calculus integration trigonometry geometry perimeter conics ellipse

Can anyone help me simplify this?

[tex]$\int \sqrt{1\\-a^2\sin{x^2} }\; dx$[/tex]

It's part of the formula for calculating the perimeter of an ellipse.
Thanks.

It is well known that elliptic integrals do not have a closed-form antiderivative.

The area of an ellipse, however, is  :pi: ab

This is probably best seen by use of Green's Theorem in the plane:

Integrate (xdy-ydx) around the ellipse using the parametrization x=acost, y=bsint

So there's no way to calculate the perimeter of an ellipse exactly?
I know there are certain ways to approximate the perimeter.

[quote="Ravi B"]There is no closed-form formula for the perimeter of an ellipse.

For the area of an ellipse, it is easier to derive the formula by stretching the ellipse into a circle.[/quote]

Easier to understand, perhaps - Green's Theorem in the plane is extremely computationally simple.  Unless you leave out the 1/2, as I did in the post above.

Thanks everyone for your replies.  I just tumbled on the perimeter problem last couple of days, I thought it was very interesting and quite hard to derive as well unlike the area formula.  Then I researched the formula and tried to verify it with a simple equation, which I obviously couldn't accomplish because of the elliptical integral which I didn't know before.

I just looked up Green Theorem and it's a quite nice theorem.  At least it gives a generalization to the area of plane curves.

[quote="gauss202"]Most people on this forum probably haven't seen Green's Theorem yet. So if you mention a theorem like that you have to take the time to explain it in terms that people can understand. Always remember who your audience is when answering a question.

As for the original integral, it is in fact an elliptic integral. It can't be evaluated in terms of what we call elementary functions, except when a = 0 or 1. For other values of a, the definite integral can be either looked up or evaluated on a calculator or a computer.

But don't think that's very strange. Elliptic functions are functions just like any other - like sine or cosine or ex. Each of them can only be precisely evaluated at certain points, and at any other values you have to approximate it or look it up. Its just take the time to study sines and cosines and ex's in high school because they come up pretty often.

If you'd like to know more about elliptic integrals you can look in a book called [u]Advanced Calculus[/u] in the Schaum's Outline Series. It has a nice little chapter on them.[/quote]

My bad, forgot that this wasn't Advanced.

Green's Theorem in the Plane states that if a closed plane curve can be parametrized by x=f(t), y=g(t) then the area enclosed by the curve can be given by:

A = int(xdy), where the integral is taken around the curve, or

A = -int(ydx) (as before)

or (sometimes more convieniently)

A = 1/2*int(xdy-ydx)

So we can parametrize an ellipse by x=acost, y=bsint

Then dx = -asint, dy = bcost

A = 1/2*int(ab(cost)^2+ab(sint)^2) over 0, 2pi

A = 1/2 [abt]0,2pi
A = 1/2 [2pi*ab-0] =  :pi: ab

[quote="blahblahblah"][quote="anipoh"][quote="blahblahblah"]
A = int(xdy), where the integral is taken around the curve, or

A = -int(ydx) (as before)
[/quote]

I don't really see how those two formulas are used or when they are used.  Can you explain further?  Thanks.
I know how to use the third one though, it seemed quite straightforward.[/quote]

Again, in the case of the ellipse, the same area is given by:

int(ydx) over [0, 2pi], using the same parametrization as before.

We get int(ab*cost) = [ab/2*(sinxcosx + x)] 0, 2pi =  :pi: ab[/quote]

I think you meant int(xdy) over [0, 2pi] and int(ab*cos^2(t)) = .....  But overall, I get what you mean.  Those two formulas can derive the third one, which is more convenient sometimes.

Correct.  In general, int(xdy) = -int(ydx).

[quote="Magnara"]Are you sure there's no way to calculate the perimeter exactly?  A friend of mine told me that it was (a+b)* :pi:, which seems reasonable enough.  He gave me that one and the  :pi:a*b one at the same time, both without proof.[/quote]

I don't think (a+b) * :pi: works for the perimeter.  It would basically mean that for any a and b such that (a+b) is constant, then sqrt[ a^2 + b^2 ] = c = approximation(P/4) = (a+b)/4 is constant, which is impossible since a^2 + b^2 varies a lot in value as a and b changes.

[quote="Magnara"]Are you sure there's no way to calculate the perimeter exactly?  A friend of mine told me that it was (a+b)* :pi:, which seems reasonable enough.  He gave me that one and the  :pi:a*b one at the same time, both without proof.[/quote]

100% sure, without the use of elliptic integrals.

Also, just look at what happens as b gets very very large (or a gets very very small).  The arclength approaches 4b, not the pi*b your formula gives.

[quote="Magnara"]Are you sure there's no way to calculate the perimeter exactly?  A friend of mine told me that it was (a+b)* :pi:, which seems reasonable enough.  He gave me that one and the  :pi:a*b one at the same time, both without proof.[/quote]

Just for clarification on the topic...

The approximation of [tex]\pi (a+b)[/tex] was discovered in the 1600's by Kepler and it is still today to best closed form approximation for the perimeter of an ellipse known today. The only true way to approximate the perimeter of an ellipse is to setup the integral and then approximate with simpson's rule. Luckily, because of the characteristics of an ellipse, the approximation where n=4 will almost always give an answer within .01 of the exact perimeter... Not Bad...

Kepler was also the one that discovered [tex]A=\pi ab[/tex].

[quote="Magnara"]Ok then.

Btw, does an elliptic integral have anything to do with an elliptic curve?[/quote]

Yes it does. But in a very non-trivial way. From elliptic integrals you can define these things called elliptic functions. Elliptic functions turn out to be a lot like trigonometric functions in that they are periodic. But instead of being periodic on the real numbers (in one dimension) as the trig functions are, they are periodic on the complex numbers (in two directions). Because of this, and the fact that they satisfy a certain differential equation, you can use them to parameterize elliptic curves (sort of like you use trig. functions to parameterize circles, or hyperbolic trig functions to parameterize hyperboli).

It's really a quite beautiful theory. You'll learn about it in a second semester or advanced course in Complex Analysis. For those of you who already know Complex Analysis you can look in a book called [u]A Course in Modern Analysis[/u] by Whittaker and Watson for more details.

[quote="JSRosen3"]What exactly does "closed-form" mean? Does it mean that there is no formula for the perimeter of an ellipse, or just no simple formula (i.e. 3x, 5x :^2: , etc.)?[/quote]

Well, we can obviously write the equation for the perimeter of an ellipse in integral form, and use a suitable series expansion to integrate it termwise.

However, this is not 'closed-form', because it refers to an quantity with an infinite number of terms which has no value like, say, sin(x), ln(x), or x^2.

[quote="gauss202"][quote="JSRosen3"]What exactly does "closed-form" mean? Does it mean that there is no formula for the perimeter of an ellipse, or just no simple formula (i.e. 3x, 5x :^2: , etc.)?[/quote]

Closed form means in terms of elementary functions like polynomials, sines, cosines, exponentials, logarithms, etc. (basically all the functions you studied in pre-calculus), and elementary constants like pi and e.[/quote]

You also need to specify that there be only a finite number of elementary functions involved in the closed-form representation.

[quote="ComplexZeta"]What is the actual definition of elementary function?[/quote]

Here is Mathworld's rather terse and cursory view on the topic:

http://mathworld.wolfram.com/Closed-FormSolution.html

[quote="ComplexZeta"]Ha they use exact and closed to mean the same thing. The region is probably star-shaped then. :)[/quote]

http://mathworld.wolfram.com/ExactSolution.html

"exact solutions therefore need not be closed-form"


Let the function $f(x)=\sin(\omega x+\varphi)\,(\omega>0, 0\leq\varphi\leq\pi)$ be an even function on $\mathbb{R}$. Suppose that the graph of $f$ is symmetrical about the point $M\left(\frac{3\pi}4,0\right)$, and $f$ is monotonic on $\left[0,\frac{\pi}2\right]$. Find the value of $\omega$ and $\varphi$.

Is the problem too easy or too hard to be posted on here? Or has it been discussed somewhere?

Well, still no one did it...
Ok, Let me post my solution so that you can tell me whether it is correct.

[hide]As $f$ is even, we get $\sin(\omega(-x)+\varphi)=\sin(\omega x+\varphi)$. Thus
\[\sin\varphi\cos\omega x-\cos\varphi\sin\omega x=\sin\varphi\cos\omega x+\cos\varphi\sin\omega x \]
for all $x$ in $\mathbb R$. Hence $\cos\varphi=0$. But $0\leq\varphi\leq\pi$, so
\[\boxed{\varphi=\frac{\pi}2}. \]
The graph of $f$ is symmetrical about the point $M(3\pi/4,0)$, so
\[f\left(-x+\frac34\pi\right)=-f\left(x+\frac34\pi\right). \]
Set $x=0$, we get $f(3\pi/4)=-f(3\pi/4)$ and so $f(3\pi/4)=0$. Moreover,
\[f\left(\frac34\pi\right)=\sin\left(\frac{3\pi\omega}4+\frac{\pi}2\right)=\cos\frac{3\pi\omega}4. \]
Hence $\cos(3\pi\omega/4)=0$. But $\omega>0$, therefore
\[\frac{3\pi\omega}4=\frac{\pi}2+k\pi, \]
where $k$ is a non-negative integer. Consequently,
\[\omega=\frac23(2k+1). \]
(a) If $k=0$, then $\omega=2/3$. In this case $\sin\left(\frac23x+\frac{\pi}2\right)$ is decreasing on $[0,\pi/2]$;
(b) If $k=1$, then $\omega=2$. In this case $\sin\left(2x+\frac{\pi}2\right)$ is decreasing on $[0,\pi/2]$;
(c) If $k\geq2$, then $\omega\geq10/3$. In this case $\sin\left(\omega x+\frac{\pi}2\right)$ is not monotonic on $[0,\pi/2]$.
Therefore,
\[\boxed{\omega=\frac23\text{ or }\omega =2}. \]
[/hide]


Let Segment $AB$ be a diameter of a circle and $C$ be a point on segment $AB$ with $2 \times AC = BC.$

Let $D$ and $E$be points on the circle such that segment $DC$ is perpendicular to segment $AB$ and $DE$ is a second diameter. What is the ratio of the area of triangle $DCE$ to the area of triangle $ABD$?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3

I am not quite sure if this is correct
[hide] Sorry, I don't have any pictures, but I will try to explain it with words.  Say the radius of this circle is $r$ and the center is O.  Since 2AC=BC, we have that AC=AB/3.  This implies that AC=AB/2-(AB/2)/3=OA+OA/3=2r/3.  This means that OC=OA-AC=r-2r/3=r/3.  Now, from E, drop a perpindicular to DC extended and say it hits the extension at P.  Since Angle P=Angle DCO=90 and Angle ODC=Angle EDP, Triangle DCO~Triangle DPE by AA.  This means that DO/DE=CO/PE.  This means that r/2r=(r/3)/PE.  So, PE=2r/3.  This impllies that [DCE] is EP(DC)/2=(2r/3)(DC)/2=DCr/3.  [ADB] is DC(AB)/2=DC(2r)/2=DCr.  This means that $\frac{[DCE]}{[ABD]}=\frac{\frac{DCr}{3}}{DCr}=\frac{1}{3}.$ [/hide]

Well i did one example and approximated the areas of the triangles using heron's formula and i got that the answer was E.

[hide]
$CE = D(\frac{2}{3}-\frac{1}{3}) = D(\frac{1}{6})$
Key to note is: $[ADE] = [AEB]$, so the ratio we're looking for is $\frac{EC}{EA}= \frac{\frac{1}{6}}{\frac{1}{2}}= \frac{1}{3}$
[/hide]

[hide]Call the center O. [DOC]=[EOC] because they have the same height. So [DCE]=2[DOC]. [DOC]/[ABC]=1/6, because they have the same height, and bases of ratio 1/6. Thus, [DCE]/[ABD]=1/3, $\boxed{C}$.[/hide]


trigonometry
algebra
system of equations

List all solutions $ (x, y, z)$ of the following system of equations with x, y, and z positive real numbers:
$ x^2 \plus{} y^2 \equal{} 16$
$ x^2 \plus{} z^2 \equal{} 4 \plus{} xz$
$ y^2 \plus{} z^2 \equal{} 4 \plus{} yz\sqrt{3}$.

[hide]The first equation implies $ x,y,4$ are the sides of a right triangle with $ 4$ the hypotenuse.

The second equation is equivalent to $ x^2\plus{}z^2\minus{}2xz\cos60\equal{}2^2$.  Thus, $ x,z,2$ form the sides of a triangle with $ 2$ opposite a $ 60$ degree angle.

The third equation is equivalent to $ y^2\plus{}z^2\minus{}2xz\cos30\equal{}2^2$.  Thus, $ y,z,2$ form the sides of a triangle with $ 2$ opposite a $ 30$ degree angle.

Now, align lengths $ x,y,z$ as shown in the picture, with $ AD\equal{}x,BD\equal{}y,CD\equal{}z$ and let $ \angle ADC\equal{}60$ so that $ AC\equal{}2$.  Also let $ \angle BDC\equal{}30$ so that $ BC\equal{}2$.  Thus, we have $ \angle ADB\equal{}60\plus{}30\equal{}90$.  Thus, $ AB\equal{}4$.  Since $ AC\plus{}CB\equal{}2\plus{}2\equal{}4\equal{}AB$, we find that $ A,C,B$ are collinear.  Thus, $ DC$ is a median of right triangle $ ABD$.  Since the length of the median in a right triangle is half the hypotenuse, we find that $ z\equal{}DC\equal{}2$.  Thus, $ x\equal{}2,y\equal{}2\sqrt{3}$ by plugging into the equations.
[/hide]

[hide="Like this"][color=blue]We can rewrite the first equation as: $ (x,y) \equal{} 4 ( \cos \theta , \; \sin \theta ) \equal{} (4c, 4s)$

We can solve the second equation for $ z$, giving $ z \equal{} 2c \pm 2\sqrt {1 \minus{} 3c^2}$

We can also solve the third equation for $ z$, giving $ z \equal{} 2 \sqrt {3} s \pm 2 \sqrt {1 \minus{} s^2} \equal{} 2 \sqrt {3} s \pm 2c$
[/color]
$ 2c \pm 2\sqrt {1 \minus{} 3c^2} \equal{} 2 \sqrt {3} s \pm 2c$ $ \iff (2 \pm 2)c \equal{} 2 \sqrt {3} s \pm 2 \sqrt {1 \minus{} 3c^2}$ [color=blue], where the $ \pm$'s are independent.[/color]

[b][color=red]Case 1:[/color][/b] $ 0 \equal{} 2 \sqrt {3} s \pm 2 \sqrt {1 \minus{} 3 c^2} \implies s \sqrt {3} \equal{} \pm \sqrt {1 \minus{} 3 c^2}$

[color=blue]Square to get[/color] $ 3s^2 \equal{} 1 \minus{} 3c^2 \implies 3 (s^2 \plus{} c^2) \equal{} 1 \implies 3 \equal{} 1$. [color=blue][i]Contradiction.[/i][/color]

[b][color=red]Case 2:[/color][/b] $ 2c \minus{} s\sqrt {3} \equal{} \pm \sqrt {1 \minus{} 3 c^2}$

[color=blue]Square:[/color] $ 4c^2 \plus{} 3s^2 \minus{} 4 \sqrt {3} cs \equal{} 1 \minus{} 3 c^2 \implies 1 \plus{} 2 c^2 \equal{} 2 \sqrt {3} cs$

[color=blue]Square:[/color] $ 4c^2 \plus{} 4c^2 \plus{} 1 \equal{} 12 c^2 \minus{} 12 c^4$

$ \iff 16 c^4 \minus{} 8 c^2 \plus{} 1 \equal{} (4c^2 \plus{} 1)^2 \equal{} 0$

$ \implies c^2 \equal{} 1/4 \implies c \equal{} \frac {1}{2} \implies s \equal{} \frac {\sqrt {3}}{2}$

$ \boxed{(x,y,z) \equal{} (2, 2 \sqrt {3} , 2)}$[/hide]


I have been trying to solve this for a while, I keep getting closer to the answer but I don't quite have it.

Here's the question:

Five friends have a small shop that opens from monday to friday. Since it is sufficient for 2 people to run the shop, the friends decide to make a work plan for each week, that specifies who will work each day and that satisfies these 2 conditions:

1. Each friend works exactly 2 days of the week.
2. All 5 pairs have to be different.

How many different plans of work can they make the work plan?

Here's what I have:

This is the number of different pairs $ \binom{5}{2}\equal{}\frac{5!}{(2!)(5\minus{}2)!}\equal{}\frac{5*4}{2!}\equal{}5*2\equal{}10$

Since there are five days the possibilities would be 10*9*8*7*6, but not all of those satisfies the condition number 1. Can anyone help me finish?

Interesting problem. Where did you find it?

*I have confused myself*

Let the 5 friends be $ A$, $ B$, $ C$, $ D$, and $ E$. $ A$ works on exactly two of the 5 days. There are $ \binom{5}{2} \equal{} 10$ different pairs of days on which $ A$ could be working. On each of those days, $ A$ works with one of the other friends. There are $ 4\cdot3 \equal{} 12$ ways to do this. So $ A$'s work schedule has $ 10\cdot12\equal{}\boxed{120}$ possibilities. Assume without loss of generality that the pairs are $ AB$ and $ AC$.

There are 3 other days we have left to fill in. Note that on none of these days will $ B$ and $ C$ be working together, because then $ D$ and $ E$ will have to work together on both of the remaining two days, which is against the criteria. So on $ B$'s other workday, s/he works with $ D$ or $ E$. That's two possible partners on three possible days, for a total of $ \boxed{6}$ possibilities for $ B$. Assume without loss of generality that $ B$ works with $ D$.

Now on the remaining 2 days to fill in, the pairs must be $ CD$ and $ DE$. That's $ 2! \equal{} \boxed{2}$ ways.

Thus the total number of ways is $ 120 \cdot 6 \cdot 2 \equal{} \boxed{1440}$.

I am not entirely sure that my answer is correct.  :maybe:

We let each day have a first slot and a second slot, such that the people in the first slots of each day are all distinct, and the people in the second are also distinct. There are $ 5!$ ways to permute the five people into the first slots for each day.

We let the operation "$ X \rightarrow$" yield the person in the second slot of the day in which the first slot is occupied by $ X$ (and by $ X \leadsto$ we indicate that a person could not be in the second slot). For example, if $ A$ works in slot one of Tuesday, and $ B$ works in slot two of Tuesday, then we write $ A \rightarrow B$. Then we note that $ X_1 \leadsto X_1$, and also $ X_1 \rightarrow X_2 \leadsto X_1$ (since all pairs are distinct).

We select an arbitrary person and repeat this operation until it becomes periodic, e.g. we have a cycle, which as we showed above must have at least $ 3$ distinct members. (and because we have a finite set it must become periodic) Clearly a person can only belong to one cycle. But since we have only $ 5$ members, then the only possibility is for one cycle with all $ 5$ members; and there are $ (5 \minus{} 1)! \equal{} 4!$ cyclic permutations of such.
However, since the order of slots in a single day does not matter, we've counted each arrangement twice. Thus $ \frac {5! \times 4!}{2} \equal{} 1440$.

[b][i]So,[/i][/b] what happens if we have $ 6$ people on $ 6$ days, distinct pairs each day, then?


[quote="Lovasz"]Let $ab \ge 1$ with $a$ and $b$ positive. Prove that: \[\frac{1}{(a+2)^{2}}+\frac{1}{(b+2)^{2}}\ge \frac{2}{3}\cdot\frac{1}{2+bc}.\] [/quote]
Let $ab \ge 1$ with $a$ and $b$ real. Prove that: \[\frac{1}{(a+2)^{2}}+\frac{1}{(b+2)^{2}}\ge \frac{2}{3}\cdot\frac{1}{2+ab}.\] This is true and nice! ;)

[hide]Multiply both sides by $3(2+ab)(a+2)^{2}(b+2)^{2}$ to reformulate as $3(2+ab)[(b+2)^{2}+(a+2)^{2}] \geq (a+2)^{2}(b+2)^{2}$

By AM-GM $(b+2)^{2}+(a+2)^{2}\geq (a+2)(b+2)$ so it remains to prove that $3(2+ab) \geq (a+2)(b+2)$

Expanding $3(2+ab) \geq (a+2)(b+2) \Leftrightarrow 1+ab \geq a+b$

Cauchy: $(a+b) \leq \sqrt{(1+b)(a+1)}= \sqrt{ab+a+b+1}\leq (ab+a+b+1) \Leftrightarrow ab+a+b\geq 0$

But with Cauchy again $(a+b)^{2}\leq (1+b)(a+1) \Rightarrow  a^{2}+b^{2}+2ab \leq ab+a+b+1 \Rightarrow  ab+a+b\geq 0$ is true iff $a^{2}+b^{2}+2ab-1 \geq 0$ which is true as $ab \geq 1$[/hide]

[quote="vulgarfraction"]

Expanding $3(2+ab) \geq (a+2)(b+2) \Leftrightarrow 1+ab \geq a+b$

[/quote]
It is wrong: $a=4,$ $b=\frac{1}{2}.$ :wink:


Problem 1:

Given that $ p$ is not a factor of $ n$ for all primes $ p \leq \sqrt[3]{n}$, show that $ n$ is either a prime or a product of two primes.

Problem 2:

Prove that Goldbach's Conjecture that every even integer greater than 2 is the sum of two primes is equivalent to the statement that every integer greater than 5 is the sum of three primes.

[hide="Problem 1"]Assume that $ n$ has three distinct prime factors $ p_{1},p_{2}$ and $ p_{3}$. Then we get $ \max\{p_{1},p_{2},p_{3}\}> \sqrt[3]{n}$. However, this contradicts our assumption. In a similar way, we can show that $ n\neq p_{1}^{a}p_{2}^{b}$ with $ a,b>1$. Then the conclusion easily follows.[/hide]

[hide="Problem 2"]For any even number $ 2n$, $ 2n-2=p+q$, where $ p,q$ are primes, is equivalent to $ 2n=p+q+2$, a sum of three primes.

For any odd number $ 2n+1$, $ 2n-2=p+q$ is equivalent to $ 2n+1=p+q+3$, a sum of three primes.[/hide]


Prove:

As \[n \rightarrow \infty\]

\[\frac{1}{\sqrt{n}}(\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+...+\frac{1}{\sqrt{n+n}}) \rightarrow 2(\sqrt2-1)\]

Hint [hide]Write this sum so that its limit appears as an integral[/hide]

Hmm...

I can see how to write this as sigma notation, but I'm too stupid to think of how to express this as an integral... *scratches head*


Prove that $ \displaystyle\sum_{k\equal{}1}^n (\minus{}1)^k(\frac{k^2}{k!}) \equal{} \frac{(\minus{}1)^n}{(n \minus{} 1)!}$.

[quote="Farnak"]Prove that $ \displaystyle\sum_{k \equal{} 1}^n ( \minus{} 1)^k(\frac {k^2}{k!}) \equal{} \frac {( \minus{} 1)^n}{(n \minus{} 1)!}$.[/quote]

just wondering, why no induction?
it is the most elegant method.

[quote="CircleSquared"]
just wondering, why no induction?
it is the most elegant method.[/quote]
I think he might be looking for an elegant combinatorial argument for it...

Oh, originally this problem had k = 4000 and it just asked you to evaluate for the LS and I found a pattern that turned out to be the RS.
However you can't use induction for the original problem so I was wondering how you would start from knowing only the LS and eventually getting to the RS.

[hide="Giveaway Hint"]
Note that: $ \frac{k^2}{k!} \equal{} \frac{k}{(k\minus{}1)!} \equal{} \frac{(k\minus{}1)\plus{}1}{(k\minus{}1)!} \equal{} \frac{1}{(k\minus{}2)!} \plus{} \frac{1}{(k\minus{}1)!}$

Now the sum telescopes.
[/hide]

[quote="Farnak"]you can't use induction for the original problem[/quote]

Why not?  :)

I think he ran across $ \displaystyle\sum_{k\equal{}1}^{4000} (\minus{}1)^k(\frac{k^2}{k!})$ on a math contest or something.

In a case like that, it would probably be faster to see the telescoping solution than to try out numbers, look for a pattern, guess the answer, and then verify by induction.

I don't see the idea of not using induction...

When you do the inductive step, you get that the last term is f(n)-f(n-1)

you might as well just replace every term in the sum with that and you get something that telescopes.


Here is this problem I recently found:

[quote="Kleppner+Kolenkow"]The acceleration of gravity can be measured by projecting a body upwards and measuring the time it takes to pass two given points in two directions
Show that if the time the body takes to pass a horizontal line $ A$ in both directions is $ T_A$, and the time to go by a second horizontal line $ B$ is $ T_A$, then, assuming that the acceleration is constant, its magnitude is:
\[ g \equal{} \frac {8h}{T_A^2 \minus{} T_B^2},
\]
where $ h$ is the height of $ B$ above $ A$.
[/quote]

This problem is pretty simple if you bust in equations, but I was looking for a more elegant solution. My work was based on the fact that $ \frac {T_A \minus{} T_B}{2}$ is the time it takes to initially pass from line $ A$ to $ B$ and that $ \frac {T_A \plus{} T_B}{2}$ is the rest of the time to hit line $ A$. However, I am unable to do anything with this. Help would be appreciated.

Since the acceleration is constant, the path from any level to the top of the trajectory and the path back to that level are "symmetrical."
Let the up speed as the body moves up through $ A$ and $ B$ be $ x$ and $ y$ respectively.
As the body comes back down, the up speed on passing $ A$ and $ B$ will be $ \minus{}x$ and $ \minus{}y$ respectively.
As the body went up from $ B$ to the top of its trajectory and back down to $ A$, its up speed changed from $ y$ to $ \minus{}x$. The change in speed is:
$ g\frac {T_A \plus{} T_B}{2}\equal{}y\plus{}x$
As the body went up from $ A$ to $ B$, the average up speed is $ \frac {x\plus{}y}{2}$, which multiplied by the time gives you the vertical distance:
$ \frac {T_A \minus{} T_B}{2}\frac {x\plus{}y}{2}\equal{}h$
Solve that equation for $ x\plus{}y$ and replace in the other equation above.

Thanks a lot man! That was good and was exactly what I was looking for :). I had gotten about 3/4s of it; I should have gotten the last part :(


Prove that
$3\sqrt[3]{(a^{3}+1)(b^{3}+1)(c^{3}+1)}+1 \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2\sqrt{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$
where a;b;c are positive real numbers satisfying $abc=1$


[quote="wailam010"]The inscribed circle of △ABC touched BC at D.
Show that the in-centre of △ABC lies on the straight line joining the mid-points of BC and AD.

Pls explain to me with steps.. THZ A LOT!![/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=5811
http://www.mathlinks.ro/Forum/viewtopic.php?t=112509
http://www.mathlinks.ro/Forum/viewtopic.php?t=106337 (thanks Pontios for this link)

  darij


[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {ab + ac - bc}{b^2 + c^2} + \frac {bc + ba - ca}{c^2 + a^2} + \frac {ca + cb - ab}{a^2 + b^2} \ge \frac {3}{2}
\]
I would like to see a solution without the modern methods such as SOS,... Also please dont use the direct expanding :)[/quote]
I cant find my mistake
$ LHS=sum \frac{(ab + ac - bc)^2}{(ab + ac - bc)(b^2+c^2)}\ge \frac{(ab+bc+ca)^2}{2abc(a+b+c)}$ since CS ineq
and $ (ab+bc+ca)^2\3abc(a+b+c)$ is obivious. :maybe:
I think my proof is wrong,but I cant find my mistake  :(

[quote="Allnames"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {ab + ac - bc}{b^2 + c^2} + \frac {bc + ba - ca}{c^2 + a^2} + \frac {ca + cb - ab}{a^2 + b^2} \ge \frac {3}{2}
\]
I would like to see a solution without the modern methods such as SOS,... Also please dont use the direct expanding :)[/quote]
I cant find my mistake
$ LHS = sum \frac {(ab + ac - bc)^2}{(ab + ac - bc)(b^2 + c^2)}\ge \frac {(ab + bc + ca)^2}{2abc(a + b + c)}$ since CS ineq
and $ (ab + bc + ca)^2\3abc(a + b + c)$ is obivious. :maybe:
I think my proof is wrong,but I cant find my mistake  :([/quote]

Oh, It 's wrong , my friend :( .

Because $ ab + ac - bc;bc + ba - ca;ca + cb - ab$ may be negative so denominator can be negative .

But when apply C-S deniminator must be non-negative

So It 's your mistake :wink:

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {ab \plus{} ac \minus{} bc}{b^2 \plus{} c^2} \plus{} \frac {bc \plus{} ba \minus{} ca}{c^2 \plus{} a^2} \plus{} \frac {ca \plus{} cb \minus{} ab}{a^2 \plus{} b^2} \ge \frac {3}{2}
\]
I would like to see a solution without the modern methods such as SOS,... Also please dont use the direct expanding :)[/quote]
Let $ a\plus{}b\plus{}c\equal{}p\equal{}1,ab\plus{}bc\plus{}ca\equal{}q;abc\equal{}r.$

$ Ineq \leftrightarrow 12q^3\minus{}11q^2\plus{}2q\plus{}2r(2q\plus{}1)\minus{}9r^2 \ge\ 0$

Let $ f(r)\equal{}2r(2q\plus{}1)\minus{}9r^2\plus{}12q^3\minus{}11q^2\plus{}2q$

$ \rightarrow f '(r)\equal{}4q\plus{}2\minus{}18r^2 \ge\ 0$

.Case1: $ q \le\ \frac{1}{4}$ we have $ r \ge\ 0$

$ \rightarrow f(r) \ge\ f(0)\equal{} 12q^3\minus{}11q^2\plus{}2q\equal{}q(4q\minus{}1)(3q\minus{}2) \ge\ 0$

.Case1: $ q \ge\ \frac{1}{4}$. Follow schur we have $ f(r) \ge\ \frac{4q\minus{}1}{9}$

$ f(r) \ge\ f(\frac{4q\minus{}1}{9})\equal{}36q^3\minus{}33q^2\plus{}10q\minus{}1\equal{}(4q\minus{}1)(3q\minus{}1)^2 \ge\ 0$

We have done! :)

[quote="onlylove_math"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {ab \plus{} ac \minus{} bc}{b^2 \plus{} c^2} \plus{} \frac {bc \plus{} ba \minus{} ca}{c^2 \plus{} a^2} \plus{} \frac {ca \plus{} cb \minus{} ab}{a^2 \plus{} b^2} \ge \frac {3}{2}
\]
I would like to see a solution without the modern methods such as SOS,... Also please dont use the direct expanding :)[/quote]
Let $ a \plus{} b \plus{} c \equal{} p \equal{} 1,ab \plus{} bc \plus{} ca \equal{} q;abc \equal{} r.$

$ Ineq \leftrightarrow 12q^3 \minus{} 11q^2 \plus{} 2q \plus{} 2r(2q \plus{} 1) \minus{} 9r^2 \ge\ 0$

Let $ f(r) \equal{} 2r(2q \plus{} 1) \minus{} 9r^2 \plus{} 12q^3 \minus{} 11q^2 \plus{} 2q$

$ \rightarrow f '(r) \equal{} 4q \plus{} 2 \minus{} 18r^2 \ge\ 0$

.Case1: $ q \le\ \frac {1}{4}$ we have $ r \ge\ 0$

$ \rightarrow f(r) \ge\ f(0) \equal{} 12q^3 \minus{} 11q^2 \plus{} 2q \equal{} q(4q \minus{} 1)(3q \minus{} 2) \ge\ 0$

.Case1: $ q \ge\ \frac {1}{4}$. Follow schur we have $ f(r) \ge\ \frac {4q \minus{} 1}{9}$

$ f(r) \ge\ f(\frac {4q \minus{} 1}{9}) \equal{} 36q^3 \minus{} 33q^2 \plus{} 10q \minus{} 1 \equal{} (4q \minus{} 1)(3q \minus{} 1)^2 \ge\ 0$

We have done! :)[/quote]
Read at my request, no modern method please!

$a,b,c,d>0,$

\[{\frac {acd+abd+abc-3\,bcd}{ \left( b+c+d \right) ^{3}}}+{\frac {abd+a
bc+bcd-3\,acd}{ \left( c+d+a \right) ^{3}}}+{\frac {abc+bcd+acd-3\,abd
}{ \left( d+a+b \right) ^{3}}}+{\frac {bcd+acd+abd-3\,abc}{ \left( a+b
+c \right) ^{3}}}\geq 0\]

[quote=can_hang2007]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac{ab\plus{}ac\minus{}bc}{b^2\plus{}c^2}\plus{}\frac{bc\plus{}ba\minus{}ca}{c^2\plus{}a^2}\plus{}\frac{ca\plus{}cb\minus{}ab}{a^2\plus{}b^2} \ge \frac{3}{2}\]
I would like to see a solution without the modern methods such as SOS,... Also please dont use the direct expanding :)[/quote]

We have
\[LHS-RHS=\sum{\frac{ab(a-b)^2}{(a^2+c^2)(b^2+c^2)}}+\frac{(a-b)^2(b-c)^2(c-a)^2}{2(a^2+b^2)(b^2+c^2)(c^2+a^2)}\ge{0}\]


search
number theory unsolved
number theory

Calculate $ n,m, k$
$ n \plus{} k \equal{} ( GCD(n,k))^2$
$ k \plus{} m \equal{} (GCD(k,m))^2$
$ n \plus{} m \equal{} (GCD(m,n))^2$

it is PEN (A113) search before posting... :)


(1)$ f: Q \implies Q$
(2) $ f(0) \equal{} 2 , f(1) \equal{} 3$
(2) for all real number x and all integer n,   $ f(x \plus{} n) \minus{} f(x) \equal{} n\{(f(x \plus{} 1) \minus{} f(x)\}$
(3) $ f(x) \equal{} f(\frac {1}{x})$
What is the possible $ x$   if $ f(x) \equal{} 2002$  ?

Answer is $ 2000 , \frac {125}{16} , \frac {16}{125}, \frac {1}{2000}$

But I don't know that reason. Can I find f(x) for all x? :(

It seems you are solving $ f(x)\equal{}2002$ not $ f(x)\equal{}2009$.

Actually $ f(\frac mn)\equal{}mn\plus{}2$ where $ (m,n)\equal{}1$. It can be shown by induction in $ k\equal{}\min(|m|,|n|)$.

Then $ mn\equal{}2000,(m,n)\equal{}1$ only has solution $ (2000,1)$ or $ (125,16)$.

[quote="xxp2000"]It seems you are solving $ f(x) \equal{} 2002$ not $ f(x) \equal{} 2009$.

Actually $ f(\frac mn) \equal{} mn \plus{} 2$ where $ (m,n) \equal{} 1$. It can be shown by induction in $ k \equal{} \min(|m|,|n|)$.

Then $ mn \equal{} 2000,(m,n) \equal{} 1$ only has solution $ (2000,1)$ or $ (125,16)$.[/quote]

You're right. my mistake. so I corrected.
 and Thank you  :lol:


Let $ ABCD$ be a quadrilateral with an incircle of center $ I$ such that $ <BAD \plus{} <ADC > 180$. Let $ l$ be a line through $ I$ and $ X$ and $ Y$ the points of intersection of $ AB$ and $ CD$ with $ l$, respectively. Prove that if $ IX \equal{} IY$ then $ (AX)(DY)\equal{}(BX)(CY)$

See [url]http://www.mathlinks.ro/viewtopic.php?t=149151[/url]
and [url]http://www.mathlinks.ro/viewtopic.php?t=146483[/url] for a short solution of equivalent problem.


algebra
polynomial
calculus
integration
algebra unsolved

find all polynomials which have -1,0,1 as their coefficients and possess all integral roots

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots+a_{1}x+a_{0}$ where $a_{i}$ are equal with 1 or -1. If $P(x)$ has an integer root $x_{0}$ then $x_{0}$ must divide $a_{0}$ , so $x_{0}$ must be 1 or -1.
Then $P(x)=(x-1)^{n_{1}}(x+1)^{m_{1}}$ where $m_{1}, n_{1}$ are natural and $m_{1}+n_{1}=n$.
The coefficient of $x^{k}$ is $(-1)^{p}C_{n_{1}}^{p}C_{m_{1}}^{q}, p+q=k$ and must be 1 or -1 for all k. So $m_{1}$ and $n_{1}$ must be 0 or 1. The polonomials are $x-1, x+1, x^{2}-1$

Any of the previous solutions can be multiplied by $x^{k}$ for natural number k.  The case of the constant term being equal to zero was overlooked.  Also the monomial $x^{k}$ is a solution.  Finally, signs can be interchanged throughout each solution.

Nothing, except you can't have $k=0$ and both minus signs simultaneously.

(nitpicking continued :wink: )
That's why you excluded $P(x)=\pm1$, I agree. But if you say $k\neq0$, then you exlude e.g. $P(x)=x-1$. Or is this just a n(ot)ational issue : my interpretation of $\mathbb{Z}^{+}_{0}$ is the set {1,2,3,...}=$\mathbb{N}_{0}=\mathbb{N}\setminus\{0\}$.

It's just a n(ot)ational issue. In my practice:

$\mathbb{Z}^{+}=\{1,2,3,\dots\}\ (=\mathbb{N})$

$\mathbb{Z}_{0}^{+}=\{0,1,2,3,\dots\}\ (=\mathbb{N}_{0})$

$\mathbb{Z}^{-}=\{-1,-2,-3,\dots\}$

$\mathbb{Z}_{0}^{-}=\{0,-1,-2,-3,\dots\}$

$\mathbb{Z}^{\pm}=\mathbb{Z}\setminus\{0\}$

It's interesting that our interpretations of $\mathbb{N}$ and $\mathbb{N}_{0}$ are directly opposite... Long live the non-ambiguity of math language ;)

TANK YOU TANK YOU .
SHEER AMBGUITY. :rotfl:  :mad:  WHO SUGGESTS me a method to solve the same when all roots are real ????


Given $ f(r)=a^{r}(b-c)+b^{r}(c-a)+c^{r}(a-b)$ in which $ a,b,c$ are nonzero and distinct. If $ n$ is a positive integer and $ f(n+1)=0$, prove that $ f(n)f(n+2)=a^{n}b^{n}c^{n}(a-b)(b-c)(c-a)f(-n)$.

[quote="N.T.TUAN"]Given $ f(r) \equal{} a^{r}(b\minus{}c)\plus{}b^{r}(c\minus{}a)\plus{}c^{r}(a\minus{}b)$ in which $ a,b,c$ are nonzero and distinct. If $ n$ is a positive integer and $ f(n\plus{}1) \equal{} 0$, prove that $ f(n)f(n\plus{}2) \equal{} a^{n}b^{n}c^{n}(a\minus{}b)(b\minus{}c)(c\minus{}a)f(\minus{}n)$.[/quote]
Let $ x_{a}\equal{} a^{n\plus{}1}(b\minus{}c)$; $ x_{b}\equal{} b^{n\plus{}1}(c\minus{}a)$ and $ x_{c}\equal{} c^{n\plus{}1}(a\minus{}b)$, then $ f(n\plus{}1) \equal{} 0$ is equivalent to $ x_{a}\plus{}x_{b}\plus{}x_{c}\equal{} 0$ so $ (x_{a}\plus{}x_{b}\plus{}x_{c})^{2}\equal{} 0$ then $ x_{a}^{2}\plus{}x_{b}^{2}\plus{}x_{c}^{2}\equal{}\minus{}2x_{a}x_{b}\minus{}2x_{b}x_{c}\minus{}2x_{c}x_{a}$
on the other hand
$ f(n)f(n\plus{}2) \equal{} (\frac{x_{a}}{a}\plus{}\frac{x_{b}}{b}\plus{}\frac{x_{c}}{c})(ax_{a}\plus{}bx_{b}\plus{}cx_{c})$
$ \equal{} x_{a}^{2}\plus{}x_{b}^{2}\plus{}x_{c}^{2}\plus{}x_{a}x_{b}(\frac{a}{b}\plus{}\frac{b}{a})\plus{}x_{b}x_{c}(\frac{b}{c}\plus{}\frac{c}{b})\plus{}x_{c}x_{a}(\frac{c}{a}\plus{}\frac{a}{c})$
$ \equal{} x_{a}x_{b}(\frac{a}{b}\plus{}\frac{b}{a}\minus{}2)\plus{}x_{b}x_{c}(\frac{b}{c}\plus{}\frac{c}{b}\minus{}2)\plus{}x_{c}x_{a}(\frac{c}{a}\plus{}\frac{a}{c}\minus{}2)$
$ \equal{} x_{a}x_{b}\frac{(a\minus{}b)^{2}}{ab}\plus{}x_{b}x_{c}\frac{(b\minus{}c)^{2}}{bc}\plus{}x_{c}x_{a}\frac{(c\minus{}a)^{2}}{ca}$
$ \equal{}\frac{x_{a}x_{b}}{ab}\left(\frac{x_{c}}{c^{n\plus{}1}}\right)^{2}\plus{}\frac{x_{b}x_{c}}{bc}\left(\frac{x_{a}}{a^{n\plus{}1}}\right)^{2}\plus{}\frac{x_{c}x_{a}}{ca}\left(\frac{x_{b}}{b^{n\plus{}1}}\right)^{2}$
$ \equal{}\frac{x_{a}x_{b}x_{c}}{abc}\left(\frac{x_{a}}{a^{2n\plus{}1}}\plus{}\frac{x_{b}}{b^{2n\plus{}1}}\plus{}\frac{x_{c}}{c^{2n\plus{}1}}\right)$
$ \equal{} a^{n}b^{n}c^{n}(a\minus{}b)(b\minus{}c)(c\minus{}a)\left(\frac{(b\minus{}c)}{a^{n}}\plus{}\frac{(c\minus{}a)}{b^{n}}\plus{}\frac{(a\minus{}b)}{c^{n}}\right)$
$ \equal{} a^{n}b^{n}c^{n}(a\minus{}b)(b\minus{}c)(c\minus{}a)f(\minus{}n)$ Q.E.P.D

This is the another solution: By $ \left|\begin{array}{ccc}f(n\plus{}2) & f(n\plus{}1) & c^{n\plus{}1}\\ f(n\plus{}1) & f(n) & c^{n}\\ 0 & 0 & 1\end{array}\right| \equal{}\left|\begin{array}{ccc}a^{n\plus{}1}& b^{n\plus{}1}& c^{n\plus{}1}\\ a^{n}& b^{n}& c^{n}\\ 1 & 1 & 1\end{array}\right|\cdot\left|\begin{array}{ccc}a(b\minus{}c) & (b\minus{}c) & 0\\ b(c\minus{}a) & (c\minus{}a) & 0\\ c(a\minus{}b) & (a\minus{}b) & 1\end{array}\right|$ we have $ f(n\plus{}2)f(n)\minus{}f^{2}(n\plus{}1) \equal{} (a^{n}b^{n}(a\minus{}b)\plus{}b^{n}c^{n}(b\minus{}c)\plus{}c^{n}a^{n}(c\minus{}a))(a\minus{}b)(b\minus{}c)(c\minus{}a) \equal{} a^{n}b^{n}c^{n}(a\minus{}b)(b\minus{}c)(c\minus{}a)f(\minus{}n)$ and we're done!


Show that every number of the form $2^{p}3^{q}$ , where $p,q$ are nonnegative integers, divides some number of the form $a_{2k}10^{2k}+a_{2k-2}10^{2k-2}+...+a_{2}10^{2}+a_{0}$, where $a_{2i}\in\{1,2,...,9\}$

[quote=cuopbientcd]$X$ is set of number of form $a_{2k}10^{2k}+...+a_{o}$ as problem.
lemma:For all k integer we have 2^k is divise a number belong to X.
By induction.
Return problem.
Let  $n=2^{p}.k$ (k is not divise by 5)and k odd.we will find a number in X which divise n.
I see there is m in X so $2^{p}|m$ and supose m has d digits.Choose $f=10^{d+1}-1$
Since $10^{\phi(fk)}=1$$mod$ $fk$
then $\frac{m(10^{(d-1)\phi(fk)})}{10^{d+1}-1}$ is number which we need find.[/quote]

Why  $\frac{m(10^{(d-1)\phi(fk)})}{10^{d+1}-1}$ is number which we need ? Can you explain more ?


Determine all $f: \mathbb{R}\rightarrow\mathbb{R}$ for which
\[ 2\cdot f(x)-g(x)=f(y)-y \textrm{ and } f(x)\cdot g(x) \geq x+1. \]

The lhs of the first condition is independant of $y$, so $f(y) = y + c$, where $c$ is a constant.
Clearly $c=f(0)$. Thus $g(x) = 2x+c$.
The second condition may be rewritten as $(x+c)(x+2c) \geq x+1$ for all $x$. Expanding, we get a quadratic expression in $x$ which has to be non-negative, and whose disriminant is $\Delta = (c-3)^2$. But, if $\Delta > 0$ then the quadratic takes positive and negative values...then $c=3$, and $f(x) = x+3$, which is a solution.

Pierre.


MATHCOUNTS
probability
geometry
AoPS Books
number theory
\/closed

I of heard that there are several new AoPS books comming out? What are they about? What will they cover? Will they still be helpfull if I've already read Aops vol 1? vol2?  When, why,  who, what, how?

thanks!

We are publishing several new books.  The first set of books will be at the Introductory level, meaning they are good for students preparing for MATHCOUNTS and/or the AMC 10/12.  Unlike the current AoPS books, our new books will each focus on a particular subject.

The first book, [i]Introduction to Counting & Probability[/i], will hopefully be submitted to the printers tomorrow  :)  and should be available in late November or early December.  The next two books, [i]Introduction to Number Theory[/i] and [i]Introduction to Geometry[/i], will hopefully be available in the next 2-6 months.  Each book will also have an accompanying solutions manual.  We also expect to have [i]Introduction to Algebra[/i] sometime later next year, and books at the Intermediate level in the next 1-2 years.

If you have [i]completely[/i] mastered AoPS Vols 1 and 2, then you won't need any of our new Introductory-level books.  If you've [i]completely[/i] mastered AoPS Vol 1, then it becomes a judgement call, and probably depends on the subject.  In particular, Counting & Probability is very lightly covered in Vol 1, so probably most students who have read Vol 1 would still benefit from our new [i]Introduction to Counting & Probability[/i] book.  If you have not completely mastered a subject in AoPS Vol 1, then you'll most likely find that our new subject books cover that subject in much greater depth.  Also our new subject books have completely new exercises than those in AoPS (the Counting & Probability book, for example, has over 400 exercises and problems, and that number is probably an underestimate, since many of the problems have multiple parts).

In the coming weeks, sometime between now and when the first book is published, we'll have a web page on our site which describes our new books in much greater detail.

Thanks for the information, I look forward to reading some of them.


inequalities
geometry
circumcircle
incenter
geometry open

Given triangle ABC with R radius of circumcenter,ra,rb,rc radius of extangent circle,r radius of incenter and a,b,c sides.Prove that
$ \frac{2R\minus{}r_{a}}{bc}\plus{}\frac{2R\minus{}r_{b}}{ac}\plus{}\frac{2R\minus{}r_{c}}{ab}\le \frac{1}{4r}$

$ \sum \frac {2R \minus{} r_{a}}{bc}\le \frac {1}{4r} \\
 \\
\sum \frac {2R}{bc} \minus{} \sum \frac {r_a}{bc}\le \frac {1}{4r} \\
 \\
\sum \frac {2\frac {abc}{4S}}{bc} \minus{} \sum \frac {\frac {S}{p \minus{} a}}{bc}\le \frac {p}{4S} \\
 \\
\sum \frac {a}{2S} \minus{} \sum \frac {S}{(p \minus{} a)bc}\le \frac {p}{4S} \\
 \\
4p \minus{} \sum \frac {4S^2}{(p \minus{} a)bc}\le p \\
 \\
3p \le \sum \frac {4S^2}{(p \minus{} a)bc} \\
 \\
3 \le \sum \frac {4(p \minus{} b)(p \minus{} c)}{bc} \\
 \\
\mathrm{Let} \begin{array}{cc} a \equal{} x \plus{} y \\
b \equal{} y \plus{} z \\
c \equal{} x \plus{} z \end{array} \\
 \\
\sum \frac {xy}{(x \plus{} z)(y \plus{} z)} \ge \frac {3}{4} \\
 \\ \sum \frac {xy(x\plus{}y)}{(x\plus{}y)(x \plus{} z)(y \plus{} z)} \ge \frac {3}{4}
\\\\
\sum xy(x \plus{} y) \ge \frac {3}{4}(x \plus{} z)(y \plus{} z)(x \plus{} y) \\
 \\
x^2 y \plus{} x y^2 \plus{} x^2 z \plus{} y^2 z \plus{} x z^2 \plus{} y z^2 \ge 6xyz$
 Which is true by $ AM \minus{} GM$.

[color=darkblue]Observe that $ S\equal{}(p\minus{}a)r_a\equal{}pr\ \implies\ ar_a\equal{}p(r_a\minus{}r)$ .

I use the relations $ abc\equal{}4Rpr$ and $ r_a\plus{}r_b\plus{}r_c\equal{}4R\plus{}r$ .

Therefore, $ \boxed {\ \begin{array}{c}
\\\
\boxed {\ \sum\frac {2R \minus{} r_{a}}{bc}\le \frac {1}{4r}\ }\\\\
\sum a(2R\minus{}r_a)\le pR\\\\
3pR\le \sum ar_a\\\\
3pR\le \sum p(r_a\minus{}r)\\\\
3(R\plus{}r)\le \sum r_a\\\\
3(R\plus{}r)\le 4R\plus{}r\\\\
\boxed {\ 2r\le R\ }\\\
\end{array}\ }$ .[/color]


In a game, $A$ flips a fair coin two times. B then flips coins until B's last two flips form a different sequence than A's. C then flips until A's sequence or B's sequence occurs. The player whose sequence occurs is the winner. Compute the probability that A wins.

[hide="comment"] I think it should be a 1/2 chance of a certain sequence of given length occurring before another sequence of the same length, but this seems too obvious...[/hide]

Hmm, are you sure the wording is correct? I don't quite understand it...

It's $1/2$... are you sure that was the question? Unfortunately, my ARML book only goes up to 2003 =\

Original problem at http://www.arml.com/2004%20Contest/2004%20ARML%20Individuals.pdf

You have to be careful, the pairs of sequences of A and B do not occur with equal probability. Also note that there are solutions on the ARML site too. You just have to get the url from the 2006 contest and replace the "2006" s with "2004" s

[hide]There are 4 different sequences, HH,TT,HT,TH.
For A, each occurs with probability 1/4.

For B,
If A is HH, the probability of B being
- TH or TT is 1/4 each
- HT, since it can follow any number of H's, prob is 1/4+1/8+1/16=1/2.
Similarly for A is TT, B is HT, HH, TH, with respective probabilities

If A is HT, sequence for B for B being
- HH is 1/4
- TH or HTH is 1/4+1/8=3/8
- Simlarly, TT or HTT, = 3/8
Similarly for A is TH, B is TT, HT or THT, HH or THH, with respective probabilities.

Total prob of B for
- TH is 1/4(1/4+1/2+3/8)=9/32 = prob of HT
- TT is 1/4(1/4+3/8+1/4)=7/32 = prob of HH

Umm, I think that the probability of any particular sequence first occurring for C is the same for all sequences, since the coin is fair...umm, so the prob of winning is related to the probs of each match up or pair of A and B's sequence.

Probability of A is...[/hide]


[quote="Big Bolzano"]On the plane we have finite set of triangles. Figure F is sum of all this triangles. $P_f = 16$, where P is an area. Prove, that we can choose triangles, which are separable and sum of their areas is $\geqslant 1$.[/quote]

I don't understand, are the triangles allowed to overlap? If the triangles are distinct, can't you just choose any triangle? i don't see the difficulty.

I think I don't get the mean.if the triangel are all in the some place?

Sorry, i forgot to write that all triangles are equilateral.  Yes, they can overlap. All triangles are on one plane.

what if all the triangles are in the same place? Or let say we choose 32 equi triangles of area $\frac{1}{2}$ with the same circumcentre?

[quote="Big Bolzano"]but in $P_f$ you have to sum all areas of triangles and than substract intersetion of every two.[/quote]

Mm I see. Okay, why can't we take equilateral triangles, with side length arbitarily close to $\frac{4}{\sqrt{3}}$ all sharing a point only, and take enough triangles such that we approach the circle of radius $\frac{4}{\sqrt{3}}$? . Furthermore, the area of such a circle is $\frac{16\pi}{3}$, so taking enough, we can exceed 16?

[quote="Big Bolzano"]$P_f = 16$, not $P_f > 16$.
What do u mean, sharing one point? Sharing one point with every triangle, or sharing one point with one of triangles?[/code][/quote]

We select a common point. All triangles have that common point. The limit would be a circle of area>16. Thus, we take enough triangles until we reach area 16. Surely we cannot find triangles such that the condition holds.

[quote="Big Bolzano"]What about taking just one triangle?[/quote]

Arbitarily close, and not equal. $\frac{4}{\sqrt{3}} -$:varepsilon: where :varepsilon: is as small as we choose.

[quote="Big Bolzano"]But we can choose such $\epsilon$ which satisfies $(\frac{4}{\sqrt{3}}-\epsilon)^2*\sqrt{3}*\frac{1}{4} > 1$[/quote]

Sorry my bad, I miscalculated Heron's.

yes.I think I come up the idea when I was sleeping. :)
assume $a_{1}$  is the largest side of the equilateral triangle.
we consider the area where there have  equilateral triangles intersect with $a_{1}$
let the area is $S_{1}$,with not had work we can know(rememeber $a_{1}$ is the largest side):
$S_{1}\leq(\pi+\frac{\sqrt{3}}{4}+3)(a_{1})^{2}  $
if $S_{1} \geq16$, we can know $\frac{\sqrt{3}}{4}(a_{1})^{2}>1 $
if $S_{1} <16$  we can chose the largest whicn does not intersect with $a_{1}$,assume its  side is intersect with $a_{2}$,so record $S_{2}$ similar.we can get $S_{2}\leq(\pi+\frac{\sqrt{3}}{4}+3)(a_{2})^{2}$,too。(if $S_{2}$ intersect with $S_{1}$,throw away these area)
wo do this again and again.finally we will get  $S_{1}+S_{2}......+S_{n}\geq16  $
so we can get $\frac{\sqrt{3}}{4}(a_{1}^{2}+a_{2}^{2}+......+a_{n}^{2})>1 $
we are done. :)    the chart.doc.is why $S_{1}\leq(\pi+\frac{\sqrt{3}}{4}+3)(a_{1})^{2}  $


I feel like I have hit a barrier in my problem solving education. This barrier is placed right at AMC 10/12 question 15. Almost anything before that, and many mathcounts questions, I do with reletive ease. Some hard mathcounts questions stump me, but I at least have a general idea of what to do(count up the areas and add, etc). But every single hard AMC or AIME question I try, I just can't do it.

Advice?

Moved to AMC from F&G.

Although, there've been plenty of discussions about this before. You may want to try searching for advice from old threads as well.

Memorize the solution to every problem you can't solve.

[quote="PenguinIntegral"][quote="white_horse_king88"]Wow, I would actually advise that you read the solution and do the problems you couldn't do before again and again until the solution comes out of your writing implement instinctively.[/quote]
That is actually.......a good idea!

Well, not good, but the best I have :([/quote]

I think it's very good advice. Always understand why you weren't able to solve a problem so you can get it when you try a similar problem.

If you can't solve a problem, play with it for a while. Try different approaches and stuff. Of course, that's only if you're training. But if there's no time limit, you might as well just work with it a little and see what you can figure out about the problem.

If you feel like you can't solve the problem much because you feel you lack skills, go through AoPS book and review your memories. Although it is good idea to solve many problems, it won't hurt you to stop for minute and develop your skill through book and then go back to problems.

This is pretty good idea for now since you have several months before AMC-12. Another thing is to take AoPS class as well. That's best I can say.


Actually I am a bit confused about something. Today I was just trying to post a heavy post (a solution to a geometry problem with two images total size around 200 KB). But I often got a blank screen when trying to submit it.

Does the server get overloaded with a bit heavier posts?

BTW I saw Darij or Yetti posting really long posts. I wonder if everyone faces the same problem or is there anything wrong with my connection/PC.  :maybe:

Sometimes, I believe. I had this happen to me before and I just put it in 2 different posts.

Still there is no solution from the admins. :(

I am trying to post a solution to a geometry problem with a 150 KB diagram. But what I get is a blank screen :(

Please look into this problem.

Can you email me the .txt file containing exactly what you put in the posting window?

Okay, so the problem was the image you were attaching: it had too much information in it :) I've now corrected for that, and also imposed a limit on the size of the images to be attached (5000 x 5000 pixels -- that should be enough :) )

It had nothing to do with the [b]length[/b] of the post.

[quote="Moonmathpi496"]Thanks. Actually by default I print the pictures with a virtual printer as a letter size page. That's why the diagrams are so big. I shall resize diagrams from now.[/quote]You can use imagemagick to trim the pictures, it's quite easy actually.

Or you can learn Geogebra, it's a visual program ;)


proof this:
$ Df_{(x_0,y_0)}(u,v)\equal{} \dfrac{\delta f}{\delta x}(x_0,y_0)(u)\plus{}\dfrac{\delta f}{\delta y}(x_0,y_0)(v)$.x and y are vectors.

in case f is differentiable, it is the chain rule


calculus
integration
LaTeX
calculus computations

Hi there. Having trouble with this problem on my Calculus final exam review sheet.
Any help would be greatly appreciated.

Integral of 1/2(2x+4)^4 xdx

What you said is badly ambiguous. You need to do one of two things: either learn how to post in LaTeX, or put in many more sets of parentheses to remove the ambiguities. I'm going to play mindreader here, and make a guess as to what I think you meant to say:

$\int\frac1{2(2x+4)^4}\,x\,dx = \frac1{32}\int\frac{x\,dx}{(x+2)^4}.$

Let's make the substitution $u=x+2,\,x=u-2,\,dx=du$:

$=\frac1{32}\int\frac{u-2}{u^4}\,du=\frac1{32}\int u^{-4}-2u^{-3}\,du.$

You should be able to finish it from there. One can also do this problem by integration by parts.

Sorry for the ambiguity. What I meant to say was:

Integral of (1/2) * (2x+4)^4 xdx

For

$\frac{1}{2}\int x(2x+4)^4 \ dx$

Just use integration by parts.

The same substitution and same idea as in my post #3 also works here.

Warning: the answer you get doing it "my" way and the answer you would get doing it Jimmy's way will look different from each other. They will actually only differ by a constant, and thus both be correct - but you may have to do a little algebra to convince yourself of that.


For every integer $n\geq 3$ determine the number of sequences $(c_{1},...,c_{n})$ with terms in $S=\{1,2,3,...,9\}$ such that for every three consecutive terms, at least two are equal.

How many different elements can the sequence have, though? If it is unrestricted, for $n = 3$ we can just have

$(0,0,1); (0,0,2); (0,0,3); \ldots$

and it wouldn't be a very interesting problem.

[hide]
We will use induction.  First, look at sequences where the last two terms are different.  For $n=3$ there are 9 ways to choose the last term, 8 ways to choose the previous term, and 2 ways to choose the first term (one of the last two) for 144 possibilities.

Now say that we have a sequence of length $r-1$ satisfying the requirements, and the last two terms are different.  To adjoin a term and form a sequence of length $r$, we need to choose one of the last two terms, so there are 2 possibilities for each initial sequence.  So clearly there are $144*2^{n-3}= 9*2^{n+1}$ sequences of length $n$ with the last two terms different.

Now, look at sequences with the last two terms the same. For $n=3$ there are 9 ways to choose the last two digits, and 9 ways to choose the first digit, for 81 possibilities.

Now say that we have a sequence of length $r-1$ satisfying the requirements, and the last two terms are the same.  The adjoined digit could be any digit, so there are 9 possibilities.  So there are $81*9^{n-3}= 9^{n-1}$ sequences of length $n$ with the last two terms the same.

So the total number of sequences of length $n$ is $9*2^{n+1}+9^{n-1}$.
[/hide]

It is almost the same problem as [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=112226[/url].


$ a,b,c$ are positive numbers,prove that
$ \frac{a(a\minus{}b)(a^2\plus{}ab\plus{}2b^2)}{a^3\plus{}b^3}\plus{}\frac{b(b\minus{}c)(b^2\plus{}bc\plus{}2c^2)}{b^3\plus{}c^3}\plus{}\frac{c(c\minus{}a)(c^2\plus{}ca\plus{}2a^2)}{c^3\plus{}a^3}\geq0$

:( I eager to know the solution to this inequlity,anyone to help?

[quote="Jianxing113725"]$ a,b,c$ are positive numbers,prove that
$ \frac {a(a \minus{} b)(a^2 \plus{} ab \plus{} 2b^2)}{a^3 \plus{} b^3} \plus{} \frac {b(b \minus{} c)(b^2 \plus{} bc \plus{} 2c^2)}{b^3 \plus{} c^3} \plus{} \frac {c(c \minus{} a)(c^2 \plus{} ca \plus{} 2a^2)}{c^3 \plus{} a^3}\geq0$[/quote]
 design square is
 $ \sum_{cyc}\frac{(a\minus{}b)^2(2b^2\minus{}a^2)}{a^3\plus{}b^3} \geq 0$

[quote="tuantam1lan"][quote="Jianxing113725"]$ a,b,c$ are positive numbers,prove that
$ \frac {a(a \minus{} b)(a^2 \plus{} ab \plus{} 2b^2)}{a^3 \plus{} b^3} \plus{} \frac {b(b \minus{} c)(b^2 \plus{} bc \plus{} 2c^2)}{b^3 \plus{} c^3} \plus{} \frac {c(c \minus{} a)(c^2 \plus{} ca \plus{} 2a^2)}{c^3 \plus{} a^3}\geq0$[/quote]
 design square is
 $ \sum_{cyc}\frac {(a \minus{} b)^2(2b^2 \minus{} a^2)}{a^3 \plus{} b^3} \geq 0$[/quote]

After expanding,AM-GM solve it easily.... :blush:I do that a fews days ago....


Please help me to make $ {x}$ the subject of the two equations. \[ {x}^{2}\minus{}x\equal{}y\] and \[ {x}^{3}\minus{}x\equal{}y\]

[hide="First one"]
Complete the square.
$ x^2\minus{}x\plus{}\frac{1}{4}\equal{}y\plus{}\frac{1}{4}$.

$ \left(x\minus{}\frac{1}{2}\right)^2\equal{}y\plus{}\frac{1}{4}$.

$ \boxed{x\equal{}\frac{1}{2}\pm\sqrt{y\plus{}\frac{1}{4}}}$.
[/hide]

Not sure if the second one is even possible (from the looks of the cubic formula, it won't be pretty).

well done grn_trtl for the first and for the second maybe we should try the cubic form  :wink:
Let's try !

Obviously that is not something to expect someone to do by hand. The following is the output from Mathematica:

$ x \equal{} \minus{} \frac {\left(\frac {2}{3}\right)^{1/3}}{\left( \minus{} 9 y \plus{} \sqrt {3} \sqrt { \minus{} 4 \plus{} 27 y^2}\right)^{1/3}} \minus{} \frac {\left( \minus{} 9 y \plus{} \sqrt {3} \sqrt { \minus{} 4 \plus{} 27 y^2}\right)^{1/3}}{2^{1/3} 3^{2/3}}$

$ x \equal{} \frac {1 \plus{} i \sqrt {3}}{2^{2/3} 3^{1/3} \left( \minus{} 9 y \plus{} \sqrt {3} \sqrt { \minus{} 4 \plus{} 27 y^2}\right)^{1/3}}$ $ \plus{} \frac {\left(1 \minus{} i \sqrt {3}\right) \left( \minus{} 9 y \plus{} \sqrt {3} \sqrt { \minus{} 4 \plus{} 27 y^2}\right)^{1/3}}{2 2^{1/3} 3^{2/3}}$

$ x \equal{} \frac {1 \minus{} i \sqrt {3}}{2^{2/3} 3^{1/3} \left( \minus{} 9 y \plus{} \sqrt {3} \sqrt { \minus{} 4 \plus{} 27 y^2}\right)^{1/3}}$ $ \plus{} \frac {\left(1 \plus{} i \sqrt {3}\right) \left( \minus{} 9 y \plus{} \sqrt {3} \sqrt { \minus{} 4 \plus{} 27 y^2}\right)^{1/3}}{2 2^{1/3} 3^{2/3}}$


Given $a,b,c>0$ are const and $x,y,z>0$ such that $x+y+z=1$ find min value of
$(\frac{x}{b+c})^{3}+(\frac{y}{c+a})^{3}+(\frac{z}{a+b})^{3}$
 :D

ok maybe it is too easy with algebraic method :D, but actually I found it intricately by geometric method, therefore can you post your solution? :)


linear algebra
matrix
topology
function
real analysis
real analysis unsolved

Let f: Rm -> Rn be a continuous mapping with m > n.  Does f have to be non-injective?  Or is this only true if f is a differentiable mapping.

It's injective because of the Borsuk-Ulam Theorem: any continuous mapping from $\mathbb S^n$ to $\mathbb R^n$ takes two antipodes to the same point. Since the mapping from $\mathbb R^m$ to $\mathbb R^n$ induces a continuous mapping from $\mathbb S^{m-1}$ to $\mathbb R^{m-1}$, that's that. The theorem is rather nasty to prove, I think. I certainly can't prove it :).

Google it and see what you find.

hm...is it really so hard, it is only one question on an analysis final exam.  Well, it is princeton's but still it is not a contest problem.  I would like to know answer...

ok, i understand that using this complicated theorem (borsuk-ulam), that a continuous mapping is good enough to guarantee that f is not injective.  But is there a much more elementary proof if we assume f is differentiable?

Suppose that f was a infinitely differentiable function.  Then if f were injective its jacobian matrix J would have to have constant rank on some open subset U.  Thus f restricted to U would been an immersion, that is the rank of J at every point of U is equal to dim U = m.  This is clearly a contradiction since rank J is at most n.

it makes sense intuitively, but rigorously, why must J have maximal rank on an open set if it is injective?  Thanks a lot for the help.

Let p be a point at which J has maximal rank r.  Then there exists a r by r submatrix of J call it A which has nonzero determinant at p.  Since the entries of J are continuous there exists a neighborhood U of p on which A has nonzero determinant.  Thus on U the rank of J is at least r.  But r is the maximal rank that J takes on and thus we have that J has rank r on U.  Now from the theory of smooth manifolds we know that an injective smooth function is an immersion.  (see theorm 7.15 in John M. Lee's introduction to smooth manifolds in the GTM series)

Thanks for the quick reply.  But I guess my real question then is that last statement.  "I mean that a smooth injective function that is of constant rank is an immersion."  Is this hard to prove?

Hm, actually does this have anything to do with the rank theorem?

If you use the Rank theorem then this is almost an immediate corollary.  However, the rank theorem is a difficult theorem to prove.  The proof is an involved construction which uses the inverse function theorem somewhere along the lines.


Why $\overline{\mathbb{Q}^k} = \mathbb{R}^k$  :?:

I know this question maybe is stupid, but could someone explain it?
Thanks.

This is simply saying that given any point $(x_1,x_2,\dots,x_k)\in\mathbb{R}^k,$ we can approximate it arbitrarily closely by a point in $\mathbb{Q}^k.$

Assume we have such a point, and let $\epsilon>0.$ Since $\mathbb{Q}$ is dense in $\mathbb{R},$ for each $j,\,1\le j\le k,$ we can find a rational number $r_j$ such that $|x_j-r_j|<\frac{\epsilon}{\sqrt{k}}.$

Then $\|(x_1,x_2,\dots,x_k)-(r_1,r_2,\dots,r_k)\|=\sqrt{(x_1-r_1)^2+\cdots+(x_k-r_k)^2}$

$<\sqrt{k\left(\frac{\epsilon}{\sqrt{k}}\right)^2}=\epsilon.$


Given $ x,y > 0$, find the minimum value of this expression:

$ \frac{1}{x^2\plus{}y^2}\plus{}\frac{x^2}{1\plus{}x^2}\plus{}\frac{y^2}{1\plus{}y^2}$

hello, we have $ \frac{1}{x^2\plus{}y^2}\plus{}\frac{x^2}{1\plus{}x^2}\plus{}\frac{y^2}{1\plus{}y^2}>1 \forall x,y \in \mathbb{R}$ with $ x,y>0$.
Sonnhard.

Does the following inequality seem right to you?
$ \frac{x}{x\plus{}1} \plus{} \frac{y}{y\plus{}1} \ge \frac{x\plus{}y}{x\plus{}y\plus{}1}$ for all $ x, y >0$


How many positive integral solutions are there for

         a+b+c+d+e=36          with 0<_ a,b,c,d,e < 12

what do u mean by combinatorics...

u mean we solve it by combinatorics or not?

Since $0 < a, b, c, d < 12$ we have $a = 12 - a'$, $b = 12 - b'$...
where $a', b'$... are all integers.
We rewrite the first as $60 - a' - b' - c' - d' - e' = 36$ and we get $a' + b' + c' + d' + e' = 24$.
So the number of positive integral solutions is $\binom{24 - 1}{5 - 1} = 8855$.

I'm not 100% sure  :)

I’m assuming that you meant to say $0 \le a,b,c,d,e < 12$

I know one way to solve it.
[hide="Solution"]
The number of possibilities is the coefficient of $x^{36}$ in the following:

$(1+x^1+x^2+\cdot\cdot\cdot+x^11)^5$
And that comes to be $\binom{40}{36} -5\binom{28}{24} +10\binom{16}{12} -10$[/hide]
Hope I’m correct.
 :!:

[hide]Using generating functions, the answer should be the coefficient of the $x^{36}$ term of $\left( \sum_{k=0}^{11} x^k\right)^5$ which can be rewritten $\frac{(1-x^{12})^5}{(1-x)^5}$. Using binomial series, this becomes $(1-5x^{12}+10x^{24}-10x^{36}+5x^{48}-x^{60})\sum_{k=0}^\infty \binom{-5}{k} (-x)^k$. Expanding out $\binom{-5}{k}$ and removing the negative signs, this becomes $(-1)^k\binom{k+4}{k}=(-1)^k\binom{k+4}{4}$, making the function equal to $(1-5x^{12}+10x^{24}-10x^{36}+5x^{48}-x^{60})\sum_{k=0}^\infty \binom{k+4}{4}x^k$. The $x^{36}$ coefficient is equal to $\binom{40}{4}-5\binom{28}{4}+10\binom{16}{4}-10=7205$.[/hide]


Identify the units on the variable x in each of the following situations.

1. x=fp	both f and p are measured in inches.

um...

before you post read some of the Getting Started stickies


$a$ is a real number such that $a>1$. $x$ and $y$ are real numbers. Show that $a^x = a^y$ implies that $x=y$, using only very basic axioms, definitions, and results (as if this were a problem in the first week of an analysis class).

Thank you for your help.

We need $x,y$ to be rational, if we're talking in 'first week of analysis class' terms.

The condition implies $a^{x-y}=1$.  Clearly $x=y$ is a solution.  Otherwise, WLOG $x>y$.  Then $x-y>0$.  I claim that $a^{x-y}>1$.

If instead we had $0\leq a^{x-y}\leq 1$, we would have, since $x-y$ is rational,

$(a^{x-y})^p=a^q\leq 1$ for integers $p,q$ which contradicts $a>1$.

Thanks for your post. Sorry about the confusion, but perhaps I should have set "first few weeks of an analysis class," since I am looking for the solution when $x, y$ are real. The problem is part of a series of problems that define the existence and uniqueness of logarithms. I have already managed to show that $x =$ sup $S$, where $S$ is the set of all real $n$ satisfying $a^n<m$, is a solution of the equation $a^x=m$. Now I just need to show that it is the unique solution.

Thus, I assumed that there exists another solution $y$ and put $a^x = m = a^y$, so I come upon the dilemma of showing that $x=y$.


A linear regression equation of best fit between a student’s attendance and the degree of success in school is h = 0.5x + 68.5. The correlation coefficient, r, for these data would be

(1) 0 <r <1         (3)r = 0
(2) –1 <r <0       (4)r = –1

[hide]If a linear regression equation has the form $y = ax+b$, the a-value tells whether the relation is direct or indirect.  Since a is positive, there is a direct correlation, and r must also be positive.  Therefore, 0<r<1.[/hide]

You said:

"Since a is positive, there is a direct correlation, and r must also be positive. Therefore, 0<r<1."

Questions:

What about if a in y = ax + b is negative?

Can r also be negative?

If so, what happens if r is not positive?

Is the equation y = ax + b the same in meaning as y = mx + b, where m is the slope?

In other words, does m = a = SLOPE?

[quote="Interval"]You said:

"Since a is positive, there is a direct correlation, and r must also be positive. Therefore, 0<r<1."

Questions:

What about if a in y = ax + b is negative? [b]Then r is negative.[/b]

Can r also be negative? [b]Yes.[/b]

If so, what happens if r is not positive? [b]If r is 0, there is no linear relation between the two variables (student's attendence and degree of success) in question. If r is negative, they are inversely related, that is, when one increases, the other decreases.[/b]

Is the equation y = ax + b the same in meaning as y = mx + b, where m is the slope? [b]Pretty much, yea.[/b]

In other words, does m = a = SLOPE? [b]Yep, it doesn't matter what you call it since they both tell you the same thing.[/b][/quote]

Bictor717,

Your answers to my questions are great for my collection of notes for math study time.


How many miles per hour is 1298 feet per second?


Tare wrote:I apologize for the lack of problems, but now I've got a couple:

The difference between the squares of two consecutive odd number is 136. Which of the two numbers is prime?
Generalize the above. (The difference between the squares of two consecutive odd number is x. What is the two numbers?; There can be cases where both is composite, ex. 25 and 27)

[hide]sigh, heres one way to do it. so then you have two numbers x and x+2. sure to be technically correct it would be 2n-1 and 2n+1, or something along those lines. but this works. so then square those to get x^2+4x+4 and x^2. subtract those for 4x+4=136. then solving x=33. so then the two numbers are 33 and 35. with none of them being prime.  i too lazy to generalize.[/hide]

Yup, that's the algebraic way to attack this problem, how about an answer from series point of view?

...wait a minute, [quote="You"]so then the two numbers are 33 and 35. with none of them being prime.[/quote]
!?!?:!::?::)

i just checked again. and 33 and 35 works. maybe there is another pair that works also. i dunno. im too tired to think right now

No no, you said that 33 and 35 were both prime :) Technically you're not done with the problem since the problem asks for only one of the numbers.

i said both are NOT prime. read it again

Oh whoops, sorry :oops:

...guess I picked a bad number, should've chosen 144 or somat or just left it the way it was (in the original problem the difference was 64)

Anyways, there's a couple more ways to do this if anyone want to give it a go, although the answer won't come out good...should I change the problem?


Instead of having one big database of questions on the forum why not split it up into sections.  That would make it a lot easier to browse and then people like me don't have to browse through Calculus questions. :D

I disagree.  First off, you might learn something from looking at threads covering topics you don't know much about.   Besides, it doesn't take that much more time to look at a problem, and sometimes one that may be too hard for you will still get you thinking.

Besides, sometimes problems fall under more than one category, there's the danger that someone will miscategorize something and get it all mixed up...  having specific categories often make things more complicated and hard to find.

However, we COULD have a field in the posting.php form for "category" - such as Algebra, trig, etc... that way people looking would immediately know what category the problem falls under.

I disagree as well. Frequently, topics may cover more than one discipline; furthermore, topics in the "easier" subjects such as algebra might not necessarily be easier than calculus problems; it's simply a different field of study.

Mmm, each person only posting once..."those were the days"... :)

I agree and disagree. We'll need to discuss long and hard how we split the forums into sub-forums since it could cause problems, but at the same time we'll be able to separate out the competitions and such.

See, now we've got all kinds of crazy spam. Just check out the Etc. forum :-)

Keep in mind that changing the forums around requires more and more work on the part of Messrs. (sp?) Rusczyk and Crawford.  I personally think the forums are fine as they are right now.

They're not really bad or anything. And yes, you did spell rusczyk right i believe


A triangle has integer sides of 12, 20, and [i]x[/i]. What is the probability that [i]x[/i] is a prime number?

[hide]We know that $x<12+20$ and $x>0.$ 31 different possibilities for the third side. Of these, 11 are primes (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31). Therefore, the probability that $x$ is prime is $\frac{11}{31}.$[/hide]

So it is possible for [i]x[/i] to equal 2?

:wallbash:

[hide]We know that $x+12>20$ and $x<12+20.$ There are 23 possibilities for the third side (9-31). Of these, 7 are prime (11, 13, 17, 19, 23, 29, 31). Therefore, the probability that $x$ is prime is $\frac{7}{23}.$[/hide]

hmmm...? Kevin, your post count is not going up...even though you have 2 posts in this thread?

You just noticed?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118586[/url]


function
limit
calculus
calculus computations

Let $ f(x,y,z)$ be differentiable function on $ \mathbb{R}^3$ such that for every real number $ t$ and every $ (x,y,z)\in \mathbb{R}^3$ the equality \[ f(tx,ty,tz)\equal{}tf(x,y,z)\] holds. Show that \[ f(x,y,z)\equal{}ax\plus{}by\plus{}cz\] for some $ a$, $ b$, $ c\in \mathbb{R}$.

$ f$ is homogeneity of degree $ 1$.
I think that $ f$ resembles linear mapping.

[quote="Jutaro"]Let $ f(x,y,z)$ be differentiable function on $ \mathbb{R}^3$ such that for every real number $ t$ and every $ (x,y,z)\in \mathbb{R}^3$ the equality
\[ f(tx,ty,tz) \equal{} tf(x,y,z)\]
holds. Show that
\[ f(x,y,z) \equal{} ax \plus{} by \plus{} cz\]
for some $ a$, $ b$, $ c\in \mathbb{R}$.[/quote]
What about $ f(x,y,z) \equal{} \frac{x^3\plus{}y^3\plus{}z^3}{x^2\plus{}y^2\plus{}z^2}$ and $ f(0,0,0)\equal{}0$. Are there any problem with that?

The problem with that is that it's not differentiable at zero. In fact, we could relax the "$ f$ differentiable" condition to $ f$ being differentiable at the origin.

Here's a proof:
The definition of differentiability at zero is that $ f(\mathbf{v})\equal{}f(0)\plus{}L\mathbf{v}\plus{}o(|\mathbf{v}|)$, as $ \mathbf{v}\to 0$, where $ L$ is a linear map we call the derivative.
By homogeneity, $ f(t\mathbf{v})\equal{}tf(\mathbf{v})\equal{}f(0)\plus{}tL\mathbf{v}\plus{}o(t|\mathbf{v}|)\equal{}tf(0)\plus{}tL\mathbf{v}\plus{}t\cdot o(|\mathbf{v}|)$. Clearly, $ f(0)\equal{}0$, which leaves us with only the $ o(t|\mathbf{v}|)\equal{}t\cdot o(|\mathbf{v}|)$ part. Divide by $ t$, and let $ t\to 0$ holding $ v$ constant; the left-hand side $ \frac1t o(t|\mathbf{v}|)$ goes to zero, while the right-hand side is constant and must therefore be equal to zero. The "error" is identically zero, and $ f(\mathbf{v})\equal{}L\mathbf{v}$ everywhere.

[quote="jmerry"]The problem with that is that it's not differentiable at zero. In fact, we could relax the "$ f$ differentiable" condition to $ f$ being differentiable at the origin.[/quote]
We can just define $ \frac{df(x,0,0)}{dx} \equal{} 1$, can't we? (and similar with $ y,z$ of course) Where lies the problem?

[quote="Mathias_DK"][quote="jmerry"]The problem with that is that it's not differentiable at zero. In fact, we could relax the "$ f$ differentiable" condition to $ f$ being differentiable at the origin.[/quote]
We can just define $ \frac {df(x,0,0)}{dx} \equal{} 1$, can't we? (and similar with $ y,z$ of course) Where lies the problem?[/quote]

Differentiable means differentiable with respect to $ (x,y,z)$ as a whole, instead of a partial derivative. If you work from the definition, you can see that as the limit $ \lim_{(x,y,z)\rightarrow (0,0,0)} \frac{(x^3\plus{}y^3\plus{}z^3)\minus{}(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2)}{(x^2\plus{}y^2\plus{}z^2)^{3/2}}$ does not exist, so it is not differentiable at 0.

[quote="Soarer"][quote="Mathias_DK"][quote="jmerry"]The problem with that is that it's not differentiable at zero. In fact, we could relax the "$ f$ differentiable" condition to $ f$ being differentiable at the origin.[/quote]
We can just define $ \frac {df(x,0,0)}{dx} \equal{} 1$, can't we? (and similar with $ y,z$ of course) Where lies the problem?[/quote]

Differentiable means differentiable with respect to $ (x,y,z)$ as a whole, instead of a partial derivative. If you work from the definition, you can see that as the limit $ \lim_{(x,y,z)\rightarrow (0,0,0)} \frac {(x^3 \plus{} y^3 \plus{} z^3) \minus{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2)}{(x^2 \plus{} y^2 \plus{} z^2)^{3/2}}$ does not exist, so it is not differentiable at 0.[/quote]
I don't know the definitions then. How would one differentiate $ f(x,y,z)$ with respect to $ (x,y,z)$ as a whole? I know that in order to differentiate $ g(x)$ we should look at $ g'(x) \equal{} \lim_{x_1 \to x} \frac {g(x) \minus{} g(x_1)}{x \minus{} x_1}$, but how do you do when you have more than one variable?

[quote="jmerry"]The definition of differentiability at zero is that $ f(\mathbf{v}) \equal{} f(0) \plus{} L\mathbf{v} \plus{} o(|\mathbf{v}|)$, as $ \mathbf{v}\to 0$, where $ L$ is a linear map we call the derivative.[/quote]

[quote="Mathias_DK"]
I don't know the definitions then. How would one differentiate $ f(x,y,z)$ with respect to $ (x,y,z)$ as a whole. I know that in order to differentiate $ g(x)$ we should look at $ g'(x) \equal{} \lim_{x_1 \to x} \frac {g(x) \minus{} g(x_1)}{x \minus{} x_1}$, but how do you do when you have more than one variable?[/quote]

Please refer to any multivariable calculus/analysis textbooks. In brief the idea is: In the 1 variable case we want to approximate a function $ f(x)$ by a linear one near a point, e.g. at point $ x_0$ the approximation is $ f(x_0) \plus{} f'(x_0)(x\minus{}x_0)$.

In multivariable case, say $ f : \mathbb{R}^m \rightarrow \mathbb{R}$, we also want to approximate it by a linear one near a point, say at point $ \vec{a} \equal{} (a_1,\cdots,a_m)$, we want an approximation of the form $ f(a_1,\cdots,a_m) \plus{} \sum c_i (x_i \minus{} a_i)$. (Approximate along each direction) If you write down the definition it would look like $ f$ is differentiable at $ \vec{a}$ if there exists constants $ c_i$ such that
$ \lim_{(x_1,\cdots,x_m) \rightarrow (a_1,\cdots,a_m)} \frac{f(x_1,\cdots,x_m) \minus{} f(a_1,\cdots,a_m) \minus{} \sum c_i (x_i \minus{} a_i)}{\sqrt{\sum (x_i\minus{}a_i)^2}} \equal{} 0$
Note that this multivariable limit requires the "limit-taking process" to occur at every direction.

Of course this is all very sketchy, so just look up some textbooks if you want more details.

[quote="Soarer"][quote="Mathias_DK"]
I don't know the definitions then. How would one differentiate $ f(x,y,z)$ with respect to $ (x,y,z)$ as a whole. I know that in order to differentiate $ g(x)$ we should look at $ g'(x) \equal{} \lim_{x_1 \to x} \frac {g(x) \minus{} g(x_1)}{x \minus{} x_1}$, but how do you do when you have more than one variable?[/quote]

Please refer to any multivariable calculus/analysis textbooks. In brief the idea is: In the 1 variable case we want to approximate a function $ f(x)$ by a linear one near a point, e.g. at point $ x_0$ the approximation is $ f(x_0) \plus{} f'(x_0)(x \minus{} x_0)$.

In multivariable case, say $ f : \mathbb{R}^m \rightarrow \mathbb{R}$, we also want to approximate it by a linear one near a point, say at point $ \vec{a} \equal{} (a_1,\cdots,a_m)$, we want an approximation of the form $ f(a_1,\cdots,a_m) \plus{} \sum c_i (x_i \minus{} a_i)$. (Approximate along each direction) If you write down the definition it would look like $ f$ is differentiable at $ \vec{a}$ if there exists constants $ c_i$ such that
$ \lim_{(x_1,\cdots,x_m) \rightarrow (a_1,\cdots,a_m)} \frac {f(x_1,\cdots,x_m) \minus{} f(a_1,\cdots,a_m) \minus{} \sum c_i (x_i \minus{} a_i)}{\sqrt {\sum (x_i \minus{} a_i)^2}} \equal{} 0$
Note that this multivariable limit requires the "limit-taking process" to occur at every direction.

Of course this is all very sketchy, so just look up some textbooks if you want more details.[/quote]
Thank you!


algebra
difference of squares
special factorizations

Pick two consecutive positive integers whose sum is less than 100. Square both of those integers and then find the difference of the squares. Which of the following could be the difference?

(a) 2
(b) 64
(c) 79
(d) 96
(e) 131

The answer was c. I'm a bit confused as to why, could you please help? Thanks.

Suppose the numbers were $ x$ and $ x \plus{} 1$. We know that $ 2x \plus{} 1 < 100$, and we want to find the possible values of $ (x \plus{} 1)^2 \minus{} x^2$. Remember that for a difference of squares $ a^2 \minus{} b^2$, it's equal to $ (a \plus{} b)(a \minus{} b)$. Thus, we have our expression as $ (2x \plus{} 1)(1) \equal{} 2x \plus{} 1$. So, we just have to look at the answer choices that can be written in that form.

It doesn't work for 2 or 64, but it does for 79. As a final check, we see it doesn't work for 96 or 131 (because 131 > 100), so the answer is 79, or $ \boxed{C}$.


This question was given to me by someone else, and it is bothering me mainly because it seems to be missing information:

Let $ f(x,y) \equal{} x^3 \plus{} 3y^2 \minus{} 2x^2y \plus{} 6$

In what direction is the function $ f$ decreasing the most, and what is the rate of decrease in that direction?

Also, what is the rate of decrease of $ f$ in the direction $ (\frac {1}{\sqrt {2}}, \minus{} \frac {1}{\sqrt {2}})?$

I feel as though a specific point should have been provided in order to solve this problem.

These are actually the last two parts of a question about that function. The first two parts used the point $ (2,3)$, so I am thinking that they want to use this point specifically? If not, I am not sure how to even solve this problem generally.

[quote="bsmith2000"]These are actually the last two parts of a question about that function. The first two parts used the point $ (2,3)$, so I am thinking that they want to use this point specifically? If not, I am not sure how to even solve this problem generally.[/quote]

You are supposed to use that point.

[quote="t0rajir0u"]You are supposed to use that point.[/quote]

Okay, cool - that's what I figured. However, since they never explicitly stated to use that point in those parts, would you happen to know of a way that this can be generally solved (that is, the way the question was asked without giving a point)?

Thanks again for your help!

edit: I also had a question about the problem if it was indeed asking about the point $ (2,3)$.

To find the fastest decrease, do we just take the negative gradient evaluated at the point? Then to find the rate of decrease, does one take the norm of this negative gradient (at the point)?  Thanks!

[quote="bsmith2000"]
edit: I also had a question about the problem if it was indeed asking about the point $ (2,3)$.

To find the fastest decrease, do we just take the negative gradient evaluated at the point? Then to find the rate of decrease, does one take the norm of this negative gradient (at the point)?  Thanks![/quote]

Yes  :)

Just to ensure I am doing this correctly, I have:

The gradient is $ (3x^2\minus{}4xy, 6y\minus{}2x^2)$, so $ f$ is decreasing the most at $ (2,3)$ in the direction $ \minus{}(3(2)^2\minus{}4(2)(3), 6(3)\minus{}2(2)^2) \equal{} 12i \minus{} 10j$.

This just sort of seems like an odd question and made me wonder if the problem was really asking for a general answer to the question, because the first question asked for the gradient at $ (2,3).$ This is merely the negative of that answer.

The rate of decrease is $ \minus{}\sqrt{12^2\plus{}10^2} \equal{} \minus{}\sqrt{244} \equal{} \minus{}2\sqrt{61}$

The rate of decrease at $ (\frac{1}{\sqrt{2}}, \minus{}\frac{1}{\sqrt{2}})$ is the dot product of the negative gradient at $ (2,3)$ with $ (\frac{1}{\sqrt{2}}, \minus{}\frac{1}{\sqrt{2}})$, which equals:

$ 12(\frac{1}{\sqrt{2}}) \minus{} 10(\minus{}\frac{1}{\sqrt{2}}) \equal{} \frac{22}{\sqrt{2}} \equal{} 11\sqrt{2}$


[quote="Gibbenergy"]Let $a,b,c$ are positive number such that $a^{2}+b^{2}+c^{2}=3$. Prove that
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq a^{3}+b^{3}+c^{3}$[/quote][hide][color=darkblue]By [b]Quadratic Mean-Arithmetic Mean Inequality[/b], \[\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\ge\frac{a+b+c}{3},\]  \[\sqrt{\frac{3}{3}}\ge\frac{a+b+c}{3},\]  \[\boxed{3\ge a+b+c}.\] By the [b]Quadratic Mean-Harmonic Mean Inequality[/b], \[\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}},\]  \[\sqrt{\frac{3}{3}}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}},\]  \[\boxed{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3}.\] Hence, by transitivity, \[\boxed{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge a+b+c}.\] By the [b]Chebyshev Sum Inequality[/b], \[\frac{a^{2}+b^{2}+c^{2}}{a^{3}+b^{3}+c^{3}}\ge\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).\] By transitivity, \[\frac{a^{2}+b^{2}+c^{2}}{a^{3}+b^{3}+c^{3}}\ge\frac{1}{3}\left(a+b+c\right)=\frac{a+b+c}{3}.\] From the [b]Arithmetic Mean-Harmonic Mean Inequality[/b], \[\frac{a+b+c}{3}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}.\] By transitivity, \[\boxed{\frac{a^{2}+b^{2}+c^{2}}{a^{3}+b^{3}+c^{3}}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}}.\] Since $a^{2}+b^{2}+c^{2}=3$, then \[\frac{3}{a^{3}+b^{3}+c^{3}}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}},\] and consequently, \[\boxed{ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge a^{3}+b^{3}+c^{3}},\] since everything is positive.[/color][/hide]

Dang, I'm bad at this inequality stuff. Stupid ordered sequences. >_<

Hmm, using the notation $(x,y,z) = \sum_{sym}a^{x}b^{y}c^{z}$, the problem reduces down to

$(5,1,0)+2 \cdot (3,2,1)+(3,3,0) \ge 4 \cdot (4,1,1)$,

which seems to be pretty strong. Definitely stronger than AM-GM; not sure how it compares with Schur, though.

[b]arqady[/b] has a counterexample. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=604270#604270

[quote="MysticTerminator"]I think the step where you cited Chebyshev is wrong.[/quote]

Yes.  The inequality sign is in the other direction.


geometry
circumcircle
conics
hyperbola
geometry solved

Find point $X$ on the circumcircle of triangle $ABC$ that its simson line passes through the circumcenter of triangle $ABC$.

[i]Edited by Myth[/i]

My dynamic sketches show that there are three such points X for an acute triangle ABC and one such point X for an obtuse triangle ABC. Hmm, this looks like a cubic equation but not Morley-related...

  Darij

Look at
http://groups.yahoo.com/group/Hyacinthos/message/2651
for a construction of the Simson lines going through a given point P using the common points of a circle and a rectangular hyperbola (the points whose Simson lines go through a given point P are generally not ruler and compass constructible, even when P = O)
Kind regards. Jean-Pierre


Let p be a prime and a is a interger such that : gcd(a,p)=1
Prove that :exist x and y such that $0<x^{2},y^{2}<p$ and x=ay (mod p)

Lemma of Thue: http://www.mathlinks.ro/Forum/viewtopic.php?t=46300 , post 5, Lemma 2.

I don't mean this.This problem is not Thue's lemma .It's difficult than it /

Would you bother to explain why it is more difficult¿
It's up to notation and to exception of the case $p|a$ the same.


i can't figure this one out.  :(

if the reciprocal of (x-2) is (x+2), what is the greatest possible value of x?  Express your answer in simplest radical form.

i found this when i was digging through my old papers, and i found i left this one unanswered.

[hide]I got sqrt5
1/(x-2)=(x+2)
1=(x-2)(x+2)
1=x^2-4
0=x^2-5
...quadratic formula...
positive or negative sqrt5
greatest value gives positive sqrt5
[/hide]

[hide]
You can set up the following equation:
$\frac{1}{x-2}=x+2$
Multiply by $x-2$ to get
$1=(x-2)(x+2)$
Simplify:
$1=x^2-4$
$x^2=5$
$x=\pm\sqrt{5}$
The greatest value is thus $\sqrt{5}$[/hide]

[hide]Write an equation like this: 1/x-2=x+2. Now cross multiply to get $x^2-4=1$. Add 4 to both sides and take the square root to get x=rt5.[/hide]

thanks that was easier than i thought  :oops:


Without a calculator, evaluate $40000+999800\cdot1000200$.

[hide]Call x 1,000,000. We have (x-200)(x+200) which is x^2-40000. When this is added to 40000, we get x^2, which is [b]1 trillion[/b][/hide]

I think you're missing a few zeros, DarkKnight, but the right idea though


function
integration
complex analysis
complex analysis unsolved

Let $ f$ be an entire function which satisfies $ |f(z)|<e^{|z|}$. prove that there exists $ K>0$ such that for every $ z \in \mathbb{C}$ Prove that there exists $ K>0$ such that for every $ z \in \mathbb{C} \  |f'(z)|<Ke^{|z|}$.

Try the Cauchy formula on the unit circle around $ z$.

Thanks. I think this works : since $ f$ is entire due to Cauchy formula for any complex $ z$ we have that ${ f'(z) = \frac{1}{2\pi i} \int _{|\psi-z|=1|}  \frac{ f(\psi) d\psi }{(\psi - z)^2}}$, hence taking modulus $ |f'(z)| \leq e^{|\psi|} \leq e^{|z|+1}$ for $ |\psi-z|=1$.


Let $n\in N$ $n\geq2$ and the set $X$ with $n+1$ elements. The ordered sequences $(a_{1}, a_{2},\ldots,a_{n})$ and $(b_{1},b_{2},\ldots b_{n})$ of distinct elements of $X$ are said to be $\textit{separated}$ if there exists $i\neq j$ such that $a_{i}=b_{j}$. Determine the maximal number of ordered sequences of $n$ elements from $X$ such that any two of them are $\textit{separated}$.
Note: ordered means that, for example $(1,2,3)\neq(2,3,1)$.

Answer $n!$.
Denote elements of $X$ by $1,2,3...,n+1$. Define transformation $T$ such that : $T$ maps  $(a_1,a_2,...,a_n)$ to   the sequence obtained by taking $a_i=X\setminus \{a_1,...,a_n\}$,if $\exists a_i$ such that $n+1=a_i$; and leave it unchanged if $\{a_1,a_2,...,a_n\}=\{1,2,...n\}$.

It is easy to check that $a,b$ are separated iff $T(a)$ and $T(b)$ are separated. the maximum number of pairwise separated sequencs with terms in the set $\{1,...,n\}$ is $n!$. We can also find $n!$ sequences pairwise separated: the last element is $n+1$, and the first $n$ are a permutation of $1,...,n$.

At first sight, the above answer is definitely [color=red]wrong[/color] (the last step counts sequences with $ n\plus{}1$ elements, anyways).
For $ n\equal{}2$ the answer can be checked by hand to be $ 3$, with e.g. the set of sequences $ (1,2)$, $ (2,3)$ and $ (3,1)$.
(Let's see what the above does: $ T(1,2) \equal{} (1,2)$, $ T(2,3) \equal{} (2,1)$, $ T(3,1) \equal{} (2,1)$. The effect is that two of the sequences are now the same (both are now $ (2,1)$). Sure enough, the set of transformed sequences are separated: $ (1,2)$ and $ (2,1)$ are separated, but they are less than than the original ones).
The problem is intriguing enough so I will try find the time to come with a proper full solution!

The answer is $ \frac {1} {2} (n\plus{}1)!$. We will do two seemingly inconsequential things that will open the way.

1. Fill an extra $ (n\plus{}1)^{\textrm{th}}$ position in a sequence with the missing element (clearly this can be done in a unique way). Now we can look at the set of ordered sequences as being $ \mathcal{S}_{n\plus{}1}$, the permutations on $ n\plus{}1$ elements.

2. Build a graph $ G\equal{}(V,E)$ having as vertices the elements $ \sigma \in \mathcal{S}_{n\plus{}1}$ (hence, according to the above, the ordered sequences), and as edges those pairs of sequences that [b]are not[/b] [i]separated[/i].

Consider an edge $ \sigma_1\sigma_2$ in this graph; it is easy to see that there must exist a unique $ 1\leq i\leq n$ such that $ \sigma_2 \equal{} \tau(i, n\plus{}1)\circ \sigma_1$, where $ \tau(i, n\plus{}1)$ is the transposition between the elements on positions $ i$ and $ n\plus{}1$.
Consider now a cycle $ \gamma$ in this graph; then, going around the edges, a vertex $ \sigma$ will be reached again in a number of steps equal to the length $ \lambda(\gamma)$ of the cycle, hence after applying $ \lambda(\gamma)$ transpositions. But transpositions have odd signature, so $ \lambda(\gamma)$ needs be even. This shows the graph contains no odd cycles, therefore is bipartite (classical result, easy to prove by splitting the vertices according to parity, on walking over a spanning tree of $ G$).

Now, each class of the bipartite $ G$ graph is a suitable set of pairwise [i]separated[/i] sequences, so the most numerous of the classes has at least $ \frac {1} {2} (n\plus{}1)!$ elements. On the other hand, assume it has strictly more; then if we look at the first $ n\minus{}1$ elements in each of the $ (n\plus{}1)!$ sequences, the number of these $ (n\minus{}1)$-prefixes is exactly  $ \frac {1} {2} (n\plus{}1)!$, so by pigeon-hole principle, there will exist two sequences with same $ (n\minus{}1)$-prefix in our set. However, these two sequences cannot be [i]separated[/i], contradiction.


A [i]palindrome[/i] is a word that is read from beginning to end and from end to beginning same. Ex: $ABABA=ABABA$, so $ABABA$ is a palindrome, but $ABAB \neq BABA$, so $ABAB$ is not a palindrome. Hope you understand. Sum of words is their concatenation. Ex: $AB+BA=ABBA$.
We are given a word of 2003 letters, all of them only $A$ and $B$.
Prove that the word can be written as a sum of at most 802 palindromes.

Let's prove that a 5-lettered word can be written as the sum of 2 palindromes. Assume WLOG that the word begins with A. The other 4 letters can be chosen in $2^4=16$ ways, so we have 16 cases:

01) AAAAA = palindrome
02) AAAAB = AAAA + B
03) AAABA = AA + ABA
04) AAABB = AAA + BB
05) AABAA = palindrome
06) AABAB = AA + BAB
07) AABBA = A + ABBA
08) AABBB = AA + BBB
09) ABAAA = ABA + AA
10) ABAAB = A + BAAB
11) ABABA = palindrome
12) ABABB = ABA + BB
13) ABBAA = ABBA + A
14) ABBAB = ABBA + B
15) ABBBA = palindrome
16) ABBBB = A + BBBB

From the 2003-lettered word, take out the last 3 letters. The word formed by them can be written as the sum of at most 2 palindromes (easy to check). Divide the rest of 2000 letters into 400 consecutive words of 5 letters. Each of the words can be written as the sum of at most 2 palindromes. This all adds up to a total of at most $2 \times 400+2 = 802$ palindromes.


What is the answer to #23 on the AMC 10B? Is it 3/8?

Nope, It can either have 0 faces of other color painted, 1 face, or a pair of opposite faces.

To have 0 faces, there is only 1 way.
To have 1 face, there is only 6 ways, 1 per face
To have 2 faces, there is only 3 ways, 6 faces/2 for pairs

That adds up to 10, then multiply by 2 cuz it can be any color, so 20.
Total possibilities = 2^6 = 64

20/64 = 5/16.


induction
geometry
parallelogram
combinatorics unsolved
combinatorics

suppos that d1,d2,...,dn are n lines so that for any two of them like di and dj which intersect on for example A there is another line which passes from A.(and any two of the lines intersect each other).prove that all the lines pass from one point.

Extremal Principle.  It's a variation of the Sylvester Problem.  You can also solve it using induction.

Masoud Zargar

I am interested in solution by induction??

Assume that's right for $n=k$, we'll proof for $n=k+1$
WE can make $k$ line concurrent at a point $A$, while assuming that the remaining line don't intersect any other line at $A$. Let this line intersects any other line, say it $d_{x}, x \in [1,2,\cdots,k]$ at $B, B \neq A$, then there is another line different than these two lines intersects $d_{x}$ at $B$ which is impossible because they've alreadymet at $A$...
Contradiction and we finish the induction...

Am I right ?

Well, that's what I did too.  I've never seen an induction proof for this problem other than yours; however, it seems perfectly logical other than the fact that you forgot to mention that it works for the initial state(s).

Yah, I think it's quite clear for the Mathlinkers  :D

What about inversion + helley's theorem or whatever that thing says about ovals with three passing through common points => all pass through common points?

I think we just use a simple answer for this one...
But that maybe useful in other hard question ,so JSteinhardt ,can u post the theorem ???  :)

[quote="RDeepMath91"]Assume that's right for $n=k$, we'll proof for $n=k+1$
WE can make $k$ line concurrent at a point $A$[/quote]
it's not obvious you must tell something more about that

What ?? I think it's clear because the statement is true for $n = k$, so what still I need to write down ??
I'm sorry but I'm really bad at combin....

yes, it's true for k, but your situation is different, let $l_{1},...,l_{k},l_{k+1}$ be those lines, you know that if you take any two lines from this set there will be another line, which intersect them , but $l_{k+1}$ may be that line, and then the assumption doesn't have to be true in the set $l_{1},...l_{k}$

I say we can make $k$ line concurrent....
So no matter what are the $k$ lines, the remaining one will intersect at the same point.... (in my proof)
Yah, maybe I just forget to write some little explanation...

I said that I am interested in solution by induction, because I know that it can't be solved that way.

Just like [b]Zorro[/b] said, your solution isn't true. It does stand for any $k$ line, but only when for every two line that intersect, there is another one that goes through that intersection point [b]BETWEEN[/b] that $k$ line.

When you have $k+1$ line, $l_{1},l_{2},...,l_{k+1}$, and you look at $l_{1},l_{2},..., l_{k}$ then, maybe, intersection point of $l_{1}$ and $l_{2}$ isn't on any other line from $l_{1},l_{2},...,l_{k}$, because it can be on $l_{k+1}$.

Just like: http://www.mathlinks.ro/Forum/viewtopic.php?t=82810

Bye

yes, I heard that there is no proof known which uses induction
I don't know how to solve it, do you know any solution??

I have come up with a solution using the extremal principal along with contradiction.  Assume that there is more than one intersection point.  Then we select an intersection point with the minimal distance to a line.  Then the rest is simple by using traingles.  We conclude that there is even a smaller distance.  And then we're done. :)  (edit: hopefully, it is right) I find this solution neater than an inductive proof which seems to be wrong.  I'll think of an inductive proof.

Masoud Zargar

let S be the set of all intersections, it's finite, so we can take the polygon with vertices from S with maximal area, since it has maximal area it have to be paralelogram and then we get a contradiction


superior algebra
superior algebra solved

You might want to take a look at this:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=automorphism&t=16615[/url].

[quote="liyi"]Show that $\mbox{Aut}(S_n)\cong S_n$ for $n\geq 3$.[/quote]

It is a famous fact that $S_6$ has outer automorphisms, so this is false.
For $n>6$ it is true.


superior algebra
superior algebra theorems

Let $G$ be a finite group acting on a set $M$. Let $k$ be the number of orbits in that $M$ is divided by this action and let $\mathrm{fix}(g), g \in G$ be the number of Fixpoints of $g$. Then it follows:
\[ \sum_{g \in G} \mathrm{fix}(g) = k \cdot |G|  \]

We count the pairs $(g,x)\in G\times M$ with $x\in\mbox{Fix}(g)$. On the one hand, there are $\sum_{g\in G}|\mbox{Fix}(g)|$ such pairs, since for every $g\in G$ there are $\mbox{Fix}(g)$ elements $x$ which form such pairs with $g$. On the other hand, there are $\sum_{x\in M}|\mbox{Stab}(x)|$, where $\mbox{Stab}(x)$ is the stabilizer group of $x$ under our action. Hence $\sum_{g\in G}|\mbox{Fix}(g)| = \sum_{x\in M}|\mbox{Stab}(x)|$.

If for each $x\in M$ we denote the orbit of $x$ by $Gx$, we have $|\mbox{Stab}(x)|=\frac{|G|}{|Gx|}$. We thus get $\sum_{g\in G}|\mbox{Fix}(g)|=|G|\cdot\sum_{x\in M}\frac 1{|Gx|}\ (*)$, which is obviously equal to $|G|\cdot k$, since for every orbit $Gx$, the term $\frac 1{|Gx|}$ appears precisely $|Gx|$ times in the RHS of $(*)$.


induction
number theory
combinatorics open
combinatorics

Let $a_1,a_2,...,a_n$be $n$ distinct integers s.t.$|A|=\frac{(n-1)n}{2}$,where $A=\{a_i+a_j|1\le i<j \le n\}$.Find the minimum of $\max|a_i-a_j|(1\le i<j \le n)$.

It's Fn - 1, where Fn denotes the nth Fibonacci number.

Sorry,but $a_1,a_2,...a_n$ should be distinct!

ya, just consider a1 = 1, a2 = 2, a3 = 3, a4 = 5, a5 = 8, ...  and i guess it's easy to prove by induction

I guess it is not easy to prove by induction ;) . The question is also on the Number Theory Forum. Look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39212]here[/url].

Can we prove that $\min(\max|a_i-a_j|)(i\neq j) \le n^{2}$ ?


Write the function as a composite of two functions, f and g. (more than one answer is correct).

f(g(x)) = (sq root of X) /(divided by) 3(sq root of x)  + 2

f(x)= ??
g(x)= (Sq root of x) right?

sorry i dont know how to use the symbols

Let $f(x)$ be an arbitrary one-to-one function; then $g(x) = f^{-1}\left( \frac{ \sqrt{x}}{3 \sqrt{x}+2}\right)$

;)

Do you want the simplest expression?

yeah basically i just want to know what f(x) is. I have a test coming tomorrow morning, so would appreciate the help. and yes thats the equation, how do i access those math symbols?  :oops:

$f(x)$ is [b]arbitrary.[/b]  It all depends on how simple you want $g(x)$ to look.

[hide="A possibility"] Let $f(x) = \frac{x}{3x+2}$. [/hide]

You use dollar-signs to enter and exit math mode.  Some help with $\LaTeX$ syntax:

http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php

As t0rajir0u already confirmed, there are an infinte number of ways you can do function composition.

For the specific problem you specified that you wanted $g(x) = \sqrt{x}$.

This works nicely.

Look in the final form. See where all the $\sqrt{x}$'s occur.
Replace those with just plain ole' $x$. Now we have two functions in X.

$f(x)=\frac{x}{3x+2}$
$g(x)=\sqrt{x}$
Now make the composition:
$(f \circ g)(x) = \frac{\sqrt{x}}{3\sqrt{x}+2}$

alright i kinda get the hang of it, i feel stupid now haha. thanks guys

Followup:
Construct two functions, F and G, that share no common domain values.


Βρειτε το ελαχιστο της παραστασης   :jump:                                                                                S=xy/z+yz/x+zx/y oπου το αθροισμα των τετραγωνων των x,y,z einai 1.

Αρχιμήδης Μικρών 2004
http://www.hms.gr/gr/activities/index_competitions.html

Ε Σιλουανε μου κανεις χαλαστρα......  :D  :D οκ ειχα ξεχασει οτι επεσε εκει.για μενα εχει ενα μηνυμα αυτη η ασκηση:Αν δε μπορεις να βρεις το ελαχιστο S tote βρες το ελαχιστο της ριζας του τετραγωνου.ειναι το ιδιο αποτελεσμα αλλα πολυ πολυ πολυ πιο ευκολα και απλα. :)

Είχα πάρει μέρος στο διαγωνισμό όντας Β Γυμνασίου τότε  και προβλήματα που έχεις διαγωνιστεί δεν ξεχνιούνται εύκολα . ;)

Κωστα βρηκα και μια αλλη λυση.Αν θεσω x^2=ab,y^2=bc,z^2=ca τοτε αρκει να βρουμε το ελαχιστο του α+b+c με την συνθηκη οτι ab+bc+ca=1 Ομως (a+b+c)^2>=3(ab+bc+ca) αρα ΜΙΝ(a+b+c)=3^1/2 :gleam:


Έστω A,B,C,D  σημεία στο χώρο και AB η απόσταση μεταξύ του A και Β , και ουτω καθεξής.Δείξτε ότι

$ {{AC}^2 \plus{} {BD}^2 \plus{} {AD}^2 \plus{} {BC}^2}\geq{{AB}^2 \plus{} {CD}^2}$

[color=red]Mod: edited[/color]

Μήπως έχει γίνει τυπογραφικό ? Γράφεις και στα δυο μέλη $ AB^2$

Όντωσ εχεις δικιο στο πρωτο μελος ειναι $ {AC}^2$

Τίποτα ... :|  γράψε λάθος

Σίγουρα είναι έτσι η εκφώνηση;

Προσωπικά νομίζω ότι είναι με μέτρα διανυσμάτων ή αποστάσεις χωρίς τετράγωνα:

$ AC \plus{} BD \plus{} AD \plus{} BC \equal{} \frac {AC \plus{} AD}{2} \plus{} \frac {BC \plus{} BD}{2} \plus{} \frac {CA \plus{} CB}{2} \plus{} \frac {DA \plus{} DB}{2} \ge$

(από τριγωνική ανισότητα) $ \frac {CD}{2} \plus{} \frac {CD}{2} \plus{} \frac {AB}{2} \plus{} \frac {AB}{2} \equal{} AB \plus{} CD$

Αν είχε και τετράγωνα, πρέπει να θέσεις ότι τα σημεία δε σχηματίζουν ανά τρία αμβλεία γωνία, επομένως θα ισχύει η ιδιότητα $ a^2 \le b^2 \plus{} c^2$ και όμοια με παραπάνω η άσκηση.

Given that the points lie in 3-dimentional space, I asummed above that either the squares are not needed or that we should use vectors instead and gave a proof in case it was as I stated it.

We can set without loss of generality A(0,0,0), B(0,1,0) C( a,b,c) and D(x,y,z)  then

$ AC^2\equal{}a^2\plus{}b^2\plus{}c^2$
$ BD^2\equal{}x^2\plus{}(y\minus{}1)^2\plus{}z^2$
$ AD^2\equal{}x^2\plus{}y^2\plus{}z^2$
$ BC^2\equal{}a^2\plus{}(b\minus{}1)^2\plus{}c^2$
$ CD^2\equal{}(x\minus{}a)^2\plus{}(y\minus{}b)^2\plus{}(z\minus{}c)^2$
$ AB^2\equal{}1$

The inequality ( if the calculus is correct ) is equivalent to

$ (b\plus{}y\minus{}1)^2\plus{}(x\plus{}a)^2\plus{}(z\plus{}c)^2 \ge0$ that is obviously true

 :)


Show that the function is continuous on the interval
[-4,4].

f(x) = x(sqrt{16-x^2})

How do I answer such questions for ANY interval?

For example, how do I answer the above question for the interval ( 3, +infinity)?

Since $f$ is only defined on $[-4,4]$, there's no point even talking about continuity elsewhere.

I'm sorry but I don't understand your answer.  Can you show me how to solve this question?

sqrt(16-x^2) is defined only for x belonging to [-4,4].
Range of sqrt(16-x^2) in [-4,4] is [0,4]
Thus its continuity has to be checked in this interval only.
Now x is continuous in [-4,4].
16-x^2 is continuous in [-4,4].
sqrt(x) is continuous in [0,4]
thus sqrt{16-x^2} is continuous in [-4,4].
thus xsqrt{16-x^2} is continuous in [-4,4].

Does this help.