How do you calculate fractional exponents? Like for example 10^1/10
Thanks.

Edit: Can't figure out how to put in the code so that it looks correct.

First of all, to use $ \text{\LaTeX}$ on exponents with more than one digit, you have to enclose the entire exponent in curly braces, for example:

[code]a^{1326598746}[/code] will get you $ a^{1326598746}$, while
[code]a^1326598746[/code] will get you $ a^1326598746$.

Now, fractional exponents can be rewritten as:

$ a^{\frac{1}{10}}=\sqrt[10]{a}$.

Basically, if you have an exponent in the form:

$ a^{\frac{n}{m}}=\sqrt[m]{a^n}$

[quote="happyme"]First of all, to use $ \text{\LaTeX}$ on exponents with more than one digit, you have to enclose the entire exponent in curly braces, for example:

[code]a^{1326598746}[/code] will get you $ a^{1326598746}$, while
[code]a^1326598746[/code] will get you $ a^1326598746$.

Now, fractional exponents can be rewritten as:

$ a^{\frac {1}{10}} = \sqrt [10]{a}$.

Basically, if you have an exponent in the form:

$ a^{\frac {n}{m}} = \sqrt [m]{a^n}$[/quote]
What does it mean when the number is on the left of a square root? And how do you do an exponent for a square root? Do you multiply it and then do the square root?

Thanks for the response.

$ \sqrt [n]{a}$ means the $ n$-th root of $ a$, and is the same as raising $ a$ to the $ \frac{1}{n}$ power.

Usually, $ a^{\frac {1}{n}}$ is used in mathematics instead of $ \sqrt [n]{a}$ because it's simpler to work with when you have many fractional exponents.

[quote="i_like_pie"]$ \sqrt[n]{a}$ means the $ n$-th root of $ a$, and is the same as raising $ a$ to the $ \frac{1}{3}$ power.[/quote]

That should read "the same as raising $ a$ to the $ \frac{1}{n}$ power".

Another reasons fractions are used instead of radicals is because sometimes multiplying fractions in the exponent simplifies the problem whereas multiplication under a radical is harder to see/do.

@ cf249 - Thanks. :blush:

@ leroyjenkens - The value is $ \approx1.2589$, but $ 10^{\frac{1}{10}}$ cannot be simplified.

[quote="leroyjenkens"]That makes sense. Thanks. But how do you calculate a number to a fractional exponent? Like for example, how do you calculate $ 10^{1/10}$?

Thanks for the responses.[/quote]

I'm pretty sure there has to be some manual way that I don't know of, but usually on your calculator (if its at least a scientific one) there should be a button with something like: $ \sqrt[x]{y}$ on it, and to calculate $ \sqrt[10]{10}$ you would just type in "10" then the "$ \sqrt[x]{y}$" button, and finally "$ 10$" again.

you could use Newton's method to find the root of x^10-10=0.
it's not great because it takes forever to converge if the first guess is off, but it's one way of finding the root manually...

[quote="leroyjenkens"][quote="i_like_pie"]@ cf249 - Thanks. :blush:

@ leroyjenkens - The value is $ \approx1.2589$, but $ 10^{\frac {1}{10}}$ cannot be simplified.[/quote]

Thanks, but I was wondering how you reach your answer of 1.2589.[/quote]

Plug it into  a calculator...

[quote="leroyjenkens"][quote="i_like_pie"]@ cf249 - Thanks. :blush:

@ leroyjenkens - The value is $ \approx1.2589$, but $ 10^{\frac {1}{10}}$ cannot be simplified.[/quote]

Thanks, but I was wondering how you reach your answer of 1.2589.[/quote]
You reach the answer by using $ x^{\frac{y}{z}=^z\sqrt{x^y}}$  X is 10, Y is 1, and Z is 10.  So:
$ ^{10}\sqrt{10^1}=^{10}\sqrt{10}\approx1.258925412.......$

[quote]Plug it into a calculator...[/quote]
How did people figure it out before they had calculators? What I'm asking is how you calculate it manually. Sorry if I wasn't clear.
[quote="ajain"][quote="leroyjenkens"][quote="i_like_pie"]@ cf249 - Thanks. :blush:

@ leroyjenkens - The value is $ \approx1.2589$, but $ 10^{\frac {1}{10}}$ cannot be simplified.[/quote]

Thanks, but I was wondering how you reach your answer of 1.2589.[/quote]
You reach the answer by using $ x^{\frac {y}{z} = ^z\sqrt {x^y}}$  X is 10, Y is 1, and Z is 10.  So:
$ ^{10}\sqrt {10^1} = ^{10}\sqrt {10}\approx1.258925412.......$[/quote]
So it would be 10 to the 1st power and then what do you do with the 10 on the left side?

example: $ 8^{2/3}$

do $ 8^2\equal{}64$
then do $ 64^{1/3}$, or the $ 3^{rd}$ root of 64
answer=4

To generalize: you raise the base to the numerator, then raise to the $ 1/x$, or take the $ x^{th}$ root, where x is the denominator.

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

That pretty much sums up everything here

[quote="jackdillon"]example: $ 8^{2/3}$

do $ 8^2 \equal{} 64$
then do $ 64^{1/3}$, or the $ 3^{rd}$ root of 64
answer=4

To generalize: you raise the base to the numerator, then raise to the $ 1/x$, or take the $ x^{th}$ root, where x is the denominator.[/quote]
Ok thanks. Is there an easy way to find the root of the number?

use a calculator. (that's about the only easy way to do it)

10^(1/10)=x (where x is the answer) can also be written as 10=x^10

There are plenty of manual ways to find roots, but are extremely tedious, depending on the power of the root, and the numbers involved.

Leroy: $ x^{\frac{m}{n}}\equal{}\sqrt[n]{x^m}\equal{}\sqrt[n]{x}^m$

You'll just have to memorize that, but here's an example:

$ 8^\frac{2}{3}$

$ \equal{}\sqrt[3]{8}^2$
$ \equal{}2^2$
$ \equal{}4$

I did not square 8 and then take the root, because 8 is a perfect cube. Note that if the number is not a perfect root, I would recommend exponentiating first.

This is the law of indices. I jz learned this part last weak =.=  Its when A^n*A^b= A^n+b. A^1/6= power root 6  A to the power of 1

shaun, everyone here already knows that. Not to mention that this thread is over a week old.

everyone does NOT know that. what is that indices or whatever thing shaun was talking about? i hav no clue. ( i dont think sixth graders learn about it, or at least i havent)

Well, you weren't involved in the thread at the time, so everyone on the thread did know that. And stop reviving old threads. Almost every one of your posts is an old thread.

they r not. the last post was less that a month ago.

[quote="jackdillon"]example: $ 8^{2/3}$
.[/quote]

You could also do this:

$ 8^{\frac{2}{3}} \equal{} 8^{\frac{1}{3}*2}$

The cube root of $ 8$ is $ 2$, and $ 2^2$ is 4.  :)

That's exactly what happens anyways...

You raise the base to the numerator(th) power, and take the denominator(th) root.

I must remind fishythefish not to revive old threads :P

Except for the part where it wasn't me... *looks at dreambold and isabella*

If you look at the dates, I never posted anything more than 1 day after the previous post. (I just added to this thread because I can see that this thread was going to continue whether we told people to stop posting or not.) :ninja:


Please describe the Gradient Vector of the Surface f(x,y,z)=0 for me.

Thanks in advance...


inequalities
rearrangement inequality
inequalities theorems

can we prove classic cauchy-schwarz inequality by using re-arrengement inequality? I couldnt find it on internet and i couldnt do it:S

Yes we can. You can use the following corollary of the rearrangement inequality:
Let $ a_1,\ldots ,a_n$ be real numbers and $ (a_1^\prime ,\ldots,a_n^\prime )$ be a permutation of $ (a_1,\ldots,a_n)$. Then $ \sum_{i \equal{} 1}^n a_i^2\geq \sum_{i \equal{} 1}^n a_ia_i^\prime$.

^^   i've found it after 5 hours i posted this topic:) anyway thanks for help puuhikki


number theory
least common multiple
divisibility tests
number theory unsolved

Prove that 1989|n^n^n^n-n^n^n for every odd n.

Can someone write down the solution, please? I need it.

Easy mod problem. You can check that for any odd $n$ that
\[n^{n^{n^{n}}}\equiv n^{n^{n}}\mod 17 \]
. Similarly, you get
\[n^{n^{n^{n}}}\equiv n^{n^{n}}\mod 13 \]
and
\[n^{n^{n^{n}}}\equiv n^{n^{n}}\mod 9 \]

ok but $1989$ doesnt divide it or am i missing something ?

the problem is equivalent to proving
$3 |$  $n^{n^{n}}-n^{n}$
and
$4 |$  $n^{n}-n$
for all odd $n$.

It's easy to show that. But can you explain me how problem is equivalent to that? You can also show that 24|n^n-n for all odd n.

for divisibility by $9,13,17$ its sufficient to show that
$48 | n^{n^{n}}-n^{n}$ for all odd $n$
because
$n^{n^{n^{n}}}-n^{n^{n}}= n^{n^{n}}[ n^{n^{n^{n}}-n^{n}}-1]$
and $\phi (13) = 12 , \phi (17) = 16, \phi(9) = 6$ $(LCM = 48)$
also, if $n$ is not coprime to $3,13,17$ then the expression is divisible by $9,13,17$ respectively.
for proving $48 | n^{n^{n}}-n^{n}$ its sufficient to prove
$3 | n^{n^{n}}-n^{n}$
and
$4 | n^{n}-n$
because $n^{n^{n}}-n^{n}=n^{n}[ n^{n^{n}-n}-1$
and $n^{4}\equiv 1(mod 16)$ for all odd $n$.


Michael has never taken a foreign language class, but is doing a story on them for the school newspaper.  The school offers French and Spanish.  Michael has a list of all 25 kids in the school enrolled in at least one foreign language class. He also knows that 18 kids are in the French class and 21 kids are in the Spanish class.  If Michael chooses two kids at random off his list and interviews them, what is the probability that he will be able to write something about both the French and Spanish classes after he is finished with the interviews?

There are $ 18\plus{}21\minus{}25\equal{}14$ students that take both French and Spanish. Michael can have a student who only takes Spanish and a student who only takes French, of which there are $ (18\minus{}14)(21\minus{}14)\equal{}28$ ways, or a French-and-Spanish student among his two, which there are $ \binom{25}2\minus{}\binom{25\minus{}14}2\equal{}245$ ways for, and the probability is $ \frac{273}{\binom{25}2}\equal{}\boxed{\frac{91}{100}}$.

[hide="solution"]When we put this into a venn diagram, we have 7 in the spanish, 4 in the french, and 14 in both. The only way Michael will not be able to write down both something about french and spanish is if both of the interviewed people are in french or both or in spanish. The probability that both of the interviewed students are in French is $\frac{4}{25}*\frac{1}{8}=\frac{1}{50}$, and the probability that both are in spanish is $\frac{7}{25}*\frac{1}{4}=\frac{7}{100}$, so the complementary probability is $\frac{9}{100}$, so the probability that we want is $\boxed{\frac{91}{100}}$.[/hide]

Another way to calculate this is to do:
(7/25)*(6/24)+(4/25)*(3/24)
which means: the probability that the first kid only enrolled French*the probability the second kid only enrolled French + probability that the first kid only enrolled Spanish* probability the second kid only enrolled Spanish.


geometry
parallelogram
geometric transformation
reflection
circumcircle
geometry proposed

Two circles with centers $K , L$  intersect at points $A , B$ . The tangents to the circles at $A$ meet them again at $C$  and $D$ .If $O$  is the center of the circumscribed circle  of triangle $ACD$ , prove that $OB$ is perpendicular to $AB$.

[u]Babis[/u]

Perform an inversion with positive power and pole $A$. If in general, for a point $X$, we denote its image by $X'$, then $AC'B'D'$ will be a parallelogram, and $O'$ will be the reflection of $A$ in the diagonal $CD$ of this parallelogram. The assertion we want to prove is equivalent to $\angle AO'B'=\frac\pi 2$, which is obvious from the observations above.

[quote]and O' will be the reflection of A in the diagonal CD of this parallelogram.[/quote] Why ??

with simples angle chasing we find that $\angle{CBD} = 2\angle{CAD} = \angle{COD}$.
so, $BODC$ is ciclyc. this implies that $\angle{OBA} = \angle{ABD} - \angle{OBD} = \angle{ABD} - \angle{OCD}$
$= 180 - (\angle{BAD}+\angle{BDA}) - (90 - \angle{CAD}) = 90$  :)

This ([b]Your solution )[/b] is the simplest elementary solution for this problem. My solution - based on the original problem and the official solution - is longer.

 << If  $F , G$ are the antidiametrical points of $A$ to $K , L$, then $F , B , G$are on the same line and the perpendiculars from $K , L$to $AC , AD$ pass through the midpoint $M$ of $FG$  , since$KM // AG$( as perp. to $AC$...) and $LM // AF$ . But then $O$ and $M$ are the same point .... >>.

 [u]Babis[/u]


inequalities
geometry
inequalities unsolved

Well-known, so certainly posted before. (btw, this should be an $ac$ in the RHS and that's a $\geq$)
There's a quick solution if you interprete it geometrically, it simply reduces to the triangle inequality.

The inequality is not true: $a=5, b=1,c=1/3$.

There was some misteake in the last square root. It must be a^2 + ac + c^2.

And this ineq really can be proved by using geometry: Let take OA = a, OB = b, OC = c and angle AOB = angle BOC = 60 degree!

The inequality holds <=> 1/a + 1/c = 1/b.


you just use a list

So first

With 3 4-pieces

so 3 4s, 1 2, 1 1

3 4s, 3 1s

Then with 2 4 pieces

2 4s 3 2s 1 1.... etc.

[quote="reve"]A customer ordered 15 pieces of gourmet chocolate. The order can be packaged in small boxes that contain 1, 2 or 4 pieces of chocolate. Any box that is used must be full. How many different combinations of boxes can be used for the customer's 15 chocolate pieces? One such combination to be included is to use seven 2-piece boxes and one 1-piece box.[/quote]

You make a chart basically.

[code]
4 piece    2 piece
3            1
              0
2            3
              2
              1
              0
1            5
              4
              3
              2
              1
              0
0            7
              6
              5
              4
              3
              2
              1
              0
[/code]

Which gives us a total of 20 different ways.


Because my computer has broken, please help me draw it on the computer:

The function is y=(x^3+4x+6)/(x^2-1)


I'm looking for proof of medians in the traingle.
Proof of than they divide each oter 2/3 and 1/3

getting started is not really for proofs, but doing the proof with points on a cartesian system is pretty straight forward, and a proof with mass points or vectors is pretty easy too

[hide]If you let A=(x1,y1), B=(x2,y2), C=(x3,y3), then ((x1+x2+x3)/3,(y1+y2+y3)/3) will be on all three medians. That point is 2/3 of the way from (x1,y1) to ((x2+x3)/2,(y2+y3)/2). Similar with the other two medians.[/hide]

Source :  Coxeter's [i]Introduction to Geometry[/i]

In triangle $ABC$, take the medians from $A$ and $B$.  Let us say they intersect their respective sides at $A'$ and $B'$, and that they intersect each other at $O$.  In triangle $AOB$, take the line connecting the bisection points of sides $AO$ and $BO$, and call it $KL$, with $K$ on $AO$ and $L$ on $BO$.  Do the same for triangle $ACB$ and the sides $AC$ and $BC$ (that line is already called $B'A'$).  Clearly, $KL$ and $B'A'$ are parallel and of the same length (half that of $AC$), so $KLA'B'$ is a parallelogram.  The diagonals of a parallelogram bisect each other, so $A'O\equiv OK$, and by definition, $OK\equiv AK$.  Thus we have proven that any two medians (and thus all of them) intersect at their trisection points, which nicely answers the question of concurrence as well.
(It is much more clear with a diagram.)

This is by far the most elegant proof I have seen, and I have been surprised not to have seen it anywhere else.  (Then again, I am not particularly well-read.)


calculus
derivative
function
calculus computations

Suppose that $f(x)-f(y)\le(x-y)^{2},\ \forall x,y\in\mathbb R$. Prove that $f$ is constant.

[quote="Kent Merryfield"]What is the derivative of $f?$[/quote]
The function $f$ is constant $\Rightarrow\frac{df}{dx}\equiv0$.

But look at the title. :roll:
For any $a,b\in\mathbb R$ and any integer $n\ge 1$ we have
$|f(a)-f(b)|\le \sum_{i=1}^{n}\left|f\left(a+i\frac{b-a}{n}\right)-f\left(a+(i-1)\frac{b-a}{n}\right)\right|\le \sum_{i=1}^{n}\frac{(b-a)^{2}}{n^{2}}=\frac{(b-a)^{2}}{n}$


Two distinct numbers are selected simultaneously and at
random from the set $ \{1, 2, 3, 4, 5\}$. What is the probability that
the smaller one divides the larger one? Express your answer as a
common fraction.

I suppose we can list out:

Smallest number 1: 4 cases, all work

Smallest number 2: 3 cases, just 4 works

Smallest number 3: 2 cases, none work

Smallest number 4: 1 case, none work.

So we have $ \frac{4\plus{}1}{1\plus{}2\plus{}3\plus{}4}\equal{}\boxed{\frac{1}{2}}$.

when they say "smaller one divides the larger one", do they mean that the number dividing into the larger one can leave remainders? Initially, when I did the problem I thought it meant it could leave remainders...

Generally, if $a$ divides into $b$, then $a$ is a factor of $b$.

[hide=sol]There are 5c2 ways to pick 2 integers from the set. Four pairs of these are 1 and another integer which automatically qualifies and the other is 2 and 4, giving us 5 in total. Thus 5/10=1/2.[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $R$ be a commutative ring with identity $1\neq 0$. Let $P$ be a projective $R$-module, $r\in R$ and $C: =P\diagup rP$. Prove that
$\mbox{Ext}_{R}(C,G)\cong \mbox{Hom}_{R}(C,G\diagup rG)$
for every $R$-module $G$.

[quote="owk"]Why is $rP$ a submodule of $P$? owk[/quote]
You are right... $R$ need to be commutative. I edit my post.

I found a mistake in my proof. So, i'm not sure that the proposition is true.
It's surely true the following:
Let $C$ be an abelian group such that $pC=0$ for a prime number $p\in \mathbb{Z}$. Prove that
$\mbox{Ext}\left( C,G\right) \cong \mbox{Hom}\left( C,G\diagup pG\right)$
for any abelian group $G$.

I define an homomorphism $\eta : \mbox{Ext}\left( C,G\right) \to \mbox{Hom}\left( C,G\diagup pG\right)$, but i can't prove that $\eta$ is an isomorphism.


trigonometry
analytic geometry
function
trig identities
Law of Cosines
geometry
Law of Sines

Convert $ \frac{7\pi}{12}$ radians to degrees and fine the reference angle in degrees.

Convert 210 degrees to radians and find the reference angle in radians.

[quote="math92"]Convert $ \frac {7\pi}{12}$ radians to degrees and fine the reference angle in degrees.

Convert 210 degrees to radians and find the reference angle in radians.[/quote]

I can't answer the first one cuz I forgot trig (literally, forgot :().  But $ \frac{7\pi}{12}$ radians would be equal to $ \frac{7 \cdot 180}{12}$.  So the angle in degrees is $ 105$.  The reference angle would be $ 180 \minus{} x$, so it is $ 65$.

$ 210^o \equal{} \frac{210\pi}{180}^r \equal{} \frac{7\pi}{6}^r$

I accidently replied to my thread rather than creating a new topic. Sorry!  :blush:

[quote="Poincare"][quote="math92"]Given that (-5,-12) is on the terminal side of an angle $ \theta$ in Standard Position, find the six trig values of $ \theta$.[/quote]
geogebra won't work so use the distance formula and make a triangle then you can use Law of Sines/Cosines i am not very sure though.[/quote]

I think you could use the Law of Cosines to find the angles.

You cannot use the Law of Sines because you have no angles.

EDIT: Nvm, you have the $ 90^\text{o}$ angle.

you can use whatever you like, i haven't had MUCH experience with trig so you should try both and see if you get the same answer :)

Actually, this is what you should do:

Draw the triangle. The legs should be -5 and -12 because of the coordinates. (Yes, I know that distances are only positive, but doing this is helpful in trig to keep track of the signs.) The hypotenuse is then 13. $ \theta$ is opposite -12.

$ \sin \theta\equal{}\frac{\minus{}12}{13}\equal{}\minus{}\frac{12}{13}$

$ \cos \theta\equal{}\frac{\minus{}5}{13}\equal{}\minus{}\frac{5}{13}$

$ \tan \theta\equal{}\frac{\minus{}12}{\minus{}5}\equal{}\frac{12}{5}$

$ \csc \theta\equal{}\frac{13}{\minus{}12}\equal{}\minus{}\frac{13}{12}$

$ \sec \theta\equal{}\frac{13}{\minus{}5}\equal{}\minus{}\frac{13}{5}$

$ \cot \theta\equal{}\frac{\minus{}5}{\minus{}12}\equal{}\frac{5}{12}$

 [geogebra]0bea224c72a76c5eecc92386f883309d6415ddd7[/geogebra]

What is the reference of an angle and what is Standard Position (referring to the questions)?

This may help.

http://www.mathwords.com/r/reference_angle.htm

Standard Position means that one side (the initial side) of the angle is the positive x axis, and the vertex is at the origin. It's the simplest angle placement. (From the origin extending to the direct right is one side, and the other can be any ray coming out of the origin).

but fishy doesn't the question say to find the angle?

I was referring to the first question, which asks you to find the 6 trig values. (sine, cosine, tangent, cosecant, secant, cotangent)

The second questions ask you to find the reference angle. (Which, succinctly, is the smallest angle between the terminal side of an angle in Standard Position, and the x-axis.)

It seems that ernie did not fully answer the second question. The reference angle of 210 degrees is 30 degrees, which converts to $ \pi/6$ radians.

Remember that:

$ (i)$ if $ P(x,y)$ is on the terminal side of $ \theta$ in standard position, then the coordinates of $ P$ in general is $ (r\cos \theta, r\sin \theta).$
$ (ii)$ There must be some circle of radius $ r \equal{} \sqrt{x^2 \plus{} y^2}$ that contains the point $ P$

So, the six trigonometric functions are still intact when we are dealing with different circles, but the angle/s must be in standard position. fishythefish's answer is correct.


Each edge of a regular octahedron is colored orange or black.  If every face of the octahedron has at least one orange edge, what is the smallest possible number of orange edges?

I got $ \fbox{7}$, did anybody get anything lower?

I get $ \boxed4$. There are $ 8$ faces, and each orange edge is part of at most $ 2$ unique faces, so we need $ \frac82$ edges at least. It isn't too difficult to show that this is achievable.

[quote="GameBot"]Each edge of a regular [b]octahedron[/b] is colored orange or black.  If every face of the octahedron has at least one orange edge, what is the smallest possible number of orange edges?[/quote]

Where I come from, octa means eight...

Oh whoops! My bad! I was thinking of an octagonal pyramid.  :rotfl:

Doesn't an octagonal pyramid have 9 faces?

Yeah, but let's just ignore that now... :P

Arggggggggg...my mind is SO phailing today...octagonal PRISM. I need my brain checked.


geometry
rectangle
calculus
ratio
LaTeX
similar triangles

[hide="Answer."]Your main concern isn't the hypotenuse. So if you graph it in an x-y plane, make the y-intercept $5$ and the x-intercept $12$. The origin will be the vertex with the right angle. The equation of the line is $y=-\frac{5}{12}x+5$. Now you didn't specify which square so I will find the square that shares the right angle. The opposite angle should have $x=y$ so plug it in and you get $(\frac{60}{17},\frac{60}{17})$. So the length of the side is $\frac{60}{17}$. If you want to make it more interesting, make it so that you have to find the minimum. :lol: [/hide]

[quote="NeverOddOrEven"]hmm i got a different answer.

[hide="My sorution"]
So say the square had a length of $x$. When you inscribe the square into the triangle, you form two other similar triangles. So you can get
\[\frac{5-x}{x}=\frac{x}{12-x}\]
So solving for $x$ yields
\[x^{2}=60-17x+x^{2}\Rightarrow x=\frac{60}{17}\]
[/hide][/quote]

Thats what I got.

Oops, I put in the wrong slope. I will edit that shortly. :blush:

[quote="NeverOddOrEven"][quote="cincodemayo5590"]Yup.

[b][i]Follow-up Question:[/i][/b]

Find and prove the maximum area that a rectangle, inscribed in a $5-12-13$ triangle, can have.[/quote]

[hide="My sorution"]
Let the dimensions of the rectangle be $x$ and $y$. When the rectangle is inscribed in the triangle, two similar triangles are formed that give the following proportion:
\[\frac{12-x}{y}=\frac{x}{5-y}\]
So we can find $x=-\frac{12}{5}y+12$
So we have
$\text{Area}=xy$
$A(y)=-\frac{12}{5}y^{2}+12y$
$A'(y)=-\frac{24}{5}y+12$
$A'(y)$ is never undefined and $A'(y)=0 \Rightarrow y=2.5$
So we find $y=2.5$ to be a local maximum for $A(y)$. Therefore, the maximum area a rectangle can have inside a $5-12-13$ triangle  is
\[A(y)=-\frac{12}{5}y^{2}+12y=-\frac{12}{5}(2.5)^{2}+12(2.5)=\boxed{27.5}\]
[/hide][/quote]

Let's try not to use calculus on this forum  :wink:.
[hide] Instead of using calculus, you can use the algebraic formula $\frac{-b}{2a}$ for the maximum (which can be proved using algebra), and you get the same thing. Also, check your arithmetic - you should get $\boxed{15}$ at the end... [/hide]

[quote="now a ranger"][color=darkred]Find the length of the side of the square $MNPR$inscribed in right $A$- triangle $ABC$ if $\{M,N\}\subset (BC)$, $P\in (CA)$ , $R\in (AB)$ and $AB=c=5$ , $AC=b=12\ .$[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote the point $D\in BC\ ,\ AD\perp BC\ .$

Therefore, $BC=a=13$ and $AD=h_{a}=\frac{bc}{a}=\frac{60}{13}\ .$ Thus,

$\left\{\begin{array}{c}\frac{PR}{BC}=\frac{AR}{AB}\\\\ \frac{RM}{AD}=\frac{BR}{BA}\end{array}\right\|\Longrightarrow\left\{\begin{array}{c}\frac{PR}{BC}+\frac{RM}{AD}=1\\\\ PR=RM=MN=NP=l\end{array}\right\|\Longrightarrow\boxed{\ \frac{1}{l}=\frac{1}{a}+\frac{1}{h_{a}}}$

In conclusion, $l=\frac{780}{229}\ .$

[b][u]Generally[/u][/b], the area $PR\cdot RM$ of the rectangle $MNPR$ is maximum if and only if $\frac{PR}{BC}\cdot\frac{RM}{AD}$ is maximum. But the sum $\frac{PR}{BC}+\frac{RM}{AD}=1\ ,$ i.e. constant.
Therefore, $\frac{PR}{BC}=\frac{RM}{AD}=\frac{1}{2}$ a.s.o.[/color]

[quote="now a ranger"][quote="NeverOddOrEven"]hmm i got a different answer.

[hide="My sorution"]
So say the square had a length of $x$. When you inscribe the square into the triangle, you form two other similar triangles. So you can get
\[\frac{5-x}{x}=\frac{x}{12-x}\]
So solving for $x$ yields
\[x^{2}=60-17x+x^{2}\Rightarrow x=\frac{60}{17}\]
[/hide][/quote]

There isn't a choice on the test that this question came from  (Grissom 2006). The choices are $\frac{3}{2}$,$\frac{43}{13}$, $\frac{85}{27}$, $\frac{253}{64}$, and $\frac{780}{229}$.

hm...

for the question that was first asked. finding the length of the square, not the maximum or anything[/quote]
[hide="I think I got it."]They were asking for one side of the square to be collinear to the hypotenuse and for the other two points to touch the legs, so use the properties of similar triangles. Let's denote the length of the triangle to be $s$. The smaller triangle that is formed by the square and shares the original right angle the original big triangle had, a picture would help, has hypotenuse $s$. The sides are in ratios, so let's find the ratio of $s$ to the little part of the leg cut out of the leg of length $12$. The ratio of these are $13: 12$(that's 13:12 in case this doesn't work in latex). Plug it in, and the length of the little part, with respect to $s$, is $\frac{12s}{13}$. But there's another triangle cut out. The legs of this triangle use a side of the square and a part of the hypotenuse as legs and the bigger part of the cut out $12$ side as the hypotenuse. Once again, the ratios should be the same, so the length of this hypotenuse is, with respect to $s$, is $\frac{5s}{13}$. Wait! If you add these together, they should make $12$! Let's simplify the sum first, then set it equal to twelve. The sum is $\frac{229s}{65}$. Set it equal to $12$ and you get $s=\boxed{\frac{780}{229}}$. :D [/hide]

If H and D are on AD and AC are respectively. H and are vertices of square, than
triangle AHD is similar to triangle ABC
so
$5x=60-12x$
$x=\frac{60}{17}$

Well I think BadBoy's answer was right since it was one of the choices, and he provided explanation.

 :lol:

[i]Mod note: please don't use LaTeX for plain text.[/i]

how do you know that hte maximum does not occur when two of the verticies of the square are on the hypotenuse?

[quote="now a ranger"]Find the length of the side of the square inscibed in right triangle ABC if AB=5 and BC=12.[/quote]

Is <B a right angle, cause from the information you can't tell?
Also, there are two ways the square can be inscribed, one with the side on a leg, and one on a hypotenuse.

[hide]If it is on a leg, then

$\frac{5-s}{s}=\frac{s}{12-s}$, by similar triangles.

So $60-17s=0$, so $s=\frac{60}{17}$

But if it is on the hypotenuse, then
, we get that

$s+\frac{5s}{12}+\frac{12s}{5}=13$, by ratios

So solving for s we get $s=\frac{780}{289}$.[/hide]

[quote="mathgeniuse^ln(x)"][quote="now a ranger"]Find the length of the side of the square inscibed in right triangle ABC if AB=5 and BC=12.[/quote]

Is <B a right angle, cause from the information you can't tell?[/quote]

Well, the problem seems to me to be looking at a $5-12-13$ triangle, although technically, we could instead be looking at a $5-\sqrt{119}-12$ triangle.  Certainly the answer would be a little messier.

the square is sitting on the hypotenuse.

Couldn't right triangle ratios be used here?

[quote="mathgeniuse^ln(x)"]Yep so the answer would be $\frac{780}{289}$.[/quote]
You mean $\frac{780}{229}$

so I can conclude that you just use similar triangles to see what the lengths are.


for ellipse the only given information is the foci (-2,0) and (2,0) and the vertices (-6,0) and (6,0)

for the hyperbola the only given information is the foci (-8,0) and (8,0) and the vertices (-6,0) and (6,0)

find the standard equation for both

help and EXPLANATION pleeease?

Recall that the standard equation of an ellipse is $ \frac{(x\minus{}h)^2}{a^2}\plus{}\frac{(y\minus{}k)^2}{b^2}\equal{}1$, where $ a$ and $ b$ are axes (major or minor depending on the ellipse's orientation) and $ (h,k)$ is the center.

Using your information, you can do a couple of things:
1. Find the center
2. Find the orientation
3. Find the major axis

You then have enough information to solve for the minor axis and then you're done.

Similar thing for the hyperbola.


[quote="Interval"]What exactly is Non-Euclidean geometry?[/quote]
Please try googling it or go to http://www.mathworld.com to find more about this.

Euclidean geometry uses 5 basic concepts as given truths.  All of these can be proved, except for one:

Given a line and a point not on that line, there is exactly one line that runs through the given point that is parallel to the given line.

This seems simple, but cannot be proven.  It leads to all sorts of interesting questions, such as:

What if we were on a sphere, like the Earth?  The sum of the degrees of a triangle would NEVER add up to 180 degrees.  Now things don't work out so well.

Different types of non-Euclidean geometries on different surfaces (i.e. hyperbolic geometry) work out differently.  Most types on non-Euclidean geometry are well beyond high school level.

These non-Euclidean geometries were all developed by mathematicians unsuccessfully trying to prove Euclid's 5th Postulate.

If you are interested in learning more, and are really good at geometry, I would recommend Kedlaya's free book Geometry Unbound, available as a PDF download somewhere on the internet.  I am currently struggling through this, and learning a lot in the process.

[quote="mysmartmouth"]Euclidean geometry uses 5 basic concepts as given truths.  [b]All of these can be proved[/b], except for one:

Given a line and a point not on that line, there is exactly one line that runs through the given point that is parallel to the given line.

This seems simple, but cannot be proven.
[/quote]First of all, you don't prove axioms. They are like 'the rules' of the game in any geometry world. You will have to either accept them or convince yourself they are true. Second, by altering the axioms generate other geometries, and hence the name non-Euclidean. Euclid set forth 5 postulates as the basic structure of his geometry. Nowadays, when we study in our school, that geometry has been revised plenty of times, most notably by Hilbert and others.

I thank you for the reply.

In terms of mathworld.com, the site is not easy for people with little math background.  The author of the site is NO DIFFERENT than any college math textbook.


I have followed the usual steps to set up TeXnic Center to compile my Asymptote file. But in the last step, after the press New Shotrtcut Key block and Ctrl+A, the Assign button does'nt light and Crtl+A is good only for selecting the code. Any suggestion? Thanks

When I tried to assign Ctrl+A as the shortcut, it said it was already assigned.  Perhaps this is the problem you are having?  If so, you should make something else the shortcut, such as, for example, Alt+A

Nené: if roadnottaken's suggestion works could you let us know here then the instructions could be changed.

I have followed the roadnottaken's suggestion that works, but also now appears the page, C:\Program Files\Asympotote\asy.exe, in black without the design. The code is:
draw((0,0)--(100,100));
draw((0,100)--(100,0));
dot((50,50));
label("$P$",(50,50),S);

That's the DOS window which should close after asymptote.exe has finished. Is there a pdf created - if your file was called file.asy it should have created file.pdf. Has that happened?

What did you change (this may be helpful to others)?

I have made three changes using your suggestions:
1) From Crtl+A to Alt+A in the New Shotrcut Key block.
2) In the argument line (Tools menu-Customize-Tools tab in the Tecnic Center) change latex to pdflatex if you want result in pdf file otherwise you get ps file.
3) clear the dollar signs near the letter P in the code.


This has probably been posted before, but I can't find it. Where can I get all of the past AMC, AIME, and USAMO Problems. Solutions are preferred, although I can always use the forum.  :D

I'm preparing to do some rigorous studying of these problems, since I'm almost done with AoPS Volume 2.

Search up Kalva on Google and go to the first result.

Would anybody happen to know an archive for past AMC problems?

I don't think that an archive with a substantial portion of previous AMC's and AHSME's exists.  But [url=http://www.math.uncc.edu/~hbreiter/]this site[/url] has a few.

Don't forget this site

http://goodlymaths.0catch.com/contests/amc/amc_archive.htm

or this one

http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm

Don't forget to check out the AMC forum, where the previous AIME's and AMC's made by AOPSers are archived.

You can always order AMC tests.

As far as AIME go, Kalva is pretty great except that their solutions aren't that great. For USAMO, the best way is build up your proof skill and solve couple problems with full solution. Post in AMC forum and see what score you can get.

The score you will get is not the exact but it will help you a lot.

P.S. In future, please ask this type of question in AMC forum.


How did everybody do on the AMC B exams?

I heard David Harris got a 143.  That doesn't suck.

Any other ludicrously high scores in Alabama?

So far we've heard of no perfects, though AAST had a few in the 140s.  We haven't talked to a lot of the MOPpers however.

From what I know of the Sophomore class:

Evan got a 146.5 on the 10B.
Ryan, Zhenya, and Kevin got 144 on the 10B (I believe).
I got a 122 on the 12B (I really thought this one was [i]much[/i] harder than the 12A).

From what it seemed like the 12B was really easy to qualify for AIME (the first 11 questions were cake), but it had far more challenging questions after that; our third highest 12B score was a 124 (out of the entire school -- Sophomore class my 122 was the highest on 12B).

For AMC10B,
Kevin got 146.5
Evan, John, and I got 144.

Evan and I both made ridiculous careless mistakes.
I had a = 2b, a/b = 1 in my work.
Evan counted 2 through 9 as 9 instead of 8.

Smack whoever got the 146.5.

When you think you've solved all but one, you might as well guess at the last one, right?  Isn't a 1 in 5 shot at a 150 cooler than the flat 146.5??

Heheh. Kevin.  :roll: I think he's convinced that he can't do probability (#21).

I think that was clearly the hardest problem on the AMC 10's this year.  A lot of our students who got high 130's to mid 140's skipped or missed that problem.


Find the number of different five-digits numbers obtain using 0,0,0,1,1,1,2,2,2

Counting problem .... best way to get wrong without really understanding why  ;)

My guess : $140$

I place the $9$ numbers this way :  xxxxXxxxx
with X = $1$ or $2$

this way I obtain $4 \times$ the wanted number (left + right for $1$, left + right for $2$).

So it is $\frac{1}{4} \cdot \frac{8!}{3! \cdot 2! \cdot 2!} = 140$

I won't bet anything on it :P

i got a totally different answer
[hide]
well the first digit can not be 0 so there are 6 choices.  the rest of the digits can be whatever so there are a total of 8! ways to arrange them.
however, since there are repeats we myst divide by the number of repeats.
so my answer is $\frac{(6)8!}{3!3!3!}=1120$[/hide]

I think $1120$ is too much :

Numbers are between $1000$ and $1222$ or between $2000$ and $2221$ which makes at most $445$ numbers.

>> i got a totally different answer

By pigeon hole one (or more probably both) solution is wrong  ;)

o woops didnt read that we were only using 5 digits.  :oops:

First, let us overcount and assume that there are available to us an unlimited number of $0, 1, 2$s.  The answer is then $2 \times 3 \times 3 \times 3 \times 3 = 162$.

Next, let us subtract the numbers where we have used more than three of any of the numbers.  For more than three $0$s, we have $10000, 20000$.  For four $1$s, the remaining digit is $0$ or $2$ and it can be in one of five places, so there are $10 - 1$ (because $0$ cannot be the first digit).  There is only one number that uses five $1$s.  Repeating similar logic gives us $10 - 1 + 1$ numbers that use four or more $2$s, so our final answer is $162 - 2 - 10 - 10 = \boxed{140}$.


function
inequalities unsolved
inequalities

Let n be a positive integer and the function: $f(x)=1+cosx+\frac{1}{2}cos2x+\frac{1}{3}cos3x+...+\frac{1}{n}cosnx$
Solve that $f(x) \geq 0$ $\forall x \in R$

[quote="frejer"]Let $n$ be a positive integer, prove that: \[f(x)=1+\cos x+\frac{1}{2}\cos 2x+\frac{1}{3}\cos 3x+\cdots+\frac{1}{n}\cos nx \ge 0 \qquad \forall x \in \mathbb R.\] [/quote]
It's Feier's ineq. Also posted but no solution.


r: radius of the small semicircle
R: radius of the big semicircle
Find X
[img]http://img46.imageshack.us/img46/7421/prob1jp5.png[/img]

Do you want $ x$ in terms of $ R$ and $ r$, or do you want the exact value?

You shoud post full problem. If you give me this picture, I don't understand?

[hide="Quick observation"]
Probably not right
$ x \propto x\plus{}8$.
[/hide]

[hide="A start"]Drawing $ JB$, we see $ \triangle AFO\sim \triangle ABJ$. Using this fact, we know $ \frac {x}{2R} = \frac {R}{x + 8}$. Solving for $ x$,
\begin{align*}x^2 + 8x & = 2R^2 \\
 \\
x^2 + 8x - 2R^2 & = 0 \\
 \\
x & = \frac { - 8\pm\sqrt {64 + 8R^2}}{2} \\
 \\
x & = \frac { - 8\pm2\sqrt {16 + 2R^2}}{2} \\
 \\
x & = - 4\pm\sqrt {16 + 2R^2}\end{align*}
$ x$ cannot be negative, so $ x = - 4 + \sqrt {16 + 2R^2}$.

Also, $ x = \sqrt {2Rr}$ by other similar triangle relations (I'll post the proof later if anybody wants it.).[/hide]


Prove that the equation $x^{2}=y^{2}+z^{2}$ does not have solutions $(x,y,z)$ such $x$,$y$,$z$ are prime numbers.

It's quite obvious.. but I can't make a good solution... help?

[hide]If $x,y,z$ are prime such that $x^{2}=y^{2}+z^{2}$
+)$y>2$ and $z>2$ then $x$ is even prime and great than $2$ , contraction!

+)$y=2$ we have $(x-z)(x+z)=4$ therefore $x+z$ is even and $\leq 4$, contraction![/hide]


You select any 2 and do simultaneous on them. This way you find the vertex of the triangle. If I haven't made any errors, I think the answer is:

[hide="Answer"] $x^{2}+y^{2}=13$[/hide]

[quote="ckck"]The vertices of a triangle of the lines $y=-x-1, x=2$, and $y=\frac{1}{5}x+\frac{13}{5}$. Find an equation of the circle passing through all three vertices.
How do you do this problem?[/quote]
solve the 3 sets of systems of 2 equations (they are the vertices), then locate the center by making 3 expressions out of distance formula for distance between x and the x value of one vertex, distance between y and y value of that same vertex, and do that for all 3 vertices. then set expressions equal to each other 2 at a time, and solve for the center. then solve for the radius with the distance formula. there u have ur circle.

however, my way is probably the longest possible way...

The easiest way to do this would be to find the circumcenter, and then go from there to one of the points and then figure out the equation.


trigonometry
integration
logarithms
real analysis
real analysis unsolved

[quote="spix"]Solve $f(x)=\sin F(x)$ where $F\in\int f(x)dx$ and $f: \Re\to\Re,F: \Re\to\Re$[/quote]That's the ordinary differential equation
\[F'(x) = \sin F(x)\]
whose general solution may be found by separation of variables, using the fact that $\int \frac{dx}{\sin x}= \log \tan \frac{x}{2}+C$.


Let $ m \in \mathbb{N}$ such that $ 4 | (m \minus{} 1)$ and $ m \equal{} s^2 \plus{} 4t^2$ (m not being a square) for some $ s,t \in \mathbb{N}$. Assume that $ x^2 \minus{} my^2 \equal{} \minus{} 1$ has integer solutions $ (x \equal{} x_1,y \equal{} y_1)$. Show that there exist natural numbers $ p,q$ such that $ m \equal{} pq$[b](corrected)[/b] and $ px^2 \minus{} qy^2 \equal{} s$ has integer solutions.

$ pq\equal{}n$ (wath $ n$?)
you want say that $ pq\equal{}m$ or $ pq\equal{}t$ or what?


in the derivation of wavelength=h/p where p is momentum

why the scientists have equalled

e=hn(n ius frequency)
and
e=mc^2

for deriving the above one
can anyone explain clearly

Check out [url]http://stedjee1.infinology.net/De_Broglie/Quantum%20Mechanics.htm[/url].  This site seems to present a very good derivation, although some of the details are absent.  Keep in mind that the E = mc^2 comes from relativity: it relates mass to its energy equivalent.  Also, E = h$\nu$ is the energy of a photon (by definition?).  Hope that helps.


Two dogs, each traveling 10 ftsec, run toward each other from 500 feet apart. As they run, a flea flies from the nose of one dog to the nose of the other at 25 ft/sec. The flea flies between the dogs in this manner until it is crushed when the dogs collide. How far did the flea fly?

Isn't this in AoPS v1? :lol:
[hide]d=rt
r=25
d=?

then we see how long it takes the dogs 500/20
so we multiply this by 25
*gets calculator out*
=[b]625[/b]
[/hide]

oh shoot.  :wallbash_red:

I thought it was way more complicated than that. Oops. Yeah its in AoPS Vol. 1

I thought since it had the sign next to it that said you might need help signaled it wasn't simple.. eh

I know I was stumped when I saw that signal for the first time. It actually makes the problem harder to be told it is harder...eh? :(


algebra
polynomial
algorithm
advanced fields
advanced fields unsolved

[quote="vinny919"]I didn't know where else to post this, so I put it here...
I am in high school, but I am doing a comp sci project on a currently open problem (minimizing total weighted tardiness on a single machine with release dates and equal processing times). It is in Industrial Engineering, specifically: scheduling. However, I was wondering if anyone could define a few words for me in realllly simple terms (I read the things on mathworld and on other parts of the internet but I couldn't understnad them):

NP-
NP-Easy-
NP-Hard-
NP-Complete
P-
P vs NP-

Thats bout it...If anyone could help me out that would be great...thanks![/quote]

I am not an expert in the topic, not even close, but let's see what can we do for you.

[b]P-[/b] This is the collection of "Problems" that can be solved in polynomial time. This mean, that you have some way to solve it, such that the number of operations (usually consider binary operations, like binary sum or multiplication) depends polynomially in the length of the imput to the problem. For example, if I give you two positive integer numbers, then you can find the maximum comon divisor to both of them in polynomial time with respect to the length of the numbers (Euclides Extended Algorithm)

[b]NP-[/b] This is the collection of "problems" that can be veryfied in polynomial time. Then, for example, factoring. It is believed that factoring is a hard problem, this is, people think that there is no algorithm to factor any number in polynomial time with respect to the length of the number. Nevertheless, if I give you two numbers and I tell you that the first is a factor of the second, you can easily find out if I am saying the true or not, in polynomial time (you just have to do the division to find out if the remainder is zero or not), this is an NP problem, because a solution can be verified in polynomial time.

[b]NP-Easy, NP-Hard-[/b] I have heard the names, and I now that it has something to do with problem reduction, but probably someone else can explain that a lot better than me.

[b]NP-Complete[/b] A problem X is said to be NP-Complete if every NP problem can be reduced to X, or equivalently, that means that if you find a solution (polynomial time algorithm) for the problem X, you have found a solution for every NP problem.

[b]P vs NP-[/b] Well, it is easy to prove that if a problem can be solved in polynomial time, it can be verified in polynomial time, that is $P\subseteq NP$, nevertheless, nobody knows if the other direction is true. Most of the cryptographic comunity believes that NP is a lot bigger than P, and that, for example, Discrete Log Problem and Factoring problems are NP but are not P. But nobody knows for sure, and the problems is open. That is what is called P vs NP.

I hope it helps. If I redacted to complicated, let me know and I will try to reexplain!

NP hard means that the problem is "at least NP."  So, the NP Complete problems are those which are both NP and NP hard.  "NP easy" presumably means "at worst NP," although I've never heard the terminology before.

Vinny,

You can learn a lot about NP-hard problems from the yahoo newsgroup "algorithm-forge", where NP-hard problems are all that is discussed.

Caligula


Let $n$ be a positive integer, such that $\frac{2^{n}+2}{n}$ is an integer.
Prove that $n$ divisible by $2$

Of course, $n>1$. The answer is yes, but the proof is far from being obvious (although very short). Apparently, it was found for the first time by Schinzel. I think you can find a solution by searching it on the forum. If you don't succeed, I will post a solution again.

I think not mentioning $n>1$ is just a little mistake.  :P

The interesting thing is that we can generalize the problem to $\frac{k^{n-1}+1}{n}$ instead of $\frac{2^{n-1}+1}{n}$, of course if my proof is correct.

I don't have the courage to follow all those indices, but this is the idea, so yes, the proof should be correct.

I have a strong feeling that instead of writing $p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{k-1}^{\alpha_{k-1}}p_{k+1}^{\alpha_{k+1}}...p_{m}^{\alpha_{m}}$ all the time, I could've just write $\frac{n}{p_{k}^{\alpha_{k}}}$...


So what did you get on round 1? i got a 4/4/3/5/5 = 21.

if you had taken it, you'd have gotten a perfect....  :P

dudes i got like a 19, i suck now

i got 45451 or something like that...

i barely tried 5, maybe i shouldve spent more time on it...

but ill make it up tomorrow...ownage............lol dang i suck at math now............depressed............

[quote="undefined117"]Hmm I couldn't take it, but danggg---congratulations to MathAnd! Only perfect from Indiana. :)[/quote]
Thanks, but I think I was somewhat lucky.  Some people complained about losing points because their solutions were too wordy.  My solutions took up 19 pages, which is about 15 more more than they should have taken up.  Yet I got a 25 and only nice comments from the graders?
Oh well.  I'll just take the score. :lol:

you beat me on #2 (just like everyone else  :( )

Hello people from Indiana (and gauss and moogra and undefined).
I phailed with 3 5 2 5 2 :(


Hi,
   Can you provide me with the information about free online classes?.
thanks
kishore

[quote="funcia"]umm if you click "online school" which is on the bar on the side you should get to the information..... :lol:[/quote]
I don't think it is free! :huh:

you are right but math jams are free...oh wait a minut better instructions i just realised if you click "math jams" above you you get to them instead of finding them weird funcia-ish ways :wink:


Let $ ABC$ be a triangle with sides $ a$, $ b$ and $ c$, and area $ K$.

Prove that:

$ \frac{K}{3}(cot(A)\plus{}cot(B)\plus{}cot(C))\leq\sqrt{\frac{1}{3}\frac{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}{4}\minus{}K}$

 :)

Hint

consider power of point and the incircle

 :)


Solve of the equation:
$\sqrt{x^{2}-2x+5}+\sqrt{x^{2}+2x+10}=\sqrt{29}$


I have tested the go to the top of the page buttom in several forums, including Intermediate counting, Getting Started, Moderator's Lounge, etc, and it does not work in any of them.  Here is one of the urls:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=91633#top[/url]

Dude, the topic you mention simply doesn't exist, no wonder it doesn't work. ;)
It works perfect for all existing posts and topics.

Sorry, the link is to the forum for Intermediate Counting and Probability, so you can't use it unless you are enrolled.

I don't know why, but the buttons work now.  Maybe I was on the other skin, or something  :maybe:


The sequence {a_n} is defined by a_0 = 9 and a_(n+1) = 3*(a_n)^4+4*(a_n)^3 for n = 0,1,2,... Prove that the decimal representation of a_10 contains more than 1000 9's.

Let b_n  = a_n +1.

Then,
b_(n+1) = 3(b_n - 1) 4  - 4(b_n - 1) 3  + 1 = b 2_n(3b 2  - 8b + 6)

Now it is easy to prove by induction that b_n is divisible by 10 t , where t = 2  n .
Then b_n ends by at least  2  n digits '0'. It follows that a_n has at least 2  n - 1 digits '9', and in particular a_10 has at least 1023 digits '9'.

Pierre.


inequalities
function
inequalities solved

If $a \geq b \geq |x|$ then $ \sqrt{a-b}+ \sqrt{a+b} \leq \sqrt{a-x} + \sqrt{a+x}\leq 2 \sqrt{a}$

cheers! :D :D :cool:

This is exactly the same as one you posted before (do not remember the n).
The right inequality is a direct consequence of the concavity of the function f(t) =  \sqrt t.
Tle left one is easy to prove since the function g(x) =  \sqrt (a-x) +  \sqrt (a+x) is even and decreasing on [0,b], thus f(x)  \geq f(b), hence result.

Pierre.


function
algebra
domain
inequalities
polynomial
Cauchy Inequality
real analysis

I'm trying for some time to find all continuous functions f:(0,infinity) with values in the same set such that f(1/2(x+y))=max{  \sqrt f(x)f(y), 1/2(f(x)+f(y)}. Can you help me? I cannot solve this one.

your function is pretty sad harazy :D:D... maybe "she" doesn't want to be solved :P

It's sad! I don't know what I do, but whenever I want to specify the domain of the function, that sad face appears. Anyway, find it and she will be happy.

I think it's (x+y)/2. I'm not sure, but the shape of the problem seems to have something to do with the arithmetic and geometric mean.

You are right, Grobber. It is the arithmetic and geometric mean.

hum hum... In this case it's not very interesting !
It is well known (especially for two variables...) that we have the Cauchy inequality :  \sqrt (xy)<=(x+y)/2 (You just need to write that (\sqrt x-\sqrt y)²>=0). So f((x+y)/2)=(f(x)+f(y))/2 and f is continous, hence f is affine.

Sorry, I'm the stupid one here. The corect statement is:
  f((x+y)/2)= max{ f( \sqrt xy), (f(x)+f(y))/2}.

First of all, from f((x+y)/2)>=(f(x)+f(y))/2 and the continuity of f we find f to be concave. I think I can show that the only funtions satisfying the condition are the increasing (not necessarily strictly increasing) 1'st degree polynomials.

Something that's going to be very helpful: If the graph of f passes through 3 colinear points A, B, C (B in (AC)) then the entire segment [AC] is part of the graph of f. This is easy to prove by some geometric arguments, using the fact that a part of f's graph is always "above" the chord having the same endpoints.

I'll be back to post more. Right now, my mom is claiming some PC time to play Minesweeper or something, so I have to go :D.

Well, waiting for your solution, I just saw that we need not ask for continuity of f. It can be deduced from the those relations!

[b]Case 1: f((x+y)/2)=(f(x)+f(y))/2[/b]

This means the function is a 1'st degree polynomial and it must not be strictly decrasing because it takes values in (0,00), so the answer is f(x)=ax+b with a,b>=0 and a2+b2>0.

[b]Case 2: there exist a,b in (0,00), b(f(a+b)+f(a-b))/2[/b]

In this case f(a)>(f(a+x)+f(a-x))/2 for all x in (b,a) because otherwise we could use the lemma in my previous message, the one about 3 collinear points on the graph of f and we would contradict the fact that f(a)>(f(a+b)+f(a-b))/2. This means that f(a)=f(\sqrt(a2-x2)) for all x in (b,a).

This shows there is a T>0 s.t. f is constant on (0,T]. Take T to be maximal with this property (assume it's not 00). For all x>T f(x)<f(T) (assume f(x)>f(T), take points (T/2,f(T/2)), (T,f(T) and (x,f(x)) and they don't satisfy the condition of concavity). Now take a>0, a<T s.t. \sqrt(ax)<T and (a+x)/2>T (x>T).

f((a+x)/2)>(f(a)+f(x))/2 (otherwise we use the lemma and we get the entire segment with endpoints (a,f(a)) and (x,f(x)) to belong to the graph of f, which would mean f(x)=f(a)=f(T), contradiction), so f((a+x)/2)=f(\sqrt(ax))=f(T) (*)(because \sqrt(ax)<T). At the same time, (a+x)/2>T, so f((a+x)/2)<f(T) (**). From (*) and (**) we have our contradiction, so we see that there is no maximal T, so f is constant on (0,00), so f(a)=(f(a+b)+f(a-b))/2 for all a,b>0, a>b, so case 2 can't exist.

I think we're done, right?


What is the sum of the digits of the first 25 natural numbers?

What is the sum of the first 25 odd numbers?

sum of the digits of the first 25 natural numbers= sum of the digits of (1+2+.....+25)
=sum of the digits of (1+2+.....+9)+sum of the digits of (10+11+.....+19)+sum of the digits of (20+21+.....+25)
45+sum of digit of(9*10+1+2+....+9)+sum of digit of (5*20+1+2+....+5)
=145

No, this one is also incorrect :( due to
$S=\sum_{i=1}^{n}(2i-1)$
$=2\sum_{i=1}^{n}i-\sum_{i=1}^{n}1$
$=n(n+1)-n=\boxed{n^2}$.
As a result, you do
$n^2=25^2=625$

Masoud Zargar

i think you guys should keep the level of math to middle school level becuase most middle schoolers do not understand that. ;)  :)

Hey math92, are you also math92_?
Because you both have 'fear the bunny!' and almost identical names.  :huh:  :D
[hide="a"]
$1+2+...9=45$
$S_{digits}{10+11+12+13+...19}=45+10=55$
$S_{digits}{20+21+22+23+24+25}=12+15=27$
$45+55+27=\boxed{127}$[/hide]
I don't get how to do B without adding them all up... which I don't want to do. :?


Hint: This has nothing to do with geometry. You only have to know how many faces and vertices are on a cube. Let's hope you can count.  ;)

On a cube, a positive integer is written on each face. A corner of a cube is the intersection of a certain set of three faces of the cube. On each corner, the number that is the product of the three integers written on the intersecting faces is written on each vertex. The numbers on the vertices add up to 70. What do the numbers on the faces add up to?

Solution  :arrow:  :arrow:  :arrow:  :arrow:  :arrow:  :arrow:  :arrow:  :arrow:

[hide]14. Substitute each number on the faces with a letter (A, B, C, D, E, and F). Thus, the numbers on the corners will be in the form XYZ, where X, Y, and Z are the numbers on the faces. Factor until you reach the conclusion in the form (G+H)(I+J)(K+L) = 70. Since all of the numbers on the faces are positive integers, G+H, I+J, and K+L will also be integers. You only have to find the three factors of 70 that do not include 1 (because the sum of any two positive integers cannot be one; convince yourself of this  :)  ) The factors, which are 2, 5, and 7, add up to 14.[/hide]

[hide]The numbers on the faces are a, b, c, d, e, and f. The numbers on the vertices are abc, acf, afe, aeb, dbc, dcf, dfe, and deb.
abc+acf+afe+aeb+dbc+dcf+dfe+deb = 70
(a+d)(bc+cf+fe+eb) = 70
(a+d)((b+f)c+(f+b)e) = 70
(a+d)(b+f)(c+e) = 70
70 = 2*5*7
So the sum is 2+5+7, or [color=green]14[/color].
[/hide]


geometry
3D geometry
sphere
trigonometry
ratio
angle bisector
similar triangles

Let O be the center of the excircle of triangle ABC opposite A. Let M be the midpoint of AC, and P be the intersection of MO and BC. Prove that if angle BAC = 2 x angle ACB, then AB=BP.

Tell me if these qs are too easy or too hard, I basically can't solve any geometry and can't tell them apart :)

[color=cyan]I'd say your geometry problems tend to the more difficult side.  On the other hand, that's what this forum is for, no?

So, while I struggle hopelessly on your problem, here's one of my own, in several parts.

Part 1 (doable):  Given 2 circles of different radii tangent to each other and their common external tangent.  How many circles are tangent to both circles and to the tangent line?  What are their radii?

Part 2 (also doable): Given 3 circles in a plane such that none intersect nor is any contained in any other, how many circles are tangent to all three?  This has a name, but I don't remember it.

Part 3 (very very hard): Given three circles each tangent to the other two -- EITHER all 3 externally tangent OR 2 internally tangent to the third and externally tangent to each other -- how many circles are tangent to all 3?  What are their radii?

Part 4 (doable again, but hard): Given three spherical balls of different radii sitting on a table such that they all are touching each other.  How many spheres are tangent to all 3 and the table?  What are their radii?

Part 5 (even more difficult than part 3): Given 4 spheres, each externally tangent to the other 3.  How many spheres are tangent to all 4?  What are their radii?

Part 6 (the hardest edge of doable): Given n n-spheres, each externally tangent to every other, and all sharing a common (n-1) tangent plane.  How many other n-spheres are tangent to all of them and to the (n-1) plane?  What are their radii?

Part 7 (don't even try): Given (n+1) n-spheres, each externally tangent to every other.  How many n-spheres are there tangent to all of them?  What are their radii?

I should note that part 3 is essentially the problem of the arbelos, which I don't care to explain right now because I'm very tired, but maybe someone else will do me the favor of explaining.  Anyhow, I'm convinced someone must have solved all of these.  I just don't know who, or what the asnwers are (except for parts 1 and 2).[/color]

The couple of questions I've posted were from national selection tests / olympiads, so they shouldn't be IMO level. And I tried to pick ones where the solutions were very short and didn't look too difficult.. of course that means nothing..

[color=cyan]Oh, I'm so pleased with myself!  :).  I did this last night after getting in to bed, 4 different ways (sort of).  So, without further ado:

Lemma 1:
a^2 = b^2 + bc

Proof 1:
Use Law of Sines on angles A and C of triangle ABC:
a/sin(2C) = c/sin(C)
a/(2c) = cos(C)
Use Law of Cosines on angle C of triangle ABC:
c^2 = a^2 + b^2 - 2ab*cos(C)
c^2 = a^2 + b^2 - 2b*a^2/(2c)
c^3 = c*a^2 + c*b^2 - b*a^2
c^3 - c*b^2 = c*a^2 - b*a^2
c*(c-b)*(c+b) = a^2*(c-b)
Assume c != b
c*(c+b) = a^2
If c = b, it gives you a specific triangle (45-45-90, in fact) and you can work out yourself that it works.

Proof 2: Draw AO.  This is the angle bisector of angle A in ABC.  Let it intersect BC at N.  Note that angle BAN = 1/2 BAC = ACB.
Also, angle ABN = angle CBA.  Therefore
triangle ANB ~ triangle CAB.

Using the angle bisector theorem, BN = ac/(b+c).  Using similar triangles, BN = c^2/a.  Setting these two expessions equal,
a^2 = bc + c^2

Lemma 2: BP = (a^2 + ac)/(a + b + c)

Proof 1:  Okay, we will use barycentric coordinates.  Define
A = [1, 0, 0], B = [0, 1, 0], and C = [0, 0, 1].  Then N = [0, b, c].  Then we can find the angle bisector from A is given by
[(b+c)(1-t), bt, ct].

Similarly, (by proving the External Angle Bisector theorem, which I did to get this), we can see that the external angle bisector to angle C has equation [-as, bs, (b-a)(1-s)].

We solve for the intersection to find O = [-a, b, c]
We know M = [1, 0, 1]
We can solve for the line between them and then find its intersection with BC (which has equation [0, t, 1-t] of course), and we get P = [0, b, c+a].  This means BP : PC = c+a : b, so
BP = (a^2 + ac)/(a+b+c).

Proof 2: We will use areas of triangles.  Let the radius of the excircle be r_a.  Let the altitude of ABC from A be h_a.
Then
[PCM] = 1/2 * (h_a / 2) * PC = h_a * PC / 4
[PCM] = [OCM] - [OCP] = (1/2 * r_a * MC) - (1/2 * r_a * PC) =
= r_a * (b/4 - PC/2)
So, setting these two equal:
h_a * PC / 4 = r_a * (b/4 - PC/2)
h_a / r_a = (b - 2PC) / PC

Now, go to triangle ABC.
[ABC] = 1/2 * h_a * a
[ABC] = 1/2 * r_a * (-a + b + c)
Setting equal,
1/2 * h_a * a = 1/2 * r_a * (-a + b + c)
h_a / r_a = (-a + b + c) / a

Substituting this in above,
(-a + b + c) / a = (b - 2PC) / PC
Adding 2,
(a + b + c) / a = b / PC
PC = ab / (a + b + c)
BP = a - PC = (a^2 + ac)/(a + b + c).

So, lets take our two lemmas.
BP = (a^2 + ac)/(a + b + c).
a^2 = bc + c^2
Substitute in for a^2:
BP = (bc + c^2 + ac)/(a + b + c).
BP = a.

Q.E.D.[/color]

Joel, wats a "lemma"?

ANd can you give us some real geometry problems? (meaning ones i can actually do?) :D

A lemma is a theorem you use to prove a more difficult theorem.

My proof is kind of similar, so I'm not going to post it all. Here are the main ideas: Let D be the foot of the angle bisector of A. Since O is the intersection of the angle bisector of A and the external angle bisectors of B and C, we can easily find (by using Menelaus and the bisector theorem) the ratio OD/OA. By applying Menelaus to the triangle ADC with transversal line MPO we get DP/PC=OD/OA, which we know (it's a/(b+c) where a, b, c are the usual notations of the sides of the triangle). We know DP/PC so we can now calculate BP in terms of a, b, c (this part isn't hard, it's just long and boring). By using the fact that a^2=c(b+c) we get exactly what we want.

My opinion is that it's not a HARD problem, it just might look that way because of the long calculations.

[color=cyan]What makes it hard isn't that there are tough calculations -- in fact, they're mostly pretty nice.  A few ugly expressions, maybe.  It's hard because it requires several distinct logical steps.  That's really quite tough, especially in geometry where there are so many available steps you could take.

One comment on your solution -- you can also find OA/OD (well, really OD/AD) using the area formulae r_a*(s-a) and 1/2*h_a*a and similar triangles.
I think I had tried to do something like what you did, but hadn't been able to make it work out right.  At least, I know I was trying to find DP.

I'm mostly pleased with myself because I got the barycentrics to work, which was fun.  [/color]

Avoiding boring calculations:
$BP=AB\iff AP$ is bisector of the angle $\angle DAC, AD$ bisector of $\angle BAC, D\in \overline{BC}$. Because $CO$ is the external bisector of $\angle ACB\implies\frac{AO}{DO}=\frac{AC}{CD}=\frac{AC}{AD}\ (\ 1\ )$.
Now apply Menelaos to $\triangle ACD$ with transversal $\overline{MPO}$ and finally get $\frac{PC}{DP}=\frac{AO}{DO}$; with $(1)$ we get $\frac{PC}{DP}=\frac{AC}{AD}$ and, indeed, $AP$ is the angle bisector of $\angle DAC$, consequently $AP\parallel DO$ and we are done.

Best regards,
sunken rock


Prove that for every natural number like $n$ we have $\frac{n!}{n^{n}}\leq\frac{1}{2^{[\frac{n}{2}]}}$.

It is rather rough inequality.
If $n=1$ or $2$ we get equality.
If $n\geq 3$, then $3n+3\leq 4n$ :arrow: $\frac{n+1}{2n}\leq \frac{2}{3}$ :arrow: for each $i=1, 2,..., n$
 $\frac{i}{n}\frac{n+1-i}{n}\leq \frac{(\frac{n+1}{2})^2}{n^2}=(\frac{n+1}{2n})^2\leq \frac{4}{9}<\frac{1}{2}$.
Multiplied for $i=1,2,...,n$ we get
 $\frac{(n!)^2}{n^{2n}}\leq \frac{1}{2^n}$. Hence
$\frac{n!}{n^n}\leq \frac{1}{2^{[\frac{n}{2}]}}$, q.e.d.

[b]Lemma 1.[/b] $\left(\forall\right)n\in N^*\ ,\ \left[\frac{n+1}{2}\right]-\left[\frac n2\right]\le 1\ .$
[u]Proof.[/u] Indeed, $\frac{n+1}{2}<1+\frac n2\Longrightarrow \left[\frac{n+1}{2}\right]\le \left[1+\frac n2\right]=1+\left[\frac n2\right]\ .$

[b]Lemma 2.[/b] $\left(\forall\right)n\in N^*\ ,\ \left(\frac{1}{n+1}\right)^{n+1}>\left(1+\frac 1n\right)^n\ge 2\ .$
[u]Proof.[/u] We will apply the A.M.-G.M. to the positive numbers
$x_k=1+\frac 1n,\ k\in \overline {1,n}$ and $x_{n+1}=1$ a.s.o.

[u]Proof of the proposed inequality.[/u] Denote $a_n=\frac{n!}{n^n}\cdot 2^{\left[\frac n2\right]} \ ,\ n\in N^*.$
We observe that: $a_1=1$ and for $n\in N^*$ we have
$\frac{a_{n+1}}{a_n}=2^{\left[\frac{n+1}{2}\right]-\left[\frac n2\right]}\cdot \frac{1}{\left( 1+\frac 1n\right)^n}<2\cdot \frac 12=1\Longrightarrow a_{n+1}<a_n\ .$
Therefore, for any $n\in N^*, n>1,\ a_{n+1}<a_n<a_1=1\ .$


Prove that $ \sin x \plus{} \cos x \equal{} \sqrt{2}\sin(\theta \plus{} \pi/4)$

[quote="veezbo"]Prove that $ \sin x + \cos x = \sqrt {2}\sin(\theta + \pi/4)$[/quote]
Assuming you meant $ \sin\theta+\cos\theta$ :P

[hide]From the LHS factor out $ {\sqrt{2}}$ to get $ \sqrt{2}(\frac{\sqrt{2}}{2}\sin\theta+\frac{\sqrt{2}}{2}\cos\theta)$
$ \sqrt{2}(\cos\frac{\pi}{4}\sin\theta+\sin\frac{\pi}{4}\cos\theta)$
$ \sqrt{2}\sin(\theta+\frac{\pi}{4})$[/hide]

[quote="abcak"][quote="veezbo"]Prove that $ \sin x \plus{} \cos x \equal{} \sqrt {2}\sin(\theta \plus{} \pi/4)$[/quote]
I didn't get your proof ckck.

Don't you just replace sin($ \theta$+$ \pi$/4) with $ cos(\theta)$, square both sides, and use a few identities?[/quote]
Uh what don't you get about my proof :maybe:
I just factored out $ \sqrt{2}$, and used the fact that $ \frac{\sqrt{2}}{2}\equal{}\sin\frac{\pi}{4}\equal{}\cos\frac{\pi}{4}$ to simplify the expression into what we needed it to be.

He's using the sum of angles formula between steps 2 and 3. That might be what is confusing.

[quote="abcak"][quote="veezbo"]Prove that $ \sin x \plus{} \cos x \equal{} \sqrt {2}\sin(\theta \plus{} \pi/4)$[/quote]
I didn't get your proof ckck.

Don't you just replace sin($ \theta$+$ \pi$/4) with $ cos(\theta)$, square both sides, and use a few identities?[/quote]

This is what tripped me up when I first read this question. $ \pi/4 \equal{} 45$ degrees not $ 90$ degrees

So $ \sin(\theta\plus{}\pi/4) \neq \cos(\theta)$

ckck's way of doing it is probably the easiest

You could also expand out the right side.

$ \sqrt{2}\left(\sin\theta\frac{1}{\sqrt{2}}\plus{}\frac{1}{\sqrt{2}}\cos\theta\right)\equal{}\sin\theta\plus{}\cos\theta$.

This is a bit more obvious than factoring $ \sqrt{2}$ out of $ \sin\theta\plus{}\cos\theta$.

hello, you can also write $ \sin(x)\plus{}\cos(x)\equal{}\sqrt{2}\cos(\frac{\pi}{4}\minus{}x)$
Sonnhard.


geometry
3D geometry
tetrahedron
analytic geometry
geometry solved

Let G be a point inside the trihedron (is this the word?) formed by the lines in space Ox, Oy, Oz. Construct the plane through G which, together with Ox, Oy, Oz forms the tetrahedron of minimum volume.

That's very easy problem.
Let G have coordinates (a,b,c).
Let a plane  \pi containing G intersects Ox, Oy, Oz at
P =(p,0,0), Q = (0,q,0), R= (0,0,r) respectively. Then the equation of \pi is
x/p+y/q+z/r = 1, so a/p+b/q+c/r = 1.
1 = a/p+b/q+c/r \geq 3 (abc/pqr)^(1/3),
so the volume of OPQR = pqr/3 \geq 9abc with equality only when a/p=b/q=c/r=1/3. (G will be the centroid of OPQR).


In one of my documents, I am creating a numbered list.  Now, in between two of the items, I have a short paragraph.  However, this paragraph is indented and I need it flush up against the left margin.  I would begin a new list, which would fix the problem, except I need the list to continue where the old one stopped (i.e. 8, 9, paragraph, 10,... instead of 8, 9, paragraph, 1, 2,...).

The resume option of the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=enumitem]enumitem package[/url] (almost certainly part of your distro) will do this for you - see page 4 of the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/enumitem/enumitem.pdf]documentation[/url].

[quote="fedja"]You can also use a brute force approach adding \setcounter{enumi}{7} right after the \begin{enumerate} command to start numbering with 8.[/quote]
That would only work if no subsequent changes were made to the previous list. Otherwise using the brute force way you would have to save the counter after paragraph 9, then load that counter into the enumi counter when starting again. But enumitem does all that for you.

[quote="JRav"]Thanks to both of you.  Just for the record, I took fedja's advice because it is simpler.[/quote]
Not if you have to make changes. And I'm not sure what you mean by 'simpler '- enumitem requires only the line
\usepackage{enumitem}
and [resume] when you start the enumeration.

You seem to be taking offense to this.  I used his method, and it worked without a problem or an error message so I have no reason to fix it.

[quote="JRav"]You seem to be taking offense to this.  [/quote]
Don't be daft! fedja clearly knows what he is talking about so his idea is fine. I'm just saying that enumitem is more in line with LaTeX's philosophy of numbering things automatically (like sections, theorems etc etc) and can cope with subsequent changes. If enumitem was complicated to use I would agree that you may want to trade simplicity for possible changes, but enumitem is about as simple as it could be. That's why I wondered about saying the setcounter method was 'simpler'.

[quote="JRav"]I used his method, and it worked without a problem or an error message so I have no reason to fix it.[/quote]I agree with you if the first list is fixed and you never want to change it in the future.

Yes I often use such brute force methods myself but really I should use packages where authors have thought about other effects it may have.

I tend to not like adding extra packages if I don't need to.  I would rather just write a command myself than try to figure out which package I need, then find the syntax.

I'm pretty much of the same type. I used to stick to AMS-TeX as long as journals accepted it. My switch to LaTeX was rather reluctant and I still prefer to avoid using many of its options. Also, rather than learning about various packages, I prefer to learn about programming possibilities, so the way I would really do this interrupted list if I were intending to change it later is something like
[code]
\newcounter{aa}
\begin{enumerate}
\item item 1
\item item 2
\setcounter{aa}{\theenumi}
\end{enumerate}
The paragraph
\begin{enumerate}
\setcounter{enumi}{\theaa}
\item item 3
\end{enumerate}
[/code]
(it is by no means the only option, just the first that came into my head) but stevem has a valid point here: there is a certain set of "standards" or "philosophy" in $ \text{\LaTeX}$ typesetting and if you want to share your TeX-file with other people, they'll be much happier to see usage of familiar packages than unfamiliar assembler-style code embedded into the text. Of course, if you just type your homework, it doesn't matter and learning basic programming in $ \text{\LaTeX}$ by writing your own macros for homeworks may have its advantages.


integration
function
real analysis
calculus
Support
real analysis unsolved

I have the simple question, it's interesting to see some different proof of this wellknown statement .
suppose $ f\in$ $ L_{1}(\mathbb{R})$ and $ L_{2}(\mathbb{R})$ (I mean in intersection of $ L_{1}$ and $ L_{2}$)
consider $ Hf: \equal{} F^{ \minus{} 1}((Ff)(z)( \minus{} isign(z)))$
where $ F$ is fourier transform .
 show that if $ Hf$ is in $ L_{1}$ then $ \int_{\mathbb{R}}f \equal{} 0$

That's [b]Hilbert[/b] transform, not Gilbert!

We can prove that if $ f\in L^1(\mathbb{R}),$ then $ \widehat{f}$ is continuous. (We can actually prove that $ \widehat{f}\in C_0(\mathbb{R}),$ the set of continuous functions that tend to zero at infinity.)

Therefore, to have $ Hf\in L^1$ with $ \widehat{Hf}(\xi) \equal{} \minus{} i\,\text{sgn}\,(\xi)\widehat{f}(\xi),$ we must have $ \widehat{Hf}$ continuous at zero, which can only happen if $ \widehat{f}(0) \equal{} 0.$

But $ \widehat{f}(0) \equal{} \int_{\mathbb{R}} f.$

Please note that this is a necessary condition for $ HF\in L^1$ but it is [b]not[/b] a sufficient condition.

A second argument: $ Hf$ is given by convolution of $ f$ with (a constant times) $ \frac1x.$ (Technically, the principal value of that, but that won't matter in the argument I'm about to make.)

Suppose $ f(x)$ is zero outside of $ [\minus{}M,M]$ and $ \int_{\minus{}M}^Mf\equal{}c\ne 0.$ Let $ x>2M.$

$ Hf(x)\equal{}\int_{\minus{}M}^M\frac{f(y)}{x\minus{}y}\,dy \equal{} \frac1x\int_{\minus{}M}^Mf(y)\,dy\plus{}\int_{\minus{}M}^M\frac{yf(y)}{x(x\minus{}y)}\,dy.$

The first integral on the right hand side above is simply $ \frac cx.$ The second integral is bounded by $ \frac{CM}{x^2}\,\|f\|_1$ for some $ C$ and is hence $ O(x^{\minus{}2})$ as $ x\to\infty.$

That means that $ Hf(x)$ is asymptotic to $ \frac cx$ as $ x\to\infty,$ and as such cannot be integrable.

To finish the argument: if $ f\in L^1$ and $ \int_{\mathbb{R}}f\ne 0,$ then we can approximate $ f$ arbitrarily closely by a function with bounded support and nonzero integral.


$A+B+C=\pi$ prove that :

$(\frac{1}{cosA}+\frac{1}{cosB}+\frac{1}{cosC})^2 \geq \frac{4}{cosAcosBcosC}$


factorial
integration
calculus
algebra proposed
algebra

\[ \sum\limits_{k \equal{} 0}^n {\left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right)\frac{{\left( { \minus{} 1} \right)^k }}
{{2k \plus{} 1}}}\]

[quote="vazk"]
\[ \sum\limits_{k \equal{} 0}^n {\left( {\begin{array}{*{20}c} n \\
k \\
\end{array} } \right)\frac {{\left( { \minus{} 1} \right)^k }} {{2k \plus{} 1}}}\]
[/quote]

More generally, compute
\[ \sum\limits_{k \equal{} 0}^n {\left( {\begin{array}{*{20}c} n \\
k \\
\end{array} } \right)\frac {{\left( { \minus{} 1} \right)^k } x^{2k\plus{}1}} {{2k \plus{} 1}}}\]
then plug in $ x\equal{}1$.  This is an improvement, because you can take its derivative.

[quote="transseries"][quote="vazk"]
\[ \sum\limits_{k \equal{} 0}^n {\left( {\begin{array}{*{20}c} n \\
k \\
\end{array} } \right)\frac {{\left( { \minus{} 1} \right)^k }} {{2k \plus{} 1}}}\]
[/quote]

More generally, compute
\[ \sum\limits_{k \equal{} 0}^n {\left( {\begin{array}{*{20}c} n \\
k \\
\end{array} } \right)\frac {{\left( { \minus{} 1} \right)^k } x^{2k \plus{} 1}} {{2k \plus{} 1}}}\]
then plug in $ x \equal{} 1$.  This is an improvement, because you can take its derivative.[/quote]7

There is another form to do that without derivate?
thanks

hello, we have $ \sum_{k\equal{}0}^n\binom{n}{k}\frac{(\minus{}1)^k}{2k\plus{}1}\equal{}\frac{(2n)!!}{(2n\plus{}1)!!}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have $ \sum_{k \equal{} 0}^n\binom{n}{k}\frac {( \minus{} 1)^k}{2k \plus{} 1} \equal{} \frac {(2n)!!}{(2n \plus{} 1)!!}$.
Sonnhard.[/quote]
What's proof?
We don't have $ \sum_{k\equal{}0}^n\binom{n}{k}\frac{(\minus{}1)^k}{2k\plus{}1}\equal{}\frac{(2n)!!}{(2n\plus{}1)!!}$. Because we [b]don't yet prove[/b] this.

[quote="latios315"][quote="Dr Sonnhard Graubner"]hello, we have $ \sum_{k \equal{} 0}^n\binom{n}{k}\frac {( \minus{} 1)^k}{2k \plus{} 1} \equal{} \frac {(2n)!!}{(2n \plus{} 1)!!}$.
Sonnhard.[/quote]
I have no idea what the right answer is but this is definitely wrong, plugging in n=3 in the original sum gives 16/35, but your formula gives 720!/5040![/quote]
I think the double factorial is $ (2n)!!\equal{}(2n)\cdot(2n\minus{}2)\cdot(2n\minus{}4)\cdots2$, not $ ((2n)!)!$.

[quote="latios315"]thank you, it makes much more sense now, I would definitely like to see how he found that[/quote]

Using $ f_n(x)\equal{}\sum_{k\equal{}0}^n\binom nk\frac{(\minus{}1)^kx^{2k\plus{}1}}{2k\plus{}1}$, you get $ f_n'(x)\equal{}\sum_{k\equal{}0}^n\binom nk(\minus{}1)^kx^{2k}$ $ \equal{}(1\minus{}x^2)^n$

So $ f_n(x)\equal{}\int_0^x(1\minus{}t^2)^ndt$ and the required value is $ S_n\equal{}\int_0^1(1\minus{}t^2)^ndt$

A basic integration by parts implies $ S_n\equal{}2n\int_0^1t^2(1\minus{}t^2)^{n\minus{}1}dt$ and so $ S_n\equal{}2n(S_{n\minus{}1}\minus{}S_n)$

So $ S_n\equal{}\frac{2n}{2n\plus{}1}S_{n\minus{}1}$ with starting value $ S_0\equal{}1$

So $ S_n\equal{}\frac{\prod_{k\equal{}1}^n2k}{\prod_{k\equal{}1}^n(2k\plus{}1)}$ $ \equal{}\frac{(\prod_{k\equal{}1}^n2k)^2}{\prod_{k\equal{}1}^{2n\plus{}1}k}$

So $ S_n\equal{}\frac{(2^nn!)^2}{(2n\plus{}1)!}$

(which indeed is another form of the result previously given by Dr Sonnhard)

[quote="pco"][quote="latios315"]thank you, it makes much more sense now, I would definitely like to see how he found that[/quote]

Using $ f_n(x) = \sum_{k = 0}^n\binom nk\frac {( - 1)^kx^{2k + 1}}{2k + 1}$, you get $ f_n'(x) = \sum_{k = 0}^n\binom nk( - 1)^kx^{2k}$ $ = (1 - x^2)^n$

So $ f_n(x) = \int_0^x(1 - t^2)^ndt$ and the required value is $ S_n = \int_0^1(1 - t^2)^ndt$

A basic integration by parts implies $ S_n = 2n\int_0^1t^2(1 - t^2)^{n - 1}dt$ and so $ S_n = 2n(S_{n - 1} - S_n)$

So $ S_n = \frac {2n}{2n + 1}S_{n - 1}$ with starting value $ S_0 = 1$

So $ S_n = \frac {\prod_{k = 1}^n2k}{\prod_{k = 1}^n(2k + 1)}$ $ = \frac {(\prod_{k = 1}^n2k)^2}{\prod_{k = 1}^{2n + 1}k}$

So $ S_n = \frac {(2^nn!)^2}{(2n + 1)!}$

(which indeed is another form of the result previously given by Dr Sonnhard)[/quote]

\begin{eqnarray*}
S_n
& = & \int_0^1(1 - t^2)^ndt \\
& = & \frac{1}{2}B\left(\frac{1}{2}, n+1\right) \\
& = & \frac{1}{2}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(n+1\right)}{\Gamma\left(\frac{1}{2}+(n+1)\right)} \\
& = & \frac{1}{2}\frac{\sqrt{\pi} n!}{\frac{(2n+1)!!}{2^{n+1}}\sqrt{\pi}} \\
& = & \frac{2^n n!}{(2n+1)!!} \ \ \left(= \frac{(2^n n!)^2}{(2n+1)!}\right)\\
& = & \frac{(2n)!!}{(2n+1)!!} \\
\end{eqnarray*}


How do I draw an a quadrant sector with a line bisecting it?


geometry
geometric transformation
induction
algebra
polynomial
binomial coefficients
algebra solved

here is a nice problem:
let S0 be an arbitrary finite set of positive integers. let Sn be defined for each n>=1,as the set consisting of all k such that just one of k,and k-1 belongs to Sn-1. Prove that for infinitely many n, Sn is the union of S0 and a translation of S0.
I was able to prove this by induction. But I also noticed that if Sm be union of S0 and a translate of itself, then m=2^k for some k. I couldn't prove this by my induction, so does any one have any idea for proving this?

To each set S_n we associate the polynomial p(x)=sum a_iX^i, where a_i is 1 if i belongs to S_n and 0 otherwise. If we restrict to polynomials over Z_2, the it is easy to find that p_n(X)=(1+X)^np_0(X). Hence, if we are in the condition of the problem, we must have (1+X)^n=1+X^n, identity true in Z_2. This means that all the binomial coefficients are even and hence that n is a power of 2.


AMC USA(J)MO USAMO function calculus derivative inequalities

let $ f\left(a,b,c\right) \equal{} \frac {a}{b \plus{} c \plus{} 1} \plus{} \frac {b}{a \plus{} c \plus{} 1} \plus{} \frac {c}{a \plus{} b \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$ is a convexe function so $ max(f) \equal{} f\left(1,1,1\right)$. and pôsted before. :wink:
Also see here
http://www.iis.it-hiroshima.ac.jp/~ohkawa/www.kalva.demon.co.uk/usa/usoln/usol805.html

How do you know it's a convex function?
And after I have known it's a convex function, why it's f(1,1,1)?

the second derivative is positive for either of the three variables so it's convex. For a convex function, the maximum appears at the extremes, so you'll compare f(0,0,0) f(0,0,1) f(0,1,1) and f(1,1,1) and choose the largest value of f from them.
Hope this helps

it's convex only if $f''(a)>0$ [b][color=red]and[/color][/b] $f''(b)>0$ [b][color=red]and[/color][/b] $f''(c)>0$?

[quote="Twins"]Show that for all non-negative reals $a,b,c\leq 1$
$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c)\leq 1$[/quote]
We have
$(1-c)(1-a)(1-b)\leq \frac{1-c}{1+a+b}$
$a\leq b \leq c$
$\to \frac{a}{1+b+c}\leq \frac{a}{a+b+1}$
...

[quote="Twins"]Show that for all non-negative reals $a,b,c\leq 1$
$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c)\leq 1$[/quote]

its a linear function on each of the variables so the maximum can be attained at the endpoints. the rest is for you :D

[quote="Twins"]Show that for all non-negative reals $a,b,c\leq 1$
$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c)\leq 1$[/quote]

Here is a "classical" solution via the arithmetic-geometric mean inequality:

1. Rewrite the right hand side 1 as $\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}$.

2. Move the three fractions from the left hand to the right hand side.
Merge every pair with identical numerator into a single fraction.
For instance $\frac{a}{a+b+c}-\frac{a}{b+c+1}$ becomes $\frac{a(1-a)}{(a+b+c)(b+c+1)}$.

3. Multiply the entire inequality by $(a+b+c) / [(1-a)(1-b)(1-c)]$.
Now the new right hand side RHS reads:
$RHS: = \frac{a}{(b+c+1)(1-b)(1-c)}+\frac{b}{(a+c+1)(1-a)(1-c)}+\frac{c}{(a+b+1)(1-b)(1-a)}$
And the entire inequality reads: $a+b+c \leq RHS$.
Up to this moment, we have only rewritten the inequality into an equivalent inequality.

4. Consider the right hand side RHS. The arithmetic-geometric mean inequality yields that the denominator of every fraction is at most 1.
For instance $(b+c+1)(1-b)(1-c) \leq 1$.
This completes the proof.

[quote="gjw"]
3. Multiply the entire inequality by $(a+b+c) / [(1-a)(1-b)(1-c)]$.

[...]

This completes the proof.[/quote]

Well...
Of course the cases with a+b+c=0, a=1, b=1, and c=1
must be analyzed separately.

[quote="Twins"]How to get the  the second derivative of [color=red]three[/color] variables?
Can anyone post it?[/quote]
You may consider the inequality of only one variable, say $a$, keeping $b$ and $c$ fixed for now [sort of "parameters"]. The functions (of $a$) $\frac{a}{b+c+1}$, $\frac{b}{a+c+1}$ ,$\frac{c}{a+b+1}$ and $(1-a)(1-b)(1-c)$ are convex, thus their sum is convex. It is known that a convex function on a closed interval $[x,y]$ attains its maximum either in $x$ or in $y$.
That is $f(a,b,c)\leq Min(f(0,b,c), f(1,b,c))$.
Repeating for $b$ and $c$ we get that $f(a,b,c)$ is maximal when $a,b,c\in\{0,1\}$. A direct verification shows that any such triple $(a,b,c)$ satisfies the equality case.

i think that to get the seconde derivation
u have just to [b]fix[/b] two nubers and working with with one, as a function


One way to guess the correct solution to this problem is to guess the correct limit $ L$ (which isn't $ \frac {1}{2}$) and then to look at the behavior of $ x_n \minus{} L$.  If you're like me and hate guessing, the [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]characteristic equations method[/url] is your friend, although understanding why it works is a little involved.

[b]Generalization:[/b]  What is $ L$ when the sequence averages its last $ k$ terms to find its new term, in terms of the first $ k$ terms?  (This is not trivial.  You first need to show that the limit exists.)  Can you generalize to an arbitrary weighted sum in terms of weights $ u_1, u_2, ... u_k$?  (The average case is $ u_i \equal{} \frac{1}{k}$.)

This is an interesting problem.

Usually in problems of this type the limit of the sequence is either independent of the initial values or it is loosely affected by the initial values (it may diverge for certain initial values).

But in this case, the limit is very dependent on the initial values of the sequence (the limit of a sequence given that recursion can be anything, and it depends on the initial values). Any limit is possible!

Of course there is the other case when the sequence almost always diverges based on the initial values.

The sequence can also oscillate.

These cases are based on the values of the roots of the characteristic equation.

As long as the weights are non-negative and add to $ 1$ (which are both implicit conditions in a "weighted average"), oscillation (and it can only be periodic) is the only non-convergent behavior (and again this must be proven).

Rewrite the reccursion as $ x_{n \plus{} 1} \plus{} \frac {1}{2}x_n \equal{} x_n \plus{} \frac {1}{2}x_{n \minus{} 1} \equal{} \cdots \equal{} x_1 \plus{} \frac {1}{2}x_0 \equal{} 1$.

$ \therefore x_{n \plus{} 1} \minus{} \frac {2}{3} \equal{} \minus{} \frac {1}{2}\left(x_n \minus{} \frac {2}{3}\right)$.

$ \therefore x_n \minus{} \frac {2}{3} \equal{} \left( \minus{} \frac {1}{2}\right)^{n}\left(x_0 \minus{} \frac {2}{3}\right)$, yielding $ x_n \equal{} \frac {2}{3} \minus{} \frac {2}{3}\left( \minus{} \frac {1}{2}\right)^n\ (n\geq 0)$.

Rider: $ x_0 \equal{} 0,\ x_1 \equal{} 1,\ x_{n \plus{} 1} \equal{} (1 \minus{} p)x_n \plus{} px_{n \minus{} 1}\ (p\neq \minus{} 1)$.

Rewite the reccursion as  $ x_{n \plus{} 1} \plus{} px_n \equal{} x_n \plus{} px_{n \minus{} 1} \equal{} \cdots \equal{} x_1 \plus{} px_0 \equal{} 1$. Since $ p\neq \minus{} 1$, we have $ x_{n \plus{} 1} \minus{} \frac {1}{1 \plus{} p} \equal{} \minus{} p\left(x_n \minus{} \frac {1}{1 \plus{} p}\right)$. $ \therefore x_n \minus{} \frac {1}{1 \plus{} p} \equal{} ( \minus{} p)^n\left(x_0 \minus{} \frac {1}{1 \plus{} p}\right) \equal{} \minus{} \frac {1}{1 \plus{} p}( \minus{} p)^n$, yielding $ \boxed{x_n \equal{} \frac {1}{1 \plus{} p}\{1 \minus{} ( \minus{} p)^n\}\ (n\geq 0)}$

[quote="mathwizarddude"]$ x_0 \equal{} 0$
$ x_1 \equal{} 1$
$ x_{n \plus{} 1} \equal{} \frac {x_n \plus{} x_{n \minus{} 1}}2$

Find $ x_n$.[/quote]
$ x_n\equal{}\frac{2}{3}\minus{}\frac{2}{3}.(\minus{}\frac{1}{2})^n$

We can classify 3 cases according to the sign of Discriminant.

By the way, From what do $ x^2\minus{}ax\minus{}b\equal{}0$ come out? :D

But no one wants to try the generalization?  :)  Try at least

$ a_{n\plus{}4} \equal{} \frac{a_{n\plus{}3} \plus{} a_{n\plus{}2} \plus{} a_{n\plus{}1} \plus{} a_n}{4}$

with initial conditions $ a_0, a_1, a_2, a_3$.  Note that it is no longer feasible to solve the characteristic equation; nevertheless, the limit can be shown to exist and can be determined.

hello, also in this case it gives an explicit formula
$ \displaystyle{a(n)\equal{}\left(\frac{1}{4}
   \left(\minus{}1\minus{}\frac{5^{2/3}}{\sqrt[3]{3 \left(\minus{}9\plus{}4
   \sqrt{6}\right)}}\plus{}\frac{\sqrt[3]{5 \left(\minus{}9\plus{}4
   \sqrt{6}\right)}}{3^{2/3}}\right)\right)^n
   c_1\plus{}\left(\minus{}\frac{1}{4}\plus{}\frac{5^{2/3} \left(1\minus{}i \sqrt{3}\right)}{8
   \sqrt[3]{3 \left(\minus{}9\plus{}4 \sqrt{6}\right)}}\minus{}\frac{\left(1\plus{}i
   \sqrt{3}\right) \sqrt[3]{5 \left(\minus{}9\plus{}4 \sqrt{6}\right)}}{8
   3^{2/3}}\right)^n c_2\plus{}\left(\minus{}\frac{1}{4}\plus{}\frac{5^{2/3} \left(1\plus{}i
   \sqrt{3}\right)}{8 \sqrt[3]{3 \left(\minus{}9\plus{}4
   \sqrt{6}\right)}}\minus{}\frac{\left(1\minus{}i \sqrt{3}\right) \sqrt[3]{5
   \left(\minus{}9\plus{}4 \sqrt{6}\right)}}{8 3^{2/3}}\right)^n
   c_3\plus{}c_4}$
Sonnhard.

But presumably not in the $ k \equal{} 6$ case!  In any case, your computation does not reveal the value of $ c_4$ :)

Let $ x_1$, $ x_2$,..., $ x_k$ be the first given terms and let the recurrence be

$ x_{n\plus{}k\plus{}1}\equal{}\frac{1}{k}\cdot (x_{n\plus{}1}\plus{}x_{n\plus{}2}\plus{}...\plus{}x_{n\plus{}k})$

It suffices to consider the case where for some $ 1\le j\le k$ we have $ a_k\equal{}1$ and $ a_n\equal{}0$ for $ x\equal{}1,2,...,j\minus{}1,j\plus{}1,...,k$.

To see this, note that we can just take a linear combination of these recurrences for each $ j$ to get the original one.

From example, it seems that the sequence converges to $ \frac{j}{\frac{(k)(k\plus{}1)}{2}  }$. So

Taking the necessary linear combination gives us $ \frac{\sum_{j\equal{}1}^k j\cdot a_j}{ \frac{(k)(k\plus{}1)}{2}}$

In the particular case, $ (k,a_1,a_2)\equal{}(2,0,1)$ so the answer is $ \frac{2}{3}$.

Of course, I didn't actually prove anything yet. This is an interesting problem...I always think that the topic of linear recurrences has been pounded into the ground but then something interesting always comes up.

[quote="chien than"]We can solve problem
$ x_0 \equal{} \alpha;x_1 \equal{} \beta$
$ x_n \equal{} ax_{n \minus{} 1} \plus{} bx_{n \minus{} 2}$
$ \alpha;\beta;a;b \in R$
Find $ x_n$
[hide]Use equation $ x^2 \minus{} ax \minus{} b \equal{} 0$[/hide][/quote]

yes , i have read a book about this kind of equation <a Vietnamese book > . And the book basically it says that we can solve :
$ x_0 \equal{} \alpha_1 , ... x_i \equal{} \alpha_i$
$ x_n\equal{} \sum^{i}_{k\equal{}1} a_kx_{n\minus{}k}$
by solving $ f(x) \equal{} x^i \minus{} \sum^{i\minus{}1}_{k\equal{}0} a_kx^{k}$


geometry
circumcircle
inequalities
inequalities proposed

Let ABC be a triangle. Prove that:

$2(m_al_c+m_bl_b+m_cl_a)\leq3\sqrt{3}Rp$

Could you clarify your notation? The only thing I'm sure of here is that R is the circumradius.

$m_a$median
$l_a$bisector
$R$circumradius
$p$semiperimeter

Very easy, but nice. This follows from Cauchy-Schwarz inequality, the inequality $ 9R^2\geq a^2+b^2+c^2 $, the identity $ m_a^2+m_b^2+m_c^2=\frac{3}{4}\cdot (a^2+b^2+c^2) $  and the fact that $ l_a\leq\sqrt{p(p-a)} $. I dislike geometric inequalities, but this one was so nice

Thanks. It also could have been done with Cebasev


Hello,

I am having some trouble with the following questions and I would appreciate any help or hints. Also, my work is shown below.

Thank you very much.

---

3. For what values of $ x$, do we have $ \log_x 5 > \log_x 10$ ?

[u]My Work :[/u]

3.

$ \log_x 5 > \log_x 10$

$ \log_x 5 \minus{} \log_x 10 > 0$

$ \log_x (1/2) > 0$

Is there anything significant that the final step provides? Or would I have to solve this problem with logic?

---

[b]ANSWERED Questions:[/b]

1. There are two radioactive isotopes of iron. Isotope A decays so that after every 6 minutes, the number of atoms remaining is halved. Initially, there are twice as many atoms of isotope A as of isotope B, and after 24 minutes there are the same number of atoms of each isotope. How long does it take the number of atoms of isotope B to halve?

2. If $ \log_8 3 \equal{} k$, express $ \log_8 18$ in terms of k.

[u]My Work :[/u]

1. I find this to be the most challenging question. Could someone please provide me with a hint as to getting the equation?

I tried something like this: $ a \equal{} a/2$ but I do not know how to incorporate the half-life of 6 minutes into the equation.

2. $ \log_8 3 \equal{} k$

$ 8^k \equal{} 3$

$ k \equal{} \frac {\log 8}{\log 3}$

How would this enable me to express $ \log_8 18$  in terms of k?

4. Determine all values of $ x$ for which $ ((\sqrt {x})^{\log_{10} x} \equal{} 100$.

4. $ ((\sqrt {x})^{\log_{10} x} \equal{} 100$

$ ((x^{1/2})^{\log_{10} x} \equal{} 100$

Which of the two forms below should I express this as?

$ (x^{\frac {log_{10} x}{2}}) \equal{} 100$ or $ {(x)^{log_{10} \sqrt {x}}} \equal{} 100$

[hide="Two."]
$ \log_{8} \left( 3 \cdot 3 \cdot 2 \right) \equal{} \log_{8} 3 \plus{} \log_{8} 3 \plus{} \log_{8} 2 \equal{} 2k \plus{} \frac {1}{3}$.
[/hide]

Solution for #1:

[hide]Isotope A has gone through 4 half-lives in 24 minutes, so there is 1/16th (2^4=16) of the original amount left.  Since the amounts of isotopes A and B are equal after 24 minutes (and there originally was twice as much of A as B), then there must be 1/8th of the original amount of Isotope B left.  Therefore, B has gone through 3 half-lives (2^3=8) and each half-life is 8 minutes long.[/hide]

Thank you for your replies, AstroPhys and cfs249. I appreciate your help.

I also have rearranged my order of questions so that the solved ones are now at the bottom.

[hide="4"]
$ \sqrt{x}^{\log x}\equal{}100$
Taking log of both sides
$ \log \sqrt{x}^{\log x}\equal{}\log 100$
$ (\log x)(\log \sqrt{x})\equal{}2$
$ (\log x)(\log x^{\frac{1}{2}})\equal{}2$
$ (\log x)(\frac{1}{2}\log x)\equal{}2$
$ \frac{1}{2}(\log x)^{2}\equal{}2$
$ (\log x)^2\equal{}4$
$ \log x\equal{}2, \minus{}2$
$ x\equal{}\frac{1}{100}, 100$
[/hide]


The sum of $ 49$ consecutive integers is $ 7^5$. What is their median?

$ \textbf{(A)}\  7 \qquad \textbf{(B)}\  7^2\qquad \textbf{(C)}\  7^3\qquad \textbf{(D)}\  7^4\qquad \textbf{(E)}\ 7^5$

[hide="Solution"]C.

Notice that because the terms are consecutive integers, the median is simply $\frac{7^5}{7^2}$.[/hide]

[hide]suppose our sequence is n-24,...,n,...n+24, sum is 49n, and median is n, so 7^5/7^2=7^3[/hide]

[hide]You can write it the sum like this:
$(x-24)+(x-23)+\ldots+(x-1)+x+(x+1)+\ldots+(x+23)+(x+24)=7^5$
If you group the first term and last term; second term and second to last term; and so on (there are 24 pairs of these, and each pair adds up to 2x), you get:
$24 \cdot 2x+x=49x=(7^2)x=7^5$
$\Rightarrow x=7^3=\boxed{343}$[/hide]

[quote="Tiago Guerreiro"]Hello  :D

[hide="My Solution"] Let $M$ be the median:
$x+(x+1)+(x+2)+\ldots+(x+48)=7^5 \longrightarrow 49x+1225=7^5 \longrightarrow x=318$
$M=\frac{318+(318+49)}{2}=342,5$ Which is approximatly $343=7^3$[/hide][/quote]

I'm confused, are you allowed calculators on the AMC ??

[quote="perfectnumber628"]yeah, they let you use calculators for the AMC.  not for the AIME though.[/quote]

Um, What grade do you write the AIME???


factorial
floor function
algorithm
search

The problem is simple:
find TOTAL number of zeros in factorial of number N
(not calculating N! of course)

I think there is some kind of method using mods but unfortunately, I don't remember it.

The problem isn't simple at all.  While the expression for the number of [i]trailing[/i] zeroes is well-known and fairly straightforward, the rest of the digits are much more difficult to analyze.

To reinterpret t0r's "more difficult to analyze," there is no elementary solution to this problem or probably even any known solution at all.

[hide="Trailing zeros for base 10"]$ \left\lfloor \frac{n}{5} \right\rfloor \plus{} \left\lfloor \frac{n}{5^2} \right\rfloor \plus{} \left\lfloor \frac{n}{5^3} \right\rfloor \plus{} \ldots$[/hide]

[quote="MellowMelon"]To reinterpret t0r's "more difficult to analyze," there is no elementary solution to this problem or probably even any known solution at all.
[/quote]

The solution should exist cause this problem was taken from some International Math Olympiad.
Calculating number of trailing zeros of N! is well known and simple problem.

[quote="Igort75"] The solution should exist cause this problem was taken from some International Math Olympiad. [/quote]

Could you give reference of the "some" International Math Olympiad? (I don't know any solution if we include inside zeroes in the computation)

[quote]Could you give reference of the "some" International Math Olympiad? (I don't know any solution if we include inside zeroes in the computation)[/quote]

I'll try to find source but I'm afraid it will not be easy... :(
But I'm absolutely sured that I've seen this problem on some math olymp..

Forget to mention that this task was also offered on Google interview.

A google interview? Perhaps it was meant to be done by a computer algorithm? That might work somehow, but I strongly doubt there is a solution up to mathematics' standards. Maybe you saw this problem in a computing olympiad.

[quote="JRav"]Could somebody provide an explanation as to why the trailing zeroes method works?[/quote]

The number of zeros which a number has following it, is the number of $ 10$s there are in its factors. Since there are enough $ 2$s to balance out the $ 5$s which arise from the factorial, we only need to consider how many of them there are. This is done by dividing by $ 5$ to find all the factors which are $ 5$ then by $ 5^{2}$ to get all of the ones which are missed the first time etc...

Well in genl there's a formula .
For the max value of x which satisfies $ \ce{a^x}$ such that $ \ce{a^x}$ is a factor of n! :
x= $ gif(n!/a) \plus{} gif(n!/\ce{a^2}) \plus{} .....\plus{} gif(n!/\ce{a^p})$
such that the final term is the the smallest term greater than 0 .
Search for the proof  :D

Not quite.  The expression is only valid for $ a$ prime, and should take the form

$ \left\lfloor \frac {n}{a} \right\rfloor \plus{} \left\lfloor \frac {n}{a^2} \right\rfloor \plus{} \left\lfloor \frac {n}{a^3} \right\rfloor \plus{} ...$

(It's perfectly legitimate to allow the terms to continue without end, since all the terms after a certain point are just $ 0$ anyway.)

Igort75:  If this was an Olympiad problem, it asked about a specific $ N$, not a general expression for all $ N$.


Έστω $ x,y,z$ θετικοί πραγματικοί ώστε $ xyz\equal{}8$. Αποδείξτε ότι $ \frac{x^2}{\sqrt{(x^3\plus{}1)(y^3\plus{}1)}}\plus{}\frac{y^2}{\sqrt{(y^3\plus{}1)(z^3\plus{}1)}}\plus{}\frac{z^2}{\sqrt{(z^3\plus{}1)(x^3\plus{}1)}}\geq\frac{4}{3}$.

[quote="NickNafplio"]Ισχυει:

$ \sqrt {x^3 \plus{} 1} \leq \frac {x^2}{2} \plus{} 1$

οποτε αρκει:

$ \sum_{cyclic}{\frac {x^2}{(\frac {x^2}{2} \plus{} 1)(\frac {y^2}{2} \plus{} 1)}} \geq \frac {4}{3}$

η

$ \sum_{cyclic}{x^2(\frac {z^2}{2} \plus{} 1)} \geq \frac {4}{3}(\frac {x^2}{2} \plus{} 1)(\frac {y^2}{2} \plus{} 1)(\frac {z^2}{2} \plus{} 1) \equal{} \frac {4}{3}(9 \plus{} \frac {1}{2}(x^2 \plus{} y^2 \plus{} z^2) \plus{} \frac {1}{4}(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2))$

Το οποιο μετα τις πραξεις ισοδυναμει με:

$ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \plus{} 2(x^2 \plus{} y^2 \plus{} z^2) \geq 72$
που ειναι προφανες απο ΑΜ-ΓΜ.[/quote]

Α ρε Νίκο! και συ τέτοια ώρα ξύπνιος?
Τελοςπαντων, δεν μπορω να ελενξω τη λυση σου τωρα γιατι ειμαι κουρασμενος, αλλα ειμαι σιγουρος οτι ειναι σωστη. Καληνυχτα :D

Πράγματι, είσαι πολύ σωστος! Δείτε και εδω:
http://gbas2010.wordpress.com/2010/01/18/inequality-30apmo-2005/


Let ABC  such that A$ \geq 120^0$
FIND  MIN
P=$ tan\frac{A}{2}\plus{}tan\frac{B}{2}\plus{}tan\frac{C}{2}$

[quote="tranquoc"]Let ABC  such that A$ \geq 120^0$
FIND  MIN
P=$ tan\frac {A}{2} \plus{} tan\frac {B}{2} \plus{} tan\frac {C}{2}$[/quote]
Let $ \tan\frac{A}{4} \equal{} x$, $ \frac {1}{\sqrt {3}}\le x < 1$
$ P\ge\tan\frac {A}{2} \plus{} 2\tan\frac {B \plus{} C}{4} \equal{} \tan\frac {A}{2} \plus{} 2\tan(\frac {\pi}{4} \minus{} \frac {A}{4}) \equal{} \frac {2x}{1 \minus{} x^2} \plus{} \frac {2(1 \minus{} x)}{1 \plus{} x} \equal{} \frac {2x^2 \minus{} 2x \plus{} 2}{1 \minus{} x^2}$
With $ \frac {1}{\sqrt {3}}\le x < 1$


Can some one tell me the solution to the problem below?

(IMO 1974/5) Determine all possible values of

                      S= a/(a+b+d) + b/(a+b+c) + c/(b+c+d) + d/(a+c+d)

where a, b, c, d are arbitary positive numbers.

It's obvious that $S > \sum \frac{a}{a + b + c + d} = 1$ and $S < a/(a + d) + b/(b + c) + c/(b + c) + d/(a + d) = 2$.
Now try to show that all numbers in $(1,2)$ are attainable using a continuity argument.
Do you see how to do this?

I am sure this one was posted before... also check:

http://www.kalva.demon.co.uk/imo/isoln/isoln745.html