Factor the following polynomial into  factors with integer coefficients.
\[ x^8\plus{}98x^4\plus{}1\]

After a bunch of guess and check (I used symmetry to cancel stuff),
[hide]$ (x^4 \plus{} 4x^3 \plus{} 8x^2 \minus{} 4x \plus{} 1)(x^4 \minus{} 4x^3 \plus{} 8x^2 \plus{} 4x \plus{} 1)$
[/hide]

How is it known to make that substitution?

Warning: Ridiculously long post ahead  :lol:

If we reverse the coefficients of a polynomial, we get a new polynomial whose roots are the reciprocals of the respective roots of the original polynomial.  Therefore, in a case like this where the coefficients of our polynomial are symmetric, we have the interesting property that if $ x$ is a root (where we allow $ x$ to be complex), then $ 1/x$ is also a root.  Now look at all possible values of $ x \plus{} 1/x$ where $ x$ is a root.  Since $ x \plus{} 1/x \equal{} (1/x) \plus{} 1/(1/x)$, we can pair up our roots with their reciprocals based on which values of $ x \plus{} 1/x$ they produce.

Still, some weird stuff can happen.  What if we have a double root at $ k$ but only a single root at $ 1/k$?  Well, then we can divide through by $ (x \minus{} k)(x \minus{} 1/k) \equal{} x^{2} \minus{} (k \plus{} 1/k)x \plus{} 1$, a polynomial with symmetric coefficients, to get another polynomial with symmetric coefficients, and we get a contradiction: our new polynomail has symmetric coefficients, a root at $ k$, but no root at $ 1/k$.  Since we can do this repeatedly, we know that the multiplicities of the roots $ k$ and $ 1/k$ must be the same.

This relies on the fact that if $ P(x)|Q(x)$ and $ P$ and $ Q$ have symmetric coefficients, then $ Q(x)/P(x) \equal{} R(x)$ has symmetric coefficients.  To prove this, note that $ P(x)R(x) \equal{} Q(x)$ and therefore $ P(1/x)R(1/x) \equal{} Q(1/x)$.  Multiplying both sides by $ x^{\text{deg}P \plus{} \text{deg}R} \equal{} x^{\text{deg}Q}$, we get $ x^{\text{deg}P}P(1/x)x^{\text{deg}R}R(1/x) \equal{} x^{\text{deg}Q}Q(1/x)$  But $ P(x) \equal{} x^{\text{deg}P}P(1/x)$ and $ Q(x) \equal{} x^{\text{deg}Q}Q(1/x)$, so $ R(x) \equal{} x^{\text{deg}R}R(1/x)$ also, which implies that $ R$ has symmetric coefficients.  Note that this same proof shows that the product of two polynomials with symmetric coefficients also has symmetric coefficients.

Now, what about 1 and -1, the two numbers that are their own reciprocals?  First of all, -1 can be a root with any multiplicity, since we can always multiply our polynomial by $ x \plus{} 1$ or divide our polynomial by $ x \plus{} 1$ if -1 is a root and get another polynomial with symmetric coefficients, and in the process increase or decrease the number of times -1 occurs as a root.  Because of this, let's only look at polynomials such that -1 is not a root.

If 1 is a root, then $ x \minus{} 1$ is a factor.  But $ x \minus{} 1$ doesn't have symmetric coefficients!  All our other roots are already paired off, and we're assuming -1 is not a root.  Therefore we must have another $ x \minus{} 1$ factor.  Note that $ (x \minus{} 1)^{2} \equal{} x^{2} \minus{} 2x \plus{} 1$, which has symmetric coefficents, so 1 can be a root, but only with even multiplicity.  Therefore, even if 1 is a root/$ x \minus{} 1$ is a factor, we can pair it off with another factor $ x \minus{} 1$.

What's the point of all this?  Well, assuming -1 is not a root, we've shown that any polynomial with symmetric coefficients has an even number of roots, say $ 2n$, that can be paired off such that each pair consists of numbers that are reciprocals of each other.  Each pair of roots $ x$ and $ 1/x$ produces one value of $ x \plus{} 1/x$.  Let the possible values of $ x \plus{} 1/x$ be $ a_{1},a_{2},\ldots ,a_{n}$, where we count a value twice if it appears for two different pairs and so on.  Now let $ P(y)$ be the polynomial with roots $ a_{1},a_{2},\ldots ,a_{n}$.  But $ P(x \plus{} 1/x)$ now has exactly the same roots (even down to multiplicity) as our original polynomial, and $ x^{n}P(x \plus{} 1/x)$ is the same as our original polynomial down to scaling.

Why is this useful?  It means that any polynomial with symmetric coefficients, assuming -1 is not a root, can be divided through by an appropriate power of $ x$ and then rewritten as a polynomial in $ x \plus{} 1/x$ in a unique way.

Let's readdress the -1 issue.  Clearly if we have a single root at -1, or -1 is a root with any odd multiplicity, then our polynomial has odd degree, which prevents us from dividing through by a power of $ x$ in a symmetric way.  For example, consider the polynomial with symmetric coefficients $ x \plus{} 1$.  Clearly $ x \plus{} 1$ can't be written as a polynomial in $ x \plus{} 1/x$, but dividing by $ x$ doesn't help either, since $ 1 \plus{} 1/x$ still can't be written as a polynomial in $ x \plus{} 1/x$.  This same problem arises for every polynomial with odd degree.  But if -1 is a root with an even multiplicity, then we don't actually have any problems, since $ (x \plus{} 1)^{2} \equal{} x^{2} \plus{} 2x \plus{} 1 \equal{} x((x \plus{} 1/x) \plus{} 2)$.

So we can actually revise our conclusion: [b]Any even-degree polynomial[/b] with symmetric coefficients can be divided through by an appropriate power of $ x$ and then rewritten as a polynomial in $ x \plus{} 1/x$ in a unique way.

Shortened version: If stuff is symmetric, the substitution $ u \equal{} x \plus{} 1/x$ helps a lot.  arqady's solution used $ u\equal{}x\minus{}1/x$, but you could just as well use $ u\equal{}x\plus{}1/x$.  The substitution $ u\equal{}x\minus{}1/x$ works well in some cases where you have symmetric coefficients down to sign, but some of the signs are slightly off.

Also: You can prove what I proved above fairly easily by induction, so this is fairly pointless.  But, Po-Shen taught me never to use induction, so this is what you get.  :D

Long, but very educational and well motivated. Nicely done!

That discussion can be phrased without talking about roots.  The underlying structure is that of a symmetric polynomial in two variables; make the substitution $ x \equal{} \frac {a}{b}$, clear denominators, cite the fundamental theorem of symmetric polynomials, and note that $ x \plus{} \frac{1}{x} \equal{} \frac{a^2 \plus{} b^2}{ab} \equal{} \frac{(a \plus{} b)^2}{ab} \minus{} 2$.

[quote="t0rajir0u"]That discussion can be phrased without talking about roots.  The underlying structure is that of a symmetric polynomial in two variables; make the substitution $ x \equal{} \frac {a}{b}$, clear denominators, cite the fundamental theorem of symmetric polynomials, and note that $ x \plus{} \frac {1}{x} \equal{} \frac {a^2 \plus{} b^2}{ab} \equal{} \frac {(a \plus{} b)^2}{ab} \minus{} 2$.[/quote]

Ah, that's very nice.  Although it remains to prove that the resulting polynomial can be expressed in terms of $ \frac{(a\plus{}b)^{2}}{ab}$, that's not hard to do at all  :)

[quote="matt276eagles"]
Shortened version: If stuff is symmetric, the substitution $ u \equal{} x \plus{} 1/x$ helps a lot.  arqady's solution used $ u \equal{} x \minus{} 1/x$, but you could just as well use $ u \equal{} x \plus{} 1/x$.
[/quote]
But without success in our case.  :wink:
For $ x^8 \plus{} 34x^4 \plus{} 1$ it helps, but it's not interesting.


Fie $O$,$A$,$B$,$C$ patru puncte necoplanare a.i. $OA \perp OB \perp OC \perp OA$. Ducem $MP \perp OB$ si $MQ \perp (AOC)$, unde $M$ se afla in interiorul triunghiului $ABC$.
Care este locul geometric al lui $M$, daca $PQ$ are valoare constanta?

Dacă înţeleg eu bine care-s $P$ şi $Q$, atunci $OM$ şi $PQ$ sunt diagonale mari în acelaşi paralelipiped, deci sunt egale. Locul geometric căutat este intersecţia unei sfere de rază $r$ cu interiorul triunghiului $ABC$ (adică ori $\emptyset$, ori un cerc ori nişte bucăţi de cerc - cu centrul în proiecţia lui $O$ pe $(ABC)$).


Hello everyone,

1. Two perpendicular lines with x-intercepts -2 and 8 intersect at (0, b). Determine both possible values for b.

2. [ (x - 1)/(x + 1) + (x + 1)/(x - 1) / ((x - 1)/(x + 1)) - ((x + 1)/(x - 1))]

I got the answer for #2 a while ago and I would just like to know the rules for the orders of operations. Can I simplify the numerator and denominator and then divide or do I have to divide right away?

Thank you in advance.

--

[u]My work:[/u]

1. I drew a graph for #1 and I figured out that -4 was one of the possible values for b. I denoted the two lines as:

y = 2x + b
y = -8x + b

2x + b = -8x + b
2x = -8x = -4

However, how do I find out the other possibility? Is my method here mathematically correct?

1. Let $A(-2,\ 0),\ B(8,\ 0)$ we have a circle with diameter $AB$. Thus the point $(0,\ b)$ is the intersection point of  the circle and $y$-axis,
so power theorem gives $|b|^{2}=2\cdot 8$ , yielding $b=\pm 4$.

per'aps we can use the $TeXeR$ for numero dos.

[hide="#2"]
Heres my rule of the thumb.

$ \frac{\left(\frac{x-1}{x+1}\right)+\left(\frac{x+1}{x-1}\right)}{\left(\frac{x-1}{x+1}\right)-\left(\frac{x+1}{x-1}\right)}$

Four layers. T[op]B[ottom]B[ottom]T[op] from the top or bottom. Thats where all the terms end up. I would simplify first though.
[/hide]


Let ${sinx}^{2}=a$
${cosx}^{2}+1=b$
We have $\sqrt{1+a^{2}}+\sqrt{1+b^{2}}=1+\sqrt{5}$
And $a+b=2,0\le a\le 1,1\le b\le 2$
Then it's easy to get the answer
a=0 and b=2
So $x=k\pi$ with $k\in Z$

[quote="hikaru123"]Let ${sinx}^{2}=a$
${cosx}^{2}=1=b$
We have $\sqrt{1+a^{2}}+\sqrt{1+b^{2}}=1+\sqrt{5}$
And $a+b=2,0\le a\le 1,1\le b\le 2$
Then it's easy to get the answer
a=0 and b=2
So $x=k\pi$ with $k\in Z$[/hide][/quote]

[quote="vmsvl"]Solve the equal
$\sqrt{1+\sin^{4}x}+\sqrt{\cos^{4}x+2\cos^{2}x+2}= 1+\sqrt 5$[/quote]

[hide]
$\sqrt{1+(1-cos^{2}x)^{2}}+\sqrt{1+(1+cos^{2}x)^{2}}= 1+\sqrt5$

let $u = cos^{2}x$

$\sqrt{1+1-2u+u^{2}}+\sqrt{1+1+2u+u^{2}}= 1+\sqrt5$

$\sqrt{u^{2}-2u+2}+\sqrt{u^{2}+2u+2}= 1+\sqrt5$

$u\in [0,1]$, so the only solution is $u=1$, so I'm getting $x=k\pi$ as well.
[/hide]

Care to explain why hikaru's was wrong? It wasn't formatted perfectly, but it seems to work.

Sorry because I didn't type the solution correctly. I edited it so my solution isn't false.

[quote="hikaru123"]Let ${sinx}^{2}=a$
${cosx}^{2}+1=b$
[/quote]

just to avoid confusion u mean $\cos^{2}{x}$ and $\sin^{2}{x}$


x,y,z are reals
 solve the equation x+ 2/x=2y
                                y+2/y=2z
                                z+2/z=2x

Naming a topic "Algebra" and posting it in "Number Theory" even if it has nothing to do with number theory is considered a serious crime. You and your thread are thus sent to the Algebra forum for all eternity. The thread shall now infamely be known as "x+2/x=2y and cyclic".

Firstly, it's easy to see that $ x,y,z$ are all positive or all negative.
[b]Case 1[/b]:all of $ x,y,z$ positive
As $ x,y,z \neq 0$, we rewrite the first equation into $ x^2\minus{}2yx\plus{}2\equal{}0$ , to have x real , we know $ (\minus{}2y)^2\minus{}4(1)(2) \ge 0$, so $ y^2 \ge 2$, similarly as $ z^2$ and $ x^2$.
Add the 3 original equation together we get $ \frac{x\plus{}y\plus{}z}{2}\equal{}\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}$
But from $ x^2, y^2, z^2 \ge 2$ we get $ \frac{y}{2} \ge \frac{1}{y}$,etc
Because equality holds, $ x^2\equal{}y^2\equal{}x^2\equal{}2$
$ (\sqrt {2},\sqrt {2}, \sqrt {2})$ are the only solution
[b]Case 2[/b]:all of $ x,y,z$ negative
Symmetrically $ (\minus{}\sqrt {2},\minus{}\sqrt {2}, \minus{}\sqrt {2})$ are the only solution.


function
real analysis
real analysis unsolved

Let $f_n$ be a familly of real increasing functions defined over an open interval such that the sequence $f_n(x)$ is bounded for every $x$.Show that there exists a subsequence $k_n$ such that $f_{k_n}(x)$ converges for every $x$.

Basically you need to use the "diagonal process" to create a subsequence which converges on a countable dense subset $D$ of numbers in your interval (easy); you can also assume this sequence to be increasing [b]for each fixed $x \in D$[/b] when $n$ is large enough (depending on $x$) (*). Then you say that the set $E$ of the points of discontinuity of all the $f_{n}$ is at most countable, whence without loss of generality one can assume $E \subseteq D$. Now using the continuity of $f_{n}$ at every point $x \notin D$, the density of $D$, the monotonicity of the $f_{n}$ and (*), it becomes easy to show that the given subsequence converges everywhere.

Yeap,that's the idea.I'll wait some more,maybe someone will post a complete solution.If not I'll post it.


calculus
integration
trigonometry
calculus computations

Prove that $ \int_{0}^{\frac{\pi}{3}} (1\plus{} \tan^{6}\theta)\ d\theta \equal{} \frac{9\sqrt{3}}{5}$.

Let $ I_n\equal{}\int_{0}^{\frac{\pi}{3}}\tan^n{\theta}d\theta$.

$ I_n\equal{}\int_{0}^{\frac{\pi}{3}}\tan^n{\theta}d\theta\equal{}\int_{0}^{\frac{\pi}{3}}\tan^{n\minus{}2}\theta\cdot\tan^2{\theta}d\theta$
$ \equal{}\int_{0}^{\frac{\pi}{3}}\tan^{n\minus{}2}\theta\cdot\sec^2{\theta}d\theta\minus{}\int_{0}^{\frac{\pi}{3}}\tan^{n\minus{}2}\theta{d\theta}\equal{}\frac{1}{n\minus{}1}(\sqrt{3})^{n\minus{}1}\minus{}I_{n\minus{}2}$.

$ I_{0}\equal{}\frac{\pi}{3}$, so $ I_2\equal{}\sqrt{3}\minus{}\frac{\pi}{3}$.
$ I_4\equal{}\frac{1}{3}(\sqrt{3})^3\minus{}I_2\equal{}\frac{\pi}{3}$, so we get $ I_6\equal{}\frac{1}{5}(\sqrt{3})^5\minus{}I_4\equal{}\frac{9\sqrt{3}}{5}\minus{}\frac{\pi}{3}$ for finally.

$ \therefore \int_{0}^{\frac{\pi}{3}}(1\plus{}\tan^{6}\theta){d\theta}\equal{}\frac{\pi}{3}\plus{}\frac{9\sqrt{3}}{5}\minus{}\frac{\pi}{3}\equal{}\frac{9\sqrt{3}}{5}$.

[quote="AndrewTom"]Prove that $ \int_{0}^{\frac {\pi}{3}} (1 + \tan^{6}\theta)\ d\theta = \frac {9\sqrt {3}}{5}$.[/quote]

${ \int_{0}^{\frac {\pi}{3}} (1 + \tan^{6}\theta)\ d\theta =[\tan \theta -\frac 13 \tan ^ 3 \theta +\frac 15\tan ^ 5 \theta}]_0^{\frac{\pi}{3}}=\frac 95 \sqrt{3}$

How did you compute the definite integral of $ 1\plus{}\tan^6{\theta}$ so quickly? Is it similar as my method?

Just remark that $ 1 \plus{} \tan ^ 6 \theta \equal{} (1 \plus{} \tan ^ 2 \theta)(1 \minus{} \tan ^ 2 \theta \plus{} \tan ^ 4 \theta)$

Looks to me like kunny used $ (1 \plus{} x^{3}) \equal{} (1 \plus{} x)(1 \minus{} x \plus{} x^{2})$ with $ x \equal{} \tan^{2}\theta$, so $ (1 \plus{} x) \equal{} \sec^{2}\theta$ and the result follows immediately.

Neatly done, kunny!

Oops, our posts crossed; sorry kunny.


factorial
floor function
search
combinatorics unsolved
combinatorics

Show that there is a natural number $n$ such that $n!$ when written in decimal notation ends exactly in 1993 zeros.

The number of trailing zeros in decimal representation of n! equals the power of 5 dividing n! that is
$d(n) = \left\lfloor\dfrac{n}{5}\right\rfloor + \left\lfloor\dfrac{n}{5^2}\right\rfloor + \left\lfloor\dfrac{n}{5^3}\right\rfloor + \dots$
We need to find such n that $d(n)=1993$.

Note that
$1993 = d(n) \leq \dfrac{n}{5} + \dfrac{n}{5^2} + \dfrac{n}{5^3} + \dots = \dfrac{n}{4}$
implying $n\geq 4\times 1993 = 7972$.

It is useful to note that $d(m) = d\left(5 \left\lfloor\dfrac{m}{5}\right\rfloor\right)$ for any m. Therefore, we may search for n being a multiple of 5.

Subsequently testing $7975, 7980, 7985$ we find that $d(7985)=1993$.


F is a bijective transformation of the plane into itself such that it maps rectangles into rectangles (rectangles are closed and continuity is not assumed).
Prove that F maps squares into squares !

By rectangle do you mean the whole surface of a rectangle or just its sides?

F maps line segments into line segments, sides of rectangles into sides of rectangles, vertices of rectangles into vertices of rectangles and perpendicular line segments into perpendicular line segments.

According to the last post of Orl, the conclusion holds trivially :
If ABCD is a square, then F(A)F(B)FC)F(D) is a rectangle, whose diagonals are the images with respect to F of the diagonals of ABCD, and then are perpendicular. Thus F(A)F(B)F(C)F(D) is a square.

I think the problem would be more interseting if the statement was :
Let F a bijection of the plane into itself which maps closed rectangles (that is a rectangle with its interior) into rectangles. Then F maps squares into squares.

Thus, you have to PROVE that F maps line segments into line segments, sides into sides vertices into vertices and perpendicular line segments into perpendicular line segments.

This may be done as follows :

Let M' = F(M) for every point M.

Let R = ABCD be a rectangle, with center O. Let X,Y,Z,T be the midpoints of sides AB,BC,CD and DA respectively.

Since F is a bijection, we have F(YT) = F(ABYT  \cap CDTY) = F(ABYT) \cap F(CDYT), where F(ABYT) and F(CDYT) are rectangles.
Suppose that F(YT) is not a side of F(ABYT) : Then, since F(YT) is contained in F(CDYT) and that F(ABYT \cup CDYT) = F(R) = R' is a rectangle, it follows that F(CDYT) is a rectangle which intersects F(ABYT) at some other points than F(YT), contradicting the relation above.
Thus, F(YT) is a side of F(ABYT), and in the same way, is a side of F(CDYT).

Since R' is a rectangle, two parallel sides of R' are sides of F(ABYT) and F(CDYT) respectively. Then :
a) F(YT) is a segment parallel to the sides of R', and whose vertices belong to the sides of R'.
or
b) F(YT) is a rectangle whose vertices belong to the sides of R', and whose sides are parallel to the sides of R'.

With the same reasoning, we deduce that F(XZ) satisfies the same conclusion as above.

Since YT \cap ZX = {O}, it follows that F(YT) \cap F(ZX) = {F(O)} is a singleton. But, from the above conditions, such a situation is possible only if F(YT) and F(XZ) are perpendicular line segments.

It easily follows that F maps segments into line segments, sides into sides vertices into vertices and perpendicular line segments into perpendicular line segments, and we are done.

Pierre.

I think that's what orl intended. I think all that stuff in his last post wasn't given, but it had to be proven, and orl just wanted to help people out (I think).

You are right, grobber. These were the feature which have to be discovered. And if you want to use it to solve the problem, you have to prove it. Well done, Pierre.


Normally, this game is for 2 people, but I'll just take everyone's submissions and do a round-robin.
I'll accept around 15-20 submissions before running the game.

[u]Rules:[/u]
1. Both you and your opponent have 100 troops that can be assigned to 10 battlefields that you want to capture (for example, you can assign 10 troops to each field, or 20 troops to each of the first 5 fields and 0 troops to each of the last 5 fields). You will PM your choices to me.
1b. (note that the order of the fields matters!)

2. For each battlefield, the person who assigns more troops to it wins the field.

3. After the winner of each field is decided, the person who wins more fields wins the match (or it can be a tie).

4. There will be two different games that will be scored in different ways, to make it more interesting:
First method (game 1): If you win a match, you get 2 points, a draw is 1 point, and a loss is 0 points, regardless of the margin of victory.
Second method (game 2): The number of points you score in a match is the number of fields you win.

5. To join, just send me a PM with your submissions. Make the title "Colonel Blotto" and make the message body in the form [code]
Game 1: a1-a2-a3-a4-a5-a6-a7-a8-a9-a10
Game 2: b1-b2-b3-b4-b5-b6-b7-b8-b9-b10[/code]
where $ ai$ or $ bi$ is the number of troops you assign to field $ i$.

6. The winners will receive bragging rights.

[u]Examples:[/u]
Player 1 decides to assign the troops in the order 10-10-10-10-10-10-10-10-10-10
Player 2 assigns his troops in the order 12-11-11-11-11-11-11-11-11-0
Player 2 wins 9 out of 10 fields and wins the match, pwning player 1 who only wins 1 field.

Player 1 does 25-25-25-25-0-0-0-0-0-0
Player 2 does 0-0-0-0-0-20-20-20-20-20
Player 2 wins 5 fields compared to player 1's 4 and wins the match, although by a narrow margin.

$ \mathcal{IN}$

When will you give the results?

[quote="zapi2007"]haha, this is like a tiebreaker I remember for ARML local[/quote]
Oh god, I failed that one.

/in

Don't understand the rules completely, but sounds fun.


[quote="Myth"]Myth is out of here. Bye [/quote]
This topic is about Myth, one of our MathLinks prolific user. I really don't know what happened to Myth because:

1) His Avatar has a funny circle prohibition meaning "No Myth" like those "No Smoking" signs

2) His Signature says "Myth is out of here. Bye "

I'm just curious as to what happened to the 'real' Myth... :rotfl:

PS: Myth, if you see this topic, please respond!!!  :lol:

lol...right when you said that he started visiting the forums again and he's posted in some threads.


Euclid 93
10.
a changing point of a sequence is defined by a term (other than the first term) such that this term differs from the previous term.  find the average number of changing points for all possible binary sequences of p zeros and q ones and p is less than or equal to q.

I think answer is 2(p+q-1)*C(p+q-2,p-1)/C(p+q,p)=2pq/(p+q).

Is that correct?

I wrote a simulation and it matched your formula.  how ever did you come up with the total number of changing points?

Let us calculate number of sequences ...10... where '10' on k-th place (k=1,2,...,p+q-1). It is obvious this number equals to C(p+q-2,p-1). The same for ...01...  So total number of changing points in all sequences is 2(p+q-1)*C(p+q-2,p-1).


Krista put 1 cent into her new bank on a Sunday morning.  On Monday she put 2 cents into her bank.  On Tuesday she put 4 cents into her bank, and she continued to double the amount of money she put into her bank each day for two weeks.  On what day of the week did the total amount of money in her bank first exceed $ \$2$?

Monday: 2 cents
Tuesday: 4 + 2 = 6 cents
Wednesday: 8 + 6 = 14
Thursday: 16 + 14 = 30
Friday: 32 + 30 = 62
Saturday: 64 + 62 = 126
Sunday: 128 + 126 = 254

Thus, our answer is $ \boxed{\text{Sunday}}$.

For int n, $ 2^0\plus{}2^1\plus{}\cdots \plus{}2^{n\minus{}1} \plus{} 2^{n}$, the sum would be $ 2^{n\plus{}1}\minus{}1$. The smallest power of 2 that is over 200 is 256, or $ 2^8$. Thus, It would take 8-1=7 days to have over $ \$2.00$ in the bank, which would be a Sunday.


The ratios are R:W=2:3 and W:B=4:5. How can I combine them? I could find, R/B=8/15, but then how can I find R:W:B=??

thanks.

hint:
ratio means divide
$ r: w\equal{}\frac{r}{w}$

$ \frac{R}{W}\equal{}\frac{2}{3}$, $ \frac{W}{B}\equal{}\frac{4}{5}$

Convert the fractions so that $ W$ is the same in both fractions

$ \frac{R}{W}\equal{}\frac{8}{12}$, $ \frac{W}{B}\equal{}\frac{12}{15}$

$ R: W: B\equal{}8: 12: 15$

[quote="Carbon57"]$ \frac {R}{W} \equal{} \frac {2}{3}$, $ \frac {W}{B} \equal{} \frac {4}{5}$

Convert the fractions so that $ W$ is the same in both fractions

$ \frac {R}{W} \equal{} \frac {8}{12}$, $ \frac {W}{B} \equal{} \frac {12}{15}$

$ R: W: B \equal{} 8: 12: 15$[/quote]

ok,I see.Thank you.


Express as a common fraction: $ \frac{. \overline{7} }{. \overline{8} }$

We get
\begin{align*}
\frac{.\overline{7}}{.\overline{8}} &= \frac{\frac{7}{9}}{\frac{8}{9}} \\
&= \boxed{\frac{7}{8}} \\
\end{align*}

Also, if you have no clue how to do this, it's not very hard to guess :wink:


algebra
polynomial
number theory solved
number theory

Let $p>2$ be a prime number. Determine all $a$ if we take: $ C_{p+k}^{k} \equiv a (mod p)$ with $1 \leq k \leq p.$

How did you reach to that answer? Can you please detail your solution!

Yes, I will detail it. For $k<p$ it suffices to observe that $ C(p+k,k)=(p+1)(p+2)...(p+k)/k!$ which is congruent to 1 because at numerator we have modulo p exactly $k!$ and $ (k!,p)=1$. For $k=p$ it follows from the fact that $ C(2p,p)=2+(C(p,1))^2+...+(C(p,p-1))^2$ and the fact that $ C(p,u)$ is divisible by $p$ for $u$ between 1 and p-1/

could you please tell where the formula comes from? I can't find at the moment a way to approach it :D.
thx

Ikap, are you kidding? Or I don't know even the easiest combinatorial identity?

I'm not kidding. i didn't knew it and just thinking for a minute I couldn't manage to work it out.
I really don't know combinatorial identities... all the combinatorial identities i have encountered over years where too complicated to keep them in mind so I thought they wouldn't help, i thought that if I will need them somewhere i will just find them at the moment :D
Allthough, it seems there are easier identities i should learn. Maybe u could suggest me where to start from :D?

There are many ways to prove the identity
\[
\binom{2n}{n} = \binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2,
\]
the easiest of which is by a direct combinatorial argument: consider the collection $\Omega$ of $n$-element subsets of $[1,2n]$ (clearly, $|\Omega| = \binom{2n}{n}$). Let $\Omega_k \subset \Omega$ be the subcollection consisting of those sets $S \in \Omega$ for which $|S \cap [1,n]| = k$. Clearly, there are $\binom{n}{k}$ choices for the $k$-element set $S \cap [1,n]$ and $\binom{n}{n-k} = \binom{n}{k}$ choices for the $(n-k)$-element set $S \cap [n+1,n]$, so $|\Omega_k| = \binom{n}{k}^2$. Since obviously $\Omega$ is the disjoint partition of the $\Omega_k$, the identity follows.

This can also be done (in half a line  ;) ) with creative telescoping of the hypergeometric summands  :D .

And by the combinatorial Sieve we can prove the more general polynomial identity
\[
\sum_{j \geq 0} \binom{n}{j}^2t^j = \sum_{k \geq 0} \binom{n}{k}\binom{2n-k}{n}(t-1)^k.
\]
[Of course, $0^0 = 1$  ;) ].

And, just a remark: the 'Littlewood Teaser,' the sum
\[
\sum_k \binom{n}{k}^3,
\]
that has long been the subject of extensive efforts, has been proven to be inexpressible in closed form.  :P

thx
i am having a headache, as a matter of fact, harazi, i surely met this before and the solution.

Yes, ikap, it is in the book for the 10-th grade!  ;)


Does anyone else take part in the "mathfax" contests that take place once a quarter in high schools?

High school?????? I took it, but I'm only in middle school. Took the geometry honors test and i believe I failed miserably like last time.
:(

A quick google search turned up the website for 'mathfax': [url]http://www.educontest.edu[/url].

The contest is 3rd through 8th grade, plus some of high school (up through advanced math, whatever that is).

It looks like it's too late to register.  :cry:  But how much does it cost?

I know very little about it.  My junior high doesn't offer it, but the high school where I take my math classes offers it to freshmen through seniors.  I really like the format.

I know the company that runs it, simply faxes the test & answers to the school.  You must take the test on the day it is faxed.

I wish my school had more of a math focus.  I've never even heard of most of the competitions that everyone is taking.

i took mathfax when i was in 8th grade. i ended up in a 3 way tie for 1st in our division. the 2 other people were also from the school i went to.
my high school doesn't offer it though... :cry:

Does anyone know where to find Mathfax tests for practice?


How many positive integers of n digits exist such that each digit is 1,2, or 3? How many of these contain all 3 digits 1,2, and 3 at least once?

This question needs refining, please put a limit on $ n$.
[hide="solution as it is now"]
The first question:
[quote="mdk"]How many positive integers of n digits exist such that each digit is 1,2, or 3?[/quote]
Is $ \infty$, as there are infinite numbers.

The second question:
[quote="mdk"]How many of these contain all 3 digits 1,2, and 3 at least once?[/quote]
Is also $ \infty$, even though there [i]are[/i] $ x$ numbers that do not contain a $ 1, 2,\text{ or }3$ there are infinite numbers that do. And, $ \infty\minus{}x\equal{}\infty$.[/hide]

[hide="solution"]$ 3^{n}$;

Numbers with 1 and 2 only: $ 2^{n}$
Numbers with 1 and 3 only: $ 2^{n}$
Numbers with 2 and 3 only: $ 2^{n}$
Numbers with 1 only: 1
Numbers with 2 only: 1
Numbers with 3 only: 1

Numbers with 1, 2, and 3: $ 3^{n}\minus{}3*2^{n}\plus{}3$[/hide]

[quote="FightTheTide"]I think mdk was asking for an answer in terms of $ n$.[/quote]
Oh..... In that case:
[hide="solution"]
For the first question:
[quote="mdk"]How many positive integers of n digits exist such that each digit is 1,2, or 3?[/quote]
If $ n\equal{}1$, then there are $ 3$ numbers which satisfy the condition, $ 1,2,\text{ and }3$.
If $ n\equal{}2$, then there are $ 9$ numbers which satisfy the condition, $ 11,12,13,21,22,23,31,32,\text{ and }33$.
If $ n\equal{}3$, then there are $ 27$ numbers which satisfy the condition, $ 111,112,113,\text{ ...},331,332,\text{ and }333$.
We can see that this follows a geometric sequence $ 3,9,27,81, ...$ so the answer is $ \boxed{3^{n}}$.

The second question:
[quote="mdk"]How many of these contain all 3 digits 1,2, and 3 at least once?[/quote]
Obviously, the $ n\equal{}1$ and $ n\equal{}2$ cases will not work for this so we start at $ n\equal{}3$.
If $ n\equal{}3$, then there are $ 6$ numbers that satisfy the condition, $ 123,132,213,231,312,\text{ and }321$.
If $ n\equal{}4$, then there are $ 36$ numbers that satisfy the condition, $ 1123,1132,1213,\text{ ...},3231,3312,\text{ and }3321$.
If $ n\equal{}5$, then there are $ 216$ numbers that satisfy the condition, $ 11123,11132,\text{ ...},33312,\text{ and }33321$
This is also a geometric sequence $ 6,36,216,1296, ...$ so the answer is $ \boxed{6^{n\minus{}2}}$.[/hide]

[quote="Kyron"]The second question:
[quote="mdk"]How many of these contain all 3 digits 1,2, and 3 at least once?[/quote]
Obviously, the $ n \equal{} 1$ and $ n \equal{} 2$ cases will not work for this so we start at $ n \equal{} 3$.
If $ n \equal{} 3$, then there are $ 6$ numbers that satisfy the condition, $ 123,132,213,231,312,\text{ and }321$.
If $ n \equal{} 4$, then there are $ 36$ numbers that satisfy the condition, $ 1123,1132,1213,\text{ ...},3231,3312,\text{ and }3321$.
If $ n \equal{} 5$, then there are $ 216$ numbers that satisfy the condition, $ 11123,11132,\text{ ...},33312,\text{ and }33321$
This is also a geometric sequence $ 6,36,216,1296, ...$ so the answer is $ \boxed{6^{n\minus{}2}}$.[/hide][/quote]

But mdk is implying that the second answer is less than or equal to the first, and $ 6^{n\minus{}2}> 3^{n}$ for lots of n.


integration
calculus
logarithms
calculus computations

$ \int_0^1 \frac {\arctan x}{1\plus{}x} \, dx$

do you rebember it?

Tks

The integral equals $ \frac{\pi\ln 2}{8}$. Integrate by parts and you get that

$ I\equal{}\frac{\pi}{4}\ln 2\minus{}\int\limits_{0}^{1}\frac{\ln(1\plus{}x)}{1\plus{}x^{2}}dx$.

For calculating the last integral use the substitution $ x\equal{}\frac{1\minus{}t}{1\plus{}t}$, and you should get it.

And here is collections of proofs.
http://www.mathlinks.ro/viewtopic.php?t=23498


Show that the function $f\left(x\right)=x^{2}$ is continuous, but not uniformly continuous.

You mean $x^{2}$ on $\mathbb{R}$?

[quote="mlok"]$f(x+\delta)-f(x)=2x\delta+\delta^{2}$[/quote]
Can you explain what you mean by that?

You can use this trick:

Let $x_{n}=\sqrt{n+1}$ and $y_{n}=\sqrt{n}$. Then

$x_{n}-y_{n}\rightarrow 0$ and $f(x_{n})-f(y_{n})=1$ which does not converge to 0, and hence the function is not uniformly convergent.


If $ a,b,c>0$ such that $ abc\equal{}1$ prove that :

$ \sum_{cyc}\frac{1}{a^3(b\plus{}c)}\ge \frac{3}{2}\sum_{cyc}\frac{ab}{a^5\plus{}b^5\plus{}ab}$

[quote="alex2008"]If $ a,b,c > 0$ such that $ abc \equal{} 1$ prove that :

$ \sum_{cyc}\frac {1}{a^3(b \plus{} c)}\ge \frac {3}{2}\sum_{cyc}\frac {ab}{a^5 \plus{} b^5 \plus{} ab}$[/quote]
Very nice :lol:
$ a^5\plus{}b^5\plus{}ab \geq a^2b^2(a\plus{}b)\plus{}ab\equal{}ab[ab(a\plus{}b)\plus{}1]\equal{}ab[ab(a\plus{}b)\plus{}abc]$
So $ \sum \frac {ab}{a^5\plus{}b^5\plus{}ab} \leq \sum \frac{ab}{ab[ab(a\plus{}b\plus{}c)]}$
Which is equivalent to $ \sum \frac {1}{ab(a\plus{}b\plus{}c)} \equal{} \sum \frac{c}{a\plus{}b\plus{}c}\equal{}1$ So the $ RHS \leq \frac{3}{2}$ So we have to prove that:
$ \sum \frac{1}{a^3(b\plus{}c)} \geq \frac{3}{2}$
Which i think was an IMO problem....
We got $ \sum \frac{a^2b^2c^2}{a^3(b\plus{}c)}\equal{}\sum \frac {(bc)^2}{ab\plus{}ac}$
So by Adreescu inequality we have that
$ \sum \frac{(bc)^2}{ab\plus{}ac} \geq \frac{(ab\plus{}bc\plus{}ca)^2}{2(ab\plus{}bc\plus{}ca)}\equal{}\frac{ab\plus{}bc\plus{}ca}{2}\geq \frac{3 \sqrt[3]{a^2b^2c^2}}{2}\equal{} \frac{3}{2}$  :)

Yes ! That is exact the solution i was thinking . Very nice

I dont' think this problem is nice,your way  is start with old problem..

try this substitution,
$ a \equal{} \frac {1}{x}$, $ b \equal{} \frac {1}{y}$, $ c \equal{} \frac {1}{z}$


Let    $ a,b,c$ positive real numbers, such that  $ a^2\equal{}11,b^2\equal{}13,c^2\equal{}48.$ It is he  true  $ a\plus{}b\equal{}c$

No. They are nearly the same: $ \sqrt{11} \plus{} \sqrt{13}$ is approximately $ 6.922176066$, whereas $ \sqrt {48}$ is approximately $ 6.92820323$, a difference of about $ 0.006027$.

[quote="AndrewTom"]No. They are nearly the same: $ \sqrt {11} \plus{} \sqrt {13}$ is approximately $ 6.922176066$, whereas $ \sqrt {48}$ is approximately $ 6.92820323$, a difference of about $ 0.006027$.[/quote]He probably wants a non-calculator proof. So square both sides to get $ 11\plus{}13\plus{}2\sqrt{143}\equal{}48$. Square again to see that they are very close. In fact, $ \sqrt{n\minus{}1}\plus{}\sqrt{n\plus{}1}\approx\sqrt{4n}$.

[quote="AndrewTom"]No. They are nearly the same: $ \sqrt {11} \plus{} \sqrt {13}$ is approximately $ 6.922176066$, whereas $ \sqrt {48}$ is approximately $ 6.92820323$, a difference of about $ 0.006027$.[/quote]

10000th User,
I was already working on this before you posted.

Original poster,

whether or not Andrew Tom used a calculator or a computer to do that, it is a good demonstration.
(I'm not meaning it was the type you wanted, though.)

Also, you could have:

$ \sqrt {11} \plus{} \sqrt {13} \equal{} ? \sqrt {48}$
$ (\sqrt {11} \plus{} \sqrt {13})^2 \equal{} ? (\sqrt {48})^2$
$ 11 \plus{} 2\sqrt {143} \plus{} 13 \equal{} ? 48$
$ 2\sqrt {143} \plus{} 24 \equal{} ? 48$
$ \sqrt {143} \plus{} 12 \equal{} ? 24$
$ \sqrt {143} \equal{} 12$ . . . False.

Yes, I used a calculator. I was going to square both sides of $ \sqrt{11} \plus{} \sqrt{13} \equal{} \sqrt{48}$ to get $ 48\equal{} 11 \plus{} 13 \plus{} 2\sqrt{143}$ and then use my calculator to get $ 2\sqrt{143}$ being about $ 23.916$ which is not quite $ 24$.

I must admit that I had not thought of squaring $ \sqrt{n\minus{}1} \plus{} \sqrt{n\plus{}1}$ to get $ n\minus{}1 \plus{} n\plus{} 1 \plus{} 2\sqrt{n^{2}\minus{}1}$ which is about $ 4n$.

Meh, or note that $ \mathbb Q(\sqrt{11},\sqrt{13},\sqrt{48})$ has dimension $ 8$.

[quote="math154"]Meh, or note that $ \mathbb Q(\sqrt {11},\sqrt {13},\sqrt {48})$ has dimension $ 8$.[/quote]

And how do you prove it without Galois and Fields Theory?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1641950#1641950]I didn't say it wasn't overkill[/url], but it doesn't need any legit field theory... and for the case of just addition of the roots, there is a solution in a link in [b]t0rajir0u[/b]'s post there I think, which just considers the polynomial $ \prod(x\minus{}\sqrt{a_1}\pm\sqrt{a_2}\pm\cdots\pm\sqrt{a_n})$.


[quote="nsato"]Can you clarify the definition of $ A_2$?[/quote]
Sorry....... :)
$ A_2\equal{}d_a\cap{BC}(d_a//AI,O\in{d_a}$)
Thank you

[quote="jayme"]Dear Mathlinkers,
The line A1A2 go through the midpoint of the side passing through A of the tangentiel triangle of ABC.
This result is not easy to prove synthetically.
This problem is then very known on Mathlinks but actually I haven't see a syntically proof.
Sincerely
Jean-Louis[/quote]
any  body?

[quote="TSTUSA12345"][quote="TSTUSA12345"]Let $ \triangle{ABC},O$the circumcenter,$ I$ the incenter.$ A_1$ the intersection $ AI$ with $ O$.$ A_2$ the point of intersection of parallel to $ AI$through $ O$.Cycially we have $ B_1,B_2,C_1,C_2$.Prove that $ A_1A_2,B_1B_2,C_1C_2$concurrent[/quote]
Yetti.Where are you?
Any body  :?:[/quote]
Please! :(


modular arithmetic
logarithms
number theory
prime factorization
number theory unsolved

Let $p$ be a prime of form $4k+1$ such that $2^{p}= 2(\mod p^{2})$. Show that there exists a prime $q$, divisor of $2^{p}-1$, such that $2^{q}>(6p)^{p}$

Harazi gave me this problem 2 years ago in order to check how I will be quick in solving, so whether it is possible to propose it for IMO selection.

I remember I solved it and there are no unsufficient conditions, but now... I can't reproduce my old solution, and I invented new one which doesn't require $p=1\pmod 4$.

Let $2^{p}-1=q_{1}q_{2}...q_{k}$, where $q_{i}$ are prime numbers. We have $q_{i}\mid 2^{p}-1$ and $q_{i}\mid 2^{q_{i}-1}-1$. It implies $p\mid q_{i}-1$, so $q_{i}=pn_{i}+1$. Note that $n_{i}$ is even. Thus \[p^{2}\mid 2^{p}-2=(pn_{1}+1)(pn_{2}+1)...(pn_{k}+1)-1.\] It follows that $2p\mid n_{1}+n_{2}+...+n_{k}$, hence $n_{1}+n_{2}+...+n_{k}\geq 2p$. We see also $2^{p}>p^{k}n_{1}...n_{k}\geq p^{k}$.
We will show there are $n_{i}$ s.t. $n_{i}\geq \log_{2}{6p}$ (in this case $2^{q_{i}}>2^{pn_{i}}\geq (6p)^{p}$). Suppose the contrary: $n_{i}<\log_{2}6p$ for all $i$. Then $k>\frac{2p}{\log_{2}6p}$, therefore \[p^{k}>p^{\frac{2p}{\log_{2}6p}},\] but $2^{p}>p^{k}$. Thus $2^{p}>p^{\frac{2p}{\log_{2}6p}}$, consequently $6p>p^{2}$ -- it is true only for $p=5,3$, but it is not a case.

Strange... Only very weak estimates.

I apologize if I misunderstand the above solution, but I think that it assumes that $ 2^p \minus{} 2$ is squarefree, causing the bounds to be off.  However, it seems to be relatively easily fixed, but I still seem to get that the condition $ p\equiv 1\bmod 4$ is irrelevant.

[hide="Solution"]
If $ p \equal{} 2$, then $ 2^p \minus{} 2$ is not a multiple of $ p^2$, so we may assume that $ p$ is odd.  Assume that no prime factor of $ 2^p \minus{} 1$ is greater than $ p\cdot (1 \plus{} \log_2 3 \plus{} \log_2 p)$.  Any prime factor of $ 2^p \minus{} 1$, say $ r$, is congruent to $ 1\bmod p$ since $ 2^p\equiv 1\bmod r$, so the order of $ 2\bmod r$ is $ p$.  Hence, the prime factorization of $ 2^p \minus{} 1$ can be written as $ \prod_{i \equal{} 1}^x {(k_ip \plus{} 1)}^{a_i}$ for some integers $ x$, $ k_i$, and $ a_i$.  Notice that\[ \prod_{i \equal{} 1}^x (k_ip \plus{} 1)^{a_i} \equiv 2^p \minus{} 1\equiv 1\bmod 2p^2\]and since\[ (pz \plus{} 1)^y\equiv 1 \plus{} pz\cdot \binom {y}{1} \plus{} p^2z^2 \cdot \binom {y}{2} \plus{} \cdots \plus{} p^yz^y\equiv 1 \plus{} pzy \bmod p^2\]we have that\[ p\sum_{i \equal{} 1}^x {k_ia_i} \plus{} 1\equiv 1\bmod 2p^2\implies \sum_{i \equal{} 1}^x {k_ia_i}\equiv 0\bmod 2p\]Since $ k_i < 1 \plus{} \log_2 3 \plus{} \log_2 p$, we get that\[ (1 \plus{} \log_2 3 \plus{} \log_2 p)\cdot \sum_{i \equal{} 1}^x a_i > \sum_{i \equal{} 1}^x {a_ik_i} \ge 2p\]

Now, we also get that\[ 2^p > 2^p \minus{} 1 \equal{} \prod_{i \equal{} 1}^x (k_ip \plus{} 1)^{a_i} \ge (p \plus{} 1)^{\sum_{i \equal{} 1}^x a_i} > p^{\sum_{i \equal{} 1}^x a_i}\implies\]\[ p > \sum_{i \equal{} 1}^x {a_i}\cdot \log_2 p\]Therefore, we have that\[ 2\sum_{i \equal{} 1}^x {a_i}\cdot \log_2 p < 2p < (1 \plus{} \log_2 3 \plus{} \log_2 p)\cdot \sum_{i \equal{} 1}^x a_i\implies \log_2 p < 1 \plus{} \log_2 3
\]which is may only happen when $ p < 6$. However, checking $ p \equal{} 3, 5$ yields contradictions since $ 2^p \minus{} 2$ would not be a multiple of $ p^2$ in these cases.  Hence, there exists a prime factor, $ q$, of $ 2^p \minus{} 1$ so that $ q > p\cdot (\log_2 3 \plus{} 1 \plus{} \log_2 p) \implies 2^q > (6p)^p$, as wanted.  [/hide]


So I'm trying to use counter-options to buffer losses in both directions, and I'm like, on something and I keep screwing up. Sorry if this is straightforward :oops:

Say a person buys $ \$ x$ in stocks and insures it with $ \$ M - x$ of options. $ M$ is the original amount.

[b]Definitions:[/b]

$ P_i$=initial price
$ P_f$ =final price
$ P_o$=option price
$ P_f'$ = option put price
$ F(x)$ = profit/loss function

$ F_{stock}(x) = \frac {x}{P_i} (P_f - P_i)$

$ F_{option}(x) = \left \{\begin{array}{cc} - (M - x), & P_f > P_f' & & & - \frac {M - x}{P_o} (P_f' - P_f) - (M - x) & P_f' \leq P_f \end{array} \right \;$

So, $ P_f > P_i \implies F_{stock} > 0$ and $ P_f' < P_f \implies F_{option} > 0$

Find $ x$ such that the person minimizes losses. If too ugly, assume $ P_f' = P_i$

I helped co-write this problem and there is a mistake.

For the second equation, the second part of the piecewise should say:

pf<=pf' not pf'<=pf


trigonometry
calculus
integration
calculus computations

[quote="SwappedDimension"]Hey guys. We just started doing this stuff the other day and I ran into this question that's a little troubling. If you guys could help me through this, I would greatly appreciate it. It's new stuff to me, so lol, I apologize if it seems easy.

Evaluate

int sinxcosxdx by four methods: a) the substitution u = cosx b) substitution u = sinx c) identity sin3x = 2sinxcosx d) integration by parts.

The last two weeks or so we've had a sub lecturer so I've been half getting it, half not.  Thanks again in advance!![/quote]
I prefer to guide you rather than solve completely since this is an exercise rather than a challenge:

a) If you let $u=\cos x$, $du=-\sin x\,dx$. Plug in values and you should be able to integrate

b) Same method as in part a except that you let $u=\sin x$ and $du=\cos x\,dx$

c) You got the identity wrong. It should be $\sin 2x=2\sin x\cos x$. Should I say more on how to integrate this: $\int \frac12\sin 2x\,dx$?

d) This part depends what you choose for $u$ and $dv$. Remember that integration by parts comes in the form:
\[ \int u dv=uv-\int v du  \]
You can do either:

1) $u=\cos x$, $dv=\sin x\,dx$. So $du=-\sin x\,dx$ and $v=-\cos x$ and you should get the key integration
\[ \int \sin x\cos x\,dx=-\cos^2x-\int (-\cos x)(-\sin x)\,dx  \]

2) $u=\sin x$, $dv=\cos x\,dx$. So $du=\cos x\,dx$ and $v=\sin x$ and you should get the key integration
\[ \int \sin x\cos x\,dx=\sin^2x-\int (\sin x)(\cos x)\,dx  \]

PS: Don't be surprised if you are getting different answers from the different parts. This explains that antiderivatives are unique up to a constant term.

I prefer that you try to do them first, when you finish you can check your answers.
[hide="Click here to see all solutions"]
a. $\int sin(x) cos(x) dx$

 let $u=cos(x) \implies du = -sin(x)dx$

$= \int sin(x) \cdot u \cdot \frac{1}{-sin(x)} du = -\int u du = \frac{-u^2}{2} = \boxed{\frac{-1}{2} cos^2(x) + C}$

b. let $u=sin(x) \implies du = cos(x)dx$

$= \int u \cdot cos(x) \cdot \frac{1}{cos(x)}du = \frac{1}{2}u^2 = \boxed{\frac{1}{2}sin^2(x) +C}$

c.  $\int sin(x) cos(x) dx = \int \frac{1}{2}sin(2x)dx = \frac{1}{2} \int sin(2x)dx$
    let $u = 2x \implies du = 2dx$
    $= \frac{1}{2} \int sin(u) \cdot \frac{1}{2}du = \boxed{\frac{-1}{4}cos(2x) +C}$

d. let $u = sin(x) \ \ \ \ \ \ \ \ \ \ dv = cos(x) dx$
         $du = cos(x)dx \ \ \ \ \ \ \ \ \ \ v = sin(x)$

$= sin^2(x) - \int sin(x) cos(x) dx = sin^2(x) - \frac{1}{2}sin^2(x) +C = \boxed{\frac{1}{2}sin^2(x)+C}$
[/hide]

Well, I was able to get the first two with your help, but somehow I still managed to screw up the last two parts of the question. I guess that's what the practice is for. I'm gonna give those last two antoher try in a bit before I check the answers.

Thanks for that help again. I reallllly appreciate it.

after thought
i did this same problem recently, except i only did it two ways, i did substitution, $\int_0^x u du$ sin(x) = u , answer is $\frac{sin(x)^2}{2}$.  then i used the double angle identity for the other way $\frac{1}{2} \int_0^x sin(2x) dx$, which is $\frac{1 - cos(2x)}{4}$.
these are both equivelent so:
$\frac{sin(x)^2}{2} = \frac{1 - cos(2x)}{4}$
or
$sin(x)^2 = \frac{1 - cos(2x)}{2}$
which is basically a "proof" for a power reduction formula, i thought it was kinda cool.
(the quotes are to avoid arguements)


1. Let $ r$, $ s$, and $ t$ be the solutions to the equation $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$.  What is the value of $ (rs)^2 \plus{} (st)^2 \plus{} (rt)^2$ in terms of $ a$, $ b$, and $ c$?

2. Let $ x,y,z$ be nonzero real numbers with $ x \plus{} y \plus{} z \equal{} 0$.  Compute $ x \left (\frac {1}{y} \plus{} \frac {1}{z} \right ) \plus{} y \left (\frac {1}{z} \plus{} \frac {1}{x} \right ) \plus{} z\left (\frac {1}{x} \plus{} \frac {1}{y} \right )$.

3. A student awoke at the end of the algebra class just in time to hear the teacher say, "...and as a hint, the solutions to this polynomial form an arithmetic progression."  Looking at the board, the student found a fifth-degree equation to be solved for homework, but he only had time to write down half the polynomail before it was erased:

$ x^5 \minus{} 5x^4 \minus{} 35x^3 \plus{} \ldots$

However, the student was still able to find the roots!  What are they?

4. Calculate the sum of the coefficients of $ P(x)$ if $ (20x^{27} \plus{} 2x^2 \plus{} 1)P(x) \equal{} 2001x^{2001}$.

5. If $ abc \equal{} 1$, prove that $ \frac {a}{ab \plus{} a \plus{} 1} \plus{} \frac {b}{bc \plus{} b \plus{} 1} \plus{} \frac {c}{ca \plus{} c \plus{} 1} \equal{} 1$.

6. The polynomial $ p(x) \equal{} x^3 \plus{} 2x^2 \minus{} 5x \plus{} 1$ has three different roots $ a$, $ b$, and $ c$.  Find $ a^3 \plus{} b^3 \plus{} c^3$.

[hide="1"]$ (rs)^2 \plus{} (st)^2 \plus{} (tr)^2 \equal{} (rs \plus{} st \plus{} tr)^2 \minus{} 2(s^2 rt \plus{} r^2 st \plus{} t^2 rs)$. [/hide]
[hide="2"]We could be cheap and plug numbers in... or we could use math154's solution.[/hide]
[hide="4"]What does $ P(1)$ represent?[/hide]
[hide="6"][url=http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums]Use Newton's sums[/url][/hide]

For 5, do you mean $ \frac {1}{ca \plus{} c \plus{} 1}$?

Hints:

[hide="1"]
$ rs\cdot st \plus{} st\cdot rt \plus{} rt\cdot rs$
[hide]
$ (rs \plus{} st \plus{} rt)^2$
[/hide]
[/hide]
[hide="2"]
$ \frac1y \plus{} \frac1z \equal{} \frac {y \plus{} z}{yz}$
[hide]
$ y \plus{} z \equal{} \minus{} x$
[/hide]
[hide]
$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} xz)$
[/hide]
[/hide]
[hide="3"]
Vieta's and $ x \equal{} \{a \minus{} 2d,a \minus{} d,a,a \plus{} d,a \plus{} 2d\}$
[/hide]
[hide="Alternate Direction for 6"]
$ (a\plus{}b\plus{}c)^3\equal{}a^3\plus{}b^3\plus{}c^3\plus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)$
[hide]
$ a\plus{}b\plus{}c\equal{}\minus{}2$, so $ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}(\minus{}2\minus{}a)(\minus{}2\minus{}b)(\minus{}2\minus{}c)\equal{}p(\minus{}2)$
[/hide]
[/hide]

[hide="1"]manipulations using Vieta's formulas

[hide="big hint!"]$ (rs \plus{} st \plus{} rt)^2$[/hide][/hide]

EDIT: BLARFGH BEAT TO IT

Zhero, those aren't so much hints as giveaways of the main tactic...

Given that you already recognize that many of these problems are just about Vieta's:
[hide="2"] "Symmetric questions have symmetric answers."  Pick some values and find out what the sum is in one case, guess that it's always that, and prove it. [/hide]
[hide="3"] There are only two things you need to solve for:  the common difference and the initial value.  You are given two coefficients.  That looks promising!  What kind of equations can you set up?  (You may find it faster to change the second variable to "the middle term.") [/hide]
[hide="5"] If I remember correctly, this problem has some interesting symmetries.  What happens when you write $ ab \equal{} \frac{1}{c}$ and so forth in the denominators and simplify? [/hide]
[hide="6"] See the hints to #1; modify the tactic.  Alternately, look up Newton's sums, but I don't recommend this approach since you'll learn more trying to figure it out yourself. [/hide]

[quote="Zhero"]For 5, do you mean $ \frac {1}{ca \plus{} c \plus{} 1}$?[/quote]

No, the given relation is true.

My bad, and thanks. I've changed it around a bit.

And math154, thanks for the heads-up. There goes the pseudo-symmetry. :\

Actually, it was supposed to be symmetrical.  Thanks to everyone who's helped!

[quote="dgreenb801"][hide="5"]
Break the symmetry, change the other fractions to suit one of the fractions.[/hide][/quote]
Hmm.  I'm not sure what you have in mind, but here's the [hide="symmetrical solution"] Write $ f(a, b, c) \equal{} \frac{a}{ab\plus{}a\plus{}1} \plus{} \frac{b}{bc\plus{}b\plus{}1} \plus{} \frac{c}{ca\plus{}c\plus{}1}$.  Writing $ ab \equal{} \frac{1}{c}, bc \equal{} \frac{1}{a}, ca \equal{} \frac{1}{b}$ and simplifying gives

$ f(a, b, c) \equal{} \frac{ca}{ca\plus{}c\plus{}1} \plus{} \frac{ab}{ab\plus{}a\plus{}1} \plus{} \frac{bc}{bc\plus{}b\plus{}1}.$

Hmm.  Multiplying the fractions by $ b, c, a$ respectively and simplifying a second time gives

$ f(a, b, c) \equal{} \frac{1}{bc\plus{}b\plus{}1} \plus{} \frac{1}{ca\plus{}c\plus{}1} \plus{} \frac{1}{ab\plus{}a\plus{}1}$.

But now we readily see that $ f(a, b, c) \plus{} f(a, b, c) \plus{} f(a, b, c) \equal{} 3$. [/hide].
Offhand, the asymmetrical solution (while it's an important tactic to keep in mind) requires more computation.

Okay, here's my solution:
[hide="Not much work"]Divide the numerator of the left fraction by a, multiply the numerator of the right fraction by b. You get
$ \frac {1}{b \plus{} 1 \plus{} \frac {1}{b}} \plus{} \frac {b}{bc \plus{} b \plus{} 1} \plus{} \frac {bc}{abc \plus{} bc \plus{} b} \equal{} \frac {1}{bc \plus{} b \plus{} 1} \plus{} \frac {b}{bc \plus{} b \plus{} 1} \plus{} \frac {bc}{bc \plus{} b \plus{} 1} \equal{} 1$  :D[/hide]
Although at first I found the given equivalent to the three fractions, but never thought of adding them. I decided to use one of each, you combined all three. Nice solution!


geometry
geometric transformation
reflection
inequalities
inequalities theorems

I think it should be in section Theorem and Formulas

[quote="Vasc"]I called the AC-Theorem the following statement:
[b]Arithmetic Compensation Theorem (AC-Theorem).[/b]Let $s>0$, and let $F(x_1,x_2,...,x_n)$ be a symmetrical continuous function on the compact set in $R^n$,
$S=\left((x_1,x_2,...,x_n): x_1+x_2+...+x_n=s, x_1 \geq 0,..., x_n \ge 0\right)$.
If $F(x_1,x_2,...,x_n) \leq \max{\left(F\left(\frac{x_1+x_2}{2},\frac{x_1+x_2}{2}, x_3,...,x_n\right),F\left(0,x_1+x_2,x_3,...,x_n\right)\right)}$
for all $(x_1,x_2,...,x_n)\in S$ with   $x_1>x_2>0$, then
$F(x_1,x_2,...,x_n) \leq \max_{1\leq k\leq n}{F\left(\frac{s}{k},..., \frac{s}{k},0,...,0\right)}$  for all $(x_1,x_2,...,x_n)\in S$.[/quote]

It's also belong of my friend, Dinh Ngoc An. He call $UMV$ theorem.

[quote="Vasc"]You can find a paper about AC-Theorem in Mathematical Reflections Site (edited by Titu Andreescu), issue 3.
http://reflections.awesomemath.org/currentissue.html
AC-Theorem and GC-Theorem (Geometric Compensation Theorem) are presented together with many applications in "Algebraic Inequalities-New and Old Methods" book.[/quote]

how could I obtain that book?


My clock chimes two times 15 minutes after the hour, four times 30 minutes after the hour, and six times 45 minutes after the hour. The clock also chimes eight times on each hour in addition to chiming the number of times equal to the hour. (So at 2:00 p.m., the clock chimes $ 8 \plus{} 2 \equal{} 10$ times.) Starting at 12:05 a.m., how many times does the clock chime in a 24-hour period?

But isn't their solution constructive counting?  :cursing:

I did it a different way.  Here it is:

2 (times it chimes at 15 after)+4 (times it chimes at 30 after)+ 6 (times it chimes at 45 after)= 12. 12 times 12 (hours from 12:05 am to 12:00 pm) equals 144.  1+8=9.  2+8=10.  (I'm doing this to find how many times it chimes on each hour.)  For each hour, the times it chimes goes up by 1 until 12 noon.  At 1:00 PM, it chimes 9 times again, and goes up by 1 until 12 midnight, so I can use the same list for the PMs.  In 12 hours, it chimes 174 times.  144 plus 174 equals 318.  Then I multiply that by 2 for the other 12 hours.  318x2=636.


geometry
topology
advanced fields
advanced fields unsolved

This appeared on one of my tutorials for diff geo. I just can't see the solution. Help, anyone?

A Riemannian Manifold, $M$, satisfies $R(u,v) = \frac{Rg(u,v)}{n}$ where $R(u,v)$ is the Ricci curvature, $R$ is the scalar curvature, $n$ is the dimension and $g(u,v)$ is the Riemannian metric. Prove that if $n \geq 3$, then $M$ has constant scalar curvature and parallel Ricci tensor.

I've worked out how to get the parallel Ricci condition from the constant scalar curvature and vice-versa, but I can't derive either one independently.

Cheers,
Mr_Happy

Use the contracted second Bianchi identity.


inequalities
vector
real analysis
real analysis unsolved

Prove that for $x,y$ in $R^{n}$
absolute value $(\parallel x\parallel -\parallel y\parallel ) \leq \parallel x-y \parallel $.

Use the triangle inequality on the vectors $x-y$ and $y.$

This, as well as your last few problems, should have gone in the 'Computations and Tutorials' section instead.


We have n odd integers,not necessarily all different, such that the sum the squares of these integers equals n^4. If n is known to be a positive odd integer, which values can n take?

Ans: n is any +ve integer.
We know that
1+3+...+(2n-1)=n^2
making it square.
Left side will have n^2 terms.
If n is odd then we may take (n-1) odd numbers unchanged then adding all the numbers (that sum is even) with another odd will give another odd numb.
If n is even then taking (n-2) terms unchanged and spiling the sum of other numbers into two part that will be even then adding them with last 2 number will give us 2 odd numbers.
So the proof is complete here.

I don't understand or believe your argument at all... for $n = 3$ I'm pretty sure we can't find $3$ odd integers whose squares add up to $3^{4}= 81$.

Ya ... I can that are 9,27,45
bcz u asked to add 3 odd integers to get 3^4
[ok mention if i have made a mistake]

...three odd integers whose [i]squares[/i]...

:( oups...I am making stupid mistakes.
I didnt notice that it need to be square...I proved it for odd numbers.
Then using the fact that every odd number is 8q+1 it can be said that if n=8q+1=x^2 i mean if n is a square then it can be true.Just add n x^3 s.Example-n=9 x^3=27 just add 9,27.
But i dont know is there any other solutions or not.

$(2k+1)^{2}\equiv 1 \pmod{4}\implies \sum \equiv 3 \pmod{4}$ so there are no solutions for this case.

So anyway, by this logic, $n \equiv 0,1 \pmod{4}$

If $n \equiv 0$, then $n^{4}\equiv 0 \pmod{256}$ and etc., the thing fails by infinite descent.

Left to prove is if $n \equiv 1 \pmod{4}$

[quote="The Zuton Force"]$(2k+1)^{2}\equiv 1 \pmod{4}\implies \sum \equiv 3 \pmod{4}$ so there are no solutions for this case.[/quote]
using this method, we can probably put some restriction on n cuz n^4 is cong to 1...
have to go now sry


Yes, invite yourself to the chatroom MathChat

please excusse my stupidity, but what would 6:30 EST be in Texas?

sorry I am really brainwashed right now.

No, I believe the south is 1 hr behind us, so when its 6:30 here, its 5:30 in TX.

Here's a website that will tell you the times all over the US, in case you get confused.

http://www.onlineconversion.com/timezone.htm


Consider the following question:(this is not a homework problem,but it is one I want to discuss regarding its feature)

A ball is falling onto a ground from a certain height and collides elastically and reaches the same height and the process continues.What is the averages force on the ground?

There are plenty of methods available to do this.What I like most is to use the ROCKET EQUATION!!!

m(dv/dt)=F-u(dm/dt) F is the external force to the system.

It first may look strange.But if you think about it,there is nothing wrong in it.After all,it is meant for a system of particle and is quite general in application.

[If you doubt,you apply it correctly to the famous problems asking the average force in water jet impinging on a wall/chain hanging on a flat platform suddenly starts to fall freely and falls a distance x/conveyer belt prblems etc...Just you will have to consider that LHS m(dv/dt) denotes the force on the concerned body...may be it is not accelerating---like rain falls on a roof and we want to find the average force exerted on the roof.It will begiven by the m(dv/dt) term but individually (dv/dt) will have no meaning here]

Applying that in this problem you will see that the required average force is the external force Mg.And this is the correct answer.

Now,the current problem that is haunting me.I refer to the same problem I quoted first in this thread,by assuming not perfectly elastic collision.That is I am given a co-eff. of restitution e<1.Suppose,I try to find the average force.My method still predicts it is Mg.But,the calculation works are not easy.Because,each time you have to find (del p/del t) where each of these are changing.Then,you have to take the average...and it is by no meanss easy to see what is the result...

Can anyone tell me what is the correct result?By my tool,I am almost sure that it is Mg.But,is there any other means to verify that?

People do not get interested!!!

What I am getting:Force at a time on the floor,F(t)=dp(t)/dt+Mg
Supposing that the initial time moment is t=0, the average force on the ground can be found by

Average force <F>=Limit(T-->0) (1/T) integration(0 to T) F(t) dt
=Limit(T-->0)[{p(t)-p(0)}/T +Mg]

Since the motion of the ball is confined in a finite region, p(T)-p(0) is always bounded, and so the limit is zero


Hey, I'm new here and I can't find a section for Quardratic equations and... I need an answer urgently.. so I am sorry if I got it wrong  :)

..2
x  (xsquare/x power 2)

Here is my problem..
(x-2)(x+2) = 12
xsquare - 2square - 12 = 0
xsquare - 4 - 12 = 0
xsquare - 16 = 0

did I do the correct thing? because i'm stuck right here

that is correct

after that add 16 to both sides and you are left with
$x^2=16$
or x=+4,-4

[quote=PizzaPerson]woowwww a long time ago[/quote]

howww did you find that post
also
[quote=K0NSTANTiNE_hacker]bro x^2 = 16 now take square root from both side we get x = 4, or x = -4. bro why did you posted this in high school?[/quote]

yup

lol what is it with bumping old posts

HSM back then wasn't as sophisticated as it is now.

[quote=K0NSTANTiNE_hacker]bro x^2 = 16 now take square root from both side we get x = 4, or x = -4. bro why did you posted this in high school?[/quote]

Dude the question was asked 15 years ago... literally before some of us were born


Consider the sequence $ a_{0}=a_{1}=1$ and we have $ (n+3)a_{n+2}=(2n+3)a_{n}+3na_{n-1}\forall n\in N$
Prove that $ a_{n}\in Z \forall n\in N$

[quote="s.tringali"]Well, I guess it must be $ 3a_{2}= 3a_{0}+0 \cdot a_{-1}$, i.e. $ a_{2}= 1$, independently of the actual value of $ a_{-1}$ (just suppose $ a_{-1}\in \mathbb{C}$). Maybe...?! :maybe:[/quote]
Maybe there should be $ a_{n+1}$ instead of $ a_{n-1}$ :roll:...
Hm.....then should be n=0,1,2... instead of n=1,2,3...

:(   I'm so sorry.I'm wrong.The true problem is
 $ a_{0}=a_{1}=1$ and  $ (n+1)a_{n+1}=(2n+1)a_{n}+3na_{n-1}$ . Prove that $ a_{n}\in Z$

This problem is very nice.I tried to solved but I couldn't.Can anyone give me a solution?I have two ways to think
1.find a general term of this sequence but I think that it's a hard work.
2. We'll prove by induction.But it's seem that not simple.
3.Can it's a recurrence of a combinatoric problem ? ( such as a permutation satisfy some property,...)
Last, I post problem again.
  Consider a sequence $ a_{0}\equal{}a_{1}\equal{}1$ and $ (n\plus{}1)a_{n\plus{}1}\equal{}(2n\plus{}1)a_{n}\plus{}3na_{n\minus{}1}$ .
Prove that $ a_{n}\in Z$ for all $ n\in N$


geometry
3D geometry
tetrahedron
circumcircle

Find the volume of the tetrahedon $ABCD$ such that $AB=6,\ BC=\sqrt{13},\ AD=BD=CD=CA=5.$

[hide="Solution"]If we take $\triangle ABC$ as basis, then $DA=DB=DC$ means that $D$ projects into the circumcenter of $\triangle ABC$.

By Heron's

\begin{eqnarray*}[ABC]^{2}&=& \frac{11+\sqrt{13}}{2}\cdot\frac{1+\sqrt{13}}{2}\cdot\frac{-1+\sqrt{13}}{2}\cdot\frac{11-\sqrt{13}}{2}\\ &=& \frac{(121-13)(13-1)}{16}\\ &=& \frac{108\cdot 12}{16}\\ &=& 81\end{eqnarray*}

hence $[ABC]=9$. Then $R=\frac{5\cdot 6\cdot\sqrt{13}}{4\cdot 9}=\frac{5\sqrt{13}}{6}$, giving $H^{2}=5^{2}-R^{2}=25\left(1-{13\over 36}\right)=\frac{25\cdot 23}{36}\implies H=\frac{5\sqrt{23}}{6}$

Then $V={1\over 3}[ABC]H=\frac{5\sqrt{23}}{2}$[/hide]


Here are a couple of good problems for the South Carolinians (especially Part II).

Answers to come in about a week.

[hide="II-6"]
Substraction gives $x=5z$ and substituting into the first equation gives $y=3z$. Thus the ratio is \[ \frac{(5z)^2+3(5z)(3z)}{(3z)^2+z^2}=\boxed {7}.  \]
[/hide]

[hide="II-8"]
Take $\tan$ of both sides gives \[ \frac{8}{31} = \tan\left(\arctan (x-1)+\arctan (x+1)\right) = \frac{2x}{2-x^2}.  \] Solving the quadratic gives $x=\frac{1}{4}\vee -8$ but $-8$ doesn't work so $x=\boxed {\frac{1}{4}}$.
[/hide]

[hide="II-10"]
From equation 1 we see that $z=y^2$ and from equation 2 we see that $z=2x+1$, or $x=\frac{1}{2}(y^2-1)$. Substituting into equation 3, we see that \[ \frac{1}{2}(y^2-1)+y+y^2 = 16\iff y=3\vee\frac{-11}{3}.  \]. This gives $z=9\vee\frac{121}{9}$ and $x=4\vee \frac{20}{3}$.

EDIT: OK the second one randomly doesn't work so maybe it's just

$\therefore (x,y,z)=\left(4,3,9\right)$.
[/hide]

[hide="II-11"]
We have $x^2<1$ and $y^2<1$. This means that $(x^2-1)(y^2-1)>0$, or $x^2y^2-x^2-y^2>-1$. Rearranging and adding $2xy$ to both sides, we have $(x+y)^2<(1+xy)^2$. This is equivalent to $|x+y|<|1+xy|$, as desired. $\boxed {}$
[/hide]

[hide]Perfecto! on numbers 6 and 11, but small problem in the details on numbers 8 and 10.[/hide]

I'm a bit tired to do these right now, but probably will do them later, here are hints to some of the remaining ones i glanced at.

[hide="1"]Harmonic Mean[/hide]

[hide="2"]Difference of Two Squares
[/hide]

[hide="4"]Use vectors, distance formula, etc.[/hide]

[hide="7"]Try visualizing it, they're the parts of circles![/hide]

[hide="12"]How many digits do the first 9 pages use?, the next 90 pages?, the next 900?, etc. [/hide]

and for number 10, you substituted into the third equation wrong, it should be $\frac{1}{2}(y^2-1)$, but other than that, it's fine.

[quote="Iversonfan2005"]Note to Chess64. For number 8, the denominator should be $1-(x+1)(x-1)=1-x^2-1=-x^2$, other than that, your methodology is correct

and for number 10, you substituted into the third equation wrong, it should be $\frac{1}{2}(y^2-1)$, but other than that, it's fine.[/quote]

#8: i'm really confused isn't it $1-(x+1)(x-1)=1-(x^2-1)=1-x^2+1=2-2x^2$

#10: i don't think that was the problem, just that one of the solutions was extraneous (maybe).

or maybe my brain just clicked off after (or maybe during :roll:) AP exams?

[quote="chess64"]

or maybe my brain just clicked off after (or maybe during :roll:) AP exams?[/quote]

no, i think my brain is clicked off from studying for them. I don't know how to add, but i see now, it's $x^2$ not $2x^2$, that's what you messed up on.

EDIT: Wait, that's what you have on your original solution, i think you had a typo on the post above, um, ok, i don't see a problem with your solution :?. (It's been a long week :( )

ok i'm uber confused now but I just typed in the expression into [url=http://www.google.com/search?hl=en&q=tan%28arctan%28-8-1%29%2Barctan%28-8%2B1%29%29-8%2F31&safe=active]google[/url] and it says that i'm right so that makes me a little less confused.

Chess64

[hide]Your answer to number 8 is still not right. If you take the $\tan^{-1}$ of a negative number it is negative, and the $\tan^{-1}$ of a positive number is positive. The sum of two negative numbers cannot be positive, so one of your solutions is extraneous.

For number 10 there are actually 3 sets of solutions. Your first one is correct, and your second one just has a miscomputation of the x-value. You haven't noticed the third one yet.[/hide]


Additive Number Theory
modular arithmetic
Sequence
number theory
IMO
IMO 1975

Let $a_{1}, \ldots, a_{n}$ be an infinite sequence of strictly positive integers, so that $a_{k} < a_{k+1}$ for any $k.$ Prove that there exists an infinity of terms $ a_{m},$ which can be written like $a_m = x \cdot a_p + y \cdot a_q$ with $x,y$ strictly positive integers and $p \neq q.$

I recall solving this once before, but since I like the problem.. :)

If we can find $p\ne q$ such that $(a_p,a_q)=1$, we're done: every sufficiently large positive integer $n$ can be written in the form $xa_p+ya_q,\ x,y\in\mathbb N$. We can thus assume there are no two such $p\ne q$. We now prove the assertion by induction on the first term of the sequence, $a_1$. The base step is basically proven, since if $a_1=1$ we can take $p=1$ and any $q>1$ we want. There must be a prime divisor $u|a_1$ which divides infinitely many terms of the sequence, which form some subsequence $(a_{k_n})_{n\ge 1},\ k_1=1$. Now apply the induction hypothesis to the sequence $\left(\frac{a_{k_n}}u\right)_{n\ge 1}$.

[quote="grobber"]
If we can find $ p\ne q$ such that $ (a_p,a_q) \equal{} 1$, we're done: every sufficiently large positive integer $ n$ can be written in the form $ xa_p \plus{} ya_q,\ x,y\in\mathbb N$[/quote]
Please can you explain why? I don't know this fact.

I believe its called the "Chicken McNugget Theorem". Its easier to give an example to show why this is intuitively true:
[b][i]Chicken McNugget Theorem:[/i][/b]
\[ \text{If }(a,b)\equal{}1\text{ then all }n>(a\minus{}1)(b\minus{}1)\text{ can be written as }xa\plus{}yb\]

Take $ 3,7$. Beyond $ 12$, we can write 13=2(3)+7, 14=2(7), and 15=5(3). Now all numbers can be written mod 3, so just scale these up by three to hit all positive integers beyond a point.

EDIT: I just saw i got beaten to this post...

[hide="my solution"]Consider modulo $ a_1$. Each of them is an element of $ \{0,1,2,...,a_1\minus{}1\}$. Since they are infinitely many of them, at least one of $ \{0,1,2,...,a_1\minus{}1\}$ has an infinite number of elements. Assume that there are infinitely many $ a_k$ such that $ a_k\equiv r\pmod{a_1}$. So we can take $ m>n$ such that $ a_m\equal{}a_1\plus{}ya_n$. Evidently, there are infinitely many such $ m$.[/hide]

If $(a_p, a_q) \neq 1$ for all $p, q$, simply scale down.

Pick $a_p, a_q$ such that $(a_p, a_q) = 1$; we simply need $a_q \mid a_m - xa_p$. But for any $a_m > a_pa_q$, there exists some $1 \leq a_p \leq a_q$ such that the conclusion is true due to the relatively prime condition. Infinitely many $a_m$ exist because the sequence is strictly increasing.


trigonometry
analytic geometry
graphing lines
slope
geometry

Let ABC be an isosceles triangle with AB=AC, and let D e the midpoint of BC.  Let E be the foot of the perpendicular from D to AB and F the midpoint of DE.  Prove that AF is perpendicular to CE.

First of all, I drew a random isoceles triangle.
I drew the point D on the triangle and joined the point A with D, which is perpendicular to line BC.  Next, I drew the point E and connected it with D so that DE is perpendicular to AB.  Next, I drew the point F and connected it with point A.  Next I joined C with E.

After drawing the diagram, I labelled BCE as X.  Then I labelled BDE as 2X.  Then Labelled ACE as 90-3X.  ABD=90-2X.  AEG=90-X.  AGE=90-X.  EAG=2X.  CAD=2X.  CDG=90-X.  EDG=90-2X.  I think I need to find AEF or FAG or AFD or AFE to prove that Af is perpendicular to EG.

Could you please explain to me what I missed or the angles I don't need please?
Also, could you help me form this into a proof.  Such as should I include a diagram in the proof?

[color=darkblue][b]Proof 1 (synthetic).[/b]Denote the intersection $I\in CE\cap AF$ and the projection $G$ of the point $C$ to the line $AB$. Thus, $DE\parallel CG$ and $EG=EB$, i.e. the ray $[CE$ is the $C$- median in the triangle $CGB$. Prove easily that $ADE\sim CBG$. The $A$-median of the triangle $ADE$ is similarly with the $C$- median of the triangle $CGB$ and the side $[AD]$ of the triangle $ADE$ is similarly with the side $[CB]$ of the triangle $CBG$. Therefore, $\widehat{DAF}\equiv\widehat{BCE}$, i.e. $\widehat{DAI}\equiv\widehat{DCI}$, what means that the quadrilateral $ACDI$ is cyclically, i.e. $\widehat{ADC}\equiv\widehat{AIC}$. But $DA\perp DC$. In conclusion, and $IA\perp IC$, i.e. $AF\perp CE\ .$

[b]Proof 2 (metric).[/b] $AC^{2}-AE^{2}=AC^{2}-(AD^{2}-DE^{2})=$ $(AC^{2}-AD^{2})+DE^{2}=DC^{2}+DE^{2}$ $\Longrightarrow$ $\boxed{\ AC^{2}-AE^{2}=DC^{2}+DE^{2}\ }\ \ (1)\ .$
$4\cdot (FC^{2}-FE^{2})=$ $2\cdot (CE^{2}+CD^{2})-DE^{2}-DE^{2}=$ $2\cdot (CE^{2}+DB^{2}-DE^{2})=$ $2\cdot (CE^{2}+BE^{2})=$ $4\cdot DE^{2}+CB^{2}=$ $4\cdot (DC^{2}+DE^{2})$ $\Longrightarrow$ $\boxed{\ FC^{2}-FE^{2}=DC^{2}+DE^{2}\ }\ \ (2)\ .$ From the relations $(1)$ and $(2)$ obtain $AC^{2}-AE^{2}=FC^{2}-FE^{2}$, i.e. $\boxed{\ AF\perp CE\ }\ .$

[b]Proof 3 (trigonometric).[/b] Denote $\phi =m(\widehat{DAF})$ and $\psi =m(\widehat{BCE})$. Therefore, $1=\frac{FD}{FE}=\frac{AD}{AE}\cdot\frac{\sin\phi}{\sin\left(\frac{A}{2}-\phi\right)}=\frac{\sin\phi}{\cos\frac{A}{2}\sin\left(\frac{A}{2}-\phi\right)}$ $\Longrightarrow$
$2\sin\phi =\sin (A-\phi )-\sin \phi$ $\Longrightarrow$ $3\sin \phi =\sin (A-\phi )$ $\Longrightarrow$ $3\tan\phi =\sin A-\cos A\tan\phi$ $\Longrightarrow$ $\boxed{\ \tan\phi =\frac{\sin A}{3+\cos A}\ }\ \ (1)\ .$
$\frac{EB}{EA}=\left(\frac{DB}{DA}\right)^{2}=\tan^{2}\frac{A}{2}\ ,\ \frac{EB}{EA}=\frac{CB}{CA}\cdot\frac{\sin\psi }{\sin \left(90-\frac{A}{2}-\psi \right)}=\frac{2\sin\frac{A}{2}\sin\psi }{\cos\left(\frac{A}{2}+\psi\right)}\ .$ Therefore, $\sin\frac{A}{2}\cos\left(\frac{A}{2}+\psi \right)=2\cos^{2}\frac{A}{2}\sin\psi$ $\Longrightarrow$ $\sin (A+\psi )-\sin \psi =2\sin\psi (1+\cos A)$ $\Longrightarrow$ $\sin A\cos\psi=3\sin\psi+\sin\psi \cos A$ $\Longrightarrow$ $\sin A=\cos A\tan \psi+3\tan\psi$ $\Longrightarrow$ $\boxed{\ \tan\psi =\frac{\sin A}{3+\cos A}\ }\ \ (2)\ .$ From the relations $(1)$ and $(2)$ obtain $\phi =\psi$ $\Longrightarrow$  the quadrilateral $ACDI$ is cyclically $\Longrightarrow$  $AF\perp CE\ .$

[b]Proof 4 (analytic).[/b] $D(0,0)\ ,\ B(0,b)\ ,\ C(-b,0)\ ,\ A(0,a)$, where $a>0$, $b>0$. The slope of the line $AB$ is $s(AB)=-\frac{a}{b}$, the slope of the line $DE$ is $s(DE)=-\frac{1}{s(AB)}=\frac{b}{a}$ and the point $E$ is the intersection between the line $AB$ with the equation $\frac{x}{b}+\frac{y}{a}=1$ and the line $DE$ with the equation $y=\frac{b}{a}x$. Thus, the point $E$ has the coordinates $x_{E}=\frac{a^{2}b}{a^{2}+b^{2}}$ and $y_{E}=\frac{b^{2}a}{a^{2}+b^{2}}$. The middlepoint $F$ of the segment $DE$ has the coordinates $x_{F}=\frac{1}{2}\cdot \frac{a^{2}b}{a^{2}+b^{2}}$ and $y_{F}=\frac{1}{2}\cdot \frac{b^{2}a}{a^{2}+b^{2}}$. Therefore, the slope of the line $CE$ is $s(CE)=\frac{\frac{b^{2}a}{a^{2}+b^{2}}}{\frac{a^{2}b}{a^{2}+b^{2}}+b}=\frac{ab}{2a^{2}+b^{2}}$ and the slope of the line $AF$ is $s(AF)=\frac{\frac{b^{2}a}{2(a^{2}+b^{2})}-a}{\frac{a^{2}b}{2(a^{2}+b^{2})}}=-\frac{2a^{2}+b^{2}}{ab}\ .$ In conclusion, $s(CE)\cdot s(AF)=-1$, i.e. $CE\perp AF\ .$

[b]Remark.[/b] I'll prove similarly the following [size=117][b][u]"two problems of the butterfly[/u][/b]"[/size] :[/color]

[color=darkred]$\blacksquare\ P.B.1$ Given are a circle $w(O)$, a line $l$ and the point $F\in l$. Let $\{M,N,P,R\}\subset w$ be four points so that $F\in MN\cap PR$.
Denote the points $A\in MR\cap l$, $B\in NP\cap l$. Prove that $\boxed{\ OF\perp l\Longleftrightarrow FA=FB\ }\ .$[/color]

[color=darkblue][b]Proof.[/b] Denote the projections $U$, $V$ of the center $O$ to the lines $MR$, $PN$ respectively. Thus, $\triangle FNP\sim\triangle FRM$ $\Longrightarrow$ $\widehat{FUM}\equiv\widehat{FVP}$ because the sides $[FM]$, $[RM]$ of the triangle $FRM$ are similarly with the sides $[FP]$, $[NP]$ of the triangle $FNP$ $\Longrightarrow$ $\widehat{FUA}\equiv\widehat{FVB}\ \ (1)$.
$\triangleright$ [b][u]Suppose that[/u][/b] $OF\perp l\ .$ The quadrilaterals $AFUO$, $BFVO$ are cycclically with the diameters $[AO]$, $[BO]$ respectively $\Longrightarrow$ $\widehat{FUA}\equiv\widehat{FOA}$ and $\widehat{FVB}\equiv\widehat{FOB}$. From the relation $(1)$ obtain $\widehat{FOA}\equiv\widehat{FOB}$. From $OF\perp l$ results that the triangle $AOB$ is isosceles, i.e. $FA=FB\ .$
$\triangleleft$ [b][u]Suppose that[/u][/b] $FA=FB\ .$ Denote the intersections $X$, $Y$ of the bisector line of the segment $[AB]$ with the lines $OU$, $OV$ respectively. The quadrilaterals $AFXU$, $BFYV$ are cyclically with the diameters $AX$, $BY$ respectively. Thus, $\widehat{FXA}\equiv\widehat{FUA}$ and $\widehat{FYB}\equiv\widehat{FVB}$. From the relation $(1)$ obtain $\widehat{FXA}\equiv\widehat{FYB}$, i.e. $X\equiv Y\equiv O$ $\Longrightarrow$ $OF\perp l\ .$[/color]

[color=darkred]$\blacksquare\ P.B.2$ Given are the circle $w(O)$, the points $\{A,B,C,D\}\subset w$ and a line $l$ for which $O\in l$.
Denote : $M_{1}\in AB\cap l$ ; $M_{2}\in AD\cap l$ ; $M_{3}\in AC\cap l$ ; $N_{1}\in CD\cap l$ ; $N_{2}\in BC\cap l$ ; $N_{3}\in BD\cap l$.
Prove that the following chain - equivalence : $\boxed{\ OM_{1}=ON_{1}\Longleftrightarrow OM_{2}=ON_{2}\Longleftrightarrow OM_{3}=ON_{3}\ }\ .$[/color]

[color=darkblue][b][u]Proof (metric).[/u][/b] Denote the intersection $R\in AC\cap BD$. The powers of the points $R\ ,\ N_{3}\ ,\ M_{3}$
w.r.t. the circle $w$ are : $p_{w}(R)=RA\cdot RC=RB\cdot RD$ : $p_{w}(N_{3})=N_{3}D\cdot N_{3}B$ ; $p_{w}(M_{3})=M_{3}A\cdot M_{3}C$.
$\blacktriangleright$ Apply the Menelaus' theorem to the transversals $\overline{AM_{2}D}$, $\overline{BN_{2}C}$ for the triangle $M_{3}RN_{3}$ :
$\frac{M_{2}M_{3}}{M_{2}N_{3}}\cdot \frac{DN_{3}}{DR}\cdot\frac{AR}{AM_{3}}=1$ and $\frac{N_{2}N_{3}}{N_{2}M_{3}}\cdot\frac{CM_{3}}{CR}\cdot\frac{BR}{BN_{3}}=1$ $\Longrightarrow$ $\frac{N_{2}M_{3}}{N_{2}N_{3}}\cdot \frac{M_{2}M_{3}}{M_{2}N_{3}}\cdot\frac{RA}{RD}\cdot\frac{RC}{RB}\cdot\frac{N_{3}D}{M_{3}C}\cdot\frac{N_{3}B}{M_{3}A}=1$, i.e. $\boxed{\ \frac{M_{2}M_{3}}{M_{2}N_{3}}\cdot\frac{N_{2}M_{3}}{N_{2}N_{3}}=\frac{p_{w}(M_{3})}{p_{w}(N_{3})}\ }\ \ (1)\ .$
$\blacktriangleright$ Apply the Menelaus' theorem to the transversals $\overline{M_{1}AB}$, $\overline{N_{1}CD}$ for the triangle $M_{3}RN_{3}$ :
$\frac{M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac{BN_{3}}{BR}\cdot\frac{AR}{AM_{3}}=1$ and $\frac{N_{1}M_{3}}{N_{1}N_{3}}\cdot\frac{DN_{3}}{DR}\cdot\frac{CR}{CM_{3}}=1$ $\Longrightarrow$ $\frac{M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac{N_{1}M_{3}}{N_{1}N_{3}}\cdot\frac{RA}{RB}\cdot\frac{RC}{RD}\cdot\frac{N_{3}B}{M_{3}A}\cdot\frac{N_{3}D}{M_{3}C}=1$, i.e. $\boxed{\ \frac{M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac{N_{1}M_{3}}{N_{1}N_{3}}=\frac{p_{w}(M_{3})}{p_{w}(N_{3})}\ }\ \ (2)\ .$
From the relations $(1)$, $(2)$ obtain the conclusion of the our problem.

[b][u]A very interesting particular case, when [/u][/b] $A\equiv B$.[/color] [color=darkred] Let $ABC$ be an acute triangle inscribed in the circle $w=C(O,R)$.
Denote the point $A'\in w\cap (AO$ and the intersections $P\in BC\cap A'A'$, $M\in AB\cap PO$, $N\in AC\cap PO$. Then $OM=ON\ .$[/color]

if we fix $BD=1$ and $\angle ABC=\theta$, then trig easily shows that $AE^{2}+CF^{2}=EF^{2}+AC^{2}$ which implies the conclusion

[b]Altheman[/b], how calculate (with "trigo") you the length of the segment $FC$ ? Thanks.

Drop a perpendicular from $E$ to $BD$ then from $F$ to $BD$, consider $\measuredangle ABD = \alpha$, it follows that $FC^{2}= (\frac{sin\alpha cos\alpha}{2})^{2}+(1+\frac{sin^{2}\alpha}{2})^{2}$

Also:
$FE^{2}=(\frac{sin\alpha}{2})^{2}$
$AC^{2}=sec^{2}\alpha$
$AE^{2}=sec\alpha-cos\alpha$

Then: $FC^{2}-FE^{2}=AC^{2}-AE^{2}=1+sin^{2}\alpha$  $\Rightarrow$  $AF \perp CE$  $\mathbb{QED}$

Yes, but in this case the trig methods are ugly because appear medians !

[quote="max1546"]Let ABC be an isosceles triangle with AB=AC, and let D e the midpoint of BC.  Let E be the foot of the perpendicular from D to AB and F the midpoint of DE.  Prove that AF is perpendicular to CE.

[/quote]

 [b]The more classical solution[/b]

 If $M$ is the midpoint of $BE$, then in triangle $ADM$ point $F$ is the orthocenter, because $MF // BD$ , so $MF \perp AD$ and $DE\perp AB$. This means that $AF \perp DM$ . But $DM //CE$

[u] Babis[/u]

I got a question.
What is FC?
What is the importance of that number.  I noticed that nearly all the people used this length in their proofs.
Therefore, on nearly all workings I'm stuck there

This is a very nice problem I saw, I knew it was posted some time ago, I searched google.
I have a question about Virgil Nicula's metric proof. Why is it true that 4(FC^2-FE^2)=2(CE^2+CD^2)-DE^2-DE^2?
Nice fast proof using the orthocenter, stergiu.


Evaluate $ \sum_{k \equal{} 1}^n (k^2 \plus{} 3k \plus{} 2)$.

Could someone also list out the properties of sigma notation manipulation or redirect me to a link?

Thanks.

Yes the properties for sigma notation can be easily derived. Here are the basic ones:

$ \sum_{i \equal{} 1}^n (a_1b_1 \plus{} a_2b_2 \plus{} ... \plus{} a_nb_n) \equal{} \sum_{i \equal{} 1}^n (a_1b_2) \plus{} \sum_{i \equal{} 1}^n (a_2b_2) \plus{} ... \plus{} \sum_{i \equal{} 1}^n (a_nb_n)$

$ \sum_{i \equal{} 1}^n ca_i \equal{} ca_1 \plus{} ca_2 \plus{} ... \plus{} ca_n \equal{} c(a_1 \plus{} a_2 \plus{} ... \plus{} a_n) \equal{} c \sum_{i \equal{} 1}^n a_i$

$ \sum_{i \equal{} a}^n c \equal{} c \plus{} c \plus{} ..(n \minus{} a \plus{} 1 times).. \plus{} c \equal{} (n \minus{} a \plus{} 1)c$

[hide="The problem"]$ \sum_{k \equal{} 1}^n (k^2 \plus{} 3k \plus{} 2) \equal{} \sum_{k \equal{} 1}^n (k^2) \plus{} 3 \sum_{k \equal{} 1}^n (k) \plus{} \sum_{k \equal{} 1}^n 2 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \plus{} \frac {3n(n \plus{} 1)}{2} \plus{} 2n$[/hide]

For a full list of properties see here

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

If you understand sigma notation, you should find the properties very intuitive.

$ \sum cf(x) \equal{} c\sum f(x)$

$ \sum f(x) \plus{} g(x) \equal{} \sum f(x) \plus{} \sum g(x)$

$ \sum_{k \equal{} a}^{b}f(k) \equal{} \sum_{k \equal{} 0}^{b \minus{} a}f(k \plus{} a)$

Anyway, for your question:

Split it into $ \sum_{k \equal{} 1}^{n}k^2 \plus{} 3\sum_{k \equal{} 1}^{n}k \plus{} \sum_{k \equal{} 1}^{n}2$

$ \boxed{\frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \plus{} \frac {3n(n \plus{} 1)}{2} \plus{} 2n}$

EDIT: Darn, beaten.

EDIT 2: I'll throw in proofs for $ \sum k$ and $ \sum k^2$, just for fun.

[hide="k"]
$ S\equal{}1\plus{}2\plus{}3\plus{}\cdots\plus{}n$
$ S\equal{}n\plus{}\cdots\plus{}3\plus{}2\plus{}1$

$ 2S \equal{}\underbrace{(n\plus{}1)\plus{}\cdots\plus{}(n\plus{}1)}_{n\ \text{times}}\equal{}n(n\plus{}1)$

$ S\equal{}\boxed{\frac{n(n\plus{}1)}{2}}$
[/hide]

[hide="k^2"]
Notice that $ (n\plus{}1)^3\minus{}n^3\equal{}3n^2\plus{}3n\plus{}1$

$ \sum_{k\equal{}1}^{n}(k\plus{}1)^3\minus{}k^3$ telescopes into $ (n\plus{}1)^3\minus{}1$

Let $ S\equal{}\sum_{k\equal{}1}^{n}k^2$

$ 3S\plus{}\frac{3n(n\plus{}1)}{2}\plus{}n\equal{}(n\plus{}1)^3\minus{}1$

Solve for $ S$.
[/hide]

$ a_k \equal{} f(k) \minus{} f(k \plus{} 1)\Longrightarrow \sum_{k \equal{} 1}^n a_k \equal{} \sum_{k \equal{} 1}^n \{f(k) \minus{} f(k \plus{} 1)\} \equal{} f(1) \minus{} f(n \plus{} 1).$

$ a_k \equal{} f(k \plus{} 1) \minus{} f(k)\Longrightarrow \sum_{k \equal{} 1}^n a_k \equal{} \sum_{k \equal{} 1}^n \{f(k \plus{} 1) \minus{} f(k)\} \equal{} f(n \plus{} 1) \minus{} f(1).$

$ \sum_{k \equal{} 1}^n (k^2 \plus{} 3k \plus{} 2) \equal{} \sum_{k \equal{} 1}^n (k \plus{} 1)(k \plus{} 2) \equal{} \sum_{k \equal{} 1}^n \frac 13 \{(k \plus{} 1)(k \plus{} 2)(k \plus{} 3) \minus{} k(k \plus{} 1)(k \plus{} 2)\}$

$ \sum_{k\equal{}1}^n (sa_k\plus{}tb_k)\equal{}(sa_1\plus{}tb_1)\plus{}(sa_2\plus{}tb_2)\plus{}\cdots \plus{}(sa_n\plus{}tb_n)$

$ \equal{}s(a_1\plus{}a_2\plus{}\cdots \plus{}a_n)\plus{}t(b_1\plus{}b_2\plus{}\cdots \plus{}b_n)$

\[ \boxed{\boxed{\sum_{k\equal{}1}^n (sa_k\plus{}tb_k)\equal{}s\sum_{k\equal{}1}^n a_k\plus{}t\sum_{k\equal{}1}^n b_k}}\]

[quote="grn_trtle"][b]If you understand sigma notation,[/b] you should find the properties very intuitive.

[/quote]

NOW I understand sigma notation - I wasn't sure about it earlier.  Now I also understand why they work.

Thanks.  :)


If $ x,y\in \mathbb{R}$ find the minimum value of :

$ E\equal{}x^2\plus{}6y^2\minus{}2xy\plus{}6x\minus{}16y\plus{}18$

[quote="alex2008"]If $ x,y\in \mathbb{R}$ find the minimum value of :

$ E \equal{} x^2 \plus{} 6y^2 \minus{} 2xy \plus{} 6x \minus{} 16y \plus{} 18$[/quote]
It's so easy  :)
$ E \equal{} x^2 \plus{} 6y^2 \minus{} 2xy \plus{} 6x \minus{} 16y \plus{} 18\equal{}(x\minus{}y)^2\plus{}6(x\minus{}y)\plus{}9\plus{}5(y^2\minus{}2y\plus{}1)\plus{}4$
$ \equal{}(x\minus{}y\plus{}3)^2\plus{}5(y\minus{}1)^2\plus{}4 \ge 4$
$ \Rightarrow E_{min}\equal{}4 \Longleftrightarrow x\equal{}\minus{}2;y\equal{}1$


calculus
integration
function
real analysis
real analysis unsolved

Is there any way to calculate

$ \int_{a}^{b} \frac{e^{\minus{}\frac{1}{2}x}}{cx^2\plus{}1} dx$

for some constants $ a$, $ b$, $ c$?

hello, i think it can not be expressed by an elemantary function,Maple 13 gives the following answer
$ \minus{}1/2\,\sqrt {\minus{}c} \left( {{\rm e}^{\minus{}1/2\,{\frac {1}{\sqrt {\minus{}c}}}}}{\it
Ei} \left( 1,\minus{}1/2\,{\frac {\minus{}x\sqrt {\minus{}c}\plus{}1}{\sqrt {\minus{}c}}} \right) \minus{}{
{\rm e}^{1/2\,{\frac {1}{\sqrt {\minus{}c}}}}}{\it Ei} \left( 1,\minus{}1/2\,{\frac
{\minus{}x\sqrt {\minus{}c}\minus{}1}{\sqrt {\minus{}c}}} \right)  \right) {c}^{\minus{}1}$
Sonnhard.


AMC
AIME
USA(J)MO
USAMO
articles
AMC 10
geometry

Hard work... definitely hard work!!!  I bet that you can not go through all the USAMO qualifiers and find one who hasn't spent a ton of time on math.  People aren't born with knowledge.  They have to go and get it.  Now some people might be born with a bigger capacity, quicker learning ability, or more desire to learn but they still have to put effort in to get smart.  As for going from a 3 to a 7, that is very possible.  I have only taken the AIME once so I can't say if I have improved by 4 or more but I hope to.  I think that that is a very realistic goal.

One person on this forum got a 4 on AIME one year and a 13 on AIME the next. I think that result speaks for itself.

Your goal sounds doable to me. Besides getting the official [url=http://www.unl.edu/amc/d-publication/d6-books/testsoln.html#13]old AIMEs and solutions[/url] from AMC, you should read the books in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8674]readable books thread[/url] to get more background. And be sure to read all the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php]articles[/url] in this site's resources section.

Good luck!

If you don't like spending money to get old AIMEs, http://www.kalva.demon.co.uk/aime.html has all of the AIMEs with answers and short solutions.  And there's always the practice problems on the official website...

[quote="nr1337"]If you don't like spending money to get old AIMEs, http://www.kalva.demon.co.uk/aime.html has all of the AIMEs with answers and short solutions.[/quote]

But the Kalva site, useful as it is, doesn't pretend to provide [b]verbatim[/b] AMC problems or complete solutions, does it? I wonder what the highest score ever obtained on the AIME by someone who never referred to the official AMC solutions is. (Yes, I'm asking anyone who passes by the thread who happens to know a particular score.)

Without putting an emphasis on studying AMC material, I went from 5 to 7 (really, really close to 9). So that goal sounds doable to me.

Thanks to you all, for all the encouraging advice. I guess then I should get to work knowing that such an improvement is indeed possible. BTW, ComplexZeta, how much did that person with AIME 13 study?

well, i went from a 113 on amc10 in 8th grade (screwed up big time) to a 134.5 this year and a 6 on aime. thanks to changing school districts and practicing hard. when you are in an ambitious environment, you will definitely do better. i am hoping to get 9+ on the aime next year so im practicing too.

Nearly every student taking the USAMO had a year where they went from being able to work a small number of AIME problems to making the USAMO.

The key is working problems and developing your strengths into near "I'll get any of this type of problem" strengths.  Once you have done that you can develop your weak points into strengths and build on your success.  This is important because the AIME has many problems that draw from more than one element of mathematics: (combinatorics and algebra, algebra and geometry, geometry and number theory, etc.).

[quote="keta"]Is it possible for one to score a 3 on the AIME on one year, study hard for another year, and achieve a 7 or above on the AIME the following year? I plan to seriously put effort into doing well on the AIME and making the USAMO, but I need to know is improvement more based on hard work or natural mathematical abilities? Advice is greatly appreciated![/quote]

I went from 2 - 8 between my grade 10 and 11 years.

(and then to 3 my grade 12 year, but let's not talk about that too much).


Dear Members,
                      Can I "CHANGE"my email address for more,better
and secured email address? If so,how? How about listing my interests? I've not done when i first register.

                                                  Sincerly,
                                                 Mikhail S.

You can change your options and information through the "profile" link at the top of the page.

Err.. you don't have to "quote" everything.


ratio
geometry
trigonometry
rectangle
angle bisector

The lengths of the legs of a right triangle are 3 and 4. Two congruent circles, externally tangent to each other, are drawn inside the triangle, with each circle tangent to the hypotenuse and one of the legs. What is the distance between the centers of the two circles?

[quote="Aznwithabrain"][hide]
$2.5$[/hide]

Not sure though[/quote]

I got 10/7 but Im not sure on that..

[hide]I got 1 as the answer but I'm unsure.[/hide]

I would like to see someone work it with the steps laid out.

ehghegh wow I just noticed it said CONGRUENT circles..  :|

nevermind about my answer then >_>

[quote="Aznwithabrain"][hide]
$2.5$[/hide]

Not sure though[/quote]

I get 2.5 also....

My approach was to just set up a right triangle inside the triangle such that they are similar with the distance between the two circles being the hypotenuse....

i didnt use a calculator but who cares........i just want to know the right answer now....i'm very curious

[quote="Art of Owna"]The line joining the center of the two circles-is that basically the diameter of one of the circles-and also is that line parallel to the hypotenuse

becuase if thats true ill still go with 10/7..[/quote]

can u demonstrate? because i did the same thing as you and i got $\frac{5}{2}$

alright call the triangle ABC
so AB =4
BC=3
AC=5

call the circle tangent to AC and BC C1, the center X
call the circle tangent to AC and AB C2, the center Y

call the point where C1 is tangent with AC O
call the point where C2 is tangent with AC P

So we basically are trying to fined CO+OP+AP, now I dont know what exactly OP is, is that just 2r where r is the radius of one of the circles?  Or is there a shorter line connecting the centers that isnt consisted of the two radii?
If that is 2r, we basically are trying to find AP and OC, so we have AP, we know that AY is an angle bisector of angle A, CX is an angle bisector of OC, so we can find tan(1/2A) and tan(1/2C)
well, tan(1/2A)=1/3
and tan(1/2C)=1/2,
PY and XO are both r, the radius
so we can use ratios as PY (r)/AP=1/3
and XO (r)/OC=1/2
so we have that AP=3r, XO=2r, now if OP is 2r, we have 7r=5 (the hyupotenuse) so r=5/7, and we are trying to find OP, so that is 10/7, but I dont know if 2r is the shortest distance betwene the center of the circles because I dont know if 2r is actually a straight line.

[hide="Hint"]Use area to find the length of one of the radius of the circle, then everything should be easy from there. [/hide]
Edit: I also get what Art Of Owna gets.

[quote="Art of Owna"][quote="ChrisA"]You're allowed to use calculator.[/quote]
So is the answer 10/7 (the only thing Im confused about is that is the shortest distance between the two centers equal to twice the radius, or is there a shorter path-I wish I had a picture to show what I mean).  If the two times the radius is the shortest distance then it is 10/7..but if it isnt it isnt 10/7[/quote]

I dont know what you mean

umm after you use trig to get ratios.. those are the ratios of parts of the triangle (sorry it's REALLY off scale)

and since the distance between the two centers is 2/7 of AC, $5\cdot\frac{2}{7}= \frac{10}{7}$

I get 10/7 too.

Art of owna, it IS a strait line between the two circles centers since they are tangent to each other.

Basicly what I did was notice taht the centers of the circles lie on the angle bisector of the angles they subtend. Using the angle bisector theory we can find the ratios between the half angles. I draw radii from the centers of the circles to the hypotenuse, and label these points p and q. The distance between p and q is 2r, and I label the distance from q to the vertice of the triangle x, and from p to the vertice 5 - 2r - x.

Using the ratios we can create 2 equations with r and x.. then its a matter of solving for r.

EDIT:
[hide="full solution"]
Label the 90 degree angle A, the smaller acute angle B and the larger, C. Draw radii from the center of the circles to the hypotenuse. Label the intersection of these lines P for the one subtending angle C, and Q for the circle subtending angle B. Lets call the circle subtending angle C, n and the other circle m.

Draw aline through the center of circle n to through vertex C so that it intersects side BA at S. Note that this bisects angle C. Do the same with circle m and angle B so that it intersects side AC at T.

By angle bisector theory we find SA to be $\frac{3}{2}$ and TA to be $\frac{4}{3}$.
Thus the tangent ratio of angle SCA is $\frac{1}{2}$ and the tangent ratio of angle TBA is $\frac{1}{3}$.

This means that
$\frac{r}{PC}= \frac{1}{2}\Rightarrow PC = 2r$
and since $QB = 5-2r-PC$:
$\frac{r}{5-2r-PC}= \frac{1}{3}\Rightarrow PC = 5-5r$

Now we can solve for R
$2r = 5-5r$
$7r = 5$
$r = \frac{5}{7}$

Since the distance between the centers of both circles is 2r, the distance is $\boxed{\frac{10}{7}}$
[/hide]

hmm I got the right answer but I want to ask one more question:

So lets say two circles are tangent, does the shortest distance between the centers go through that tangent point and thus making the shortest distance the sum of the two circles radii (if congurent just the diameter of one circle)?

yes.

Draw a line tangent to a circle, and a radii to the tangent line. This will form a 90 degree angle.
Now put a circle tangent to the line on the other side at the exact same point and draw a radius, this will form a 90 degree angle, meaning that the angle between the two radii is 180 degrees, or in other words, is a straight line.

[quote="ChrisA"]I had said before that a calculator was allowed, but I'm actually not sure because contest is from 1981. However, I think it is safe to assume that Texas Instrument graphing calculators weren't around then.[/quote]
You dont need a caculator-just look at what me and everyone else did with tangents and angle bisectors.  All you need to know is the $\tan(\alpha/2)$ formula

[quote="lkryptonicl"]glentheman....how do u know that the centers of the circles lie on the angle bisector?[/quote]

it has to, try drawing a circle and two lines tangent to it from a point outside of the circle, by the triangles HL congruency, they are congruent meaning the angles are congruent, meaning it is the angle bisector.

[hide="Way without trig or whatever you you guys did before"]Call the right triangle ABC, with AB equaling 3, BC=4, and AC=5. Call the circles, Q and P, with Q next to vertex A, and P next to vertex C. From Q, draw perpendicular lines to the side AC and AB, and call the points X and Y respectively. Then draw one line from Q straight to A. Then from P, draw perpendicular lines to AC and BC and call the points M and N respectively. Then draw form P to C. Then connect the centers of the two circles. And finally, from both centers, draw a line to B. QX=r, PM=r, QY=r, and PN=r. And PQ is 2r. We can find the area of the rectangle, which is $2r^{2}$. The area of triangle AQB is $\frac{3r}{2}$. The area of triangle CPB is $2r$. Now we need to find the area of triangle PQB. We already have the base, and we just need to find the height. The height of that triangle equals the height of triangle ABC minus the radius. The height of triangle ABC is $\frac{12}{5}$. So the height is $\frac{12}{5}-2r$. So that triangle area is $\frac{12}{5}r-r^{2}$. Finally we have the last two triangles, AQX and CPM. Combine these two triangles because they have the same height. And they're base is 5-2r. So the area of this combined triangle is ${\frac{5r}{2}-r^{2}}$. Adding up all these area, we get $6r+\frac{12}{5}r=6$. Notice how all the $r^{2}$ cancels out.  :D  So now solving for the radius, we get that the radius is $\frac{5}{7}$. The distance between the centers is thus $\frac{10}{7}$.[/hide]
Finally done. It may look long, but really, it can be done in 3-5 minutes.

My way was without any trig at all really, just a side length ratio which I referred to as the tangent ratio since it concerned the opposite and adjacent sides to the angle.

I like your method though, its very different from the way I went about it  :thumbup:

An Entirely Different Solution Using Basic Geometry

[hide]

First, the distance between the two centers of the circle is the sum of the radii of the two circles because the circles are tangent.

Make the right triangle shown below in red with its hypotenuse being the distance between the two centers of the circles such that the sides of the drawn triangles are parallel to the sides of the original 3-4-5 triangle. Calculate the lengths of the sides of the new triangle using similar triangles.

[img]http://img.photobucket.com/albums/v735/morebadcomix/1.png[/img]

Make another similar right triangle this time using the center of the circle shown below and the radius of the circle as a leg. Calculate the lengths of the sides using similar triangles.

Make one last similar right triangle as shown below. You know the length of the side of the new triangle opposite the angle opposite the side with length 3 of the original triangle because it is also a side of the x by (5/3)x rectangle shown. This may not seem like it is so because the picture is drawn not to scale. Calculate the lengths of the sides using similar triangles.

[img]http://img.photobucket.com/albums/v735/morebadcomix/3-1.png[/img]

Using the lengths we have labeled, observe that x + (8/5)x + (4/3)x + (5/3)x = 4.

Solve for x and double the result.

[img]http://img.photobucket.com/albums/v735/morebadcomix/5.png[/img]

[/hide]


hey guyz can any one tell what is the speciality of the number 6174 and who is it named after?
thank you

6174 is Kaprekar number.  whatever 4 digit no u start with, if u arrange the no first in ascending then in descending order and then subtract bigger no from smaller one , after a series of such steps(max 7) u will get 6174


Find all positive integers$n$  for which$n^{10}+1$  is divisible by $10!$

Suppose that there exist a number $n$ such that $10! \mid n^{10}+1$.
Then also $7 \mid n^{10}+1 \Rightarrow$

$n^{10}\equiv-1 \pmod 7$

Checking all the possibilities in [b]modulo 7[/b] we will find that this equivalence has no solution

Assertion: There are no $n$ satisfying these conditions
Proof: For it to be divisible by 10! it must be divisible by 4. When we take $n^{10}+1 \pmod 4$, we must have this equal to 0 for $n^{10}+1$ to be divisible by 4.

$n^{10}+1 \pmod{4}\equiv 0\rightarrow n^{1}0\equiv 3 \pmod 4$. However, $n\pmod 4 \in\{ 0,1,2,3\}\rightarrow n^{10 }\pmod 4 \in \{0,1\}$. Thus there are no solutions for $n$ in integers. (Which implies no solutions in positives either.)

You are right!
We can also work in modulo 3, for less work

$0^{10}\equiv 0 \pmod 3$
$(\pm 1)^{10}\equiv 1 \pmod 3$

So there's no $n$ such that $n^{10}\equiv-1 \pmod 3$