calculus
derivative
trigonometry
function
integration
geometry
analytic geometry

So, the AP test is coming up in two weeks and since most of the problems on the AP test are really predictable and repetitive, i was wondering if anyone had a nice need to know list of formulas and theorems I could look over couple nights before the actual test to cram?

Well I'm sure we'll be glad to help you when you have any problems with the material.
(We don't do your drudge work here; I'm sure you can review your notes from class)

Good luck!

If I can recall correctly, here's everything you need to know.  Or at least what I think you need to know.

1) Limits-basically computing them, figuring out if they exist, continuity
2) Derivatives-definition of the derivative, power rule, product rule, quotient rule, chain rule, trig and exponential functions
3) Applications of the Derivative-optimization, max/mins/critical points/inflections, inverse functions, related rates, MVT/Rolle's Theorem, logarithmic differentiation, parametrized curves, polar curves, position/velocity/acceleration
4) Integrals-Riemann sums, indefinite integrals, FTC (both parts), trig functions, exponential functions, u-substitution, integration by parts, trig substitutions, improper integrals
5) Applications of the Integral-area under a curve, area between curves, volumes spanned by revolving curves around lines and cross-sectional shape, polar curves, arc length, partial fractions
6) Differential Equations-separable equations, slope fields
7) Series and Sequences-arithmetic and geometric sequences, tests for convergence and sums, ratio test, integral test, alternating series test, comparison test, power series, Taylor and Maclaurin Series, Lagrange error

I hope this is comprehensive enough.  CB provides a list as well.

Hyperbolic integration and differentiation is a must have also I think.

Hyperbolic funtions aren't on the AP BC exam. Although I really don't see why not. (The exam also got rid of important stuff like the rigorous definition of a limit.)

Like all standardized exams, this isn't all that bad if you look at old tests. They don't really like coming up with new problems, so they will just be variants on existing ones.

The thing that hurt me the most on the exam was probably series. Know the standard ones!! Also, make sure you can tell if they converge quickly without having to do tests (this is useful for multiple choice where you don't show any work). Make sure you understand how to compute a Maclaurin series (has problem 6 ever been something different than this?)

Outside of that, formulas given on a standard textbook front cover are a must along with things like arclength (for BC know the polar form as well) or really anything on a Calc I/Calc II formula sheet is fair game (I think these are made by Barnes and Noble or Borders?)

Like I said, though, it is standardized, so doing a couple old tests will probably tell you absolutely everything you will encounter on the actual test.

don't forget the logistic growth equation

There is ALWAYS a question about that. Usually, they ask about the related equation y'=y(k-x) as well

Logistic growth is an easy separable equation:
$ \frac{dy}{dx}\equal{}y(k\minus{}x)\Rightarrow \int\frac{dy}{y}\equal{}\int k\minus{}x\ dx\Rightarrow\ln{y}\equal{}kx\minus{}\frac{x^{2}}{2}\plus{}C\Rightarrow y\equal{}Ce^{kx\minus{}x^{2}/2}$


geometry
area of a triangle
geometry unsolved

Let $ ABCD$ be a circumscriptible quadrilateral, let $ \{O\} \equal{} AC \cap BD$, and let $ K$, $ L$, $ M$, and $ N$ be the feet of the perpendiculars from the point $ O$ to the sides $ AB$, $ BC$, $ CD$, and $ DA$. Prove that: $ \frac {1}{\left|OK\right|} \plus{} \frac {1}{\left|OM\right|} \equal{} \frac {1}{\left|OL\right|} \plus{} \frac {1}{\left|ON\right|}$.

we have $ \frac {1}{|OM|} \equal{} \frac {DC}{OD.OCsinDOC}$and...
so $ \frac {1}{|OK|} \plus{} \frac {1}{|OM|} \equal{} \frac {1}{|OL|} \plus{} \frac {1}{|ON}\Longleftrightarrow
 \frac {DC}{OD.OC}\plus{}\frac {AB}{OA.OB}\equal{}\frac {AD}{OA.OD}\plus{}\frac {CB}{OC.OB}$
but since $ OA\equal{}\frac {AB.AD.BD}{BC.AB\plus{}AD.CD}$and....
we must have   $ \frac {1}{|OK|} \plus{} \frac {1}{|OM|} \equal{} \frac {1}{|OL|} \plus{} \frac {1}{|ON}\Longleftrightarrow AB\plus{}CD\equal{}BC\plus{}AD$ and so  there exists an incircle of $ ABCD$.

$ OCD \sim OBA$ so $ \frac {OA}{OD} \equal{} \frac {AB}{CD}$ ie $ OD \equal{} \frac {OA.CD}{AB} (1)$
 and $ ODA \sim OBC$ so $ \frac {OA}{OB} \equal{} \frac {AD}{CB}$ so $ OB \equal{} \frac {OA.BC}{AD}(2)$
$ (1) \plus{} (2)\Longrightarrow BD \equal{} OA(\frac {CD}{AB} \plus{} \frac {BC}{AD})$ and so
$ |OA| \equal{} \frac {|AB|.|AD|.|BD|}{|BC|.|AB| \plus{} |AD|.|CD|}$
and we replace in $ \frac {DC}{OD.OC}\plus{}\frac {AB}{OA.OB}\equal{}\frac {AD}{OA.OD}\plus{}\frac {CB}{OC.OB}$

Tchebytchev, a [i]circumscriptible[/i] quadrilateral is a quadrilateral with an incircle, not one with a circumcircle.

  darij

See Theorem 17 in the just-updated [url=http://www.cip.ifi.lmu.de/~grinberg/CircumRev.pdf]"Circumscribed quadrilaterals revisited"[/url]. (You don't have to read the whole note; you'll only need the text up to Theorem 3, and Theorems 16-17 with proofs; and you need to know that I use $ \alpha$, $ \beta$, $ \gamma$, $ \delta$ to denote the angles $ \measuredangle DAB$, $ \measuredangle ABC$, $ \measuredangle BCD$, $ \measuredangle CDA$ of the quadrilateral.)

[hide="Old version of this post"]Before I wrote up the proof and posted it on my webpage, this post consisted of a hint to the problem. Here is this hint, for the sake of completeness:

Read my note [url=http://www.cip.ifi.lmu.de/~grinberg/CircumRev.pdf]"Circumscribed quadrilaterals revisited"[/url] until Theorem 3. Your point $ O$ is called $ P$ in this note. Theorem 3 yields $ \frac {AO}{CO} \equal{} \frac {a}{c}$. But $ \frac {AO}{CO} \equal{} \frac {\left|AOB\right|}{\left|BOC\right|}$, where the notation $ \left|P_1P_2P_3\right|$ means the area of a triangle $ P_1P_2P_3$. Hence, $ \frac {\left|AOB\right|}{\left|BOC\right|} \equal{} \frac {a}{c}$. This and similar equations yield the proportionality

[color=blue][b](1)[/b][/color] $ \left|AOB\right|: \left|BOC\right|: \left|COD\right|: \left|DOA\right| \equal{} \left(ab\right): \left(bc\right): \left(cd\right): \left(da\right)$.

On the other hand, $ AB \equal{} a \plus{} b$.

But $ \left|AOB\right| \equal{} \frac12\cdot AB\cdot OK$, so that $ OK \equal{} \frac {2\cdot \left|AOB\right|}{AB} \equal{} \frac {2\cdot\left|AOB\right|}{a \plus{} b}$. By [color=blue][b](1)[/b][/color], this yields that $ OK$ is proportional to $ \frac {2ab}{a \plus{} b} \equal{} \frac {2}{a} \plus{} \frac {2}{b}$. Similarly for $ OM$, $ OL$ and $ ON$, and the equation $ \frac {1}{OK} \plus{} \frac {1}{OM} \equal{} \frac {1}{OL} \plus{} \frac {1}{ON}$ becomes trivial (since it rewrites as $ \left(\frac {2}{a} \plus{} \frac {2}{b}\right) \plus{} \left(\frac {2}{c} \plus{} \frac {2}{d}\right) \equal{} \left(\frac {2}{b} \plus{} \frac {2}{c}\right) \plus{} \left(\frac {2}{d} \plus{} \frac {2}{a}\right)$).

Sorry for being vague, I hope someone else finds a simpler proof anyway...[/hide]

  darij

This is one of the improved theories of Butterfly Th..


trigonometry
inequalities
induction
calculus
analytic geometry
graphing lines
slope

Prove that |sin nx| :le: n|sin x|, for n = 0, 1, 2, ..., and any real number x.

Wow, that's really great!  It looks good to me!

using some kind of calculus argument, couldn't you show it is true for any real number n (instead of just integers).  I mean the result is obvious if you just look at the graph y = sin(x) and y = x.  The slope of sin(x) is always less than or equal to 1.

It's not true for any n. Try n = 1/2 and x = :pi:.

By the way, well done Beta! :-)

beta wrote:I am not sure if this is right

[hide]Using induction,when n=0, we have the inequality true.
Suppose
|sin nx| :le: n|sin x| is true,
then
|sin(nx)cos(x)| :le: n|sin(x)|  because |cosx| :le: 1
|sin(x)cos(nx)| :le: |sinx| because of same reason
add them up we have
|sin(nx+x)| :le: (n+1)|sinx|   as desired.[/hide]

This proof needs one more step.

By the triangle inequality,

|sin(nx)cos(x) + sin(x)cos(nx)| :le: |sin(nx)cos(x)| + |sin(x)cos(nx)|.

You couldn't simply add them straight up since the left hand sides are different functions within the absolute values.


abstract algebra
algebra
polynomial
induction
group theory
geometry
geometric transformation

I was wondering if anyone here could point me to a proof of Abel's Impossibility Theorem. Thank you...

Isn't this easy? Every abelian group is solvable thus theorem follows from Galois theorem.

What have abelian groups to do with the question¿  :huh:

As far as I know, Abel's Impossibility theorem is the following:

If char $(K)=0$ and $f\in K[x],f\not=0$ is a nonzero polynomial with abelian Galois group, then $f(x)=0$ is solvable with radicals. And Galois theorem is the following: Let $f\in K[X],f\not =0$ and char$(K)=0$ or char$(K)>$deg$(f)$.Then $f(x)=0$ is solvable with radicals iff the Galois group of $f$ is solvable.

Correct me if I'm wrong. By the way: do we say in English soluble or solvable?

I know Abel's Impossibility theorem as given in http://mathworld.wolfram.com/AbelsImpossibilityTheorem.html  :?

OK. Let's try to prove that theorem.

Let $K$ be a field and $N$ be the splitting field of $f$ with respect to $K$. We will assume first that $f(x)=0$ is solvable with roots and that $E$ is some iterated radical extension of $K$.

Then we have sequence $(K_1)_{0\leq i \leq m}$ where $K_0=K,K_m=E$ and for all $i\in [1,m]$ we have
\[ K_i=K_{i-1}(x_i),  \]
where $x_i^{n_i}=a_i\in K_{i-1}$ for some $n_i>0$. Let $L_i$ be the normal extension of $K$ spanned by $K_i$ $(1\leq i\leq m)$. It is separable thus the Galois extension of $K$.

We will show by induction with respect to the $m$ that the Galois group of $L=L_m$ is solvable. Let $L'=L_{m-1}$ and $Gal(L'/K)$ solvable.

The normal extension spanned by $K_m=K(x_1,\ldots ,x_m)$ is an extension of $K$ spanned by conjugates of $x_i$ $(1\leq i\leq m)$. Let
\[ B=\{y_1,\cdots,y_n\}\subset L=L'(y_1,\cdots,y_n)  \]
be the set of conjugates of $x_m$ with respect to $K$. Because $L'$ consists the conjugates of $x_1,\cdots,x_{m-1}$, we have
\[ L=K(L'\cup B)  \]
Furthermore every $y_j^{n_m}$ is the conjugate of $x_m^{n_m}=a_m\in K_{m-1}$ with respect to the $K$ thus $y_j^{n_m}=b_j\in L'$. This means that $L$ is the splitting field of
\[ \prod_{j=1}^n (X^{n_j}-b_j)\in L'[X]  \]
with respect to the $L'$. Therefore Gal$(L/L')$ is solvable.

Because Gal$(L'/K)$ is isomorphic to the quotient group
\[ Gal(L/K)/Gal(L/L')  \]
is solvable we have by induction assumption that Gal$(L/K)$ is solvable and
\[ Gal(L/K)/Gal (L/N)\cong Gal (L/N)  \]
is solvable too.

Assume now that Gal$(N/K)$ is solvable. We have to prove that $f(x)$ is solvable by radicals. Let
\[ n=[N: K]=(Gal (N/K): 1)  \]
and $R$ a field of $n$th a roots of of unity in $K$. Because $N$ consists the conjugates of its elements it spans an extension $E=R(N)$ which is normal and thus a Galois extension.

Every $R$-automorphism $\sigma: E\to E$ is $K$-automorphism and thus $\sigma (N)=N$ for every normal extension $N$ of $K$.Therefore homomorphism
\[ Gal(E/R)\to Gal(N/K),\sigma\mapsto\sigma |N  \]
is injective because $N$ spans $E$.Therefore Gal$(E/R)$ is solvable and its order divides $n$.

Group $G=Gal(E/R)$ has thus a composition series $(G_i)_{0\leq i \leq m}$ where every quotient $G_i/G_{i+1}$ $(1\leq i\leq m-1)$ is cyclic and its order $d_i$ is a factor of $n$.

Let $E_i=E^{G_i}$ be the subextensions of $E$ corresponds to $G_i$. Now $E_i$s form a sequence of extensions
\[ R=E_0\subset E_1\subset\cdots\subset E_m=E.  \]
Therefore Gal$(E/E_i)=G_i$ for every $i\in[0,m]$ and because $G_{i+1}$ is the normal subgroup of $G_i$ as $0\leq i\leq m-1$ we have that $E_{i+1}$ is the Galois extension of $E_i$ and Gal$(E_{i+1}/E_i)\cong G_i/G_{i+1}$ thus cyclic group with $d_i$ elements.

Because $d_i|n$, $R$ consists the root of units of order $d_i$, and we have that $E_{i+1}$ is a radical extension of $E_i$. Furthermore $R$ is a radical extension of $K$ spanned by some primitive $n$th root of unity. From this it follows that $E$ is an iterated radical extension of $K$, and because it consists the splitting field $N$, Thus $f(x)=0$ is solvable by radicals.

I'm not sure if radical extension is a good word in English since I just know that term in Finnish.

[quote="puuhikki"]I'm not sure if radical extension is a good word in English since I just know that term in Finnish.[/quote]
I think it is- my algebra text does have a section titled "Solvable and Radical Extensions" (which leaves out the term "radical extension" in favor of "solvable by radicals").
On the other hand, each instance of "soluble" should be "solvable". In normal usages, those two words are interchangeable, but this isn't a normal usage.

Most of the mathematical translation problems I see come from French.

Thanks jmerry. I just like to learn to write more fluent English and maths so it is good if someone shows me mistakes I make in my language and maths too!

Its easier than that proof... just note that S_n is not a solvable group for n>=5, or that A_4 is simple, or whatever.

Any Galois theory book [even the one for begginers] will contain a proof that just need Zassenhaus lemma.


conics
parabola
analytic geometry
graphing lines
slope
calculus
derivative

Given two points $ P,\ Q$ on the parabola $ y \equal{} x^2$. Let $ h$ be the $ y$ coordinate of the midpoint of the segment $ PQ$.

(1) Let $ L$ and $ m$ be the length and slope of the segment $ PQ$ respectively. Express $ h$ in terms of $ L,\ m$.

(2) When $ L$ is fixed, find the minimum value of $ h$.

[color=green]Corrected by moderator.[/color]

[quote]Let  $ {h}$ be the  coordinate of the segment .
[/quote] What dose it mean?

Sorry, I made a typo. Correctly, as follows.

[quote="kunny"]Given two points $ P,\ Q$ on the parabola $ y \equal{} x^2$. Let $ h$ be the $ y$ coordinate of [color=blue]the mid point[/color] of the segment $ PQ$.

(1) Let $ L$ and $ m$ be the length and slope of the segment $ PQ$ respectively. Express $ h$ in terms of $ L,\ m$.

(2) When $ L$ is fixed, find the minimum value of $ h$.[/quote]

To moderator, could you correct the context such a blue line?

I can't edit it.

Thanks in advance.

Japanese Communities Moderator

kunny

Kunny, the answer for question b is

$ \ Min(h)\equal{}  \frac{1}{4}\left(2L\minus{}1\right)$ , isn't it? Thanks for checking  :)

Oh, I see your point, my mistake when I take the whole thing into consideration

When $ 0 \le  L\le 1$, the minimum happens if and only if $ |m|\equal{}0$ (take the derivative and let the first derivative be 0). Hence, $ min(h)\equal{}L^2$

 When $ L> 1$, the minimum happens if and only if $ |m|$ different from 0, $ min(h)\equal{}2L\minus{}1$

thanks for checking, I may start again
For question a

$ h\equal{}\frac{1}{4}\left(\frac{L^2}{m^2\plus{}1}\plus{}m^2\right)$

and therefore, we have $ \min(h) \equal{} \frac{2L\minus{}1}{4}$

Dear kunny, Does question (b) need calculus ?

My solution doesn't need to use calculus.

here is my solution for part(a):DENOTE the angle $ {\alpha}$;and let the $ {Q\minus{}x\equal{}x}$;then we get the equation :$ {(x\minus{}cos\alpha L)^2\plus{}sin\alpha L\equal{}x^2}$;solve it $ {\Rightarrow}$ $ {x\equal{}\frac  {cos\alpha L\plus{}tan\alpha}{2}}$;we also know :$ {h\equal{}M\minus{}y\equal{}\frac {(x\minus{}cos\alpha L)^2\plus{}x^2}{2}}$;put in $ {x}$ $ {\Rightarrow}$:$ {h\equal{} \frac{cos^2\alpha L^2\plus{}m^2}{4}}$;
we also have :1$ {sin^2\alpha\plus{}cos^2\alpha \equal{} 1}$ and 2.$ {m*cos\alpha\equal{}tan\alpha*cos\alpha\equal{}sin\alpha}$;sovle the equation system we get :$ {cos^2\alpha\equal{}\frac {1}{m^2\plus{}1}}$;put in $ {cos^2\alpha}$ we get :$ {h\equal{}\frac{\frac{L^2}{m^2\plus{}1}\plus{}m^2}{4}}$
and for (b) I think my solution must be wrong ,I am still in senior one for  the first term  and I have no idea about this question ,can you give me some hints,kunny?Thanks a lot ! :oops:
[hide]my solution(must be wrong :( )$ {4h\le 2*L*\sqrt{\frac{m^2}{m^2\plus{}1}}\le 2L}$;so $ {h\le \frac{L}{2}}$,so the mininum value is $ {\frac{L}{2}}$[/hide]

Ok, kunny, could you please post your full solution so that everybody can understand what goes beyond the problem?. I was thinking of using Cauchy inequality for the last expression and I derived my solution based on Cauchy inequality.

Thanks a lot! :)

great math, CCMath, castigioni, your answers to question (1) are correct. :)

For question (2), setting $ t\equal{}1\plus{}m^2$ gives $ h\equal{}\frac{1}{4}\left(t\plus{}\frac{L^2}{t}\right)\minus{}\frac{1}{4}\ (t\geq 1)$.

Now consider the graph $ y\equal{}t\plus{}\frac{L^2}{t}\ (t\geq 1)$.

Let $ P(p,p^2)$, $ Q(q,q^2)$.

Then $ m \equal{} p \plus{} q$, $ h \equal{} \frac {1}{2}\cdot (p^2 \plus{} q^2)$.

$ L^2 \equal{} PQ^2 \equal{} (p \minus{} q)^2(1 \plus{} (p \plus{} q)^2)$

So easily, $ 4h \equal{} \frac {L^2}{1 \plus{} m^2} \plus{} m^2$.

If $ L < 1$, then $ 4h \minus{} L^2 \equal{} \frac {m^2}{1 \plus{} m^2}(m^2 \plus{} 1 \minus{} L^2)\ge 0$.

Equality occurs at $ p \equal{} \frac {L}{2}$, $ q \equal{} \minus{} \frac {L}{2}$.

If $ L > 1$, then $ 4h \equal{} \frac {L^2}{1 \plus{} m^2} \plus{} m^2 \plus{} 1 \minus{} 1\ge 2L \minus{} 1$ by AM-GM.

Equality occurs if $ p,q \equal{} \frac {\sqrt {L \minus{} 1}\pm\sqrt {L}}{2}$.

Somehow I think we should determine a nice geometric property of a parabola from this discussion because fixing $ L$ is related to having a circle centered at $ P$ and intersecting the parabola again at $ Q$, $ Q'$ and somehow we have a nice case when $ L\equal{}1$...


Let $a$, $b$, $c$ be three positive reals. Prove the inequality

$\frac{a\left(b^2+c^2\right)}{\left(b+c\right)\left(2a^2+b^2+c^2\right)}+\frac{b\left(c^2+a^2\right)}{\left(c+a\right)\left(2b^2+c^2+a^2\right)}+\frac{c\left(a^2+b^2\right)}{\left(a+b\right)\left(2c^2+a^2+b^2\right)}$
$\leq\frac34$.

  Darij

Very Very Nice inequality.And it took me about 2 hours to come out an unnice solution :) .(Much shorter than half a year :D ).Hope it is right.And Here it is.
$Solution$:
$\sum$ means cyclic sums.
So:
$\frac{a\left(b^2+c^2\right)}{\left(b+c\right)\left(2a^2+b^2+c^2\right)}+\frac{b\left(c^2+a^2\right)}{\left(c+a\right)\left(2b^2+c^2+a^2\right)}+\frac{c\left(a^2+b^2\right)}{\left(a+b\right)\left(2c^2+a^2+b^2\right)} \le \frac 3 4$
$\iff \sum \frac{a\left(b^2+c^2\right)}{\left(b+c\right)\left(2a^2+b^2+c^2\right)} \le \frac 3 4$
$\iff \sum \frac{4a\left(b^2+c^2\right)}{\left(b+c\right)\left(2a^2+b^2+c^2\right)} \le 3$
$\iff \sum \frac{4a\left(b^2+c^2\right)}{\left(b+c\right)\left(2a^2+b^2+c^2\right)}-\sum \frac{2a\left(b+c\right)^2}{\left(b+c\right)\left(2a^2+b^2+c^2\right)}\le 3-\sum \frac{2a\left(b+c\right)^2}{\left(b+c\right)\left(2a^2+b^2+c^2\right)}$
$\iff \sum \frac{2a}{b+c} \cdot \frac{\left(b-c\right)^2}{\left(b+c\right)\left(2a^2+b^2+c^2\right)} \le \sum \frac{\left(a-b\right)^2+\left(a-c\right)^2}{2a^2+b^2+c^2}$
$\iff \sum \left( \frac{1}{2a^2+b^2+c^2}+ \frac{1}{a^2+2b^2+c^2}- \frac{1}{a^2+b^2+2c^2} \cdot \frac{2c}{a+b} \right)(a-b)^2 \ge 0$
$\iff \sum \left( \frac{1}{c^2(2a^2+b^2+c^2)}+ \frac{1}{c^2(a^2+2b^2+c^2)}- \frac{1}{a^2+b^2+2c^2} \cdot \frac{2}{c(a+b)} \right)c^2(a-b)^2 \ge 0$
let $S_c=\frac{1}{c^2(2a^2+b^2+c^2)}+ \frac{1}{c^2(a^2+2b^2+c^2)}- \frac{1}{a^2+b^2+2c^2} \cdot \frac{2}{c(a+b)}$
and $S_a,S_b$ similar.
Wlog $a \ge b \ge c$ :)
We have:
$b^2 S_b=\frac 1{2a^2+b^2+c^2}+\frac 1{a^2+b^2+2c^2}-\frac{2b}{a+c} \cdot \frac 1{a^2+2b^2+c^2}$
$\ge \frac 1{2a^2+b^2+c^2}+\frac 1{a^2+b^2+2c^2}-\frac{2b}{a} \cdot \frac 1{a^2+2b^2+c^2}$
$=\left(\frac 1{2a^2+b^2+c^2}-\frac{b}{a} \cdot \frac 1{a^2+2b^2+c^2} \right)+\left(\frac 1{a^2+b^2+2c^2}-\frac{b}{a} \cdot \frac 1{a^2+2b^2+c^2}\right)$
$=\left(\frac{a(a-b)^2+a(a+b)(a-b)+(a-b)c^2}{(2a^2+b^2+c^2)(a^2+2b^2+c^2)} \right)+\left(\frac 1{a^2+b^2+2c^2}-\frac{b}{a} \cdot \frac 1{a^2+2b^2+c^2}\right)$
$\ge \left(\frac{a(a-b)^2+a(a+b)(a-b)+(a-b)c^2}{(2a^2+b^2+c^2)(a^2+2b^2+c^2)} \right)+\left(\frac 1{a^2+b^2+2c^2}-\frac 1{a^2+2b^2+c^2}\right) \ge 0$
And
$c^2S_c=\frac{1}{2a^2+b^2+c^2}+ \frac{1}{a^2+2b^2+c^2}- \frac{1}{a^2+b^2+2c^2} \cdot \frac{2c}{a+b}$
$\ge \frac{1}{2a^2+b^2+c^2}+ \frac{1}{a^2+2b^2+c^2}- \frac{1}{a^2+b^2+2c^2}$
$= \left(\frac{1}{2a^2+b^2+c^2}-\frac{1}{2(a^2+b^2+2c^2)} \right)+ \left(\frac{1}{a^2+2b^2+c^2}- \frac{1}{2(a^2+b^2+2c^2)}\right)$
then we get $S_b,S_c \ge 0$
and also
$S_b+S_a=\frac{1}{b^2(2a^2+b^2+c^2)}+ \frac{1}{b^2(a^2+b^2+2c^2)}- \frac{1}{a^2+2b^2+c^2} \cdot \frac{2}{b(a+c)}-\frac{1}{a^2(a^2+2b^2+c^2)}+ \frac{1}{a^2(a^2+b^2+2c^2)}- \frac{1}{2a^2+b^2+c^2} \cdot \frac{2}{a(b+c)}$
$\ge \frac{1}{b^2(2a^2+b^2+c^2)}+ \frac{1}{b^2(a^2+b^2+2c^2)}- \frac{1}{a^2+2b^2+c^2} \cdot \frac{2}{ab}+\frac{1}{a^2(a^2+2b^2+c^2)}+ \frac{1}{a^2(a^2+b^2+2c^2)}- \frac{1}{2a^2+b^2+c^2} \cdot \frac{2}{ab}$
$= \left( \frac{1}{b^2(a^2+b^2+2c^2)}+ \frac{1}{a^2(a^2+2b^2+c^2)}-\frac{1}{a^2+2b^2+c^2} \cdot \frac{2}{ab} \right)$
$+\left( \frac{1}{b^2(2a^2+b^2+c^2)}+ \frac{1}{a^2(a^2+b^2+2c^2)}-\frac{1}{2a^2+b^2+c^2} \cdot \frac{2}{ab} \right)$
$\ge 0$
Then use the method of SOS.We get our result.
Well,what is your solution,Darij?

Zhaobin, your solution is correct, and not that unnice. Mine also uses degree-3 calculations:

Using the notation $\sum$ for cyclic sums, the inequality in question, $\sum\frac{a\left( b^2+c^2\right) }{\left(b+c\right) \left( 2a^2+b^2+c^2\right) }\leq\frac{3}{4}$, rewrites as $\frac{3}{4}-\sum\frac{a\left( b^2+c^2\right) }{\left( b+c\right)\left( 2a^2+b^2+c^2\right) }\geq 0$. Now,

$\frac{3}{4}-\sum\frac{a\left( b^2+c^2\right) }{\left( b+c\right)\left( 2a^2+b^2+c^2\right) }=\frac{3}{4}\sum\frac{a}{a+b+c}-\sum\frac{a\left( b^2+c^2\right) }{\left( b+c\right) \left( 2a^2+b^2+c^2\right) }$
$=\sum\left( \frac{3}{4}\cdot\frac{a}{a+b+c}-\frac{a\left( b^2+c^2\right) }{\left( b+c\right) \left( 2a^2+b^2+c^2\right) }\right)$
$=\sum\frac{a\left( 5ab+bc+ca+b^2\right) \left( a-b\right) -a\left(5ca+ab+bc+c^2\right) \left( c-a\right) }{4\left( a+b+c\right) \left(b+c\right) \left( 2a^2+b^2+c^2\right) }$
$=\frac{1}{4\left( a+b+c\right) }\cdot\sum\left( \frac{a\left(5ab+bc+ca+b^2\right) \left( a-b\right) }{\left( b+c\right) \left(2a^2+b^2+c^2\right) }-\frac{a\left( 5ca+ab+bc+c^2\right) \left(c-a\right) }{\left( b+c\right) \left( 2a^2+b^2+c^2\right) }\right)$
$=\frac{1}{4\left( a+b+c\right) }\cdot\left( \sum\frac{a\left(5ab+bc+ca+b^2\right) \left( a-b\right) }{\left( b+c\right) \left(2a^2+b^2+c^2\right) }-\sum\frac{a\left( 5ca+ab+bc+c^2\right)\left( c-a\right) }{\left( b+c\right) \left( 2a^2+b^2+c^2\right)}\right)$
$=\frac{1}{4\left( a+b+c\right) }\cdot\left( \sum\frac{b\left(5bc+ca+ab+c^2\right) \left( b-c\right) }{\left( c+a\right) \left(2b^2+c^2+a^2\right) }-\sum\frac{c\left( 5bc+ca+ab+b^2\right)\left( b-c\right) }{\left( a+b\right) \left( 2c^2+a^2+b^2\right)}\right)$
$=\frac{1}{4\left( a+b+c\right) }\cdot\sum\left( \frac{b\left(5bc+ca+ab+c^2\right) \left( b-c\right) }{\left( c+a\right) \left(2b^2+c^2+a^2\right) }-\frac{c\left( 5bc+ca+ab+b^2\right) \left(b-c\right) }{\left( a+b\right) \left( 2c^2+a^2+b^2\right) }\right)$
$=\frac{1}{4\left( a+b+c\right) }\cdot\sum\left( b-c\right) \left(\frac{b\left( 5bc+ca+ab+c^2\right) }{\left( c+a\right) \left(2b^2+c^2+a^2\right) }-\frac{c\left( 5bc+ca+ab+b^2\right) }{\left(a+b\right) \left( 2c^2+a^2+b^2\right) }\right)$.

Now, we want to show that this is $\geq 0$; we will do this by proving that the sum $\sum\left( b-c\right) \left( \frac{b\left( 5bc+ca+ab+c^2\right)}{\left( c+a\right) \left( 2b^2+c^2+a^2\right) }-\frac{c\left(5bc+ca+ab+b^2\right) }{\left( a+b\right) \left( 2c^2+a^2+b^2\right) }\right)$ consists of nonnegative terms only. In fact, let's show that the term $\left( b-c\right) \left( \frac{b\left( 5bc+ca+ab+c^2\right) }{\left(c+a\right) \left( 2b^2+c^2+a^2\right) }-\frac{c\left(5bc+ca+ab+b^2\right) }{\left( a+b\right) \left( 2c^2+a^2+b^2\right) }\right)$ is nonnegative (the other two terms are analogous).

In fact, two cases are possible: $b\geq c$ and $b\leq c$. We consider the case $b\geq c$ first.

In this case, $b-c\geq 0$, so that in order to prove that $\left( b-c\right) \left( \frac{b\left( 5bc+ca+ab+c^2\right) }{\left(c+a\right) \left( 2b^2+c^2+a^2\right) }-\frac{c\left(5bc+ca+ab+b^2\right) }{\left( a+b\right) \left( 2c^2+a^2+b^2\right) }\right) \geq 0$, it will be enough to show that $\frac{b\left( 5bc+ca+ab+c^2\right) }{\left( c+a\right) \left(2b^2+c^2+a^2\right) }-\frac{c\left( 5bc+ca+ab+b^2\right) }{\left(a+b\right) \left( 2c^2+a^2+b^2\right) }\geq 0$, or, equivalently, that $\frac{b\left( 5bc+ca+ab+c^2\right) }{\left( c+a\right) \left(2b^2+c^2+a^2\right) }\geq\frac{c\left( 5bc+ca+ab+b^2\right)}{\left( a+b\right) \left( 2c^2+a^2+b^2\right) }$. This rewrites as

$b\left( 5bc+ca+ab+c^2\right) \left( a+b\right) \left( 2c^2+a^2+b^2\right) \geq c\left( 5bc+ca+ab+b^2\right) \left( c+a\right)\left( 2b^2+c^2+a^2\right)$.

First, note that $a+b\geq c+a$ (since $b\geq c$), so that it remains to show the inequality

$b\left( 5bc+ca+ab+c^2\right) \left( 2c^2+a^2+b^2\right) \geq c\left( 5bc+ca+ab+b^2\right) \left( 2b^2+c^2+a^2\right)$.

Now, this inequality is the sum of the inequalities

$b\left( ca+ab\right) \left( 2c^2+a^2+b^2\right) \geq c\left(ca+ab\right) \left( 2b^2+c^2+a^2\right)$        and
$b\left( 5bc+c^2\right) \left( 2c^2+a^2+b^2\right) \geq c\left( 5bc+b^2\right) \left( 2b^2+c^2+a^2\right)$.

The first of these inequalities simplifies to $b\left( 2c^2+a^2+b^2\right) \geq c\left( 2b^2+c^2+a^2\right)$ and follows from

$b\left( 2c^2+a^2+b^2\right) -c\left( 2b^2+c^2+a^2\right) =\left( b-c\right) \left( a^2+bc+\left( b-c\right) ^2\right) \geq 0$

(since $b-c\geq 0$). The second inequality,

$b\left( 5bc+c^2\right) \left( 2c^2+a^2+b^2\right) \geq c\left( 5bc+b^2\right) \left( 2b^2+c^2+a^2\right)$,

rewrites as

$bc\left( 5b+c\right) \left( 2c^2+a^2+b^2\right) \geq bc\left( 5c+b\right) \left( 2b^2+c^2+a^2\right)$,

and simplifies to

$\left( 5b+c\right) \left( 2c^2+a^2+b^2\right) \geq\left( 5c+b\right) \left( 2b^2+c^2+a^2\right)$.

But this is the sum of the inequalities

$\left( 5b+c\right) a^2 \geq\left( 5c+b\right) a^2$            and
$\left( 5b+c\right) \left( 2c^2+b^2\right) \geq\left( 5c+b\right) \left( 2b^2+c^2\right)$.

The first of these inequalities follows from $5b+c\geq 5c+b$ (what is equivalent to $4b\geq 4c$ and thus to $b\geq c$). The second inequality follows from

$\left( 5b+c\right) \left( 2c^2+b^2\right) -\left( 5c+b\right) \left( 2b^2+c^2\right) =3\left( b-c\right) ^3\geq 0$

(using $b-c\geq 0$ again). This completes the proof for the case $b\geq c$.

Similarly, in the case $b\leq c$, we get $b-c\leq 0$ and $\frac{b\left( 5bc+ca+ab+c^2\right) }{\left( c+a\right) \left(2b^2+c^2+a^2\right) }-\frac{c\left( 5bc+ca+ab+b^2\right) }{\left(a+b\right) \left( 2c^2+a^2+b^2\right) }\leq 0$, so that $\left( b-c\right) \left( \frac{b\left( 5bc+ca+ab+c^2\right) }{\left(c+a\right) \left( 2b^2+c^2+a^2\right) }-\frac{c\left(5bc+ca+ab+b^2\right) }{\left( a+b\right) \left( 2c^2+a^2+b^2\right) }\right) \geq 0$.

And the problem is solved.

  Darij

[quote="perfect_radio"]I'll never understand how you all make that $(a-b)^2$ appear in these situations. How do you do this?[/quote]
In this one,I want to prove:
$\frac{4a\left(b^2+c^2\right)}{\left(b+c\right)\left(2a^2+b^2+c^2\right)} \le \sum \frac{2a\left(b+c\right)^2}{\left(b+c\right)\left(2a^2+b^2+c^2\right)}$
But I find I reversed,then the solution is come out.
Anyway,Darij your solution is more elementary :) .Mine uses Sos.


Suppose that river moves with velocity V along  (i^) direction.(plz. read it as 'i cap' as I do not know the code language.)
let the boat make an angle 90+x with the direction of flow of river.(i.e. x with normal to yhe bank)

hence,the boat tries to have velocity,
-V/2sin(x) (i^) +V/2cos(x) (j^)

But its actual velocity due to the flow of the river (V(i^)) becomes,
[V-V/2sin(x)] (i^) +V/2cos(x) (j^)

let drift=D and W =width of river
then,
W=V/2cos(x)t     where t=time taken to cross the river
&
D=[V-V/2sin(x)]t

hence,
D=W[2sec(x)-tan(x)]
d[D]/dx=W[2sec(x)tan(x)-sec^2(x)]=0 for maxima or minima.

i.e.    x=30 deg.    (clearly for minima)

required angle=90+30=120 degrees.

[quote="bkunchanapalli"]A  boat moves relative to the water with a velocity which is n=2 times less than the river flow velocity.At what angle to the stream direction must the boat move to minimize drifting

can anyone please clearly solve this problem[/quote]
you shoi


function
logarithms
algebra
rational function
real analysis
real analysis unsolved

[b]Theorem. [/b] Let $b_1,b_2,\ldots$ be integers such that the function $\sum_{n \geq 1} nb_nx^n$ is rational. Then, the function $\sum_{n \geq 1}b_nx^n$ is also rational.

[b]Alternative Formulation: [/b] If $\{a_n\}_{n \geq 1}$ is a linear recursive sequence of integers such that $n | a_n$ for all $n$, then the sequence $\big\{\frac{a_n}{n}\big\}$ also satisfies a linear recurrence.

--Vesselin

We shall use some well-known facts in the proof:
1. A rational function $\frac {P(X)}{Q(X)}$ can be written as $\sum \frac {r_{i,n}}{(x-x_i}^n+\sum {a_{i,n}x+b_{i,n}}{(x-y_i)^n(x_\overline{y_i})^n}$ where $x_i$ are real solutions and $y_i$ are non-real solutions to $Q(X)$, $a_{i,n}, b_{i,n}, t_{i,n}$ are integers (or rationals, I don't remember).
2. The terms of a linear recurrence (possibly multiplied by a suitable constant) are all integers iff the generating function determined by them is a rational function.
3. Every term $\frac 1{(x-x_i)^n}$ or $\frac 1{(x^2+t)^n}$ can be integrated as a sum of terms of form $\frac 1{x-x_i}^n$, $\ln (x-x_i)$   or $\frac 1{(x^2+t)^n}$ , $t \arctan {\frac xt}$.
4. from 1.) and 3.) we derive that every rational fuction can be integrated similarly as in 3.)

Now from here we can prove the assertion.
Let $P(X)$ be the generating function of the sequence.
Then $Q(X)=(XP(X))'$ is a rational function.
Hence integrating $Q$ as in 4.) we dedcue that $P(X)=H(X)+t(X)$ where $H$ is a rational function and $T$ is the sum of terms of form $\frac 1{x-x_i}^n$ or $t \arctan {\frac xt}$.
Thus all terms $c_n$ where $\sum_{i=0}^{\infty} a_i X^i=t(X)=P(X)-H(X)$ are integer (or become after a suitable multiplication).
However, expanding $t(X)$ by Taylor we get that the $c_i=\frac 1{i!} t^{(i)}(0)$ and it's clear that $t^{n}(0)$ can be written
as $\sun \alpha_i \epsilon_i^n$ hence $|c_i|\rightarrow 0$ as $i$ tends to $\infty$ (here we use Stirling).
Since $c_i$ are integers or become integers after multiplyiing them by a constant this implies that $c_i=0$ from some moment.
This means $t(X)$ is a rational function and thus so is $P(X)$.
QED


LaTeX
group theory
abstract algebra
superior algebra
superior algebra theorems

Can anyone give me the proof of Jordon-Holder theorem.I know that I have to use 1st isomorphism theorem but the problem is I dont know how to use it & the text books I have donot give the proof of the theorem..

You can post the statement here for our happiness. ;)

Hmm your last post didnot bring me any happiness as I dont know latex & typing it is impossible.I am trying to find the link for the statement

It is not an argument. You see a huge number of examples of using LaTeX on this forum.

No I know thats not an excuse.But I am too busy studing for my college exams.I will learn latex after my exams,meanwhile the theorem is there in Dummit & Foote but not the proof

No need to think about it I have at last found the proof :lol:

Jordan-Hölder theorem follows immediately from the following Schreier theorem: Let $G$ be a group. Two normal towers of subgroups ending with the trivial group have the same equivalent refinements.

It would be nice if someone gave an application of this theorem

Does it help in any way to prove statements about finite groups?

I don't really see the use yet as one series produced by the theorem does not define the group

Well a important consequence of the theorem is:
Prove that Jordon-Holder theorem implies fundamental theorem of arithmatic.


[hide]If we let $\text{cis}{x}=e^{ix}$ for any $x \in \mathbb{C}$, that's
\[\sqrt{\lfloor{(5+\frac{1}{e^{\sqrt{2}}})^{\frac{1}{1024}}\rfloor}}=1 \]
[/hide]

Did you omit the square root signs just because they were too tedious?

In any case, this problem turns out to be quite silly, if I understand it correctly.

[hide="Because, I mean..."] $5+\text{cis }(i \sqrt{2}) = 5+e^{-\sqrt{2}}< 6$, and of course, since

$6 < 2^{1024}$

We know that

$\sqrt[1024]{6}< 2$

But it's also greater than $1$, so its floor is $1$, and the square root of $1$ is $1$. [/hide]

Oh yeah...forgot to type that 1024 in...


quadratics
abstract algebra
number theory proposed
number theory

Prove that there doesn't exist integer numbers $ x$ and $ y$ such that $ x^4 \plus{} y^7 \equal{} 31^{10}$.
I would like to see a nice solution.
This problem was proposed this year on Bulgarian TST for BMO.

I know this solution but is there anything else?

Can you tell how did you come up to mod 29? Have you checked all the primes smaller than 29?

[b]Lemma:[/b]  Let $ p$ be a prime.  There are $ \frac{p\minus{}1}{\gcd(k, p\minus{}1)}$ nonzero $ k^{th}$ powers $ \bmod p$.

[i]Proof.[/i]  The $ k^{th}$ powers are precisely the roots of $ x^{ \frac{p\minus{}1}{\gcd(k, p\minus{}1)} } \minus{} 1 \bmod p$, which is a divisor of $ x^{p\minus{}1} \minus{} 1$, so the result follows by Lagrange's theorem and Fermat's little theorem.

This quantity is minimized when $ k | p\minus{}1$, which here is $ 28$.

Thanks, [b]t0rajir0u[/b]. :lol:  I didn't know this lemma but it seems really useful :wink:

calculating residues with such a big exponent is tedious without calculator in a contest...is there any way to ease this calculation ??please help someone

Your definition of big is strange :D
$ 4,7,10$ are really not big.

$ 29$ isn't a big modulus; the whole computation took maybe two minutes, and in any case it's much less than the amount of time you'd usually have to spend on an Olympiad problem.  It's helpful to know how many nonzero residues you can expect because you can stop when you have that many - I didn't need to square $ 14$ things to compute the quartic residues.

Edit:  Perhaps you aren't familiar with [url=http://en.wikipedia.org/wiki/Exponentiation_by_squaring]binary exponentiation[/url].  To compute quartic residues all you have to do is square twice.  And to compute seventh powers $ \bmod 29$ it's enough to figure out what the square root of $ \minus{}1$ is (why?).

[quote="SAPOSTO"]I agree with [b]ishfaq420haque[/b]. How am I going to calculate all the residues modulo 29. I get the idea of modulo 29 but I thought it's senseless. Even if I had start with modulo 29 I don't know if I would manage to find all these residues and to finish the solution. Is there some easy way to calculate these residues.[/quote]
Well.. $ 2^4 \equal{} 16$ is a $ 4^{th}$ power $ \bmod 29$, but not a $ k^{th}$ power for any $ k \nmid 4$. So, you could just calculate $ 16^0, 16^1, 1^2,....,16^6$, to get all $ 4^{th}$ powers $ \bmod 29$. And you can do the same with $ 2^7 \equiv 12$, to find all $ 7^{th}$ powers. ($ 2$ is not a quadratic residue, since $ 29 \equiv 5 \bmod 8$)


Determine all integer solutions of the system \[2uv-xy=16,\] \[xv-yu=12.\]

System equavalent to $ (x\plus{}u\sqrt 2 )(y\minus{}v\sqrt 2 )\equal{}\minus{}16\minus{}12\sqrt 2$.
$ 16^2\plus{}2*12^2\equal{}2^5*17$. There are only one way factorizate $ 17\equal{}(5\plus{}2\sqrt 2)(5\minus{}\sqrt 2)$.
Uniques are $ \pm 1, \pm 1\pm \sqrt 2$ and 2 ways factorizate $ 4\equal{}2*2\equal{}(2\plus{}\sqrt 2)(2\minus{}\sqrt 2 )$.
It give all ways factorizate $ \minus{}16\minus{}12\sqrt 2$ or all solutions these system.

[quote="Rust"]System equavalent to $ (x \plus{} u\sqrt 2 )(y \minus{} v\sqrt 2 ) \equal{} \minus{} 16 \minus{} 12\sqrt 2$.
$ 16^2 \plus{} 2*12^2 \equal{} 2^5*17$. There are only one way factorizate $ 17 \equal{} (5 \plus{} 2\sqrt 2)(5 \minus{} \sqrt 2)$.
Uniques are $ \pm 1, \pm 1\pm \sqrt 2$ and 2 ways factorizate $ 4 \equal{} 2*2 \equal{} (2 \plus{} \sqrt 2)(2 \minus{} \sqrt 2 )$.
It give all ways factorizate $ \minus{} 16 \minus{} 12\sqrt 2$ or all solutions these system.[/quote]
I don't think that the above proof is correct. In the UFD $ \mathbb Z[\sqrt2]$, the absolute norm of $ \minus{} 16 \minus{} 12\sqrt2$ is $ |16^2 \minus{} 2\cdot 12^2| \equal{} 2^5$, (not $ 16^2 \plus{} 2\cdot 12^2 \equal{} 17 \cdot 2^5$), and there are infinite number of units, viz. $ \pm(1 \plus{} \sqrt2)^n$, $ n\in\mathbb Z$, (not just $ \pm 1$, $ \pm 1\pm \sqrt 2$).

You are right. Solutions are
$ x \plus{} y\sqrt 2 \equal{} \pm (\sqrt 2)^m(1 \plus{} \sqrt 2)^k, y \minus{} v\sqrt 2 \equal{} \mp (\sqrt 2 \minus{} 1)^{k \minus{} 2}(\sqrt 2)^{5 \minus{} m},m \equal{} 0,1,2,3,4,5, k\in Z.$


function
real analysis
real analysis unsolved

Let $E=$(the set of lipschitz functions)$f: \mathbb{R} \to \mathbb{R}$. For all $f \in E$, define norm
$||f||=|f(0)|+sup_{x\neq y} |\frac{f(x)-f(y)}{x-y}|$. Prove that $E$ is complete space with metric induced by it's norm.

Take a Cauchy sequence $(f_{n})$ for the induced metric. Then clearly $(f_{n}(0))$ is a Cauchy sequence in $\mathbb{R}$ and hence $f_{n}(0) \rightarrow f(0) \in \mathbb{R}$. Now taking $y=0$ in the second summand shows that $(f_{n}(x))$ must be Cauchy for every $x \neq 0$ as well, whence $f_{n}(x) \rightarrow f(x) \in \mathbb{R}$ . Whence $f_{n} \rightarrow f$ pointwise, and $f$ is clearly lipschitz (write the definition of $f_{n}$ being lipschitz and take limits). Also it is easy to show that $||f_n-f|| \rightarrow 0$ for $n \rightarrow +\infty$ .


A={1,2,3,.........150}
B={x*y|x,y € A and x and y aren't equal)
how many elements of B are divisible by 3?

[hide="Answer"]
For x not multiples of 3, only those y will cause x*y to be divisible by 3, which are multiples of 3. Thus this gives us (150 - (150/3) ) * (150/3) = 50000 results. x*x will not affect the result.

For x being multiples of 3, all products will be divisible by 3 except x*x which is to be ignored. This gives us (150-1)*(150/3) = 7450 results.

Thus, the no. of elements of B divisible by 3 are 57450

[/hide]

first (150-150/3)*(150/3)=5000 its not 50000.
second i think there are too many doubles which gives the same result such as 18*1=6*3=9*2
i think answer must be smaller than 7450 because we choose one of the numbers which are divisible by 3 and than we choose any other number from A.50*149=7450.but as i said there will be doubles which give the same result.i couldn't find a way to solve this problem.


If $ AD$ is the internal bisector of the angle $ A$ of the triangle $ ABC$, prove that $ \overline{AD}^2 \equal{} \overline{AB}\cdot\overline{AC} \minus{} \overline{BD}\cdot\overline{DC}$.

[hide="Hint"]
Internal Bisector Theorem and Stewart's  Theorem
[/hide]

[quote][color=darkred]Let the ray $ [AD$ be the internal bisector of $ \widehat {BAC}$ in  $ \triangle\ ABC$, where

$ D\in (BC)$. Prove that exists the relation $ \boxed {AD^2 \equal{} bc \minus{} DB\cdot DC}\ .$[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote the second intersection $ E$ of the line $ AD$ with the circumcircle $ w$ of $ \triangle ABC$.

Observe that the power $ p_w(D)$ of $ D$ w.r.t. $ w$ is $ p_w(D) \equal{} \minus{} DA\cdot DE \equal{} \minus{} DB\cdot DC$.

Therefore, $ \triangle ABE\sim\triangle ADC$ $ \implies$ $ \frac {AB}{AD} \equal{} \frac {AE}{AC}$ $ \Longleftrightarrow$ $ bc \equal{} AD\cdot AE$ $ \Longleftrightarrow$

$ bc \equal{} AD\cdot (AD \plus{} DE)$ $ \Longleftrightarrow$  $ AD^2 \equal{} bc \minus{} DA\cdot DE$ $ \Longleftrightarrow$ $ \boxed {AD^2 \equal{} bc \minus{} DB\cdot DC}\ .$

[b][u]Remark.[/u][/b] $ \frac {DB}{c} \equal{} \frac {DC}{b} \equal{} \frac {a}{b \plus{} c}\implies$ $ AD^2 \equal{} bc \minus{} \frac {bca^2}{(b \plus{} c)^2}$ $ \implies$

$ \boxed {AD \equal{} \frac {2\sqrt {bcp(p \minus{} a)}}{b \plus{} c}}$ i.e. $ \boxed {AD \equal{} \frac {2bc\cdot\cos\frac A2}{b \plus{} c}}\ ,$ where $ 2p \equal{} a \plus{} b \plus{} c\ .$ [/color]

[quote][color=darkred][b][u] A similar problem.[/u][/b] Let the ray $ [AD$ be the external bisector of $ \widehat {BAC}$

in nonisosceles $ \triangle\ ABC$, where $ D$ belongs to the sideline $ BC$.

Prove that exists the relation $ \boxed {AD^2 \equal{} DB\cdot DC \minus{} bc}\ .$[/color][/quote]


For positive integers $n$, let $\tau (n)$ denote the number of positive integer divisors of $n$, including $1$ and $n$. For example, $\tau (1)=1$ and $\tau(6) =4$. Define $S(n)$ by \[S(n)=\tau(1)+ \tau(2) + ... + \tau(n).\] Let $a$ denote the number of positive integers $n \leq 2005$ with $S(n)$ odd, and let $b$ denote the number of positive integers $n \leq 2005$ with $S(n)$ even. Find $|a-b|$.

[hide="Solution"]Since a perfect square always has an odd number of divisors, $\tau (n)$ is odd if and only if $n$ is a perfect square. Additionally, $S(n)$ changes parity with each perfect square. Hence, all the numbers between each perfect square must be either even or odd.

Now, we can begin calculating. The largest perfect square that is less than $2005$ is $44^2$.

From $S(1)$ to $S(44^2-1)=S(1935)$, we have
$(2^2-1^2)+(4^2-3^2)+...+(44^2-43^2)=990$ integers for which $S(n)$ is odd.

We have $(3^2-2^2)+(4^2-3^2)+...+(43^2-42^2)= 945$ integers for which $S(n)$ is even. We also need to take into consideration the 70 integers from $1936$ to $2005$ which are even.

Thus, $a=1015$, $b=990$, and $|a-b|=25$ Our answer is $25$.[/hide]

[hide="Answer"]The only numbers with $\tau(1)$ odd are perfect squares, so $S(n)$ changes between odd and even only at perfect squares. We begin with three odds, and add two each time, so we have three odd, five even, seven odd, etc. Specifically, the sequence becomes even when it reaches an even perfect square, and odd for an odd perfect square. The last square before $2005$ is $44^2=1936$, so we have a series of $70$ evens, and the last full set was $87$ odds. Therefore, we have $a=3+7+11+...+83+87=990$ and $b=5+9+13+...+85+70=1015$, so $|a-b|=\boxed{25}$.[/hide]

[hide]At every perfect square, we change parity.  That's because only perfect squares can get an odd number of divisors.
1,2,3 are odds
4-8 are evens
9-15 are odds...
so a=3+7+11+...
while b=5+9+...
Since 2025=45^2, 44^2 is the last square.  Since at even squares we're even, a contains as many terms as b.
The numbers in a are 2 less than the corresponding numbers in b except for the case from 44^2 since we don't reach 45^2.  We're 20 short of 2025.  We have 22 terms each since we have the same on a and b.  Since b is normally 2 greater than a but we're short twenty, at the last one, a is greater than b by 18.  But in the 21 other places, b is greater than a by 2 each time so 21*2=42.  |18-42|=|-24|=24.[/hide]

Oops, I forgot to add one when I count how many numbers are in a series of numbers.

[quote="314.math"][quote="4everwise"]Additionally, $ S(n)$ changes parity with each perfect square. Hence, all the numbers between each perfect square must be either even or odd.[/quote]

I don't get how this is found.  Could someone please explain it to me?

Thanks in advance![/quote]

Perfect squares are the only numbers that can have an odd number of divisors.  We can see this by taking the prime factorization, and taking the product of one plus all the exponents.  Since all numbers that are not perfect squares have an even number of divisors, the parity will not change.

[quote="4everwise"]From $ S(1)$ to $ S(44^2 \minus{} 1) \equal{} S(1935)$, we have
$ (2^2 \minus{} 1^2) \plus{} (4^2 \minus{} 3^2) \plus{} ... \plus{} (44^2 \minus{} 43^2) \equal{} 990$ integers for which $ S(n)$ is odd.[/quote]

Could someone please explain the second line to me?  I don't get how that was gotten from $ S(44^2 \minus{} 1) \equal{} S(1935)$.

[b]Solution:[/b] Realize that $\tau (n)$ is odd if and only if $n$ is a square. If $n$ is not a square, it is a certainty that $\tau (n)$ is even. Noting this, the pattern for the parity of $S(n)$ follows as such: odd (1), odd (2), odd (3), even (4), even (5), even (6), even (7), even (8), odd (9), ...

Now we can just apply some arithmetic stuff to get that $b = 45 \cdot 21 + 70$ and $a =  45 \cdot 22$. So we get that $b-a = 70 - 45 = \boxed{025}$. And we are done. $\blacksquare$

Funny thing is I was practicing old aime problems and I just did this


Could someone help me with the formula, or even better, the derivation for the area of an octagon with radius "R"? I can't see how to it without trigonometry. Thanks.

This is a fairly sizable hint, I believe:[hide]Put the octagon inside a square.[/hide]

More hint from JBL:

Relationship of 4 Right Triangles and Square.

[hide]

[/hide]

This practically gives it away  .

[hide]The length of each side of the octagon can be represented by x.

So, the length of each side of the square is (x + x :sqrt: 2) .

The area is (x + x :sqrt: 2) 2 - 2x 2 .

[/hide]

Derivation. You can fill in the missing steps.

Thanks for the help guys! :)  I think I get it now.

i dont have time to test it now but by that picture it looks like...

[hide]
4 x ( .5 (diameter) )
[/hide]


vector
induction
combinatorics unsolved
combinatorics

Let $n$ be a positive integer, set $S_n = \{ (a_1,a_2,\cdots,a_{2^n}) \mid a_i=0 \ \text{or} \ 1, 1 \leq i \leq 2^n\}$. For any two elements $a=(a_1,a_2,\cdots,a_{2^n})$ and $b=(b_1,b_2,\cdots,b_{2^n})$ of $S_n$, define
\[ d(a,b)= \sum_{i=1}^{2^n} |a_i - b_i| \]
We call $A \subseteq S_n$ a $\textsl{Good Subset}$ if $d(a,b) \geq 2^{n-1}$ holds for any two distinct elements $a$ and $b$ of $A$. How many elements can the $\textsl{Good Subset}$ of $S_n$ at most have?

I think the answer is $2^{n+1}$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=vectors&t=43211]this[/url] problem. It states that the maximal number of vectors in $\mathbb R^{d}$ such that each two have non-positive scalar product is $2d$.

Our problem asks for the maximal number of binary vectors of length $2^{n}$ such that each two differ in at least $2^{n-1}$ coordinates. Of course, $0$ and $1$ play no special role here. We may as well assume we’re working with vectors with entries $\pm 1$, and in this case the condition that two vectors differ in at least $2^{n-1}$ is the same thing as saying that each two vectors have non-positive scalar product. We can thus use the problem cited above for $d=2^{n}$ and conclude that $2^{n+1}$ is an upper bound for the size of a good set.

In order to prove that, conversely, $2^{n+1}$ can be achieved, we look for an orthogonal basis $e_{1},\ldots,e_{2^{n}}$ of $\mathbb R^{2^{n}}$ formed with vectors having entries $\pm 1$, and take our good set to be $\{\pm e_{1},\ldots,\pm e_{2^{n}}\}$. We can construct such a basis $\{e_{i}\}$ by induction on $n$:

Assuming we have the basis $e_{1},\ldots,e_{2^{n}}$ in $\mathbb R^{2^{n}}$, we can take, for example, $f_{i},\ i=\overline{1,2^{n}}$ to be the vectors of size $2^{n+1}$ formed by appending $e_{i}$ after $e_{i}$ (i.e. we just duplicate the $e_{i}$), and $f_{j},\ j=\overline{2^{n}+1,2^{n+1}}$ to be the vectors obtained by appending $e_{j-2^{n}}$ after $-e_{j-2^{n}}$. $f_{1},\ldots,f_{2^{n+1}}$ will be the basis in $\mathbb R^{2^{n+1}}$, and this completes the induction step.

[quote="grobber"]I think the answer is $2^{n+1}$.
[/quote]
This is the correct answer!  :)

Just linking. See [url=https://artofproblemsolving.com/community/c6h51286p324345]this[/url] thread.


Find all the $ a,b\in \mathbb{N}$ such that :

$ \frac{1}{x}\plus{}\frac{1}{y}\in [a,b]\ ,\ (\forall)x,y\in [a,b]$

Does $ [a, b]$ mean $ b \ge a$? Anyway, assuming it does:

$ x, y \equal{} a$
$ \frac {2}{a} \le b$
$ ab \ge 2$
$ x, y \equal{} b$
$ \frac {2}{b} \ge a$
$ ab \leq 2$
$ ab \equal{} 2$
$ \{1, 2\}$


symmetry
combinatorics unsolved
combinatorics

Around the circular table there are $2005$ knights, and one of them has $z\leq 2005$ coins. He is supposed to deliver these coins to the other knights in such a way that no knight ends up with more than one coin. His intention is that , in each step, one of those knights which has more than one coin (if there is any such knight) given one coin to each of his neighbours.
a) Prove that the livery must stop if $z<2005$
b) Can the delivery be finished if $z=2005$.

First we consider the situation where there is one knight at each integer on the real line, and the knight initially at $0$ has the coins, it is reasonably easy to prove (I think) that if the knight at $k>0$ recieves a coin, then the knights from $1$ to $k-1$ must at some previous turn have been holding $k$ coins (in fact, $k$ turns before the knight at $k$ recieves a coin, each of the knights from $2$ to $k-1$ must have had one coin, and the knight at $1$ must have had two).

Now this means that if any coin travels more than half way around the table (i.e. is passed to more than $1002$ distinct knights), then the $1002$ knights it passed on it's way must at some point have been holding $1003$ or more coins.  But since the number of coins being held by the $1002$ knights adjacent to the first going clockwise (say) is obviously equal to the number held by the $1002$ knights adjacent to the first travelling anti-clockwise (by symmetry), all the knights together must have been holding at least $2006$ coins, which is impossible.

Now we number the knights by their distance from the starting knight, along whichever path is shorter, and define the weight of a coin as the square of the number of the knight holding it. Whenever a knight distributes two coins to his neighbours, he increases the sum of all the weights by two (the only knights for which this would not be true are the two numbered 1002, but we have seen that they can never pass on coins to one another).  But the sum of the weights is obviously bounded, and so the coins can only be shared a finite number of times - that is, the delivery must end.

Incidentally, I'm not sure why you separate out the case of $z=2005$, since this is exactly equivalent to the case with $z=2004$, except that the starting knight always has one more coin (but since his number of coins is odd, he doesn't make any more swaps as a result), so the delivery ending for $z=2005$ is an immediate consequence of it ending for $z=2004$.

Apologies for the messing write-up - I hope there are no mistakes.


Let n be an odd natural number. Prove that the equation 4/n = 1/x + 1/y has a
solution (x,y) with natural numbers x,y iff n has a divisor of the form 4k-1 (k \in N) !

Let's assume first that 4/n=1/x+1/y has solns. Then 4xy=n(x+y). let d=(x, y) and x=ds, y=dt with (s, t)=1. We have 4dst=n(s+t). So 4|s+t. Because s and t are coprime they can't both be even so they must both be odd. 4|s+t so s=M4+1 and t=M4+3 (or vice-versa). t|n(s+t) and (t, s+t)=1 so t|n. t has form 4k+3 so n has a divisor of the form 4k+3 (M4+3).

Now for the other part. We can observe that if there are solns for n then there are solns for kn: 4/n=1/x+1/y => 4/kn=1/kx+1/xy. So in order to prove that there are solns for each n which has a divisor of the form 4k-1 it's enough to prove that there are solns for each n of the form 4k-1:

Let n=4k-1. We can see that 4/(4k-1)=1/k+1/k(4k-1) so x=k, y=k(4k-1) is a soln for n=4k-1. This ends the proof.


[quote="reason"]hi!
calculate the following limit:

$ \lim_{x\to2}\frac {\sqrt {x \plus{} 2} \plus{} \sqrt {x^{2} \plus{} 5} \minus{} 5}{\sqrt {3x \minus{} 2} \plus{} \sqrt {4x^{2} \plus{} 5x \plus{} 23} \minus{} 9}$[/quote]

$ \lim_{x\to2}\frac {\sqrt {x \plus{} 2} \plus{} \sqrt {x^{2} \plus{} 5} \minus{} 5}{\sqrt {3x \minus{} 2} \plus{} \sqrt {4x^{2} \plus{} 5x \plus{} 23} \minus{} 9}$

$ \equal{} \lim_{x\to2}\frac { (\sqrt {x \plus{} 2}\minus{}2) \plus{} (\sqrt {x^{2} \plus{} 5} \minus{} 3)}{ (\sqrt {3x \minus{} 2}\minus{}2) \plus{} (\sqrt{4x^{2} \plus{} 5x \plus{} 23} \minus{} 7)}$

$ \equal{} \lim_{x\to2}\frac { \frac{1}{ \sqrt{x\plus{}2}\plus{}2 } \plus{} \frac{x\plus{}2}{\sqrt{x^2\plus{}5}\plus{}3} }{ \frac{3}{ \sqrt{3x\minus{}2}\plus{}2 } \plus{} \frac{ 4x\plus{}13 }{\sqrt{4x^2\plus{}5x\plus{}23}\plus{}7} }$

$ \equal{}\frac{ \frac{1}{4} \plus{} \frac{2}{3} }{ \frac{3}{4} \plus{} \frac{3}{2} }$

$ \equal{} \frac{11}{27}$


This is from the 2002 State Sprint Round problem #25:

What is the positive value of m such that the triangle bounded by the lines y=0, x=12 and y=mx has an area of 1872 square units?

Answer:
[hide]26[/hide]

Clearly, this triangle is a right triangle. The base would be the distance from (0,0) to (12,0) and the height is the distance from (12,0) to the intersection of $ y\equal{}mx$ and $ x\equal{}12$. The area is $ 1872\equal{}12h/2\equal{}6h$. So, $ h\equal{}312$. Hence, $ y\equal{}mx$ intersects $ x\equal{}12$ at $ (12,312)$. So, $ m\equal{}312/12\equal{}26$.

[quote="math154"]Graph it. Notice that the line $ y \equal{} mx$ passes through the origin and intersects $ x \equal{} 12$ at $ (12,12m)$. [geogebra]1ea461e35a70047ede3c941d4038a36ac898a478[/geogebra]

We see that this triangle has a base of length $ 12$, a height of length $ 12m$, and an area of $ \frac12\cdot12\cdot12m \equal{} 72m \equal{} 1872\implies m \equal{} \boxed{26}$.[/quote]
Did you rotate a graph $90^{\circ}$? >_<

Otherwise I don't get the $12, 12m$ thing

He didn't rotate it; he just didn't draw axes going through the origin (it's the point labeled $(0,0)$; think about it). Rather, he drew the relevant lines $y=0,x=12,y=mx$.

The $(12,12m)$ comes from finding the intersection of $x=12$ and $y=mx$. Because we already know $x=12$, we plug that into $y=mx$ to get $y=12m$. Therefore, the intersection is $(12,12m)$.


calculus
integration
algebra
polynomial
LaTeX
trigonometry
calculus computations

two integrals in the attachment doc1. , both algebraic , i was hoping for two different solutions:
one to verify my approach using just the polynomials and no special substitutions , and the other using trigonometric substituitions.
Ive written down for the 1st only and followed the same in the 2nd.

In the future, try to post your problems in Latex instead of attaching a file with them, in the sake of easy reading.

The indefinite integrals in question are:

1. $ \int \frac{1}{(1\plus{}\sqrt{x}) \sqrt{x\minus{}x^2}}\,dx$

2. $ \int \frac{1}{(\sqrt{5\plus{}2x\plus{}x^2})^3}\,dx$

[hide="Number 1"]Consider $ x \equal{} \sin^2t$, $ dx \equal{} 2 \sin t \cos t \,dt$, and so

$ P \equal{} \int \frac {1}{(1 \plus{} \sqrt {x}) \sqrt {x} \sqrt {1 \minus{} x}}\,dx \equal{} \int \frac {2 \sin t \cos t}{(1 \plus{} \sin t) \sin t \cos t}\,dt \equal{} 2 \int \frac {1}{1 \plus{} \sin t}\,dt$

$ \equal{} 2 \int \frac {1 \minus{} \sin t}{\cos^2t}\,dt \equal{} 2 \int \sec^2t \,dt \plus{} 2 \int ( \minus{} \sin t) (\cos t)^{ \minus{} 2}\,dt \equal{} 2 \tan t \minus{} \frac {2}{\cos t} \plus{} C$

$ \equal{} 2 \frac {\sin t \minus{} 1}{\cos t} \plus{} C \equal{} 2 \frac {\sqrt {x} \minus{} 1}{\sqrt {1 \minus{} x}} \plus{} C$.[/hide]

Also:
$ 2\int\frac{1\minus{}\sin t}{\cos^{2}t}\,dt$

$ \equal{}2\int(\sec^{2}t \minus{}\sec t\tan t)\,dt$

$ \equal{}2(\tan t \minus{} \sec t)$

$ \equal{} 2\frac{\sqrt x \minus{} 1}{\sqrt {1\minus{}x}}$

Number 2.

$ (\sqrt{5\plus{}2x\plus{}x^{2}})^{3}$ $ \equal{}((x\plus{}1)^{2}\plus{}4)^\frac{3}{2}$.

Let $ x\plus{}1\equal{}2\tan\theta$. Then $ dx\equal{}4\sec^{2}\theta\,d\theta$.

Therefore $ \int\frac{1}{(\sqrt{5\plus{}2x\plus{}x^{2}})^{3}}\,dx$

$ \equal{}\int\frac{4\sec^{2}\theta}{8\sec^{3}\theta}\,d\theta$

$ \equal{}\frac{1}{2}\int\cos\theta\,d\theta$

$ \equal{}\frac{1}{2}\sin \theta \plus{} c$

$ \equal{}\frac{x\plus{}1}{2\sqrt{(x\plus{}1)^{2}\plus{}4}} \plus{}c$

$ \equal{}\frac{x\plus{}1}{2\sqrt{5\plus{}2x\plus{}x^{2}}} \plus{}c$.


I read somewhere once that if you take the terms in the right order, an alternating series (e.g. the alternating harmonic series) can be shown to converge to any value.  Does anyone have a proof for this?

The correct statement is that a conditionally convergent real series can be rearranged to converge to any real number. We can have conditionally convergent series that are not, strictly speaking, alternating, and we can have alternating series that are absolutely convergent - so "alternating" was the wrong word for you to use.

A rough sketch of the proof: the postive terms sum to $+\infty$ and the negative terms sum to $-\infty.$ (If only one of these is infinite, then it's not a convergent series. If both of these are finite, then the original series is absolutely convergent.)

List only positive terms until the partial sum is greater then your target value; then list only negative terms until your partial sum is less than your target value, then switch back to positive terms, and so on. You only overshoot the target value by less than the last term you add in. Since the terms go to zero, we have convergence.

What happens if we try to rearrange the terms in, say, $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}$ ?

That is an alternating series, but it is not conditionally convergent since it is absolutely convergent. The theorem about absolutely convergent series is this: any rearrangement of this series converges to the same result as the original series.

All of this can be found in many different beginning analysis textbooks.

Thanks for the correction and the help, Kent!

Here is a proof by grobber: [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=convergent&t=67043]Link.[/url]
It is the same idea Kent mentioned.


I have sent my solutions to the USAMTS about a month and a half ago, and when I check my profile or my email, I don't know whether they've received them. Should I sent another copy?

it can't hurt. The only bad thing that could possibly happen is that they become frustrated with you.

I sent mine a week ago exactly and I still don't see the green check.

[quote="DPatrick"]We have not yet processed any of the Round 1 solutions, nor will we begin doing so until after the due date (October 10).[/quote]

so how would we know if you got our answers or not? my email account is weird and sometimes emails I send don't get through for whatever reason.

can I send it again just in case?

Please do not submit solutions more than once.

If you followed the directions on the USAMTS web site for submitting your solutions, then we almost certainly have them.

We will begin processing solutions next week.


geometry
parallelogram
geometry unsolved

[quote="georgech"]Help needed..!!..REALLY..!

1)Let ABCD be a convex equilateral and M,N the midpoints of AB and CD respectively.If CM and BN intersect at K and AN and DM at L,show that (MKNL)=(KBC)+(LAD).

2)Let ABC be a triangle and D a point of AB such that if AC is extended to E , CE=BD.Prove that ,if DE and BC intersect at M,  (MD/ME)=(AC/AB).

Any ideas ?[/quote]

1) "(Convex) equilateral ABCD" ? You meant "rhombus". Then
$AMND$ and $MBCN$ are parallelograms $\Rightarrow$
$S_{LMNK}=S_{LMN}+S_{MKN}=S_{ALD}+S_{MKC}$

2) Under the Menelaus theorem for triangle $AED$ and secant $CB$ we have
$\frac{AC}{CE}*\frac{CM}{MD}*\frac{DB}{AB}=1 \Rightarrow$
as $CE=DB \Rightarrow \frac{AC}{AB}=\frac{MD}{ME}$

[quote="georgech"]Maybe i didn't make it clear but in (1) ABCD is not a rhombus..it's just a quadrilateral and we don't know anything for its sides..[/quote]
just notice that $S_{BCM}+S_{ADM}=S_{ABN}$,and minus the public part then we get it .:)

Sorry for being so disturbing but i just don't get it!why should (BCM)+(ADN)=(AMN) and how does the conclusion follow?!

The median divides a triangle into two equivalent triangles.
$\left[DNB\right] = \left[NKC\right] + \left[KBC\right]$
$\left[MBD\right] = \left[LAD\right] + \left[LAM\right]$
Adding these $\left[MKNL\right] + \left[LDN\right] + \left[BKM\right] = \left[LAD\right] + \left[NKC\right] + \left[KBC\right] + \left[LAM\right]$. Hence the result.

And again..(BKM)+(LND)=(KCN)+(MLA) because....?!..If (MANC) AND (BMDN) were equal then ok..but how do we prove it?

$\left[DMN\right] = \left[CMN\right]$ and $\left[BNM\right] = \left[ANM\right]$.
Adding these we get $\left[DMBN\right] = \left[AMCN\right]$.


If Today is Monday, January 1 2009.  Then what day of the week will be $ 2{}^9{}^8{}^3{}^3$ days from now??

[hide="Easy."]
=====
$ 7$ is prime and $ 9833\equiv 5 \pmod{6}$ so $ 2^{9833}\equiv 2^5 \equiv 4 \pmod{7}$ by Fermat so the answer is $ \boxed{\text{Friday}}$.
=====
[/hide]

That's probably the shortest way though...
I'll Show another way which might be a longer way in finding for the answer. But, without using Modulus arithmetic...

First: Find a pattern/expression for $ 2^n$ that can be a multiple of $ 7$(so that it will be easier to know what day will it be if a number of days exceeds a week).

Notice,

$ 2^3 = 8 ; 8-1 = 7$
$ 2^6 = 64 ; 64-1=63$
$ 2^9=512 ; 512-1=511$

Implies that $ 2{}^3{}^n - 1$ is a multiple of $ 7$.

Therefore, $ 2{}^3{}^n - 1+1=2{}^3{}^n$ is one day after $ 7$ days. Since, $ 9833$ is not divisible by $ 3$, look for the highest number less than $ 9833$ that is divisible by $ 3$. That number is $ 9831$.

$ 2{}^3{}^n=2{}^9{}^8{}^3{}^3 = 2^3^({}^3{}^2{}^7{}^7^)^ +^ 2 = 2{}^9{}^8{}^3{}^1 \cdot 2^2 = 2{}^9{}^8{}^3{}^1 \cdot 4$

And since $ 2{}^3{}^n$ is one more than a multiple of $ 7$, If you multiply $ 4$ to it, we have:

$ 4(7k + 1)=28k+4=4(7k)+4$

So, $ 2{}^9{}^8{}^3{}^3$ days after a Monday is $ 4$ days after another Monday.

And that day is $ \boxed{\text{Friday}}$

[quote="jvenezuela716"]Okay, but try this problem:

If today is Monday, January 1, 2009, then what day of the week will it be $ 1000^{999^{998^{\dots^{2^1}}}}$ days from now?

It's difficult without modular arithmetic.[/quote]

Wahahahaha... You've got a point,cases like that do need modulus....

Yes, and the problem is actually easier than yours: Fermat is not needed.

$ 1000\equiv \minus{}1 \pmod{7}$ so the number of days is congruent to $ (\minus{}1)^{999^{\dots}}$. The exponent is odd so it is congruent to just $ \minus{}1$, so the answer is $ \boxed{\text{Sunday}}$.

waaaaaaaaaaaaaaaaa... hmm... Kuya [b]jvenezuela716[/b] hmm... ngayon ko lang  kasi ulit napansin ito.... hmmm.. i think $ (\minus{}1)^{999^{...}}$ is not yet congruent to $ \minus{}1$ kasi look po...

$ (\minus{}1)^{999^{998^{...^{2^{1}}}}}$ $ \Longrightarrow$ $ (\minus{}1)^{998^{...^{1}}}$

$ \Longrightarrow$ $ 1^{997^{...^{2^{1}}}}$ $ \Longrightarrow$.. So... Therefore, It'll be congruent to $ 1$ and that day will be $ \fbox{Tuesday}$

Hi

mukhang may mali..

1.  you dont consider leap year. so this is isnt a real calendar :D
2.  i think you must look at your parentheses  eg. (x^5)^4  !=  x^(5^4) so dapat mauuna evaluate yung exponents right??

in the first case its equivalent to x^20 which is actually multiplying the exponent :D and i assume its not the problem you intend right??

regards


$10$ people land on a deserted island. There they find lots of coconuts and a monkey. During their first day they gather coconuts and put them all in a community pile. After working all day they decide to sleep and divide them into two equal piles the next morning. That night one castaway wakes up hungry and decides to take his share early. After dividing up the coconuts he finds he is one coconut short of ten equal piles. He also notices the monkey holding one more coconut. So he tries to take the monkey's coconut to have a total evenly divisible by 10. However when he tries to take it the monkey conks him on the head with it and kills him. Later another castaway wakes up hungry and decides to take his share early. On the way to the coconuts he finds the body of the first castaway, which pleases him because he will now be entitled to 1/9 of the total pile. After dividing them up into nine piles he is again one coconut short and tries to take the monkey's coconut. Again, the monkey conks the man on the head and kills him. One by one each of the remaining castaways goes through the same process, until the 10th person to wake up gets the entire pile for himself. What is the smallest number of possible coconuts in the pile, not counting the monkeys? :rotfl:

[hide="Solution"]If you include the monkey's coconut, the total number of coconuts is divisible by 1,2,3,4..., and 10. The LCM of 1,2,3,4,... , and 10 is 2520. We then subtract 1 to disclude the monkey's coconut to get [b]2519[/b] as our answer.[/hide]

simple question, yes the answer is right.

It is
[hide]
less than 10!.  I know that 60 is divisible by 1,2,3,4,5,6, and 10.  Thus I multibly by 2 (so the 4 can be multiplied to 8) to get 120, by 3 (to "fill out" the 3 into a 9) to get 360, andby 7 to get 2520, far less than 10!.  Oh, and subtract the one that the monkey was killing the sailors with to get [b]2519[/b].
[/hide]


Given:  2 sets of identical 4-sided dice, 3 in each set, all dice are fair.  These dice have only positive whole numbers on their faces, and repeats of numbers are allowed.  You have 3 dice, all alike, and the opponent has 3 dice (but different from yours) all alike.

When 2 or more dice are rolled by the same person, then the numbers on their faces are added together for their "roll number."

You have just rolled 3 of your dice 30 times in a row against
the opponent's 3 dice each time.  The opponent has won the majority of these and it has put both of you tied at money from betting on the dice rolls.

You will roll one more time against the opponent, and allow you to come out with a relatively small amount more than the opponent, should you win this one roll.  However, you will be allowed to choose the number of dice rolled per person.

Which situation gives the best chance for you?  Choose one.

1) Roll 3 dice against the opponent's 3 dice.

2) Roll 2 dice against the opponent's 2 dice.

3) Roll 1 die against the opponent's 1 die.

[hide] In all three situations, you have an equal chance to 'win' as your opponent, but u also have a chance to tie.  If you tie, you don't 'win' and therefore want to reduce the chance of a tie: 3dice Vs 3dice reduces the chance of a tie by the most(we don't know by howmuch because we don't know what is one the dice; this is your best choice. [/hide]

*** Spoilers Ahead ***

Suppose the numbers on each of your dice are 4,4,4,9

Suppose the numbers on each of your opponent's dice
are 6,6,6,1

If that were so, that would explain why your opponent won
more often.  If this example is not just an aberration of the possibilities of what the numbers on the faces could be, by affecting the averages?
>
>
>
You should choose to roll two of your dice against two of the
opponent's dice.


calculus
integration
modular arithmetic
number theory proposed
number theory

Determine all integral solutions $(a,b)\in\mathbb{Z}^{2}$ such that
\[221a^{2}\pm 20=b^{2}\]

[hide] I used mod 17 instead, but same idea.[/hide]

In fact $221a^{2}+-20k=b^{2}$ have no solution for $k\leq 4$

Note: You can prove Hawk Tiger's statement using mod 13, mod 17, mod 13, and mod 13 for k equaling to 1,2,3, and 4, respectively.


Hi everyone, im just new here. If anybody knows a good sight regarding circular track problems, kindly inform me. thanks.

Here's a cute one:

A car runs laps around a circular track. Along the track there are some fuel tanks, such that the total fuel of all the tanks around the track is just enough to do one lap. A car starts at one fuel tank, then as it gets to the next one, it takes the fuel of the following tank and so on, until it completes the lap. Prove there's always a tank from which it can start so that it doesn't run out of fuel before making a lap.

Cool! This is an excellent problem to do for a warm-up in class next week. More class discussions!

[quote="obfuskate"][quote="amirhtlusa"]there was a good (easy actualy) prbolem about five runners in running a circular track, in opposite directions, while holding a light, and that every time one runner passes another, he passes the light, and the question was who has the light at the end.
i can dig it up, if you want to...[/quote]

I don't quite understand how 5 runners can run in opposite directions... are there 3 going one way and 2 going the other?[/quote]
all right, i have the original problem, it's a good one_it's easy:

in a circular track there are five runners standing apart from each other with the same distance. lets call them : $A,B,C,D,E$. runner $A$ has a piece of wood in his hand.
when the race startes, runner $A$ runs[u] counter clock wise[/u] with speed of 4 rounds per hour,(it means in one hour, he runs the track four times). runners $B,C,D,E$ run colckwise with speed of one round per hour.
when a runner that holds the piece of wood, passes another runner, then he gives the piece of wood to the runner that passed.
after one hour, which one of the runners holds the piece of wood?


Mark on the two given lines that cuts in A. the points D,E respectively such  AE=AD=1/2p. Construct the circle (O) tangent to these lines in the points D,E. The line solution is the tangent from P to the circle (O) that separates O and A in different semiplanes.


A space station consists of 2 donut-shaped living chambers, A and B, that have the radii rA=3.2*10^2m, rb=1.10*10^3m.  As the station rotates, an astronaut in chamber A is moved 2.40*10^2m along a circular arc.  How far along a circular arc is an astronaut in chamber B moving during the same time?

although I cant understand the donut-shaped meaning{please explain it }
but I think the soloution should be this:
S=2.4*$10^2$m   R=$r_{a}$=3.2*$10^2$m
s=?                      r=$r_{b}$=1.1*$10^3$m
X=the angle that the station has rotated
$X_{a}$=$X_{b}$
so:
$S/R$=$s/r$
so we can find s :oops:


[quote="worthawholebean"]Determine the number of non-degenerate rectangles whose edges lie completely on the grid lines of the following figure.
$ \begin{tabular}{|c|c|c|c|c|c|} \hline & & & & & \\
\hline & & & & & \\
\hline & & \multicolumn{1}{c}{} & & & \\
\cline{1 \minus{} 2}\cline{5 \minus{} 6} & & \multicolumn{1}{c}{} & & & \\
\hline & & & & & \\
\hline & & & & & \\
\hline \end{tabular}$[/quote]
[hide="wrong solution"]If we were given a $ 6 \times 6$ grid, then the ans would be $ \binom{7}{2}^2$. But a $ 2\times 2$ grid has $ \binom{3}{2}^2 \minus{} 1$ rectangles, which do not have all the sides on the boundary.
So the answer is $ \binom{7}{2}^2 \minus{} \binom{3}{2}^2 \plus{} 1 \equal{} 433$[/hide]
Sorry! I made a mistake :|
When we choose a rectangle, we choose four co-ordinates namely $ (a,c),(a,d),(b,c),(b,d)$.
We have to find how many rectangles have $ (3,3)$ as a vertex. As we already have $ 2$ values among $ (a,b,c,d)$. We can find other two in $ \binom{6}{1}^2 \equal{} 36$ ways.
That yields the answer $ \binom{7}{2}^2 \minus{} \binom{6}{1}^2 \equal{} 405$
I hope that it is correct now.  :ninja:

[quote="Moonmathpi496"][quote="worthawholebean"]Determine the number of non-degenerate rectangles whose edges lie completely on the grid lines of the following figure.
$ \begin{tabular}{|c|c|c|c|c|c|} \hline & & & & & \\
\hline & & & & & \\
\hline & & \multicolumn{1}{c}{} & & & \\
\cline{1 \minus{} 2}\cline{5 \minus{} 6} & & \multicolumn{1}{c}{} & & & \\
\hline & & & & & \\
\hline & & & & & \\
\hline \end{tabular}$[/quote]
If we were given a $ 6 \times 6$ grid, then the ans would be $ \binom{7}{7}^2$. But a $ 2\times 2$ grid has $ \binom{3}{2}^2 \minus{} 1$ rectangles, which do not have all the sides on the boundary.
So the answer is $ \binom{7}{7}^2 \minus{} \binom{3}{2}^2 \plus{} 1 \equal{} 217$[/quote]

First of all, I got an even number, because of the symmetry.

I got $ 222$ non-degenerate rectangles.

Rectangles that do not have the center 2-by-2 square as part of them:

1 by 2 -- 40
1 by 3 -- 32
1 by 4 -- 24
1 by 5 -- 16
1 by 6 --  8

2 by 3 -- 16
2 by 4 -- 12
2 by 5 --  8
2 by 6 --  4

Rectangles that do have the center 2-by-2 square as part of them:

2 by 3 -- 4
2 by 4 -- 6
2 by 5 -- 4
2 by 6 -- 2

3 by 4 -- 12
3 by 5 -- 8
3 by 6 -- 4

4 by 5 -- 12
4 by 6 --  6

5 by 6 -- 4
----------------
............ $ 222$

Hmm, well I went with the idea of overcounting and then correcting.

So in total there would be, if the lines WERE there, $ (\binom{7}{2})^2\equal{}441$ rectangles.

Now, of those 441 rectangles, $ 6*\binom{7}{2}\equal{}126$ of them contain the vertical missing line.

Another $ 126$ contain the horizontally missing line.

However, some of those with missing lines are being repeated, accounting for subtracting $ 6*6\equal{}36$ too many.

So...I got $ 441\minus{}252\plus{}36\equal{}\boxed{225}$ rectangles.

Hmm, three different answers...

I made a mistake. I hope that it is correct now.

[quote="Ihatepie"][quote="10000th User"]You all have to realize that you can't have 3x3, nor 3x4, nor 4x4 rectangles and so on...[/quote] Why not? The empty space can be inside the rectangle.[/quote]

10000th User,

it is true you can't have 3 X 3 and 4 X 4 rectangles, for instance,
because squares are not allowed.  However, the 3 X 4, 3 X 5, 3 X 6,
4 X 5, 4 X 6, and the 5 X 6 rectangles are possible, despite the
2 X 2 hole in the figure.

Wait, why aren't squares allowed? The problem simply said to find the rectangles, and did not put any restrictions on the types of rectangles.

I'm also confused on where this came from: $ (\binom{7}{2})^{2}\equal{}441$ to get the rectangles if teh lines were there.

I have the same question, why aren't squares  allowed? :ninja:

Yes. Now I realize that I have made mistake again. (I am so careless :mad: )
It is not necessary that the vertex is $ (3,3)$ (if we introduce co ordinate system). Rather we need to subtract the rectangles, which have sides that go through $ (3,3)$.
So we can calculate them considering two cases.
1. The one of the vertex is $ (3,3)$. There are $ \binom{6}{1}^2\equal{}36$ such rectangles.
2. Now we consider the rectangles having vertices on $ x\equal{}3,y\equal{}3$, and which has sides going through $ (3,3)$.
For $ y\equal{}3$. We have to choose 2 points on the left of $ (3,3)$, and 3 on the right, and one point on $ x\equal{}3$. Which can be done in $ 3\times 3 \times 6\equal{}54$ ways. The same reasoning works for $ x\equal{}3$.
In total we have $ 2\times 54\plus{}36\equal{}144$ rectangles, which are invalid.
So the number of valid rectangles is $ 441\minus{}144$
Can anyone check this answer  :|

[quote="Ihatepie"]Squares are allowed, and every size of rectangle (squares are rectangles) with a width and length ≤ 6 is possible.

I'm too lazy right now to solve the problem, but I'm pretty sure no one in this topic has posted the right answer.[/quote]

No, they are *not*.

Go back and read the OP's original post.

[quote="worthawholebean"]
Determine the number of non-degenerate rectangles whose edges lie completely on the grid lines of the following figure.
[/quote]

It states "non-degenerate rectangles."
A square is a non-degenerate rectangle.

I wasn't sure (at the time it was posted) of the meaning so I looked it up.

worthawholebean could have stated (alternatively) that he wants the count of all rectangles,
except those that are also squares.

Meh. I looked around online just now, and it seems there are two definitions for a degenerate rectangle. One is "a line," and the other is "a square." From what I've seen, the more common definition is the former.

Anyways, I checked the [url=http://web.mit.edu/hmmt/www/datafiles/solutions/2008/solutions-guts-2008.pdf]solution[/url] and it seems they used the first definition.


integration
trigonometry
complex analysis
complex analysis unsolved

For readers unfamiliar with this notation, $C_{n}^{k}=\binom{n}{k}$.

I got, if my calculations are correct, that the limit is 1. Here is my solution to this problem:

Let $a_{n}$ be your sum. Then, a calculation shows that

$a_{n}=\sum_{k=0}^{2n}(-1)^{k}C_{4n}^{2k}(2n+1)\int_{0}^{1}x^{k}(1-x)^{2n-k}dx$

$=(2n+1)\int_{0}^{1}(1-x)^{2n}\sum_{k=0}^{2n}C_{4n}^{2k}\left(\frac{-x}{1-x}\right)^{k}dx$

$=\frac{2n+1}{2}\int_{0}^{1}(1-x)^{2n}\left(\left(1+\sqrt{\frac{-x}{1-x}}\right)^{4n}+\left(1-\sqrt{\frac{-x}{1-x}}\right)^{4n}\right)dx$

$=\frac{2n+1}{2}\int_{0}^{1}\left(\left(\sqrt{1-x}+i\sqrt{x}\right)^{4n}+\left(\sqrt{1-x}-i\sqrt{x}\right)^{4n}\right)dx$.
Now we sue the substitution $\sqrt{x}=\sin t$ and we get after some calculations that,

$a_{n}=\frac{2n+1}{2}\int_{0}^{\frac{\pi}{2}}\left(\sin(4n+2)t+\sin(4n-2)t\right)dt=\frac{8n(2n+1)}{(4n+2)(4n-2)}\rightarrow 1$.
Therefore, the limit is 1.

Thanks Didi again for your help ,but must admit that this demonstration is to advanced for me ,and i would like someone to help me with it ,because i don't realy know so many trigonometrics

[quote]Silviu,

Here is what I have done:

I will consider only the integral, so we have the substitution $\sqrt{x}=\sin t$ or $x=\sin^{2}t$, i.e, $dx=\sin (2t) dt=\frac{e^{2ti}-e^{-2ti}}{2i}dt$,

I=$\int_{o}^{\pi/2}((\cos t+i \sin t )^{4n}+( \cot t-i \sin t )^{4n})\frac{e^{2ti}-e^{-2ti}}{2i}dt$

In other words we get that:

I=$\int_{0}^{\pi/2}(e^{4nit}+e^{-4nti})\frac{e^{2ti}-e^{-2ti}}{2i}dt$
I=$\frac{1}{2i}\int_{0}^{\pi/2}(e^{(4nt+2t)i}-e^{(4nt-2t)i}+e^{-(4nt-2t)i}-e^{-(4nt+2t)i})dt$
Now we use the fact that ${e^{ui}-e^{-ui}}{2}=\sin u$,

We get that

I=$\int_{0}^{\pi/2}(\sin(4n+2)t+\sin(4n-2)t)dt$.
I hope it helps.
Didi [/quote]


Because of the revision, will we be getting new student numbers? Or can we start filling/faxing the forms with the old numbers?

i'm guessing we keep the old numbers, but this is a good question and should be verified...
*bump*

i'm prettys ure they will just reassign the old numbers

yeah, im pretty sure most of the people won't have their numbers changed
but maybe thats cause im such a low number and 5% club people have higher numbers...


I have a question:

 i don't know how to wite iupac names of free radicals, carbocations, etc...  :oops:  :blush:
So, i'm posting image
can u help me??

i told u da....it's really beautiful...r u using morisson :?:

Yes, i'm using morrison but only sometimes for reactions...hydrocarbons and cycilc compounds are very fascinating....can u ppl post some problems wrt these topics...i'm not able to follow much of wat is given in chem forum...plz consider the level of difficulty and post...all help would ne much appreciated

hey pardesi can we direct him to the chemistry forum....

@hell ever- it's really good at the chem forum.. just check it out

i know the forum but am not able to understand much of wat is discussed there

if u have any "good " doubts post it in the IITJEE thread...

or pm me/chemrock/madness/pardesi

yes u can always pm us ...but posting in the chem forum will give u views from Carcul himself.....
also if u r following morisson thoroughly look at my posts from early april last year till date....u will have lots of doubts from morrison solved by carcul...we were then the only two people who kept the forum alive.. :)
Anyway u have to be patient in organic...maturity will come with time :)

morisson is good, but the problems are very easy, tho at the initial stage you might wanna do only that. you could also refer solomons which is also a nice book.
by the way what sort of questions would you like us to post?

Interact more with Carcul ,he is a treasure....

I have learnt organic more from hin than my school teacher... :D

I learnt Organic Only from Carcul and by reading myself :D
I odn't think Morrison Problems are that easy especially those after chapters 19,20 ...they are quite analytical and they develop ur skills much more than some 'IIT-Chemistry" book would :P
I have said this and let me tell u this again "Be Very Very Patient" :)

hell_ever ...why don't u try ...we will help u out :)

For free radical the radical postn is always no.1 irrespective of a $ \ce{\minus{}COOH}$ or a $ \ce{\minus{}SO_3H}$. Then if it is $ \ce{CH_3^0}$ then it is methyl free radical or 1-phenyl methyl radical and so on. In ur case it is cyclohexyl radical.

Well then.. should no.4 be cyclohex-2-ene radical? and no.3 be 3-phenyl cyclohex-2-ene radical?

Cyclohexenyl radical that's what I think :maybe:  :maybe:

[quote=Stringintegral]Hey you all,
Like hell_ever I am too weak in chemistry.  Please explain me what is morrison and who is carcul[/quote]

How do you people even find such an old thread ?!?! Have a look. It's 8 years old. Whenever posting anything, notice the date of the last post. :)


If a and b are positive integers such that $ 1/a \plus{} 1/2a \plus{} 1/3a \equal{} 1/ (b^2\minus{}b)$, what is the smallest possible value of
$ a \plus{}b$?

[hide]$ \frac1a\left(1\plus{}\frac12\plus{}\frac13\right)\equal{}\frac1{b^2\minus{}b}$

$ \frac1{a}\left(\frac{11}6\right)\equal{}\frac1{b^2\minus{}b}$
$ \frac{6a}{11}\equal{}b^2\minus{}b$
$ 6a\equal{}11(b^2\minus{}b)$
$ a\equiv0\pmod{11}$
Because $ a$ is positive, $ a\geq11$.  Clearly, the sum is minimized when $ a\equal{}11$, so we find that $ b^2\minus{}b\equal{}6\Rightarrow b\equal{}3$ and $ a\plus{}b\equal{}14$.[/hide]

[hide]
$ \frac {1}{a} \plus{} \frac {1}{2a} \plus{} \frac {1}{3a} \equal{} \frac {1}{b^2 \minus{} b}$

$ \frac {11}{6a} \equal{} \frac {1}{b^2 \minus{} b}$

$ \frac {6}{11}a \equal{} b^2 \minus{} b$

$ \frac {6}{11} \equal{} \frac {b(b \minus{} 1)}{a}$
For a to be as small as possible, we have $ a \equal{} 11$, and then
$ b(b \minus{} 1) \equal{} 6$
$ b \equal{} 3$
So $ a \plus{} b \equal{} 11 \plus{} 3 \equal{} \boxed{14}$
[/hide]
Edit: I was beaten by 3 minutes!


calculus
integration
function
real analysis
real analysis unsolved

Let $ (X,M,m)$ be a measure space with $ f$ a nonnegative extended real valued $ M$-measurable function on a set $ D \in M$ with $ m(D) < \infty$.  Let $ D_n \equal{} \{x \in D : f(x) \geq n \}$ for $ n \in \mathbb{N}$.
Show that $ \int_D f dm < \infty$ if and only if $ \sum_n m(D_n) < \infty$.

Hint:

How is the function $ g(x)\equal{}\sum_{n\equal{}1}^{\infty}\chi_{D_n}(x)$ related to $ f(x)?$

Hint: If $ m \{ x \in D \, | \, f(x) \equal{} \infty \} \equal{} 0$, the sum $ \sum_{n\equal{}1}^{\infty} m(D_n)$ estimates $ \int_{D} f \, dm$. To confirm this, put $ E_n \equal{} D_{n\minus{}1} \minus{} D_{n} \equal{} \{ x \in D \, | \, n > f(x) \geq n\minus{}1 \}$. Then

$ \sum_{k\equal{}1}^{n} m(D_k) \equal{} n \cdot m(D_n) \plus{} \sum_{k\equal{}1}^{n} (k\minus{}1) \cdot m(E_k).$

$ \sum_{n \equal{} 1}^{\infty}1_{D_n}(x) \le f(x) \le m(D) \plus{} \sum_{n \equal{} 1}^{\infty}1_{D_n}(x)$

thats the only thing you really need.


A regular hexagon and an equilateral triangle have equal perimeters. What is the ratio of the area of the hexagon to the area of the triangle? Express your answer as a common fraction

draw a diagram.

If the perimeter of 2 polygons are 6x, a side length of triangle is 2x and a side length of hexagon is x.

divide the triangle and hexagon by smaller triangles with side length of x.

there's total 4 small triangles in the triangle, and 6 small triangles in the hexagon.

so the ratio of the area of the hexagon to the area of the triangle is 6:4, which is 3:2.

The answer is 3:2.

[quote="GameBot"] Express your answer as a common fraction[/quote]

Nice...


I just got my SAT scores from Collegeboard yesterday and received a 600 (reading), 700 (writing) and 780 (math). I know I really need to improve on reading :blush:

I'm going to need to know how to approach that strategically so I can get at least 740+

Can anyone help>

Thanks

What grade are you in? If this was in 11th grade you might have to work extrahard on comprehension

Wow.  That's pretty darn amazing.  Congratulations!

My advice for improving your reading/comprehension scores is to just read a lot.  Read challenging material with big words...  Sign up for a word-a-day to from Dictionary.com to your email.  Read some more...  That is what there is to it.  You've got a couple years though to improve, so don't stress it.

Congrats again!  I hope that I do that well, and I'm in 11th grade...  lol

Well, i have expanded my vocabulary so I can understand a lot of the words used. Its just that I am not able to quickly complete all of the questions acurately.

You probably just need to read more, and do more critical thinking about what you read. There are some common reasons to get these sorts of problems wrong:

1. You know all the definitions of the vocabulary, but you don't know the connotations of the words, so you fall for one of the trick answers.

2. You read too quickly and miss some details or misread some things, and then you answer a question from memory instead of going back to check (to save time) and you answer wrongly because of that.

3. You read too slowly and always go back to check the passage carefully so as not to miss problems, but then you run out of time.

My strategy was always to quickly read the essay so that I would know where everything was in it, and then I could quickly go back and check for the questions whose answers weren't obvious. This helped me avoid (2) and (3). Also, as you go, mark ones that you're unsure about and just move on, and when you've finished you can go back and think about those ones more.

For (1) (and related problems), you really just need to read more, like byrdgyrl said. In your English classes, never do the Spark Notes thing. Always actually read the books. (It's worth the effort!) Read on your own time, too. It doesn't have to be fancy literature, as long as you're not reading little kids' books. Read sci-fi or fantasy if that's what catches your fancy. Read mystery books. Read history books. Whatever. Reading is fun! Just don't read math books and convince yourself that this is helping with your English skills...

Good grief way to go! You won't be having much of a problem then later on. You beat my 8th grade score, and I thought mine was pretty darn high. Try doing a little bit of freerice.com. Obviously you'll have a much higher vocab level in three years. And with the time thing, I had the same problem too. Usually i have ample time leftover on standardized tests, but not with the SAT. Maybe thats because I took practice tests on the kitchen table while multitasking. Looks like you aced the essay too. Underline key points in each passage. Did you finish the reading and math section? I mean get all they way through to reading the last question.

i got a 10/12 on the essay

Of course I finished the math :P I just messed up on one

freerice.com is trivial. Its like 4th grade level :o

Wow. Dude man. You're a genius. I only got a 1940 (in 7th grade =P). And a 1740 (in 6th grade =P). Notice some arithmetic sequence here xD. Watch me get a 1600 (800+800) =D... Freerice.com is not trivial once you get to like lvl 40 or something up there xD

You could try sparknotes.com for improving your vocabulary, they have some good suggestions :)

I'm not sure how much PSAT scores correlate to SAT scores, but I improved my reading score by 13 points from 8th grade to 11th (which I guess is like 130 points compared to the SAT) by doing nothing.  So I don't think you need to worry about your scores too much.

I need to work on my reading and comprehensino skills (according to the PSATs). Maybe it would have helped if I didn't skip every question that I was too lazy to do  :blush: .

dude I got a 740 reading and apparently that's too low for Asian parents :wink:

anyway... so... do lots of practice stuff. Reading comp is definitely way harder than vocab... (to me anyway)

and then... writing. Get all the grammar! I only got 9 essay but 800 writing... because I got all the grammar! :P

if you're lazy. Take a catnap. and then do them. xD... ahh 740 reading. I hate you. geniussss

A friend of my sister got a perfect SAT score when he was in the 6th grade without studying, so i'm not really surprised; the SAT tests topics are covered in middle school for the most part.

If you get confused or tripped up because multiple answers make sense to you, this applies to you:

SAT's aren't meant to weed out geniuses, they're meant to assess the average college-bound student. So, they want the most obvious interpretation that makes sense in context. So to prevent yourself from wandering off with multiple interpretations, consider the following [u]strategy[/u]:

1. Skim the passage briefly
2. Skim the questions, quickly answering those that you are confident of the answers to, and getting a feel of the rest
3. Go back and read the passage for real
4. Answer the remaining questions, spending very little time per question
5. If you have time, go back and see if your answers make sense

This may or may not help you, but for me it improved my CR scores from the upper 500's/lower 600's up into the 700's.

[i](when I took the SAT's for real, I was really sick and still got a 700 - in comparison, when I took them in 8th grade, I got a 580)[/i]


Without looking at the labels, Adrien placed four CDs in four cases. What is the probability that exactly two of the CDs are in the wrong cases? Express your answer as a common fraction.

There are $ \binom{4}{2}\equal{}\frac{4\cdot3}{2}\equal{}6$ ways to choose the $ 2$ CDs that are in the right case. We then have only one way to arrange the other $ 2$ CDs, so our answer is $ \boxed{6}$.

but the answer is 1/2..

how can a probability be bigger than 1?

If Adrien put two CD's in the incorrect cases, then clearly she put two in the right cases. Pick any $ 2$ of the $ 4$ to be the correct cases. She can now put the remaining CD's in the cases in exactly two ways. Of these, one of then yields an incorrect case, and hence the probability is $ \boxed{\frac12}$.


The polynomial $P(x)$ is given such that $P(x)\neq x$ and :

$Q_{1}(x)=P(x)-x,Q_{2}(x)=P(P(x))-x,...,Q_{n}(x)=\underbrace{P(P(...P}_{n}(x)))-x$
Prove that $Q_{n}(x)$ is divisible by $Q_{1}(x)$.

For every complex $\xi$ such that $P(\xi)=\xi$, we have $P^{[n]}(\xi)=P^{[n-1]}(\xi)=\dots=P(\xi)=\xi$, hence all the roots of $Q_{1}(x)$ are the roots of $Q_{n}(x)$.

[quote="harazi"]Rust, should we remind that if any root of $f$ is a root of $g$ then it is not always true that $f$ divides $g$?[/quote]
What do you mean ?
If deg $f$ is m then $f$ has $m$ complex roots , and if all of them are also roots of $g$ , then $f|g$ .What is wrong?


are you looking forward to school this year? I certainly am! :)

NO! i wish summer would last longer. :surf:

I'm definately looking forward to it. But I'm also kinda wishing summer would never end. :jump:  :agent:

it's not just that school's starting...it's the last year of high school...
the last year of home and seeing current friends a lot...
also around when school starts...all the seniors from last year are gonna leave  :(

but other than that...summer should stay...cuz it's getting nicer outside (temp-wise) and cuz i've finally fixed Diablo II exp  :D

no i hate school and i have been too busy this summer so id like to relax

i'm in the same situation as mwahahahaha (wow that name sounds weird in a sentence)
summer should never end... im actually getting so much done this summer - self studying physics and math and other fun stuff

Yes and no...

I'm going to miss all the free time, but it'll be great to see all my friends again.

whoever voted that they were going to get a crippling disease, think of your mothers and don't :D

Well, it gives me something to do for a change...
but I can't say I look foward to seeing my friends because, well, I won't be seeing them because I'm skipping a grade.

Im looking forward to school- Im starting high school, and high school is something i've been looking forward to for a long time, so i cant wait

Im glad that yes is so far the prevailing choise :)

:(and not too far behind are those who will soon be crippled

good thing only 5 people will suffer. :)

i want it to start, because i miss my friends, but then i think of all that homework...........

I'm not looking foward to the work of the school year, but I am looking foward to finishing my last year in my school, so i voted yes.

always think in a positive light :) this is my last year of middl school i cant wait until i am in high school.  Now there are 6 people coming down with malaria :o

Bah, school's already been in since last week for me.  I am definitely do not look forward to school (and homework) except this year I get to take the AMC's for the first time...that's the only think I'm looking forward to.

i wish summer would last longer, but i'm glad i'm not a freshman anymore!

High School starts tommorow for me(cant wait!)

I'm going into middle school this year, I'm looking forward to it. The only thing that I can do is go on AoPs, but school does start for me tomorrow :)

my school starts on Sept. 7th.  Which is a wednesday.  Why can't they start it on Monday?

i WAS looking foward to the start of school (as i do every year)... until it STARTED. please take me out of school! any way possible. besides, if i got a crippling disease, i'd probably see more of my friends than i do now (they all have different classes from me)- they'd come visit! and we could watch movies together! and paint! and build things out of legos! and... and... !

Not particularly.  However, what I am looking forward to is that school will end in another ten months.

hahah school ends in another 9 months for seniors here

and any productive meaning of "school" and "class" practically ends in 8 months, after the AP tests

my school still has not started... CAVA Virtual Acadamy...online homeschool... :( almost everyone else has already started school and i want to so bad :mad: maybe i will mess up their date website :idea:


Four distinct points, $A,B,C,$ and $D,$ are to be selected from 1996 points evenly spaced around a circle. All quadruples are equally likely to be chosen. What is the probability that the chord $\overline{AB}$ intersects the chord $\overline{CD}$?

$\displaystyle \text{(A)} \ \frac{1}{4} \qquad \text{(B)} \ \frac{1}{3} \qquad \text{(C)} \ \frac{1}{2} \qquad \text{(D)} \ \frac{2}{3} \qquad \text{(E)} \ \frac{3}{4}$

[hide]We can make quadrilaterals out of the points, $ABCD$, $ACBD$, $ADBC$, $ADCB$, $ABDC$, and $ACDB$, and only two $(ACDB,ADBC)$ out of the six have the intersection, so $\boxed{\frac{1}{3}}$.[/hide]


[quote="Interval"]A circular courtyard has an area of 10 – 2x^2. There are two rectangular flower beds in the courtyard. Which of the following expressions represents the green lawn area.

(A)....(10 – 2x^2) – (2x^2 + 3x^2)= 10 + 3x^2
(B)....(10 – 2x^2) – 2x^2 – 3x^2 = 10 – 7x^2
(C)....(10 – 2x^2) – 2x^2 – 3x^2 = 10 – 3x^2

What is the answer? A, B or C?

NOTE:

There is a picture of a cicular garden and inside the garden there are two rectangles.

One rectangle has the following dimensions: Length = 2x; Width = x.

The other rectangle has: Length = 3x and Width = x.[/quote]
[hide="hint"]Yo man, all you have to do, is like fine the area of that one thing. Then add that to the area of that one other thingy. Then you can take that away from that third thing. Whatever floats your boat dude.[/hide]

you already posted this.  do you still not knoow how to solve it?


This is pretty weird ... :)

Let ABC and PQR be triangles such that P is the midpoint of BC and A is the midpoint of QR, and such that QR bisects BAC and BC bisects QPR.

Prove that AB + AC = PQ + PR.

And pretty old: http://www.mathlinks.ro/Forum/viewtopic.php?t=39245 .

  darij


[quote="Farenhajt"][hide="Solution"]By Euclides' algorithm:
\begin{eqnarray*}\gcd(n-13,5n+6) &=& \gcd(n-13,4n+19)\\ &=& \gcd(n-13,3n+32)\\ &=& \gcd(n-13,2n+45)\\ &=& \gcd(n-13,n+58)\\ &=& \gcd(n-13,71)\end{eqnarray*}
Since $71$ is prime, the smallest $n$ is obtained when $n-13=71\iff n=84$[/hide][/quote]
Whats the Euclides' alogrithm, i have heard a lot about it. Also, could you please post an alternative solution?thanks

[quote="Inspired By Nature"]Whats the Euclides' alogrithm, i have heard a lot about it.[/quote]

$gcd(a, b) = gcd(b, a-b)$

More generally:  Let $a > b$ be positive integers (not necessary, but simplifies the statement).  Let $r$ be the remainder upon division by $b$.  Then

$gcd(a, b) = gcd(r, b)$

[hide="alternative soln"]
Kind of lame compared to Euclidean Alg., but here it goes:

Let $p$ be the smallest number that divides both the numerator and the denominator.  Since the numerator and denominator has to be divisible by $p$, we know that $n\equiv 13 \pmod p$ and also $5n+6\equiv0\pmod p$.  This implies that $5*13*6\equiv71\equiv0\pmod p$.  Since 71 is prime, $p=71$.  Now $n = 71+13 = 84$, we find the numerator is 71 and the denominator is 426, both of which are divisible by 71.  Furthermore we know that $n=71k+13$ will always produce reducible fractions.

[/hide]

[/hide]

[quote="Inspired By Nature"]Find the least positive integer $n$ for which $\frac{n-13}{5n+6}$ is a non-zero reducible fraction.[/quote]

[hide]If a number is reducible, then so is it's reciprocal.

Taking the reciprocal we get

$\frac{5n+6}{n-13}=5+71\frac{n-13}$.

So, the smallest value for which this is a non reducible fraction is a factor of 71 that is n>0, and that n-13 is a factor of 71.

also, n-13 can't equal 1 otherwise it will already have been reduced, so n-13=71, hence n=84.

$84$[/hide]


vector
calculus
linear algebra
calculus computations

I know there are entirely separate courses dedicated to linear algebra but are basic vectors and inner product spaces/vector spaces covered in Calc 1, 2, or 3? I'm asking because I'm currently learning calculus and I need to know a bit about those topics for learning Fourier Series and such. I need to know if it's within the scope of my teacher

Calculus 3 (multivariable calculus) has some elements of linear algebra (vectors and determinants for sure), but not necessarily the definition of an inner product space. That should not stop you from asking your teacher anyway. Math teachers know more about math than they tell their classes. :ninja:


$ ABC$ is a triangle which is not isosceles. Let the circumcenter and orthocenter of $ ABC$ be $ O$, $ H$, respectively,
and the altitudes of $ ABC$ be $ AD$, $ BC$, $ CF$. Let $ K\neq A$ be the intersection of $ AD$ and circumcircle of triangle $ ABC$, $ L$ be the intersection of $ OK$ and $ BC$, $ M$ be the midpoint of $ BC$, $ P$ be the intersection of $ AM$ and the line that passes $ L$ and perpendicular to $ BC$, $ Q$ be the intersection of $ AD$ and the line that passes $ P$ and parallel to $ MH$, $ R$ be the intersection of line $ EQ$ and $ AB$, $ S$ be the intersection of $ FD$ and $ BE$.
If $ OL \equal{} KL$, then prove that two lines $ OH$ and $ RS$ are orthogonal.

$ DH=DK=OM$ because $ L$ is the midpoint of $ OK$. Then $ OH$ is parallel to $ BC$.
Since $ OM$ is parallel to $ AK$ and $ L$ is the midpoint of $ OK$ then $ P$ is the midpoint of $ AM$.
Now, $ PQ$ parallel to $ MH$ implies that $ Q$ is midpoint of $ AH$.
Let $ E\^{A}Q=\alpha$,  $ Q\^{A}R=\beta$ so we have  $ E\^{S}D=E\^{R}F=2\alpha+\beta$.
Then the quadrilateral $ ERFS$ is cyclic and $ E\^{R}S=E\^{F}S=2\alpha=E\^{Q}H$.
So $ RS$ is parallel to $ QH$.