Let $ a,b,c$ be postive real numbers. Prove that
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} 4\left(\frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}\right) \plus{} a \plus{} b \plus{} c \ge \frac {18(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}.$

[quote="arqady"][quote="can_hang2007"]Let $ a,b,c$ be postive real numbers. Prove that
$ \frac {a^2}{b} + \frac {b^2}{c} + \frac {c^2}{a} + 4(\frac {bc}{a} + \frac {ca}{b} + \frac {ab}{c}) + a + b + c \ge \frac {18(a^2 + b^2 + c^2)}{a + b + c}.$[/quote]
Let $ a + b + c = 3u,$ $ ab + ac + bc = 3v^2,$ $ abc = w^3$ and $ u^2 = tv^2.$
Hence, $ \sum_{cyc}(\frac {a^2}{b} + \frac {4ab}{c} + a)\geq\frac {18(a^2 + b^2 + c^2)}{a + b + c}\Leftrightarrow$
$ \Leftrightarrow u\sum_{cyc}a^3c + 36uv^4 - 75u^2w^3 + 36v^2w^3\geq0\Leftrightarrow$
$ \Leftrightarrow u\sum_{cyc}(a^3b + a^3c) + 72uv^4 - 150u^2w^3 + 72v^2w^3\geq u\sum_{cyc}(a^3b - a^3c)\Leftrightarrow$
$ \Leftrightarrow3(3u^3v^2 + 6uv^4 - (17u^2 - 8v^2)w^3)\geq u^2(a - b)(a - c)(b - c).$
Let $ f(w^3) = 3u^3v^2 + 6uv^4 - (17u^2 - 8v^2)w^3.$
Hence, $ f$ is a decreasing function. Thus, $ f$ gets a minimal value when $ w^3$ gets a maximal value, which holds when two numbers from $ \{a,b,c\}$ are equal.
Let $ b = c = 1.$ Hence, $ f(w^3)\geq0\Leftrightarrow a^4 - 7a^3 + 21a^2 - 25a + 10\geq0\Leftrightarrow$
$ \Leftrightarrow(a - 1)^2(a^2 - 5a + 10)\geq0,$ which is true.
Id est, it remains to prove that
$ (3(3u^3v^2 + 6uv^4 - (17u^2 - 8v^2)w^3))^2\geq u^4\prod_{cyc}(a - b)^2.$
But $ (3(3u^3v^2 + 6uv^4 - (17u^2 - 8v^2)w^3))^2\geq u^4\prod_{cyc}(a - b)^2\Leftrightarrow$
$ \Leftrightarrow(3(3u^3v^2 + 6uv^4 - (17u^2 - 8v^2)w^3))^2\geq27u^4(3u^2v^4 - 4v^6 + 6uv^2w^3 - 4u^3w^3 - w^6)\Leftrightarrow$
$ \Leftrightarrow(73u^4 - 68u^2v^2 + 16v^4)w^6 + 3u(u^6 - 10u^4v^2 - 13u^2v^4 + 8v^6)w^3 +$
$ + 3u^2v^6(4u^2 + 3v^2)\geq0.$
Hence, for $ u^6 - 10u^4v^2 - 13u^2v^4 + 8v^6\geq0$ the original inequality is proven.
Thus, it remains to check this inequality for $ u^6 - 10u^4v^2 - 13u^2v^4 + 8v^6 < 0,$
which gives $ 1\leq t < 11.045....$
Id est, it remains to prove that
$ (3u(u^6 - 10u^4v^2 - 13u^2v^4 + 8v^6))^2 - 12(73u^4 - 68u^2v^2 + 16v^4)u^2v^6(4u^2 + 3v^2)\leq0.$
But $ (3u(u^6 - 10u^4v^2 - 13u^2v^4 + 8v^6))^2 - 12(73u^4 - 68u^2v^2 + 16v^4)u^2v^6(4u^2 + 3v^2)\leq0\Leftrightarrow$
$ 3(t^3 - 10t^2 - 13t + 8)^2 - 4(4t + 3)(73t^2 - 68t + 16)\leq0\Leftrightarrow$
$ \Leftrightarrow(t - 1)^2(3t^3 - 54t^2 + 111t - 64)\leq0,$ which is true for $ 1\leq t < 11.045....$ Done![/quote]
Thanks, arqady. But is there any simpler solution?

[quote="can_hang2007"]
Thanks, arqady. But is there any simpler solution?[/quote]
Here is simpler:
$ \sum_{cyc}\left(\frac {a^2}{b} \plus{} \frac {4ab}{c} \plus{} a\right)\geq\frac {18(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}c(a^2 \plus{} 3bc \minus{} 2ac \minus{} 2ab)^2\geq0.$
Here is  a similar one:
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} 10\left(\frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}\right) \geq\frac {33(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}\]

Ok! very nice solution, but I want to find a different solution?


[hide]The only two possibilities for x are 11 and 2, but x can't be 2 because 11 isn't even. Therefore, $x=11$ and $y=2$.  That leaves $z=3$.  $(e)$ is the correct answer.[/hide]

[hide]From xy=22, x and y are positive integers, y is even, and x is prime, x could only equal 11 and y equal 2. As xyz=66, z=3. Therefore, y<z<x, answer [b]E[/b].[/hide]

[quote="236factorial"]The product of positive integers x, y, and z is 66. If y is even, x is prime, and xy=22 which of the following must be true?

a) y<x<z
b) x<z<y
c) x<y<z
d) z<x<y
e) y<z<x[/quote]

[hide]since xy is 22, z has to be 3. since x has to be prime, it has to be either 2 or 11. since y has to be even, y is 2 and x is 11. this makes...    y<z<x[/hide]

[hide]x*y*z=66 xy=22=11*2 x is prime so x=11. y is even so y=2. 11*2*z=66 so z=3 and y<z<x(E)[/hide]

[quote="iluvpi"][hide]x+y+z=66 xy=22=11*2 x is prime so x=11. y is even so y=2. 11+2+z=66 22+z=66 so z=44 and y<x<z(A)[/hide][/quote]

sorry pi...but you made a mistake. The problem reads that the $product$ of $x$, $y$, and $z$ is $66$, but you calculated your answer thinking that the problem was $x+y+z=66$...

z=xyz/xy=66/22=3.

xy=22=2*11

y=2, because y must be even; x=11, because x is prime.

So, x=11; y=2; z=3.

Then, the correct one is E.


vector
calculus
function
analytic geometry
linear algebra
linear algebra unsolved

lol..the title rules..now for the prob:

For a triangle ABC, find a point P in the triangle such that the vectors AP^2+  BP^2+ CP^2 is at a minimum.

GOOD LUCK..  :starwars:

Knowing a little calculus can help... That or guessing the answer.

[hide="Solution 1."]
Let $A(a,b), \, B(u,v), \, C(s,t)$ and $P(x,y)$.
We have to minimize $f : \mathbb R^{2}\to \mathbb R$, $f(x,y) = 3 \left( x^{2}+y^{2}\right)-2x(s+u+a)-2y(t+v+b)+\left( s^{2}+t^{2}+u^{2}+v^{2}+a^{2}+b^{2}\right)$.
Since it has everything you'd want from such a function (for example, it is differentiable everywhere), the local extremum point can be found out by setting $\nabla f = 0$. That gives $x = \frac{u+s+a}3$ ( :D ) and $y = \frac{t+v+b}3$, which are the coordinates of the centroid.
It's pretty clear that this is a minimum point (think about what happens when we let $x,y \to \infty$).
[/hide]
So that's how I found out the answer. Using the hint from the title of the topic, it's very easy now:

[i]Solution 2.[/i]
We have that:
\begin{eqnarray*}AP^{2}+BP^{2}+CP^{2}&=& \sum_\textrm{cyc}\overrightarrow{AP}^{2}\\ \ &=& \sum_\textrm{cyc}\left( \overrightarrow{AG}+\overrightarrow{GP}\right)^{2}\\ \ &=& \sum_\textrm{cyc}\left( \overrightarrow{AG}^{2}+2 \cdot \overrightarrow{AG}\cdot \overrightarrow{GP}+\overrightarrow{GP^{2}}\right) \\ \ &=& \sum_\textrm{cyc}\overrightarrow{AG}^{2}+2 \cdot \overrightarrow{GP}\cdot \sum_\textrm{cyc}\overrightarrow{AG}+\sum_\textrm{cyc}\overrightarrow{GP}^{2}\\ \ &=& \sum_\textrm{cyc}\overrightarrow{AG}^{2}+2 \cdot \overrightarrow{GP}\cdot \overrightarrow 0+\sum_\textrm{cyc}\overrightarrow{GP}^{2}\\ \ &=& \sum_\textrm{cyc}\overrightarrow{AG}^{2}+\sum_\textrm{cyc}\overrightarrow{GP}^{2}\\ \ &=& \sum_\textrm{cyc}AG^{2}+\sum_\textrm{cyc}GP^{2}\end{eqnarray*}

Since $\sum_\textrm{cyc}AG^{2}$ is obviously constant, it suffices to minimize $\sum_\textrm{cyc}GP^{2}$. But the minimum of this expression is $0$ (it is reached only for $P=G$), so we're done.


Let n>=2 and 0<=xi<=1 for i=1,2,3,...,n.Prove that the inequality
 (x1+x2+...+xn)-(x1x2+x2x3+...+xnx1)<=[n/2].
(when is there an equality?)

Two term AM-GM can take care of this if you factor into $x_1(1-x_2)$


When I tried to access the forums page earlier today, I got:
[quote]

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Created and Maintained by Valentin Vornicu - (c) AoPS Inc. 2004-2007 - Powered by phpBB
[/quote]

[I inserted the hide because it's a *really* long line]

It didn't happen again. Was this just a random thing?

I was just trying to preview a post in the Intermediate forum and got

[quote]Home

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SQL Request: SELECT c.class_id FROM classes c, enrollments e WHERE e.user_id = 1430 AND c.start_date <= DATE_ADD(NOW(),INTERVAL 7 DAY) AND c.end_date > DATE_ADD(NOW(),INTERVAL -50 DAY) AND e.dropped = 0

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Given triangle $ ABC$. Points $ M$, $ N$ are the projections of $ B$ and $ C$ to the bisectors of angles $ C$ and $ B$ respectively. Prove that line $ MN$ intersects sides $ AC$ and $ AB$ in their points of contact with the incircle of $ ABC$.

Let $X,Y$ be the intersections of $ MN$ with $ AB,AC,$ respectively. It suffices to see that quadrilateral $ MNCB$ is cyclic, thus $ \angle NMC \equal{} \angle NBC \equal{} \angle ABN$ $ \Longrightarrow$ $ MXIB$ is cyclic. Hence, $ \angle IXB \equal{} \angle IMB \equal{} 90^{\circ}.$ Mutatis mutandis $ \angle IYC \equal{} 90^{\circ}$ $\Longrightarrow$ $ X,Y$ are the tangency points of the incircle with $ AB,AC.$

Dear Mathlinkers,
for more see for example
http://pagesperso-orange.fr/jl.ayme/vol4.html     an unlikely concurrence p. 2-5
Sincerely
Jean-Louis


inequalities
function
inequalities proposed

Given $x,y,z \ge 0 , x + y + z = 1, xy + yz + zx = \frac{2}{9}$

Find the minimal and maximal of $x^n + y^n + z^n$

What is the condition of $n$?
$n$ is real or integer?

n is an positive integer number.

From the result of of minimal and maximal of $x^n + y^n + z^n$, consider a convergence series:

$f(x) = \sum a_{n} x^{n}$

given $a_{n} > 0$

You could get the result of the minimal and maximal of

$f(x) + f(y) + f(z)$

For example: $e^x + e^y + e^z$

To people who are interested in my problem, have you solved it??  :)

I am happy to see how great your technique is, how powerful is the inequality you uses  :)

I think both solutions are false.
The signs are strict.
Equality case cannot be attained.

[quote="Chang Woo-JIn"]Hmm.. I think my proof is right, cuz $x, y, z$ can be zero.[/quote]

your first solution requires two variables to be 0.
Your second solution used inequalities with x=y=z as equality case.
Both violates with xy+yz+zx=2/9

Sorry for my mistake; I considered just one condition in each ineq;

Sorry; :(  :(

[quote="Vasc"]According to (n-1)VE Theorem, we may assume $y=z$ for any real $n$.[/quote]

Could u show me that theorem? I am a bit curious about it. I believe that you are right. Of course, you can view the problem like finding a maximal and minimal value of a 3-variables function with some conditions.   However, if you do that, you put yourself in the position of an undergraduate student, then the problem become boring.

See topic
"just two different value" principle
in this forum.


conics hyperbola symmetry geometry trapezoid circumcircle ratio

Hi, everyone.

There is a hyperbola with two foci A and B, and infinite many circles (with center C) tangent to the
hyperbola. draw two tangent lines BD, BE from a focus B.
Prove that $ \angle{DBE}$ is always constant.

[b]with synthetic proofs (no analytic, no trigonometric)[/b]

By symmetry, circle $ (C)$ is tangent to the hyperbola at 2 points $ S, T$ symmetrical WRT the hyperbola minor axis. $ ST \parallel AB$ and $ AS \equal{} BT$ $ \Longrightarrow$ $ ASTB$ is isosceles trapezoid, cyclic, let $ (P)$ be its circumcircle centered on the minor axis. The hyperbola tangent at $ T$ is also tangent of $ (C).$ This tangent bisects $ \angle ATB$ $ \Longrightarrow$ it cuts $ (P)$ again at $ X$ on the minor axis. Angles $ \angle XTC$ is right $ \Longrightarrow$ $ C \in (P).$ Then $ BC$ bisects $ \angle SBT.$ Let $ Q \equiv BC \cap ST$ $ \Longrightarrow$ $ \frac {SB}{SQ} \equal{} \frac {TB}{TQ}.$ Since the hyperbola is locus of points with equal ratios of distances from the focus $ B$ and corresponding directrix $ b,$ it follows that $ Q \in b$ $ \Longrightarrow$ $ \frac {TB}{TQ} \equal{} \epsilon \equal{} \text{const}$ is equal to the hyperbola excentricity for any $ (C).$ Triangles $ \triangle BTC \sim \triangle TQC$ are similar, having equal angles $ \Longrightarrow$ $ \frac {BC}{TC} \equal{} \frac {TB}{TQ} \equal{} \epsilon.$ This means that the figure $ (C) \cup B$ remains similar for any $ (C)$ and $ \angle DBE \equal{} \text{const}.$
[hide]$ \angle DBC \equal{} \frac {_1}{^2}\angle DBE,\ 0^\circ < DBC < 90^\circ$ and $ \cos \widehat {DBC} \equal{} \frac {TC}{BC} \equal{} \frac {1}{\epsilon}$[/hide]

Thanks, yetti

I am unable to take my eyes off your great solution!  :wow:

By symmetry $ C$ lies on the perpendicular bisector of $ AB$ and the normal line to the hyperbola at $ P.$ In other words, $ C$ lies on the circumcircle of $ \triangle PAB$ being the midpoint of the arc $ APB.$ We must prove that ratio beetwen the distance $BC$ and the radius $ CD \equal{} CP \equal{} r$ is constant. Indeed, by Ptolemy's theorem for the cyclic quadrilateral $ ABPC,$ keeping in mind that $ CB \equal{} CA,$ we have

$ r \cdot AB \plus{} PB \cdot CB \equal{} CB \cdot AP  \Longrightarrow  \frac {r}{CB} \equal{} \frac {AP \minus{} PB}{AB} \equal{} \text{const}.$

[quote="armpist"]...[/quote]
Your previous post was spam and your next post was a lie.


$ \frac 1a<\frac 4a$
$ \frac 2{a\plus{}b}<\frac 4{b}$
since $ 2<4$ and $ a\plus{}b>b$
$ \frac 3{a\plus{}b\plus{}c}<\frac 4{c}$
since $ 3<4$ and $ a\plus{}b\plus{}c>c$

However Dr. Sonnhard has posted a similar problem that froze the algebra II marathon that is much harder, and suitable for this level.

Show that for any positive real numbers  the following inequality is fulfilled

$ \frac{1}{a_{1}}\plus{}\frac{2}{a_{1}\plus{}a_{2}}\plus{}...\plus{}\frac{n}{a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}}<4\left(\frac{1}{a_{1}}\plus{}...\plus{}\frac{1}{a_{n}}\right)$

[quote=RobRoobiks]

Show that for any positive real numbers  the following inequality is fulfilled

$$ \frac{1}{a_{1}}\plus{}\frac{2}{a_{1}\plus{}a_{2}}\plus{}...\plus{}\frac{n}{a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}}<4\left(\frac{1}{a_{1}}\plus{}...\plus{}\frac{1}{a_{n}}\right)$$[/quote]
[url=https://artofproblemsolving.com/community/c6h406491p2269940]Russian olympiad 1986[/url]


calculus
integration
algebra
function
domain
abstract algebra
Ring Theory

Let $R$ be a Dedekind ring and $P$ be it's prime ideal, let $R_{p}$ be a local ring at $P$. Prove that $R_{p}$ is also a Dedekid ring.

You could, for example, characterize Dedekind rings as those integral domains which are Noetherian, integrally closed, and with Krull dimension $1$ (i.e. all non-zero prime ideals are maximal). All these properties are conserved when you pass to the ring $S^{-1}R$ for some multiplicatively closed subset $S$ of $R$ (well, except maybe for the Krull dimension: it could decrease and become $0$, in which case $S^{-1}R$ becomes a field; this is not the case for our rings $R_{P}$ though), so all rings of fractions $S^{-1}R$ (which are not fields) of a Dedekind domain $R$ are Dedekind domains. In particular, this holds when $S=R\setminus P$, the case which interests us.

In fact, if you take for granted all basic properties of Dedekind domains, it's not hard to prove that $R_{P}$ are actually discrete valuation rings: principal ideal domains with exactly one non-zero prime ideal.

Thank you, Grobber. But the thing is i'd like to have a proof using the following definition of Dedekind ring : as a commutative integral domain with $1$ such that it's fractional ideals from a group.

And what i actually interested is how to show that under the hypothesis that $R$ is Dedekind show that the fractional ideals of $S^{-1}R$ form a group.


http://www.math.umd.edu/highschool/mathcomp/2008winner.html

Congratulations to Holden, Qinxuan, and Aki!

Honorable Mention and Baltimore County Winner for me.  I don't know where I ranked yet.

Yeah, finally I'm not the 3rd place any more.

P.S. Is Holden new to Maryland?

this is the most recent one right?
i did terrible.... i suck at writing proofs
gratz top 3 =)
according to the magic piece of paper with rankings, sam zbarsky got 4th 0.o middle schoolers ftw =D

Zbarsky got fourth as an 8th grader?  Wow.

I ended up getting fifth.  If only I hadn't failed Part I . . .

least thats what i think 0.o dont quote me on it, but im pretty sure =/ im scared of what he'll be next year

what are the middle school results? can you post them?... cuz i wanna find out where i ranked... if i ranked at all. :P i did badly on part II. :blush:

I doubt that separate middle school results have been compiled, but Dr. Washington might tell you if you asked him.

By the way, it appears that an 8th grader from MD got a 12 on the AIME.  I assume this is Zbarsky?


There are 11 distinct points on a plane. You have to join pairs of these points by straight
lines of three colors; red, blue and green. What is the maximum number of lines that can
be drawn so that no triangle of sides of same color is formed?

What are you and RedAlice both cheating on?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=253940

Probably the same thing that was mentioned here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=253893


A sequence of real numbers has first term $ a_0 \equal{} 1$, and satisfies $ 2^{a_n} \equal{} 4^{a_{n \minus{} 1}} \minus{} 1$ for $ n \geq 1$.

Find, with proof, the number of integer terms in the sequence.

[hide]Assuming the irrationality of $ \log_2 3$ (I haven't actually got the foggiest)

We obtain the formula

$ a_n = \log_2 ( 4^{a_{n - 1}} - 1)$

The first few terms are $ a_0 = 1, a_1 = \log_2 3, a_2 = 3, a_3 = \log_2 63$

Now consider a general next term (taken as $ \log_2 k$)

$ a_x = \log_2 ( 4 ^{\log_2 k} - 1 )$

However $ 4^{\log_2 k} = k^2$

Therefore it can be written as

$ a_4 = \log_2 ( 63^2 - 1 )$

Now consider the inside of the $ \log$ for $ a_n$, $ n \geq 5$ $ \pmod{3}$

Clearly $ a_5 = \log_2 ( (63^2 - 1)^2 - 1 )$ Therefore $ 3|I(a_5)$, where I is the inside

If $ 3|I(a_n) \Rightarrow 3|I(a_{n + 2})$ So clearly we have $ a_{2n+1} = \log_2 ( 3 \cdot m )$ which is irrational $ \forall m$

If $ 31|(a_n) \Rightarrow 31|(a_{n+1})$ So clearly we have $ a_{2n} = \log_2 ( 31 \cdot m )$ which is irrational $ \forall m$ (Since $ 31|a_6$)

Therefore there are 2 rational cases, $ a_0$ and $ a_2$
[/hide]

[b]Edited, Thanks TZF[/b]

[quote="SimonM"]Assuming the irrationality of $ \log_2 3$ (I haven't actually got the foggiest)[/quote]

This is straightforward.  If $ \log_2 3 \equal{} \frac{p}{q}$ for integers $ p, q$ then $ 3^q \equal{} 2^p$, contradiction.

SimonM, my gripe with your solution is your line $ 3|I_n \implies 3|I_{n \plus{} 1}$.

$ a \equiv 0 \pmod{3} \implies a^2 \minus{} 1 \equiv \minus{} 1 \pmod{3}$, so it would only prove it for every other number. It's quite easy to patch that up, though :)

[quote="The Zuton Force"]SimonM, my gripe with your solution is your line $ 3|I_n \implies 3|I_{n \plus{} 1}$.

$ a \equiv 0 \pmod{3} \implies a^2 \minus{} 1 \equiv \minus{} 1 \pmod{3}$, so it would only prove it for every other number, I think.  :)[/quote]

Ah yes, I see. You can fix that since the other ones are divisible by $ 31$

Proof will follow if you so wish

[hide="Another Solution"]Define a new sequence $ b_{n}$ such that $ b_{n}\equal{}2^{a_{n}}$ for all $ n\ge 0$.  Then we have $ b_{0}\equal{}2$, and $ b_{n}\equal{}b_{n\minus{}1}^{2}\minus{}1\equal{}(b_{n\minus{}1}\minus{}1)(b_{n\minus{}1}\plus{}1)$ for $ n\ge 1$.  Clearly all the terms of $ b_{n}$ are integers.

Now suppose $ a_{n}$ is an integer for some $ n\ge 1$.  Then $ b_{n}$ must be a perfect power of 2, so $ b_{n\minus{}1}\minus{}1$ and $ b_{n\minus{}1}\plus{}1$ must both be perfect powers of 2.  But 2 and 4 is the only pair of perfect powers of 2 with difference 2, so $ b_{n\minus{}1}\equal{}3$.

We have $ b_{0}\equal{}2$, $ b_{1}\equal{}3$, and the sequence is increasing so 3 cannot appear again.  Therefore the only perfect powers of 2 in the sequence are $ b_{0}\equal{}2$ and $ b_{2}\equal{}8$, so there are two integer terms in the original sequence.[/hide]

[quote="The Zuton Force"][quote="matt276eagles"][hide="Another Solution"]... Then $ b_{n}$ must be a perfect power of 2, so $ b_{n \minus{} 1} \minus{} 1$ and $ b_{n \minus{} 1} \plus{} 1$ must both be perfect powers of 2... [/hide][/quote]

This was the intended approach. In retrospect, I should've put this in intermediate... I feel bad for contributing to the easy problems in PO  :([/quote]But I was so happy I solved a problem in PO..


What features related to hydrogen bonding give storage polysaccharides, such as starch and glycogen, very different properties from structural polysaccharides, such as cellulose and chitin?

For glucose, there are 2 kinds, an alpha type and a beta type. Basically, the difference between these two are its placement of the -OH group. For storage polysaccharides such as starch or glycogen, all the glucose molecules are alpha type while for cellulose they are all beta type. Because of this, every glucose is essentially "upside down" compared to the one next to it, giving the celluose fiber a very rigid structure.

As for chitin, they contain a nitrogen group.


geometry
analytic geometry
trigonometry
quadratics
calculus
derivative
algebra

If $O$ is a given point on the prolongation of diameter $BA$ of a semi-circle, and if $ODC$ is a secant cutting the semi-circle at $D$ and $C$, prove that quadrilateral $ABCD$ has maximum area when the orthogonal projection of $DC$ on $AB$ is equal to the radius of semi-circle.

WLOG, assume that the points $O, A, B$ and the points $O, D, C$ follow on the rays $OA, OD$ in these orders. Put the coordinate origin at the point $O$ and the positive x-axis along the ray $OA$. Denote the angle $\angle AOD \equiv BOC = \phi$. The area $\Sigma$ of the quadrilateral $ABCD$ is then equal to the difference of the areas of the triangles $\triangle OBC, \triangle OAD$:

$\Sigma = |ABCD| = |\triangle OBC| - |\triangle OAD| = \frac 1 2 (OB \cdot OC - OA \cdot OD) \sin \phi$

Let $p = OP = \frac{OA + OB}{2}$ be the x-coordinate of the circle center $P$ and $r = PA = PB = \frac{OB - OA}{2}$ the circle radius. Then $OA = p - r$ and $OB = p + r$. The equations of line $OD \equiv OC$ and the circle $(P)$ are:

$OC:\ \ y = x \tan \phi$

$(P):\ \ (x - p)^2 + y^2 = r^2$

Substituting the line equation into the circle equation and using $\cos^2{\phi} = \frac{1}{1 + \tan^2{\phi}}$, we obtain a quadratic equation for the x-coordinates of the quadrilateral vertices $C, D$:

$\frac{x^2}{\cos^2{\phi}} - 2xp + p^2 - r^2 = 0$

$x = \cos^2{\phi} (p \pm \sqrt{p^2 - \frac{p^2 - r^2}{\cos^2{\phi}}}) = \cos \phi ( p \cos{\phi} \pm \sqrt{r^2 - p^2 \sin^2{\phi}} )$

$y = x \tan \phi = \sin \phi ( p \cos{\phi} \pm \sqrt{r^2 - p^2 \sin^2{\phi}} )$

By assumption, $x_D < x_C,\ y_D < y_C$. Thus the expressions with the plus signs yield the coordinates of the vertex $C$ and the expressions with the minus signs the coordinates of the vertex $D$. Consequently,

$OC = \sqrt{x_C^2 + y_C^2} = p \cos{\phi} + \sqrt{r^2 - p^2 \sin^2{\phi}}$

$OD = \sqrt{x_D^2 + y_D^2} = p \cos{\phi} - \sqrt{r^2 - p^2 \sin^2{\phi}}$

Let $E, F$ be the orthogonal projections of the vertices $C, D$ on the diameter $AB$ and denote $q = EF$. Then

$q = EF = OE - OF = x_C - x_D =  2 \cos{\phi} \sqrt{r^2 - p^2 \sin^2{\phi}}$

Substituting the expressions for $OA, OB, OC, OD$ into the formula for the quadrilateral area, we get

$\Sigma = \frac 1 2 \sin \phi [(p + r) ( p \cos{\phi} + \sqrt{r^2 - p^2 \sin^2{\phi}} ) - (p - r) ( p \cos{\phi} - \sqrt{r^2 - p^2 \sin^2{\phi}} )] = $

$= p \sin \phi (r \cos \phi + \sqrt{r^2 - p^2 \sin^2{\phi}})$

For the quadrilateral area to be maximum, the derivative $\frac{\text d\Sigma}{\text d\phi} = 0$ has to be zero. Once we are done differentiating by $\phi$, we can substitute $q = 2 \cos \phi \sqrt{r^2 - p^2 \sin^2{\phi}}$.

$\frac{\text d\Sigma}{\text d\phi} = p \cos \phi (r \cos \phi + \sqrt{r^2 - p^2 \sin^2{\phi}}) - p \sin \phi (r \sin \phi + \frac{p^2 \sin \phi \cos \phi}{\sqrt{r^2 - p^2 \sin^2{\phi}}}) =$

$= p [r (\cos^2{\phi} - \sin^2{\phi}) + \frac{q}{2} + (\frac{q^2}{4 \cos^2{\phi}} - r^2) \frac{2 \cos^2{\phi}}{q} ] =$

$= p [r (2 \cos^2{\phi} - 1) + q - \frac{2 r^2 \cos^2{\phi}}{q} ] =$

$= \frac p q ( 2rq \cos^2{\phi} - rq + q^2 - 2r^2 \cos^2{\phi} ) =$

$= \frac p q [2r \cos^2{\phi}(q - r) + q(q - r)] =$

$= \frac p q (q - r)(2r \cos^2{\phi} + q) = 0$

The possibility of factoring the ugly derivative $\frac{\text d\Sigma}{\text d\phi}$ is rather surprising. It is clear that the only factor that can be zero is $q - r = 0$, because all terms of the other factor are positive. Hence, when the quadrilateral area is maximum, $q = EF = r$ and the maximum area angle $\phi$ is given by the following quadratic equation:

$r = 2 \cos \phi \sqrt{r^2 - p^2 \sin^2{\phi}}$

$r^2 = 4 \cos^2 \phi [r^2 - p^2 (1 - \cos^2{\phi})]$

$p^2 \cos^4{\phi} - (p^2 - r^2) \cos^2{\phi} - \frac{r^2}{4} = 0$

$\cos^2{\phi} = \frac{p^2 - r^2 \pm \sqrt{(p^2 - r^2)^2 + p^2r^2}}{2p^2}$

Since $\cos^2{\phi} \ge 0$, only the positive root (with the plus sign) is acceptable. Thus given the distance $p = OP = \frac{OA + OB}{2}$ and the circle radius $r = PA = PB = \frac{OB - OA}{2}$, the maximum area quadrilateral $ABCD$ is even constructible.

My Solution :

Let $Q$ be the center of the semicircle, $r$ is the radius of the semicircle, and $L$ is the length $OQ$, $r$ and $L$ are given constants.
We proceed to express the area of $ABCD$ in terms of variable $\alpha=\angle DOA$. Let $M$ be the foot of perpendicular from $Q$ to $CD$, then $M$ bisects $CD$.
From triangle $OMQ$, $OM=L\cos \alpha$ and $MQ=L\sin \alpha$. Applying Phytagoras theorem to triangle $DMQ$, we find $DM=\sqrt{r^2-L^2 \sin^2 \alpha}$

$[ABCD]=[OBC]-[OAD]$
$[ABCD]=\frac{1}{2}OB.OC \sin \alpha- \frac{1}{2}OA.OD \sin \alpha$
$[ABCD]=\frac{1}{2} \sin \alpha ((L+r)(L\cos \alpha +\sqrt{r^2-L^2 \sin^2 \alpha})-(L-r)(L \cos \alpha- \sqrt{r^2-L^2 \sin^2 \alpha}))$
$[ABCD]=L \sin \alpha (\sqrt{r^2-L^2 \sin^2 \alpha}+L \cos \alpha)$

When the area of $ABCD$ is maximum, the derivative of this expression with respect to $\alpha$ is equal to $0$. From this we obtain\[(2h \cos \alpha -r)(h+r \cos \alpha)=0,\]where $h=\sqrt{r^2-L^2 \sin^2 \alpha}=DM=CD/2$

Therefore, $r=2h \cos \alpha=CD \cos \alpha$

Q.E.D


Let $a$, $b$, $c$ be sides of $\triangle ABC$

Prove: $-a^3+a^2b+a^2c+ab^2-2abc+bc^2+ac^2-b^3+b^2c-c^3>0$

should it be greater or equal to 0? Because I tried an equilateral triangle with sides 8,8,8 and I got 0. And with 7,8,8 I get -71... maybe you wanted to say a   >=  b  >=  c and make it greater/equal to 0?

i think he forgot to add bc²  :?  (not sure though)
edit: and the > should indeed be  >=  :)

Could I change the answer? Maybe yes. :D
a,b,c is a side of a triangle. prove that
\[a^{3}+b^{3}+c^{3}-a^{2}b-ab^{2}-b^{2}c-bc^{2}-c^{2}a-ac^{2}+2abc<0\]
Now it's easy to prove.

[quote="mathpk"]Could I change the answer? Maybe yes. :D
a,b,c is a side of a triangle. prove that
\[a^{3}+b^{3}+c^{3}-a^{2}b-ab^{2}-b^{2}c-bc^{2}-c^{2}a-ac^{2}+2abc>0\]
Now it's easy to prove.[/quote]

Um, $(a+b-c)(b+c-a)(c+a-b)>0$, which is the reverse inequality.


MATHCOUNTS
Gauss
AMC
USA(J)MO
USAMO
USAMTS
AMC 8

Who do you guys think will win the 2009 IL MATHCOUNTS? Please make predictions here. ALso feel free to include multiple ppl and who will make the team, win CD round, top team, etc.

If you're gonna predict the TOP person, do not vote for yourself if you're gonna mention yourself.

Um.. I'm not sure because I've never done mc before lol.. :P

Ravi probably...maybe you megan/gauss
Or the other people.  I don't see any others making it, unless they sstep up their game a lot.

I don't think gauss likes people revealing her name  :wink:

Some interesting facts to consider: :D

-Ravi: Took 4th place :lol:  last year in state mathcounts, 52nd (?) place in Nats, 2007 AMC 8 24/25, 2008 AMC 8 perfect, taking WOOT this year, 7th grader this year, wants to make USAMO (?)

-Xiaoyu: Took 5th place last year in state mathcounts, 1st place chapter in 2008, 1st place chapter team :lol: , 10th (?) place in Wisconsin Chapter mathcounts as a 6th grader, top 10 in WI state, no known AMC participation, 8th grader, Female, likes  :P

-Megan: Took 6th place last year in state mathcounts, 4th place chapter in 2008, nth place chapter team in 2008 where n>12 :( , 6th place chapter in 2007, 7th place chapter team in 2007, Didn't take AMC 8, going to take AMC 10 this year, Participates in USAMTS and got an 11/25 in Round 1, likes  :D , 8th grade.

-Patrick: Took 7th place last year in state mathcounts, 1st place chapter in 2008, 12th (I think) place chapter team 2008, 2009 is his second time participating in mathcounts, 8th grade.

OTHERS:

-Kennedy JHS took 2nd place team in 2008 states, 1st place team in 2007 states
-Avery Coonley took 1st place team in 2008 states, had 2 kids in top 10 in 2008 states

Argh.
Too many smilies in your post: 12.
You can have at most 10 smilies in this message. Please remove some smilies before submitting this message.

WHY DO THOSE  :lol: s have to count as 2 smilies???????????????????????????/

muahahaaha lol i voted megan chen cuz gaussy is awesome  :D whoa why do  :lol: s count as two?!?!?!?!?!!?!?!? that is really messed up

Did you notice that most of the male AoPSers have no preferences over smilies? :D
 :lol: =worst smiley ever :D See that?
Too many smilies in your post: 5.
You can have at most 4 smilies in this message. Please remove some smilies before submitting this message.

Hmm.. I never noticed that

guineapiggies do you live in IL?

She does. She did mathcounts last year. And she's in 9th grade right now.

[quote="Stargirl240"]Interesting... lol, I did AMC 10 when I was in 6th grade. Rather not tell my score though.[/quote]
Off Topic Spam, huh stargirl? Lol thats a lot weirder too.

But anyways

There's a new kid on the block:

[b]James Tao[/b]

1st place in my chapter

If you're gonna vote for him in the predictions poll, please quote this post.

[quote="Stargirl240"]Interesting... lol, I did AMC 10 when I was in 6th grade. Rather not tell my score though.[/quote]

lucky. KJHS only offered AMC 8 for 7th/8th graders and only sometimes 10/12.

[quote="gauss1181"][quote="Stargirl240"]Interesting... lol, I did AMC 10 when I was in 6th grade. Rather not tell my score though.[/quote]
Off Topic Spam, huh stargirl? Lol thats a lot weirder too.[/quote]
Oops. Err.. nevermind peoples... ignore that.  :blush:

Update:

All of the IL chapter contests have finished! Any new predictions? Individual winners, team winners, countdown qualifiers, countdown winners, nats qualifiers, etc.

Team winner: Not my school
Winner: Dunno
Qualifying: Top four non-eighth graders from last year, I hope =P

Also as a side note, do not vote for yourself as a prediction for winners. Do not vote for your school either. If you think you're gonna win, say the next top person.

Team Winner: Kennedy JHS or Avery Coonley School
Individual: Ravi maybe...or someone who dominated...?
Countdown: Ravi Jagadeesan
Nats: Ravi Jagadeesan (definite), and three other 8th graders

patrick?

Who knows.. we all thought Ravi was going to win.

It was pretty shocking.......

So how did it go?  Nats for y'all I hope?
I wonder how much Kennedy failed...lol

[quote="Aryth"][quote="cheesymath"]Why didn't anybody have faith in me?  The competition is over and I only have one vote (and that was from me after I won).[/quote]

You need to either spam more on the AoPS forums or be spammed about more.  :wink:[/quote]
How about no spam? That'd be great. Do more math on the forums instead. :D

lol manhattanative
Teams:
1. Kennedy Junior High School
2. Avery Coonley School
3/4. University Laboratory High School or UOC Lab Schools (can't remember which was which)
5. Plum Grove Junior High School
6. (don't remember)
7. Lincoln Junior High School

Individuals:

Nats Team:
1. Patrick Liscio, Lincoln Junior High School
2. Megan Chen, Wood Oaks Junior High School
3. James Tao, Plum Grove Junior High School
4. David Wang, Hinsdale Middle School

Also noted: stargirl got 15th place, ravi jagadeesan 5th place, suma_milli 60something place, helagha was 13th

Countdown:

1. Eric Lullo, Kennedy Junior High School
2. Ravi Jagadeesan, Avery Coonley School
3. Megan Chen, Wood Oaks Junior High School
4. Thomas Wu, Kennedy Junior High School

EDITS: I found out Eric's last name.
It's DAVID Wang, not Kevin. Sorry for mixing up the names. And he's from Hinsdale MS.
helagha took 13th.

LOLOLOL THOMAS WU
that's hilarious...how he got 4th in countdown...wow I almost died laughing. xD
I also lol at how Kennedy is once again saved by a newcomer, in this case I'm assuming Kevin Wang, but I congratulate them nonetheless
Congrats to you all too! =P

ANd....how did Ravi not make nats??

i have no idea. :maybe:

BTW, I saw Josh Dement and Adam Kalinich at states today. :D

:D  :D  :D
well at least i pwned him. lol xD no offense ravi if you're reading this

Josh is my slave.

wow how does KJHS survive?  :huh:

[quote="Aryth"]Josh is my slave.[/quote]

 :rotfl:  :rotfl:  :rotfl:

[quote="Aryth"]wow how does KJHS survive?[/quote]
IDK! MAYBE THEY JUST BEAST

i got 13th  :(
also, i think kenedy changed their team cause they got 4th in chapter and only like one person got from the top ten
 (i think 7th or something around that) :huh:

Lol Kennedy is just weird. They send all 8th graders to compete each year, and each year these first-timers somehow win.  Well, I guess that's a generalization, as there have been 7th graders who went to state before...
Except for last year.  We failzed and got 2nd >.<

[quote="cheesymath"]How can any team end up with five people in the to 16 at state?[/quote]

How can any team have five people?

Guys, there's no score discussion allowed on AoPS until the state competitions are all done and the tests are up on the MATHCOUNTS website. This is a general yearly AoPS rule.

[quote="ManHatTaNatiVe94"]That's pretty sad...he totally let us all down.  We thought he was gonna keep pwning more and more until he was done with mathcounts. D:
And congrats on meeting these math celebrities; I used to see Josh on a daily basis back in 8th grade lol. :D  I've seen Adam before but I've never really talked to him.[/quote]
who are these "math celebrities"?
i see kevin chen often. i see bobby shen REALLY often
mark zhang went to my school but then got recruited to fort settlement

I'm using Illinois standards here.  I'm sure you know more than I do. :P

[quote="ManHatTaNatiVe94"]Lol Kennedy is just weird. They send all 8th graders to compete each year [/quote]

not true.

[quote="ManHatTaNatiVe94"], and each year these first-timers somehow win. [/quote]

not true.

[quote="ManHatTaNatiVe94"] Well, I guess that's a generalization, as there have been 7th graders who went to state before... [/quote]

and sixth graders.

[quote="ManHatTaNatiVe94"] Except for last year.  We failzed and got 2nd >.<[/quote]

true  :wink:

Lol but Kennedy still beasts overall. The school always like gets a team trophy every year at states :lol:

Just joined to clear up some misinformation about kennedy.
First, they didn't switch members from chapter to state.
Second, they won huge at chapter - 3rd, 4th, 5th, 6th individually.  And Eric Lullo won the countdown.
Third, they don't have 5 members - Taylor Herr qualified for state as an individual  by taking first at chapter in the written part.
And finally, they have new members each year because it is incredibly competitive to get on the team.  Kennedy's top rejects would get on any other team in the state (including Avery C.).  Just look at what Taylor did - 1st in chapter and 12th-ish in state and he didn't even make the team - he was an alternate!
So... Kennedy isn't weird, they are just brilliant.

[quote="helagha"]Pat, what was your score?
and Megan, what was yours?[/quote]

Sorry, I can't post my score now.

:o trainer are you by chance related to rebecca?  :!:

Dude READ THE STICKIES IN THE MATHCOUNTS FORUM. DO NOT DISCUSS STATE SCORES UNTIL THE ADMINS SAY IT'S OKAY. Please edit/delete your post.

zOMG I GOTS A 1337 SPRINT AND A 42 TARGET
and I got 1729 of the CD problems.

y'all n00bs lolzpwned.

wow i'm feeling really out of it today.


geometry
geometric transformation
reflection

Why can't I link to the website of mathematical reflections?
Is it open to the public?

[quote="SUPERMAN2"]http://www.reflections.awesomemath.org/[/quote]
thank you very much :D


In the rectangle ABCD, the side AB has length srqt(2) and the side AD has length 1. Let the circle with centre B and passing through C meet AB at X.
Find angle ADX (in degrees).

I cheated on this one :oops:.

[hide="My Solution"]
Because the radius of the circle is $ 1$, $ AX \equal{} AB \minus{} BX \equal{} \sqrt {2} \minus{} 1$. By the given information, $ AD \equal{} 1$. Let the foot of the perpendicular from $ X$ to $ \overline{CD}$ be $ Y$. $ AXYD$ is a rectangle because $ \overline{XY}$ is a tangent to the circle. Drawing $ \overline{DX}$ we have two right triangles.

$ \tan{\angle ADX} \equal{} \sqrt {2} \minus{} 1 \implies \angle ADX \equal{} 22.5^\circ$

(I have no clue how you'd find the arctan of that value without a calculator unless you recognized it/memorized it.)
[/hide]

I have a simpler solution for this Sunehra:

By Pythagoras, XC =sqrt(2). Therefore angle XCB = 45 degrees. Therefore angle DCX = 45 degrees and XC = CD implies triangle DCX is isosceles. Therefore angle CDX = (180 - 45)/2 = 67.5. Therefore angle ADX = 90 - 67.5 = 22.5.

Any ideas on the problem posted as "circular arcs and a right-angled-triangle"?


algebra
polynomial
number theory unsolved
number theory

I have two problem about irreducible and reducible:
First problem:let [tex]p[/tex] be a prime number.Show that polynomial:[tex]P(x)=x^{p-1}+2x^{p-2}+...+(p-1)x+p[/tex] is irreducible in Z[x]
Second one:Let polynomial [tex]Q(x)=x^4+x^3+x^2+x+1[/tex].Show that:[tex]Q(5y^2)[/tex] is recible in Z[y]

For the first one, look at $\tilde P(x)=P(x+1)$ instead.

If we reduce $\tilde P$ modulo $p$, we get $(x+1)x^{p-2}$. If we assume that $\tilde P=fg$, then $f$ will be equivalent to $(x+1)x^a$ modulo $p$, while $g$ will be equivalent to $x^b$ modulo $p$, where $a+b=p-2$ and $a,b>0$. This means that the free terms of both $f$ and $g$ are divisible by $p$, so the free term of $\tilde P$ must be divisible by $p$. However, its free term is $\frac{p(p+1)}2$, contradiction.

Could someone write why this polynomial looks like that after reduction of $ P(x\plus{}1)$ modulo $ p$?

For the second one, if $ Q(x) \equal{} x^4\plus{}x^3 \plus{} x^2 \plus{} x\plus{}1$ then:

$ Q(5y^2) \equal{} (25y^4 \minus{} 25y^3 \plus{} 15y^2 \minus{} 5y \plus{} 1)(25y^4 \plus{} 25y^3 \plus{} 15y^2 \plus{} 5y \plus{} 1)$

To see it better, $ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1 \equal{} (x^2 \plus{} 3x \plus{} 1)^2 \minus{} 5x(x\plus{}1)^2$. From this we easily see that if we plug instead of $ x$ the expression $ 5x^2$, the second term becomes a perfect square, and so we can factorize.


linear algebra
matrix
algebra
polynomial
vector
linear algebra unsolved

Please pay attention to the fact that my math knowledge wasn't gained in English so I may not be familiar with some of the expressions here.

my question is: (it's taken from a test)
I have a matrix  A=Matrix(3X3), which is symmetrical and has REAL values (belongs to R). It's characterial polynomial is (x-1)(x-2)(x-3). Does exist a vector x such that |x|=1  and |Ax|=4 ?

thanks, Tom.

Your matrix is diagonalisable with diagonal $(1;2;3)$

You are looking if exist a vector $w=(a,b,c)$ with norm $1$ such that $||Aw||=4$
$a^2+b^2+c^2=1$
$a^2+4b^2+9c^2=16$
or
$a^2+b^2+c^2=1$
$3b^2+8c^2=16$
from now not difficult to finish

if I choose my matrix A to be diag(1,2,3)
then:
a^2+b^2+c^2=1^2=1
a^2+4b^2+9c^2=4^2=16

it's easy to see that there isn't a solution to this equation...

Anybody else wants to try to solve ?

regards, Tom.

How complicated! Just write $ 16=a^2+4b^2+9c^2\leq 9(a^2+b^2+c^2)=9$.

"harazi rusé comme un renard"   ;) I don't know how to say in english

I understand "harazi", "comme", "un", "renard", but "ruse" I don't know. Anyway, I don't think it's good... :P .

please pay ATTENTION to the first message in this thread!
MY QUESTION STILL WASN'T ANSWERED... :(

anybody ? help?

TOMTOMMMM.

Your question has been answered- it's impossible.

A key point which should have been mentioned: your matrix is orthogonally diagonalizable because it is symmetric. Transform to an orthogonal basis of eigenvectors, and the result is clear using the Euclidean norm.

apparently the result is not so clear to me...
I know that there exists P , orthogonal matrice, such that : A=P^-1 Diag(1,2,3) P.
also, I know that |v|=1
what's next ? please try to explain the levels of your proof, step by step.

tomtommmm.

Suppose $A$ is a real symmetric matrix with eigenvalues $\lambda_1,\dots,\lambda_n.$ Then if $||x||=1,\,||Ax||\le \max(|\lambda_1|,\dots,|\lambda_n|).$ Write that last maximum as $|\lambda|_{\text{max}}.$

The exists an orthogonal matrix $Q$ such that $Q^TAQ=D=\text{diag}(\lambda_1,\dots,\lambda_n).$

$||Ax||^2=(Ax)^T(Ax)=x^TA^TAx=x^TQ^TD^TQQ^TDQx=x^TQ^TD^2Qx$

This can be written as $y^TD^2y$ where $y=Qx$ and $||y||^2=x^TQ^TQx=||x||^2=1.$

$y^TD^2y=\sum_{j=1}^n\lambda_j^2y_j^2\le \sum_{j=1}^n\lambda_{\text{max}}^2y_j^2=|\lambda|_{\text{max}}^2||y||^2= |\lambda|_{\text{max}}^2.$

So $||Ax||^2\le |\lambda|_{\text{max}}^2$, as claimed.

More generally, for a matrix $A$ which is not necessarily symmetric, $\max_{||x||=1}||Ax||^2$ is the largest eigenvalue of $A^TA.$

In your particular example, $A$ is not merely symmetric but also positive definite. With some slight attention to detail in the above proof, we could prove that for $||x||=1,$ $1\le ||Ax||\le 3.$


Find all positive integer solutions x, y, z, p with p a prime, of the equation x^p + y^p = p^z.

I found some solutions but I can't prove that I found all.
Thanks.

There are some obvious solutions when $p=2$ ($z=2k+1,x=y=2^k,\ k\ge 0$), and it's easy to prove that they're the only ones, so let's assume that $p$ is prime.

Asume WLOG that $p\not|x,y$. We have $x+y|p^z$, so $y=p^a-x$ for some $1\le a\le z$. Now write $x^p+(p^a-x)^p=p^z$, and expand $(p^a-x)^p$ to get $(p^a)^p-\binom p1(p^a)^{p-1}x+\binom p2(p^a)^{p-2}x^2-\ldots+\binom p{p-1}p^ax^{p-1}=p^z\ (*)$. From $(*)$, we conclude that $z$ must be equal to $a+1$ (the smallest exponent of $p$ in one of the terms of $(*)$ is $a+1$, occuring only in $\binom p{p-1}p^ax^{p-1}$), so $x^p+y^p=p^{a+1}=p(x+y)$. Since $p\ge 3,\ x,y\ge 1$, some simple estimates show that $p=3,\{x,y\}=\{1,2\}$ (and $z=2$).

P.S.

I might actually have missed some small solutions; sorry if this is the case :).

[b]Case 1:[/b] $x=y$
Since thr LHS is even we have that $p=2$ and the diophane becomes $x^2=2^{z-1}$ thus $x=2^{\frac{z-1}{2}}$ and we see that $z$ is odd.
Now the solution on this case is $(x,y,z,p)=(2^{\frac{z-1}{2}}, 2^{\frac{z-1}{2}},z,2)$
[b]Case 2:[/b] $x \ne y$
W.L.O.G $x>y$ then assume that $(x,y,p) \ne (2,1,3)$ thus by Zsimondy we have that there exists a prime $q$ such that $q \mid x^p+y^p$ and $q \not \; \mid a+b$ and with this we have that $q \mid p^z$ meaning that $q=p$.
Assume that $kp^{\alpha}=x$ and $lp^{\beta}=y$ where $\gcd(k,p)=\gcd(l,p)=1$ then replacing we have $p \beta<z$ and $k^pp^{\alpha-\beta}+l^p=p^{z-p \beta}$ thus $p \mid l$ and we have a contradiction!!.
Thus we have $p \not \; \mid x,y$ thus by FLT: $x+y \equiv 0 \pmod p$ and we get contradiction!!.
Thus we have $(x,y,p)=(2,1,3)$ meaning that $z=2$.
Thus all the solutions are: $(x,y,z,p)=(2^{\frac{z-1}{2}}, 2^{\frac{z-1}{2}},z,2), (2,1,2,3), (1,2,2,3)$
And we are done  :D  :blush:


I'm not very clear with the bondings,   e.g. covalent bond , etc.
Can anyone tell me the meaning of each bond and the difference between them?
 :?
 :?

In brief, covalent bond = sharing of electrons
ionic bond = transfer of electrons.

But there are rarely pure covalent/ionic bonds.


geometry
circumcircle
incenter
geometry proposed

Let P be an arbitrary point on the side  of triangle ABC.Circle O1 tangent to AP,BP and intouches Circle O simultanously at D,where Circle O is the circumcircle of triangle ABC.The mixtilinear circle of angle A incircles AB at E. I1 is the incenter of  triangle ABP
Prove that I1,E,D ars collinear

Dear plane geometry:
I think your problem is false! please you check and fix it!

Hi,dear Mr.Fang!
Why my problem is false?My Geometer's Sketchpad told me it is so :(

There must be something wrong,I am sorry,guys :blush:
Thank  you for your interests!  :lol:


function
algebra
polynomial
quadratics
sum of roots

Determine all real numbers $ p$ such that the three roots of \[ 5x^3 \minus{} 5(p \plus{} 1)x^2 \plus{} (71p \minus{} 1)x \minus{} (66p \minus{} 1) \equal{} 0\] are positive integers.

since all the roots are integers then the sum, product two at a time and product also have to be an integer
sum of roots = 5(p+1)/5 = p+1 is an integer for all p
71p-1 = 0 mod 5
so p = 1 mod 5
66p-1 = 0 mod 5
p = 1 mod 5
sum of roots = p+1 so p >= 2
because each root >= 1
so p = every fifth integer starting from 6

$ (x \minus{} 1)(5x^2 \minus{} 5px \minus{} 66p \minus{} 1) \equal{} 0$
Let $ a,b$ be roots of polynomial $ 5x^2 \minus{} 5px \plus{} (66p \minus{} 1)$
Therefore $ a \plus{} b \equal{} p , 5ab \equal{} (66p \minus{} 1)$
$ \therefore 5ab \minus{} 66(a \plus{} b) \equal{} \minus{}1$
$ \therefore (5a \minus{} 66)(5b \minus{} 66) \equal{} 4351 \equal{} (19)(229)$
Therefore $ a \equal{} 17,b \equal{} 59 \rightarrow p \equal{} 76$

[hide]It factors to $ (x-1)(5x^2-5px+66p-1)$
The roots of the quadratic are $ \frac{5p \pm \sqrt{(5p-132)^2-132^2+20}}{10}$.  Clearly $ p$ is a positive integer.
So it suffices to find all $ p$ such that that thing under the square root is a perfect square.  Let that thing be $ n^2$.  Then
\[ (5p-132-n)(5p-132+n)=132^2-20\]

The thing on the right is $ 2^2 \cdot 19 \cdot 229$.  So then it can be written as a product of two things of the same parity in exactly two ways: as $ (2)(2 \cdot 19 \cdot 229)$ or $ (2 \cdot 19)(2 \cdot 229)$.  We then get
$ 5p-132=19 \cdot 229+1$ or $ 19 + 229$
$ 5p-132=4352$ or $ 248$
The first case dies.  So we get $ p=\boxed{76}$[/hide]


At the state competition for my state, I saw so many different types of calculators!

There are so many varieties.

Discuss the calculators here!  :lol:

Does anyone have a preference on a  specific calc?

I have a TI- 84 plus, and it is pretty awesome!
But I heard the TI-89 Titanium is the best.

My family has a couple TI-29 (or something like that), a Sharp calculator at the same level, and I have a TI-89 Titanium that I got this summer. Last year, I used the Sharp and the TI-89 as backup, and it worked okay. But graphing calculators can do more things like find prime factorizations and such.

Also, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=287024]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=271903]here[/url] are a couple of related topics.

[color=red][size=67]AIME15: Fixed your tags...use the Preview function please.[/size][/color]

well, I use TI-84 plus, but i saw a lot of TI-30X IIS's...

haha i didn't know you could use graphing calcs. i used a TI-30  :P

had a ti-34 last year, but at actual competitions i just used one of my math team teachers TI-84 silver editions...now i have one of my own :)

I like my casio fx-260 solar; bring it everywhere.

For MC (did in 7th and 8th), I owned a TI-Nspire, so I used that for all rounds. Then at nats 2008, they gave out another Nspire (with a case!). And then at nats 2009, they gave out another Nspire (with a shirt!).

[quote="nikeballa96"]Hah, guys....calculators don't matter. I got until Nationals last year (with a 44 at States) with a TI-34, and used an N-spire at nationals, getting a 31. It's about how well you can use what you do have.[/quote]

So true, I used a TI-30X.

I used a ti-89, though I think it should be removed from the competition as it makes some of the problems too easy.

[quote="Thunder365"]lol, what a strange topic :)

My team member who got first at states used a Casio, im pretty sure. Also, some other dude on my team used a crappy Casio which prevented him from getting one of the target questions right (even though you really didnt need it)[/quote]

yeah that was me  :)  i just feel that graphers arn't needed for mathcounts that much.....most of the problems can be done with a simple scientific

personally, i feel that graphers are more inconvenient than scientifics on tests like mathcounts

of course, if all depends on what test it is

I use a Casio FX-300ES Scientific Calculator. I didn't know you were allowed to use graphing calculators on Math Counts?

Interesting...

[quote="$ LaTeX$"]I agree with chetjan, after all TI-89 Titanium is banned on a lot of major exams. Personally, I use my TI-84 Plus Silver.[/quote]

I have never used a Ti-89, but why does it make MATHCOUNTS too easy?

Are you allowed to  progrm the calcultors.

[quote="AwesomeToad"]Yes. But the programs and memory should be cleared before the competition.[/quote]

So before every competition, someone has to check your calc or is it the Honor system?

It is the honor system. I doubt they have time to check everyone's calculator.  :)

There are proctors, so they might check it. It isn't required, so usually it isn't checked.

[quote="$ LaTeX$"]I agree with chetjan, after all TI-89 Titanium is banned on a lot of major exams. Personally, I use my TI-84 Plus Silver.[/quote]

Which examinations is it banned from?

TI-89T FTW! Makes the TI-83 and Ti-84 look like something from 80's. This thing's got a REAL UI, with serious performance upgrades with not only the processor but the OS itself. Considering overclocking it over the summer, but it would be pretty much pointless....

But, I agree. The rules are pretty lax and very easy to bend. Most graphing calculators have memory enough to quite a few formulas. I'm pretty sure I saw a HOWTO somewhere on the internet to add a EEPROM chip to regular calculators (without modifying firmware).

[quote="AwesomeToad"]Yes. But the programs and memory should be cleared before the competition.[/quote]

I don't know where you got this factoid from, but as far as I know, its false. And if it was true, then at least half the people at nationals cheat.

Yes, I was wondering about that. First, how would a proctor find out? Second, I know a few people who have simple programs on their calculators but nevertheless participate in MATHCOUNTS.  :gleam:

Does anyone have any information on the next calculatr by TI?
 :wow:

Highly programmed calculators make practically no difference at nats though, i find the best way to prepare to do things without it to develop speed.

Most calc-problems at nats involve complex arithmetic though... :wink:

I prefer TI's recent multiple line display scientific calculator, the TI-30XS Multiview.

Even thought MathCounts allows powerhouse calculators like the TI-89 and Nspire with CAS, I find the TI-30XS Multiview is adequate for nearly every MathCounts problem and much easier and quicker for most to use than the graphing calculators (with or without CAS).

The 30XS is basically an update and upgrade from the 30xiis.  Its multi-line display, natural textbook-like display, memory interface, and table features are the most useful upgrades.

As for graphing calculators, I still prefer the HP's, such as the 50g.

HP advantages:
[list]Most people find the RPN and stack protocol of the HP's much faster to work with than the TI's once they get used to it.  If not, they can set their HP to use the standard TI protocol.
The HP is more powerful.
The HP is less expensive.
Less people know the HPs' capabilities and are less likely to prohibit it as they might the TI-89 or Nspire CAS.
[/list]

TI advantages:
[list]TI gives free calculators and trainings to promote use of their calculators in schools
Consequently teachers know them, can better instruct students on how to use them, and promote them.
(Of course is why they cost more and teachers know to prohibit CAS equipped TI's.)
The promotion by teachers leads to a bigger user base and greater availability of software.
[/list]

In summary, TI has long dominated the educational market by supporting teacher's use of calculators in schools.  HP has dominated the professional market by building better calculators at more reasonable prices.

Of course, in the last 10-20 years, the professional market has been disappearing.  Few adults have a compelling reason to use a powerful scientific calculator rather than software on a computer.  Now that high powered calculators cost half as much as [b]much [/b]more powerful netbooks or desktops running free software such as maxima and openoffice.org.  Furthermore any adult who might use a scientific calculator already has use of a computer -- there's no marginal cost.  I'm not an Iphone user but I wouldn't be surprised if with the right aps, an ipod touch or iphone can outgun any calculator in most ways except speed of data entry.  Of course you can't use a computer or ipod at mathcounts.  Calculators are becoming a niche product for schools and exams, rather than professional tools.

Hmmm..there are far more TI-84's than anything else. Is this because the TI-84 is specially made for MATH?

I use TI-83 Plus because that's what my mathcounts coach provides :P


The forum moderators may already have it on their to-do list, but again, I would like to respectfully ask that there be no discussion of the 2007 AMC 10 B and 2007 AMC 12 B problems or solutions until Thursday, February 22, 2007.  Remember that other students in other time zones will be taking the B-date contests throughout the day, after school, etc.  You don't want to spoil the fun by discussing problems before others get a chance to see them.

Thanks for your cooperation, and happy problem solving to all.

Steve Dunbar
AMC Director

Yes, it is already Wednesday morning in the Far East, so we have locked the AMC forum.

Do not discuss any aspect of the AMC 10B/12B on any part of the website.  This includes the forum, blogs, and the wiki.

Good luck on the contest, and we'll reopen this forum Thursday morning.

It's now OK to discuss the AMC 10B/12B.

Remember, our Math Jam is Friday (Feb 23) at 7 PM Eastern / 4 PM Pacific, where we will discuss the harder problems on both contests.

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Click here for the complete Math Jam schedule.[/url]


calculus
integration
limit
calculus computations

(1) For $ a>0,\ b\geq 0$, Compare

$ \int_b^{b\plus{}1} \frac{dx}{\sqrt{x\plus{}a}},\ \frac{1}{\sqrt{a\plus{}b}},\ \frac{1}{\sqrt{a\plus{}b\plus{}1}}$.

(2) Find $ \lim_{n\to\infty} \sum_{k\equal{}1}^n \frac{1}{\sqrt{n^2\plus{}k}}$.

we have :
 $ \frac {1}{\sqrt {a \plus{} b \plus{} 1}} < \int_b^{b \plus{} 1} \frac {dx}{\sqrt {x \plus{} a}} \equal{} \frac {2}{\sqrt {a \plus{}b } \plus{} \sqrt {a \plus{} b \plus{} 1}} < \frac {1}{\sqrt {a \plus{} b}}$

$ \lim_{n\to\infty} \sum_{k \equal{} 1}^n \frac {1}{\sqrt {n^2 \plus{} k}} \equal{} \int_{0}^{1} \frac {1}{\sqrt {1 \plus{} x}} \equal{} 2\sqrt {2} \minus{} 2$

hello kunny, the integral $ \int_{b}^{b\plus{}1} \frac{dx}{\sqrt{x\plus{}a}}$ is to compare with the terms $ \frac{1}{\sqrt{a\plus{}b}}$ or $ \frac{1}{\sqrt{a\plus{}b\plus{}1}}$. Is this correct?
Sonnhard.

@Dr Sonnhard Graubner, yes, it is. How would you sai the context in plain English?

@Hong Quy

For first part your answer is correct, but the latter part is incorrect.

hello, i think
$ \lim_{n\to \infty}\sum_{k\equal{}1}^n\frac{1}{\sqrt{n^2\plus{}k}}\equal{}1$.
Sonnhard.

That's right. :)

See more harder problem posted by Virgil Nicula:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1287168#1287168[/url]

2 has been posted before. It is squeezed


integration
trigonometry
calculus
calculus computations

[color=darkred] Find the recurrent relations for  $ A_n \equal{} \int (\sin x \plus{} \cos x)^n\ \mathrm {dx}$

and $ a_n \equal{} \int_0^{\frac {\pi}{2}}(\sin x \plus{} \cos x)^n\ \mathrm {dx}$ , where $ n\in \mathcal N$ .

See http://www.mathlinks.ro/Forum/viewtopic.php?p=1025505#1025505  [/color]

$ \sin x\plus{}\cos x\equal{}\sqrt{2}\sin(x\plus{}\pi/4)$, therefore...

[quote="Virgil Nicula"][color=darkred] Find the recurrent relations for  $ A_n \equal{} \int (\sin x \plus{} \cos x)^n\ \mathrm {dx}$

and $ a_n \equal{} \int_0^{\frac {\pi}{2}}(\sin x \plus{} \cos x)^n\ \mathrm {dx}$ , where $ n\in \mathcal N$ . [/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] I"ll use the identity $ (\sin x \plus{} \cos x)^2 \plus{} (\cos x \minus{} \sin x)^2 \equal{} 2$ . Thus, for $ n\in\mathcal N$ ,

$ A_{n \plus{} 2} \equal{} \int (\sin x \plus{} \cos x)^n\left[2 \minus{} (\cos x \minus{} \sin x)^2\right]\ \mathrm {dx}\implies$ $ \boxed{A_{n \plus{} 2} \equal{} 2A_n \minus{} H_n}$ ,

where $ H_n \equal{} \int (\sin x \plus{} \cos x)^n(\cos x \minus{} \sin x)^2\ \mathrm {dx}$ . Apply the integration by parts :

$ \overline {\underline {\left\|\begin{array}{ccc} u(x) \equal{} \cos x \minus{} \sin x & \implies & u'(x) \equal{} \minus{} (\sin x \plus{} \cos x) \\
 \\
v'(x) \equal{} (\sin x \plus{} \cos x)^n(\cos x \minus{} \sin x) & \implies & v(x) \equal{} \frac {1}{n \plus{} 1}\cdot (\sin x \plus{} \cos x)^{n \plus{} 1}\end{array}\right\|}}$

Therefore, $ H_n \equal{} \frac {1}{n \plus{} 1}\cdot (\sin x \plus{} \cos x)^{n \plus{} 1}(\cos x \minus{} \sin x) \plus{} \frac {1}{n \plus{} 1}\cdot A_{n \plus{} 2}$ and

$ \boxed {\begin{array}{c} A_0 \equal{} x \plus{} \mathcal C\ \ ,\ \ A_1 \equal{} \sin x \minus{} \cos x \plus{} \mathcal C \\
 \\
(n \plus{} 2)\cdot A_{n \plus{} 2} \equal{} 2(n \plus{} 1)\cdot A_n \minus{} (\sin x \plus{} \cos x)^{n \plus{} 1}(\cos x \minus{} \sin x)\ ,\ n\in\mathcal N\end{array}}$ .

Analogously, $ \boxed {\begin{array}{c} a_0 \equal{} \frac {\pi}{2}\ \ ,\ \ a_1 \equal{} 2 \\
 \\
(n \plus{} 2)\cdot a_{n \plus{} 2} \equal{} 2(n \plus{} 1)\cdot a_n \plus{} 2\end{array}}$ .[/color]


geometry
AMC
AIME
algebra
polynomial
probability
AMC 10

I'm just curious. Usually, whenever I do mock tests on here, I get around 70 to if lucky, 90. I'm planning to take AMC-10 this year. Does those score mean my score will be around 70-90? I hope not :( And the questions on the mock AMC's, the geometry is not sticking to me.

I've been studying geometry really hard but none of the things I used came to me on the mock tests. The application questions are like "totally different"

I don't know the reason "why" I'm not so good at geometry.

I might really need to take the class...

 :(

The level of the problems you post is very low in relation to actual AMCs.  If you struggle with those, it is extremely unlikely that you will score over 100 on any level of the AMC.

The main thing to do the first time you take an AMC 10 or AMC 12 test, and this is a tip that JBL shared last year, is to MAKE SURE NOT TO ANSWER TOO MANY QUESTIONS. There is a penalty for guessing in the official scoring, and you are much better off answering, for example, only the first eighteen or so of the questions on the AMC 10, or only the first fifteen or so on the AMC 12, rather than trying to answer all of them. Leaving a question blank is much better than guessing wrong, the way the scoring works.

Each year, the level of difficulty of the official tests varies a little bit. And what's an easy question for YOU may be a hard question for someone else, but in general on all AMC tests the easier questions appear earlier in the test than the hard questions. Work on the early problems in your test, think about them, make sure you choose the correct answer, and then just leave the last several questions blank when your time runs out. Really. I mean just don't even look at the last several questions, so you have more time to think about the first questions. This worked very well for a sixth-grader on my math team, who was our state's champion on the AMC 10B last year. Last year he only answered eighteen questions, but he got all eighteen correct. That meant he qualified for AIME, which few sixth-graders do. This year, he may try answering a few more questions, but not all of them, on the AMC 10. He expects to try the AMC 12 for the first time this year (his seventh grade year). Don't take my word for it: I got this good advice from JBL. Make sure you answer the questions you are sure about, and don't guess on an AMC 10 or AMC 12 test.

Good luck on the tests. Enjoy your practicing until then.

Thanks to tokenadult for great reply.

I figured that mock AMC's were harder but not too much than AMC's. I remember of three mock I did, the highest score I got was 96 and I guess I got that because I recognized the two problems.

I probably will take geometry class. So, geometry is covered. I figured I need to work on polynomials and probability little bit more but other than, I don't seem to have problem unless I keep doing it.

P.S. I did some questions on AMC-10 2002 on the AMC site and except the last one, I got 4 right out of 5  :lol: .

Thanks everyone!


calculus integration algebra polynomial AMC AIME trigonometry

So... what was your favorite question from round 4?  I personally liked 3 - it took me an incredibly long time to get the solution, but it was really elegant when I found it.  Question 4 was a close second.

Same here, I thought #3 was really cool.  Problems #4,5 were ok I guess, and #1,2 were just blah.  :P

What answer did you get, by the way?  I used the identity that

[tex](x + y +z)(x^2 + y^2 + z^2 - (xy + yz + xz)) = x^3 + y^3 + z^3 - 3xyz[/tex]

and proved that the second part of the left side is always positive for integral x,y, z to find the requested polynomial, which was

[tex]f(x,y,z) = x^3 + 2y^3 + 4z^3 - 6xyz[/tex]

[quote="mathfanatic"][quote="zanttrang"]What answer did you get, by the way?  I used the identity that

[tex](x + y +z)(x^2 + y^2 + z^2 - (xy + yz + xz)) = x^3 + y^3 + z^3 - 3xyz[/tex]

and proved that the second part of the left side is always positive for integral x,y, z to find the requested polynomial.[/quote]

I got $x^3 + y^3 + z^3 - 6xyz$; not sure if this is correct, but it held up to a lot of trials in Excel.[/quote]
LOL

I want to see a solution for problem 3! I didn't solve it . . . sort of like problem 2 or whatever it was on round 4 last year, which also had to do with constructing a polynomial.

3 was probably my favorite (yes, I've changed my mind since posting in the other topic)
It was certainly the most frustrating-I had good solutions for the other 4 in the first week or so, then spent the last 3 weeks trying to find a better answer than the one that I ended up with (which I had immediately convinced myself was not right).

\[x^3+2y^3+4z^3-6xyz\]

I've attached my round four solutions, for anybody who's interested.  My explanation of the solution to #3 is pretty easy to understand (if not exactly intuitive), so take a look if you want some insight into that problem.

#5 was definitely my favorite.  It took me a while to prove, and my solution wasn't very elegant, but I enjoyed solving it.  I think I just like geometric proofs, I've had fun with all of them this year.

Now, I hated #3.  I couldn't solve it!  It was the only question I left blank this year.  I spent a good amount of time trying to figure it out but nothing worked for me.

The funny part is, I solved both #1 and #2 in the first five minutes of looking at the Round 4 problems.  That was misleading because I thought the rest of the problems would be as easy.

I think I'll make that "Silver" List.  I did a lot better last year, but I believe the grading is tougher this year.

Of the 5 problems on this round, #3 was probably the most interesting (although #4 was more challenging for me). #1 was nice and a good difficulty for a #1. #5 was entertaining, although I had a lot of algebra in my solution...

Yes, the solution to #3 was sooo elegant...

$f(x,y,z) = x^3+2y^3+4z^3-6xyz$ works perfectly with a really easy proof too.

Yes, the solution is elegant once you find it, but that problem is like many on this year's AIME: Those of us who knew the trick could get it in a few seconds, whereas it was hard for those who didn't know it. In a good problem, those with advanced knowledge wouldn't have such an advantage over those in earlier math classes. I preferred number 5, which didn't require much more than the laws of sines, cosines, and area.

-Adam Hesterberg

Question #1 was easy, except I didn't know what the theorem that I used to find the number of factors was called :blush: so I'm hoping that it's commonly known...


I have trouble multiplying in different bases.

Take 88 in base 9 X 87 in base 9...
I know that when you multiply 8*7 (the 8 is the last digit of 88), you get 56 but you have to convert it to base 9 to get 62. Then you write down "2" underneath the 7 and you carry the 6. Now here comes my question. Now you multiply the first 8 in 88 * 7 to get 56. Now my question is: [i]do you add 56 + the carried 6 = 62 and then you convert the whole thing to base 9, or 68; or do you first convert 56 to base 9, or 62, then add the carried 6 to get 69.[/i] I know that it doesn't really matter for this particular question. But what would the correct step be?

Now, having answered that question, does it really matter which step you do first? I mean, in really high bases, would it matter which you did first????

Now, for a couple of subquestions...
why do we have to convert to base 9 (in this example as described above) after we multiply the 8*7? would you do the same in low bases? Why? How about in super high bases? Why?

could someone show me how to do the multiplication in the problem above?????

Could someone explain to me how to do multiplication in different bases????????

It's good practice to do everything in base $ 9$ and [b]also[/b] to clearly denote what numbers are and aren't in base $ 9$.  You can get very confused trying to figure out what $ 56$ is and what $ 62$ is but if you keep in mind that you're writing $ 8_9 \cdot 7_9 \equal{} 56_{10} \equal{} 62_{9}$ and $ 56_{9} \plus{} 6_{9} \equal{} 63_{9}$ (not $ 62$ or $ 69$!) you won't have any problems.

thanks  :lol:

I tried that, but I still get mixed up...
could someone try to answer the questions???

Convert to base 10 - multiply - then convert the answer back to base 9.

So $ 88_9 \times 87_9 \equal{} (8*9 \plus{} 8)_{10} \times (8*9 \plus{} 7)_{10} \equal{} 80_{10} \times 79_{10} \equal{} 6320_{10} \equal{} 8602_9$

The difficulty in multiplying outright in other bases is that we don't have multiplication tables memorized in other bases, only in base 10. And that's a key element to being able to multiply efficiently in any base.

To convert $ 6320_{10}$ to base 9, I divided by $ 9^3 \equal{} 729$ (the largest power of 9 less than 6320) to get 8 with a remainder of 488. So the digit 8 is the first digit (of four) in the base 9 representation of 6320.

Then I divided the remainder, 488, by $ 9^2 \equal{} 81$ to get 6 with a remainder of 2. So the digit 6 is the second digit in the base 9 rep. of 6320. Now since the remainder is 2, which is less than 9, the third digit (the 9's place) must be zero and the fourth digit (the 1's place) must be 2.

Alternatively, to convert $ \overline{a_na_{n\minus{}1}a_{n\minus{}2}\ldots a_2a_1a_0}_{10}$ to an arbitrary base $ b$, we can divide the base ten integer by $ b$, find the remainder, divide the quotient by $ b$, find the remainder, etc. until we get to a quotient of $ 0$, and then reverse the remainders obtained to find the base $ b$ representation.

This works because if we let the converted number be $ \overline{b_kb_{k\minus{}1}b_{k\minus{}2}\ldots b_2b_1b_0}_b$, then this equals

$ (((\cdots(((b_k)b\plus{}b_{k\minus{}1})b\plus{}b_{k\minus{}2})b\plus{}\cdots)b\plus{}b_2)b\plus{}b_1)b\plus{}b_0)_{10}$,

and the process above is simply the reverse of this (which can also be used to convert from base $ b$ to base $ 10$).

For example, to convert $ 6320_{10}$ to base $ 9$, we can continually divide, etc., so

$ \frac{6320}9\equal{}702\cdot9\plus{}{\color{blue}2}\implies\frac{702}9\equal{}78\cdot9\plus{}{\color{blue}0}\implies\frac{78}9\equal{}8\cdot9\plus{}{\color{blue}6}\implies\frac89\equal{}0\cdot9\plus{}{\color{blue}8}$ (everything here is in base $ 10$),

and our base $ 9$ representation is $ 8602_9$.

[quote="gauss202"]The difficulty in multiplying outright in other bases is that we don't have multiplication tables memorized in other bases, only in base 10. And that's a key element to being able to multiply efficiently in any base.[/quote]
Multiplying efficiently is less important than understanding that the basic idea behind multiplication doesn't depend on what base you're in.  You'll gain a better understanding of multiplying in general if you understand how it works in bases other than $ 10$.

I really appreciate everyone's answers.  :lol:

However, can someone please answer the question(s)???????????

I think you should break down base multiplication into the simplest step: repeated addition. Suppose you have $ 2_9 \cdot 3_9$. Simplifying yields:

$ 2_9 \cdot 3_9 \equal{} 2_9 \plus{} 2_9 \plus{} 2_9 \equal{} 6_9$

Obviously that was one of the simple multiplication problems. Now you can try: $ 7_9 \cdot 5_9 \equal{} 7_9 \plus{} 7_9 \plus{} 7_9 \plus{} 7_ 9$. Remember when adding, when you get to 9, it's the same as getting to 10 in base ten and starting over. (Remember that if you are in base $ n$ you can never have $ n$ as a digit.) So: $ 7_9 \plus{} 7_9 \plus{} 7_9 \plus{} 7_ 9 \equal{} 15_9 \plus{} 15_9 \equal{} 32_9$. Hope that helps. It's best to think of it in small steps and then work your way up to the multiplication algorithm that we commonly use in base 10 and instead use in base $ n$.

cool! thanks  :lol:
I think I have down that step now...

although I still don't think that my original question(s) were answered  :(

Is your question how to multiply numbers like $ 27_9\cdot25_7$. Where the bases are different.

If so, the only have that I know of is to change the numbers to common bases then multiple.

Thanks Nandubetax

[quote][b]Now my question is: do you add 56 + the carried 6 = 62 and then you convert the whole thing to base 9, or 68; or do you first convert 56 to base 9, or 62, then add the carried 6 to get 69. I know that it doesn't really matter for this particular question. But what would the correct step be?

Now, having answered that question, does it really matter which step you do first? I mean, in really high bases, would it matter which you did first????

could someone show me how to do the multiplication in the problem above?????

Could someone explain to me how to do multiplication in different bases????????[/b][/quote]

The commutativity of addition and multiplication appears as a symmetry of the tables with respect to the main diagonal. One may discern other features of the tables that have algebraic interpretation. So looking at the whole table has definite advantages over looking at it a line at a time. Even more regularity can be observed when several tables one for each different base system are studied together.
----------
smithsan
[url=http://paid.cashsurveys.net]Paid Surveys[/url]

[quote="theax"]thanks, but...

[quote]Quote from t0rajir0u's post:

Multiplying efficiently is less important than understanding that the basic idea behind multiplication doesn't depend on what base you're in. You'll gain a better understanding of multiplying in general if you understand how it works in bases other than. [/quote][/quote]

Basically there are tables out there that help you multiple in other bases. He is saying that instead of memorizing the tables, you should understand how the tables work (basically understand, don't memorize).


What is the largest digit which can replace $ b$ to make the number $ 437,b32$ divisible by 3?

2 but on the answer thing it said 8. Because if u add all the digits u get 19. Add 2 and u get 21. 21/3=7 so its divisible. So i think im right

BUt if you add 8 you get 27, which also works. As a matter of fact, 2,5, [b]and[/b] 8 all work! So which one do we do? Rereading the question, we see it asks for the largest possible, so the answer is 8.

[quote="GameBot"]What is the[b][size=200][color=red] largest digit[/color][/size][/b] which can replace $ b$ to make the number $ 437,b32$ divisible by 3?[/quote]

Always read  :P


[quote="EndlessEquationz"]Okay so I'm new (this is my first post :ninja: ) and i looked around and have decided to post a logic problem.
Dan, Amy, Billy, Bob, And Linda were playing a videogame. Determine what order they ranked in.

Dan scored higher than Amy, but Less than Bob.
Amy was less than Billy and higher than Linda.
Bob's score was next to Linda's on the High scores list.
Linda had a higher score than Bobby.

Oh and i made up this problem :o .[/quote]

nice problemo:
[hide="soltution"] Lets see, Amy < Dan < bob, Linda < Amy < Billy
Bobby < Linda < Amy <Dan < Billy < Bob wait, this makes no sense if amy < bob but bob is next to Linda.

THIS IS IMPOSSIBLE!!!![/hide]

Your problem makes no sense. We can't solve it.

yeah, u should change up the prob
 :wink:

I'm Sorry i didn't have anough time to check it.. Okay a book problem i wanna see if anyone will get.

(a-x)(b-x)(c-x)...(z-x)=???

I got that too but my Mom didn't get it for like an hour so i was just wondering whether i had a brain spark sheesh! :(

this problem is impossible bobs score can not be next to Linda while Linda is higher than bob.

 :rotfl:  :stretcher:  :starwars:

Useless but:
Haha,
the (x-x) looks like a face

So did u not check the logic prob or something?

Oh yeah, and does MOEMS typically have logic probs?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Characterize those finite groups having the property that every two non-trivial subgroups have non-trivial intersection.

I'd say which problem this is related to, but that other topic actually contains the answer, and I don't want to spoil the fun :).

I think the group you have mentioned is generlized quaternion group, since minimal subgroup P is unique with |P|=p a pime divisor of |G|.

That's almost the whole answer, but it needs proving (most mathematical statements do :)).

Suppose G satisfies the condition and G has non-trivial subgroup. Let p and q are prime divisors of |G|, obviously, G has a subgroup of order p and q, denote P and Q, respectively. Then P and Q has only common element 1. And if p=q and P is not equel to Q, then P and Q has also common element 1. Therefore G has only minimal subgroup for every prime divisor of |G|, so that G is generalized quaternion group by book of THE GROUPPEN, written by Huppert, and |G| is 2^n. Suppose G has only trivial subgrooup, then |G| is prime .

That's not exactly the answer; every cyclic group of order $p^{n}$ works.

ommit a little.

Suppose G satisfies the condition and G has non-trivial subgroup. Let p and q are prime divisors of |G|, obviously, G has a subgroup of order p and q, denote P and Q, respectively. Then P and Q has only common element 1. And if p=q and P is not equel to Q, then P and Q has also common element 1. Therefore G has only minimal subgroup for every prime divisor of |G|, so that G is generalized quaternion group and |G| is 2^n, and cylic group order p^n by book of THE GROUPPEN, written by Huppert, . Suppose G has only trivial subgrooup, then |G| is prime .
 :)

Why did you just repeat your earlier post exactly, including typos and bad English?

I am from China, indeed bad english, but better Chinese than you!!

That wasn't really my point- my point was that you copied your previous post rather than saying anything new. Your argument is incomplete; it misses an answer.


I have used several lemmas in different problems on the Round 1 USAMTS solutions sheet.  Lemma is a theorem-like environment defined by:

\newtheorem{lemma}{Lemma}

The problem that I have is that the lemmas are numbered in order without respect to each problem's solution.  The consequence is that problem one contains Lemma 1, and the first lemma in problem two is Lemma 2.  I am also encountering this difficulty with numbered equations.

Is there a way to reset these counters at the begining of each problem?

[quote="dts"][quote="jhaussmann5"]Try the command \setcounter{lemma}{0}.[/quote]
Does there exist an analogous command for eqnarray?[/quote]
Yes:  \setcounter{equation}{0}

By the way, for the USAMTS, if you don't want to bother resetting the equation numbers, lemma numbers, etc., then don't bother; the graders don't care one bit.

Your replies have been very helpful!  Thanky you

[quote="DPatrick"][quote="dts"][quote="jhaussmann5"]Try the command \setcounter{lemma}{0}.[/quote]
Does there exist an analogous command for eqnarray?[/quote]
Yes:  \setcounter{equation}{0}[/quote]This may sound ridiculous, but how do I refer to a numbered equation. I mean, suppose that in the first page of my souce I have this

[code]\begin{eqnarray}
\sum_{k=1}^{n} & = & 1^2+2^2+\cdots+n^2\nonumber\\
               & = & \frac{n(n+1)(2n+1)}{6}
\end{eqnarray}[/code]
Then, after the sumatory, probably I would have a (1). How do I refer to that same equation, say, three pages latter (suppose that we are withing the same chapter, or in an article documentstyle), of course, without directly writting (1), for obvius reasons! I suppose it should be something obvius, but I still don't figure it out.

Thanks :first:

[quote="ComplexZeta"]In your equation, put something like \label{eqnname}, and then to refer to it later, type \ref{eqnname}.[/quote]OK, that works wonderful, thank you ComplexZeta, but now I have a small problem: Within a chapter I have several sections, and on each section I have several forumulas. Now, I have formulas (1), (2), (3) in the section one, and formulas (4), (5), ... in section two. I want a numbering that works by section. If I do \setconunter{eq}{0} at the begining of each section, then the numbering resets to (1), (2), (3) in first section, and (1), (2), ... in the second. What should I do if  want them to go (1.1), (1.2), (1.3), (2.1), (2.2), ...

[quote="Ledurt"]put the command

\numberwithin{equation}{section}

in the preambule of your document. The rest stays the same.  And in appendix, you'll have A.1 ...

for more info, try this page

[url=http://www.texnik.de/mathstuff/numbering.phtml]www.texnik.de/mathstuff/numbering.phtml[/url]

btw, the \label + \ref commands work for equations but also for figures, tables and sections but be careful where you put the command \label to make sure it's associted with the right equation, section or figure.  I'm not sure though where exactly you need to put it

ledurt[/quote]

Thank you very much! it works perfectly!


Just like the geometric mean of ellipse.I just want to know that what's the the geometric mean of ellipsoid.
[img]http://mathworld.wolfram.com/images/equations/Ellipsoid/equation1.gif[/img]

It seem that my question is so easy or silly,because I'm a High-School boy,I'm very interested in math.I'll appreciate you if you can tell me the answer.

Thanks!

[quote="kg2100"]what's the relation between the points of the ellipsoid surface and the a,b,c(when a b c are not equal)[/quote]

I'm not sure what you mean.


Let $ P(x)$ be a polynomial with degree 2008 and leading coefficient 1 such that
\[ P(0) \equal{} 2007, P(1) \equal{} 2006, P(2) \equal{} 2005, \dots, P(2007) \equal{} 0.
\]Determine the value of $ P(2008)$.  You may use factorials in your answer.

[quote="worthawholebean"]Let $ P(x)$ be a polynomial with degree 2008 and leading coefficient 1 such that
\[ P(0) \equal{} 2007, P(1) \equal{} 2006, P(2) \equal{} 2005, \dots, P(2007) \equal{} 0.
\]
Determine the value of $ P(2008)$.  You may use factorials in your answer.[/quote]
[hide="Solution"]
Consider the polynomial $ P(n) \minus{} 2007 \plus{} n \equal{} Q(n)$.  Notice that $ 0, 1, 2\cdots 2007$ are all roots of it.  Hence, $ Q(n) \equal{} x(x \minus{} 1)(x \minus{} 2)\cdots (x \minus{} 2007)$.  Thus, \[ Q(2008) \equal{} 2008!\implies P(2008) \minus{} 2007 \plus{} 2008 \equal{} 2008!\implies P(2008) \equal{} \boxed{2008! \minus{} 1}\] [/hide]

I understand your solution,  but what if $ Q(x)$ was multiplied by some constant?  Wouldn't that still work, but totally change the answer?

The leading coefficient must be 1, so we can't multiply by a constant.


$ a$ and $ b$ are real numbers and satisfy $ ab^2\equal{}\frac{27}{5}$ and $ a^2b\equal{}135$. Compute $ a\plus{}5b$.

hello, solving the given system we get $ a\equal{}15$, $ b\equal{}\frac{3}{5}$ as the solution, so we have $ a\plus{}5b\equal{}18$.
Sonnhard.

$ \frac{ab^2}{a^2b}\equal{}\frac{b}{a}\equal{}\frac{1}{25} \implies b\equal{}\frac{a}{25}$.

Thus, $ a^2b\equal{}\frac{a^3}{25}\equal{}135 \implies a\equal{}15$. Plugging this in, we get that $ b\equal{}\frac35$, and then adding,

$ 15\plus{}3\equal{}\boxed{18}$.


[quote="pdah120290"]Hi!

I am looking for Integer Solutions of

A^2 = 16(stv)^4+2sut(ut^2-vs^2)^2

where A,s,t,u,v are Integers and s,t are coprime and u,v are coprime.  :?

Is the number of Solutions infinite?

Best Regards Philipp[/quote]

Yes, it's infinite taking $u=0$, $v=1$ and obviously $16(st)^4$ with $(s,t)=1$ is always a square.

RX the Crazy Math


The 18th (2004-2005) Korean MO, Round 1
May 23th, 2004

This material is only in Korean. sorry.

[url=http://www.msquare.or.kr/sp2004/board.cgi?id=moset&action=download&gul=15]www.msquare.or.kr/sp2004/board.cgi?id=moset&action=download&gul=15[/url] (pdf format)

제18회(2004-2005) 한국 수학올림피아드 1차시험
2004년 5월 23?

Can someone please translate it into English? Thanks! :)

i will have it posted after texing.. or just write in here... either way... i can translate it

The original test papers are here.
Round 1 problems are quite easy ones.
Some of the problems are multiple choice type.

Problem 1 and 18 in high school paper is the same with problem 2 and 19 in middle school paper each,
so these problems are omitted in the above solution material.

In Korea,
middle school = 7th to 9th year and
high school = 10th to 12th year.
But 12th year students cannot participate in KMO Round 1.

I can supply the figures in the above solution material in eps format.
Or I can do texing with those figures after your translation.
(For me, translating English to Korean is easy but Korean to English is hard, sorry)

can you send them to my email address?

it is es7abl1sh3d@gmail.com.

Also, I am posting some of those problems in the Pre-Olympiad Section. (although they are easier than i thought). I have actually read the problems, and translated into ENglish for some of those problems.

I am too lazy to do actual TeXing on my computer right now (the reason why I am posting it on AoPS Pre-Olympiad Section instead), but if you can send those eps files or even the original paper on pdf (i cannot download your attached file) through email, I will be glad to translate them for you.

BTW, i havent seen Korean Math Olympiad problem for quite a while, not since i moved here when i was in 6th grade. Thanks for the problems.

?사합니다. 그리고 제 소개를 깜빡했네요. 저는 워싱턴주 시애틀? 사는 정????고 합니다. 옛날? 경시문제를 많?했었어요. 만약? ? 어려운 올림피아드 문제가 있으면 꼭 제 ?메??나 여기 AoPS? ?주세요. ?사합니다.

Wow... I have forgotten all the mathematical terms in Korean... I moved to US when I was 10-year-old and I have been practicing math in Korean texts books but it's been while... I seriously feel that I am way behind...


Prove that :
|{n/1} - {n/2} + {n/3} - ....+(-1)n-1{n/n}| < (2n)1/2 for all natural n. < {x}:the fractional part of x>

Can you post the soln for this one??? :?

i don't exactly get the bound, but at least my proof should give an idea how to proceed:
we look first at the terms ${\frac{n}{k}}$ with $k>\sqrt{n}$. these are at most $\sqrt{n}$ terms, each is at most 1 and at least 0, but half of them are positive and half of them negative, thus their sum is at most $\frac{\sqrt{n}}{2}$.
now look at the smaller terms. so, let $k$ be an integer, $2\leq k\leq \sqrt{n}$. we look at the terms ${\frac{n}{m}}$ where $k-1\leq \frac{n}{m}\leq k$, so ${\frac{n}{m}}=\frac{n}{m}-(k-1)$. but since their fractional parts are increasing as m decreases and one alternately sums them positive and negative, that sum is at most the first term which is smaller than 1. thus, for every k we get a sum smaller than 1, but there are at most $\sqrt{n}$ k's we have to look at, so the smaller terms sum to at most $\sqrt{n}$. all in all, we get at most $3/2\sqrt{n}$. this is not exactly the desired bound, but i'm sure one can improve my bounds.

Peter


Find all prime numbers p,q such that $ (p\plus{}1)^{q} \equal{}k^{2}$ where k is an integer.

Case 1 : $ p\plus{}1$ is a perfect square. $ \Rightarrow p\plus{}1\equal{}a^2 \Rightarrow p\equal{}(a\minus{}1)(a\plus{}1) \Rightarrow p\equal{}3$. Hence, $ q$ is all primes
Case 2 : $ p\plus{}1$ is not a perfect square $ \Rightarrow q$ is even $ \Rightarrow q\equal{}2$ and $ p$ is all primes

I am not sure that it's correct but i hope. :lol: If q=2 than every prime number p , satisfies the condition. If q doesn't equal to 2, than q must be an odd number. Now we can say that p+1 must be a perfect square. Because k^2 is a perfect square and all prime factors of k^2 has even power.Now let p+1 =r(1)^a(1)  . r(2)^a(2) ... r(n)^a(n). Here r(i)'s are different primes and a(i)'s are their powers. Multiplicating this with q we get that q.a(i) 's are all even numbers. So a(i)'s are even numbers. p+1=x^2 then (x-1)(x+1)=p. If x-1=1 than p=3. There is no more solution.


Please answer these questions as quickly as possible
I finished all of the questions so check my work and if I got some wrong then please write your solution
5 Questions as usual:
My answers:
 1. 49
 2. 99/100
 3. 6/7
 4. 30 degrees
 5. 5/13

i got
1. 49
2. 99/100
3. 6/5
4. 15
5. 1/4
[hide]
1. 2+rtx=9, rtx=7, x=49
2. 1-(1/2)+(1/2)-(1/3)+...+(1/99)-(1/100)=99/100
3. let the sidewalk have length x, then x=r(mariela's speed)/20 and x= s(sidewalk's speed)/30, so 30x=s and 20x=r and r+s = 50x, so (r+s)/50 = x and the answer is 60/50 = 6/5
4. let bac = x, then cbd = 2x and cdb = 2x and dce = 3x and dec = 3x and edc = 180-6x and (180-6x)+(cdb) = 120, and we know cdb = cbd = 2x, so 180-4x = 120 and x=15
5. there are 24 with 1 face, 24 with 2 face, 8 with 3 faces and 8 with 0 faces painted. therefore our probability is (24*1+24*2+8*3+8*0)/(64*6) = 1/4[/hide]

1 and 2 are right
for #3, assume the sidewalk is 6 feet long
she walks 2 feet in 1 minute and the sidewalk moves at 3 feet in 1 minute.
together, they move at 5 feet in 1 minute
so the answer is $ \frac{6}{5}$
#4 looks like itll take alot of time for me 2 explain....
#5, there are 384 faces, 96 painted so $ \frac{96}{384}\equal{}\frac{1}{4}$
i think

kay, we got the same answers and solutions. =)

May I ask what or who is setting this deadline for you?

Sorry guys for not replying earlier but these were questions that were on the 2005 Math Awareness Month Comptetition held every year. The main reason I only had two hours was because I had to go to the math comptetion by then. Yeah I know, sorry for putting these questions in the MC forum  :oops:. Anyways I found this years questions extremely easy.
Here is the 2008 test, which I just took 5 hours ago.
My answers:
1. 1400 ways
2. 8 sq. units
3. 5/32
4. Andy spent 60 minutes and Beth spent 45 minutes
5  62

No seriously, these problems appeared in MathCounts before, I think mostly on Chapter Team Rounds and such.
Aren't MC problems copyrighted or something?

maybe its like how MC lets FTW use their problems

Just make sure that when you post them, the competition deadline is way over (i.e., nobody else is taking it).  We want to make sure to preserve the integrity of the competitions :) Thanks.


integration
projective geometry
calculus
calculus computations

Hello:

I am wondering what one must do in order to revolve about y=x, as opposed to the usual revolving about the x or y axis.

Suppose we had to find the volume of the solid obtained when we revolve
$y=x$ and $y=\frac{x^{2}}{2}$ about y=x.

Would we have to translate the graph of $\frac{x^{2}}{2}$ and integrate from 0 to $2\sqrt{2}$?.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109728=]Take a look the last my post save one[/url]

kunny

I like the general derivation. I was working on a rotation.

$x=\frac{\sqrt{2}}{2}(x'-y')$
$y=\frac{\sqrt{2}}{2}(x'+y')$

This leads to: ${\pi}\int_{0}^{2\sqrt{2}}(-\sqrt{4\sqrt{2}x'+2}+x'+\sqrt{2})^{2}dx'$

This gives: $\frac{2\sqrt{2}{\pi}}{15}$

Which agrees with the your method. The Centroid Pappus method.

with a=1/2 and b=1


Find the sum of the roots of $x^{2001} + \left(\frac{1}{2} - x\right)^{2001}$

[hide]The $x^{2001}$ out front will cancel with a $-x^{2001}$ from the expansion of $\left(\frac{1}2-x \right)^{2001}$, so we have something of degree $2000$.  We need to find the coefficients of $x^{2000}$ and $x^{1999}$ in order to compute sum.  The general term of the binomial expansion is $\binom{2001}{n} \cdot \frac{1}{2^n} \cdot (-x)^{2001-n}$, so when $n=1$ we can calculate the coefficient of $x^{2000}$ as $\frac{2001}2$.  When $n=2$ we find the coefficient of $x^{1999}$ as $-2001 \cdot 250$.  Therefore the sum of the roots is $\frac{2001 \cdot 250}{2001 \cdot \frac{1}2}=500$[/hide]

[quote="curl"][quote="d343seven"] We need to find the coefficients of $x^{2000}$ and $x^{1999}$ in order to compute sum. [/quote]

Why will these coefficients tell us the sum of the roots? (P.S. I am somewhat of a newbie problem solver.)[/quote]
if you have (where a[k] represents a variable) (x-a[1])(x-a[2])(x-a[3])... (x-a[n]) (the roots of this polynomial is a[1], a[2], etc.), by the distributive property, then n-1th term's negative is going to be the sum of all the roots (a[1] has to be paired up with the n-1 other x's, the a[2] has to be paired up with the n-1 other x's etc.

it's pretty much the same thing as -b/a is sum for quadratics and c/a is product for quadratics, except generalized.  But the sum never changes and is always -b/a, however the product changes depending on a negative sign for polynomials with even or odd degree.

[quote="curl"][quote="d343seven"] We need to find the coefficients of $x^{2000}$ and $x^{1999}$ in order to compute sum. [/quote]

Why will these coefficients tell us the sum of the roots? (P.S. I am somewhat of a newbie problem solver.)[/quote]
Viets formula for $i=1$


Given 9 points in a plane,no three of them being collinear.Prove that you can always find a convex pentagon among the 9 points.
Can you show me the proof of the famous theorem?

Klein's theorem is for $5$ points and a convex quadrilateral.
Anyway, have a look here :

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=szekeres&t=21981

Pierre.

I know the proof of  n=5,anyway,thank you.


calculus
derivative
trigonometry
calculus computations

Find all solutions to the following differential equation:

$ \frac{d^3y}{dx^3}\equal{}y$

well, ummm...  :( i hope you meant to say the characteristic equation is
$ r^3 \minus{} 1 \equal{} 0\qquad\Rightarrow\qquad (r \minus{} 1)(r^2 \plus{} r \plus{} 1) \equal{} 0$
$ \Rightarrow\qquad r \equal{} 1,\;\;r \equal{} \frac {1}{2}\left( \minus{} 1\pm\imath \sqrt 3\right)$

hence, $ y \equal{} C_1e^x \plus{} C_2e^{ \minus{} 1/2\,x}\cos\left(\frac {\sqrt 3}{2}\,x\right) \plus{} C_3e^{ \minus{} 1/2\,x}\sin\left(\frac {\sqrt 3}{2}\,x\right)$


calculus
integration
trigonometry
logarithms
limit
calculus computations

$ \int_0^{\infty} \frac {(sinx)^2}{x}\ dx$

hm.?!

the integral don't converges.

$ \int \frac{\sin ^2(x)}{x} \, dx\equal{}\int \left(\frac{1}{2 x}\minus{}\frac{\cos (2 x)}{2 x}\right) \,dx\equal{}\frac{\log (x)}{2}\minus{}\int \frac{\cos (2 x)}{2 x} \, dx$ $ \equal{}$ $ \frac{1}{2} (\log (x)\minus{}\text{Ci}(2 x))$

'cause $ \lim_{x\to \infty } \, \text{Ci}(2 x)\equal{}0$