trigonometry quadratics function inequalities conics parabola algebra

If we are to model the arrow motion as the usual projectile motion, then we could find values for the arrow's initial speed and angle of shooting  in order for the shooting angle to be $ 0^{o}\leq\theta < 90^{o}$. Targeting bellow the aim wouldn't make for sure the arrow hit it. Unless I am not understanding what you mean by "aim above the target to hit it or below".

Well, when you're shooting at something that is in the same horizontal line as you, you have to aim a bit higher in order to hit it. Similarly, if you're shooting something in the same line as you, but uphill, you have to aim higher up to hit it. My teacher says that to hit something downhill, you have to aim below the target to hit it. Although I think what he might be doing is setting his aiming system on his bow a little higher than what he's aiming at, so what it looks like to him is that he's aiming below the target. I don't know..

Consider hitting below the target, and ignore the effects of gravity at the moment such that the horizontal and vertical speeds remain constant. The arrow would then travel in a straight line and hit below the target. If we now consider gravitational forces acting, then it would be dragged even further down, hitting below the point we aimed at (which in turn is below the target). So this definetely won't work.

What your teacher may mean, is that you have to aim below the horizontal (but still above the target), but this is not a requirement (as I will show later). Your teacher may be talking about "reality", where an updraft may act on the arrow, bringing it up into horizontal position. Perhaps you have to aim lower to compensate for such an updraft.

Anyway, consider an archer standing at the origin, and the target is at a point $ (X,Y)$ in the $ (x,y)$ plane. The axes are chosen such that $ X\geq 0$. The $ x$-axis is horizontal so we will use a small $ x$ to represent horizontal components of vectors. A similar notation is used for the vertical $ y$-axis, with the upwards direction chosen as positive. The arrow is fired at a velocity $ u$ at an anticlockwise angle of $ \theta$ from positive $ x$-axis. Because $ X$ is positive, we need only consider $ -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$. Let the time of flight be $ t$. Our aim is to find the angle of release that the projectile requires to hit the target.

The initial horizontal component of velocity is $ u_{x}= u\cos{\theta}$. The horizontal component of velocity remains constant throughout its motion.

$ s_{x}= u_{x}t\Rightarrow t =\frac{s_{x}}{u_{x}}=\frac{X}{u\cos{\theta}}$

The initial vertical component of velocity is $ u_{y}= u\sin{\theta}$. The vertical acceleration is $ a_{y}=-g$.

$ s_{y}= u_{y}t+\frac{1}{2}a_{y}t^{2}$

$ \Rightarrow Y = u\sin{\theta}\,\left(\frac{X}{u\cos{\theta}}\right)-\frac{1}{2}g\left(\frac{X^{2}}{u^{2}\cos^{2}{\theta}}\right)$

$ \Rightarrow Y = X\tan{\theta}-\frac{gX^{2}}{2u^{2}}\sec^{2}{\theta}$

$ \Rightarrow Y = X\tan{\theta}-\frac{gX^{2}}{2u^{2}}(\tan^{2}{\theta}+1)$

$ \Rightarrow\frac{gX^{2}}{2u^{2}}\tan^{2}{\theta}-X\tan{\theta}+\frac{gX^{2}}{2u^{2}}+Y = 0$

This is a quadratic equation in $ \tan{\theta}$.

Lets break off at a tangent (both literally and metephorically!) justnow and show that this equation indeed implies that you cannot hit the target, if you aim lower than the target.

$ -1\leq\cos{\theta}\leq 1\Rightarrow 0\leq\cos^{2}{\theta}\leq 1\Rightarrow\sec^{2}{\theta}\geq 1$

$ \Rightarrow\frac{gX^{2}}{2u^{2}}\sec^{2}{\theta}> 0$ assuming $ X > 0$

$ \Rightarrow Y < X\tan{\theta}$

$ \Rightarrow\frac{Y}{X}<\tan{\theta}$

But $ \frac{Y}{X}=\tan{\alpha}$ where $ \alpha$ is the anticlockwise angle that the line joining the origin to the point $ (X,Y)$ makes with the positive $ x$-axis.

$ \Rightarrow\tan{\alpha}<\tan{\theta}$

Since the tangent function is strictly increasing in the range $ -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, then $ \alpha <\theta$. Hence the angle that we fire at to hit the target should be greater (or more anticlockwise) than the angle that the arrow makes if we aim for (or below) the target. The exception to this is when $ X = 0$: the target is directly above or below, so you must fire in the direction of the target. If you fire at an angle more anticlockwise than the bearing of the target, then you will miss.

Now we find the discriminant of the quadratic in $ \tan{\theta}$, $ \Delta$:

$ \Delta = X^{2}-4\left(\frac{gX^{2}}{2u^{2}}\right)\left(\frac{gX^{2}}{2u^{2}}+Y\right)$

$ = X^{2}\left(1-4\left(\frac{g}{2u^{2}}\right)\left(\frac{gX^{2}}{2u^{2}}+Y\right)\right)$

$ = X^{2}\left(1-\left(\frac{g^{2}X^{2}}{u^{4}}+\frac{2gY}{u^{2}}\right)\right)$

$ =\frac{X^{2}}{u^{4}}(u^{4}-g^{2}X^{2}-2gYu^{2})$

$ =\frac{X^{2}}{u^{4}}(u^{4}-2gYu^{2}-g^{2}X^{2})$

So using the quadratic formula:

$ \tan{\theta}=\frac{X\pm\sqrt{\frac{X^{2}}{u^{4}}(u^{4}-2gYu^{2}-g^{2}X^{2})}}{\frac{gX^{2}}{u^{2}}}$

$ \Rightarrow\tan{\theta}=\frac{X\pm\frac{X}{u^{2}}\sqrt{u^{4}-2gYu^{2}-g^{2}X^{2}}}{\frac{gX^{2}}{u^{2}}}$

$ \Rightarrow\tan{\theta}=\frac{1}{gX}\left({u^{2}\pm\sqrt{u^{4}-2gYu^{2}-g^{2}X^{2}}}\right)$

$ \Rightarrow\theta =\tan^{-1}{\left(\frac{1}{gX}\left({u^{2}\pm\sqrt{u^{4}-2gYu^{2}-g^{2}X^{2}}}\right)\right)}$

Completing the square under the square root gives a - slightly - "nicer" expression:

$ \theta =\tan^{-1}{\left(\frac{1}{X}\left(\frac{u^{2}}{g}\pm\sqrt{\left(\frac{u^{2}}{g}-Y\right)^{2}-(X^{2}+Y^{2})}\right)\right)}$

Now lets get down to the interesting part of analysing what this equation means.

Firstly, if we we want to hit the target the angle must be real. So the what is under the square root, or the discriminant in the messier expression (which we shall use) is greater than or equal to 0.

$ \Rightarrow u^{4}-2gYu^{2}-g^{2}X^{2}\geq 0$

If this inequality is not satisfied then the arrow will miss the target.

If $ u^{4}-2gYu^{2}-g^{2}X^{2}= 0$ then there is only one "input". So only one angle which you can use to hit the target at. This angle is $ \tan^{-1}{\left(\frac{u^{2}}{Xg}\right)}$

The $ tan^{-1}$ function is one to one. So no two different inputs will give the same angle.

Hence if $ u^{4}-2gYu^{2}-g^{2}X^{2}> 0$ then there will be two possible angles which you use to hit the target. The tan of these two angles (or the gradients of the tangents to the parabola just after release) always average to $ \frac{u^{2}}{Xg}$. Note the independance of $ Y$ here.

It we want a positive angle, we need a positive input. But $ u^{2}+\sqrt{u^{4}-2gYu^{2}-g^{2}X^{2}}> 0$ so there is always a positive angle solution. So every target below the horizontal can be hit, by just firing above the horizontal. So it is never a necessity to aim below the horizontal.

Wow. Thanks so much, that was amazing  :o

[quote="Icy"]If you're standing on a cliff, and there is a target below and in front of you, do you have to aim above the target to hit it or below? My teacher says you have to aim below the target to hit it, and I don't know about that. That doesn't make sense.[/quote]

You would have to shoot slightly higher than the target because of acceleration due to gravity (projectile motion). If you shoot below, the arrow will got even farther below your target.


When $x+y+z=1$, $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\geq\frac{3}{2}$ by Nesbitt Inequality. :?

[quote="silouan"]If $x,y,z>0$ with $x+y+z< 1$ prove that

$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}\leq\frac{3}{2}$[/quote]
sorry.I think it is wrong....

I want from the moderators to delete all the posts . Wrong problem, sorry :blush:

[quote="silouan"]If $x,y,z>0$ with $x+y+z< 1$ prove that

$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}\leq\frac{3}{2}$[/quote]
What if when $x\geq\frac{3}{5}$? :?


$ P$ is inside triangle $ ABC$. Prove that at least one of $ \angle PAB, \angle PBC, \angle PCA$ has measure less than or equal to $ 30^\circ$.

[hide="Solution"]

Define $ A^\prime, B^\prime, C^\prime$  as the projections of $ P$ onto $ BC, CA, AB$, respectively.

Assume for sake of contradiction that $ \angle PAB, \angle PBC, \angle PCA > 30^\circ$.

Notice that $ \angle ABC > \angle PBC > 30^\circ$ and $ \angle BCA > \angle PCA > 30^\circ$. Then $ \angle ABC \plus{} \angle BCA > 60^\circ \Rightarrow \angle CAB < 120^\circ$. But since $ \angle PAB < \angle CAB$, we have $ \angle PAB < 120^\circ$.

Similarly, $ \angle PBC < 120^\circ$ and $ \angle PCA < 120^\circ$.

Since the angles have measures between $ 30^\circ$ and $ 120^\circ$, we have $ \sin \angle PAB, \sin \angle PBC, \sin \angle PCA > \frac{1}{2}$.

This implies that $ \frac{PC^\prime}{PA}, \frac{PA^\prime}{PB}, \frac{PB^\prime}{PC} > \frac{1}{2}$.

Then $ 2PC^\prime > PA, 2PA^\prime > PB, 2PB^\prime > PC$.

$ \Rightarrow 2(PA^\prime \plus{} PB^\prime \plus{} PC^\prime) > PA \plus{} PB \plus{} PC$

But by the Erdös-Mordell Inequality, since $ P$ is inside $ ABC$, we must have

$ PA \plus{} PB \plus{} PC \ge 2(PA^\prime \plus{} PB^\prime \plus{} PC^\prime)$

so we have reached a contradiction.

Therefore, at least one of $ \angle PAB, \angle PBC, \angle PCA$ has a measure that is at most [hide="thirty degrees"]

"Be forte, grumpy lemon!" is an anagram of "Fun geometry problem!"

[img]http://sebrinaa.ca/wp-content/uploads/lemon-300x297.jpg[/img]

[/hide]

[/hide]

Man that is your favorite inequality isn't it...

One of them :D
I would compare it to another one but karel would probably destroy me for it.

[hide="Solution Outlline"]
Ceva and Jensen QED.[/hide]

This was an IMO #5, wasn't it? And wasn't like #4 the hardest one on the second day of that imo?


1)Let a and b real numbers.Find x values satisfy that
 a.|x-3|+b.|3-x|=2003
2)Find x natural numbers that satify
  x^2+72=n^2 where n integer.

For number 2, any x in the form:

$x= \frac{2^{2n}3^{2m}-18}{2^{n}3^m}$

with $0 \leq n,m \leq 2$    with $n,m$ integers

will work.

$x^2+72=n^2$
$(n+x)(n-x)=72$

Pick m such that m divides 72.... ie $2*2*2*3*3$

$n+x=m$

$n-x= \frac {72}{m}$     Subtracting these:

$2x= m- \frac{72}{m}=\frac{m^2-72}{m}$

$x= \frac{m^2-72}{2m}$

Therefore m is even, ie, since m divides $2*2*2*3*3$
it is in the form

$2.2^n.3^m=2^{n+1}.3^m$ with $0 \leq m,n \leq 2$

Then $x= \frac{2^{2n}.3^{2m}-18}{2^n.3^m}$ with $0 \leq n,m \leq 2$

Subbing in all the values for n,m, we get the answers:

$x=3, 17, 10.5, 35.5$ all of which give a rational n, with the halves creeping in from when we divided by 2 back up the way. So the solutions are x=3, 17 (i hope)

2
[hide]
I got $x = {3, 7, 17}$

First, I notice the difference between two squares is even.  Knowing the difference between the squares of any two consecutive integers is always odd, the two numbers cannot be consecutive.  Furthermore, the difference between the two numbers must be even.  This is because the difference between their squares is the same as the sum of the differences of consecutive squares between them.  Since the difference of consecutive squares is odd, for the difference between the squares of two integers to be even, their difference must be even.  (Yeah, I couldn't think of how to explain that well.)

So basically I found 72 had to be the sum of an even amount of consecutive odd integers.  Once I found a set of those, I would look at the smallest one, and find out which two consecutive squares would give me that number, and that would give me a value of x.

First I found two consecutive odd integers whose sum was 72.  35 and 37.  Now I need to derive a formula for how to find $a$ and I know $(a+1)^2 - a^2$.  Looking at lower numbers, such as 1 and 2, and 2 and 3, I get $a = \frac{d-1}{2}$ where d is the difference between the consecutive squares.  Using this for 35, a working value of x is 17.

Now I find four consecutive odd integers: 15, 17, 19, 21.  Using 15, another value of x is 7.

Six consecutive odds: 7, 9... 17.  Plugging in 7, a third value of x is 3.

There are no eight consecutive odd integers whose sum is 72.  But we know 3 is the smallest value of x possible.  Plugging 3 into the original equation, $9 + 72 = n^2 = 81$  There are no perfect squares less than 81 but greater than 72 thus 3 is the smallest possible value (well, integer) for x.

So the solution is ${3, 7, 17}$
[/hide]

Please see first question
1)Let a and b real numbers.Find x values satisfy that
a.|x-3|+b.|3-x|=2003

[quote="mathmaniack"]1)Let a and b real numbers.Find x values satisfy that
 a.|x-3|+b.|3-x|=2003.[/quote]

For $a+b\neq 0$,  $x=3\pm \frac{2003}{a+b}$

For $a+b=0$, we have no roots.


calculus
integration
real analysis
real analysis unsolved

Evaluate $ \displaystyle\int^\infty_0 \frac{x}{e^x\plus{}1}\,dx$ if it is known that  $ \displaystyle\sum_{n\equal{}1}^{\infty}\frac{1}{n^2}\equal{}\frac{\pi^2}{6}$.

$ \frac{x}{1\plus{}e^x}\equal{}e^{\minus{}x} \frac{x}{1\plus{}e^{\minus{}x}}\equal{}e^{\minus{}x} x \sum ^ \infty_{n\equal{}0} (\minus{}1)^n e^{\minus{}n x}\equal{} \sum^\infty _ {n\equal{}0} (\minus{}1)^n \frac{x}{e^{x(n\plus{}1)}}$

We have:

$ \int^\infty_0\frac{x}{1\plus{}e^x} \,dx \equal{} \sum^\infty _ {n\equal{}0} (\minus{}1)^n \int^\infty_0 \frac{x}{e^{x(n\plus{}1)}} \,dx$

Using substitution ($ x(n\plus{}1) \to t$) and integration by parts:

$ \int^\infty_0 \frac{x}{e^{x(n\plus{}1)}} \,dx \equal{} \frac{1}{(n\plus{}1)^2}\int^\infty_0 \frac{t}{e^t} \,dt \equal{} \frac{1}{n^2\plus{}1}$

So:

$ \int^\infty_0\frac{x}{1\plus{}e^x} \,dx \equal{} \minus{} \sum^\infty _ {n\equal{}1} \frac{(\minus{}1)^n}{n^2}$

Also, we have:

$ \sum^\infty_{n\equal{}1}\frac{1}{n^2}\plus{}\sum^\infty_{n\equal{}1}\frac{(\minus{}1)^n}{n^2} \equal{}$

$ \equal{}\sum^\infty_{n\equal{}1}\frac{1\plus{}(\minus{}1)^n}{n^2} \equal{}$

$ \equal{}\sum^\infty_{n\equal{}1}\frac{2}{(2n)^2} \equal{}$

$ \equal{}\frac{1}{2} \frac{\pi^2}{6} \Rightarrow \sum^\infty_{n\equal{}1}\frac{(\minus{}1)^n}{n^2}  \equal{} \minus{}\frac{\pi^2}{12}$

And finally:

$ \int^\infty_0\frac{x}{1\plus{}e^x} \,dx \equal{} \frac{\pi^2}{12}$


inequalities
geometry
function
circumcircle
inequalities solved

We have a quadrilateral ABCD with AB=a,BC=b,CD=c,AD=d, S is its area
and p=(a+b+c+d)/2.
Do we have or not the inequality

3*sqrt(S)<=p+sqrt4(a*b*c*d) ?

(where sqrt(S) is the square root, and sqrt4 the order 4 root).

just a quick notice: you can use the notation abcd for the product of a,b,c and d rather than a*b*c*d. :)

Just another quick notice:

http://www.mathlinks.ro/phpBB/viewtopic.php?t=58

......

cheers! :D :D

IMHO the ineq. should be not true.

If not, it should be valid also for d-->0, id est when the quadrilateral becam a triangle.
In this limit case we should have that

3 \sqrt S  \leq p.

And this is not generally true.

Remark- If the quadrilateral has overall the angles 90, the inequality is trivially true.

it's curious because the inequality is NOT homogen ...

I'm sorry, but don't understand the observation! Could you explain please what do you mean? Exactly which step of the reasoning do you think is not valid?

Try to be more rigorous (maybe an epsilon-delta proof)

The basis of my reasoning is that "S", "p" and "abcd" are continous functions of, for istance, d. Then f(d)=3*sqrt(S)-p+(a*b*c*d)^(1/4) is a continous function of d. If f(0)>0 then f(d)>0 at least in a "little" interval of 0.
To find a case where f(0)>0 considere a triangle whith a=b=c.

Is this what do you mean whith "delta-epsilon proof"?

My conclusion is that, for istance, a=b=c=1 and d= 0,01 should be a counterexample.

[quote="xxxxtt"]That "SHOULD BE" a counterexample or "IS" a counterexample?[/quote]

Shoul be IS :) !

Actually i think you may be wrong.
The problem is with the area of the quadrilateral.First of all, it cannot depend exclusively on the sides a,b,c,d it must be a function of the sides and also angles (like in the formula with cos B+D/2).
And another thing-if you have a,b,c and B+D constant how can you make
d->0 ? you must prove that there exist quadrilaterals with that constant elements and d->0.
I think the problem remains open.

<< Actually i think you may be wrong. >>
May be. But I don't see where.

<<The problem is with the area of the quadrilateral.First of all, it cannot depend exclusively on the sides a,b,c,d it must be a function of the sides and also angles (like in the formula with cos B+D/2). >>

I know this, but doesn't matter.

<< And another thing-if you have a,b,c and B+D constant how can you make d->0 ? you must prove that there exist quadrilaterals with that constant elements and d->0.  >>
This is, of course, non possible. But I never said that. I only said that f(.) is a continous function whith respect to d.
Let me tray to make in a different (I hope more convincing) way my idea(I have very big difficulty whith english, but I hope it will works).  Given trhee fixed point A,B and C and given an other point D on the arc(AC) of the circumcircle c(ABC). Let's pose AB=a, BC=b, CD=c and DA=d. In this case, given that a=b=constant, we have that f(.) is funcion of c and d but c is continuos function of d then so is S, p and a*b*c*d.

If the main statment  be true it should be true for each position of D on the above given construction.

But we can easily see that for AB=10^4AD the thesis is not valid.

<< I think the problem remains open. >>

Please, try to calcute the value of f(.) when AB=BC=CA=1 and D is on the parallel from A to BC so that AB=10^4AD.


LaTeX
number theory proposed
number theory

Please post your problems directly into the Forum, using $ \text{\LaTeX}$ (see http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 ).


Find all pairs of positive integers $(a, b)$ such that $5a^{b} - b = 2004$.

when $b=1$ , $a=401$

clearly $a=1$ yields no solution . So when $a,b\geq 2$ , by Bernoulli Ineq

$2004=5a^b-b\geq 5\cdot 2^b-b=5\cdot (1+1)^b-b\geq 5(1+b)-b=5+4b$

$\Longleftrightarrow 499.75\geq b$ so $499\geq b\geq 2$

Hence this also gives us $5a^b=2004+b\leq 2004+499=2503$ $\Longleftrightarrow a^b\leq 500$

So $500\geq a^b\geq 2^b$ gives us $b\leq 8$

Now taking mod 5 at the original equation , we see that $b\equiv 1\mod 5$ .So we only have $b=6$ (since $2\leq b\leq 8$) . But checking this value yields no integer solution for $a$ . Hence the only solution are

$(a,b)=(401,1)$

 :)


$H(x) = \frac{x^3-1}{x^2-9}$

Step 5.  Locate the horizontal or oblique asymptotes, if any, using the procedure given earlier.  Determine points, if any, at which the graph of H intersects these asymptotes.

The answer that the book gives is:
5.  Oblique asymptote: $y=x$, intersected at $(0, \frac{1}{9})$

I agree with the equation of the asymptote, but if the intersection is on $y=x$, shouldn't the point be at $(\frac{1}{9}, \frac{1}{9})$?

i would agree that the intersection of the oblique asymtote and the function is at $(\frac{1}{9},\frac{1}{9})$.  i guess it was a misprint in the book. who knows.


I'm trying to configure the TeXnicCenter thing and it says "Enter the full path of the directory, where the executables (latex, tex, etc.) of your TeX distribution are located:"
I searched previous topics with this and they said to type in "C:\texmf\MiKTeX\bin\", and then it said "The application 'LaTeX' has not been found in the directory of your TeX-distribution."
What do I do?  :huh:

It looks like you have installed LaTeX in a different directory. Look for the directory containing the file latex.exe.


algebra
polynomial
number theory unsolved
number theory

Is it true that any polynomial $ f\in\mathbb{Z}[x]$ that has a zero modulo $ n$ for any positive integer $ n$ must have a rational zero?

Nice! :) I'm proving there exists a polynomial $ f(x)\in\mathbb{Z}[x]$ such that, for all primes $ p\in\mathbb{N}$, $ f(x) \equal{} 0$ is solvable in $ \mathbb{F}_{p}$, but $ f(x) \equal{} 0$ isn't possible over the rational field. So admit $ p\in\mathbb{N}$ is an arbitrary prime. Then $ x^{2}\equal{}\minus{}1$ is solvable in $ \mathbb{F}_{p}$ iff $ p\equiv 1\bmod p$ or $ p \equal{} 2$. Hence suppose $ p\equiv 3\bmod 4$. Then there is $ x_{0}\in\mathbb{F}_{p}$ such that either $ x_{0}^{2}\equal{}\minus{}2$ or $ x_{0}^{2}\equal{} 2$, since $ \left(\frac{\pm 2}{p}\right) \equal{}\pm\left(\frac{2}{p}\right)$. Thus take $ f(x) \equal{} (x^{2}\plus{}1)(x^{4}\minus{}2)$.

No, try $ (x^{2}\minus{}2)(x^{2}\minus{}17)(x^{2}\minus{}34)$ or $ (x^{2}\plus{}3)(x^{3}\minus{}7)$. I proved that the latter is minimal in respect of the degree in http://www.mathlinks.ro/Forum/viewtopic.php?t=110231 .

Edit: @s.tringali: there is no solution $ \mod 4$ (you just considered prime moduli, but we are asked for all positive integer moduli).

[quote="ZetaX"](you just considered prime moduli, but we are asked for all positive integer moduli).[/quote]
but if $ f$ has a zero modulo $ p$ for any prime $ p$ then if has a zero modulo $ n$ for any positive integer $ n$.

No it has not, I already said that his example has no solution $ \mod 4$.
You miss some prime powers.


function
analytic geometry
graphing lines
slope

Let $ f$ be a linear function for which $ f(6)\minus{}f(2)\equal{}12$. What is $ f(12)\minus{}f(2)$?

$ \textbf{(A)}\ 12 \qquad
\textbf{(B)}\ 18 \qquad
\textbf{(C)}\ 24 \qquad
\textbf{(D)}\ 30 \qquad
\textbf{(E)}\ 36$

[hide="Solution"]
Because $ f$ is linear, its slope is constant.  Thus, $ \frac{f(6)\minus{}f(2)}{6\minus{}2}\equal{}\frac{f(12)\minus{}f(2)}{12\minus{}2}$, so $ 3\equal{}\frac{f(12)\minus{}f(2)}{10}$, so $ f(12)\minus{}f(2)\equal{}30$, corresponding to answer choice $ \textbf{D}$.
[/hide]

I think you have to find a linear function which satisfies f(6)-f(2)=12.
My linear function is y=3x.
Let x=6.
f(6)=18
Let x=2.
f(2)=6.
Thus, f(6)-f(2)=12.
Hence; f(12)-f(2)=30

Hello Shan remember that a linear function can have a constant term

so the function f can be $ 3x \plus{} c$ where c is any constant .

It does not aaffect ur solution but i justed wanted to clarify with u
Thanks,

Befuddlers

Sorry Befuddler,
I was going to put in the calculations to get the slope of the function but didn't as I found it unnecessary to the final solution.
Thanks for the input anyway :)

[hide="Another solution"]Let $ f(x) \equal{} ax \plus{} b$ for $ a,b\in\mathbb{R}$.
$ f(6) \minus{} f(2) \equal{} 6a \minus{} 2a \plus{} b \minus{} b \equal{} 4a \equal{} 12\implies a \equal{} 3$
$ f(x) \equal{} 3x \plus{} b$
$ f(12) \minus{} f(2) \equal{} 36 \minus{} 6 \plus{} b \minus{} b \equal{} 30\implies\boxed{\text{D}}$[/hide]

[hide=solution]Because we know that f is a linear function, we say that $f(x)=ax+b,$
So, $$f(6)-f(2)=12$$
$$6a+b-2a+b=12$$
$$4a=12$$
$$a=3$$
So, $f(12)-f(2)=12a+b-2a+b=10a=\boxed{30}$[/hide]

[quote=EWenslawski][hide]30[/quote]

FTFY

Btw, this was a question in 2009 lol

[quote=franzliszt]This was actually from a AMC test before 2006[/quote]

I mean asked and also for the “official” AoPS solution is [url=https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_9]here


solve this equation in $ R$:  $ x^4\minus{}3x^3\minus{}1\equal{}0$

[quote="thanhnam2902"]solve this equation in $ R$:  $ x^4 \minus{} 3x^3 \minus{} 1 \equal{} 0$[/quote]

Rewrite the given equation as

$ x^4\plus{}2\left(\minus{}\frac32x\plus{}r\right)x^2\plus{}\left(\minus{}\frac32x\plus{}r\right)^2\equal{}\left(\minus{}\frac32x\plus{}r\right)^2 \plus{}2rx^2\plus{}1$

$ \implies \left(x^2\minus{}\frac32x\plus{}r\right)^2\equal{}\left(\frac94\plus{}2r\right)x^2 \minus{}3rx\plus{}r^2\plus{}1$

Now select the real $ \ r$ such that RHS is a perfect square$ \ \implies$ Discriminant of RHS is zero

$ \implies 9r^2\minus{}4\left(\frac94\plus{}2r\right)\left(r^2\plus{}1\right)\equal{}0$

$ \implies 8r^3\plus{}8r\plus{}9\equal{}0$

Now we have to solve this cubic resolvent.

hello, the explicit formulas are very complicated, see here
http://www.wolframalpha.com/input/?i=x%5E4-3x%5E3-1%3D0
Sonnhard.


group theory
abstract algebra
superior algebra
superior algebra open

Given subgroup G of group of permutations of n elements.
What is a minimum power of set U, such that there exists an element X of U^n ( that is X=(x_1, x_2, ... , x_n) ):
for every element g (g!=e) from G: g(X)!=X.

P.S.: sorry, if there are some mistakes in problem specification.

What do you know about $ G$ (if anything?)


AMC
AIME
USA(J)MO
USAMO
geometry
3D geometry
AMC 10

Well, apparently I think this year will be lower than last year.  Last year it was lower than usual because they switched to 1.5 points for blank answers.  So I would assume without calculators, the top 5% should drop a bit more.  I've heard as low as 93, but I was thinking 96 or 96.5.  Obviously since 96.5 isn't even a viable score, 96 is the next best.  What's your guys' take?

As for AMC 10?  I don't really know.  I heard it was fairly easy from some, while others struggled.

Either way, thanks.

Yeah, I wouldn't be surprised to see it at 96, or even 94.5.  People seemed to struggle with this one quite a bit.

My friend was top 6th 11th grader in my county.  He got a 99 on the AMC 12A.  This is probably a fairly good indicator that, since top 6 11th graders is certainly top 5%, at least in my county, that the cutoff score is at most a 99.

My guess is a 96.  Last year, I think the floor was 97.5 for the 12A (somebody check that?), and I thought this one was harder.

And no, although you can take both AMCs, you cannot take both AIMEs.

Yea, the cutoff last year was 97.5 for AMC 12.

ok thx krsattack. I really hope the calculator rule will "lower" the USAMO cutoff. :roll:

i thought this test had some really nice problems. I made a stupid mistake and the cube one, but i still managed to get a 99. considering the problems i managed to do in like 5 minutes when i got home i should have gotten a 114.  :mad:
i guess this is an ok score for my first time ever taking an AMC, considering im in middle school. i don't think the 12 was too difficult, except the last 2 or so problems were beast.

[quote="rem"]By the way, can we take both AIMEs if we make it through?[/quote]

No, you may take only one and no more.


[Moderator edit: This is a current contest problem. Cheating!]

For positive reals a,b,c and natural n proof that

$ \frac {a^{n \plus{} 1}}{b \plus{} c} \plus{} \frac {b^{n \plus{} 1}}{c \plus{} a} \plus{} \frac {c^{n \plus{} 1}}{a \plus{} b} \ge ( \frac {a^{n}}{b \plus{} c} \plus{} \frac {b^{n}}{c \plus{} a} \plus{} \frac {c^{n}}{a \plus{} b})(\sqrt [n] {\frac {a^{n} \plus{} b^{n} \plus{} c^{n}}{3}})$

STOP!
it's the problem of 61st Polish National Mathematical Olimpiad!

don't chelp contest cheater


trigonometry
geometry
pigeonhole principle

I was just wondering if people could post so prove-it type problems at the beginner level, I feel pretty good with my algebra and trig (intermediate) but my proof writing skills are weak, and it doesnt seem that there are many challenges at this level to prove stuff.

Hmm... this one's fairly easy, once you know the technique:

Nine points are placed randomly inside a square with side length one.  Prove that a triangle exists with area  :le: 1/8 where the vertices are all among the nine points.

Hint: [hide]Pigeonhole Principle[/hide]

Here's one that has half solving, half proof.

You have some stamps (enough so that you won't run out).  They are of two values: 7 cents and 13 cents.  What is the largest integer amount of cents that you cannot make, and prove that that is the largest.

[quote="NAS"]I was just wondering if people could post so prove-it type problems at the beginner level.[/quote]

There isn't a lot of proof to be found at the Getting Started level, the way the mathematics curriculum is set up in nearly all the countries in which participants in this forum reside. In my mind, when the problem starts saying "Prove . . . " instead of saying "How many . . . " or "Solve . . . " is when you go from Getting Started to Intermediate in this site's set of forums. Of course, one can never get too many problems that involve proof, but I think it's remarkably hard to design a problem that really asks for a proof that really isn't a higher level problem.

Tip: one of the best places to look for problems that ask for proofs that really, truly are at the beginning level is in the popular books for children by Raymond Smullyan.

Take any high-school Geometry textbook and attempt to prove some of the easier theorems without peeking at the solutions.

I'm not sure you got that right - take a = b = c = 1.

I think it's

prove a2 + b2 + c2 :ge: ab + bc + ca

instead.

Oh, and a, b, c must be reals...

Yep.  That's the one.  Just ab, ac, bc, no two's

Beta, you shouldn't be working getting started problems.  And use spoiler, too.

1. Let a and b be two distinct positive real numbers, n a positive integer, and x0, x1, x2,... , xn integers. Suppose that (ax0)(ax1) ... (axn) = (bx0)(bx1) ... (bxn), and x1 + x2 + ... + xn = 1. Find x0.

(This may look hard but try it, it's really not)

2. Prove that :rt2: cannot be written as a fraction of integers, i.e. a/b.

3. Prove that |x+y| :le: |x| + |y|

4a. Prove that for any set of 3 integers, there is a subset of them whose sum is divisible by 3.

4b. Can you prove a similar result for sets of more integers? <-- this is a little harder than beginner's level

(If you've seen these before leave them for someone else)


search
number theory unsolved
number theory

prove that  (xy+1)(xz+1)(zy+1) is a perfect  square if and only if xy+1,xz+1,yz+1 are  perfect squares


Kunny, how much is ${1\over 4}\left( 1+{1\over 2}\right)$? :)


trigonometry
algebra
polynomial
arithmetic sequence
number theory
relatively prime
algebra unsolved

Does there exist an angle $ \alpha\in(0,\pi/2)$ such that $ \sin\alpha$, $ \cos\alpha$, $ \tan\alpha$ and $ \cot\alpha$, taken in some order, are consecutive terms of an arithmetic progression?

[hide="I think this works"]Let $ \sin \alpha \equal{} a, \cos \alpha \equal{} b$, so $ a^2 \plus{} b^2 \equal{} 1$. WLOG $ a < b$. Then the four terms are $ a, b, a/b, b/a$, with possible orders
1. $ a, b, a/b, b/a$
2. $ a, a/b, b, b/a$

In the second case, we have $ a/b \plus{} b/a \equal{} 2b$. But that implies $ 2b \geq 2$ and thus $ b \equal{} 1$, which is not in the domain.

In the first case, $ a\plus{}a/b \equal{} 2b$. So $ b \plus{} 1 \equal{} 2bb/a$ and $ b \equal{} 2/a \minus{} 1 \minus{} 2a$. Then $ b^2 \equal{} 4/a^2 \minus{} 4/a \minus{} 7 \plus{} 4a \plus{} 4a^2$, or $ 0 \equal{} 5a^4 \plus{} 4a^3 \minus{} 8a^2 \minus{} 4a \plus{} 4$.
We also have $ 2a/b \equal{} b \plus{} b/a$. Then $ 2a \equal{} 1/a \minus{} a \plus{} 1 \minus{} aa$ and $ 0 \equal{} a^3 \plus{} 3a^2 \minus{} a \minus{} 1$.

But these two polynomials are relatively prime, so no $ a$ exist.[/hide]

[hide]Since $ \sin (\alpha) \equal{} \cos (\frac{\pi}{2}\minus{}\alpha)$ and $ \tan (\alpha) \equal{} \cot (\frac{\pi}{2}\minus{}\alpha)$, WLOG, let $ \alpha \in (0, \frac{\pi}{4}]$.

It is easy to see that $ \alpha \equal{} \frac{\pi}{4}$ does not yield an arithmetic progression, so $ \alpha \in (0, \frac{\pi}{4})$.

Also, it is easy to see that the sequence cannot be constant, so WLOG, let the sequence be increasing.

For all $ \alpha \in (0, \frac{\pi}{4})$, we have:

$ \sin \alpha < \cos \alpha$

$ \sin \alpha < \tan \alpha$

$ \tan \alpha < \cot \alpha$

$ \cos \alpha < \cot \alpha$

So, the sequence in order is either:
$ \sin \alpha$, $ \cos \alpha$, $ \tan \alpha$, $ \cot \alpha$
OR
$ \sin \alpha$, $ \tan \alpha$, $ \cos \alpha$, $ \cot \alpha$

Let $ A \equal{} \sin \alpha$, and $ B \equal{} \cos \alpha$.

Then, the terms in order are either $ A, B, \frac{A}{B}, \frac{B}{A}$ OR $ A, \frac{A}{B}, B, \frac{B}{A}$.

In the 1st sequence $ B\minus{}A$ and $ \frac{B}{A} \minus{} \frac{A}{B}$ are both equal to the common difference.

In the 2nd sequence $ B\minus{}A$ and $ \frac{B}{A} \minus{} \frac{A}{B}$ are both equal to twice common difference.

Therefore:
$ B \minus{} A \equal{} \frac{B}{A} \minus{} \frac{A}{B}$

$ AB^2 \minus{} A^2B \equal{} B^2 \minus{} A^2$

$ (B\minus{}A)(AB) \equal{} (B\minus{}A)(B\plus{}A)$

$ B \equal{} A$ OR $ AB \equal{} A\plus{}B$

$ B \equal{} A$:
$ \cos \alpha \equal{} \sin \alpha$, so $ \alpha \equal{} \frac{\pi}{4}$, which, as previously stated, does not yield an arithmetic progression.

$ AB \equal{} A\plus{}B$:

Since $ 0< AB <1$: $ AB < \sqrt{AB}$

Since $ A\plus{}B > 0$: $ \frac{A\plus{}B}{2} < A\plus{}B$

And, by AM-GM: $ \sqrt{AB} \le \frac{A\plus{}B}{2}$

So, $ AB < \sqrt{AB} \le \frac{A\plus{}B}{2} < A\plus{}B$. Thus $ AB \neq A\plus{}B$.

Therefore, there are no such angles $ \alpha$ which yields an arithmetic progression.[/hide]

Oh, that's nicer than mine. Though a faster way to finish from [hide]$ AB \equal{} A \plus{} B$ is to reduce it to $ A(1\minus{}B) \plus{} B \equal{} 0$ and note that two positive numbers can't add to $ 0$.[/hide]

if there exists,let us suppose $0<a<45$
then $sina<cosa,tana<cota$
so $sina+cota=cosa+tana$
hence $a=45$
contradiction!


Find all n that satisfies the condition:
First digits of 2^n and 5^n are the same.

It is easy to show that if $2^n$ and $5^n$ have the same first digit then this digit is 3 ($2^5=32$, $5^5=3125$). But I don't know at the moment whether there are other appropriate values of $n$.


Suppose that the monic polynomial $ p(x) \equal{} x_n \plus{} a_{n \minus{} 1}x^{n\minus{}1}\plus{} ... \plus{} a_1x \plus{} 1$
has non-negative coefficients and $ n$ real roots. Show that $ p(2) \ge 3^n$. When
does equality occur?

I'm assuming you meant $ p(x) \equal{} x^n \plus{} a_{n\minus{}1} x^{n\minus{}1} \plus{} \ldots \plus{} a_1 x \plus{} 1$


Let $a,b\in N$ and $ab\neq 1$, find all integer values of $\frac{a^2+ab+b^2}{ab-1}$

I believe this was a problem in the Romanian selection tests 2004 or 2005. So, use the search button, please :)

Use infinite descent to bash it out to 3 o r 7 I think,.....

Bomb

The problems in TST's are many. So Arne could you give to us the exact link?Or post the solution here please?

thank you but the solution grbber gave is strange . can anyne explain it.Thank you


what u does it mean when a string is slack. Guys could u explain the last example prob from hc verma 8th chapter.(Particularly resolving of forces)

slacking means see when u try to move a stone tied to a thread with little force a goes in an incomplete circle and falls without completing it... the stretched condition of the thread changes and it is in a relaxed condition now that is slacking of rope or thread....

when string slacks does tension become zero. :D
[hide]
plz answer my doubt in hc verma
[/hide]

i think it shud become zero

[quote="befuddlers"]answer my doubt in hc verma
[/quote] :rotfl: what does that mean??

heys thats cool... give me solution to hcv can anyone ????
 :rotfl:  :rotfl:


does anyone know any where that i can find the USA unit of measurement of table?...like a site or somethin

Hey!!!

hm... not sure what you are asking... are you asking for like how many ounce is in a pound or do you want to know how many inches is in a meter?

o well, i found a site about both... (I just went to google):

http://convert.french-property.co.uk/#top


geometry
inequalities
trigonometry
inradius
circumcircle
area of a triangle
Heron\u0027s formula

[quote="conceive"]Why not to have all basics geometry theorems here? E.g. in section Inequalities there are all famous inequalities... Please post here basics geometry theorems, you know.[/quote]

OK, I can start. In right angled triangle with hypotenuse c is true, that $a^2+b^2=c^2$. This is calles The Pythagorean theorem. And its converse is also true.

[quote="conceive"]Why not to have all basics geometry theorems here? E.g. in section Inequalities there are all famous inequalities... Please post here basics geometry theorems, you know.[/quote]

I'll continue.

$HI^{2}=2R^{2}-4R^{2}\cos A\cos B\cos C$

[quote="conceive"]Why not to have all basics geometry theorems here? E.g. in section Inequalities there are all famous inequalities... Please post here basics geometry theorems, you know.[/quote]

Come on, people.

I don't want my posts to be alone in this forum...

[quote="conceive"]Why not to have all basics geometry theorems here? E.g. in section Inequalities there are all famous inequalities... Please post here basics geometry theorems, you know.[/quote]

Come on, people.

So, I continue. Let $[ABC]$ denote the area of given triangle ABC. Than it is true, that: $[ABC]=\frac{1}{2} bc.sin(\alpha)=\frac{1}{2}a.v_a=\sqrt{s(s-a)(s-b)(s-c)}$.

The last formula is called Heron's formula and it is valid for $s=\frac{a+b+c}{2}$.

I feel kinda dumb the theorems I know aren't anywhere near as complicated...  :?

C.P.C.T.C.-- Corresponding parts of congruent triangles are congruent.

Halves of equals are equal.

[quote="Davron"]The sum of angles of a triangle equals to 180 degrees and try to prove it please ...

Davron[/quote]

Davron, I don't want people to prove this here, my original intention was just a quick geometry survey.

Anyway, thanks for posting here..

Or some ineqaulities

$\tan A+\tan B+\tan C\geq 3\sqrt3$

dear conceive...

u did it good. ;)

but should the answers be also basic?

because i want to prove this theorem by inversion, and even projection. :!:

$OI^2=R^2-2Rr$
$HI^2=2R^2-4R^2\cos A\cos B\cos C$
$\tan A+\tan B + \tan C=\tan A\tan B\tan C$ for A,B,C angles in a triangle
Stewart's theorem: If P lies on BC such that $\frac{BP}{PC}=\frac{l}{m}$
then $mAB^2+lAC^2=mPB^2+lPC^2+(l+m)AP^2$

Ceva's theorem: if AD,BE,CF are concurrent cevians (from vertex to opposite side(possibly extended?)) then
$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1$

Menelaus: If D lies on BC, E on AC, F on AB then DEF is a line iff:
$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = -1$

$\cot {\omega} = \cot A + \cot B + \cot C$ where ${\omega}$ is the brocard angle

$Area ABC = \frac{a^2 + b^2 +c^2}{4} \tan {\omega}$

$Area ABC = rs$ where r is the inradius and s the semiperimeter

$OH^2 = 9R^2 - (a^2 + b^2 + c^2)$

Erdos-Mordell inequality: let ABC be a trianle and P be a point in the plane of the triangle. Let the sum of the distances from P
to the sides be S and the sum of the distances from P to the vertices be V, then
$2S \leq V$ with equality iff the triangle is equalitaeral.

(Sorry for any mistakes or repetitions of other's posts)

Tons of geometry formulas and tidbits here

http://www.cut-the-knot.org/geometry.shtml

Wow, I and Arne and Darij have been discussed it many times ago.

In that time, I called Pythagorean theorem,  fisaghoures.

As all of us know, Pythagorean theorem is very very well known ,I think

It will be good if I prove it with a strange way, wont it?

Maybe more than $1000$ways prepared for this theorem…

Ok , let us use $INVERSION$.

Suppose $(C)$ is the circumcircle of triangle $ABC$, ($\angle A=90$)

I choose a point $O$ as the center of inversion , in which $AO$ is diameter.

With a little thinking we can conclude :

$AC=OB$

$AB=OC$

$BC=OA$

I choose $BC^2$ as the radius of inversion.

Ok, lets start to invert the shape .

Circumcircle of $ABC$, changes to a line which is tangent to the circle,

In point $A$.

$A=A'$      and     $B,C$ are the intersections of $OB,OC$ with the line

respectively .

before continuing I should say a lemma:

lemma:

suppose $AB$ is a segment and $A'B'$ is its inversive form after doing

and inversion with center$O$ , and radius $K$.

we have :    $A'B'=\frac{K.AB}{OA.OB}$

…………………………………………………………………………………

I like to know what are the lengths of $A'B',A'C',B'C'$ ?

According to the lemma we have:
............................................................................................................................

$A'B'=\frac{AB. BC^2}{OA.OB}=\frac{AB.BC^2}{BC.AC}$

$=\frac{AB.BC}{AC}$

............................................................................................................................

$A'C'=\frac{AC.BC^2}{OA.OC}=\frac{AC.BC^2}{AB.BC}$

=$\frac{AC.BC}{AB}$

…………………………………………………………………………………

$B'C'=\frac{BC^2.BC}{OB.OC}=\frac{BC^3}{AB.AC}$

…………………………………………………………………………………

as u know $A',B',C'$ are collinear, and we have :$A'B'+A'C'=B'C'$

then we should have:

$\frac{AB.BC}{AC}$+$\frac{AC.BC}{AB}$=$\frac{BC^3}{AB.AC}$

know multiple the equation at $\frac{AB.AC}{BC}$

and finally:  $AB^2+AC^2=BC^2$

and every thing is ok…

wasn’t it enjoyable?  How ever I like it.

(Note: consider that to prove the basically theorems and rules of inversion ,

Didn’t use Pythagorean theorem)

i don't see sine law  :o ...
a triangle with the sides a,b,c would have the law applied as:
a/sinA=b/sinB=c/sinC=2R
also,
http://en.wikipedia.org/wiki/Category:Mathematical_theorems
this has a huge list of therems :lol:. Enjoy!

it is good to prove easy theorems, but its alittle boring. :?

why dont we prove hard theorems?

[quote="José"]emmm.... what about the feuerbach's nine points circle?[/quote]

u did it good , my dear jose.(Argentina 10  Mexico 0 ;) )

any way im curies to see the solutions for this one.

u know i have a solution with inversion for it.
 :D

hurry up dears...

[quote="Ashegh"][quote="José"]emmm.... what about the feuerbach's nine points circle?[/quote]

u did it good , my dear jose.(Argentina 10  Mexico 0 ;) )[/quote]

I'd be glad if......... any solution?

You can also gain many theorems by solving old IMO problems to see what is useful in solving them.

[color=darkred][b]I recomend you place here that properties of the geometrical figures which are used more often by you.[/b][/color]

For example, apply the property  -- [color=darkblue] Let $ABC$ be a triangle and let $M$ be a point which belongs to the side-line $BC$. Then $\boxed{\ \frac{MB}{MC}=\frac{AB}{AC}\cdot \frac{\sin \widehat{MAB}}{\sin\widehat{MAC}}\ }$[/color]-- to the following problem :

[color=green][b]Problem.[/b] Let $ABC$ be a triangle inscribed in the circle $w$. Denote the intersection $T\in BB\cap CC$.
Prove that the ray $[AT$ is the $A$- symmedian for the given triangle, i.e. $\frac{\delta_{AB}(T)}{\delta_{AC}(T)}=\frac{c}{b}$, where $\delta_{d}(X)$- the distance of the point $X$ to the line $d\ .$[/color]

ptolemy's theorem
simsons line
power of a point


Asymptote
trigonometry
logarithms
limit
calculus
calculus computations

I have an exam soon and i'd really appreciate it if someone would upload solutions of the following:

1) lim [ (sin(x)/x)^(1/x^2) ]
    x->0

2) ln( |1-1/(x^2)| )
    find when f''(x) and f'(x) >,<0
    asymptotes

thanks!

(1) $\frac{\sin x}x=1-\frac{x^2}6+O(x^4).$

Since $\ln(1+u)=u+O(u^2)$ we have $\ln\left(\frac{\sin x}x\right)=-\frac{x^2}6+O(x^4).$

Then $\frac1{x^2}\ln\left(\frac{\sin x}x\right)=-\frac16+O(x^2).$

Taking the exponential, we have $\lim_{x\to0}\left(\frac{\sin x}x\right)^{\frac1{x^2}}=e^{-\frac16}.$


x,y,z are positive and real s.t. x+y+z=3

prove that:

1/x^0.5+1/y^0.5+1/z^0.5 is greater than or equal to 3

i cant solve it using AM-GM but im pretty sure you can

btw can anyone tell me how to do the greater than or equal to sign using latex. thanx

happy solving!!

[this problem comes from an australian source, and is part of an enrichment program for year 10, i.e. 15-16 year olds]

Hello random_mathematician!! :)

In order to learn Latex math typesetting program  view
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5261[/url]
read carefully the guide (in pdf format).

Anyway the ' greater than or equal to ' sign is   \geq
that is $\geq$....
eg.  $e^x\geq x+1$

Now to the problem...:

From AM-HM (arithmetic-harmonic mean) we get
$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\geq\frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}}$

In addition we have $3(x+y+z)\geq(\sqrt{x}+\sqrt{y}+\sqrt{z})^2$
or $(\sqrt{x}+\sqrt{y}+\sqrt{z})\leq3$ and the conclusion follows!!

ok??
 :lol:  :)

hey

i figured it out using jensen's inequality - but your way is better known!

i was thinking 1/root x is convex when x is defined over the positive reals

so using the 3 term jensen （unweighted), the answer comes out in 2 lines

let f(s)=1/ root x

then: f((x+y+z)/3) is less than or equal to 1/3(f(x)+f(y)+f(z))

i.e. 3f(1) is less than or equal to f(x)+f(y)+f(z)

then the result follows from f(s)=1/ root x

ok. That's also nice!!
Note that jensen's ineq is widely used and works out
many problems.

For more inequalities check the " inequalities lessons and maybe a marathon "
in intermediate topics.!
( [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=40857&start=160[/url] )


The Wonder Island is inhabited by Hedgehogs. Each Hedgehog consists of three co-plane segments of unit length having a common endpoint, with all three angles between them equal to 120 degree. Given that all Hedgehogs are lying at on the island and no two of them touch each other, prove that there is a finite number of Hedgehogs on Wonder Island.

I suppose the island is a bounded plane region, right? In that case, you can basicly choose a disk of radius $r$ around the "center" of each hedgehog. If $r$ is small enough, no two such disks will be able to touch each other because of the shapes of the hedgehogs, I think. There can only be finitely many disjoint disks of radius $r$ in a bounded region of the plane, so that's that.

I'm not sure it's Ok though.

Yes, the island is a bounded plane region.


inequalities
IMO Shortlist
inequalities proposed

Find the least real number $ k$ such that, for all positive real numbers $ a,b,c,d$, $ \sum_{cyc}\frac{ka^2\plus{}(3\minus{}k)bc}{a\plus{}b\plus{}c}\geq a\plus{}b\plus{}c\plus{}d$

[quote="bakerbakura"]Find the least real number $ k$ such that, for all positive real numbers $ a,b,c,d$, $ \sum_{cyc}\frac {ka^2 \plus{} (3 \minus{} k)bc}{a \plus{} b \plus{} c}\geq a \plus{} b \plus{} c \plus{} d$[/quote]
Take $ a \equal{} c$ and $ b \equal{} d,$ the inequality becomes
$ \frac {ka^2 \plus{} (3 \minus{} k)ab}{2a \plus{} b} \plus{} \frac {kb^2 \plus{} (3 \minus{} k)ab}{a \plus{} 2b}\ge a \plus{} b,$
which is equivalent to
$ \frac {(a \plus{} b)(a \minus{} b)^2(k \minus{} 2) }{(2a \plus{} b)(a \plus{} 2b)} \ge 0.$
The last inequality holds if and only if $ k \ge 2.$

By the way, the constant $ k \equal{} 2$ satisfies the original inequality. Because for $ k \equal{} 2,$ the inequality is equivalent to
$ \sum \frac {(a \minus{} b)(a \minus{} c)}{a \plus{} b \plus{} c} \ge 0,$
which is the IMO Shortlist 2008 problem.

There is a nice and simple solution for this inequality. See [b]Problem 3.18 - page 178[/b] in new book:

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL Publishing House 2009.


Let C is the intersection point of the circle c1 with diameter AB and the perpendicular from a point D on the segment AB.c2 touches c1 internally and touches CD AB at K and L respectively. Prove that angleKCL=angleLCB

It is solved in the forum again. I can not find the link(I do not know how to search). I have the solution. If nobody answers, I' ll send the solution.
 I hope that someone will send the link. It is a nice problem !

 [u]Babis[/u]

[quote="stergiu"]It is solved in the forum again. I can not find the link(I do not know how to search).[/quote]

Searching in general is rather easy: Click on the "Search" link (which is in the upper right part of every MathLinks page) and enter the keywords you want.

However, searching for this particular problem was not that easy. First, I searched with the keywords <<[b]*circle diameter* tange* bisect*[/b]>> (note that I wrote "*circle", not "circle", since sometimes it is referred to as "semicircle" only, so that the word "circle" doesn't appear in the post) and got a lot of topics, from which I found

http://www.mathlinks.ro/Forum/viewtopic.php?t=23606 ;
http://www.mathlinks.ro/Forum/viewtopic.php?t=17446 (here, a slightly different problem is discussed, but your problem gets solved too).

A subsequent search for the keyword <<[b]*t=23606[/b]>> (the intention was to find threads linking to the first of the above two threads!) yielded

http://www.mathlinks.ro/Forum/viewtopic.php?t=51954 ;
http://www.mathlinks.ro/Forum/viewtopic.php?t=46339 .

  Darij

Darij , thank you very much: for the links and the searching methods , too!
Really , can we search a word (or symbol ) written in LATEX and how ?

 [u]Babis[/u]


[quote="Aryth"]Suppose your number has $ d$ digits. Then

$ 10^d > \text{your number} \ge 10^{d \minus{} 1}$

We can take the logarithm with base 10 of all sides to obtain:

$ d > \log (\text{your number}) \ge d \minus{} 1$

So if you know (very approximately) the value of $ \log (\text{your number})$, then you can figure out $ d$, which is how many digits your number has.

For example, if we wanted to find how many digits are in $ 5^{44}$, we would do

$ \log 5^{44} \equal{} 30.754680190785...$

And, thus, since $ 31 > 30.754680190785... \ge 30$, we know that $ d \equal{} 31$, so $ 5^{44}$ has 31 digits.[/quote]

How is it possible to approximate such large logarithms without help of a calculator? Is there a formula, perhaps?

This problem was from AoPS Vol 2. and a majority of their problems don't require the use of calculators. This problem could be solved by using the properties of logarithms. So, $ \log 5^{44} \equal{} 44(\log 5) \equal{} 44(\log 10 \minus{} \log 2) \equal{} 44(1 \minus{} 0.3010) \equal{} 30.756...$ In this, the only thing that you had to "know" was $ \log 2 \equal{} 0.3010$ but that information was given in the problem.

[quote="themorninglighttt"][quote="Poincare"]Calculus to the rescue! That's what my brother said when I asked him this question. But, he doesn't really know any math aside from the stuff they teach in school so I'd double check...[/quote]

I see no calculus there... or am I missing something? I don't think so, though.[/quote]

Not in that answer, but Taylor Series can be used to approximate many different functions such as logarithms by merely adding/multiplying.


linear algebra
matrix
linear algebra unsolved

Prove that if $A_n(R)$ is the set of antisymetric real matrices then $ SO_n(R)=e^{A_n(R)}$.

Let's first prove that every matrix of the form $U=e^A$, where $A$ is antisymmetric is orthogonal and has determinant $1$.Indeed, if $UU^{T}=exp(A)exp(A^T)=exp(A)exp(-A)=I, detU=e^{trA}=e^0=1.$
Now, we'll begin with another part of our problem,stating that every orthogonal matrix with determinant $1$ can be represented in form $e^X$, where $X$ is antisymmetric:
Firstly,let's prove this result for the simpliest form of matrix in $SO_n(\mathbb{R})$: $A=\left( \begin{array} {cc} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha  \end{array} \right)$. Really it is not difficult to verify that $A=e^X$,where $X=\left( \begin{array} {cc} 0 & -\alpha \\ \alpha & 0 \end{array} \right)$.
From now our result immediately follows from the fact that every matrix $A$ from $SO_n(\mathbb{R})$ can be represented in from $A=S^{-1}diag(A_1,...,A_k,I)S$, where $A_i$ is of the form $\left( \begin{array} {cc} cos\alpha_i & -sin\alpha_i \\ sin\alpha_i & cos\alpha_i  \end{array} \right)$.


let a+b+c=3, a,b,c positive
prove that:
$ \frac{1}{2\plus{}a^2\plus{}b^2}\plus{}\frac{1}{2\plus{}b^2\plus{}c^2}\plus{}\frac{1}{2\plus{}c^2\plus{}a^2}\le\frac{3}{4}$


Prove that there are infinite values of n for which $ n^2\plus{}1$ is prime

In fact it is an open problem which is thought to be true
I think it will be great that someone could answer your post but don't expect too much because it's a very hard problem, and, moreover, if they find and answer they will not probably post it here, first they would publish it :P

Daniel


[hide]There are 12 exposed sides, and each is $\displaystyle\frac {72}{12} = 6$.  So each square has an area of $6^2=36$, and the total area is 36*5=180.[/hide]

[quote="Barnacle"][hide]36*6 = 180[/hide][/quote]
you mean 36*5=180

yes i did, haha, I made a typo.

i hope that my carelessness has not persuaded anyone to believe that 36*6 = 180, as there is clearly no factor of 10 in 36*6  :rotfl:

side question: how could you see that the product is wrong just by looking at the last digits?  can you generalize to multiplying 2 numbers that both end in 6?

even though the problem was easy you should still use spoiler tags

[quote="math92"]Five congruent squares are put together without gaps or overlap to form a figure shaped like a plus sign. The perimeter of the figure is 72 cm. Find the area of the figure, in sq cm.[/quote]

[hide] If you put the side length of each side as $x$, then $12x=72$, so $x=6$

Each area of the square would then be $6^2$ or $36$

So the combined area of the plus sign figure would be $36*5$ which is $180 sq. cm.$[/hide]


ratio
percent
geometry
3D geometry
LaTeX

4. tennis coaches high lob and low smash decided to share with their assistants, love and vantage, the money they earned from tennis lessons. they agreed on the following ratios: lob : love = 17 : 12, love : smash = 3 : 4, and smash : vantage = 32 : 15.if their earnings totaled $3,150, how much did love receive?

5. the cost of living in each quarter (3 months) incrased by 2% over the previous quarter. to the nearest tenth of a percent, to what annual percentage rate of increase does this correspond.

6. if x varies as the cube or y, and y varies as the fifth root of z, then x varies as the nth power of x. what is n?

7. it is now between 10:00 and 11:00. six minutes from now, the minute hand of a watch will be exactly opposte the place where the hour hand was three minutes ago. what is the exact time now?

8. from time t=0 to time t=1, a popultion incrased by i% and from time t=1 to time t=2, the popultion increased by j%. by what percent, in terms of i and j, did the popultion increase from time t=0 to time t=2?

9. (what is the quickest way) if for three distinct positive number x, y, and z, y/(x-z) = (x+y)/z = x/y, then find the numerical value of x/y.

10. two men starting at a point on a circular 1-mile race track walk in opposite directions with uniform speeds and meet in 6 minutes, but if they walk in the same direction, it requires one hour for the faster walker to gain a lap. what is the rate of the slower walker?

thanks!!

By the way, do you have the AoPS solutions manual?  It's worth every penny.

ya i have it but i dont get the solutions.

[hide]4)  love -- $720

5)  6.12%[/hide]

I might be wrong...

I haven't had a chance to learn Tex, so this isn't real pretty, but here goes:

4)

[hide] I prefer to work with single letter variables, so:

high lob = a    ,     low smash = b     ,   love = c      ,    vantage = d

we know:

a/c = 17/12       ,       c/b = 3/4    ,       b/d = 32/15

a + b + c + d = 3,150

we want to know what c= , so we get all of the variables in terms of c:

a and b are easy enough, just solve the ratios they give us:

a = 17c/12              b= 4c/3

now for d:

b = 32d/15           b = 4c/3      thus      32d/15 = 4c/3

d = 5c/8

now plugging these in:

17c/12 + 4c/3 + 5c/8 +c = 3,150

I'm going to assume you can use the above to solve for c, so c = 720 which means that love makes $720.

[/hide]

5)

[hide] I think I remember learning a more elegant way of doing this, but I can't remember it right now...

I'm assuming you want the % increase after 4 quarters:

c = initial cost of living

after 1 quarter:  c1 = c+c(.02) = 1.02c

after 2 quarters: c2 = 1.02c + 1.02c(.02) = 1.0404c

after 3 quarters: c3 = 1.0404c + 1.0404c(.02) = 1.061208c

after 4 quarters: c4 = 1.061208c + 1.061208c(.02) = 1.08243216c

so the annual % increase is 8.2%

[/hide]

6) if you want x in terms of x^n then n=1  , I'm guessing that the question is really asking for x in terms of z^n

[hide]

x =y^3          y=the fifth root of z = z^(1/5)

substituting for y:       x = [z^(1/5)]^3

combining the exponents:       x = z^(3/5)

so n = 3/5

[/hide]

well its getting late and I'm going to go to bed, if no one else does the others I'll see if I can help more tomorrow.  Hope these help, if I am not clear please let me know  !

[hide]4)

lob:love=17:12, love:smash=3:4, smash:vantage=32:15.

34:24, 24:32, 32:15

34+24+32+15=105

3150/105=30

So, 24*30=720.

5)

Let's say the cost starts at 100

2nd quarter -- 104.04

4th quarter -- $106.12

So annual percentage is 6.12%. To nearest tenth is 6.1%...[/hide]

That's how I solved it.

eqn1190 wrote:6. if x varies as the cube or y, and y varies as the fifth root of z, then x varies as the nth power of x. what is n?

i suppose you mean "x varies as the cube of y", right? and also, i'll suppose you meant x varies as the nth power of z.

6. [hide]x=y :^3:

y=z^(1/5)

y :^3: =z^(3/5)

x=z^(3/5)[/hide]

8. [hide] if the initial population is x, then the population at T1 is (1+i/100)x. the population after T2 is (1+i/100)x+(j/100)(1+i/100)x

so it would be (1+j/100)(1+i/100)x[/hide]

10. [hide]if it takes the faster person 60 minutes to gain a lap when they're walking in the same direction, then the faster person walks 1 mph faster.

if s=the speed of the slower walker, then s+s+1=10mph (6 minutes a mile)

2s=9

s=4.5 mph[/hide]

For #9, earlier in the chapter it stated that a/b = c/d = (a+c)/(b+d). This can be generalized to any number of equal fractions, since if a/b = c/d = e/f=....,
then a/b=(c+e+g+...)/(b+d+f+...) is equivalent to a/b=(c+ka)/(d+kb), and after multiplying that out, it reduces to bc=ad, or a/b = c/d.

So all you have to do is to write a fraction whose numerator is the sum of the numerators of the three fractions and whose denominator is the sum of the denominators of the three fractions. Adding those up, you get
2(x+y)/(x+y), or 2, which equals the original fraction, x/y.

[quote="philomath"]I haven't had a chance to learn Tex, so this isn't real pretty, but here goes:

[/quote]

It seems many people in here are having trouble with Latex so this is a topic I posted and I learned everything from here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12230&start=0

This was once very popular topic in Latex forum.

it looks like nobody's done number 7, so ill do it lol

[hide]7. in order for the minute hand to be opposite where the hour hand was, it would have to be around the 20-25 minute range after 10:00.  After 9 minutes, the hour hand travels 4.5 degrees around the clock.  So because the minute hand is opposite where the hour hand was 3 minutes ago, the angle created by the two hands after 6 minutes is 180 degrees plus 4.5 degrees for the hour hand.

Every minute the hour hand travels 0.5 degrees and the minute hand travels 6 degrees.  At 10:00, the hands form a 60 degree angle.  So every minute, the angle created by the two hands increases by 5.5 degrees because while the minute hand travels 6 degrees, the hour hand catches up by 0.5 degrees.  Thus the equation solving for the time after 6 minutes from now is:

[tex]60+5.5x=184.5[/tex]

[tex]5.5x=124.5[/tex]

[tex]x=22\frac{7}{11}[/tex]

Therefore six minutes from now it will be [tex]22\frac{7}{11}[/tex] minutes past ten, so the time right now is [tex]16\frac{7}{11}[/tex] minutes past ten.[/hide]

goldenfire, you might want to put your answer in spoiler, in case other people also want to answer the problem.

On number 9, i also got another answer. I cross multiplied  (x+y)/z = x/y, yielding xz = xy+y :^2: .   Then i cross multiplied y/(x-z) = x/y, yielding x :^2:  -xz = y :^2: .

Simplify and what not, then subract the equations:

      xz = y :^2:  + xy
 -    xz = -y :^2: + x :^2:
----------------------------------
       0 =  2y :^2:  + xy - x :^2:

Then you get (2y-x)(y+x) = 0

2y = x
[b]x/y = 2 [/b]

y = -x
[b]x/y = -1[/b]

Is this a legitimate solution?

pkothari, you subtracted your equations wrong -- the y :^2: terms should have gone away.  Check that over.

I forgot about the positive integer restriction, but i still don't see how i subracted incorrectly. These were the equations i got:

xz = y :^2:   + xy
xz = -y :^2:   + x :^2:

the minus sign in front of the second equation was intended to subract the whole equation, not to make the xz negative. I think that's where you might have gotten confused. So from that, i subracted the 2 equations which yielded

0 = 2y :^2:  + xy - x :^2:

From there i factored and got my answer.

It was pointed out that all integers had to be positive, so 2 would be the only correct solution.

Oh, I see.  Sorry, I misread that, you are correct.


everybody post your bmat scores and percentiles when they become available

bmat is brilliant's mock all india test series for iit jee.

at sagar vaze, what test series you had joined and  how were your marks last year. also i need suggestions in inorganic chemistry.

[hide]cming here after a long time. surprised to see someone invoke me out of the blue[/hide]
me and test series  :rotfl:
i hadn't joined any. just relied on our m prakash tests  :oops:

for inorganic, i wud say use j d lee.
it is much better than ncert.

if u r gud in inorganic, then do cotton willikinson.
but that is too high level.

can u plz post the shcedule.........someone.......plz........


Let A = 1!2!3!...100!
Prove that there exists a unique integer k such that S/k! is a perfect square.
Find the value of k.

Arne wrote:Let A = 1!2!3!...100!
Prove that there exists a unique integer k such that A/k! is a perfect square.
Find the value of k.

[hide]I shall actually go about this problem in reverse order.  First, I'll find k, then prove uniqueness.

A definition of notation:  P(f(i), i, a, b) = f(a)*f(a + 1)*f(a + 2)*...*f(b)

So, lets look at P(n!, n, 1, 100) = P((2n - 1)!*(2n)!, n, 1, 50) =

= P((2n - 1)!:^2:*(2n), n, 1, 50) = P((2n - 1)!, n, 1, 50):^2: * P(2n, n, 1, 50) =

= P((2n - 1)!, n, 1, 50):^2: * 2^50 * 50! = 50! * z:^2: where z is some (very large) integer.  Thus, the answer to the question is 50.

Now, to prove uniqueness.  In our origional product, we have a factor of 47 for i! for every i from 47 to 93, and 2 factors of 47 for every i from 94 to 100.  This gives us an odd number of factors of 47 in the entire expression.  Thus, k! must contain an odd number of factors of 47, so 47 :le: k :le: 93.  Similarly, by analyzing powers of 37, we see that 37 :le: k :le: 73.  Thus  47 :le: k :le: 73.  Now let's check out 19.  We have and even number of 19's in our original product (just go check it out -- it works  ) .  So that means k! must have an even number of 19's.  So 0 :le: k :le: 18 or 38 :le: k :le: 56 or 76 :le: k :le: 94.  Since we already know that 47 :le: k :le: 73, we get that 47 :le: k :le: 56.  Doing the same thing with 17 narrows it down further to 47 :le: k :le: 51.  We know 50 works.  Thus, we have a perfect square in S/50!, and we want to see whether S/47!, S/48!, S/49!, or S/51! are also perfect squares.

S/51! = m^2/51 is not a perfect square.

S/49! = m^2 * 50 is not a perfect square

S/48! = m^2 * 50 * 49 is not a perfect square

S/47! = m^2 * 50 * 49 * 48 is not a perfect square.  Okay.  All done.  Unique.[/hide]


geometry
3D geometry
angle bisector
geometry unsolved

[quote="math_07"]In triangle $ ABC$, with $ AB > BC$, $ BM$ is a median and $ BL$ an angle bisector. The line through $ M$ parallel to $ AB$ meets $ BL$ at $ D$ and the line through $ L$ parallel to $ BC$ meets $ BP$ at $ E$.
Prove that $ ED \perp BL$.[/quote]
Let, $ MD \cap BC\equal{}N, LE \cap MD\equal{}P$.
We have, $ \angle PLD\equal{}\frac B2\equal{}\angle LDP \iff LP\equal{}PD$ and $ N$ is the midpoint of $ BC$. So, $ LP\equal{}PE\equal{}PD$. So, $ D$ lies on a circle centered at $ P$ and with $ LE$ as diameter. So, $ ED \perp BL$. :)

Dear Mathlinkers,
I take the problem in this way.
1. let Y, Z be the points of contact of the incircle of ABC wrt CA, AB
and A' be the midpoint of BC.
2. It is well knowed that YZ, BL, MA' and the perpendicular from C to BL are concurrent ; D is therefore this meetpoint.
3. Let E be the point of intersection of BM and CD.
4. DMA' being the D-median of DBC, LE//BC.

For more about 1.
see : http://perso.orange.fr/jl.ayme   vol. 4  An unlikely concurrence revisited and generalized p. 4.
Sincerely
Jean-Louis

It is known for quite some time that for an isosceles triangle

   BCC' (C' is on AB  and <C = <C' ) when BM is median then

   line // to AB thru M cuts CC' in a point on B-angle bisector BL,

  the line  thru L  and // to AB cuts CC' in a point on B-symmedian BS,

  the line thru S and  // to AB cuts CC' in a point on a line that divides the

  base AC proportionate to cubes of the sides  AB  and BC  of ABC, etc...

  Once we know that in our problem points D and E are on B-angle bisector and

  B-median it means that  they are on line CC'  orthogonal to B- bisector.

   I want to dedicate this note to all victims of San Francisco earthquake 100 years ago.

  M.T.


We define the set u(n) as follows:
1)u(0)=x, where x belongs to N
2)u(n+1)=u(n)/2 if u(n) is even else u(n+1)=3*u(n)+1.

Prove that there is m such that u(m)=1.
(We get that after this number u(n) becomes periodic)

ah, ah, ah...really funny. :maybe:

Pierre.

Should I get that there's no solution(yet discovered) for this problem?

Have a look here :
http://mathworld.wolfram.com/CollatzProblem.html

Pierre.


An urn contains 2009!+1 black balls and 2009!+1 red balls . In every step we take out 2009 balls and discard all of them if they are all of the same color or keep the red ones and discard the black ones if they are of different colors . Given that 1 ball remains at the end after a finite number of steps find the probability that it is a red ball .

Sorry Thanks to rahul

The question already exists http://www.mathlinks.ro/viewtopic.php?t=250398


[quote="Peter"]Show that for all primes $p$, \[\sum^{p-1}_{k=1}\left \lfloor \frac{k^{3}}{p}\right \rfloor =\frac{(p+1)(p-1)(p-2)}{4}.\][/quote]

This is also a problem from the German National Olympiad (DeMO, 4th round) 2002, and I have written a solution for the op02 project. Since op02 does not seem to have survived, I am publishing the solution here:

We start by showing something really trivial:

[b]Lemma 1.[/b] For any integer $n$ and any non-integer real number $a$, we have $\lfloor a \rfloor+\lfloor n-a \rfloor = n-1$.

[i]Proof.[/i] Since $a$ is not an integer, we have $a=\lfloor a \rfloor+q$ for some real $q$ with $0<q<1$. Hence, $n-a = n-\left(\lfloor a \rfloor+q\right) = \left(n-1-\lfloor a \rfloor\right)+\left(1-q\right)$. Hereby, $0<1-q<1$; thus, $\lfloor n-a \rfloor = n-1-\lfloor a \rfloor$, what yields Lemma 1.

For every $k$ such that $1\leq k\leq p-1$, define the number $a = \frac{k^{3}}{p}$. This number $a$ is trivially non-integer, since $k^{3}$ is not divisible by $p$ because $p$ is a prime and $k<p$. Also, set $n = p^{2}-3pk+3k^{2}$. Then, Lemma 1 yields

$\left\lfloor \frac{k^{3}}{p}\right\rfloor+\left\lfloor \left(p^{2}-3pk+3k^{2}\right)-\frac{k^{3}}{p}\right\rfloor = \left(p^{2}-3pk+3k^{2}\right)-1$.

This simplifies to

$\left\lfloor \frac{k^{3}}{p}\right\rfloor+\left\lfloor \frac{\left(p-k\right)^{3}}{p}\right\rfloor = p^{2}-3pk+3k^{2}-1$.

Now, the problem is straightforward:

$2\sum^{p-1}_{k=1}\left \lfloor \frac{k^{3}}{p}\right \rfloor = \sum^{p-1}_{k=1}\left \lfloor \frac{k^{3}}{p}\right \rfloor+\sum^{p-1}_{k=1}\left \lfloor \frac{\left(p-k\right)^{3}}{p}\right \rfloor$
$= \sum^{p-1}_{k=1}\left( \left\lfloor \frac{k^{3}}{p}\right\rfloor+\left\lfloor \frac{\left(p-k\right)^{3}}{p}\right\rfloor \right) = \sum^{p-1}_{k=1}\left(p^{2}-3pk+3k^{2}-1\right)$
$= \left(p-1\right)p^{2}-3p \sum^{p-1}_{k=1}k+3 \sum^{p-1}_{k=1}k^{2}-\left(p-1\right)$
$= \left(p-1\right)p^{2}-3p \frac{\left(p-1\right)p}{2}+3 \frac{\left(p-1\right)p\left(2p-1\right)}{6}-\left(p-1\right) = \frac{\left(p-2\right)\left(p-1\right)\left(p+1\right)}{2}$,

so that

$\sum^{p-1}_{k=1}\left \lfloor \frac{k^{3}}{p}\right \rfloor = \frac{\left(p-2\right)\left(p-1\right)\left(p+1\right)}{4}$,

completing the solution.

  darij

[quote="Peter"]Show that for all primes $ p$,
\[ \sum^{p\minus{}1}_{k\equal{}1}\left\lfloor\frac{k^{3}}{p}\right\rfloor \equal{}\frac{(p\plus{}1)(p\minus{}1)(p\minus{}2)}{4}.\]
[/quote]

The same idea, but with different style.

Lemma. Let $ \alpha,\beta\in\mathbb{R}$ with $ \alpha\plus{}\beta\equal{}n\in\mathbb{Z}$. Then, we have
  \[ \lfloor\alpha\rfloor\plus{}\lfloor\beta\rfloor \equal{}\begin{cases}\alpha\plus{}\beta &\left(\alpha\in\mathbb{Z}\right),\\ \alpha\plus{}\beta\minus{}1 &\left(\alpha\not\in\mathbb{Z}\right).\\ \end{cases}\]

Proof. In case when $ \alpha\in\mathbb{Z}$, we also have $ \beta\in\mathbb{Z}$. Then, we obtain $ \lfloor\alpha\rfloor \equal{}\alpha$  and $ \lfloor\beta\rfloor \equal{}\beta$.  Hence,  $ \lfloor\alpha\rfloor\plus{}\lfloor\beta\rfloor \equal{}\alpha\plus{}\beta$.  Now, we consider the case when  $ \alpha\not\in\mathbb{Z}$. Since $ \alpha\plus{}\beta\in\mathbb{Z}$,  we also have $ \beta\not\in\mathbb{Z}$.  Since $ \alpha$ is not an integer, one may write $ a \equal{} k\plus{}q$,  where $ k \equal{}\lfloor\alpha\rfloor\in\mathbb{Z}$ and  $ q\in (0,1)$. Observer that $ \beta \equal{} n\minus{}\alpha \equal{} (n\minus{}k\minus{}1)\plus{}(1\minus{}q)$. Since $ n\minus{}k\minus{}1\in\mathbb{Z}$ and  $ 1\minus{}q\in (0,1)$, this means that $ \lfloor\beta\rfloor \equal{} n\minus{}k\minus{}1$. It follows that
 \[ \lfloor\alpha\rfloor\plus{}\lfloor\beta\rfloor \equal{} k\plus{}(n\minus{}k\minus{}1) \equal{} n\minus{}1 \equal{}\alpha\plus{}\beta\minus{}1 .\]

Lemma. Whenever $ k\in\{ 1,\cdots, p\minus{}1\}$, we obtain
\[ \left\lfloor\frac{k^{3}}{p}\right\rfloor\plus{}\left\lfloor\frac{\left(p\minus{}k\right)^{3}}{p}\right\rfloor \equal{}\frac{k^{3}}{p}\plus{}\frac{\left(p\minus{}k\right)^{3}}{p}\minus{}1\]

 Proof. Let $ k\in\{ 1,\cdots, p\minus{}1\}$.  Since $ p$ is prime, we see that $ k^{3}$ is not divisible by $ p$ so that $ \frac{k^{3}}{p}$ is not an integer.  Furthermore, from the congruence $ k^{3}\plus{}(p\minus{}k)^{3}\equiv k^{3}\minus{}k^{3}\equiv 0\; (mod\; p)$, we see that
 \[ \frac{k^{3}}{p}\plus{}\frac{\left(p\minus{}k\right)^{3}}{p}\in\mathbb{Z}.\]
 The above proposition yields the desired result.

 Now, we compute the summation. By symmetry, we obtain
\[ \sum^{p\minus{}1}_{k \equal{} 1}\frac{k^{3}}{p}\equal{}\sum^{p\minus{}1}_{k \equal{} 1}\frac{(p\minus{}k)^{3}}{p}.\]
and
\[ \sum^{p\minus{}1}_{k \equal{} 1}\left\lfloor\frac{k^{3}}{p}\right\rfloor \equal{}\sum^{p\minus{}1}_{k \equal{} 1}\left\lfloor\frac{(p\minus{}k)^{3}}{p}\right\rfloor.\]
Applying the above results we proved, we compute

 \[ \sum^{p\minus{}1}_{k \equal{} 1}\left\lfloor\frac{k^{3}}{p}\right\rfloor\\ \equal{}\frac{1}{2}\left(\sum^{p\minus{}1}_{k \equal{} 1}\left\lfloor\frac{k^{3}}{p}\right\rfloor\plus{}\sum^{p\minus{}1}_{k \equal{} 1}\left\lfloor\frac{(p\minus{}k)^{3}}{p}\right\rfloor\right)\\ \equal{}\frac{1}{2}\sum^{p\minus{}1}_{k \equal{} 1}\left(\left\lfloor\frac{k^{3}}{p}\right\rfloor\plus{}\left\lfloor\frac{(p\minus{}k)^{3}}{p}\right\rfloor\right)\\ \equal{}\frac{1}{2}\sum^{p\minus{}1}_{k \equal{} 1}\left(\frac{k^{3}}{p}\plus{}\frac{\left(p\minus{}k\right)^{3}}{p}\minus{}1\right)\\ \equal{}\minus{}\frac{p\minus{}1}{2}\plus{}\frac{1}{2}\sum^{p\minus{}1}_{k \equal{} 1}\left(\frac{k^{3}}{p}\plus{}\frac{\left(p\minus{}k\right)^{3}}{p}\right)\\ \equal{}\minus{}\frac{p\minus{}1}{2}\plus{}\sum^{p\minus{}1}_{k \equal{} 1}\frac{k^{3}}{p}\\ \equal{}\minus{}\frac{p\minus{}1}{2}\plus{}\frac{1}{p}\left(\frac{(p\minus{}1)p}{2}\right)^{2}\\ \equal{}\frac{(p\plus{}1)(p\minus{}1)(p\minus{}2)}{4}.\\ \]


vector
calculus
real analysis
real analysis unsolved

$ \vec{\nabla} \times \vec{F}  \equal{} [x,y,z]$
Find [i]F[/i]

Yes, and so can you - using one of those [url=http://en.wikipedia.org/wiki/Vector_calculus_identities]vector calculus identities[/url].

Do you mean
$ \nabla \cdot ( \nabla \times \mathbf{F} ) \equal{} 0$
which isn't satisfied in this case?


Edit: Moved to College Playground. My bad

[quote="Numbnut247"]I have this question which I've been struck on for a long time:

A spotlight on the ground shines on the outside wall of a parking garage 12m away. If a 2-m tall man walks toward the garage at a speed of 0.75m/s, how fast is the height of the man's shadow on the garage wall decreasing when he is 4m from the building?

I have no clue how to solve it. :oops:

Edit: I don't know if I posted this in the right forum. It's a calculus question but I'm in high school so :maybe:[/quote]

calc doesnt belong here

[quote="jli"]calc doesnt belong here[/quote]

To clarify: the "Calculus Computations and Tutorials" forum is in the "College Playground" section, but we do take calculus questions from anyone, even if they're in high school.


Found this in one of the older threads.

In the school band 5 children each own their own trumpet. In how many different ways can exactly 3 of the 5 children take home the wrong trumpet while the other 2 take home the right trumpet?

I am not sure although i ended up with two different solutions, cant convince myself.

[hide="Solution"]
1 - 5c2 (way of choosing the children who take home the right trumpet) *  3! (1-1/1!+1/2!-1/3!)Number of derangements of 3 objects

Gives 10 :(

Any flaw with the logic?

The answer is 5c3+5c2 = 20.. Not quite sure how though..any help?[/hide]

5c3 because you are choosing 3 people to take home the wrong trumpet. 5c2 because you are choosing two people to take home the right trumpet. Since both are dependent, you add to get 5c3+5c2=20.

Ah got it. Even the derangement stuff gives out the answer

What do you mean by "derangement", exactly? It sounds more like a mental health issue to me. :|

By derangement, i think he/she means arrangements.

Oh, okay. That made me wonder, because it also looked a little like rearrangements. :lol:

A derangement of a set is a permutation of the set in which none of the elements are in their original positions.
Example: 4321 is a derangement of 1234. 4231 is not a derangement of 1234 because the 2s are in the same position.
It's on AoPSWiki somewhere.

Its definitely not a mental health issue  :)

Here is the proof [url]http://www.artofproblemsolving.com/Wiki/index.php/Derangement[/url]

Oh, I get it now. That was a term I had not encountered before.

There's a more detailed explanation on my blog, in the entry titled derangments. The table of contents is on my website, since you probably don't want to comb through 127 entries... :P

[quote="solveitall"]........I am not sure although i ended up with two different solutions, cant convince myself.
[hide="Solution"]
1 - 5c2 (way of choosing the children who take home the right trumpet) *  3! (1-1/1!+1/2!-1/3!)Number of derangements of 3 objects

Gives 10 :(

Any flaw with the logic?

The answer is 5c3+5c2 = 20.. Not quite sure how though..any help?[/hide][/quote]
I like your logic. It makes sense to me, and it leads to the accepted solution. I think you made an error in transcribing or calculating somewhere.
5c2=5X4/2=10
the number of derangements of 3 is clearly 2. (No need for complicated formulas. You have two (2) choices for first place, then the other two places are already decided).
10X2=20


hello,

is there a way to solve for x analytically in x=5(1-e^(-x) )?


quadratics
induction
number theory
prime numbers
number theory unsolved

If $ x,y,z\in N$ and $ xy\equal{}z^2\plus{}1$, prove that there exists integers $ a,b,c,d$ such that $ x\equal{}a^2\plus{}b^2,y\equal{}c^2\plus{}d^2,z\equal{}ac\plus{}bd$.

$ xy \equal{} (z \plus{} i)(z \minus{} i)$ and $ \gcd(z \plus{} i, z \minus{} i) | 2i$.  The result follows.

Alternately:  any prime factor of $ xy$ has $ \minus{}1$ as a quadratic residue, hence is congruent to $ 1 \bmod 4$.

Quite interestingely, you can prove this directly (e.g. by induction) and then deduce even the formula for the number of representations as a sum of two squares from it.

Well? I'd like to see the details. In fact, I was always curious to see an elementary proof of this formula (that is, without Gaussian numbers). The only obstacle is the possibility that two representations obtained by multiplication "stick together". How do you get over it?

[quote="t0rajir0u"]

Alternately:  any prime factor of $ xy$ has $ \minus{} 1$ as a quadratic residue, hence is congruent to $ 1 \bmod 4$.[/quote]

How does this yield the result?

Am I missing something here?

$ xy \equal{} (a^2 \plus{} b^2)(c^2 \plus{} d^2) \equal{} z^2 \plus{} 1 \equal{} (ac \plus{} bd)^2 \plus{} 1 \\
\Leftrightarrow a^2d^2 \plus{} b^2c^2 \equal{} 2abcd \plus{} 1 \Leftrightarrow (ad \minus{} bc)^2 \equal{} 1$. Clearly there exist $ a, b, c, d$ such that $ |ad \minus{} bc| \equal{} 1$ and $ ac \plus{} bd > 0$. For example, take $ (a, b, c, d) \equal{} (3, 4, 2, 3)$. Then $ (3^2 \plus{} 4^2)(2^2 \plus{} 3^2) \equal{} 25*13 \equal{} 325 \equal{} 18^2 \plus{} 1$.

[quote="radio"][quote="t0rajir0u"]

Alternately:  any prime factor of $ xy$ has $ - 1$ as a quadratic residue, hence is congruent to $ 1 \bmod 4$.[/quote]

How does this yield the result?[/quote]

It's a famous result that every prime $ \equiv 1 \bmod 4$ is a sum of two squares, and that a product of two sums-of-squares is again a sum-of-squares, hence $ \exists a, b, c, d$ such that $ x = a^2 + b^2, y = c^2 + d^2$.

On the other hand, to show from here that $ xy = (ac + bd)^2 + (ad - bc)^2$ where $ |ad - bc| = 1$ is not trivial without the use of some lemmas in the Gaussian integers, so I suppose this is not a big improvement over the other solution.

[quote="t0rajir0u"]$ xy \equal{} (z \plus{} i)(z \minus{} i)$ and $ \gcd(z \plus{} i, z \minus{} i) | 2i$.  The result follows.

Alternately:  any prime factor of $ xy$ has $ \minus{} 1$ as a quadratic residue, hence is congruent to $ 1 \bmod 4$.[/quote]

I have a few questions regarding this problem:

1. Why can't $ xy$ be even?

2. You say the result follows but I can't seem to find out how. Can you tell me?

I see that $ x$ and $ y$ can be written both as the sum of squares but why $ z\equal{}ac\plus{}bd$ or the equivalent expression $ |ad\minus{}bc|\equal{}1$.

Thanks in advance.

[quote="Lightning_17"][quote="t0rajir0u"]$ xy \equal{} (z \plus{} i)(z \minus{} i)$ and $ \gcd(z \plus{} i, z \minus{} i) | 2i$.  The result follows.

Alternately:  any prime factor of $ xy$ has $ \minus{} 1$ as a quadratic residue, hence is congruent to $ 1 \bmod 4$.[/quote]

I have a few questions regarding this problem:

1. Why can't $ xy$ be even?

2. You say the result follows but I can't seem to find out how. Can you tell me?

I see that $ x$ and $ y$ can be written both as the sum of squares but why $ z \equal{} ac \plus{} bd$ or the equivalent expression $ |ad \minus{} bc| \equal{} 1$.

Thanks in advance.[/quote]
The proof was given in PEN section,but it is not elementary and uses Gaussian prime numbers :(

[quote="radio"]So, is there a totally elementary solution?  :D[/quote]
yup,you can see it here:
http://www.mathlinks.ro/viewtopic.php?t=150421


[color=darkred]Let $ABCD$ be a convex quadrilateral. Define the points: $E\in CD,\ AE\parallel BC$;
$F\in CD,\ BF\parallel AD$. Prove that the following statements are equivalently:
$1^{\circ}.\blacktriangleright$ the quadrilateral $ABCD$ is cyclically;
$2^{\circ}.\blacktriangleright$ the quadrilateral $ABEF$ is cyclically;
$3^{\circ}.\blacktriangleright$ there is the relation $AB^2=CE\cdot DF\ .$[/color]

Who are you ? Send me a private message. know you I ?

intr-adevar e usoara problema... se extinde o linie, se observa unghiuri egale...triunghiuri asemenea...se scrie proportiile...si mi-e prea lene sa scriu amanuntele.

yeah, man, you know me, we used to throw potatoes off my grandmother's balcony back in romania.

baftos!

I don't know you. For these words there are and private messages, please ! Here is  only math-chat. I wait from you and another [color=red]"pleasant"[/color] remembrances.


Given k segments in a plane, show that the number of triangles all of
whose sides belong to the given set of segments is less than
C.k^(3/2), for some positive constant C wich is independent of k.

We will prove this statement for C=2.

Suppose there exists segment which concern at most to \sqrt k triangles. If we delete this segment then there will be k-1 segments with at most 2(k-1)\sqrt (k-1) triangles (by induction) ==> initially there were at most \sqrt k + 2(k-1)\sqrt (k-1) \leq 2k\sqrt k triangles.

Now suppose each segment concern at least to \sqrt k triangles ==> degree of each end of segment is at least \sqrt k. Let V be the number of all ends of segments. We have V.\sqrt k \leq 2k ==> V \leq 2\sqrt k ==> number of triangles is at most C(2\sqrt k, 3) \leq 2k\sqrt k.


How many odd numbers between 0 and 100 are not multiples of 3?

A.17	B.32	C.33	D.34	E.35

[hide=" click to reveal hidden content"]

($ 0$ is not odd.)

There are $ 100$ numbers between $ 1$ and $ 100$ and $ 50$ of these are divisible by $ 2$; $ 33$ of the numbers are divisible by $ 3$. This gives $ 83$ numbers but they are not all distinct: we must subtract the number of those which are divisible by $ 6$, that is $ 16$ of them to get $ 83\minus{}16\equal{}67$ which are divisible by $ 2$ or $ 3$ or both and so $ 100\minus{} 67 \equal{} 33$ which are odd and are not divisible by $ 3$.

[/hide]


Prove that $\forall\ n\in\mathbb{N},n\ge 3,\ \exists a,b\in\mathbb{N},a,b\neq 0$ such that $n=[a\sqrt{2}+b\sqrt{3}]$
[x] is the greatest integer $\le x$

Let m is greates number suth that $m\sqrt 3 <n+1-\sqrt 2$. We have $m\sqrt 3+\sqrt 2<n+1.$ If $m\sqrt 3+\sqrt 2>n$,
then a=1,b=m. Else $n+1-\sqrt 2-\sqrt 3<m\sqrt 3<n$. In this case a=2 or 3 b=m-1 or m-2.


proof this progression

\[\begin{array}{l}a_{n}= \left\{ \begin{array}{l}\begin{array}{*{20}c}0 &{if}&{n = 0}\\ \end{array}\\ \begin{array}{*{20}c}{2 \times a_{\left({n-1}\right)}+1}&{if}&{n > 0}\\ \end{array}\\ \end{array}\right. \\ proof \\ a_{n}= 2^{n}-1 \\ \end{array}\]

Find the value of $\alpha$ such that $a_{n}-\alpha =2(a_{n-1}-\alpha )\ (n\geq 1)$.

Your solution is correct, but it takes time a little. :(

:lol:
can you show to me other solution better than that please

I've never really had any experience with induction, so see if I make any unreal assumptions:

[hide]Base case: when n = 1, we have $a_{1}=2(a_{0})+1=1=2^{1}-1$.
So let's assume that this holds for $n=k$, and then for $n=k+1$

$a_{k+1}=2{a_{k}}+1=2(2^{k}-1)+1=2^{k+1}-1$,

and we are done. [/hide]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108627]Please take a look[/url]


If you know calculus, that makes it easier.

khob midoonim $ \sin \alpha\equal{}\minus{}\sin (\alpha\minus{}\pi)$ pas $ \sin^2 \alpha\equal{}\sin^2 (\alpha\minus{}\pi)$ pas darim:

$ \sin^2 \frac{\pi x}8\equal{}\sin^2 \frac{\pi(x\minus{}8)}8$

khob avval biaid soorate kasro dar mozdavajesh zarbo taghsim konid yeni age gharar bedim $ A\equal{}\sqrt{2x}\plus{}\sqrt[3]{x^2}$ darim:

$ (\sqrt{2x}\minus{}\sqrt[3]{x^2})^2\equal{}(\sqrt{2x}\minus{}\sqrt[3]{x^2})^2\times \frac{A^2}{A^2}\equal{}\frac{(2x\minus{}x\sqrt[3]{x})^2}{A^2}$

hala biaiad in akhariaro dar chagho laghaz zarb konid yeni gharar bedid $ B\equal{}4x^2\plus{}2x^2\sqrt[3]{x}\plus{}x^2\sqrt[3]{x^2}$ hala darim:

$ (\sqrt{2x}\minus{}\sqrt[3]{x^2})^2\equal{}\frac{(2x\minus{}x\sqrt[3]{x})^2}{A^2}\times \frac{B^2}{B^2}\equal{}\frac{(8x^3\minus{}x^4)^2}{A^2B^2}$

khob hala haddi ke bayad hesab konim be shekle zir dar miad:

$ S\equal{}\lim_{x\to 8} \frac{(\sqrt{2x}\minus{}\sqrt[3]{x})2}{\sin^2 \frac{\pi x}8}$

$ \equal{}\lim_{x\to 8} \frac{(8x^3\minus{}x^4)^2}{A^2B^2.\sin^2 \frac{\pi(x\minus{}8)}8}$

$ \equal{}\lim_{x\to 8}\left(\frac{x^6}{A^2B^2}.\frac{(8\minus{}x)^2}{\sin^2 \frac{\pi(x\minus{}8)}8}\right)$

$ \equal{}\lim_{x\to 8}\left(\frac{x^6}{A^2B^2}\right).\lim_{x\to 8}\left(\frac{(8\minus{}x)^2}{\sin^2 \frac{\pi(x\minus{}8)}8}\right)$

amma midoonim $ \lim_{t\to 0}\frac{t}{\sin t}\equal{}1$ pas age gharar bedim $ t\equal{}x\minus{}8$ oonmoghe vaghti $ x\to 8$ darim $ t\to 0$ pas:

$ \lim_{x\to 8}\frac{(x\minus{}8)^2}{\sin^2 \frac{\pi(x\minus{}8)}8}\equal{}\frac1{\frac{\pi^2}{64}}$

pas haddi ke mikhaim be shekle zir dar miad:

$ S\equal{}\lim_{x\to 8}\left(\frac{x^6}{A^2B^2}\right).\frac{64}{\pi^2}$

amma darim:

$ \lim_{x\to 8} \frac{x^6}{A^2B^2}$

$ \equal{}\lim_{x\to 8} \frac{x^6}{(\sqrt{2x}\plus{}\sqrt[3]{x^2})^2(4x^2\plus{}2x^2\sqrt[3]{x}\plus{}x^2\sqrt[3]{x^2})^2}$

$ \equal{}\frac{8^6}{(4\plus{}4)^2(4.8^2\plus{}4.8^2\plus{}4.8^2)^2}\equal{}\frac1{9.16}$

pas:

$ S\equal{}\frac1{9.16}.\frac{64}{\pi^2}$

$ \Rightarrow \boxed{S\equal{}\frac{4}{9\pi^2}}$

[quote="sepehr ahmadzadeh"]hala ke bahse had sho in soalam ghashange :$ \lim_{x\to0}\frac {tan(nx) \minus{} ntan(x)}{nsin(x) \minus{} sin(nx)}$ ghashange :D[/quote]

 :D  Fekr konam beshe $ \frac{{2n^2  \minus{} 1}}{{n^2  \minus{} 1}}$ albate motmaen nistam  :maybe:

[quote="sepehr ahmadzadeh"]na  :D  jalebish ine ke aslan be $ n$ rabti nadare :D albate ino goftaaam fek konam dg tablo shod..[/quote]

Sepehr khan javabesh chi mishe  :blush:

[quote="sepehr ahmadzadeh"]hala ke bahse had sho in soalam ghashange :$ \lim_{x\to0}\frac {tan(nx) \minus{} ntan(x)}{nsin(x) \minus{} sin(nx)}$ ghashange :D[/quote]

[hide]=2[/hide]


geometry
circumcircle
projective geometry
geometry unsolved

Let AOD be a diameter of the circumcircle with center of O of triangle ABC. The tangent at D meets the extension of BC at E and the line EO intersects AB at P and AC at Q with P and Q being two different points. Prove that AD bisects PQ.

$CO$ intersects $(O)$ at $F$, $DF$ intersects $AB$ at $P'$. Use Pascal's theorem with the hexagon $DDABCF$ , we have $DD \cap BC = \{ E \}, DA \cap CF = \{ O \}, AB \cap FD = \{ P' \}$ so $P',O,E$ are collinear. Thus, $P \equiv P'$, or $DP \equiv DF \parallel AC$. Similarly, $DQ \parallel AB$, then $APDQ$ is a parallelogram. So $AD$ bisects $PQ$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=110996
http://www.mathlinks.ro/Forum/viewtopic.php?t=108346

  darij


geometry
geometric transformation
rotation
search
abstract algebra
advanced fields
advanced fields unsolved

Suppose that $ E$ is an affine euclidean space of dimension $ 2$ or $ 3$ and $ G$ is a group of affine isometries such that any of its elements has at least a fixed point. Then there is a point which is fixed by all elements of the group. Moreover, the conclusion does no longer hold in dimension at least $ 4$ (even though, Grobber showed on the forum-with a very difficult proof- that the conclusion holds in any dimension if one assumes the group closed).

Are you sure, harazi? Let $ G$ be the group generated by $ x\mapsto R_{1,2}x\plus{}a_i$ where $ R_j$ are rotations by $ \alpha_j$ on the plane and the equation $ m_1\alpha_1\plus{}m_2\alpha_2\equal{}2\pi \ell$ has no non-trivial solutions in integer $ m_1,m_2,\ell$. Then, clearly, each element of the group has a unique fixed point but it is easy to choose $ a_1$ and $ a_2$ so that the fixed points are different for the 2 generators :? Of course, this group is very far from being closed, but, since you are aware of that issue, it makes me wonder what the missing condition is...

Why is it clear that every element of the group has a fixed point? :? You can choose your generators such that they have fixed points, but can you keep track of what happens when you multiply those generators with each other in various ways?

Actually, I think the problem is easy in dimension $ 2$. There is a proof in [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=91772428&t=85482]the topic mentioned by harazi[/url] of the fact that [i]abelian[/i] groups of isometries all of whose elements have fixed points must have a global fixed point (regardless of whether the group is closed or not); it's the piece of the proof labelled as Step 1. That proof extends easily to solvable groups, and since the group of isometries of Euclidean $ 2$-space is solvable, that's that. In fact, this argument ca be extended in the following way. Say that a group $ G$ is [i]good[/i] if it has the property that whenever it acts on a Euclidean space such that its elements all have fixed points, $ G$ must have a fixed point. Then a group having a [url=http://planetmath.org/encyclopedia/NormalSeries.html]subnormal series[/url] with good quotients is good.

P.S.

The proof in that topic is not 100% correct; if the Lie algebra breaks up into factors, the group itself need not do so. It can easily be repaired though :). Sorry about that.

[quote="grobber"]Why is it clear that every element of the group has a fixed point? [/quote]
Oops, I was wrong: I overlooked the possibility that commutators may give you shifts, not just identity mappings. :blush:. So, I withdraw the claim.


Let $ a,b,n $ be possitive integers , $ gcd(a,b) =1 $ . $ (x, y) $ is the non-negative solutions of the equation : $ ax+by=n $ .
  Prove that :
 a) If $ n \geq (a-1)(b-1) $ then the equation has nonnegative solutions
 b) If $ n = ab-a-b $ then the equation hasn't nonnegative solutions
 c) There are exactly $ (a-1)(b-1)/2 $ numbers n such that the equation has non-negative solutions

You may find it here (p.30-33) among many other things :
http://www.animath.fr/cours/arithm.pdf

Pierre.

I don't think you have to know french to understand the solution  ;)

Pierre.


there are two pics here with 11 true/false questions
answers are included with them
i just need help understanding each one

Just tell us what problems you encounter; it's undoable (and unuseful) to explain all 11 questions in full detail without knowing what you do and do not understand.


inequalities
trigonometry
inequalities unsolved

Prove that for every $x,y,z(x\ne y,y \ne z,z\ne x)$ we have:
$\frac{|x-y|}{\sqrt{1+x^2}.\sqrt{1+y^2}}+\frac{|y-z|}{\sqrt{1+y^2}.\sqrt{1+z^2}}>\frac{|x-z|}{\sqrt{1+x^2}.\sqrt{1+z^2}}$

Let $x=\tan \alpha,\ y=\tan \beta,\ z=\tan \gamma,$ we are to prove that $|\sin (\alpha-\beta)|+|\sin (\beta-\gamma)|>|\sin (\gamma-\alpha)|.$


Hey guys,

Just say I have $f(a,b,c)$ and I notice that $f(a,b,c)=0$ automatically when $a=b=c$.  Is there any kind of remainder theorem that tells me what I can now factor $f(a,b,c)$ into?  Because it works for 2 variables ie. you can factor out $(a-b)$ if $f(a,b)=0$ when $a=b$.  Is there a similar trick for functions in 3 or more variables?

Let me know thanks.

[quote="liangchene"]maybe you can factor out (a-b)(b-c)(c-a) something like this.[/quote]

I don't think it works because $f(a,b,c)$ will be 0 when $a=b \not= c$, which isn't necessarily true.

EDIT: Latex didn't show up properly.

There is no remainder theorem for the multivariable case, at least not that I'm aware of.

For example, let $f(a,b,c) = (a - b)^2 + (a - c)^2 + (b - c)^2$.  Then $f(a,a,a) = 0$, but $f$ itself has no factors.  You can still work with variables individually, but you can't work with more than one at a time.


Let $ A\equal{}\left(\begin{array}{cc}a & b\\c & d\end{array}\right),\ E\equal{}\left(\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right), P\equal{}\left(\begin{array}{c}p\\ q\end{array}\right)\neq\left(\begin{array}{c}0\\ 0\end{array}\right)$.

Supposed that $ AP\equal{}P,\ a\plus{}d\equal{}3$.

(1) Find the value of $ ad\minus{}bc$.

(2) Express $ A^{n}$ in terms of $ A$ and $ E$.

1) From $ AP\equal{}1P$, we know that $ 1$ is an eigenvalue corresponding to the eigenvector $ P$. But we are given that the $ \text{tr}(A)\equal{}a\plus{}d\equal{}3$ which is the sum of the eigenvalues, so the other eigenvalue must be $ 2$.

We want to know $ \det(A)\equal{}ad\minus{}bc$, but the determinate is the product of the eigenvalues, which is $ 1\cdot 2\equal{}2$, so $ ad\minus{}bc\equal{}2$.

For part 2, we know by the Cayley-Hamilton theorem that
\[ A^{2}\minus{}3A\plus{}2I\equal{}0,\]
which means
\[ A^{2}\equal{}3A\minus{}2I.\]
By induction on $ n$, we can show that
\[ A^{n}\equal{}(2^{n}\minus{}1)A\minus{}(2^{n}\minus{}2) I.\]

Here I have used $ I$ for the identity matrix, instead of $ E$.


probability
linear algebra
expected value
research

This is a population dynamics problem I'd played around with on and off since high school.  I think it's a cute problem with potential for some fairly rich mathematics, but at the same time lots of low-hanging results easily obtained by a lower level undergrad.   Depending on how you approach it and how deeply you choose to analyze it, it can take you anywhere from linear algebra and systems of difference equations to ODEs and analysis of nonlinear dynamical sytems (and likely some simulation).

[b]Some brief background[/b]: Human blood is commonly classified into four groups: O, A, B, and AB (also called, respectively, Blood Group I, II, III, and IV in some countries).  The group to which an individual belongs is determined by two alleles -- genes inherited from ones parents, one from the father and one from the mother.  There are three possible alleles: $ I_a$, $ I_b$, and $ I_o$.  $ I_a$ produces antigen A, $ I_b$ produces antigen B, and $ I_a$ is a recessive gene that produces neither antigen A nor B.  Each allele is selected from a parent's two alleles with equal likelihood.  Thus, for example, a child with parents having alleles $ I_aI_o$ and $ I_aI_b$ will have genotype $ I_aI_a$ with probability $ \frac {1}{4}$, $ I_aI_b$ with probability $ \frac {1}{4}$, $ I_aI_o$ with probability $ \frac {1}{4}$, and $ I_bI_o$ with probability $ \frac {1}{4}$.  The way in which these two alleles determine which blood group one belongs to is shown below:
\begin{tabular}{ | l | l | l | l |} \hline - & I_a & I_b & I_o \\
\hline I_a & A & AB & A \\
\hline I_b & AB & B & B \\
\hline I_o & A & B & O \\
\hline \end{tabular}
So, in the previous example, the parents will have a child with blood type A with probability $ \frac {1}{2}$, a child with blood type AB with probability $ \frac {1}{4}$, and a child with blood type B with probability $ \frac {1}{4}$.

[b]Objective[/b]: We would like to construct a mathematical model that, given initial data describing the distribution of genotypes in a population, predicts the distribution of genotypes at a future time, or at least in the very distant future.

[b]Some assumptions[/b]: To get a handle on a mathematical model of the situation, we begin with some simplifying assumptions:
(1) The population mates uniformly, i.e. all other things being equal, a person with any given genotype is equally likely to select a person with any other genotype as a partner.
(2) Life expectancy and the expected number of children is the same across all genotypes.

Neither of these assumptions are very good, in fact, but they are a starting point.

[b]Mathematical model[/b]: Under these assumptions, it is common in population dynamics modeling to assume that the rate of growth of a population is roughly proportional to the product of the sizes of the populations.  In other words, if X and Y produce Z then $ \frac {dZ}{dt} = \alpha X\cdot Y$.

Some questions to consider:
- How many populations do we need to be modeling?  (While there are 4 blood types, there are not necessarily just 4 significant populations to keep track of).
- How do we model the population growth, as a continuous process or at discrete time steps? What might be some advantages and disadvantages of either approach?
- How do we account for death?  Should we account for death at all?  After all, we're only interested in the proportion of the population that is a given genotype, not what the actual size of the population is (how is assumption (2) relevant?).

[b]Steady-state[/b]: Having considered these questions, try first to construct equations describing only the steady-state solutions (since the system is time-homogeneous (why?), the steady state condition is $ \frac {dX}{dt} = 0$, or if you are working in discrete steps then it is when $ X_{n + 1} = X_n$).  Can you classify which of these equilibria are stable or unstable?  Can you say anything about how the steady-state might depend on the initial conditions?  Try some phase plane analysis on a few interesting 2-dimensional surfaces within the 6-dimensional space.

[b]Time-dependent problem[/b]: This is a more ambitious problem yielding a rich variety of nonlinear phenomena.  One neat thing to do might be to determine numerically some basins of attraction for the various equilibria -- might not be an easy task, since the problem is (at least on the surface) at least 6-dimensional.

Though the problem is simple on the surface and immediately applicable to a very real-life problem with a slew of available data, it yields some elegant mathematics and a couple of surprising results that might even be of interest to some biologists. ;)

mathematical genetics is a very active subject. Have you checked: http://mathgen.stats.ox.ac.uk/

It's certainly a very active subject, but I don't think this particular project would fall necessarily along the lines of things that are popular right now in that field.  It's more a population dynamics problem really, with genetics only appearing superficially.  Which isn't to say it's a bad problem -- I think it's perfect for an undergrad interested in modeling and nonlinear systems analysis.


[(a/c) + (a/b) +1] / [(b/a) + (b/c) +1] = 11
a,b,c are positive integers. If a+2b+c is less or equal to 40, how many different values for (a,b,c) are there?

My friend just gave me this to try out, and I really can't figure it out. I'd like to know how to do this thing, can someone help me PLEASE? Much thanks.

errr its kinda late so i didn't check my work

but this is what i got:

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getting common denominators, you have:

$\displaystyle\frac{\frac{ab+ac}{bc} + 1}{\frac{bc+ba}{ac} + 1} \implies \displaystyle\frac{\frac{ab+ac+bc}{bc}}{\frac{bc+ab+ac}{ac}} \implies \frac{a}{b} = 11$

so, we now have $a+2b+c \le 40$ and $\displaystyle\frac{a}{b} = 11$

now, we can set values for $a$ and $b$, so $a = 11, b=1$, $a=22,b=2$, $a=33,b=3$. Now we plug that in and find values for $c$, so we have:

$11+2+c \le 40 \implies c \le 27$ giving you $27$ possible solutions

$22+4+c \le 40 \implies c \le 14$ giving you $14$ possible solutions

$33 + 6 + c \le 40 \implies c \le 1$ giving you $1$ solution

so total there are $42$ possible in the end

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Wow. I can't believe it was that simple. Much thanks man.  :blush:


Was the Middle School Tournaments Forum deleted?

EDIT: Never mind, I found it in the Classroom Math forum  :blush: