have 3S \geq (a+b+c)2=27/4, so S \geq 9/4. [b](2)[/b]

From (1) we get 3-S=\sum x2 \geq 3(xyz) 2/3=3((S-2)/2) 2/3, so
3-S \geq 3(1-S/2) 2/3 <=> (1-S/3) 3 \geq (1-S/2) 2 <=> 9/4 \geq S [b](3)[/b]

(2) and (3) show that S=9/4 and all the inequlaities are equalities, so
a=b=c and x=y=z. This means a=b=c= (\sqrt 3)/2 and x=y=z=1/2.

How about this system of equations ?

a+b+c+x+y+z=(3+3 \sqrt 3)/2
a  2  +b  2  +c  2  = 2xyz + 2
a   2  +b  2  +c  2  +x  2 +y  2  +z  2  = 3

Can you solve it algebraically ?

Of course we can. The main idea is to show that from x^2+y^2+z^2+2xyz=1 => xyz <=1/8 (it is trivial, AM-GM for x^2+y^2+z^2) and x+y+z<=3/2. The second one was discussed on forum, I rewrite: it enough to show that if x+y+z=3/2 then x^2+y^2+z^2+2xyz>=1. It is equivalent to (2/3)(x+y+z)(x^2+y^2+z^2)+2xyz>=(8/27)(x+y+z)^3 <=> 5 Sum x^3 + 3xyz >= 3 Sum (x^2y+xy^2) (sum are cyclic) which is true by Schur and AM-GM.


Suppose that $ \{a_n\}$ is an arithmetic sequence with \[a_1 \plus{} a_2 \plus{} \cdots \plus{} a_{100} \equal{} 100\quad\text{and}\quad a_{101} \plus{} a_{102} \plus{} \cdots \plus{} a_{200} \equal{} 200.\] What is the value of $ a_2 \minus{} a_1$?

$ \textbf{(A)}\ 0.0001 \qquad \textbf{(B)}\ 0.001 \qquad \textbf{(C)}\ 0.01 \qquad \textbf{(D)}\ 0.1 \qquad \textbf{(E)}\ 1$

[hide]$ 200\minus{}100\equal{}100\equal{}(a_{101} \plus{} a_{102} \plus{} \cdots \plus{} a_{200} ) \minus{} (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_{100})\equal{}100\cdot100d$, where $ d$ is the common difference. So $ d\equal{}.01\equal{}a_2\minus{}a_1$. $ \boxed C$[/hide]


$ x\ge 0$ be the last square root. Then obviously the LHS is an increasing function so there can be at most one solution. $ x\equal{}2$ obviously works.

[quote="Altheman"]$ x\ge 0$ be the last square root. Then obviously the LHS is an increasing function so there can be at most one solution. $ x \equal{} 2$ obviously works.[/quote]
[color=darkblue]Yes thank! I want see your solution.[/color]


[quote="Raja Oktovin"]In triangle $ ABC$, given points $ D,E,F$ on $ BC,CA,AB$ such that side $ AD,BE,CF$ meet at point $ O$. Prove that $ \frac {AO}{AD} \plus{} \frac {BO}{BE} \plus{} \frac {CO}{CF} \equal{} 2$. :D[/quote]

$ \frac{AF}{FB}\equal{}n$
$ \frac{BD}{DC}\equal{}k$---> Chev th --> $ mnk\equal{}1$
$ \frac{CE}{EA}\equal{}k$

Ban-Obel th---->

$ \frac{AO}{OD}\equal{}\frac{AF}{FB}\plus{}\frac{AE}{EC}\equal{}n\plus{}\frac{1}{k}\equal{}n\plus{}nm$
$ \frac{BO}{OE}\equal{}\frac{BD}{DC}\plus{}\frac{BF}{FA}\equal{}m\plus{}\frac{1}{n}\equal{}m\plus{}mk$
$ \frac{CO}{OF}\equal{}\frac{CE}{EA}\plus{}\frac{CD}{DB}\equal{}k\plus{}\frac{1}{m}\equal{}k\plus{}nk$

$ \frac{AO}{AD}\equal{}\frac{1}{1\plus{}\frac{OD}{AD}}\equal{}\frac{1}{1\plus{}\frac{1}{n\plus{}nm}}\equal{}\frac{nm\plus{}n}{nm\plus{}n\plus{}1}$

$ \frac{BO}{BE}\equal{}\frac{1}{1\plus{}\frac{OE}{OB}}\equal{}\frac{1}{1\plus{}\frac{1}{m\plus{}mk}}\equal{}\frac{m\plus{}mk}{m\plus{}mk\plus{}1}\equal{}\frac{nm\plus{}nmk}{nm\plus{}nmk\plus{}n}\equal{}\frac{nm\plus{}1}{nm\plus{}n\plus{}1}$

$ \frac{CO}{CF}\equal{}\frac{1}{1\plus{}\frac{OF}{OC}}\equal{}\frac{1}{1\plus{}\frac{1}{nk\plus{}k}}\equal{}\frac{nk\plus{}k}{nk\plus{}k\plus{}1}\equal{}\frac{n^{2}mk\plus{}nmk}{n^{2}mk\plus{}nmk\plus{}nm}\equal{}\frac{n\plus{}1}{n\plus{}1\plus{}nm}$

$ \frac{AO}{AD}\plus{}\frac{BO}{BE}\plus{}\frac{CO}{CF}\equal{}\frac{nm\plus{}n}{nm\plus{}n\plus{}1}\plus{}\frac{nm\plus{}1}{nm\plus{}n\plus{}1}\plus{}\frac{n\plus{}1}{n\plus{}1\plus{}nm}\equal{}2$

When posting, please give better titles than "---" (I've changed it to make it a little better).  Also, when posting solutions, please use hide tags to hide them.

Thanks.

[hide]Using mass points:
Let the weights of $ A,B,C$ be $ a,b,c$ respectively. Then the weights of $ D,E,F$ are $ b\plus{}c,c\plus{}a,a\plus{}b$ respectively, and the weight of $ O$ is $ a\plus{}b\plus{}c$.

Thus:
$ \frac{AO}{AD}\equal{}\frac{b\plus{}c}{a\plus{}b\plus{}c}$

$ \frac{BO}{BE}\equal{}\frac{c\plus{}a}{a\plus{}b\plus{}c}$

$ \frac{CO}{CF}\equal{}\frac{a\plus{}b}{a\plus{}b\plus{}c}$

And thus, $ \frac {AO}{AD} \plus{} \frac {BO}{BE} \plus{} \frac {CO}{CF} \equal{} \frac{2(a\plus{}b\plus{}c)}{a\plus{}b\plus{}c}\equal{}2$ as desired.[/hide]

This is an easy problem. It has a really easy proof.
Observation: This is Gergonne`s theorem  :D .
Proof:
Let us take $ X,Y \in BC$ such that $ OX \perp BC$ and $ AY \perp BC$.
Because $ OX \parallel AY \Rightarrow \frac{OD}{AD}\equal{}\frac{OX}{AY}.(1)$.
Then, $ \frac{area[BOC]}{area[ABC]}\equal{}\frac{OX}{AY}.(2).$.
From $ (1)$ and $ (2)$ we get that $ \frac{area[BOC]}{area[ABC]}\equal{}\frac{OD}{AD}$ which means $ \sum \frac{OD}{AD}\equal{}\frac{area[BOC]\plus{}area[AOB]\plus{}area[AOC]}{area[ABC]}\equal{}1$.
We reach at $ \sum \frac{AO}{AD}\equal{}\sum \frac{AD\minus{}OD}{AD}\equal{}\sum(1\minus{}\frac{OD}{AD})\equal{}3\minus{}1\equal{}2$.

[quote="wangsacl"][quote="Xantos C. Guin"][hide]Using mass points:
Let the weights of $ A,B,C$ be $ a,b,c$ respectively. Then the weights of $ D,E,F$ are $ b \plus{} c,c \plus{} a,a \plus{} b$ respectively, and the weight of $ O$ is $ a \plus{} b \plus{} c$.

Thus:
$ \frac {AO}{AD} \equal{} \frac {b \plus{} c}{a \plus{} b \plus{} c}$

$ \frac {BO}{BE} \equal{} \frac {c \plus{} a}{a \plus{} b \plus{} c}$

$ \frac {CO}{CF} \equal{} \frac {a \plus{} b}{a \plus{} b \plus{} c}$

And thus, $ \frac {AO}{AD} \plus{} \frac {BO}{BE} \plus{} \frac {CO}{CF} \equal{} \frac {2(a \plus{} b \plus{} c)}{a \plus{} b \plus{} c} \equal{} 2$ as desired.[/hide][/quote]

What is mass points, actually?[/quote]

This [url=http://mathcircle.berkeley.edu/mpgeo.pdf]article[/url] should explain what it is. I'm not familiar with its use, however.


After expanding, it is
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \sum (\frac {a}{b} \plus{} \frac {a}{c}) \plus{} 3 \ge 2(a \plus{} b \plus{} c) \plus{} 6$
Since $ \frac {a}{b} \plus{} \frac {b}{a} \ge 2$,
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3 \ge 2(a \plus{} b \plus{} c)$
Let $ a \plus{} b \plus{} c \equal{} k$, then $ a^2 \plus{} b^2 \plus{} c^2 \ge \frac {k^2}{3}$, so we have to show
$ \frac {k^2}{3} \plus{} 3 \ge 2k \implies (k \minus{} 3)^2 \ge 0$. (Or by AM-GM)

Oh, come on!
All we need to do is just using Cauchy- Schwartz
We have:
$ \frac {2}{a \plus{} 1} \leq \frac {2}{2\sqrt {a}} \equal{} \frac {1}{\sqrt {a}} \equal{} \sqrt {bc}$
In fact, we also have the well known inequality
$ a \plus{} b \plus{} c \geq \sqrt {bc} \plus{} \sqrt {ca} \plus{} \sqrt {ab}$
and we have finished! :)

[quote="archimedes1"]Let $ a,b,c \ge 0$ and $ abc \equal{} 1$.
Prove that $ a \plus{} b \plus{} c \ge \frac {2}{a \plus{} 1} \plus{} \frac {2}{b \plus{} 1} \plus{} \frac {2}{c \plus{} 1}$.[/quote]
I think the following inequality is true too.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1$ and $ k_{0}$ is a real root of the equation $ 4x(x \plus{} 1)^2 \equal{} 1.$
Prove that for all $ k\geq k_{0}$ the following inequality holds:
\[ a \plus{} b \plus{} c \geq \frac {1 \plus{} k}{a \plus{} k} \plus{} \frac {1 \plus{} k}{b \plus{} k} \plus{} \frac {1 \plus{} k}{c \plus{} k}
\]
For $ k\equal{}\frac{1}{5}$ we obtain:
$ \frac{a\plus{}b\plus{}c}{6}\geq\frac{1}{5a\plus{}1}\plus{}\frac{1}{5b\plus{}1}\plus{}\frac{1}{5c\plus{}1}.$

[quote="archimedes1"]Let $ a,b,c \ge 0$ and $ abc = 1$.
Prove that $ a + b + c \ge \frac {2}{a + 1} + \frac {2}{b + 1} + \frac {2}{c + 1}$.[/quote]

let$ x = ln(a),y = ln(b),z = ln(c)$and $ x + y + z = 0$
$ f(x) = exp(x) - \frac {2}{exp(x) + 1}$
because of $ f''(x) > 0$   use Jensen inequality
so$ f(x) + f(y) + f(z)\ge \3f(0) = 0$


symmetry
inequalities unsolved
inequalities

Denote $ S(x,y,z) \equal{} \sum{x^2(y\plus{}z)} \equal{} \sum{x^2 \minus{} x^3}$.

Note that $ S$ is symmetric.

Since $ \frac{d^2}{dx^2}(x^2 \minus{} x^3) \equal{} 2 \minus{} 6x$, $ x^2 \minus{} x^3$ is convex in $ [0,\frac 13]$ and concave in $ [\frac 13,1]$.

For three number $ x,y,z$, two of them must be both greater than $ \frac 13$ or both less than or equal to $ \frac 13$. Because of symmetry, we assume that they are $ y,z$. Hence, we must have:
$ S(x,y,z) \leq S(x,y\plus{}z,0)$ or $ S(x,y,z) \leq S(x,\frac{y\plus{}z}{2},\frac{y\plus{}z}{2})$

Combining both, we have
\[ S(x,y,z) \leq \max(S(x,y\plus{}z,0), S(x,\frac{y\plus{}z}{2},\frac{y\plus{}z}{2}))\]

Arithmetic Compensation Method gives us:

$ max S \equal{} \max (S(1,0,0) , S(\frac 12, \frac 12, 0), S(\frac 13,\frac 13,\frac 13)) \equal{} \max(0,\frac 14, \frac 59) \equal{} \frac 59$.

The maximum is attained at $ x \equal{} y \equal{} z \equal{}\frac 13$.

[quote="April"]If $ x$, $ y$, $ z$ are nonnegative real numbers with the sum $ 1$, find the maximum value of $ S \equal{} x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y)$ and $ C \equal{} x^2y \plus{} y^2z \plus{} z^2x$.[/quote]
\[ a^2 b \plus{} b^2 c \plus{} c^2 a \plus{} abc \le \frac{4}{{27}}\left( {a \plus{} b \plus{} c} \right)^3
\]

$ \sum x^2(y \plus{} z)\leq\frac {1}{4}\Leftrightarrow\sum x^2(y \plus{} z)\leq\frac {1}{4}(\sum x)^3$
$ \Leftrightarrow(\sum x)^3\geq 4\sum x^2(y \plus{} z)\Leftrightarrow\sum x^3 \plus{} 6xyz\geq\sum x^2(y \plus{} z)$
It's true by Schur's ineq

[quote="Honey_S"][quote="April"]If $ x$, $ y$, $ z$ are nonnegative real numbers with the sum $ 1$,
\[ a^2 b \plus{} b^2 c \plus{} c^2 a \plus{} abc \le \frac {4}{{27}}\left( {a \plus{} b \plus{} c} \right)^3
\]
[/quote][/quote]
Find the greatest value of $ a^nb\plus{}b^nc\plus{}c^na$ where $ n\in N$ :lol:

the generalization has also been discussed
http://www.mathlinks.ro/Forum/viewtopic.php?t=2549


geometry
quadratics
algebra
geometry proposed

[color=darkred]Let $ABC$ be triangle. Denote the distance $h_{a}$ from the point $A$ to the side $BC$ and $2s=a+b+c\ .$

Prove that : $\{\begin{array}{cc}1.\blacktriangleright & \boxed{\ h_{a}=\frac{2s(s-a)x}{x^{2}+s(s-a)}\Longleftrightarrow x\in\{\ r_{b}\ ,\ r_{c}\ \}\ }\ .\\\\ 2.\blacktriangleright & \boxed{\ h_{a}=\frac{2(s-b)(s-c)x}{|x^{2}-(s-b)(s-c)|}\Longleftrightarrow r\in \{\ r\ ,\ r_{a}\ \}\ }\ .\end{array}$[/color]

Let $\triangle$ be the area of the triangle $\triangle ABC$. Substituting $s(s - a) = \frac{\triangle^2}{(s - b)(s - c)}$ into the 1st quadratic equation:

$h[x^2 + s(s - a)] = 2s(s - a)x$

$x^2 - \frac{2 \triangle^2}{h(s - b)(s - c)}\ x + \frac{\triangle^2}{(s - b)(s - c)} = 0$

Substituting $h = \frac{2 \triangle}{a}$ and then $a = 2s - (b + c) = (s - b) + (s - c)$ into the mixed term of this quadratic equation,

$x^2 - \frac{a \triangle}{(s - b)(s - c)}\ x + \frac{\triangle^2}{(s - b)(s - c)} = x^2 - \frac{[(s - b) + (s - c)] \triangle}{(s - b)(s - c)}\ x + \frac{\triangle^2}{(s - b)(s - c)}$ [color=white].[/color]$= 0$

$x^2 - \left(\frac{\triangle}{s - b} + \frac{\triangle}{s - c}\right) x + \frac{\triangle^2}{(s - b)(s - c)} = \left(x - \frac{\triangle}{s - b}\right) \left(x - \frac{\triangle}{s - c}\right) = 0$

$x = \frac{\triangle}{s - b} = r_b$ [color=white].[/color] or [color=white].[/color] $x = \frac{\triangle}{s - c} = r_c$

where $r_b, r_c$ are radii of the excircles against the vertices $B, C$. Similarly, substituting $(s - b)(s - c) = \frac{\triangle^2}{s(s - a)}$ into the 2nd quadratic equation:

$h[x^2 - (s - b)(s - c)] = 2(s - b)(s - c)x$

$x^2 - \frac{2\triangle^2}{hs(s - a)}\ x - \frac{\triangle^2}{s(s - a)} = 0$

Substituting $h = \frac{2 \triangle}{a}$ and then $a = s - (s - a)$ into the mixed term of this quadratic equation,

$x^2 - \frac{a \triangle}{s(s - a)}\ x - \frac{\triangle^2}{s(s - a)} = x^2 - \frac{[s - (s - a)] \triangle}{s(s - a)}\ x - \frac{\triangle^2}{s(s - a)}$ [color=white].[/color]$= 0$

$x^2 - \left(\frac{\triangle}{s - a} - \frac{\triangle}{s}\right) x - \frac{\triangle^2}{s(s - a)} = \left(x + \frac{\triangle}{s}\right) \left(x - \frac{\triangle}{s - a}\right) = 0$

$x = - \frac{\triangle}{s} = -r$ [color=white].[/color] or [color=white].[/color] $x = \frac{\triangle}{s - a} = r_a$

where $r, r_a$ are radii of the incircle and excircle against the vertex $A$.


[i]Κατασκευασμένη απο έναν master του είδους. Θα σας πω μετά από ποιον![/i]

Αν $ a_1,a_2,...,a_n, n\leq 81$ μη αρνητικοί πραγματικοί αριθμοί τέτοιοι ώστε

$ a_1^6\plus{}a_2^6\plus{}...\plus{}a_n^6\equal{}n$, να δείξετε ότι

$ a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2\leq a_1^5\plus{}a_2^5\plus{}...\plus{}a_n^5$

Θεωρώ ότι είναι δύσκολη και σφιχτή (έχεις elementary solution?) (και αλήθεια ποιος την κατασκεύασε?)

Θα αποδείξω την ανισότητα για $ n = 81$ και για $ n < 81$ μετά θα είναι προφανής (γιατί ? :wink: )
Έστω $ x_i = a_1^6$ τότε $ \sum_{i = 1}^{81} x_i = 81$

Θεωρούμε την συνάρτηση $ f(x) = \sqrt [3]{x} - \sqrt [6]{x^5}$ και ${ f''(x) = \frac {1}{36}x^{ - \frac {5}{3}}}(5\sqrt {x} - 8)$

άρα η $ f$ έχει unique inflection point και το άθροισμα $ \sum_{i = 1}^{81} x_i = 81$ είναι fixed άρα από  Karamata

to μέγιστο του $ \sum_{i = 1}^{81} f(x_i)$ επιτυγχάνεται όταν οι 80 μεταβλητές είναι ίσες (παρατηρείστε ότι δεν χρησιμοποιώ EV theorem  ).

Αρα telika αρκεί νδο $ 81(80 + x^5)^2\geq (80 + x^6)(80 + x^2)^2$ που ισοδυναμεί με

$ (x - 1)^2(x - 2)^2(x^6 + 6x^5 + 21x^4 + 60x^3 + 75x^2 + 60x + 20)\geq 0$

which is ... trivial  :P

oraios silouan! tou vasile einai...oxi den exo elementary solution, apla tin evala gia na thimithoume ligo to lcf ...

[quote="silouan"]
Έστω $ x_i \equal{} \sqrt [6]{a_1}$ τότε $ \sum_{i \equal{} 1}^{81} x_i \equal{} 81$

[/quote]

Ωραία λύση (όχι ότι την κατάλαβα πλήρως δηλαδή, αλλά τέσπα :P  )

Ομως εδώ προφανώς εννοείς οτι $ x_i\equal{}a_i^6$ (το λέω γιατί μπορεί αρχικά να μπερδευτεί κάποιος όταν το βλέπει...)

Ρε nick άλλη φορά μη γράφεις όχι πολύ δύσκολη γιατί έχασα 2 ώρες και δεν κατάφερα τίποτα και λέω ούτε τις εύκολες δεν μπορώ να λύσω... :rotfl:

kosta...koita genika i askisi einai dyskoli, alla an exeis sto mnualo sou to ev i to lcf, opos o silouan, tote baineis sti diadikasia epilusis. symfoneis sil?

Κοίτα και ο Μιχάλης όταν βλέπει κάτι τέτοιες λέει  trivial , trivial .
Δεν είναι ακριβώς έτσι αλλά είναι σαν να σου ζητάω να αποδείξεις ότι για $ a\plus{}b\plus{}c\equal{}1$ ndo

$ \sum\frac{a}{\sqrt{1\minus{}a}}\geq\sqrt{\frac{3}{2}}$ .

Για κάποιον που ξέρει Jensen είναι trivial για κάποιον που δεν ξέρει ...είναι κάπως πιο δύσκολη ίσως πολύ πιο δύσκολη .
Έτσι και εδώ ... Το κακό είναι ότι εδώ ο μόνος τρόπος για να  παρακάμψεις τα EV και LCF είναι το Karamata όπως δείχνω παραπάνω αλλά και πάλι ...Τέσπα η δυσκολία είναι κάτι σχετικό  :)

Νίκο προφανώς εδώ την βρίκες.....Είπα να τη βάλω έτσι πληροφοριακά......


$ a_1\equal{}\frac{1}{3}$ and $ \forall n\geq1$    $ a_{n\plus{}1}\equal{}\frac{a_n}{\sqrt{1\plus{}13a^2_n}}$

if $ a_k$ is the greatest term of sequence $ (a_n)$ such that $ a_k<\frac{1}{50}$

$ k\equal{}?$

[hide="fun problem"]
Let $ u_n\equal{}\frac{1}{a_n^2}$. Then $ u_1\equal{}9$, and $ u_{n\plus{}1}\equal{}u_n\plus{}13$, and $ u_k$ is the smallest $ u_n>2500$. So $ k\equal{}193$.[/hide]


The midpoint of each side of a convex pentagon is connected by a segment with the intersection point of the medians of the triangle formed by the remaining three vertices of the pentagon.
Prove that all five such segments intersect at one point.


Hello friends,
                                Recently I stumbled upon this curious problem. Show that 1805/1806 is the biggest proper fraction which can be written as sum of  maximum 4 different unit fractions. A unit fraction is a fraction of the form 1/n where n is a natural number.

Well I would think:
just take 1/2, then add 1/3, then add the largest unit fraction that keeps their sum smaller than 1 and repeat this another time. You will get $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}=\frac{1805}{1806}$. (Because $\ldots + \frac{1}{42}=1$)


hi, i'm wondering if there is a polar representation of the graph of y=|x|.. thanks

Yes there is. It's similar to the equation of a vertical line (x = something), or a horizontal line (y = something).

In this case, instead of an x or a y that's constant, it's :theta: that's constant. An equation would be :theta: = :pi:/4 and :theta: = 3:pi:/4, with r :ge: 0.

[hide]

:theta:=:pi:/4 or  :theta:=3:pi:/4

R :ge:0

or

:theta:= 5:pi:/4 or  :theta:=7:pi:/4

R :le: 0

[/hide]


Let A be a subset of the set {1,11,21,31,...,541,551} having the property that no two elements of A add up to 552. Prove that A cannot have more than 28 elements.

This one is very easy and a classical pigeon-hole problem type :

The given numbers are those of the form 1 + 10k, where k = 0,1,...,55.
We may divide them into 28 pairs (1+10k, 1+10(55-k)).
Any subset A with more than 28 elements has to contain at least two numbers from a same pair, so that it contaiins two elements which add up to 552, hence result.

Pierre.


[hide]83. You don't have that many options, so it's easy enough that I'd just use guess and check. Obviously, the original # is larger, so the possibilities are 92, 83, 74, and 65. Yeah..... I think people can get it from there. [/hide]

[quote="math92"]Find the 2 digit number.
The sum of the digits is 11.
If the digits are reversed, the original number is 7 more than twice than twice the new number.
What's the number?[/quote]

[hide]let the digits be x, y respectively.

x+y=11

x=11-y

10x+y-7=2(10y+x)

8x-7=19y

88-8y-7=19y

81/27=y

y=3

x=8

83[/hide]

[hide="jli"]in your work i think y is the tens digit and x is the ones digit.[/hide]


Η eκφώνηση είναι στο σχήμα. Η χορδή ΑΒ είναι τυχαία καθώς και το $ C$.Ο μικρός κύκλος εφάπτεται στα δύο τμήματα και στον κύκλο.

Κλασσική όμορφη άσκηση .Να η λύση μου ...
Έστω ότι ο κύκλος εφάπτεται στον μεγάλο κύκλο στο $ M$ και στην $ CD$ στο $ N$ και η CD τον μεγάλο κύκλο στο $ Q$  .
Από τα στοιχεία της άσκησης έχουμε ότι $ \angle{CAB}\equal{}\angle{QAB}$ δηλαδή το $ B$ είναι το μέσο του τόξου $ CQ$
Η ομοιοθεσία κέντρου $ M$ που στέλνει τον μικρό κύκλο στον μεγάλο στέλνει το $ N$ στο μέσον του τόξου $ CQ$ άρα στο $ B$
επομένως $ M,N,B$ συνευθειακά και επομένως $ BE^2\equal{}BN\cdot BM$ (1). όμως $ \angle{CAB}\equal{}\angle{BMC}\equal{}\angle{BCN}$
δηλαδή ο περίκυκλος του CMN εφάπτεται της BC στο C και επομένως $ BC^2\equal{}BN\cdot BM$  (2)
Από (1),(2) => $ BC\equal{}BE$ so QED  :)


Solve the equation:  $ 2^{\tan x}\plus{}2^{\cot x}\equal{}2\cot 2x$

hello,where can i find the problems from the Romanian N.O. shortlist 2007?
Sonnhard.

Maybe it should byt

$2^{tanx}-2^{cotx}=2cot2x$
Here my solution:

So that $cotx-tanx=2cot2x$,in this case $f(tanx)=f(cotx)$
Let $f(t)=2^t+t$ Since $f$ is strictly increasing,we get $tanx=cotx$.
Answer $x=\pm \frac{\pi}{4}+\pi k$


You are given $2000$ numbers: $10^{1}+1$, $10^{2}+1$, $\cdots$ $10^{2000}+1$.

Now prove that at least $99 percent$ of these numbers are not prime.

[hide]$a^{n}+1|a+1$ for $n \geq 3$
So all of the values excluding n=1,2 are composite.
So 99% are prime.[/hide]

[quote="Fraxia"]Doesn't that factorization only work if $n$ is odd?[/quote]

Or has an odd factor (that is ,  if n has an odd factor , $10^{n}+1$ is not prime  :)
[hide="IOW"](IOW  all n's except when n is 1,2,4,8,16,32  ...)

For example, $10^{6}+1 = 100^{3}+1$ is divisible by $100+1$
[/hide]

Of course, it so happens that,  only  $11$ and $101$ in the  above series are prime,  (bpms is right  at least for numbers less than $10^{2000}+1$  :) )

Because:

[hide]One can verify

$(73|10^{4}+1), (17|10^{8}+1), \ (353|10^{1 6}+1), \ (19841|10^{32}+1),\ (1265011073 |10^{64}+1) \dots$

etc .. you don't have to take my word for it, just verify for 128,256,512,1024
that factors are 257,10753,1514497 and 1856104284667693057 :)  [/hide]

Oh yeah, I forgot you could factor in terms of 10^2 or some other power of 10  :oops:


[quote="Ahiles"]Determine the polynomial $ P(X)$ satisfying simoultaneously the conditions:
$ a$) The remainder obtained when dividing $ P(X)$ to the polynomial $ X^3 \minus{} 2$ is equal
to the fourth power of quotient.
$ b$) $ P( \minus{} 2) \plus{} P(2) \equal{} \minus{} 34$.[/quote]

I surely misunderstood the problem : $ P(x)\equal{}(x^3\minus{}2)Q(x)\plus{}R(x)$ with $ R(x)\equal{}Q(x)^4$, so (since degree of $ R(x)<3$) : $ P(x)\equal{}a(x^3\minus{}2)\plus{}a^4$

Then the second condition is $ a((\minus{}2)^3\minus{}2)\plus{}a^4\plus{}a((2)^3\minus{}2)\plus{}a^4\equal{}\minus{}34$ and so $ a^4\minus{}2a\plus{}17\equal{}0$ and this equation has no real solution.


The harmonic mean of two numbers is the reciprocal of the average of their reciprocals. Express the harmonic mean of 5 and 7 as a mixed number.

What?

[hide]I think that the answer is:
1/5+1/7=12/35, dividing by 2 gives us 12/70, which is the average of the reciprocals.
Reciprocal of 12/70 is 70/12 or as a mixed number,

5 5/6?[/hide]

I just learned this in Geometry Class:
[hide="Solution"]The harmonic mean of two numbers is $\frac{1}{\frac{1}{2}(\frac{1}{a}+\frac{1}{b})}$. Using that, or what the problem provides (which is the same thing) you get $\frac{1}{\frac{1}{2}(\frac{1}{7}+\frac{1}{5})}$, which is $\frac{1}{\frac{1}{2}(\frac{12}{35})}$. That equals $\frac{1}{\frac{6}{35}}$, which equals $\frac{35}{6}$. The problem asks for a mixed number, so the answer is $5\frac{5}{6}$.[/hide]

I know that this is off the topic, but how to you put those mixed numbers picture-like things in your solution? It's probably easy for you because I see you do it all the time.

No prob... I just learned it a week ago. You don't really need to do this, but you can download the LaTex software by clicking on the link on the side of the screen. If you get the basics after running the program (TexNic Center) then you won't need it. To insert anything in Tex, use two dollar signs. For fractions, you type in $1\frac{2}{3}$ to get like say 1 2/3(I disabled Tex to show you.. the numbers represent numbers... for example). (The numbers can be swapped around. Don't forget to close your dollar signs, and another one: square root is $\sqrt{5}$ for the square root of five.

Question: How do you get the subtopics, like you put the source as your subtopic?

And [i]please[/i] don't forget to disable latex when you use dollar signs for a dollar amount!!

for hockey:

Is that question for me? I don't know what you're talking about. what subtopics?

I mean... How do you post the source or whatever you want under the topic name on the forum list..

In order to post a subtopic/source, just look right under the "Subject" area, and you should find the "Source" area

BTW: You dont really need to download any software to write in $%Error. "latex" is a bad command.
$ all you have to do in include the dollar signs around the latex code, and the computer automatically trasfers your code into $an image type thing$. Oh, and in classrooms(mathjams, online classes, etc.) you need to add a semicolon at the beginning as well as the dollor signs.

[hide]1/5+1/7-->(12/35)/2-->6/35-->35/6--> 5 5/6
>>5 5/6<<[/hide]

[hide="easy...."]
well (1/x)+(1/y)= (x+y)/xy
so...
one fifth plus one seventh equals .... 12/35
/2
6/35
the reciprocal of that is....

35/6

YAY!!!!  :D  :lol:
[/hide]

um... what's the difference between MiKTeX and TeXnicCenter? which one do I download?

Ps.
[hide]hey mathlover, you are a gurl too!!! yay!!! never mind, just ignore me.....[/hide]

It's best to download both in the order the instructions tell you to... but you don't need either to actually put in TeX in your posts... just the dollar signs are fine...

Also realize, downloading those softwares WILL NOT HELP YOU IN THE AoPS FORUM IN NEARLY ANY WAY
(Besides maybe helping you in learning (maybe) some code)

Dasherm... remember that it's the reciprocal of the AVERAGE of the reciprocals.. you forgot to divide 12/35 by two...


calculus
integration
geometry
geometric transformation
rotation
calculus computations

Can somebody show me how to do the Disk, Washer, and Shell Methods in terms of integration please. I just started my first day of calculus 2 and i dont wanna be lost on the first day lol

You will not be lost. Those topics will be covered in class.

we already went over it....i mean i kinda understand it, but i just want 2 know if there is a good site that explains them well

What book are you using? Most of the standard books do a good job. Here is what I did that helped. Try to understand [i]why[/i] they are called what they are. Disk, washer, etc. I could do the problems, but didn't really understand what I was doing until one day I was like, "Oh, that is why it is called that." If you understand the name you will [i]never[/i] forget how to do them (or if you do, you'll be able to figure it out without the aid of any books or websites or anything).

When I teach these methods, my exam questions don't ever specify which method you should use. We've got such-and-such an object, a solid of revolution. Find its volume by a convenient method. In practice, that does require you to have everything under command, because you don't know up front which is the more convenient method.

Generally, I'd do it in this order:

First, draw a good two-dimensional sketch of the flat object before it's rotated. Include the equations of every line or curve that forms any part of the boundary - including equations like $ x=2$ or $ y=0.$ Be prepared to flip equations around: if you're given $ y=3-2x,$ you might eventually want $ x=\frac{3-y}2.$

Decide which way it is more convenient to slice that two-dimensional object into infintesimally thin strips. Should we slice vertically at location $ x$ with width $ dx$ (and we'll have to figure out what the $ y$ values of the endpoints of that strip are in terms of $ x$)? Or would we be better off slicing horizontally at location $ y$ with width $ dy,$ finding the $ x$ values of the ends of the strip in turns of $ y$? Be prepared to answer this question either way - you have to adapt to the situation.

Now that you've done that, and you know what the axis of rotation is, visualize rotation that infinitesimally thin strip around that axis. What does it sweep out? If you started out with the strip parallel to the rotation axis, you sweep out a cylindrical shell. (If so, my next step is to get out the invisible scissors, cut the shell open, and roll it out to a rectangular slab, one dimension of which is the circumference of the shell.) If you started out with the thin strip perpendicular to the rotation axis, you sweep out a disk or washer.

You'll notice that I'd rather not start from the question of whether to use shells or disks. Instead, I start with the question of which way to slice the 2-dimensional object; whether it's shells or disks follows from that.


Let $ a,b,c>0$ . Show that :

$ \frac{a\plus{}b}{b\plus{}c}\plus{}\frac{b\plus{}c}{c\plus{}a}\plus{}\frac{c\plus{}a}{a\plus{}b}\plus{}\frac{abc(a\plus{}b\plus{}c)}{a^3b\plus{}b^3c\plus{}c^3a}\ge 4$

try $ a\equal{}2$ and  $ b\equal{}c\equal{}1$  :wink:

It's even not true that $ LHS \ge \frac{7}{2}$.

obviously  the best constant  in RHS  is : $ 3$  :wink:

If you set $ c \equal{} 0$, then the minimum is about $ 3.48$.

Is somebody able to find the minimum of $ \sum_{cyc} \frac{a\plus{}b}{b\plus{}c} \plus{} \frac{abc(a\plus{}b\plus{}c)}{a^3b\plus{}b^3c\plus{}c^3a}$?

yes you are right , my intuition  said that the second term ( $ abc..$) is  very weak  term it vanishes  when  c is close to 0 , however I think that $ 3$  is the inf , and there is no min ,probably I'm wrong  :wink:  :maybe:

[quote="aadil"]it is not true that lhs>=3[/quote]

Are you sure ?
\[ LHS > \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b}\ge 3,\]
by AM-GM

Anyway it is a nice problem to find the minimum value of the LHS , if it exists .


AMC
AIME
geometry
3D geometry
number theory
prime numbers

Hi,
I have a problem with the 2005 AIME problem 3 :
How many positive integers have exactly three proper divisors, each of which is less than 50?

It is written in the solution that "To have 3 proper divisors, we need 1 and two other divisors. The ways in which this can occur is if we have a number which is the product of two primes, or the square of a prime number. " (entire solution here : http://www.mathlinks.ro/Forum/resources.php?c=182&cid=45&year=2005&p=365485)

But I don't agree with that because if $n$ is the product of two prime numbers, then $n$ have 4 divisors not 3 ($1$,$p_{1}$,$p_{2}$, and $p_{1}* p_{2}$)
Am I missing something here ?
Thank you.

$n$ would have 4 divisors but only 3 proper divisors.

...but then case two isn't correct :huh:

It should say "the cube of a prime," because it's proper divisors.  Despite this error, the answer is correct: if the number is $p^{3}$, its proper factors are $1, p, p^{2}$ and so we must have $p^{2}\leq 50$, for which there are the 4 primes named.  Someone should fix that solution, though -- it provides really insufficient explanation for what's going on.


function
real analysis
real analysis unsolved

Find the maximum value of the function
\[ f(x,y,z,t)\equal{}\frac{\sqrt[16]{x}\sqrt[8]{y}\sqrt[4]{z}\sqrt{t}}{1\plus{}x\plus{}2y\plus{}z\plus{}t}\]
under conditions $ x\geq 0,\;y\geq 0,\;z\geq 0,\;t\geq 0.$

$ \sqrt [16]{x}\sqrt [8]{y}\sqrt [4]{z}\sqrt {t} \equal{} 4\sqrt [16]{xy^2(0.25z)^4(0.125t)^8} \leq \frac {4}{16}(1 \plus{} x \plus{} 2y \plus{} 4*(0.25z) \plus{} 8*(0.125t)) \equal{} \frac {1}{4}(1 \plus{} x \plus{} 2y \plus{} z \plus{} t)$
So, the maximum value of your function is $ 0.25$ if and only if $ x \equal{} y \equal{} 1, z \equal{} 4, t \equal{} 8$.


quadratics function conics parabola symmetry

Find the quadratic function $f(x)$ which satisfies $|f(1)|=|f(2)|=|f(3)|=|f(4)|=1$.

proof:
[hide]by sketching a rough parabola and the lines y=1 and y=-1, we see that f(1)=f(4)=1 and f(2)=f(3)=-1.  Thus, we let the quadratic function be $(x-a)(x-b)=x^2-(a+b)x+ab$, where a and b are real.  plugging in x=1, 2 we get the equations

$1-a-b+ab=1\Rightarrow a+b=ab$
$4-2a-2b+ab=-1\Rightarrow 2(a+b)-5=ab$

Solving for $ab$ and $a+b$, we get $ab=5$, and $a+b=5$.  Thus the quadratic function is $x^2-5x+5$.[/hide]

I think there are two answers...

[hide]A quadratic function of x is symmetric about a vertical line x=h.  At most two outputs of a quadratic function can be equal; hence, two of f(1), f(2), f(3), f(4) are positive and two are negative.  If f(1) = f(2) then the axis of symmetry must be x=1.5; but then f(3) $\neq$ f(4), a contradiction; if f(1) = f(3), then the axis of symmetry is x=2; but then f(2) $\neq$ f(4), a contradiction; thus, f(1) = f(4) and f(2) = f(3), and the axis of symmetry is x=5/2.

So the parabola is of the form y = a(x-5/2)^2 + k, where a and k are still unknown.

CASE ONE: f(1) = 1.  Then f(2) = -1.  Plugging these into our equation yields 1 = 9a/4 + k and -1 = a/4 + k, so that 2a = 2 or a = 1, and k = -5/4.  Thus the equation is f(x) = (x-5/2)^2 - 5/4 = x^2 - 5x + 5.  We know this will satisfy f(3) = -1 and f(4) = 1 because it has an axis of symmetry at x=5/2.

CASE TWO: f(1) = -1.  Then f(2) = 1.  Plugging yields -1 = 9a/4 + k and 1 = a/4 + k, so 2a = -2 or a = -1 and k = 5/4.  Thus the equation is f(x) = -(x-5/2)^2 + 5/4 = -x^2 + 5x - 5.  Again, this satisfies the other two.[/hide]

come to think of it, you're right.  whoops.


Pencil or pen? And how do I make one of those poll interfaces?

For USAMO, anything that will fax is fine. If you make mistakes even on final copies you should use pencil, but make sure you write dark enough so that it will be readable after it's faxed.

I personally would use a mechanical pencil because it is very precise and doesn't require sharpening.. although I only use certain types cause the others don't give you the same amount of comfort.  Pen is nice because it is usually pretty exact but you can't erase and those erasable pens are pretty lame.  I only use pen on a test I'm exceedingly confident on  :lol:

It really isn't that big of a decision, its just your matter of personal preference.  Pen usually requires a little more correction, unless you're god or Alex Zhai.

If you can write well, use a blue or black pen. If you can't and make several mistakes, Use a dark pencil. The pencil doesn't have to be #2 since a professor grades it, not a machine.

You should still have a #2 because #3 might not fax that well...

I'm too lazy to look for it right now, but I think the official handbook states that you have to use pen. It's on  [url]http://www.unl.edu/amc[/url]

err... maybe that explains last year's score?

No, the rules may have changed this year, the handbook says:

Please leave a 1” margin on all sides of the Answer Sheet and on any additional pages and write DARKLY and LEGIBLY, so your faxed copy can be easily read. A #2 lead pencil, black ink pen or marker fax best.

My bad, I remember reading something about pens in that thing...

[quote="MithsApprentice"]You could use crayons for all they cared as long as its legible and not obnoxiously large when you fax it. I personally have a blue and black pen handy for writing solutions.[/quote]

or for the AIME, you can work in whatever method you like, as long as you bubble with pencil

for example, you can scribble with crayons, write in pen, or even better, do it all in your head

Bring both and do scratch work in pencil and write your solution in pen.

Well, that's what I would do at least...

Personally I prefer yellow highlighter but w/e floats your boat.

I say you bring a pen to stab yourself, and then write it using your blood...

No, just kidding. That's just wrong.  Everyone knows pencils are better for pricking.

But really, pencils are useful if you make mistakes, but pens usually look neater.

[quote="randomdragoon"]We should all write up our USAMOs in crayon.[/quote]

White crayon. :D


limit
logarithms
calculus
calculus computations

Generalization of (i):
\[ \boxed{\boxed{\lim_{x\rightarrow 0} \frac {a^x - b^x}{c^x - d^x} = \frac {\ln \frac {a}{b}}{\ln \frac {c}{d}}}}\]
for positive real numbers $ a,\ b,\ c,\ d\neq 1.$

1) applying L'hospital rule we $ \log_{2}{\frac {4}{3}}$

2)it can be written as $ \lim_{x\to\ \infty}\frac {{1} - (3/4)^{x}}{1 - (2/4)^{x}}$

hence answer is 1.

1) Multiply and divide by $ x$ . We have the $ \lim_{x\to\ 0}\frac{a^{x}\minus{}1}{x}\equal{}\ log(a)$ also , $ \lim_{x\to\ 0} \frac{a^{x}\minus{}b^{x}}{x}\equal{}\ln \frac{a}{b}$

That's right. First of all, we should learn the definition of $ \frac{d}{dx}a^x$ from the dinition of $ e$.

I like to give explanations to my satisfaction.. :)  :lol:


function
algebra
domain
limit
advanced fields
advanced fields unsolved

The most random thought just occurred to me:

Suppose $f(x)$ is continuous and we know $f(a)$ for all $a \in S$, some set that is dense in the domain of $S$. Does that uniquely determine $f$? (This is either obviously true or I'm totally blind.)

If yes, it seems to me that would imply that (in some sense) we can say that the Dirac distributions $\delta_{a}, a \in S$ are a countable basis for the space of functions $C(\overline{S})$. I mean, consider $S = \mathbb{Q}\cap [0, 1]$. A basis for $L^{2}([0, 1])$ is $\{e^{2\pi inx}\}$, right? There, we can reconstruct $f(x)$ from all of the $\langle f, e_{n}\rangle$. Also (if my thought is correct) we can reconstruct any continuous $f$ from $\langle f, \delta_{a}\rangle$ (though not in the "normal" way).

This might be total madness, or totally useless, but sometimes useless stuff helps me see how things work anyway.  :wink:

For each $x$, you know $\lim_{t\to x}f(t)$ exists, and density in this case means you can approximate the value of $f(x)$ closely enough, so yes, $f$ is uniquely determined.

As blahblahblah pointed out, yes, it's obvious (sorry).

However, your other question is interesting because I always like to think of deltas as a basis for functions.  In a really stupid sense, they are by definition a "basis" (in some perverse sense) for the space of "all" functions, but unfortunately, we can never profitably work with the space of "all" functions.

For continuous functions, there are some problems.  First, while values of a cts fn on a dense subset S determine the fn, you cannot assign arbitrary values to the dense set.  Specifically, you can have a  sequence in Q that converges in R, but your chosen values might not converge.   So in some sense, the deltas you mention will "span," but they are not "independent."  Or a better way to say it is that the span of your delta's (in a distribution space) are dense in C(R), in some sense, which doesn't sound as exciting (especially because it's paraphrasing the fact that Q is dense in R).

It is not obvious (to me, at least), whether there are simple conditions you might place on your function on Q to guarantee the existence of a cts extension to R.  My gut tells me that asking for continuity on Q is insufficient, but I'm too lazy to figure it out.  Prove or disprove as an exercise.

By the way, if you'd like to see a rigorous explanation of how deltas might be thought of as a "basis" (for L^2 perhaps), look for "rigged Hilbert spaces."  It's a mathematical physics concept which I have never actually seen put to use, but it's nice to know that certain vague ideas can be made rigorous.

hm.. just en passant, $f: q \mapsto (q-\sqrt2)^{-1}$ is continuous on $\mathbb{Q}$ but doesn't admit a continuous extension to $\mathbb{R}$...
i was just thinking about it a couple of days ago..


The ski club has 8 women an 7 men. What is the probability that if the club elects 3 officers, all 3 of them will be women?

I know that the bottom should be 15C3 but I can't figure out what the top is. I thought it would be 8C3 but thats wrong. My book says the answer is .431. Does anyone know what im doing wrong?

.431 doesn't seem very reasonable... does it say anything about how they vote?

The simplest method would probably be this:

$ P\equal{}\frac{8}{15}\cdot\frac{7}{14}\cdot\frac{6}{13}$

$ \equal{}\frac{8}{65}$

which is approximately $ .124$. I have no idea how your book got that answer.

The problem probably is determining how they vote.

If you typed the question rightly then you aren't doing anything wrong.The probablity should be 8[b]C[/b]3/15[b]C[/b]3=.124(app.)'cause the number of ways all 3 elected will be women is equivalent to the number of ways to choose 3 women from 8 women.Either you wrote the question wrong or the answer in the book is wrong.

[b]Follow-up:[/b] If there is a man among the elected officials, there is a probability of P that the women will accuse the male voters of sexism and demand a complete revote. Find P such that the overall probability of all the elected officials being women is 28/65.

[i]Yes, there can be more than one revote; no, I don't mean to offend anyone[/i] :)

Can there be more than one revote? (Actually, I think it might not matter, but it is kinda humorous to think about a woman accusing the men of sexism 100 or so times...)

btw fishythefish, the answer without voting conditions rounds to $ 0.123$.  :wink:

With the new voting conditions by The Zuton Force, we have

$ \displaystyle \frac{28}{65} \equal{} \frac{\binom{8}{3}}{\binom{15}{3}} \plus{} P\left(\frac{\binom{8}{3}}{\binom{15}{3}} \plus{} P\left(\frac{\binom{8}{3}}{\binom{15}{3}} \plus{} P\left(\frac{\binom{8}{3}}{\binom{15}{3}} \plus{} P\left(\ldots\right)\right)\right)\right)$

That whole infinite section can be shortened:

$ \displaystyle \frac{28}{65} \equal{} \frac{\binom{8}{3}}{\binom{15}{3}} \plus{}  \frac{28}{65}P$

$ \displaystyle \frac{28}{65} \equal{} \frac{8}{65} \plus{}  \frac{28}{65}P$
$ \displaystyle P \equal{} \boxed{\frac{5}{7}}$

I got a different solution for ZutonForce's extension, but it's not a pretty answer, so I'm not sure...  :maybe:

[hide="solution"]
Let C= probability of all of the elected members are women= 28/65.
Let x= probability that all of the elected members are women on first try= x.

Thus, $ C\equal{} x \plus{} (1\minus{}x)pC$, since the probability of a recount $ \equal{} (1\minus{}x)p$, and once the recount starts, it just becomes a loop. Plugging in for C and x, then solving for p yields $ \boxed{\frac{325}{399}}$ [/hide]

[hide="Hint"]Count the complement[/hide]
[hide="More Hint"]What must happen? Let q be the natural probability, ie 8/65, r be the complement, and p be the probability of a revote, and s be the complement of that

For a man to be elected, we must have:
- man is initially elected (r) AND no revote (s) = rs
- man is initially elected (r) AND revote (p) AND man is elected again (r) AND no revote (s) = r^2ps
- man is elected, revote, elected, revote, elected, no revote = r^3 p^2 s

etc.

We see that we  have terms of the form rs(rp)^n, n ranging from 0 to infinity.[/hide]

The number probably isn't pretty, but the point of the extension was, as [b]fishy [/b]put it:

"The problem probably is determining how they vote."

I got $ \frac{325}{399}$. Same as erdogankerem123.

Hm...I also tried it another way and still got $ \frac{325}{399}$...

well? is $ \frac{325}{399}$ correct. Three people have gotten that including me...

I think it's wrong because um...Zuton Force posted the hints like 7 hours after the first person who said that.

hmm.. I don't see how its wrong. I've used infinite geometric series and recurrence and they give me the same answer.

Well, if P=1, then the R.H.S. goes to infinity when it should go to 1.

It's because he didn't weight the probability of there being one male, he simply added it on.

I might have made an arithmetic error in here somewhere, I'm doing this by hand

[hide="Solution"]Let a be the event that at least one male is elected. Let b be the event that a recount is declared. Let c be the event that a recount is not declared.

The ways we can have one male being elected are:

ac
abac
ababac
abababac
etc...

If you consider the respective probabilities (let's use the same variables), we have terms of the form:

ac(ab)^n, where n is a non-negative integer.

The total probability is then:

37/65=ac/(1-ab)

However, note that c=1-b, so we really have 37/65 = (a-ab)/(1-ab)

We know that a=57/65, and we want to solve for b. This gives me 325/469[/hide]


Euler
logarithms
parameterization
integration
function
real analysis
real analysis unsolved

$ x^2 * y'' \minus{} 8*x*y' \plus{} 14*y \equal{} 3*x^5 \minus{} 5*x^2$

LHS (the homogeneous part) is: $ y \equal{} A*x^2 \plus{} B*x^7$

$ 3*x^5 \minus{} 5*x^2 \equal{} x^2 * (3*x^3 \minus{} 5)$. RHS is of the form: $ p(y) \equal{} v*x^2*(a\plus{}b*x\plus{}c*x^2\plus{}d*x^3)$

Attempted to find $ p(y')$ and $ p(y'')$ w.r.t. both x and v of the above in order to bring

$ x^2 * p(y'') \minus{} 8*x*p(y') \plus{} 14*p(y) \equal{} 3*x^5 \minus{} 5*x^2$ to a first order linear differential equation but could not simplify the expansion (it's huge!).

The final particular solution is:

$ \minus{} \frac{1}{2}*x^5 \plus{} x^2*ln(x) \plus{} \frac{1}{5}*x^2$

Is there a simplier way to find a particular solution?

How to find a particular solution: If $ D(y)\equal{}x^2y''\minus{}8xy'\plus{}14y$, $ D(x^k)\equal{}c(k)x^k$. This suggests a solution of the form $ ax^5\plus{}bx^2$.
Now $ (5\cdot 4\minus{}8\cdot 5\plus{}14)a\equal{}3$ and $ (2\cdot 1\minus{}8\cdot 2\plus{}14)b\equal{}\minus{}5$, so $ a\equal{}\frac{3}{\minus{}6}\equal{}\minus{}\frac12$ and $ b\equal{}\frac{\minus{}5}{0}$- oops, this one doesn't work. Taking a cue from the method of undetermined coefficients, we try a constant times $ x^2\ln x$ instead, and $ x^2\ln x$ works.

If we substituted $ x\equal{}e^t$, this would become the constant-coefficient equation $ y''\minus{}9y'\plus{}14y\equal{}3e^{5t}\minus{}5e^{2t}$, and the duplicated $ e^{2t}$ means we have to go up a power of $ t$.

Firstly we solve the homogenious Euler equation. We use the subtitution $ x\equal{}e^t$ or $ t\equal{}lnx$. So we have $ u(t)\equal{}y(e^t)$.
$ y'\equal{}\frac{dy}{dx}\equal{}\frac{du}{dt}\frac{dt}{dx}\equal{}e^{\minus{}t}u'$ and $ y''\equal{}(u''\minus{}u')e^{\minus{}2t}$

Thus, $ x^2y''\minus{}8xy'\plus{}14y\equal{}0\Leftrightarrow u''\minus{}9u'\plus{}14u\equal{}0\rightarrow u(t)\equal{}c_1e^{7t}\plus{}c_2e^{2t}\Leftrightarrow y_h(x)\equal{}c_1x^7\plus{}c_2x^2,$ where $ y_1\equal{}x^7 , y_2\equal{}x^2$
We have $ y''\minus{}\frac{8}{x}y'\plus{}\frac{14}{x^2}y\equal{}3x^3\minus{}5 , g(x)\equal{}3x^3\minus{}5$,

according to the method of variation of parameters a particular solution of the equation is $ Y\equal{}v_1(x)y_1(x)\plus{}v_2(x)y_2(x)$,

where $ v_1(x)\equal{}\minus{}\int \frac{g(x)y_2(x)}{w(y_1,y_2)}dx$ and $ v_2(x)\equal{}\int\frac{g(x)y_1(x)}{w(y_1,y_2)}dx , w(y_1,y_2)\equal{}y_1y_2'\minus{}y_1'y_2$.

The general solution of the d.e. is:
$ y\equal{}y_h\plus{}Y \leftrightarrow y\equal{}c_1x^7\plus{}c_2x^2\plus{}\frac{1}{10}(2x^2\minus{}5x^5\plus{}10x^2lnx)$ :)

Thank you both!  :)

I have not thought of the Wronskian function and the method of variation of parameters. But it's good to know both methods for the future. :-)


There are 6 gold coins and one is different from others.There is balance(with 1 hand and shows the weight).You can put coins on balance only three times.Find the different one and find its weight(heavy from others or not)???
It is an amazing question!

Let $ a,b,c,d$ be four of the coins.
You may compare the weight of $ a\plus{}b$ and $ c\plus{}d$.

- If they are equals, then the fake coins is among the two other coins, say $ x,y$. In particular, you know that $ a$ is fair.
Then, you only have to compare $ a$ and $ x$, and $ a$ and $ y$ to determine which one has not the same weight than $ a$ and if it is heavier or not.

- If, say, $ a\plus{}b < c\plus{}d$. then the fake coin is among these four coins.
Now compare $ a$ and $ c$. If, say $ a<c$ then the fake one is $ a$ or $ c$. Thus compare $ a$ and $ d$. Since $ d$ is fair, you have the desired informations.
If $ a\equal{}c$ then you know that they are fair. Thus compare $ a$ and $ d$. If $ a\equal{}d$ then the fake one is $ b$ and since $ a\plus{}b < c\plus{}d$ you know that it is lighter than a fair coin. If $ a \neq d$ yoo know that $ d$ is fake and if its weight is gretare or not than a fair coin.

Subsidiary question : why is it posted in the algebra forum?

Pierre.

But You cannot compare coins you can only put some coins on balance because balance is ONE handed and it shows its weight.Thank You

goldboy, if you know the solution, then it shouldn't be in this section, it should be in the section under that one, i think.

Hmm, i have gotten some pretty good ideas.

I have a hint that might help, but it might not. i think it might have to do with the solution though.

[hide]Alright, lets say we have 'y' coins. Let's say that y = x + z , where x, y, and z are all natural numbers. Lets also say that x = mz, where m is a rational number. Lets say that the weight of one coin is w. Now let's say that we weigh 'x' coins. The weight is xw. Then we weigh the z coins and the weight is zw. We know that the weight of 'x' coins divided by 'm'  must always equal the weight of 'z' coins, if the weight of all 'y' coins are the same. I thought this might be helpful.  [/hide]

Sorry for double posting, but is anybody going to comment on what I said?


function
limit
trigonometry
calculus
real analysis
calculus computations

Prove that the function $ f(x) \equal{}sin(\frac{1}{x})$ for x in $ (0,\frac{2}{\pi})$ is not uniform continuous.

we consider two squences  $ (x_n)$ and $ (y_n)$ such that $ x_n=\frac{1}{2\pi n}$ and ${ y_n =\frac{1}{2\pi n+\frac{\pi}{2}}}$ we have $ f(y_n)-f(x_n) \to 1$ but $ y_n-x_n \to 0.$

A uniformly continuous function on a bounded set can be extended to a continuous function on the closure; uniform continuity implies that $ f(x_n)$ is Cauchy whenever $ x_n$ is Cauchy.

Since $ \lim_{x\to 0^ \plus{} }\sin\frac1x$ doesn't exist, $ \sin\frac1x$ is not uniformly continuous on $ (0,c)$.

Of course, tchebytchev's sequences are how we prove that the limit doesn't exist.

Please explain this question by using epsilon delta method.....
Actually in my calculus class the original question that has asked to me is following
Prove that following function is not uniform continuous in $ (0,\frac{2}{\pi})$
$ f(x) \equal{} sin(\frac{1}{x})$.

Thanks in Advanced.\


In this game, someone posts a question, and others try to think of song titles that answer the question. For instance, if the question was "Where do you want to be?", some possible answers could be

Heaven-The Talking Heads
Chinatown-Thin Lizzy
America-Spinal Tap
Where Eagles Have Been-Wolfmother.

For the first question, try to think of more answers for "Where do you want to be?"

Cold Pizza - m1dy
Cum On Everybody - Eminem
Gasoline - Seether
Scar Tissue - Red Hot Chili Peppers
Slim Anus - Insane Clown Posse
Spiders - System of a Down
Weenie In A Bottle - Weird Al Yankovic

Pick one.


Take a certain integer $k$ such that $k^2$ divides 60 with a remainder of 6. What is the remainder when 216000 is divided by $k^3$?

[hide="dumb way"]It looks like k=3 works. 216000/27 leaves a remainder of 0. [/hide]

[quote="JesusFreak197"]That's dumb? That's the way I would have done it... :P[/quote]

I'm wondering if there was a more algebraic way. Now come to think of it, it isn't that dumb  :P


algebra
function
domain
calculus
calculus computations

Lets say we choose $ \delta \equal{} \epsilon/2$ for a limit. Would any $ \delta$ smaller work also? Like if $ \delta \equal{} \epsilon/3$?

Yes.  (Well, it still has to be positive, of course.)  All we need is that [i]some[/i] value of $ \delta$ exists (depending on $ \varepsilon$, of course) that satisfies the necessary conditions, and if you have a single $ \delta$ then any smaller value will work, too.

To see why, let's say $ \delta_0>0$ 'works'. This means that $ |x\minus{}a|<\delta_0\Rightarrow |f(x)\minus{}\lambda|<\epsilon$ for all $ \epsilon>0$.

But for any $ 0<\delta<\delta_0$, all the $ x$ satisfying $ |x\minus{}a|<\delta$ also satisfy $ |x\minus{}a|<\delta_0$. In other words:
$ |x\minus{}a|<\delta\Rightarrow |x\minus{}a|<\delta_0\Rightarrow |f(x)\minus{}\lambda|<\epsilon$ for all $ \epsilon>0$

Geometrically, the definition of a limit says that for any ball around $ \lambda$ (radius $ \epsilon$) in the range of $ f$ we can find a ball around $ a$ in the domain of $ f$ (radius $ \delta_0$) that maps into the ball around $ \lambda$.  And clearly if that ball maps into the ball around $ \lambda$ then any ball contained in it (radius $ \delta < \delta_0$) has the same property.


calculus
integration
function
geometry
rectangle
calculus computations

The integral of the function f(t) = t from 1 to x is

x^2 - 1^2
----------
     2

The sum of the Integers from 1 to x is

x(x+1)           x^2 + x
------    =   -----------
  2......................2

this means in R+ the Sum of Integers is always bigger than the Integral

I thought the Integral is the same thing as the sum with only difference that integral also adds function terms for all numbers between the limits including nonnatural numbers

how can this happen?
im very confused :(

Have you tried drawing any pictures? Represent an integral as the area under a curve; represent a sum as the sum of areas of rectangles of width $ 1$ and varying heights. Be careful in your pictures about where sums and integrals start and stop.

Do that correctly, and you should be able to show that:

$ \int_0^nt\,dt<\sum_{k\equal{}1}^nk<\int_1^{n\plus{}1}t\,dt$

That simplifes to $ \frac{n^2}2<\frac{n^2\plus{}n}2<\frac{(n\plus{}1)^2\minus{}1}2\equal{}\frac{n^2\plus{}2n}2.$

No contradictons there.

By the same idea, for any $ m>0:$

$ \int_0^nt^m\,dt<\sum_{k\equal{}1}^nk^m<\int_1^{n\plus{}1}t^m\,dt$

$ \frac{n^{m\plus{}1}}{m\plus{}1}<\sum_{k\equal{}1}^nk^m<\frac{(n\plus{}1)^{m\plus{}1}\minus{}1}{m\plus{}1}.$


(a, b, c) are all non zero pozitive integers and a<b<c.

If (1/a + 1/b + 1/c = 1), find a, b, c

[quote="bthings_2000"][hide] (2,3,6)[/hide][/quote]

Only solution??? Got proof?

If you wanted to extend this, you could say that

\[ \sum_{k=1}^{n}\frac{1}{a_{k}}=1  \]

if $a_{n}$ is a perfect number and each $a_{k}$ is a factor of $a_{n}$ such that $\sum_{k=1}^{n-1}a_{k}=a_{n}-1$.  For instance,

\[ \frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28}=1  \]

and $2+4+7+14=27=28-1$.  The same holds true for $496$ and every other perfect number (like the original problem, $6$ certainly works).

yea, i've seen that problem before, where we're given the sum of some number of reciprocals, and they turn out to be all the factors of a perfect number...interesting little fact.

[quote="shyong"]$\frac{1}{a}>\frac{1}{b}>\frac{1}{c}$

$\implies \frac{3}{a}>\frac{1}a+\frac{1}{b}+\frac{1}{c}=1$ $\implies a<3$

$\implies a=2$

then $\frac{2}{b}>\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$ so $2=a<b<4$ , so $b=3$

then we get $c=6$ ...[/quote]

Thanks !

[quote="0714446459923"]I thought a problem similar to this came up before. I seem to remember several solutions. They were asking for the one where the sum of the reciprocals was the lowest.

Nevermind, the problem I was thinking of used 4 numbers instead of 3.
2,3,6 is the only solution to the three number problem.[/quote]

Thanks !

[quote="lingomaniac88"]Technically, it's possible to write $n$ different unit fractions that have a sum of 1, as long as $n$ is odd.

Basically, you start with $\frac12+\frac13+\frac16=1$, and you expand $\frac16$ into many fractions.  It's based on the fact that $\frac1{2x}=\frac1{3x}+\frac1{6x}$ for any nonzero value of $x$.
Expand $\frac16$ into $\frac1{12}+\frac1{12}$.  Now, let $x=6$ in the above equation, and expand the last $\frac1{12}$ to $\frac1{18}+\frac1{36}$.  You can telescope this infinitely.

Correct me if I'm wrong, but I think that it's safe to say that [hide]$\sum_{x=0}^{n}(\frac{1}{6^n\cdot2}+\frac{1}{6^n\cdot3})+\frac{1}{6^{n+1}}=1$.[/hide][/quote]

Actually, $n$ can be any integer greater than or equal to 3. Define $a_k$ by $a_1=2$ and $a_{k+1}=a_k^2-a_k+1$. The sum \[ \left( \sum_{k=1}^{n-1} \frac{1}{a_k} \right) +\frac{1}{a_n-1} \]
is always equal to 1, $a_k$ is strictly increasing, and $a_{n-1}=a_n-1$ only when $n=2$.


Let me ask you something. Is it possible to pair off the integers 1, 2, 3...50 in such a manner that the sum of each pair of numbers is a [b]different[/b] prime number?

[hide]No.

The highest number you can get from the sum of two numbers from 1 to 50, is 99.  So the question is how many primes are there less than 100?  For that, I get 25.  The number of pairs is 25 aswell.  The problem is, that it is impossible to add two different numbers from 1 to 50, and get a 2.  So there are only 24 primes that you can get, so at least one pair will not sum to a prime, or sum to a prime that another pair sums to.[/hide]

yep thats how i did it..i just used the [hide]prime number theorem[/hide]. it was quicker. if you dont know what that is, it states that p(x) [which is the number of primes less than x] can be approximated by x/ln(x).

The prime number theorem isn't quite good enough for this problem. Just because pi(x)~x/ln(x) doesn't mean that for small values of x (such as 100) the two aren't way off.

ooh you're right...by using floor[x/ln(x)] i got only 21. however, i did notice that 2 couldnt be attained, so i still might have got partial credit.
i think legendre found a better expression that was more accurate. but nothing beats direct counting.

i think (im not sure) that the logarithmic integral is the closest approximation.

a better approximation is floor[x/(ln[x]-1.08366)], but what tetra said is probably better. however, for small numbers, this still is too high. i think taking floor(average of normal prime number theorem and approximation) has even greater accuracy.

Actually of all the expressions of the form x/(ln(x)-k), the best approximation is obtained when k=1 for large values of x. However, the logarithmic integral is better than any of them.

But the stupid-proof way is to just count the darn things.  It's a proof, so you won't be hurt by spending 2 minutes doing that.

haha!

i agree, there's no need to get fancy on this problem!! :D

what if were asked to pair off 1,2,11,12,21,22,... etc into primes? There must be a solution to this problem that does not require counting.


geometry
perimeter
inequalities
conics
ellipse
parameterization
parametric equation

the answer (according to me is  a 2 (\pi- \sqrt 3)/2
note that if we fix on of the distance suppose PA=r (Note the word distance  ,not point P) then the region for which the inequality holds is the intersection  of the circle of radius "r" ,centre A and the ellipse with focus B and C and major axis =( 2a-r)/2
now prove that the intersection of the ellipse and this circle lie on the arcs of the circles with radius =a and centres B and C respectively.maximum value of r=2a and then the ellipse degenerates to a straight line.so the required convex region is the region bounded by the circles of radius =a and centres A,B,C. So the area is
area triangle ABC +3*area(minor segmentAB)= WOT I WROTE ABOVE
hope the solution is correct

sorry ,i was wrong (though approach seems right).just woke up from my slumber and realized that the solution wasnt anywhere near correct
moral of the story : dont write solutions when u r half asleep
and more importantly ,an ellipse and a circle can intersect at  4 points also :D  :D  :D

analytically

A (0,\sqrt3a/2) B (-a/2,0) C (a/2,0)  P(x,y)

so you get  \sqrt (x²+(y-\(sqrt3)a/2)²) + \sqrt ((x+a/2)²+y²) + \sqrt ((x-a/2)²+y²) = 2a

simply solve the equitation and you'll get the parameterequitation

and from there it's simple :D

(not gonna type the whole stuff - way too long)

well that 's what i also did peter ,but the thing is, this method is 3RD DEGREE TORTURE.could u please post the parametric equation u got.

[Edit: it is a wrong solution]
You missed something...

Construct three arcs of radii $a$ with centers $A,B,C$ as it is shown below. They bound a convex area. We will prove that it is a required figure. Indeed, suppose $P$ lies on the arc $AB$, then applying Ptolemei's theorem to $APBC$ we conclude $AP+BP=PC=a$ !
Suppose now that $E$ lies outside of this figure.
[b]Case 1.[/b] $E$ lies inside of angle $BAC$. Then $AE+BE+CE>AE'+BE'+CE'=2a$.
[b]Case 2.[/b] $E$ lies in the antiangle $C$. Then $ABC\subset EAB$, so $AE+BE>2a$.

Myth, are you sure about your solution?

[quote="Myth"]then applying Ptolemei's theorem to $APBC$ we conclude $AP+BP=PC=a$ ![/quote]

Actually, the quadrilateral APBC is not cyclic, is it?

  Darij

Ops... It is cyclic for another arc, but it will not help. :(

[quote="A1lqdSchool"]Let ABC be an equilateral triangle with perimeter 3a . Calculate the area of the convex region consisting of all points P such that : $PA+PB+PC \leq 2a$.[/quote]

This is the American Mathematical Monthly problem E2983, solved in American Mathematical Monthly, Vol. 96, No. 7 (Aug. - Sep., 1989), pp. 642-645. See the attached ZIP file for a scan of this solution. At the end of the solution, the editors remark: "No correct solutions were received before the customary deadline. The above solution was prepared by Mr. Fee at the request of the editors." This, and also the fact that the problem stood around unsolved on ML for a long time, is a consequence of the fact that the solution is a straightforward computational disaster. Just for fun, let me copy the answer to the problem:

The area of the region consisting of all points P such that $PA+PB+PC \leq 2a$ equals

$\left(62\frac{\pi}{9}-\frac{124}{9}\arcsin\frac{5\sqrt3}{9}-\frac{20\sqrt2}{9}\right)a^2$.

Do you like this formula? ;)

  Darij


The point $ O$ is the center of the circle circumscribed about $ \triangle ABC$, with $ \angle BOC \equal{} 120^\circ$ and $ \angle AOB \equal{} 140^\circ$, as shown. What is the degree measure of $ \angle ABC$?
[asy]unitsize(2cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));

pair B=dir(80), A=dir(220), C=dir(320), O=(0,0);

draw(unitcircle);
draw(A--B--C--O--A--C);
draw(O--B);

draw(anglemark(C,O,A,2));

label("$A$",A,SW);
label("$B$",B,NNE);
label("$C$",C,SE);
label("$O$",O,S);
label("$140^{\circ}$",O,NW,fontsize(8pt));
label("$120^{\circ}$",O,ENE,fontsize(8pt));[/asy]$ \textbf{(A)}\ 35 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 60$

[hide]angle AOC = 100 degrees

Therefore, ARC AC= 100 degrees

ANGLE ABC is half of the measure of the ARC AC, so 100/2 = 50 degrees.

which is choice D[/hide]

Solution without inscribed angles.

[hide]Each base angle in triangle AOB is $(180-140)/2=20$ degrees.  Similarly, each base angle in triangle AOC is $(180-120)/2=30$ degrees.  So angle ABC is the sum of those, or $50$.[/hide]


limit
function
calculus
geometry
integration
calculus computations

This can be written as:  $ \frac{2007! n! n^{2007}}{(n+2007)!}$.

We then just use Stirling's approximation:  $ n! \sim \sqrt{2 \pi n}\left( \frac{n}{e}\right)^{n}$.

This gives:  $ \frac{\sqrt{2 \pi n}\left( \frac{n}{e}\right)^{n}n^{2007}2007!}{\sqrt{2 \pi (n+2007)}\left( \frac{n+2007}{e}\right)^{n+2007}}$

=$ e^{2007}2007! \left( \frac{n}{n+2007}\right)^{n+2007.5}$

=$ 2007!$ when $ n\rightarrow \infty$.

This seems to smell the flavour of gamma functuion.

What happened to algebraic manipulations?
\begin{eqnarray*}\frac{n! \cdot n^{2007}}{2008 \cdot 2009 \cdot 2010 \cdot \ldots \cdot (2007+n)}&=& \frac{2007! \cdot n^{2007}}{(n+1)(n+2) \ldots (n+2007)}\\ \ &=& \frac{2007!}{\left( 1+\frac1{n}\right) \left( 1+\frac2{n}\right) \ldots \left( 1+\frac{2007}{n}\right)}\\ \ &\rightarrow& 2007! .\end{eqnarray*}

A better proof than the one I gave above can be had from realizing that one of the definitions of the Gamma function is as follows:

$ \lim_{n\rightarrow \infty}\frac{n! n^{z}}{z(z+1) \cdots (z+n)}$

That's typical of me though.   :oops:

As for avoiding algebra, well, this is the calculus forum after all.  Though I am not sure if someone asked for the area of a square here if I would be inclined to use an integral, hmm...

[quote="edoo"]As for avoiding algebra, well, this is the calculus forum after all.[/quote]
I was just implying that this is the kind of exercise that is accessible to anyone that knows just the definition of a limit plus the sum/difference/product/quotient properties. BTW, algebraic manipulations are standard in solving limits.
Nonetheless, I have nothing against using advanced techniques (I'm quite a fan of them, actually).


function
analytic geometry
real analysis
real analysis unsolved

Polar equation: r=f(t). Show that if f is even, then it is symmetric about the polar axis. And if f has period pi, show it is symmetric about the origin....

It is too obvious...I've no idea on how to show...can anyone help me???

Any function $f$ is considered even if $f(-x)=f(x)$.

So in your case $r=f(\theta)=f(-\theta)$.

This means that if your angle is negative the radius will still be the same, so the graph is symetrical where the "x-axis" would be if you were using a rectangular coordinate system.


[quote="besttate"]Find, with proof, all real solutions for $a$ to $a^{2003}\equiv 1 \bmod{17}$[/quote]

Umm 1+17k for any integer k?
Its actually $a-1=17k$ so $a=1+17k$
and $a^{2}+a+1=17k$, but I dont know how to prove there are no integral solutions to that equation for a when k can be any integer from negative infinity to infinity.

consider the fact that $a^{16}\equiv 1 \bmod{17}$ for any integer a.

[quote="besttate"]I see why that fact would allow it to be true. Why is your fact true?[/quote]

wait what, are you contradicting yourself?

[quote="besttate"]Sorry for being unclear  :blush:

What I meant to say was: I did not know that was a fact, but if it were true it would make your solution true.

Would you please explain why the statement you claimed was a fact is indeed a fact?[/quote]

O its a theorm, I think its fermats little theorm.

Thank you :) You were right on the name and everything! Thanks!

[quote="besttate"]Find, with proof, all [b]real[/b] solutions for $a$ to $a^{2003}\equiv 1 \bmod{17}$[/quote]

Er.  $a = \sqrt[2003]{1+17k}, k \in \mathbb{Z}$.  Perhaps you wanted integer solutions.

[quote="Art of Owna"]consider the fact that $a^{16}\equiv 1 \bmod{17}$ for any integer a.[/quote]

Only if $gcd(a, 17) = 1$.

obviously if $17|a$, then there is no solution...so $17\not| a$...then we apply FLT: $a^{16}\equiv 1\mod 17$, so $a^{14}\equiv 1\mod 17$...now i guess i would just test 0,1,2,...,8...blah factor...only 1 works...so $a\equiv \pm 1\mod 17$

It's a little easier than that.  $x \to x^{7}\bmod 17$ is bijective, so we can deduce directly that $a^{2}\equiv 1 \bmod 17$.

Alternately, $a^{-2}\equiv 1 \bmod 17 \Leftrightarrow a^{2}\equiv 1^{-1}\equiv 1 \bmod 17$.

How do we go from $a^{16}\equiv{1\mod{17}}$ to $a^{14}\equiv{1\mod{17}}$?

The original problem states that $a^{2003}\equiv 1 \bmod 17$.

However, Altheman seems to have taken the exponent $\bmod 17$, when properly it should be taken $\bmod 16$.  This gives

$a^{3}\equiv 1 \bmod 17$

And since $x \to x^{3}\bmod 17$ is also bijective, this has only the solution $a = \boxed{1}$.

t0rajir0u - can you show why all these bijections are true (i think it may have to do with the fact that $gcd(3,16) = gcd(7,16) = 1$, but is there any more basis than that?)

there also happen to be complex solutions...but w/e.

[quote="Teki-Teki"]t0rajir0u - can you show why all these bijections are true (i think it may have to do with the fact that $gcd(3,16) = gcd(7,16) = 1$[/quote]

You are correct.

In particular, it is known that $\mathbb{Z}_{p}$ has a primitive root, so all unique residues $\bmod p$ can be expressed as $r^{k}, k = 0, 1, 2, ... p-2$.  Hence multiplication $\bmod p$ reduces to addition $\bmod p-1$ of the exponents; the transformation $x \to x^{q}$ where $gcd(q, p-1) = 1$ is obviously a bijection now.

[quote="besttate"]I see why that fact would allow it to be true. Why is your fact true?[/quote]

You should read what Sir Tim O'Brien says on the difference between truth and fact.  Fool.


Did anyone go to All-State (for music) or is anyone here planning to go to the CT math league meet on April 10th?

[quote="piac31"]Almost made eastern region for music (I was within 10 points. Should have bought the cd). I'm not nearly good enough for all-state and clarinet is fairly competitive.

I am going to the ct math meet on Tuesday. I'm on Glastonbury's team. Do you know if this feeds into a national competition?[/quote]
CSAML does not feed into anything. For most local leagues, as far as I know, the top two teams qualify for the New England competition, NEAML. Also, the qualifiers for ARML, who took the runoff test in early March, are announced at CSAML.

I'll be there, with Ignite168, on the Hopkins (New Haven) team.

Archived NEAML competitions can be found at [url]http://www.matharchive.org/[/url]

Way to help the enemy. We should not be looking to just win, we should be looking to crush the competition and get a 126 (we're a small school I think)

Helping the enemy with past problems is not the way to go.

Thanks for the problems, but I'm too busy anyway (english paper :(  )

When you say the top two teams from each qualify for NEAML, you mean the entire A teams right? Thats so cool because I think my school is number 2 in the capital area.

Don't worry about helping the enemy. We aren't that good. My school hasn't had a USAMO qualifier since 1999. No one practices and it shows.

alex, that's gonna change really quickly (if i don't make it freshman, i will sophomore)

OMG, STILL CAN'T BELIEVE U DID NOT MAKE USAMO

jorian

It's really simple how i didn't make USAMO. I did next to no practice and I made a ton of stupid mistakes, but I will make it next year.

I'd say clinching the win by at least 2 points even if we got a 0 on the team round and every other school got an 18 is unacceptable. Only clinching a tie after round 5 is too. We should've clinched the win by the end of round 4.

Yeah, the Small division isn't too great.

Alas, with All-State and States once again approaching, let us revive this thread.

For those in the orchestra, are Rach 2.3 and 2.4 (symphony.movement) any fun?

How inconvenient--All-state starts that same day.

A decision between Rachmaninoff and donuts, I suppose.


[quote="Interval"]For each question below, do the following:

*Find an equation of the line with the given properties.
*Express your answer using EITHER the general form OR the slope-intercept form of the equation of a line, whichever you prefer.

(1) x-intecept = -4; y-intercept = 4.

(2) Parallel to the line y = -3x; passing through point Q(-1, 2).[/quote]

(1) [hide]Figuring out the slope, it would be $\frac{4-0}{0-(-4)}=1$. Then putting it in y-intercept, you get $y=x+4$ (4 for the y-intercept.)[/hide]

(2) [hide]Since the slope of $y=-3x$ is -3, you follow the point Q to the y-intercept.
$(-1,2)+(1,-3)=(0,-1)$
Then putting it together, you get $y=-3x-1$[/hide]

[hide]

(1) x-intecept = -4; y-intercept = 4.

(2) Parallel to the line y = -3x; passing through point Q(-1, 2).

1.

We have two points (-4,0) and (0,4).

The slope is 1.

I'm going to pick the second point to use in the point-slope equation.

$y-y_{1}= m(x-x_{1})$

$y-4 = 1(x-0)$

$y = x+4$

2.

The slope must be -3. Our point is (-1,2).

$y-2 = (-3) \cdot (x-(-1)$

$y-2 =-3x-3$

$y =-3x-1$

[/hide]

[quote="Interval"]I don;t get when you said:

"...you follow the point Q to the y-intercept."

How do I follow point Q?[/quote]
I meant that you would follow the line that contains Q and has the same slope. Look at erik-the-red's math - you should probably use the point-slope equation: $(y-y_{1})=m(x-x_{1})$

Using the point-slope formula is much easier amd student-friendly.

Thanks.


Let m, n be positive integers. The equation
(x+m)(x+n)=x+m+n
has an integer root (doesn't have to be positive).
Show that
$\frac 1 2 \leq \frac m n \leq\frac 2 1$.

We have $-m-n<x<0$ integer root (x-1)(x+m+n)+mn=0,
$m=\frac{-(x-1)(x+n)}{x+n-1}$. Because (x+n,x+n-1)=1, $1+k=\frac{x-1}{x+n-1} \in Z$.
It give k is divisor n (1<=k<n) and m=(1+k)(n/k-1)=n-k+n/k-1. Therefore
$\frac 12 <\frac mn <2$.


[hide]if 57 deer are tagged per 3 days, 19 will be tagged everyday, or 133 every week.[/hide]

[quote="joejia"]On the first three days of a seven-day deer population study, 57 deer were tagged. At this rate, what is the number of deer that will be tagged during the entire study.[/quote]

[hide]$\frac{3}{7} = \frac{57}{x}$

$3x = 57(7)$

$57(7) = 399$

$3x = 399$

$x = 133$[/hide]

[quote="joejia"]On the first three days of a seven-day deer population study, 57 deer were tagged. At this rate, what is the number of deer that will be tagged during the entire study.[/quote]

[hide]
19 deer a day. 19*7=133[/hide]

[hide]
3/7 is 57/x
then if you solve the proportion
3x=57*7
3x=399
x=133[/hide]

[hide]
57/3=19
19 deer are tagged each day.
19*7=133 deer[/hide]


The fluorescent light tubes made by the company Well-lit have lifetimes which are normally distributed with mean 2010 hours and standard deviation 20 hours. The company decides to promote its sales of the tubes by guaranteeing a minimum life of the tubes, replacing free of charge any tubes that fail to meet this minimum life. If the company wishes to have to replace free only 3% of the tubes sold, find the guaranteed minimum it must set.

What I did was:
Let X be the minimum number of hours,
(X - 2010) / 20 = -1.8808
X = (-1.8808 * 20) + 2010 = 1972.384
So my answer is 1972 hours. However, the answer provided by the book is 1970 hours but I only can get 1972.384.. If anyone can get 1970, how did you arrive to that answer?

Thanks in advance!!  :)

It looks like it's just roundoff. The numbers 2010 and 20 probably aren't known to a sufficient degree of accuracy that it makes all that much difference.