Vieta
AMC
AIME
algebra
polynomial
AIME I

Let $ r$, $ s$, and $ t$ be the three roots of the equation
\[ 8x^3\plus{}1001x\plus{}2008\equal{}0.\]Find $ (r\plus{}s)^3\plus{}(s\plus{}t)^3\plus{}(t\plus{}r)^3$.

[hide]By Vieta's, $ r \plus{} s \plus{} t \equal{} 0$. We can rewrite the problem as $ (r\plus{}s)^3 \plus{} (s\plus{}t)^3 \plus{} (t\plus{}r)^3 \equal{} \minus{}(t^3 \plus{} r^3 \plus{} s^3)$

Since $ r^3 \plus{} s^3 \plus{} t^3 \minus{} 3rst \equal{} (r\plus{}s\plus{}t)(r^2 \plus{} s^2 \plus{} t^2 \minus{} rs \minus{} rt \minus{} st) \equal{} 0$, $ r^3 \plus{} s^3 \plus{} t^3 \equal{} 3rst \equal{} 3(\frac{\minus{}2008}{8})$ ($ rst \equal{} \frac{\minus{}2008}{8}$ by Vieta's)

Therefore, $ \minus{}(r^3 \plus{} s^3 \plus{} t^3) \equal{} \boxed{753}$[/hide]

The one computation error I managed to make on the AIME was $ 250*3 \equal{} 775$ on this problem.

[hide]
Vieta's formula gives $ r \plus{} s \plus{} t \equal{} 0$ and $ 8r^3 \plus{} 1001r \plus{} 2008 \equal{} 0\Longleftrightarrow \minus{} 8r^3 \equal{} 1001r \plus{} 2008$, so do for $ s,\ t$.

Thus $ 8\{(r \plus{} s)^3 \plus{} (s \plus{} t)^3 \plus{} (t \plus{} r)^3\} \equal{} \minus{} 8(r^3 \plus{} s^3 \plus{} t^3)$

$ \equal{} 1001(r \plus{} s \plus{} t) \plus{} 2008\cdot 3 \equal{} 3\cdot 2008$, yielding the answer is $ 753$. [/hide]

[hide]
By Vietes, we know that our answer should be $ \minus{}(r^3 \plus{} s^3 \plus{} t^3)$.

Now, we also know, by Vietes, $ r \plus{} s \plus{} t \equal{} 0$

We have:

$ 8r^3 \plus{} 1001r \plus{} 2008 \equal{} 0$

$ 8s^3 \plus{} 1001s \plus{} 2008 \equal{} 0$

$ 8t^3 \plus{} 1001t \plus{} 2008 \equal{} 0$

Adding these, we get

$ 8(r^3 \plus{} s^3 \plus{} t^3) \equal{} \minus{}2008 \cdot 3$

So our answer is $ \fbox{753}$
[/hide]

Why didn't they have a problem like this on the AIME I... they always do!! These problems are always relatively simple if you know vieta's.

Zenith Xan's factorization is nice solution, and shake9991's is also...

I just used Vieta's to get the sum of the roots is 0, then Newton Sums.

This problem is sort of analogous to #8 on the AIME I (the arctangent sum problem) in that it's likely many people looked at the problem and immediately knew how to proceed and solve.

[hide="Slightly less nice solution"]
By Vieta's, $ r+s+t=0$ and $ rst=-\frac{2008}{8}=-251$. Define $ s_n=r^n+s^n+t^n$. Expanding out the product and factoring, we have:
\begin{eqnarray*}
(r+s)^3+(s+t)^3+(t+r)^3&=&2(r^3+s^3+t^3)+3(r^2s+rs^2+s^2t+st^2+tr^2+rt^2)\\
&=&2s_3+3((rs+st+tr)(r+s+t)-3rst)\\
&=&2s_3-9rst\\
&=&2s_3+9(251)
\end{eqnarray*}
By Newton's sums, $ 8s_3+0s_2+1004s_1+3(2008)=0$ or $ s_3=-\frac{3(2008)}{8}=-3(251)$. Substituting this into the above, we have that the desired sum is $ (-6+9)(251)=\boxed{753}$
[/hide]

[quote="shake9991"][hide]
By Vietes, we know that our answer should be $ \minus{} (r^3 \plus{} s^3 \plus{} t^3)$.

Now, we also know, by Vietes, $ r \plus{} s \plus{} t \equal{} 0$

We have:

$ 8r^3 \plus{} 1001r \plus{} 2008 \equal{} 0$

$ 8s^3 \plus{} 1001s \plus{} 2008 \equal{} 0$

$ 8t^3 \plus{} 1001t \plus{} 2008 \equal{} 0$

Adding these, we get

$ 8(r^3 \plus{} s^3 \plus{} t^3) \equal{} \minus{} 2008 \cdot 3$

So our answer is $ \fbox{753}$
[/hide][/quote]

Wow. Beautiful.
I just went ahead and did it Zenith Xan's way...

[quote="archimedes1"][quote="kunny"]shake9991's solution=My solution. :wink:

Anyway, I have just found Super Solution. :)

$ (r \plus{} s)^3 \plus{} (s \plus{} t)^3 \plus{} (t \plus{} r)^3 \equal{} \minus{} 3stu$. Thus.... :D[/quote]
Super Solution = Zenith Xan's solution... :wink: :wink:[/quote]

Super Solution $ \neq$ Zenith Xan's solution.  :wink:

Just use $ \boxed{(a \plus{} b \plus{} c)^3 \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(a \plus{} b)(b \plus{} c)(c \plus{} a)}$.  :P

kunny et al.:
i mean, you used the well-known equality:

$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} ac \minus{} bc) \equal{} \frac{1}{2} (a \plus{} b \plus{} c)((a\minus{}b)^2 \plus{} (b\minus{}c)^2 \plus{} (c\minus{}a)^2))$

which is easily derived from plugging $ a,b,c$ back into the polynomial
$ P(x) \equal{} (x\minus{}a)(x\minus{}b)(x\minus{}c)$

and summing $ P(a) \plus{} P(b) \plus{} P(c) \equal{} 0$ and doing a fair amount of algebra.

i personally like the cyclic sum of squares version (idk aesthetically pleasing i guess, and recall what Hardy said about beauty being the first test ;P)

[hide="Vietas"]
Vietas will simplify our answer to be $ \minus{}(r^3\plus{}s^3\plus{}t^3)$. Now, we use Vieta's to get $ r\plus{}s\plus{}t\equal{}0$. Since the sum is 0, we can plug in $ \minus{}8r^3\equal{}1001r\plus{}2008$, $ \minus{}8s^3\equal{}1001s\plus{}2008$, $ \minus{}8t^3\equal{}1001t\plus{}2008$. Therefore, $ 8(r^3\plus{}s^3\plus{}t^3)\equal{}\minus{}(3)(2008)$. So, we have $ 251\times 3\equal{}\boxed{753}$.[/hide]

Our solution are exactly same; kunny's=shake9991=Mersenne. :wink:

[hide=Solution]
We know that $r+s+t=0$ by Vieta's so $(r+s)^3+(s+t)^3+(t+r)^3=-(t^3+r^3+s^3)$.

We know that $r^3+s^3+t^3=(r+s+t)(r^2+s^2+t^2-rs-st-tr)+3rst$

We already know that $r+s+t$ so we just need to find $3rst$.

We know that $rst=\frac{-2008}{8}=-251$ by Vieta's so $3rst=-753$.

We take the opposite of this to get the answer of $\boxed{753}$.
[/hide]

[hide=Solution]Because the coefficient of $x^2$ is $0$, by Vieta's we know that $r+s+t=0$. Thus, $r+s=-t$, $s+t=-r$, and $t+r=-s$, so $$(r+s)^3+(s+t)^3+(t+r)^3=(-t)^3+(-r)^3+(-s)^3=-(r^3+s^3+t^3)$$ $r^3+s^3+t^3$ is the $3$rd power sum ($p_3$) of the roots, so Newton Sums tell us that $p_3=s_1p_2-s_2p_1+3s_3$, where $s_k$ is the $k$th elementary symmetric sum of the roots. $s_1 = r+s+t =0$ and $s_1=p_1$ so we have $$p_3=0 \cdot p_2-s_2\cdot0+3s_3=3s_3$$ By Vieta's we know that $s_3=rst=-\frac{2008}{8}=-251$ so to summarize, $$(r+s)^3+(s+t)^3+(t+r)^3=(-t)^3+(-r)^3+(-s)^3=-(r^3+s^3+t^3)=-p_3=-3s_3=-3(-251)=\boxed{753}$$[/hide]

Since $r+s+t=0$ (Vieta), so we just need $-r^3-s^3-t^3$, which is just $(r+s+t)(rs+st+tr-r^2-s^2-t^2)-3rst$, or $\boxed{753}$ by Vieta's again.

This is trivial like its AMC 8 level why is it AIME #7?!

[hide=Solution]From Vieta, we have $r+s+t=0$, so $r+s=-t$, $s+t=-r$, $t+r=-s$. Hence the expression is $(-t)^3+(-r)^3+(-s)^3=-(r^3+t^3+s^3)$. Note that $r^3+t^3+s^3=(r+s+t)(r^2+s^2+t^2-(rs+rt+st))+3rst$, and $r+s+t=0$, so it is just $3rst=-\frac{2008}{8} \cdot 3=-753 \implies \boxed{753}$.[/hide]

@dj: lol, you're probably right. I was half joking - I was thinking maybe as an AMC 8 #25 but I think it would be about AMC 10 #16-19 so a bit too hard as u said :)

[quote=OlympusHero]This is trivial like its AMC 8 level why is it AIME #7?!

[hide=Solution]From Vieta, we have $r+s+t=0$, so $r+s=-t$, $s+t=-r$, $t+r=-s$. Hence the expression is $(-t)^3+(-r)^3+(-s)^3=-(r^3+t^3+s^3)$. Note that $r^3+t^3+s^3=(r+s+t)(r^2+s^2+t^2-(rs+rt+st))+3rst$, and $r+s+t=0$, so it is just $3rst=-\frac{2008}{8} \cdot 3=-753 \implies \boxed{753}$.[/hide][/quote]

this aime was before i was born...

[quote=OlympusHero]This is trivial like its AMC 8 level why is it AIME #7?!

[hide=Solution]From Vieta, we have $r+s+t=0$, so $r+s=-t$, $s+t=-r$, $t+r=-s$. Hence the expression is $(-t)^3+(-r)^3+(-s)^3=-(r^3+t^3+s^3)$. Note that $r^3+t^3+s^3=(r+s+t)(r^2+s^2+t^2-(rs+rt+st))+3rst$, and $r+s+t=0$, so it is just $3rst=-\frac{2008}{8} \cdot 3=-753 \implies \boxed{753}$.[/hide][/quote]

it is AMC 8 level

[quote=OlympusHero]This is trivial like its AMC 8 level why is it AIME #7?![/quote]
While this is pretty easy for an AIME #7, I'd argue that Newton's sums/Vieta's on a cubic/the $r^3+s^3+t^3-3rst$ factorization is not AMC8 level ;)

I think it'd be around recent AMC 10s #18

I'd argue that this was appropriate for its placement. Newton's Sums is NOT a first five level problem. Personally, I think this problem would be at 5-7 for today's contests, so yes, it is a bit easy, but overall, I'd say this wasn't horribly misplaced.

[hide]The first thing to notice is that r + s + t = 0, so we're looking for -(r^3+s^3+t^3). Since r, s, and t are roots, -8(r^3+s^3+t^3) = 1001(r+s+t) + 2008*3 = 6024. dividing by 8 gives you the answer of 753.[/hide]

Note that as $\sigma_1=0$, the answer is equivalent to $-(r^3+s^3+t^3)=-\rho_3=-\sigma_1^3+3\sigma_1\sigma_2-3\sigma_3=-3(-\tfrac{2008}{8})=\boxed{754}$. Alternatively if you do not know Newton's Sums, note that $t=-r-s$, so our answer is equivalent to $(r+3)^3-r^3-s^3=3rs(r+s)=-3rst$ which is equivalent to what we got above.

By Vieta's Formula, we get,
r+s+t=0=>r+s=-t,  similarly for s & t & rst= -251
Reqd. (r+s)³+(s+t)³+(t+r)³= -(r³+s³+t³)= -(3rst)
=> -(3•-251) =>753
Nice one!!!

Note that since by Vieta's, $r+s+t=0$, giving us that
\[(r+s)^3+(s+t)^3+(t+r)^3=-(r^3+s^3+t^3),\]
and since $r+s+t=0$, we also have that
\[-(r^3+s^3+t^3)=-3rst=-3*\frac{-2008}{8}=753,\]
by Vieta's, finishing the problem.


Careful now...

$\text{What is the arithmetic mean of }1.\.{2} \text{ and } 2.\.{1}? \ (0.\.{3} = 0.333333 \ldots)$

[hide]so $1.2$ with the dot on the top would be $1.222222...$ ?

so that would be
$\frac{1\frac{2}{9}+2\frac{1}{9}}{2}$

right?... :huh:

$\frac{5}{3}$[/hide]

[hide]
Yup, that's right.

However when this was set as a UK Senior Maths Challenge question, there was a large proprotion of people answer $1.6\.{5}$.
[/hide]


For positive reals [tex]a,b,c,[/tex] which satisfy [tex]abc(a+b+c)=1[/tex], prove that

[tex]\displaystyle\frac{a}{(a+b)(a+c)}+\displaystyle\frac{b}{(b+c)(b+a)}+\displaystyle\frac{c}{(c+a)(c+b)} \leq \displaystyle\frac{1}{2abc}[/tex]

Did you make the problem yourself this time?

Yes, but I got the idea from a BMO inequality I was working on, and just modified it.  The problem was:

Find the minimum value of [tex](a+b)(b+c)[/tex], given that [tex]abc(a+b+c)=1[/tex].

[quote="ThAzN1"]Yes, but I got the idea from a BMO inequality I was working on, and just modified it.[/quote]

What is BMO?

BMO=British Mathematical Olympiad. It was from "The Mathematical Olympiad Handbook", which I think is a pretty good collection of easier olympiad problems.

Well, (a+b)(a+c) = a(a+b+c)+bc = bc+1/bc >= 2 by AM-GM. Then,

[tex]\displaystyle \frac{a}{(a+b)(a+c)}+\frac{b}{(b+c)(b+a)}+\frac{c}{(c+a)(c+b)} \le \frac{a}{2}+\frac{b}{2}+\frac{c}{2} = \frac{1}{2abc}[/tex].

I don't think equality is attainable.


Show that if $n$ is a natural number we have:

\[\sum^n_{k=1} (-1)^{k+1} \frac{1}{k} = 2 \cdot \left( \sum^{\dfrac{n}{2}}_{k=1} \frac{1}{n + 2k} \right).\]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


summer program
Mathcamp
PROMYS
geometry
symmetry
number theory
Galois Theory

Which program do you think is better? camp alumi are urged to share their experience(s)

Well this summer I chose to go to PROMYS instead of Mathcamp because I wanted to do more open-ended, research-oriented problems.  I also decided I'd benefit more from learning a lot and thinking deeply in one subject rather than seeing many different types of math.  I debated between the two a long time before choosing PROMYS.

I'm not saying PROMYS is better: perhaps my impression of Mathcamp was different from the real thing.  Neither is a bad decision: it's a win-win situation.  I will just say that I don't at all regret going to PROMYS.  It's fun, and you get to figure things out on your own.  You also, of course, think very deeply.  The focus at PROMYS is of course, depth over breadth, and complete rigor.  And yeah, it's number theory.

You also get a lot of freedom and have a lot of opportunities to do stuff.

I'm sure PROMYS is a great program, but you can do a lot of that at Mathcamp too, if you want.  Most students do a project, and if you want that project can take over most of your day the last three weeks.  Also, it is often possible to specialize in one area; during week four of Mathcamp I took three (excellent) number theory classes.  And at Mathcamp you also have a lot of freedom and opportunites.

[quote="Phelpedo"]As far as I understand though, PROMYS is ALL number theory. So if you're looking for some diversity, MathCamps offers a variety of topics and classes.[/quote]

Well, it depends what you define as number theory.  Because if number theory is all that there is a PROMYS, then number theory is a pretty large branch of mathematics.  There also was combinatorics, galois theory, and geometry and symmetry this year...not to mention you could do plenty of other stuff in research.

But the thing is, there isn't a big focus on classes on PROMYS...rather on problem sets.  Of course, I'd expect that also at mathcamp...


trigonometry
geometry
trig identities
Law of Sines

1. A fire in a building B is reported on telephone to two fire stations P and Q , 20 km apart from each other on a straight road. P observes that the fire is at an angle of 60 degrees to the road and Q observes that it is at an angle of 45 degrees to the road. Which station should send its team and how much will its team have to travel?

2. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60 degrees. After 10 seconds , its elevation is observed to be 30 degees. Find the speed of the aeroplane in km/hr.

1. Draw triangle with vertices P, Q, and F (fire). It should be obvious that the third angle is 75 degrees; from there on we apply the law of sines and find that side PF is the shortest, with distance:

$ \frac {20}{\left(\frac {\sqrt {6} \plus{} \sqrt {2}}{4}\right)} \cdot \frac {\sqrt {2}}{2} \equal{} 20\sqrt {3} \minus{} 20 \approx 14.641 \mbox{km}$

(You can do this with 30-60-90 and 45-45-90 triangles but it's slightly more complicated)

2. Using the known proportions of 30-60-90 triangles, we get that the distance traveled by the plane in $ \frac {1}{360}$ hours is:
$ \frac {\sqrt {3}}{1} \minus{} \frac {1}{\sqrt {3}} \equal{} \frac {2\sqrt {3}}{3} \mbox{km}$

Multiplying by 360 the plane's velocity is $ 240\sqrt {3} \frac {\mbox{km}}{\mbox{h}} \approx 415.692 \frac {\mbox{km}}{\mbox{h}}$, which seems reasonable (albeit somewhat low) for a plane's airspeed.


[quote="Hexaditidom"][hide="Just a guess"] $e$ ? [/hide][/quote]
 :huh: ? Are you talking about the limit of this sequence? It clearly doesn't converge to $e$.
Indeed, we have : $f(a_n) = a_{n-1}$ with $f(x) = x + 1/x$, and, $\lim_{x \to +\infty} f(x) = +\infty$, so, this sequence diverges.
But, I think that shobber's question is to find $a_n$ in terms of $n$. :)

Since it seems so close to the harmonic series in some ways, I would think that it would involve ln(n) in some way, but that's really just a guess.  However, I'm not really sure that the means actually exist to find an exact closed form for the sequence.

It might not include $ln$ as it contains rational terms ,i have an idea but i don't know if it might lead to a solution we consider two new sequences $p_n$ and $q_n$ such that $a_n=\frac{p_n}{q_n}$ and $gcd(p_n,q_n)=1$ we'll find that:
$p_n=p_{n-1}^2+q_{n-1}^2$
and$q_n=p_{n-1}^2 q_{n-1}^2$

Partial sums of the harmonic series are rational, but they are still related to ln(n).  Also, I think $q_n=p_{n-1}q_{n-1}$.  Check out http://www.research.att.com/~njas/sequences/A073833 and http://www.research.att.com/~njas/sequences/A073834 for the numerators and denominators of the sequence.  The site says that $\lim_{n\to\infty}\frac{1}{n}e^{2(a_{n}^2-2n)}$ is a constant, namely $0.57...$, which is probably irrational, if not transcendental.

[quote="kokosai"]
The site says that $\lim_{n\to\infty}\frac{1}{n}e^{2(a_{n}^2-2n)}$ is a constant, namely $0.57...$, which is probably irrational, if not transcendental.[/quote]
Isn't it Euler-Mascheroni constant?

Is that a guess, or do you have a proof?


inequalities
trigonometry
geometry
geometry solved

Let a, b, c be the sides of a triangle and S its area. Prove

$14a^2+2c^2-b^2\geq 8S\sqrt3$.

Determine all conditions for equality.

Bomb

This was at an Australian team training last year, though I don't know its source.

Writing $c$ and $S$ in terms of $a$ and $b$ and the angle $C$ opposite $c$ in the triangle, the inequality is equivalent to
\[ 16a^2+b^2-4ab\cos C\geq 4ab\sqrt{3}\sin C \]
or, dividing by 4 and rearranging,
\[ (2a)^2+(\frac{1}{2}b)^2\geq (2a)(\frac{1}{2}b)[\cos C+\sqrt{3}\sin C] =2(2a)(\frac{1}{2}b)\sin(C+30^o). \]
Now since $\sin x \leq 1$ for all x, a stronger inequality is
\[ (2a)^2+(\frac{1}{2}b)^2\geq 2(2a)(\frac{1}{2}b) \]
and this holds by AM-GM.

Equality conditions are: $C+30^o=90^o$, $2a=\frac{1}{2}b$.  This happens when
\[ a: b: c=1: 4: \sqrt{13}. \]

Let XYZ be a triangle with sides $x=YZ=1$, $y=ZX=4$ and $z=XY=\sqrt{13}$ (in fact, there exists such a triangle since the triangle inequalities for the numbers 1, 4 and $\sqrt{13}$ are fulfilled: $1+4=5>\sqrt{13}$, $4+\sqrt{13}>4>1$ and $\sqrt{13}+1>3+1=4$). Then, $y^2+z^2-x^2=4^2+\left(\sqrt{13}\right)^2-1^2=28$, $z^2+x^2-y^2=\left(\sqrt{13}\right)^2+1^2-4^2=-2$ and $x^2+y^2-z^2=1^2+4^2-\left(\sqrt{13}\right)^2=4$; also, by the Heron formula, the area T of the triangle XYZ equals

$T=\frac14\sqrt{\left(x+y+z\right)\left(y+z-x\right)\left(z+x-y\right)\left(x+y-z\right)}=\frac14\sqrt{2y^2z^2+2z^2x^2+2x^2y^2-x^4-y^4-z^4}$
$=\frac14\sqrt{2\cdot 4^2\cdot\left(\sqrt{13}\right)^2+2\cdot\left(\sqrt{13}\right)^2\cdot 1^2+2\cdot 1^2\cdot 4^2-1^4-4^4-\left(\sqrt{13}\right)^4}$
$=\frac14\sqrt{48}=\sqrt3$.

Now, by the Pedoe inequality ( http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3 Theorem [b](a)[/b]), applied to the triangles ABC and XYZ, we have

$a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$;

using the above, this becomes

$a^2\cdot 28+b^2\cdot\left(-2\right)+c^2\cdot 4\geq 16S\sqrt3$.

Division by 2 yields $a^2\cdot 14+b^2\cdot\left(-1\right)+c^2\cdot 2\geq 8S\sqrt3$, i. e. $14a^2-b^2+2c^2\geq 8S\sqrt3$. This is the inequality that we had to prove. As we have applied the Pedoe inequality to the triangles ABC and XYZ, equality will hold if and only if the triangles ABC and XYZ are similar, what means that $a: b: c=x: y: z$, or, equivalently, $a: b: c=1: 4: \sqrt{13}$.

So the problem is solved.

  darij

Indeed, I used cosine rule, area rule, auxilary form andf AMGM...

Bomb


search
number theory proposed
number theory

Is there a natural number n> (10^1000) which is not divisible by 10 and which satisfies: in its decimal representation one can exchange two distinct non-zero digits such that the set of prime divisors does not change.

Please give me a solution for this problem.

I suppose you are not allowing 555...555 either?

the digits we are exchanging are distinct! otherwise the answer is trivially yes!

please, help me.
this problem is really hard?  :(

help me.. please..  :(
i think that this problem not use hard theory in your solution..  :(

Lopes, asking for 7 times the solution for a certain problem will certainly not determine the others to solve it.

I've just remember that some time ago my friend dzeta told me he tried in vain to prove that it does not exist before finding the miraculous number 40 9999...99 ( 1038 nines) 59 . I have now idea whether this works or not, but I know that if it works, then I detest this problem.  :D

If you complete your solution, please, post.  :)

No, no, no, it is your turn to think a little bit on it.

There are simpler examples, such as
\[
n = 13 \cdot \frac{10^{360t} - 1}{9} = 144 \cdots 43.
\]
Then, changing the $1$ and the $3$, the number $344\cdots 41 = 31 \cdot \frac{10^{360t} - 1}{9}$ has the same prime divisors (because $360 = \phi (13 \cdot 31)$ and hence $10^{360t} - 1$ is divisible by both $13$ and $31$).

There might be, of course. Anyway, vess, search for another number, your number is not as large as they want.

Oh, sorry, just replace $360$ with $360t$, where $t \in \mathbb{N}$ is an arbitrary integer.  :)

Vess, why can't you switch the 3 and the 4, and you gain 2 as a prime divisor...thus altering the set of primes that divide the number (if i am understanding the problem correctly)

The problem does not require that switching [i]any[/i] pair of digits preserves the set of prime divisors (this is certainly impossible unless in the trivial cases); we are just asked to construct an $n$ for which [i]there exists[/i] a switch of different digits that preserves the prime divisors.

gotcha, i misread the problem, excuse me.


Hello!!!

Someone posted a problem in a brazilian discussion list (obm-l) and I'd like to know if there's an answer as a number not as a function.  :maybe:

Problem:

What is the sum of the products between all numbers from the set {2, 4, 6, ..., 2000} (even numbers from 2 to 2000, inclusive) and the set {1/2, 1/3, 1/4, ..., 1/2001} (natural numbers from 2 to 2001 inversed).

The sum would be something like (1/2 + 1/3 + 1/4 + ... + 1/2001)*(2 + 4 + 6 + ... + 2000) = [H(2001) - 1]*(2+2000)*1000/2 = [H(2001) - 1]*1001000, where H(N) is the sum of all terms from the harmonic series with the last term equals to 1/N. H(N) = 1 + 1/2 + 1/3 + 1/4 + ... + 1/N.

The answer should be given like that as a function of H(N) or there is a way to find a value for it.  :?:

Thanks in advance  :lol:

I take it this is what you're asking for:

$\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{2001}\right) \cdot (2+4+6+...+2000)$

[hide]There is a formula for $1+2+3+\ldots+n=\frac{n(n+1)}{2}$.  We'll use that.

$\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{2001}\right) \cdot (2+4+6+...+2000)$
$=\left[ \sum_{n=2}^{2001}\frac{1}{n}\right] \cdot 2(1+2+3+...+1000)$
$=\left[ \sum_{n=2}^{2001}\frac{1}{n}\right] \cdot 2 \left( \frac{1000(1001)}{2}\right)$; Use formula I mentioned above to get here.  Now the right side is done.
$=\left[ \sum_{n=2}^{2001}\frac{1}{n}\right] \cdot 1000(1001)$
$=1000 \cdot 1001 \cdot \sum_{n=2}^{2001}\frac{1}{n}$

I've never seen a way to cleanly sum the harmonic series, so to denote it as a function is probably the best way to answer your question.. but if you want it from $2$ to $2001$ as a decimal the calculator gives $7.1788678537352$.[/hide]

$H(N)$ does not have a closed form in terms of elementary functions.


sinx+sin2x+sin3x+sin4x= sin78*sin84 then x=?

[quote="EUCLA"]non so la risposta ma volevo salutarti xke sei italiano e invitarti nel forum (italiano, ovviamente :D )[/quote]
can you translate it into English?

sorry :blush:
i said him: i don't know the answer, but i'd like to invite u in the italian forum cos i have just posted a new topic :)


This is where we will discuss the mock ARML after 7:30PM on Thursday. There is also a poll to find out the difficulty of the ARML. If people like it, I can make one again next week. Don't vote until you have tried all the questions.

The Mock ARML Discussion will begin at 6:00 PM today. There's one general guideline for the discussion: DO NOT DISCUSS YOUR INDIVIDUAL SCORES (WHICH MEANS NO BRAGGING, NO WHINING, AND NO ASKING ABOUT WHAT EVERYONE MADE). The discussion will be about the questions themselves, the solutions, the approaches, and the overall feeling of the test.

OK, cool.  Can you post the winners and the answers?

Here are the answers to the individual round:

1. $8$
2. $12$
3. $49$
4. $\frac{17}{6}$
5. $310-320$
6. $-1$
7. $\frac{3^{59}+1}{4 \cdot 3^{59}}$
8. $0$

Here were the general statistics: Average Score of the 8 participants: $5.5$

The highest score on this was $8$ by MysticTerminator (NC) and chinagirl (NC)
The second highest score was $7$ by joml88 (SC) and Elemmenop (AL)

There were three people who took the relays, and all three of them made a perfect score on that: mysmartmouth, solafidefarms, chinagirl

And oh yeah, North Carolina won by a landslide, average score was $7.33$

FEEL FREE TO DISCUSS QUESTIONS NOW!

Congrats Chinagrl! I would say the same to Arnav, but that's taken for granted. So... I thought 1,2,3,5,6 were easy, 4 and 7 were pretty simple but took a little more time. 8 was weird, but I guessed right anyway (you shouldn't make the answer 0--it's too guessable). It was a pretty good test, though.

Can someone post a good and exact solution for number 5? I got 314 but the exact answer is 317 which a few people got :huh:

And four was simple and fast if we figured out that it was a right triangle *laughs wickedly* :diablo:

$(\frac{3ab(a+b)}{3ab})^2\le 100+2ab$ means $a^2+b^2\le 100$, so for a, there are this many values of b (positive and negative):
0-21
1-19
2-19
3-19
4-19
5-17
6-17
7-15
8-13
9-9
10-1
Since a can also be negative, we have 21+2(148)=317.

Ah, ok, what i did was when i simplified it down to $a^2+b^2\leq 100$, i found the approximate area of the circle since it was all the points within the circle, which was $314$...hmmm...lol, the test maker is asking the questions about the answers :roll:

What did you guys think of numbers 7 and 8...hehe

Eh, 7 was just recursion if you see it early enough, and then it's pretty simple from there.

8 was kinda logical, though a bit confusing to read.


Let there be a 6 digit number 576abc that is divisible by 5,7,9.  Find all 3 digit numbers abc such that it fits that category.

[hide]We have two cases, last digit is a five, or a zero.
Using the divisibility rule for 9, we come up with a list of possible last three digits for the first case.
990
090
180
270
360
450
540
630
720
810
900
Since the first number divisible by 7 in the 576,000's is 576,002, we know that abc is 2 mod 7.
Our list of numbers ending in 0 that are 2 mod 11:
30
100
170
240
310
380
450
520
590
660
730
800
870
940
The only overlap in the two lists is 450.
Now that we know that 576,450 is divisible by 315, we know that 576,135 and 576,765 are also divisible by 315.
[/hide]


Junior question paper INAO 2008 ,

1) when r1 and r2 are connected in series, the current is 2A. when they are connected in parallel, the current is 4A. find r1 and r2.

I think, the question was incomplete as u cud only derive a relation between the resistors r1 & r2 and not the values.

Can somebody comment ?

2) Is there any relation between gravity and magnetic force ?

3) Which unit is universally fundamental - kelvin/a.m.u/lightyear/none ?

Regards

just look at two topics below u posted that is [url=http://www.mathlinks.ro/viewtopic.php?t=186188]here[/url]...for the First

2)No there isn't any relation(Until unless someone discovers :D )
3)none of the above..
because 1 Lyr-Distance travelled by light in a year but the year is not fixed...
              a.m.u-Can be made with absolutely anyhing...previously it was with oxygen and now it is with                carbon so....
 Kelvin's def depend on riple point of water....firstly water may not be available evrywhere and moreover even if it is availabl;e the triple point is measurd at atm pressure hence may change with planet

Continue all ur further discussion [url=http://www.mathlinks.ro/viewtopic.php?t=186188]here[/url]
I am locking this topic


:?: Let x, y, z a positive numbers. Prove that
$\sum \frac{yz}{x}\ \ge\ \sum \frac{y^2+z^2}{y+z}.$

I proved for $\min \{ x^2, y^2, z^2\}\ge \frac{xyz}{x+y+z}.$

Write the initial ineq as
$\sum x^4(y+z)(y-z)^2 \ge 0$

With the same condition,prove that:
[tex]\sum\frac{x^2+y^2}{x+y}\leq 3.\frac{a^2+b^2+c^2}{a+b+c}[/tex]

[quote="levi"]Let x, y, z a positive numbers. Prove that
xy/z + yz/x + zx/y >= (x^2+y^2)/(x+y)+(y^2+z^2)/(y+z)+(z^2+x^2)/(z+x).[/quote]

We have to prove the inequality

$\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z}\geq\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}+\frac{x^2+y^2}{x+y}$

for any three positive reals x, y, z. In fact, using the cyclic sum notation, this inequality rewrites as $\sum\frac{yz}{x}\geq\sum\frac{y^2+z^2}{y+z}$. This is equivalent to $\sum\frac{yz}{x}-\sum\frac{y^2+z^2}{y+z}\geq 0$. But

$\sum\frac{yz}{x}-\sum\frac{y^2+z^2}{y+z}=\sum\left(\frac{yz}{x}-\frac{y^2+z^2}{y+z}\right)=\sum\frac{y^2\left(z-x\right)-z^2\left(x-y\right)}{zx+xy}$
$=\sum\left(\frac{y^2\left(z-x\right)}{zx+xy}-\frac{z^2\left(x-y\right)}{zx+xy}\right)=\sum\frac{y^2\left(z-x\right)}{zx+xy}-\sum\frac{z^2\left(x-y\right)}{zx+xy}$
$=\sum\frac{x^2\left(y-z\right)}{yz+zx}-\sum\frac{x^2\left(y-z\right)}{xy+yz}=\sum\left(\frac{x^2\left(y-z\right)}{yz+zx}-\frac{x^2\left(y-z\right)}{xy+yz}\right)$
$=\sum\left(x^2\left(y-z\right)\left(\frac{1}{yz+zx}-\frac{1}{xy+yz}\right)\right)$.

Now imagine for a while that $y\geq z$. Then, $xy\geq zx$, so that $xy+yz\geq yz+zx$, so that $\frac{1}{xy+yz}\leq\frac{1}{yz+zx}$, and thus $\frac{1}{yz+zx}-\frac{1}{xy+yz}\geq 0$, while $y\geq z$ yields $y-z\geq 0$. Hence, $\left(y-z\right)\left(\frac{1}{yz+zx}-\frac{1}{xy+yz}\right)\geq 0$. So far we have only considered the case when $y\geq z$, but in the case of $y\leq z$, we similarly find $\frac{1}{yz+zx}-\frac{1}{xy+yz}\leq 0$ and $y-z\leq 0$, so that $\left(y-z\right)\left(\frac{1}{yz+zx}-\frac{1}{xy+yz}\right)\geq 0$. Hence, in any case, we have $\left(y-z\right)\left(\frac{1}{yz+zx}-\frac{1}{xy+yz}\right)\geq 0$. Thus,

$\sum\frac{yz}{x}-\sum\frac{y^2+z^2}{y+z}=\sum\left(x^2\left(y-z\right)\left(\frac{1}{yz+zx}-\frac{1}{xy+yz}\right)\right)\geq 0$,

and the proof is complete.

@Keira_Khtn: Did you mean the inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=32967 post #3?

  darij

Levi,here's a better one: $ab/c+bc/a+ca/b>=sqrt(3(a^2+b^2+c^2))$

If $ x,y,z > 0$ then $ x^{2}\plus{}y^{2}\plus{}z^{2}> \equal{} 2yzcosA\plus{}2zxcosB\plus{}2xycosC$ where $ A,B,C$ are the angles of a triangle.
[Moderator edit: I have moved all discussion of this inequality to http://www.mathlinks.ro/Forum/viewtopic.php?t=5243 .]

[quote="cezar lupu"]If $x,y,z>0$ then $x^2+y^2+z^2>=2yzcosA+2zxcosB+2xycosC$ where $A,B,C$ are the angles of a triangle.[/quote]

A proof in my book
$(x-(z cos B + y cos C))^2 + (y sinC - z sinB)^2 \ge 0$

[quote="levi"][b][i]Thank you [/i]! [/b][b]I do not your book. What is its name ? For entertainment,  I[/b] [b]propose this conditional inequality:[/b]\[if\ \ a,b,c,x,y,z\in R\ \ such\ that\ \ x^3+y^3+z^3=3xyz\ \ \]
\[then\ \ (a^2+b^2+c^2)(x^2+y^2+z^2)+2(ab+bc+ca)(xy+yz+zx)\ge o.\][/quote]
Well, from the condition we only need to deal with two cases.
(1) $x+y+z=0$, then $2(xy+yz+zx) = -(x^2+y^2+z^2)$ So the given ineq is equivalent to $(a^2+b^2+c^2+ab+bc+ca)(x^2+y^2+z^2) \ge 0$ which is obvious.
(2) $x=y=z$ Then we need to show $(x^2+y^2+z^2)(a^2+b^2+c^2+2ab+2bc+2ca) \ge 0$ which is $(x^2+y^2+z^2)(a+b+c)^2 \ge 0$ so it's true as well.

***
For the book, it's a set of chinese mathematics book, entitled 華羅庚學校數學課本. I don't think it has any english translation.

[quote="levi"]Both the my amusing problem and the your cool solution are nicely. It get on very well with us at classroom ! I am a Romanian retired teacher and can not to sleep (here is time 1/30').[/quote]

You are now living in USA?
Anyway, thank you for your nice problem.

I think we can to find all x such that with all a,b,c>0
ab/c+bc/a+ca/b>=3(a^x++b^x+c^x)^(1/x)
   Result:x<=ln3/ln(3/2)
   Notes:2,7<ln3/ln(3/2)<2,8.


Let a,b,c reals. Find x,y,z reals such that xy(x+y-2z)=c, yz(y+z-2x)=a and zx(z+x-2y)=b.

here's wot i did
first note that ax + by + cz=0
set x=mz and y=nz then the above relation becomes
am+bn+c=0 (z=0 we can tackle the case later,easy i guess)
and substituting the values of x and y in any two equations and dividing them we get another relation linking m and n.so substitue the value of m in that relation and solve the quadratic for m
the problem's nearly finished
the solution is DIRTY.
wot is ur solution,Harazi??

I posted it in the unsolved problems. So, I didn't solved it completely, since I got stuck in a horribile 3 degree equation. My idea was the following: We must have cz/(x+y-2z)=ax/(y+z-2x)=by/(z+x-2y)=t. Thus, (x+y)/z=2+c/t and so on. It follows easily considering this as a linear system that (2+a/t)(2+b/t)(2+c/t)=8+(a+b+c)/t. And from here?

Why it is well known? It seems no one knows it after 3 years.

hello, i have solved this system with Mathematica, but the solution is ugly.
Sonnhard.


http://www.artofproblemsolving.com/Resources/AoPS_R_Websites.php

Methinks you should remove the (broken redundant) link to MathLinks on that page.

Also, on ML Skin, About and News both are errors.
About:  Not found (404)
News:   Forum you requested does not exist.
:)


calculus
integration
calculus computations

Evaluate
$ \int_{0}^{\infty} \frac{x^3}{e^x\minus{}1}\,dx$


function
calculus
derivative
trigonometry
geometry proposed
geometry

The picture is a corridor.

Find the length of the longest stick AB that can move from one of the wings to the other. (It's assumed that the thickness of the stick is negligeable and during the motion the stick stays horizontal.)

[img]http://img264.imageshack.us/img264/8341/amiensfx0.gif[/img]

Using mister Pitagoras and Mister Thales.

$ AB \equal{} \sqrt{a^2\plus{}x^2} \plus{} \sqrt{b^2\plus{}y^2}$ Pitagoras

$ \frac{b}{y}\equal{}\frac{x}{a}$  Thales

Solve for $ x$ or $ y$ in the second and put it on $ AB$.

Derive in function of the variable $ x,y$ equal to zero and solve for the variable.

You will find $ \left(a^{2/3}\plus{}b^{2/3}\right)^{3/2}$

This is the minimum length.

The maximum length don't make any sense, since it is infinity (make $ x\equal{}0$).

[quote][color=darkred][b][u]An equivalent enunciation[/u][/b]. Let $ OAFB$ be a rectangle, where $ \{\begin{array}{c} OA = a \\
\ OB = b\end{array}$ . Let $ \{\begin{array}{c} X\in OA\ ,\ A\in (OX) \\
\ Y\in OB\ ,\ B\in (OY)\end{array}$

be two points so that $ F\in XY$ . Find the [b]shortest length of the segment[/b] $ [XY]$ .[/color][/quote]
[color=darkblue][u][b]Proof I.[/b][/u] Denote $ \{\begin{array}{c} OX = x \\
\ OY = y\end{array}$ . Observe that $ \mathbb D\equiv\{\begin{array}{c} \ (x,y)\in \mathbb R^2 \\
\ x > a\ ,\ y > b\end{array}||\ \frac ax + \frac by = 1\ \}$

and $ XY^2 = x^2 + y^2$ . The our problem has the following [b]mathematical model[/b] : $ \min_{(x,y)\in \mathbb D}\ (x^2 + y^2)$ .

Using the substitutions $ \{\begin{array}{c} u = \frac ax \\
 \\
v = \frac by\end{array}$ obtain the problem $ \min_{\begin{array}{c} 0 < u,v < 1 \\
\ u + v = 1\end{array}}(\frac {a^2}{u^2} + \frac {b^2}{v^2})$ . Using derivatives obtain

$ v'(u) = - 1$ and $ \frac {a^2}{u^3} + \frac {b^2}{v^3}\cdot v'(u) = 0$ $ \implies$ $ \frac {a^2}{u^3} = \frac {b^2}{v^3}$ $ \implies$ $ \frac uv = (\frac ab)^{\frac 23}$ $ \implies$ $ \frac ax\cdot\frac yb = (\frac ab)^{\frac 23}$ $ \implies$ $ \frac xy = (\frac ab)^{\frac 13}$ .[/color]
[hide="Answer."]
[color=darkblue]$ \{\begin{array}{c} x = a + \sqrt [3]{ab^2} \\
\ y = b + \sqrt [3]{a^2b}\end{array}$ , i.e. $ \frac {x - a}{y - b} = \sqrt [3]{\frac ba}$ . Observe that $ \frac {FX}{FY} = (\frac ba)^{\frac 23}$ and $ \boxed {\frac {OX}{OY} = \sqrt [3]{\frac ab}}$ .

In conclusion, $ \boxed {\min XY = (a^{\frac 23} + b^{\frac 23})^{\frac 32}}$ , i.e. $ XY^{\frac 23} = a^{\frac 23} + b^{\frac 23}$ .[/color][/hide]
[color=darkblue][b][u]Proof II.[/u][/b] Denote $ m(\widehat {AFX})=\phi$ . Then $ XY=f(\phi )$ , where $ f(\phi )=\frac {a}{\sin\phi}+\frac {b}
{\cos \phi}$ and $ \phi\in (0,\frac {\pi}{2})$ .

Observe that $ f'(\phi )=\frac {b\sin\phi }{\cos^2\phi}-\frac {a\cos\phi }{\sin^2\phi}$ and $ f_m=\min_{(0,\frac {\pi}{2})}f(\phi )=f(\phi_m)$ , where $ \boxed {\phi_m=\arctan \sqrt [3]{\frac ab}}$ a.s.o.[/color]


A thermally insulated container is divided into two section by a thermally insulating partition. One section contain 3 mol of neon at 400 K and the other 1 mol of neon at 600K.The partition is then removed.find the equilibrium temperature of the neon , assuming that neon is ideal monatomic gas
good luck :D

Is it Advanced Physics?

 [hide="Clik to reveal hidden content"]

Let C bee the neon's molar heat capacity, $\;\;n_{1}\,=\,3\,,\,T_{1}\,=\,400\, K\;\;$ the number of mols and initial absolutte temperature in the first section, and $\;\;n_{2}\,=\,1\,,\,T_{2}\,=\,600\,K\;\;$ in the other section.

As the system is thermal insulated, the equilibrium temperature, T, is given by heat conservation ("First Law of thermodynamics"):

$\qquad n_{1}.C.(T-T_{1})\,+\,n_{2}.C.(T-T_{2})\,=\,0\;\;\Rightarrow\;\;T\,=\,{{n_{1}.T_{1}+n_{2}.T_{2}}\over{ n_{1}+n_{2}}}\,=\,450\,K\;\;$
[/hide]

Is THIS advanced PHYSICS?

also there is no fun in this problem.

GET something from FLUID MECHANICS. THATS FUN!!!

 :)

ADvance physics
Is it easy?
hhaha
A long time ago,I couldn't solve it,but i hada dream ,which lead me to the solution.So the thread is called for fun :blush:


geometry
circumcircle
trigonometry
inequalities
function
inequalities unsolved

show
sinA + sinB + sinC =< 3 (sqrt3)/2

the most straight forward one would be the one using concavity (Jensen's inequ), but this time I am allowed to use AM > GM only

i am trying the conversion sinA = a /2R, R radius of circumcircle

(sorry for the double posts, i posted it in the wrong place so i repost it here)

$ \sin A \plus{} \sin B \plus{} \sin C\leq\frac {3\sqrt3}{2}\Leftrightarrow\sum_{cyc}\frac {2S}{ab}\leq\frac {3\sqrt3}{2}\Leftrightarrow4S(a \plus{} b \plus{} c)\leq3\sqrt3abc.$
Let $ a \equal{} x \plus{} y,$ $ b \equal{} x \plus{} z$ and $ c \equal{} x \plus{} y.$ Then your inequality is equivalent to
$ 64(x \plus{} y \plus{} z)^3xyz\leq27(x \plus{} y)^2(x \plus{} z)^2(y \plus{} z)^2.$
But $ 9(x \plus{} y)(x \plus{} z)(y \plus{} z)\geq8(x \plus{} y \plus{} z)(xy \plus{} xz \plus{} yz)\Leftrightarrow\sum_{cyc}(x^2y \plus{} x^2z \minus{} 2xyz)\geq0,$
which true by AM-GM for six numbers.
Thus, it remains to prove that $ (xy \plus{} xz \plus{} yz)^2\geq3(x \plus{} y \plus{} z)xyz,$
 which true by AM-GM too since, $ \frac {x^2y^2 \plus{} x^2z^2}{2}\geq x^2yz.$

$ sinA\plus{}sinB\plus{}sinC$ $ \equal{}$ $ \frac{2}{\sqrt{3}} \displaystyle \left [ sinA.\frac{\sqrt{3}}{2}\plus{} sinB.\frac{\sqrt{3}}{2} \right ]\plus{}\sqrt{3} \displaystyle \left [\frac{sinA}{\sqrt{3}}.cosB\plus{} \frac{sinB}{\sqrt{3}}.cosA \right ]$

$ \le \frac{1}{\sqrt{3}} \displaystyle \left [(sin^2A\plus{}\frac{3}{4})\plus{}(sin^2B\plus{}\frac{3}{4}) \right ]\plus{}\frac{\sqrt{3}}{2}\displaystyle \left [(\frac{sin^2A}{3}\plus{}cos^2B)\plus{}(\frac{sin^2B}{3}\plus{}cos^2A) \right ]\equal{}\frac{3\sqrt{3}}{2}$

[quote="miff"]show
\[ \sin{A} + \sin{B} + \sin{C} \leq \frac{3\sqrt{3}}{2}\]
The most straight forward one would be the one using concavity (Jensen's inequ), but this time I am allowed to use $ AM \geq GM$ only. I am trying the conversion $ \sin{A} = \frac{a}{2R}$, $ R$ radius of circumcircle (sorry for the double posts, i posted it in the wrong place so i repost it here)[/quote]

Dear Miff,

I was not able to find the solution using $ AM\geq GM$ (respect to [b]Arqady[/b] and [b]Thanhnam2902[/b], both solutions are very glad!!). But I found another method, how to prove your inequlity:

The following identity is true for any 3 real numbers $ \alpha, \beta, \gamma$ (even if $ \alpha+\beta+\gamma \ne \pi$):

\[ {\sin{\alpha}+\sin{\beta}+\sin{\gamma}-3\sin\frac{\alpha+\beta+\gamma}{3}
=-4\sin{\frac{\alpha+\beta+\gamma}{3}}\sin^2({\frac{\alpha-2\beta+\gamma}{6})-4\sin{\frac{\alpha+\gamma}{2}}\,\sin{\frac{2\alpha-\beta-\gamma}{6}}\,\sin{\frac{\alpha+\beta-2\gamma}{6}}}
}\]

If $ \alpha, \beta, \gamma$ are the angles of some triangle, e.g $ \alpha+\beta+\gamma=\pi$, then the identity above simplifies to:

\[ {\sin{\alpha}+\sin{\beta}+\sin{\gamma}-\frac{3\,\sqrt{3}}{2}=-2\sqrt{3}\,\sin^{2}({\frac{\beta-\pi/3}{2}}})-4\cos{\frac{\beta}{2}}\,\sin\frac{\alpha-\pi/3}{2}\sin{\frac{\pi/3-\gamma}{2}}\]

Now let us assume that $ \alpha\geq\beta\geq\gamma$. Then $ \alpha\geq\frac{\pi}{3}$ and $ \gamma\leq\frac{\pi}{3}$, and the identiy above clealy shows that

\[ \sin{\alpha}+\sin{\beta}+\sin{\gamma}\leq\frac{3\sqrt{3}}{2}\]

This identity is not so nice as [b]thanhnam2902[/b] or [b]arqady[/b] solutions. However, the same method that was used for finding this identity above can be used for a lot of more inequalities. [hide="Here is the explanation."]

Imagine you need to find maximum for $ f(x)+f(y)+f(z)$ for some function $ f(x)$ under condition $ x+y+z=contant$, and you guess that maximum is equal to $ 3f(\frac{constant}{3})$ and is reached at $ x=y=z$. Let us have a look at the function $ g$:

\[ g(x,y,t)=f(t)+f(x+y-t)-f(x)-f(y)\]

From the geometric point of view, $ g(x,y,t)$ is the doubled length of the segment $ AB$, where $ A$ is the midpoint between the points $ (x,f(x))$ and $ (y,f(y))$, and $ B$ is the midpoint between the points $ (t,f(t))$ and $ (x+y-t,f(x+y-t))$.

Function $ g(x,y,t)$ has 2 properties:

[b]1.[/b] For all $ x,y,z$ and any $ f$

\[ {3f(\frac{x+y+z}{3})-f(x)-f(y)-f(z) = g(x,y,\frac{x+y+z}{3})+g(\frac{2x+2y-z}{3},z,\frac{x+y+z}{3})}\]

This can be easily proven by the removal of parentheses.

[b]2.[/b] In a lot of practical cases formula for $ g(x,y,z)$ can be simplified and some identity can be found easily so that it will be clear that $ g(x,y,z)$ is positive or negative under the assumption that $ x\geq y \geq z$. In our case $ f(x)=\sin{x}$, and

\[ g(x,y,t)=\sin{t}+\sin{(x+y-t)}-\sin{x}-\sin{y} =\\
={2\sin{\frac{x+y}{2}}\cos{\frac{2t-x-y}{2}} - 2\sin{\frac{x+y}{2}}\cos{\frac{x-y}{2}} = 2\sin{\frac{x+y}{2}}(\cos{\frac{2t-x-y}{2}}-\cos{\frac{x-y}{2})}=\\
=4\sin{\frac{x+y}{2}}\sin{\frac{x-t}{2}}\sin{\frac{t-y}{2}}}\]

From this identity it's clear that if $ x \geq t \geq y \geq 0$ and $ x+y\leq\pi$, then $ g(x,y,t)\geq0$. Now let us return to the original question of finding miminum of $ f(x)+f(y)+f(z)$ for $ f(x)=\sin{x}$. To do this, just substitute the identity

\[ g(x,y,t)=4\sin{\frac{x+y}{2}}\sin{\frac{x-t}{2}}\sin{\frac{t-y}{2}}\]

into the identity

\[ {3f(\frac{x+y+z}{3})-f(x)-f(y)-f(z)=
g(x,y,\frac{x+y+z}{3})+g(\frac{2x+2y-z}{3},z,\frac{x+y+z}{3})}\]

And you will have exactly the identity which is written at the beginning:

\[ {\sin{\alpha}+\sin{\beta}+\sin{\gamma}-3\sin\frac{\alpha+\beta+\gamma}{3}
=-4\sin{\frac{\alpha+\beta+\gamma}{3}}\sin^2({\frac{\alpha-2\beta+\gamma}{6})-4\sin{\frac{\alpha+\gamma}{2}}\,\sin{\frac{2\alpha-\beta-\gamma}{6}}\,\sin{\frac{\alpha+\beta-2\gamma}{6}}}
}\]

If you like to practice this method, here is the exercise:

For positive $ a,b,c$ it is given that $ a+b+c=1$. Prove that $ a^3+b^3+c^3\geq\frac{1}{9}$. Of course, there are a lot of ways to prove this; try to find the identity using the method described above.
[hide="Solution."]
\[ {a^3+b^3+c^3-\frac{1}{9}(a+b+c)^3 = \frac{1}{3}
 ( a+b )  ( 2\,a-b-c )  ( a+b-2\,c ) +\frac{2}{9}\, ( a+b+c )  ( a-2\,b+c ) ^{2}}\]

The idea is easy, so probably it is well-known - sorry, I am not specialist in inequalities. I have found this method when I was in school a lot of years ago.
[/hide]
[/hide]

Sergey Markelov
markelov@mccme.ru

[quote="Ronald Widjojo"]$ sin A \plus{} sin B \plus{} sin C \plus{} sin 60 \leq 2sin {\frac {A \plus{} B}{2}} \plus{} 2 sin {\frac {C \plus{} 60}{2}} \leq 4 sin {\frac {A \plus{} B \plus{} C \plus{} 60}{4}} \equal{} 4 sin 60$
So, $ sin A \plus{} sin B \plus{} sin C \leq 3 sin 60 \equal{} \frac {3\sqrt {3}}{2}$  :P  :P[/quote]

[color=darkblue]it's nice too.[/color]

[size=200][color=violet]
\[ \begin{array}{l} \sin A \plus{} \sin B \plus{} \sin C \\
\equal{} 3 \times \frac {{\sin A \plus{} \sin B \plus{} \sin C}}{3} \\
\le 3 \times \sin \frac {{A \plus{} B \plus{} C}}{3} \\
\equal{} \frac {{3\sqrt 3 }}{2} \\
\end{array}
\]
[/color][/size]
We've done. :)

Analogously[quote="miff"] [color=darkred]Show  that in any triangle $ ABC$ exists the inequality $ \sin A + \sin B + \sin C \le\frac {3\sqrt 3}{2}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] I"ll use the well-known identities $ \left\|\begin{array}{cc} \frac {a}{\sin A} = \frac {b}{\sin B} = \frac {c}{\sin C} = 2R & (1) \\
 \\
(p - a)(p - b)(p - c) = pr^2 & (2) \\
 \\
r_a + r_b + r_c = 4R + r & (3) \\
 \\
r_ar_b + r_br_c + r_cr_a = p^2 & (4) \end{array}\right\|\ .$ Thus,  $ \boxed {\ \sin A + \sin B + \sin C \le\frac {3\sqrt 3}{2}\ \ \stackrel{(1)}{\Longleftrightarrow}\ p\le \frac {3R\sqrt 3}{2}\ }$ .

$ \blacktriangleright\ \frac {(p - a) + (p - b) + (p - c)}{3}\ge \sqrt [3]{(p - a)(p - b)(p - c)}$ $ \Longleftrightarrow$ $ p^3\ge 27(p - a)(p - b)(p - c)$ $ \stackrel {(2)}{\Longleftrightarrow}$ $ p^3\ge 27pr^2$ $ \Longleftrightarrow$ $ \boxed {\ 3r\sqrt 3\le p\ }\ \ (*)$ .

$ \blacktriangleright\ 3(r_ar_b + r_br_c + r_cr_a)\le (r_a + r_b + r_c)^2\ \ \stackrel{(3)\wedge (4)}{\Longleftrightarrow}\ \ 3p^2\le (4R + r)^2$ $ \Longleftrightarrow$ $ p\sqrt 3\le 4R + r$ $ \Longleftrightarrow$ $ \boxed {\ 9p\le 3\sqrt 3(4R + r)\ }\ (**)$ .

Now add the boxed relations $ (*)$ , $ (**)$ and obtain $ 8p\le 12R\sqrt 3$ , i.e.  $ \boxed {\ p\le \frac {3R\sqrt 3}{2}\ }$ .[/color]

[hide="Remark."]
 [color=darkblue]Memorize $ \boxed {\ 3r\sqrt 3\ \le\ p\ \le\ \frac {3R\sqrt3}{2}\ }\ !$  Show analogously that $ \boxed {\ \cos A + \cos B + \cos C = 1 + \frac rR\ \le\ \frac 32\ }$ .

Indeed, $ \cos A + \cos B + \cos C - 1 =$ $ (\cos B + \cos C) - (1 - \cos A) =$ $ 2\cos \frac {B + C}{2}\cos\frac {B - C}{2} - 2\sin^2\frac A2 =$

$ 2\sin \frac A2\cos\frac {B - C}{2} - 2\sin\frac A2\cos\frac {B + C}{2} =$  $ 2\sin \frac A2\left(\cos\frac {B - C}{2} - \cos\frac {B + C}{2}\right) =$ $ 4\sin\frac A2\sin\frac B2\sin\frac C2 =$

$ \frac {4(p - a)(p - b)(p - c)}{abc} =$ $ \frac {4pr^2}{4Rpr} = \frac rR$ . Thus, $ \cos A + \cos B + \cos C = 1 + \frac rR\le \frac 32$ because $ 2r\le R$ .

I used the well-known identities $ \sin\frac A2=\sqrt {\frac {(p-b)(p-c)}{bc}}$ , $ \sin\frac B2=\sqrt {\frac {(p-c)(p-a)}{ca}}$ , $ \sin\frac C2=\sqrt {\frac {(p-a)(p-b)}{ab}}$ .[/color][/hide]

hi !

what is $ r_a$ , $ r_b$ and $ r_c$ in the prof of -Virgil Nicula-

[quote="y-a-s-s-i-n-e"]hi ! What is $ r_a$ , $ r_b$ and $ r_c$ in the prof of -Virgil Nicula-[/quote]
[b]Standard notation[/b]. Denote $ w \equal{} C(I,r)$ -  the incircle and $ \boxed {w_a \equal{} C(I_a,r_a)}$ - the $ A$-[b]excircle[/b] of $ \triangle\ ABC$ , i.e. the circle which is tangent to the

sidelines of $ \triangle\ ABC$ in the points $ X\in (BC)$ , $ Y\in CA$ , $ Z\in AB$ so that $ C\in (AY)$ and $ B\in (AZ)$ . Show easily that $ \left\|\begin{array}{c} AY \equal{} AZ \equal{} p \\
\ BZ \equal{} BX \equal{} p \minus{} c \\
\ CY \equal{} CX \equal{} p \minus{} b\end{array}\right\|$ .

You can prove easily the relations $ (2)$ , $ (3)$ , $ (4)$ from my last reply using the remarkable formulas

$ \boxed {S \equal{} \sqrt {p(p \minus{} a)(p \minus{} b)(p \minus{} c)} \equal{} pr \equal{} r_a(p \minus{} a) \equal{} r_b(p \minus{} b) \equal{} r_c(p \minus{} c)}$ .

Observe that $ ar_a \equal{} p(r_a \minus{} r)$ a.s.o and $ \sum ar_a \equal{} 2p(2R \minus{} r)$ . I"ll present you an aplication :

$ \sum\frac {a}{r_a} \equal{} \sum\frac {a^2}{ar_a}\ge \frac {\left(\sum a\right)^2}{\sum ar_a} \equal{} \frac {4p^2}{2p(2R \minus{} r)}\ \implies\ \boxed {\ \sum\frac {a}{r_a}\ge \frac {2p}{2R \minus{} r}\ }$ .


Just some possible values:
$ (a,b,c,d)\equal{}(1,1,c,c^3\plus{}2c)$ then $ f(a,b,c,d) \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{1 \plus{} abcd}\equal{}\frac{c^6\plus{}4c^4\plus{}5c^2\plus{}2}{c^4\plus{}2c^2\plus{}1}\equal{}c^2\plus{}2$

[quote="SUPERMAN2"]Let $ a,b,c,d$ be the positive integers.Denote $ f(a,b,c,d) \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{1 \plus{} abcd}$.Find all possible value of $ f(a,b,c,d)$ if $ f(a,b,c,d) \in N$[/quote]
Some other possible values :

$ a^2\plus{}b^2\plus{}c^2\plus{}((a^2\plus{}b^2\plus{}c^2)abc)^2$ $ \equal{}(a^2\plus{}b^2\plus{}c^2)(1\plus{}abc((a^2\plus{}b^2\plus{}c^2)abc))$

So $ f()$ can take at least any integer value which is a sum of three squares, that's to say at least all odd integers and most of even one.

to be continued ...


trigonometry
trig identities
cosine double angle

$\frac{\cos^8(2x)-\sin^8(2x)}{\cos^4(2x)+\sin^4(2x)}=\cos(nx)$ for all real numbers $x$.  What is the positive integer value of $n$?

[hide]Noticing that the numerator of the LHS is simply the difference of 2 squares, we can break it down:

$\frac{\cos^8(2x)-\sin^8(2x)}{\cos^4(2x)+\sin^4(2x)}=\cos(nx)$
$\frac{(\cos^4(2x)+\sin^4(2x))(\cos^4(2x)-\sin^4(2x))}{\cos^4(2x)+\sin^4(2x)}=\cos(nx)$
$\cos^4(2x)-\sin^4(2x)=\cos(nx)$

Now we have another differnece of squares, so we break that down as well:

$(\cos^2(2x)-\sin^2(2x))(\cos^2(2x)+\sin^2(2x))=\cos(nx)$

We know from a pythagorean identiy that: $\cos^2(\theta)+\sin^2(\theta)=1$ for all $\theta$.  Now we have:

$\cos^2(2x)-\sin^2(2x)=\cos(nx)$

From a cosine double angle identity, we can rewrite the LHS as:
$\cos(4x)=\cos(nx)$

Therefore, $n=\boxed{4}$
[/hide]


trigonometry
analytic geometry
circumcircle
projective geometry
geometry proposed
geometry

$ \Delta ABC$ circumscribed circle $ I$ with sides $ BC,AC$ tangent to circle $ I$ at points $ M,N$ respectively. $ E$ and $ F$ are the midpoints of $ AB$ and $ AC$ respectively. Straight line $ EF$ intersects $ BI$ at point $ D$. Prove $ M,N$ and $ D$ are collinear.

cause A,D,M,I are on the same circle $ AD \perp DB$. Let K be the intersection of AB and a parallel line from D to BC. Then $ \overline{AK}\equal{}\frac{c \sin \frac{B}{2} \cos \frac{B}{2}}{\sin B} \equal{} \frac{c}{2}$. Then $ E \equiv K$.

Using barycentric coordinates: E(1;1;0); F(1;0;1); M(0;a+b-c; a+c-b) and N(a+b-c;0;b+c-a).
The equations of line  EF: $ x \equal{} y \plus{} z$ and the line  BI:  $ \frac {x}{a} \equal{} \frac {z}{c} \Rightarrow$ D(a;a-c;c) and, the determinant of coordonnes of points
D, M an N is =0 $ \Rightarrow$ the points D, M and N colinear.

Dear Mathlinkers,
this situation is well knowed and has been studied by Lascases, a french geometer.
For more facts and a first generalisation see : http://perso.orange.fr/jl.ayme   vol. 1  Another unlikely concurrence.
This problem has been proposed on Mathlinks. Where?
Sincerely
Jean-Louis

[quote="Virgil Nicula"] [color=darkred]The incircle $ (I)$ of $ \triangle ABC$ touches it in the points $ M\in BC$ , $ N\in CA$ . Denote the midpoints

 $ E$ and $ F$ of $ AB$ and $ AC$ respectively. Prove that the lines $ MN$ , $ EF$ , $ BI$ are concurrently.[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $ X\in BI\cap MN$ . Observe that $ m(\widehat {AIX}) \equal{} m(\widehat {MNC}) \equal{} m(\widehat {ANX}) \equal{} \frac {A \plus{} B}{2}$ , i.e. the quad

$ AINX$ is cyclically. Thus, $ XA\perp XB$ and $ \widehat {EXB}\equiv\widehat {EBX}\equiv\widehat {XBC}$ , i.e. $ XE\parallel BC$ . In conclusion, $ X\in EF$ .[/color]

[quote="jayme"] [color=darkred][b]Dear Mathlinkers[/b], this situation is well knowed and has been studied by [b]Lascases[/b], a [u]french geometer[/u].

Sincerely, [b]Jean-Louis[/b][/color] [/quote]
[color=darkblue][b]Yes, it is well-known.[/b]

And yet in my opinion, here is a joke, an entertainment.

At least for me, for ... my remained mind, a little  there, indeed.

I like such problems. Sincerly, [b]Virgil Nicula.[/b] [/color]

Dear Mathlinkers,
for more details and references, see for example
http://perso.orange.fr/jl.ayme  vol. 4  An unlikely concurrence, revisited and generalized
Sincerely
Jean-Louis

Here is my solution using [b]Harmonic division[/b]:
Let $ D'\equiv BI\cap MN$, $ K\equiv MN\cap BC$. Then, it is easy to notice that $ (K,P,A,B) \equal{} \minus{} 1$, where $ P$ is the tangency point of $ (I)$ and $ AB\Longrightarrow D'(KPAB) \equal{} \minus{} 1$. In the other hand, $ D',I,B$ are collinear. Thus, $ D'$ lies on the bisector line of $ PN\Longrightarrow \angle PDB \equal{} \angle BDN \Longrightarrow D'B$ is the outward bisector of $ \angle PDN$.
Combine with $ D'(KPAB) \equal{} \minus{} 1\Longrightarrow D'A\bot D'B\Longrightarrow \angle AD'B \equal{} 90^{0}$
$ \Longrightarrow \angle AED' \equal{} 2\angle D'BA \equal{} \angle B$, which implies that $ D'E//BC$, but $ EF//BC$, hence $ D',E,F$ are collinear $ \Longrightarrow D'\equiv D\Longrightarrow D,M,N$ are collinear $ \blacksquare$
Our proof is completed. :lol:


inequalities
algebra
polynomial
calculus
derivative
inequalities unsolved

Let f(x) be a polynomical with degree n and $ f(x) \geq 0$ for all x.
Prove that:
  \[ \sum^{n}_{k=0}f^{(k)}(x) \geq 0\]

let $g(x)=\sum^{n}_{k=0}f^{(k)}(x)$. obviously, $g(x)-g'(x)=f(x)$. now, since f is everywhere nonnegative, the leading coefficient must be positive and f has even degree, so the leading coefficient of g is positive and g has even degree, since all derivatives of f smaller degree. thus g tends to positive inifinity as x tends to positive or negative inifinity. thus if $g(x_0)<0$, then there exist zeros $x_1<x_0$ and $x_2>x_0$ of g. choose $x_1,x_2$ closest to $x_0$. then g is everywhere negative between $x_1$ and $x_2$. but by rolle, there exists $x_3$ between $x_1$ and $x_2$ such that $g'(x_3)=0$. thus $0>g(x_3)=g(x_3)-g'(x_3)=f(x_3)$, contradiction.
i will now show that actually $g(x)>0$ for $f\neq 0$ everywhere, just using the fact that a zero of a polynomial which is every nonnegative has even degree. suppose $g(x_0)=0$ and let k be the degree of $x_0$, so k is even. since $g^{(m-1)}(x)-g^{(m)}(x)=f^{(m-1)}(x)$ for every $m=1,2,...$, $x_0$ is a zero of f with degree k-1, but k-1 is odd, contradiction.


function
induction
algebra proposed
algebra

Find all functions$f: N_0\to N_0$ such that $mf(n)+nf(m)=(m+n)f(m^2+n^2)$ for all $m,n\in{N_0}$.

We do $m=0: nF(0)=nF(n^2)$ , but $m \not= 0 : F(0)=F(n^2)$
Then:$F(n^2)=c..................[1]$ such that $c$ it is a constant .
Now by induction for k we will demonstrate ;$F(k)=c$
$k=1$:  $1=1^2$ of [1] $F(1)=c$
$k=2$ : $2=1+1$ of $n=m=1 \to F(2)=c$
.
.
.
$k=p$:we suppose that yes one is fulfilled $F(p)=c$
Then :$n=1;m=p \to F(p+1)=c$..............the induction this complete one

The only function that fulfills is $F(x)=c ; \forall x \in N$ ,such that $c$ it is a constant


I just graduated from school and applied for university. In my country (Australia), in the first year we do a common engineering course but in the second year we choose which engineering field we want go into.

So im just wondering which field is the most maths based?

I think by definition, engineering=math.  Mechanical engineering is pretty math intensive (if you like physics).  Chemical engineering has a lot of chemistry-related math.  It all depends on what type of math.  But anyway, choose an engineering field because you find it interesting, not because of how much math it has.


Hi, please let me know if this problem is posted inappropriately for this forum. I have solved this problem programmatically using a brute-force approach (i.e. go through all possible combinations until best one found), but I think I'm really looking for a formula. The brute-force solution quickly breaks down with larger data sets. Thanks in advance. Here's the problem:

The following shipping companies charge these per-pound rates:

[code]
			National				International
			Zone 1	Zone 2		Zone 3	Zone 4
UPS		2.00		3.00			4.00		5.00
FedEx	1.00		1.50			4.50		6.00
[/code]

A new company, Shippit, is offering a flat rate of 2.50 across the entire globe, provided that no more than 30% of the total weekly poundage ship internationally. Every internationally-shipped pound above 30% is surcharged a 3.50 penalty.

Every day, your business needs to send out shipments to various locations around the world. Write a program to determine the best way to allocate your shipping so that you are using the cheapest possible arrangement.

The problem, as I understand it:

Every day there is a certain distribution of shipments $ z_{1}, z_{2}, z_{3}, z_{4}$ that must be made.  Your job is to distribute these shipments across three shipping companies to minimize the total cost of shipment.  Is that it?  I need to clarify because it seems trivial to me:

Without Shippit, all it seems that you would need to do is to go with the cheapest company in each zone.  This is trivial.

With Shippit's constraint (which is expressed in weekly rather than daily terms; why is this?), just ship as much overseas as you can without incurring the surcharge.

So I'm afraid I don't understand the source of the difficulty.

Hi, t0rajir0u, thanks for replying.

I think the problem is as follows. Say I need to ship 10 pounds to Zone1, and 10 pounds to Zone4. Using your formula, I’d use Fedex for Zone1 and UPS for Zone4, and it would cost 60.00:

10(1.00) + 10(5.00) = 60.00

Had I used Shippit for all shipping, incurring penalty for weight above the limit, it would be cheaper at 57.00:

10(2.50) + 10(2.50) + 2(3.50) = 57.00

But depending on the distribution of weights and destinations, Shippit is not always the best choice. How do I write a formula that calculates the best choices available to me?

Er, sorry, actually my math is wrong in that last post -  :blush: - the correct total for the second choice is 64.00...

Here is a better example distribution: 10 pounds to Zone2, and 10 pounds to Zone3.

If we use Fedex's z2 rate (1.50) and UPS' z3 rate (4.00) our total will be 55.00.

If we distribute things a bit, we can lower our bill to 47.50:

7 pounds to Fedex z2 = 10.50
3 pounds to UPS z3 = 12.00
7 pounds to Shippit z2 = 17.50
3 pounds to Shippint z3 = 7.50

[quote="big maybe"]Here is a better example distribution: 10 pounds to Zone2, and 10 pounds to Zone3.

If we use Fedex's z2 rate (1.50) and UPS' z3 rate (4.00) our total will be 55.00.

If we distribute things a bit, we can lower our bill to 47.50:

7 pounds to Fedex z2 = 10.50
3 pounds to UPS z3 = 12.00
7 pounds to Shippit z2 = 17.50
3 pounds to Shippint z3 = 7.50[/quote]

This looks like $ 14$ pounds to Zone2 and $ 6$ pounds to Zone3.   :huh:

Again, the problem seems to me a matter of shipping as much overseas with Shippit as you can internationally without incurring the surcharge, since you never profit by using Shippit domestically.

If you need to send 14 pounds to Zone 2 and 6 pounds to Zone 3, then it's cheaper to not use Shippit at all (21+24<50).
[hide]If you send to Zone 3 using Shippit, you end up losing money. By not using FedEx or UPS, you save 1.50 per pound sent to Zone 3, but you lose at least 1.00 per pound sent domestically. This is bad because you need to send over twice as much domestically as you send internationally to avoid the surcharge. (You never want to pay the surcharge because the total cost per extra pound is greater than just using UPS.) Your losses overtake your gains.

So you should only use Shippit if you have shipments to Zone 4. However, if you don't have enough to send to Zone 2 to complement the shipments to Zone 4, you need to send Zone 1 shipments using Shippit also. In this case, the losses can overtake the gains. Sending 10% to Zone 1, 60% to Zone 2, and 30% to Zone 4 with Shippit is the same as not using Shippit at all. So if more than 10% of your Shippit shipments are to Zone 1, you are paying more than you would by not using Shippit.

My train of thought ends here. Someone else can build on this, or I can come back later.[/hide]


rotation
geometry
geometric transformation

A square of side length s lies in a plane perpendicular to a line  L. One vertex of the square lies on L. As this square moves a distance  H along L  the square turns one revolution about L to generate a corkscrew like column with square cross sections.

a) Find the volume of the column

b)What will the volume be if the square turns twice instead of once? Explain.

I considered an element of distance "dH", the corresponding element of angle "dx" must satisfy dx/dH = H/2pi hence dx = 2*pi*dH/h. As dH tends to zero, dx tends to zero, and the element of volume dV approximates to a cuboid of sides s, s and dH. Hence dV = (s^2)dH  and V = (s^2)H by integration.

For the second part, the volume is unchanged as cuboid approximation still works as dH tends to zero. Although the square rotates around twice, it "spends less time" on each rotation on one side of the line. The effect of this is to give the same volume. One could also use the last result as follows: consider the square rotating twice over a length of 2H. This would obviously just give a volume of 2V = 2(s^2)H. Lets now "squash" this column along L into length H. By similar volumes, this column has half the volume as it had before. i.e. its volume is V. Hence the volume is unchanged by the square rotating twice.


I stumbled across a problem that has both me and my mom (who's my teacher since I'm homeschooled) stumped. It's SQRT(z)-3 = SQRT(z-27). The stuff in parenthesis is outside of the square root. Can someone explain how to solve this problem?

                                                 -Nick

For the sake of brevity, let's let u=sqrt z.

(u-3)^2=u^2-27
u^2-6u+9=u^2-27
-6u+9=-27
6u=36
u=6
sqrt(z)=6
z=36

Edit: Fixed.

The solution that my mom showed in the solution book said that the answer was z=36. I don't understand where the 6:sqrt:z comes from either.

                                            -Nick

A round about but a fun way:

If you use shorthand  $A = \sqrt z$   and  $B = \sqrt {z-27} $
You are given:

$\sqrt  z - \sqrt {z-27} = 3 $     which is nothing but $A - B = 3 $     (eq 1)
You also know
$z - (z-27) = 27 $  which is nothing   but  $ A^2 - B^2 = 27$           (eq 2)
(Divide (eq2) by (eq 1) )
so $ A + B =  9 $  (eq 3 )  ( This is because    $ (A + B) = (A^2 - B^2) / (A - B) $)

Adding  (eq 1) and (eq 3) gives    $ 2 A = 12  \  or  \  A = 6  $  so $z=36$


Okay, really stop spamming with problems that really don't make sense (even though there is a proof that $ 1\equal{}2$).  maybe 9  :D

[quote="007math"]Okay, really stop spamming with problems that really don't make sense (even though there is a proof that $ 1 \equal{} 2$).  maybe 9  :D[/quote]
IGNORE THIS POST

[quote="jonathanchou711"]Ok tcjy. This has got to stop. You are posting random problems that are not benifiting this forum. PLEASE STOP OR I WILL REPORT YOU!!!  :mad:   :|[/quote]
OKAY!!!!!!!!
REAL PROBLEM!!!!!!!!!!:
I'll just post a problem that actually HELPS this forum:
A Natural Number N has a remainder of 3 when divided by 7 and a remainder of 4 when divided by 5. What is the smallest value N can have?

You can count up on one restriction to get the other...
lets go

$ 3,10,17,24,...$

$ 24$ is the first one that satisfies both restrictions...
so answer is $ \boxed{24}$


Factor this product completely:

$(7x^3-11x^2+25x-12)$


[quote="Apprentice"]Define the operation blah by a blah b = 1/a + 1/b.  Find all values c such that:
(1 blah 2) blah c = 1 blah ( 2 blah c )[/quote]
[hide]
Lol.
$(1+\frac{1}{2})\text{ blah }c=1\text{ blah } (\frac{1}{2}+\frac{1}{c})$
$\frac{2}{3}+\frac{1}{c}=1+\frac{2c}{c+2}$
$\frac{2c+3}{3c}=\frac{3c+2}{c+2}$
$7c^2-c-6=0$
$(7c+6)(c-1)=0$
$\boxed{c_1=-\frac{6}{7}, c_2=1}$[/hide]


Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \frac {4a^2 \minus{} b^2 \minus{} c^2}{a(b \plus{} c)} \plus{} \frac {4b^2 \minus{} a^2 \minus{} c^2}{b(a \plus{} c)} \plus{} \frac {4c^2 \minus{} a^2 \minus{} b^2}{c(a \plus{} b)}\leq3
\]

[hide="Outline of Solution"]
After using some identities, it turns out that this is equivalent to the cyclic sum of $ (\frac {1}{ab \plus{} ac} \plus{} \frac {1}{ab \plus{} bc} \minus{} \frac {3}{(a \plus{} c)(b \plus{} c)})(a \minus{} b)^2 \ge 0.$

This is provable using condition 2 of S.O.S. outlined

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127
[/hide]

I will post my full solution when I have time... don't have much time right now...

[hide="Full Solution"]
We use the identity $ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \equal{} \frac {3}{2} \plus{} \frac {1}{2}\sum \frac {(a \minus{} b)^2}{(a \plus{} c)(b \plus{} c)}$, which was cited at http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=239445157&t=248960

Then, we have
\[ \frac {4a^2 \minus{} b^2 \minus{} c^2}{a(b \plus{} c)} \plus{} \frac {4b^2 \minus{} a^2 \minus{} c^2}{b(a \plus{} c)} \plus{} \frac {4c^2 \minus{} a^2 \minus{} b^2}{c(a \plus{} b)}\leq3
\]

\[ \Longleftrightarrow \sum_{cyc} \frac {4a^2 \plus{} 2ab \plus{} 2ac \minus{} b^2 \minus{} c^2}{ab \plus{} ac} \le 9
\]

\[ \Longleftrightarrow \sum_{cyc} \frac {6a^2 \minus{} (a \minus{} b)^2 \minus{} (a \minus{} c)^2}{ab \plus{} ac} \le 9
\]

\[ \Longleftrightarrow 6 \sum_{cyc} \frac {a}{b \plus{} c} \minus{} \frac {(a \minus{} b)^2 \plus{} (a \minus{} c)^2}{ab \plus{} ac} \le 9
\]
Using the identity for $ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b}$, we get
\[ \Longleftrightarrow \sum_{cyc} (\frac {3}{(a \plus{} c)(b \plus{} c)} \minus{} \frac {1}{ab \plus{} ac} \minus{} \frac {1}{ab \plus{} bc})(a \minus{} b)^2 \le 0
\]
Negating both sides,
\[ \Longleftrightarrow (\frac {1}{ab \plus{} ac} \plus{} \frac {1}{ab \plus{} bc} \minus{} \frac {3}{(a \plus{} c)(b \plus{} c)})(a \minus{} b)^2 \ge 0
\]
We can prove this using SOS condition 2...

We have
\[ S_a \equal{} \frac {1}{bc \plus{} ba} \plus{} \frac {1}{bc \plus{} ca} \minus{} \frac {3}{(b \plus{} a)(c \plus{} a)}
\]

\[ S_b \equal{} \frac {1}{ca \plus{} cb} \plus{} \frac {1}{ca \plus{} ab} \minus{} \frac {3}{(c \plus{} b)(a \plus{} b)}
\]

\[ S_c \equal{} \frac {1}{ab \plus{} ac} \plus{} \frac {1}{ab \plus{} bc} \minus{} \frac {3}{(a \plus{} c)(b \plus{} c)}
\]
We need to prove that $ S_b, S_b \plus{} S_a, S_b \plus{} S_c \ge 0.$ First we prove $ S_b \ge 0$
\[ \frac {1}{ca \plus{} cb} \plus{} \frac {1}{ca \plus{} ab} \minus{} \frac {3}{(c \plus{} b)(a \plus{} b)} \ge 0
\]
Multiplying by $ (c \plus{} b)(a \plus{} b),$ we get:
\[ \Longleftrightarrow \frac {c \plus{} b}{c} \plus{} \frac {a \plus{} b}{a} \ge 3
\]

\[ \Longleftrightarrow \frac {b}{c} \plus{} \frac {b}{a} \ge 1
\]
Which is true since b is greater than or equal to c.

Similarly, $ S_a$ is nonegative, which implies that $ S_a \plus{} S_b \ge 0.$ Now we must prove that $ S_b \plus{} S_c \ge 0.$

Unfortunately, I haven't been able to prove this other than expanding out everything...

Expanding out everything gives us that this is equivalent to $ a^2(b^2 \plus{} c^2) \plus{}$ a few other terms $ \ge 2a^2bc$, which is true since $ b^2 \plus{} c^2 \ge 2bc$ by AM-GM...
[/hide]

$ \frac {4a^2 \minus{} (b^2 \plus{} c^2)}{a(b \plus{} c)} \equal{} \frac {4a}{b \plus{} c} \minus{} \frac {b^2 \plus{} c^2}{a(b \plus{} c)}\leq 2x \minus{} \frac {1}{x}$ with $ x \equal{} \frac {2a}{b \plus{} c}\ (0 < x\leq 1)\ , W.L.O.G. \ 0 < a\leq b\leq c)$, since $ 2x,\ \minus{} \frac {1}{x}$ are increasing fuctions, we have $ 2x \minus{} \frac {1}{x}\leq 1$ with equality $ x \equal{} 1$.

$ \therefore \sum_{cyclic} \frac {4a^2 \minus{} (b^2 \plus{} c^2)}{a(b \plus{} c)}\leq 3$ Q.E.D.

How do we know that x $ \le 1...$

$ x \le 1$ only for $ \frac {4a^2 \minus{} (b^2 \plus{} c^2)}{a(b \plus{} c)} \equal{} \frac {4a}{b \plus{} c} \minus{} \frac {b^2 \plus{} c^2}{a(b \plus{} c)}\leq 2x \minus{} \frac {1}{x}$, since you assumed $ 0 < a \le b le c,$ x might not be less than or equal to 1 for the cyclic permutations...

Oops! The inequality was not symmetric for $ a,\ b,\ c$ but cyclic. :blush:


Let a,b,c are length sides of triangel such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that
$ \sum_{cyc} \frac{c}{a}\sqrt{2\minus{}a^2} \geq \sqrt{3(\frac {1}{a^2}\plus{}\frac {1} {b^2} \plus{}\frac {1}{c^2})}$
 :D
it is not easy

[quote="thegod277"]Let a,b,c are length sides of triangel such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that
$ \sum_{cyc} \frac {c}{a}\sqrt {2 \minus{} a^2} \geq \sqrt {3(\frac {1}{a^2} \plus{} \frac {1} {b^2} \plus{} \frac {1}{c^2})}$
 :D
it is not easy[/quote]
 I think we need use 3 lemma. I will post my solution

This is your inequalities! :D  But I think your solution is long. So I hope a nice solution. Can anyone help me? :?:


Solve for x:

$ 6x^2\minus{}\sqrt{3x\plus{}1}\minus{}\sqrt{2x\minus{}1}\equal{}1$

 :o


At what time is the angle between the two hands of an analog clock $170$ degrees?

There's a formula for this kind of problems but you need to know the hour or the minutes.

Uh
I didn't get an answer. The closest i got was 170.5.

[quote="236factorial"]Maybe this?

$\frac{11}{2}m-30h=|A|$?

I made that up, but it works sometimes.  :D[/quote]

What's $|A|$

[quote="nonie"][hide]10:20 is one of the answers.[/hide]

whats that formula youre talkin about?[/quote]

[hide]yes.. i got 10:20 also[/hide]

10:20 works because $\frac{11}{2}*20-300=110-300=-190$ absolute value that and we get 190.  Either 236factorial has an incorrect formula, or 10:20 is an incorrect time.


greatest common divisor
number theory
relatively prime

Prove or disprove that there are infinite integer triples $(a,b,c)$ such that the greatest common divisor of $a$, $b$, and $c$ is $1$, and such that $a^2+b^2=c^2$

consider u^2-v^2, 2uv, and u^2+v^2. being a, b, and c respectively. Since we can set u^2-v^2 to any prime, 2uv and u^2+v^2 will be relatively prime to it.
Since we can check those values work for the formula, then there are infinitely many values that work. (Note: this is far from proving that all relatively prime triples are in the form u^2-v^2, 2uv, and u^2+b^2)

there is a simpler way :)
think of this: parametrize a circle  :)

[quote="prototypex24"][quote="Scrambled"]Note: this is far from proving that all relatively prime triples are in the form u^2-v^2, 2uv, and u^2+b^2[/quote]

Just a question: Are [i]all[/i] pythagorean triples in this form :?:[/quote]

All [i]primitive[/i] Pythagorean triples have that form, i.e. $(m^2-n^2,2mn,m^2+n^2).$  A primitive Pythagorean triple is one where GCD(a,b,c)=1 and a, b, c are the sides of the triangle.  So, for example, (3,4,5) is primitive whereas (9,12,15) is not.

But knowing all primitive triples lets you know know [i]all[/i] Pythagorean triples....all you have to do to get the rest is multiply each element of the primitive triple by the same natural number.

Wow, I never knew that!  I was asking this because my math teacher claimed it was "extremely difficult" to prove that there are infinite "primitive" pythagorean triples (I had a feeling there was a simple way like this).  This clearly proves that there are infinite.  Is it possible to prove that all can be expressed in the form $u^2-v^2, 2uv, u^2+v^2$?


geometry algebra polynomial vector function induction calculus

Let $ F(x,y,z)$ be polynomial with integer coefficients, for which it is known, that there exists real $ x_0,y_0$ and $ z_0$ so that $ F(x_0,y_0,z_0)\equal{}0$. Prove, that for some real [i]algebraic[/i] numbers $ x_1,y_1$ and $ z_1$ it is true that $ F(x_1,y_1,z_1)\equal{}0$.

Here's a possible approach: consider the Jacobian $ \left(\frac {\partial F}{\partial x}, \frac {\partial F}{\partial y}, \frac {\partial F}{\partial z}\right)$ of $ F$ at the give root $ (x_0, y_0, z_0)$. If this is not the zero vector, assume wlog that $ \frac {\partial F}{\partial z}(x_0, y_0, z_0) \neq 0$. By the implicit function theorem there's a function $ g(x,y)$ such that $ F(x,y,g(x,y)) \equal{} 0$ whenever $ (x,y)$ are in some small neighborhood of $ (x_0, y_0)$. In particular, you can certainly find a point $ (x_1,y_1)$ such that $ (x_1,y_1)$ are rational and $ F(x_1,y_1,z_1)$=0 where $ z_1 \equal{} g(x_1,y_1)$. But now $ z_1$ is a root of $ p(Z) \equal{} F(x_1,y_1,Z)$ which is rational polynomial, so $ z_1$ is algebraic and the point $ (x_1,y_1,z_1)$ is as required (with the added bonus that $ x_1$ and $ y_1$ are actually rational).

If the given solution $ P \equal{} (x_0, y_0, z_0)$ is a singular point so that the Jacobian vanishes there, then $ P$ is also a root of the polynomials $ \frac {\partial F}{\partial x}, \frac {\partial F}{\partial y}, \frac {\partial F}{\partial z}$, which cannot all vanish unless $ F \equal{} 0$. So one of those is a genuine polynomial of which $ P$ is a root, and it has a lower total degree, so we can finish off the argument by induction on the degree.

I think the problem is easier than you might think. Since there exists a real solution $ \left(x_0,y_0,z_0\right)$ of $ F\left(x_0,y_0,z_0\right) \equal{} 0$, the polynomial $ F$ is not constant (unless it is identically $ 0$, in which case the problem is trivial). Consider the polynomial $ F\left(X,Y,Z\right)$ as an element of the polynomial ring $ \mathbb{Q}\left(X,Y,Z\right)$. Its three (formal!) partial derivatives $ \frac {d}{dX}F\left(X,Y,Z\right)$, $ \frac {d}{dY}F\left(X,Y,Z\right)$ and $ \frac {d}{dZ}F\left(X,Y,Z\right)$ cannot all be zero (otherwise, a simple comparison of coefficients would show that $ F$ is constant), so we can WLOG assume that $ \frac {d}{dZ}F\left(X,Y,Z\right)\neq 0$ in the polynomial ring $ \mathbb{Q}\left(X,Y,Z\right)$. Then, there exist rationals $ \alpha$ and $ \beta$ such that $ \frac {d}{dZ}F\left(\alpha,\beta,Z\right)\neq 0$ in the (smaller) polynomial ring $ \mathbb{Q}\left(Z\right)$ (because $ \frac {d}{dZ}F\left(\alpha,\beta,Z\right) \equal{} 0$ is a polynomial identity in $ \alpha$ and $ \beta$ over the integral domain $ \mathbb{Q}\left(Z\right)$, so if it holds for all $ \left(\alpha,\beta\right)\in\mathbb{Q}\times\mathbb{Q}$, it must hold as a polynomial identity, because $ \mathbb{Q}$ is infinite). This yields that the polynomial $ F\left(\alpha,\beta,Z\right)$ is nonconstant, hence has an algebraic root for $ Z$.

The result generalizes: If $ K\subseteq L$ is a field extension, and $ I\subseteq K\left[X_1,X_2,...,X_n\right]$ is an ideal such that the polynomials from $ I$ have a common root in $ \overline{L}$, then these polynomials also have a common root in $ \overline{K}$. This can be seen by one of the following two arguments:
1) Elimination theory. This can be used to actually construct such a root, and its construction only requires finite extensions of field, so if the polynomials were defined over $ K$, they must have a common root in $ \overline{K}$.
2) Hilbert's Nullstellensatz. This doesn't yield the assertion as trivially as 1), but is far better known (unfortunately). Using the Nullstellensatz, the existence of a common root becomes equivalent to the ideal being the whole polynomial ring, and what remains is to show that an ideal $ I$ of $ K\left[X_1,X_2,...,X_n\right]$ is the whole $ K\left[X_1,X_2,...,X_n\right]$ if and only if the ideal $ I\cdot L\left[X_1,X_2,...,X_n\right]$ of $ L\left[X_1,X_2,...,X_n\right]$ is the whole $ L\left[X_1,X_2,...,X_n\right]$. While $ \Longrightarrow$ is trivial, what we need is $ \Longleftarrow$. This, again, is trivial by taking a basis of $ L$ over $ K$ (or, if we wish to avoid Hamel bases, by a basis of a part of $ L$, namely the one required to ensure that $ I\cdot L\left[X_1,X_2,...,X_n\right] \equal{} L\left[X_1,X_2,...,X_n\right]$).

  darij

I think it's doesn't have to concern any $ \frac {\partial F}{\partial X}$
By induction on $ n$ we can prove for any field $ Q\subseteq k\subseteq R$ and any n-variable polynomial $ F\in Q[X]$ the proposition holds.
n=1 is trivial.
supposing it holds for n=h-1. If $ x_1 \equal{} t$ is transcendent on $ k$, by induction the other $ x_i$ is algebra on $ k(t)$, set $ E \equal{} k(t,x_2,\cdots ,x_n) \equal{} k(t,y)$ where y is algebra on k(t), $ irr(k(t),y) \equal{} h_t(y)$. Mutiply all conjugates of $ F$ (as a k(t)[y] polynomial), we get a polynomial of $ t$, since $ t$ is transcendent this polynomial must be 0. Thus for any $ t\in k$ and any root of h_t(y) there is a corresponding solution of $ F$, choose $ t'$ so close to $ t$ that $ h_t(y)$ still has a real root, but this time $ y$ is algebra and so all $ x_i$ is algebraic on $ k$.

by the way, darij: how dose your generalization represent ……F has not only algebraic root but $ real$ algebraic root. Dose $ \bar{K}$ means algebra closed field of $ K$ ?

Ah, I see. I misread the "real" in the "real algebraic", and the title "algebraic geometry" didn't help me either, since algebraic geometry has not much to do with $ \mathbb{R}$ for me. Of course, this makes my post completely irrelevant to this topic. My apologies.

  darij

[quote="randomgraph"]

If the given solution $ P \equal{} (x_0, y_0, z_0)$ is a singular point so that the Jacobian vanishes there, then $ P$ is also a root of the polynomials $ \frac {\partial F}{\partial x}, \frac {\partial F}{\partial y}, \frac {\partial F}{\partial z}$, which cannot all vanish unless $ F \equal{} 0$. So one of those is a genuine polynomial of which $ P$ is a root, and it has a lower total degree, so we can finish off the argument by induction on the degree.[/quote]

Hello,What do you mean by genuine polynomial ?And How to use induction then?Thank you


A mass is attached to an ideal spring. At time $ t \equal{} 0$ the spring is at its natural length and the mass is given an initial velocity; the period of the ensuing (one-dimensional) simple harmonic motion is $ T$. At what time is the power delivered to the mass by the spring first a maximum?

a) $ t\equal{}0$
b) $ t\equal{}\frac{T}{8}$
c) $ t\equal{}\frac{T}{4}$
d) $ t\equal{}\frac{3T}{8}$
e) $ t\equal{}\frac{T}{2}$

I'd prefer hints first, please.

I think 123456789 posted the following equation in another thread, but here it is again:

$ \frac {d^2x}{dt^2} \equal{} \minus{}\frac {k}{m}x$

You can use it to get the harmonic motion velocity equation, and knowing that $ P \equal{} \frac{dE}{dt}$, where $ E$ is just the kinetic energy of the mass, you can do some differentiation (twice from the $ v$ equation) to find the maximum power. Compare that to the $ T$ formula and you should get $ \frac {T}{8}$. Dunno if that's the right answer.

HINTS
$ P\equal{}F \times v$
The motion is a S.H.M. $ F\equal{}\minus{}\omega^2 \times x$
$ P\equal{}max$ when $ \frac{dP}{dt}\equal{}0$

[hide="Solution"]
So, $ P\equal{}(\minus{}\omega^2 x)v$
max.power $ \implies$ $ \frac{dP}{dt}\equal{}0$
$ v^2 \plus{} xa \equal{} 0$
$ v^2 \equal{} {\omega}^2 x^2$ $ \implies$ $ \omega^2(A^2\minus{}x^2)\equal{}\omega^2 x^2$ which gives $ x\equal{}\frac{A}{\sqrt2}$ $ \implies$ $ t\equal{}\frac{T}{8}$
[/hide]

Really? Why doesn't this work?

[hide]
Assume that the displacement is given by $ x(t) \equal{} \sin(\omega t)$

Then $ v(t) \equal{} \omega \cos(\omega t)$ so $ K(t) \equal{} \frac{1}{2}m\omega^2\cos^2(\omega t)$.

Then $ P(t) \equal{} \frac{dK}{dt} \equal{} \minus{} \frac{1}{2}m \omega^3 \sin(2\omega t)$

This is maximized when $ \sin(2\omega t) \equal{} \minus{}1$ so $ 2\omega t \equal{} \frac{3\pi}{2} \implies t \equal{} \frac{3}{8}T$[/hide]

[quote="kitsune"]Really? Why doesn't this work?

[hide]
Assume that the displacement is given by $ x(t) \equal{} \sin(\omega t)$

Then $ v(t) \equal{} \omega \cos(\omega t)$ so $ K(t) \equal{} \frac {1}{2}m\omega^2\cos^2(\omega t)$.

Then $ P(t) \equal{} \frac {dK}{dt} \equal{} \minus{} \frac {1}{2}m \omega^3 \sin(2\omega t)$

This is maximized when $ \sin(2\omega t) \equal{} \minus{} 1$ so $ 2\omega t \equal{} \frac {3\pi}{2} \implies t \equal{} \frac {3}{8}T$[/hide][/quote]

Check your $ \frac {dK}{dt}$:

"Then $ P(t) \equal{} \frac {dK}{dt} \equal{} \minus{} \frac {1}{2}m \omega^3 \sin(2\omega t)$"

Perhaps the chain rule is appropriate here?

Sorry to go slightly off topic, but I took the F=ma in 2009, but don't have the problems - can any of you send the file to me? Thanks!

There's no pdf file available through the internet, but you can go to WebAssign at https://www.webassign.net/login.html?message=ended , and sign up for the F=ma exam through instructions on http://www.aapt.org/physicsteam/2010/registration.cfm (you dont need to pay to access the practice tests), and you will be able to access the 2009 exam on WebAssign

Blahhhh

$ P\equal{}dW/dt\equal{}dK/dt$.

$ K\equal{}\dfrac 12 mv^2$. $ x\equal{}A\cos(\omega t \plus{} \phi)$ and $ v\equal{}\minus{}A\omega \sin(\omega t\plus{}\phi)$. Clearly $ \phi\equal{}\pi/2$.
Then $ K\equal{}\dfrac 12 m A^2\omega \sin^2(\omega t\plus{}\pi/2)$. then plug in for A and take derivatives -> I get 3T/8...blah need to go to class now don't have time atm to finish typin

Here's how I attempted it and found $ t \equal{} \frac {T}{8}$. Can anyone point out the mistakes?

$ F \equal{} \minus{} kx \equal{} m\frac {d^2x}{dt^2}$

Let $ x \equal{} x_{max}sin(At)$, such that $ v \equal{} Ax_{max}cos(At)$. Differentiating again and substituting into the force equation gives $ A \equal{} \sqrt {\frac {k}{m}}$

So $ v \equal{} x_{max}\sqrt {\frac {k}{m}}cos(\sqrt {\frac {k}{m}}t)$

$ KE \equal{} \frac {1}{2}mx_{max}^2\frac {k}{m}cos^2(\sqrt {\frac {k}{m}}t)$

Differentiating with respect to time gives $ P \equal{} mx_{max}^2\frac {k}{m}cos(\sqrt {\frac {k}{m}}t)sin(\sqrt {\frac {k}{m}}t)\sqrt {\frac {k}{m}}$

Differentiating again with respect to time and setting it to 0 gives (I'll leave out the messy details) $ t \equal{} \frac {\pi}{4}\sqrt {\frac {m}{k}}$. Since $ T \equal{} 2\pi\sqrt {\frac {m}{k}}$, $ t \equal{} \frac {T}{8}$.

Comments?

Yeah, we understand that the answer is $ \frac{3T}{8}$, but the question is HOW?

The magnitudes at 3T/8 and T/8 are the same. However, at T/8 the force is in the opposite direction of velocity, so the spring provides negative power.


Assume $ 0 < r < 3$.  Below are five equations for $ x$.  Which equation has the largest solution $ x$?

$ \textbf{(A)}\ 3(1 \plus{} r)^x \equal{} 7\qquad \textbf{(B)}\ 3(1 \plus{} r/10)^x \equal{} 7\qquad \textbf{(C)}\ 3(1 \plus{} 2r)^x \equal{} 7$
$ \textbf{(D)}\ 3(1 \plus{} \sqrt {r})^x \equal{} 7\qquad \textbf{(E)}\ 3(1 \plus{} 1/r)^x \equal{} 7$

All the equations are of the form $ 3(1\plus{}k)^x \equal{} 7$, so we seek to minimize $ k$.

Obviously $ \dfrac{r}{10} < r < 2r$

$ \dfrac{r}{10} < \sqrt{r} \longrightarrow r < 100$, which is within the constraints.

$ \dfrac{r}{10} < \dfrac{1}{r} \longrightarrow r^2 <10$, which is true since $ 0 < r<3$

Thus $ k \equal{} \dfrac{r}{10}$

$\textbf{B}$

[hide="Observation"]
Someone who wished to guess could get this down to two possibilities without any thinking at all - if $ r\equal{}1$, within the constraint, choices A, D, and E become identical, so clearly none of them can be the unique minimum.

Not a well-thought out problem in my opinion.
 [/hide]

@above
Exactly how I did the problem... It was easy to choose between B and C after that.

[hide]
Since all of the equations have 3 as a coefficient, we want to minimize whatever is in the parentheses to maximize x.
Clearly, out of A, B, and C, B is the least. For r<10, we see that B<D,E. Since r<3, $ \boxed{B}$ is the answe.r
[/hide]


geometry
geometric transformation
reflection
trigonometry

When reflects the graph across the x-axis and when reflects the graph across the y-axis?

[quote="cdymdcool"]When reflects the graph across the x-axis ?[/quote]

When you sketch the graph of $y=|x|,\ |x^{2}-1|$ e.t.c.

[quote="cdymdcool"]When reflects the graph across the y-axis ?[/quote]

When you sketch the graph of $|y|=x+1$ e.t.c.

When you sketch the graph of $|x|+|y|=1$, you can reflect across the $x$ and $y$ axis simultaneously.

What kunny says will give you both the graph and its reflection at the same time

if you want only the reflection then with a graph $f(x)$

$f(-x)$ will give you the same graphed flipped over the y axis

$-f(x)$ will give you the same graph flipped over the x axis

$\sin (-\theta ) =-\sin \theta$
$\tan (-\theta ) =-\tan \theta$
$\csc (-\theta ) =-\csc \theta$
$\cot (-\theta ) =-\cot \theta$

take a look at the graph

$f(x) = \sqrt{x}$ - It starts at the origin and goes up and to the right

$f(-x) = \sqrt{-x}$ is flipped over the y axis

It starts at the origin and goes up and to the left

$-f(x) =-\sqrt{x}$ is flipped over the x axis

It starts at the origin and goes down and to the right

If you wanted to flip it over both axes:
$-f(-x) =-\sqrt{-x}$ would go down and left

hope this example helps


There are infinitive many:

$ n\equal{}p^9$ has 10 divisors for every prime p.
Since there are infinitive many prime numbers you have infinitive many n with 10 divisor

[quote="guile"]If d(n) is the number of postive divisors of n, find all integers such that d(n) = 10.[/quote]

[hide]Use the "prime factorize, add one to each exponent, multiply them all together" formula.

All such integers are either $ p^9$ or $ p_1^1 p_2^4$[/hide]

I know.
I just wanted to show answer is infinitive many.


Lightning strikes a point 2 miles away and you hear thunder 15 seconds later. Sixty seconds after seeing a second bolt of lightning, you hear the thunder. How many miles away did the second lightning bolt strike?

Proportions: If the time is quadrupled, the distance is also quadrupled (rate is unchanged). Therefore 8 miles is the answer.

[b]The proportion is basically 2/15 is equal to x/60. You can just cross multiply and get 120 is equal to 15x. Divide 120 by 15 to get 8[/b]