let x and y in $ R$
prove that x^{2}+y^{2}+xy = 1  =>   x^{3}y +y^{3}x \ge -2

Clearly true if $ x$ and $ y$ have the same sign. Suppose WLOG that $ x < 0, y > 0$. Let $ z \equal{}\minus{}x$. Then we can rewrite the problem with positive variables:

$ z^{2}\plus{}y^{2}\equal{} 1\plus{}z y\Rightarrow z^{3}y\plus{}z y^{3}\leq 2$

By AM-GM on the condition, we have $ 1\plus{}z y\geq 2z y$, or $ 1\geq z y$.

Notice by the condition that $ z^{3}y\plus{}z y^{3}\equal{} z y\plus{}z^{2}y^{2}$. Since $ z y\leq 1$, we also have $ z^{2}y^{2}\leq 1$. It follows that $ z y\plus{}z^{2}y^{2}\leq 2$, and we're done.

no the right exercice is
let x and y in  $ R$ :D
prove that x^{2}+y^{2}+xy = 1 => x^{3}y +y^{3}x >= -2

this problem is equivalent to:

$ xy(x^{2}\plus{}y^{2})\ge\minus{}2(x^{2}\plus{}xy\plus{}y^{2})^{2}$

If $ x,y<0$ then the inequality is weaker, so WLOG $ y\equal{}1$. Then

$ x^{3}\plus{}x\plus{}2(x^{2}\plus{}x\plus{}1)^{2}\ge 0$.

$ 2x^{4}\plus{}5x^{3}\plus{}6x^{2}\plus{}5x\plus{}2\equal{}(x\plus{}1)^{2}(...)$

which is easy to prove.


trigonometry
geometry
geometric transformation
reflection
calculus
derivative
inequalities

There is a water pipe that needs to be connected to two towns. Town X is 4 miles directly above point A, Town Y is 3 miles directly above point B. The distance from point A to B is 6 miles. There must be a pump placed in the line segment AB, and then pipes connected to the pump towns for the water to flow into the towns.

Each mile of pipe from the pump will cost $300.

Question: What should the distance of the pump from each town be in order to have the least cost?

Here's an illustration of the problem: http://i36.tinypic.com/m79vtd.gif (not drawn to scale)

Can anyone explain the solution to me (thoroughly, if possible)? Thanks.

[hide="solution 1"][geogebra]a2895c5a3943cc14498a51d8827ad5b7378d29fc[/geogebra]

let x be the distance from A to the certain point which will make the pipes to have the least cost.

let y be the distance from B to that point.

then $ x \plus{} y \equal{} 6$, so $ x \equal{} 6 \minus{} y$

and the total length of the pipes wlil be $ \sqrt {16 \plus{} x^2} \plus{} \sqrt {9 \plus{} y^2}$, which is $ \sqrt {y^2 \minus{} 12y \plus{} 52} \plus{} \sqrt {9 \plus{} y^2}$

differentiate that to find the least value of it.

$ \frac {2y}{2\sqrt {y^2 \minus{} 12y \plus{} 52}} \plus{} \frac {2y}{2\sqrt {9 \plus{} y^2}} \equal{} 0$

therefore $ 4y\sqrt {9 \plus{} y^2} \plus{} 4y\sqrt {y^2 \minus{} 12y \plus{} 52} \equal{} 0 \implies \sqrt {y^2 \plus{} 9} \equal{} \minus{} \sqrt {y^2 \minus{} 12y \plus{} 52}$

square it to obtain $ y^2 \plus{} 9 \equal{} y^2 \minus{} 12y \plus{} 52 \implies 12y \equal{} 43 \implies y \equal{} \frac {43}{12}$

therefore, the distance between the certain point and town A is $ \sqrt {\frac {29}{12}^2 \plus{} 4^2} \equal{} \sqrt {\frac {3145}{144}}$

and the distance between the point and town B is also $ \sqrt {\frac {43}{12}^2 \plus{} 3^2} \equal{} \sqrt {\frac {3145}{144}}$

answer : $ \sqrt {\frac {3145}{144}}$[/hide]

[hide="solution 2... much easier"]blah i forgot this way...

the distance becomes least when the distance between the point and the towns are equal.

so let the distance between the point and the town be x.

then $ \sqrt {x^2 \minus{} 16} \plus{} \sqrt {x^2 \minus{} 9} \equal{} 6$ so solve that to get x[/hide]

EDIT:

blah so much of a long solution

sry this is a wrong solution please see azjps's

Let $ C$ be that point on $ \overline{AB}$. Reflect $ Y$ about the line $ \overline{AB}$ to a point $ Y'$. Then the desired distance $ XC \plus{} CY \equal{} XC \plus{} CY' \ge XY'$ (shortest distance between two points). But $ XY' \equal{} \sqrt{6^2 \plus{} 7^2} \equal{} \sqrt{85}$.

FantasyLover, you miscalculated the derivative of $ \sqrt{y^2 \minus{} 12y \plus{} 52}$.

[quote="azjps"]Let $ C$ be that point on $ \overline{AB}$. Reflect $ Y$ about the line $ \overline{AB}$ to a point $ Y'$. Then the desired distance $ XC \plus{} CY \equal{} XC \plus{} CY' \ge XY'$ (shortest distance between two points). But $ XY' \equal{} \sqrt {6^2 \plus{} 7^2} \equal{} \sqrt {85}$.

FantasyLover, you miscalculated the derivative of $ \sqrt {y^2 \minus{} 12y \plus{} 52}$.[/quote]

Thanks. I asked a peer to solve it for me, he showed me this diagram:

[img]http://i33.tinypic.com/50fkhe.gif[/img]

But I couldn't get him to explain properly... I understand that those angles are vertical pairs and all that, but how can I be so sure that the $ x$ and $ 6\minus{}x$ that we get from this solution are the best distances to place the pump from points $ A$ and $ B$? Can anyone please explain it to me? Thank you!

[hide="hint"] The shortest distance between two points is a line.
Now look at what your friend showed you.[/hide]

Whoops, mis-read the question. For equality to hold in the above triangle inequality, we need $ C$ to lie on $ \overline{XY}$, which implies that $ \triangle AXC \sim \triangle BY'C \Longrightarrow \frac {AX}{XC} \equal{} \frac {4}{x} \equal{} \frac {3}{6 \minus{} x} \equal{} \frac {BY'}{Y'C} \Longrightarrow x \equal{} \frac {24}{7}$.

In case no one understood my hint, look at the diagram that I Hate Watermelons made.

It is clear that a line starting on AB will have the same distance whether it goes to the second town, or the reflection.
Therefore, the line that goes from city 1 to the reflection of city 2 will be the shortest distance.

Law of Sines/Cosines? (reflecting doesn't help much)


If $i=\sqrt{-1}$, then how many distinct complex numbers exist that are possible values of $i^{n}$ if $n$ is a member of the set of natural numbers?

complex numbers are in the form $a+bi$. but if I raised $i$ to a power then it will be either $-1$ and $-\sqrt{-1}$. I'm not sure what the question is asking  :(


[quote="mdk"]In what base b is the equation 53*15 = 732 valid? Here all three numbers are base-b
numbers, and b must be a positive integer.[/quote]

$ 53\equal{}5b\plus{}3$
$ 15\equal{}1b\plus{}5$
$ 732\equal{}7b^{2}\plus{}3b\plus{}2$ ----->

$ 53*15\equal{}732$ ---> $ (5b\plus{}3)(b\plus{}5)\equal{}7b^{2}\plus{}3b\plus{}2$ ---> $ 2b^{2}\minus{}25b\minus{}13\equal{}0$ ---> $ b\equal{}13$

[hide="meh"]Let's consider the last digits. Since 3*5 is 15 in base 10, we know that $ 15\equiv2\pmod{b}$, so, 13 is divisible by b, and b must be 13.[/hide]


calculus
integration
logarithms
calculus computations

Let $a$ be a constant number such that $0<a<1$ and $V(a)$ be the volume formed by the revolution of the figure

which is enclosed by the curve $y=\ln (x-a)$, the $x$-axis and two lines $x=1,x=3$ about the $x$-axis.

If $a$ varies in the range of  $0<a<1$,  find the minimum value of $V(a)$.

I could be wrong because it's almost 2 am here and I'm tired :lol: , but here's what I did.

The volume can be given by

$V(a)=\pi\int_1^3 \ln^2(x-a) \ dx$.

Using integration by parts, we can evaluate the integral:

$\int \ln^2 (x) \ dx$

$u = \ln^2 (x) \Rightarrow du = \frac{2\ln(x)}{x}$

$dv=dx \Rightarrow v = x$

$\int \ln^2 (x) \ dx = x\ln^2(x) - 2\int \ln x \,dx = x\ln^2 x -2x\ln x +2x$

Therefore $\int \ln^2 (x-a) \ dx = (x-a)\left(\ln^2(x-a)-2\ln(x-a)+2\right)$

$V(a) = \pi\left[(3-a)\left(\ln^2(3-a)-2\ln(3-a)+2\right)-(1-a)\left(\ln^2(1-a)-2\ln(1-a)+2\right)\right]$

$V'(a) = \ln^2(1-a)-\ln^2(3-a) = 0$

Let $x=\ln(1-a)$ and $y=\ln(3-a)$, then the above equation has a solution if $x=\pm y$.

[b]Case 1:[/b] $x=y$:  No solutions

[b]Case 2: [/b] $x=-y$:  $\ln(1-a)\ln(3-a)=1\Rightarrow a = 2\pm\sqrt{2}$

But $2+\sqrt{2}$ is extraneous, so the minimum value occurs at $a=2-\sqrt{2}$. (We know this is a minimum because $V''(a)$ is positive at that point.)

Finally,

$V\left(2-\sqrt{2}\right) =\pi\left[ (\sqrt{2}+1)\left(\ln^2(\sqrt{2}+1)-2\ln(\sqrt{2}+1)\right) - (\sqrt{2}-1)\left(\ln^2(\sqrt{2}-1)-2\ln(\sqrt{2}-1)\right) +4\right]$

Yes, you are right, you have a good power of calculation, Jimmy. :)  :)

kunny


Let be $ n\in \mathbb{N}\ ,\ n\ge 2$ and $ a_1,a_2,...,a_n\in \mathbb{N}^*$ . Prove that the roots of all the equations : $ x^2 \minus{} (a_{k \plus{} 1})x \plus{} a_{k \plus{} 1} \equal{} 0\ ,\ k \equal{} 1,2,...,n\ ,\ a_{n \plus{} 1} \equal{} a_1$ are integers if and only if $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n$ .

x^2-(a_{k+1})x+a_{k+1}=0\ ,\ k=1,2,...,n\ ,\ a_{n+1}=a_1

Check this equation please... I'm thinking there might be an error, cos there's nothing linking {a} :wink:

[quote="Q.E.D.avid"]$ x^2\minus{}(a_{k\plus{}1})x\plus{}a_{k\plus{}1}\equal{}0\ ,\ k\equal{}1,2,...,n\ ,\ a_{n\plus{}1}\equal{}a_1$

Check this equation please... I'm thinking there might be an error, cos there's nothing linking {a} :wink:[/quote]

I think the equations are correct. Check that he mentioned [b]all[/b] the equations.
[quote="alex2008"]Let be $ n\in \mathbb{N}\ ,\ n\ge 2$ and $ a_1,a_2,...,a_n\in \mathbb{N}^*$ . Prove[b] that the roots of all the equations [/b]: $ x^2 \minus{} (a_{k \plus{} 1})x \plus{} a_{k \plus{} 1} \equal{} 0\ ,\ k \equal{} 1,2,...,n\ ,\ a_{n \plus{} 1} \equal{} a_1$ are integers if and only if $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n$ .[/quote]


calculus
function
analytic geometry
graphing lines
slope

please do not use any calculus or above methods to solve it (It's not allowed :()

I need help solving several Physics problems :( I have the answers for some of them and some i'm not sure. I need help in the solution.

4.24. A large 500-kg magnet exerts a constant force of 3.00 N on a 0.250-kg bar magnet. What magnitude force does the bar magnet exert of the big magnet?

For this problem I just did   (500/3 x 0.250) which gave me 41.6667 N
but i'm not sure if this is correct.

4.32 Figure 1 shows a graph of distance along a straight course as a function of time for a 1000-kg dragster.Consturct a graph of the horizontal force between the tires and ground (no slipping) as a function of time. (Hint: First consturct a plot of velocity, taking points, say every 0.5 s)

Here figure 1 graph it shows an exponential increase graph.

for the 1st second it's 5m, the 2nd second its 25, 3 is 55, 4 is 95, 5 is 150.
So since F=ma and a=v/t, so do i have to first find the slope of every place, which will give me the velocity and then divide each one by time and multiply each point by 1000? can someone please help!

4.33 a) A 17.0-kg bucket is lowered by a rope with constant velocity of 0.500 m/s. What is the tension in the rope. b) A 17.0-kg bucket is lowered with a constant downward acceleration of 1.00 m/s^2. What is the tension in the rope? c) A 10.06-kg bucket is raised with a constant upward acceleration of 1.00m/s^2. What is the tension?

for a) since tension is m1 x  F/(m1+m2) and F = 0.500 in thise case, it should be 17 x (.500/17) which shouldgive 144.5 but it says 167N as the answer. What went wrong.

b) i dont get how it's 150N. i did  everything i can like T=ma and stuff but it never gave me 150N.

c) how is this 115N?

4.34

A 0.0050-kg bullet traveling with a speed of 200 m/s penetrates a large wooden fence post to a depth of 0.030m. What was the average resisting force exerted on the bullet.

Here i have to find frcition. since m = 0.0050 and v= 200m/s , i get f=mv which is 1. since the depth is 0.030 i thought the resisitng force to be .030. I doubt this is correct :S

4.35

A 0.0048-kg bullet traveling with a speed of 400 m/s penetrates the same thing above. If the average resisting force exerted on the bullet was 4.5 x 10^3 N, how far did the bullet penetrate?

I didn't know how to solve this.. i need a solution hhe.. the answer is 8.5 cm by the way:)

4.36

Two air-track gliders m1 and m2 are joined together with a light string. A constant horizontal force of 4 N to the right is applied to mass m2. a) If m1= 1.5 kg and m2= 0.50 kg what is the acceleration of the gliders.
(it shows m1 is in left and m2 is in the right)

I found that  delta a= F/m1+m2
so i thought it should be 4/2 which the acceleration is 2. true?

b) Tension in the cord?

noidea.

4.37

Find the accelerations and the tension for the situation of problem 4.36 given that m1.0.50 kg and m2. 1.5kg.

-> shouldnt this be the same as above?

****4.39. A simple Atwood machine composed of a single pulley and two masses m1 and m2 is on an elevator. When m1= 44.7 kg and m2 = 45.3 kg, it takes 5.00 s for mass m2 to descend exactly one meter from rest relative tothe elevator. What is the elevator's motion?

didn't get why they gave so much numbers but it's accelerating upward anyways.

4.41. A 9.75-kg lead birck rests on a level wodden table. If a force of 46.4 N is required to slide the brick across the table at a constant speed what is the coefficient of frction?

didn't know how to start this at all :(

4.43. A 5-00 g nickel coin sliding along a level table top with an initial velocity of 200 cm/s comes to a stop after traveling 50.0 cm
a) what is the accl of the coin?

first i found 200/50 which gave me 4 but the answer is -4 :S

b) what is the coefficient of friction btw the coin and the table?

no idea :S  Ff = mew (N) no?

****4.51. Two clay pots joined together by a light string rest on a table. The frictional coefficient btw the pots and the table is 0.35. The pots are also joined to a 4.0-kg mass by a string of negligible mass passed over an ideal pulley as shown in the figure.

a) calculate the acceleration of the system when the 4.0-kg is released.
b) Also find the Tensions T1( which is between the pots) and T2(which is after the 2nd pot and same x-axis as T1)

These are suppose to be basic but I'm learning physics on my own and i'm having a very hard time. Please help :(  :oops:
Thanks :)

There is a seperate physics forum, this should be moved.

ur right someone should move it

(although i find these problems to be much more easier compared to other problems in the physics problems)

First you REALLY need to sit and learn your physics. Most of your "solutions" show that you have a very vague idea (to say the least) about what's going on there. No notion of Third Newton's Law, tension, action, energy, you name it.

And I doubt anyone here or in Physics forum would volunteer to be your online tutor.

i understand

but can somebody just help me with the problems

with the ***** in the front...

obviously, I solved all the ones without the ***** while I was going.

thanks

I was trying to say that neither of your solutions is even remotely correct.


[quote="ductrung"]For any positive real numbers $a,b$ and $c$, \[\frac{a(a+b)}{b+c}+\frac{b(b+c)}{c+a}+\frac{c(c+a)}{a+b}\ge 3\ \sqrt{\frac{ab^{2}+bc^{2}+ca^{2}}{a+b+c}}\] [/quote]
$\frac{a(a+b)}{b+c}+\frac{b(b+c)}{c+a}+\frac{c(c+a)}{a+b}\ge 3\ \sqrt{\frac{ab^{2}+bc^{2}+ca^{2}}{a+b+c}}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(\frac{a(a+b)}{b+c}\right)^{2}+2\cdot\sum_{cyc}\left(\frac{a(a+b)}{b+c}\cdot\frac{b(b+c)}{c+a}\right)\geq9\cdot\frac{ab^{2}+bc^{2}+ca^{2}}{a+b+c}.$
Id est, enough to prove that $\sum_{cyc}\left(\frac{a(a+b)}{b+c}\cdot\frac{b(b+c)}{c+a}\right)\geq3\cdot\frac{ab^{2}+bc^{2}+ca^{2}}{a+b+c}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{ab(a+b)}{a+c}\geq3\cdot\frac{ab^{2}+bc^{2}+ca^{2}}{a+b+c}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^{5}c+a^{4}b^{2}-a^{3}b^{2}c-a^{2}b^{2}c^{2})\geq0\Leftrightarrow$
$\Leftrightarrow\frac{1}{6}\cdot\sum_{cyc}(4a^{4}b^{2}+b^{4}c^{2}+c^{4}a^{2}-6a^{3}b^{2}c)+\sum_{cyc}(a^{5}c-a^{2}b^{2}c^{2})\geq0.$ :)

[quote="arqady"]
Id est, enough to prove that \[\sum_{cyc}\left(\frac{a(a+b)}{b+c}\cdot\frac{b(b+c)}{c+a}\right)\geq3\cdot\frac{ab^{2}+bc^{2}+ca^{2}}{a+b+c}\] $\Leftrightarrow\sum_{cyc}\frac{ab(a+b)}{a+c}\geq3\cdot\frac{ab^{2}+bc^{2}+ca^{2}}{a+b+c}$
$\Leftrightarrow\sum_{cyc}(a^{5}c+a^{4}b^{2}-a^{3}b^{2}c-a^{2}b^{2}c^{2})\geq 0$
$\Leftrightarrow\frac{1}{6}\cdot\sum_{cyc}(4a^{4}b^{2}+b^{4}c^{2}+c^{4}a^{2}-6a^{3}b^{2}c)+\sum_{cyc}(a^{5}c-a^{2}b^{2}c^{2})\geq 0.$ :)[/quote]

Cool and bullet-proof, arqady.

[quote=ductrung]For any positive real numbers $a,b$ and $c$, \[\frac{a(a+b)}{b+c}+\frac{b(b+c)}{c+a}+\frac{c(c+a)}{a+b}\ge 3\ \sqrt{\frac{ab^{2}+bc^{2}+ca^{2}}{a+b+c}}\][/quote]

[img]http://s2.sinaimg.cn/middle/006ptkjAzy76sAYIOl3c1&690[/img]

Yes!  :-D The last it's my old identity.


topology
geometry
rectangle
3D geometry
tetrahedron
real analysis
real analysis unsolved

Prove that $R^2$ is not homeomorphic with $R^3$.

Posted before by alekk.

The removal of one point from $\mathbb R^3$ leaves it simply connected, while the removal of a point $x$ from $\mathbb R^2$ does not leave $\mathbb R^2\setminus\{x\}$ simply connected.

Edit:

I think the following (inspired by what I read in Alexandrov’s “Elementary Concepts in Topology” :)) is a sufficiently “elementary” proof, not making use of any strong reults belonging to the field of algebraic topology.

First we prove that given any compact subset $K$ of $\mathbb R^2$ and any covering of $K$ with open sets $(U_n)_n$, there is an open refinement $(V_m)_m$ which covers $K$ s.t. every point in $K$ belongs to the closure of at most $3$ sets $V_m$.

Take a covering of the plane by rectangular sets arranged in brick-tiling manner (see the picture below). Slightly enlarge the rectangles to make them open. It's clear that every point in the plane belongs to the closure of at most $3$ of these open rectangles. Now, by Lebesgue's Number Lemma, if the rectangles are small enough, then all those which intersect $K$ wil be contained in some $U_n$. We take our $V_m$'s to be precisely these slightly enlarged open rectangles which intersect $K$.

Now we prove the negation of the above for $\mathbb R^3$. To be precise, we prove that we can find a compact $K\subset\mathbb R^3$ which has an open covering $(U_n)_n$ s.t. for every open covering $(V_m)_m$ which refines $(U_n)_n$, there are $4$ sets $\overline{V_m}$ with non-empty intersection.

We take our $K$ to be a tetrahedron $\mathcal T=A_1A_2A_3A_4$, and we take the finite family of open sets $(U_n)_n$ which cover $\mathcal T$ to consist of sets so small that the closure of none of them intersects all four planes $(A_iA_jA_k)$ for $\{i,j,k\}\subset\{1,2,3,4\}$. Now let $F_1$ be the closure of the union of those $U_n$’s which do not intersect $(A_2A_3A_4)$, $F_2$ - the closure of the union of the remaining $U_n$’s which do not cut $(A_1A_3A_4)$, $F_3$ - the closure of the union of the remaining $U_n$’s which do not intersect $(A_1A_2A_4)$, and, finally, $F_4$ - the closure of the union of the remaining $U_n$’s which do not intersect $(A_1A_2A_3)$. It’s easy to see that the closed sets $F_i$ cover $\mathcal T$.

Now, for each $n$, triangulate $\mathcal T$, dividing it into tetrahedra of diameter $\le\frac 1n$. We label each vertex $x$ of this triangulation with some $i$ for which $x\in F_i$. It’s easy to see that this labeling is proper, in the sense that $A_i$ has label $i$ for $i\in\overline{1,4}$, and if, say $x\in(A_iA_jA_k)$, then $x$ bears one of the labels $i,j,k$, while if $x\in A_iA_j$, $x$ bears one of the labels $i,j$. Sperner’s lemma says that there is a small tetrahedron $\mathcal T_n$ in this triangulation whose vertices have different labels. We denote the vertex of $\mathcal T_n$ having label $i$ by $x_i^n$. For compactness reasons, we can extract convergent subsequences $(x_i^{n_k})_k,\ i=\overline{1,4}$, and since the mutual distances between the $x_i^{n_k}$ go to $0$ as $k\to\infty$, these sequences converge to a point $x$ which must be common to all the $F_i$’s and thus to some four $\overline{U_n}$’s.

I hope it's Ok.


In the Figure, $a$ is a stone meteorite with mass $m$ and the larger one is earth with mass $M$ and radius $R$. $O$ is the centre of Earth and $O'A=l$. $a$ starts moving with initial kinetic energy $E_k$. It's initial velocity is paralel to $OO'$. If the meteorite and Earth do not collide, find the minimun value of $l$.

If we minimize L, the meteorite will just barely brush by the surface of earth. When it brushes by the earth, the angular momentum
$L = Rv_f$

Our initial angular momentum
$L = lv_i$

By conservation of angular momentum
$Rv_f=lv_i$

We can also set up an energy equation relating these two points.
$KE_i + PE_i = PE_f + KE_f$

Substitute $v_f = lV_i/R$ into the energy equation.

Only $PE_i$ and $KE_f$ contain $l$ in them, and the rest of the variables are fixed, so all that's left is to solve for $l$, which requires an approximation: assuming $O'A << OO'$ we can assume $OA = OO'$. Any sharper approximation, and the algebra becomes too messy.


Prove that any prime of the form $ 3n\plus{}1$ is also of the form $ 6m\plus{}1$, and that each integer of the form $ 3n\plus{}2$ has a prime factor of this form ($ 3n\plus{}2$, that is).

[hide="hint 1"]$ 3n\plus{}1 \in \text{prime} \Rightarrow 3n \in \text{even} \ \therefore n\equal{}2m$.[/hide]
[hide="hint 2"]$ 3n\plus{}2 \in \text{prime} \Rightarrow 3n\plus{}3 \in \text{even} \ \therefore n\plus{}1\equal{}2m$.[/hide]

Solution 1

[hide]All primes except for two are odd. However, $ 2$ is not of the form $ 3n\plus{}1$. Thus any prime of the form $ 3n\plus{}1$, being odd, is one more than an even number. Thus $ 3n$ is even. $ 3$ is not even, so $ n$ is, and thus may be written as $ 2m$. Our prime now becomes $ 6m\plus{}1$, as desired.[/hide]

Solution 2

[hide]All positive integers are either $ 0$, $ \minus{}1$, or $ 1 \bmod{3}$. The number in question, $ 3n\plus{}2$, is congruent to $ \minus{}1 \bmod {3}$. Consider the prime factors of this number. They cannot all be only of $ 0 \bmod{3}$'s and $ 1 \bmod{3}$'s, for then the product will just $ 0$ or $ 1 \bmod{3}$. Therefore we have that at least one prime factor that is $ \minus{}1 \bmod{3}$, that is, expressible in the form $ 3n\plus{}2$.[/hide]


logarithms
calculus
calculus computations

The position of a particle on the x axis at time t, t>0 is ln(t). The average velocity of the particle for 1 <= t <= e is...

Let $ x(t)\equal{}\ln t\Longrightarrow v(t)\equal{}\frac{d}{dt}x(t)$.

The desired average velocity is $ \frac{v(e)\minus{}v(1)}{e\minus{}1}$.

Shouldn't it be
$ v_{avg} \equal{} \frac{x(e) \minus{} x(1)}{e \minus{} 1}$

because unit of your avg speed [b]kunny[/b] is $ m s^{\minus{}2}$.

You are right, my calculation was the average of accerelation. :blush:


calculus
integration
calculus computations

If $ \int^4_1 f(x)dx \equal{} 6$, what is the value of $ \int^4_1 f(5 \minus{} x)dx$?

[quote="kunny"]Let $ 5 \minus{} x \equal{} t$.[/quote]

I think I got it (I used u instead of t)

[hide]$ 5\minus{}x\equal{}u$
$ \frac{du}{dx}\equal{}\minus{}1$
$ \int^{4}_{1}f(5\minus{}x)dx\equal{}\minus{}\int^{1}_{4}f(u)du\equal{}\int^{4}_{1}f(u)du$
That is the same as the original equation that equaled 6, there for $ \int^{4}_{1}f(5\minus{}x)dx \equal{} 6?$[/hide]


AMC
AMC 12
AIME
AMC 8
AMC 10
percent
MATHCOUNTS

Here it is once again.

Signups begin now. (June 18 start)

This is "realtime"! So, hurry up!

/in

*sigh* there exists a lower bound for the number of characters in a non-moderator post.

im in...what exactly is this amazing ace anyway???

rd5493- wow your in 3 tournaments :!: (Mine, AIME15, and this)

JOIN

What is it going to be like? How hard are the questions?

look at the first tournament to see the details.

If you go like that, it should be Chapter ---> AMC 8 ---> State ---> AMC 10 ---> National ---> AMC 12 ---> AIME ---> Proof (easy Olympiad?).

Most consider AMC 8 to be easier than State MATHCOUNTS (at least the hardest problems on each) and National MATHCOUNTS to be a bit more difficult than AMC 10.

Who said they were in order of difficulty?  :P

I consider Nationals easier than AMC10, bar the time restriction.

Since Bugi is dropping out, can I join?
EDIT: never mind, I just remembered I'm busy on the 18th of June. Sorry.

Ok then.

Due to severe computer lagging issues I will start the competition on Monday

I would like to join if anyboy is dropping out

We're back with the first problem!

1. Given that the sum of two numbers is 7 and the product of the two numbers is 14, find the reciprocal of the sum of the reciprocals of the two numbers.

PM answer, get the next problem.

It has come to attention to PM wording. Please put as subject an indication of the problem number.

Problem 3 - PowerOfPi
Problem 2 -  rd5493, AIME15, alex960531 (wait to 11:18)
Problem 1 - everybody else

Apparently Obscuron, alex960531, and I are doing it. I can say I am on problem 5.


If you enroll for a normal class, you have until the third class starts to drop the class. For this, it's a 2-day course so can you drop it after it starts? For example, if its too hard or too easy for me after the first day, am i required to stick with it?

There is no dropping the Special AIME Problem Series once it starts.


trigonometry
ratio
integration
limit
calculus
inequalities
calculus computations

Show that the following infinite sum diverges or find its sum:
$ \arctan(\tfrac{1}{1}) \plus{} \arctan(\tfrac{1}{2}) \plus{} \arctan(\tfrac{1}{3}) \plus{} \arctan(\tfrac{1}{4}) \plus{} \cdots$

Well you could try using $ \tan^{ \minus{} 1}(x) \plus{} \tan^{ \minus{} 1}(y) \equal{} \tan\left(\frac {xy}{1 \minus{} xy}\right)$ repeatedly to find the first few partial sums. As I take the partial sums, I see no convergence what so ever so I think it diverges. You could also view this as the infinite product $ \mathrm{arg}{\left\{(1 \plus{} i)(2 \plus{} i)(3 \plus{} i)\cdots \right\}}$.

[quote="t0rajir0u"]I'd be curious if you could find an approach to this problem appropriate for this forum.  The most straightforward solution I can think of is either a limit or direct comparison to the harmonic series.[/quote]

Well, one can use the 'limit comparison test' to compare $ \arctan n^{\minus{}1}$ with $ n^{\minus{}1}$.  By L'Hopital's rule, it's easy enough to see that the limit of the ratio of the two sequences is $ 1$, so the convergence of the sum in question is determined by the convergence of the harmonic series, which diverges.

Of course, you know all this.  But it does seem appropriate for this forum.

I assume you read this after it was moved; this topic was originally in Intermediate Topics.

Since $ \arctan\left(\frac{1}{x}\right)$ is decreasing on this interval, we know that:

\[ \sum_{n\equal{}1}^{\infty} \arctan\left(\frac{1}{n}\right) > \int_1^{\infty} \arctan\left( \frac{1}{x} \right)\: dx\]

By drawing some triangles, it becomes clear that the following relationship is also true on this interval:

\[ \arctan\left( \frac{1}{x} \right) \equal{} \frac{\pi}{2} \minus{} \arctan\left(x\right)\]

Integrating this expression (and integrating $ \arctan\left(x\right)$ by parts), we arrive at:

\[ \lim_{a \to \infty} \int_1^{a} \left(\frac{\pi}{2} \minus{} \arctan\left(x\right) \right)\: dx \equal{} \lim_{a \to \infty} \left[ \frac{\pi}{2} x \minus{} \left( x \arctan\left(x\right) \minus{} \frac{1}{2} \ln(1 \plus{} x^2) \right) \right]_1^a\]

So we are really only interested in whether the following limit converges or diverges, that is, the existence of:

\[ \lim_{a \to \infty}  \left( a\left( \frac{\pi}{2} \minus{} \arctan\left(a\right) \right) \plus{} \frac{1}{2} \ln(1 \plus{} a^2) \right)\]

This limit does not exist, so our integral does not converge and as such the sum diverges.

Using the inequality $ \arctan x \ge x\minus{}\frac{x^2}{2}$ will just do the trick because the sum becomes $ \ge \frac{1}{1} \plus{} \frac{1}{2}\plus{}...\plus{}\frac{1}{n} \minus{} \frac{1}{2} \cdot (\frac{1}{1^2}\plus{}...\plus{}\frac{1}{n^2})$ which tends towards infinity.


Ramsey Theory
combinatorics unsolved
combinatorics

Show that each of the following statements is equivalent to van der
Waerden's theorem:

1. If k is a positive integer, and the set of all positive integers is colored in k
colors, one color always contains arbitrarily long arithmetic progressions.

2. If a0, a1, . . . is an infinite sequence of integers satisfying 0 < ak+1-ak < r
for some fixed r, then the sequence contains arbitrarily long arithmetic
progressions


$\triangle ABC$ is a right angle at $C$ and $\angle A = 20^\circ$. If $BD$ is the bisector of $\angle ABC$, then $\angle BDC =$
[asy]
size(200);
defaultpen(linewidth(0.8)+fontsize(11pt));
pair A= origin, B = 3 * dir(25), C = (B.x,0);
pair X = bisectorpoint(A,B,C), D = extension(B,X,A,C);
draw(B--A--C--B--D^^rightanglemark(A,C,B,4));
path g = anglemark(A,B,D,14);
path h = anglemark(D,B,C,14);
draw(g);
draw(h);
add(pathticks(g,1,0.11,6,6));
add(pathticks(h,1,0.11,6,6));
label("$A$",A,W);
label("$B$",B,NE);
label("$C$",C,E);
label("$D$",D,S);
label("$20^\circ$",A,8*dir(12.5));
[/asy]

$ \textbf{(A)}\  40^\circ   \qquad
\textbf{(B)}\   45^\circ  \qquad
\textbf{(C)}\   50^\circ  \qquad
\textbf{(D)}\    55^\circ \qquad
\textbf{(E)}\     60^\circ $

$\angle B = (\angle C - \angle A )^\circ$ so $\angle B = (90-20)^\circ = 70^\circ$. Since $BD$ - a bisector, hence $\angle DBC = DBA = 0.5$ angle $ABC = 0.5*70^\circ$, so $\angle DBC = 35^\circ$. Considering the $\triangle DBC$, that is also rectangular at $\angle C$, we have that $\angle BDC = (\angle C - \angle DBC )^\circ$ so $\angle BDC = (90-35)^\circ = 55^\circ$.

[color=orange][size=75][Dollar signs inserted by moderator solafidefarms - have to have dollars around $\LaTeX$ in the forums!][/size][/color]

[hide]Angle ABC measures 70 degrees, and DBC measures 35 degrees. Thus angle BDC measures 180 - 90 - 35 = 55 degrees. [/hide]

[quote="moogra"]this problem is fun. I like geo. I think it is 55?[/quote]
spoiler it next time

[ hide ]
text inside
[ /hide ]

without the spaces
[color=orange]
[size=75][Edited by solafidefarms because spoiler isn't how you hide any more - you use [ hide ] and [/hide ]...][/size][/color]

[quote="Silverfalcon"]$\triangle ABC$ is a right angle at $C$ and $\angle A = 20^\circ$. If $BD$ ($D$ in $\overline{AC}$) is the bisector of $\angle ABC$, then $\angle BDC =$

A. $40^\circ$
B. $45^\circ$
C. $50^\circ$
D. $55^\circ$
E. $60^\circ$[/quote]
[hide]
We have a right triangle. $\angle ABC=90-20=70^\circ$.
Drawing in DB, we get $\angle DBC=70/2=35^\circ$.
We have a new rt. Triangle, with sides CD, DB and CB. ACB is a right angle, because it was given. BDC=90-DBC=90-35=55 degrees.
[/hide]


inequalities
calculus
derivative
quadratics
algebra
polynomial
Galois Theory

Solve the equation :
     $ \sqrt{5+\sqrt{4\sqrt{9-2\sqrt{x}}}}=2\sqrt{13}(13-x)$

[quote="Harry Potter"]Solve the equation :
     $ \sqrt{5+4\sqrt{9-2\sqrt{x}}}=2\sqrt{13}(13-x)$[/quote]

Oh boy, it 's my problem and it has its name("[b][color=blue]The Unique[/color][/b]")? So why you don't give the source of the problem and the author? I don't like the way you do much. But it 's a hard and beautiful problem, I hope that you will have fun with it.
(Check again the problem, be sure the you type the right)

One more problem:
Solve the solution
$ 16x^{3}-1=\sqrt[4]{x+\frac{1}{2}}$
This problem has been posted somewhere, but there 's no complete solution for it.

[quote="timlaiminh"][quote="Harry Potter"]Solve the equation :
     $ \sqrt{5+4\sqrt{9-2\sqrt{x}}}=2\sqrt{13}(13-x)$[/quote]

One more problem:
Solve the solution
$ 16x^{3}-1=\sqrt[4]{x+\frac{1}{2}}$
This problem has been posted somewhere, but there 's no complete solution for it.[/quote]

We know that for any positive integer n, $ \sqrt[2n]{a}\geq 0$. Thus we find that:

$ \sqrt[4]{x+\frac{1}{2}}\geq 0$. For $ 16x^{3}-1$ to equal $ \sqrt[4]{x+\frac{1}{2}}$, the nonegative condition must hold for both equations. We then get:

$ 16x^{3}-1 \geq 0$
$ 16x^{3}\geq 1$
$ x^{3}\geq \frac{1}{16}$
$ x \geq \sqrt[3]{\frac{1}{16}}$

It is easy to see that $ \sqrt[3]{\frac{1}{16}}\neq \frac{-1}{2}$. Therefore, we've established that:

$ x > \frac{-1}{2}$ and $ x > \sqrt[3]{\frac{1}{16}}$. However, if the two x values are equal, they must both be greater than a certain value. Thus, we find that the only condition remaining is $ x > \sqrt[3]{\frac{1}{16}}$

That's all I can really find at the moment. Sorry if it isn't the complete solution. I found the answer to be 1/2 on my calculator, but of course that's of no use if you can't prove how it was attained.

Well, this second equation doesn't seem so difficult... The easiest way to solve it is to put in a substitution $ y=RHS{\geq}0$. After putting the substitution, we get a 12-degree equation which can be divided by $ y-1$ ($ y=1$ is the only solution of the equation, as we will later see...). The equation we get is an 11-degree equation shown under

$ 16y^{11}+16y^{10}+16y^{9}+16y^{8}-8y^{7}-8y^{6}-8y^{5}-8y^{4}+4y^{3}+4y^{2}+4y+3=0$

Now, let's show that this equation doesn't have nonnegative solutions. It can be easily proven using the $ AG$ inequality (in various ways...,), or by taking the first derivative and showing (again, with the help of $ AG$ inequality) that its minimum is greater than 0...

So, the only solution is $ x=1/2, (y=1)$.

EDIT: The easiest method of proving that the equation above has no nonnegative solutions is factorization of it, as follows:

$ (16y^{8}-8y^{4}+4)(y^{3}+y^{2}+y+1)=1$

Expressions in both brackets on LHS are bigger (or equal, in the case of the second bracket) than one ($ 16y^{8}-8y^{4}+4>1$ is a simple quadratic inequality, and the second bracket is a sum of nonnegative sumands and one, which is at most one). Because of that, we have a contradiction ($ 1<1$), i.e. the equation has no nonnegative solutions.

[quote="hsiljak"]$ 16y^{11}+16y^{10}+16y^{9}+16y^{8}-8y^{7}-8y^{6}-8y^{5}-8y^{4}+4y^{3}+4y^{2}+4y+3=0$
Now, let's show that this equation doesn't have nonnegative solutions. It can be easily proven using the $ AG$ inequality (in various ways...,), or by taking the first derivative and showing (again, with the help of $ AG$ inequality) that its minimum is greater than 0...
So, the only solution is $ x=1/2, (y=1)$.[/quote]

I don't understand much what you write above. Can you show me the clearly way of your solution?
In your solution, we have the root of equation is greater than 0, but it doesn't show that the root must and only be 1/2.

The equation we get after the division with $ (y-1)$ is an equation without nonnegative solutions (which can be proven using both $ AG$  inequality and the factorization shown in my post). Because of that, the initial equation has only one solution: $ y=1$, hence $ x=1/2$ is the only solution of the problem. If you don't understand the part with the use of $ AG$ ineq, then skip it and read the "EDIT" part...

If I understand correctly we study 2 equations:
1) That of HP with 4 root-signs in LHS; then the $ x$-minimal polynomial over $ \mathbb{Q}$ is of degree 16 and ?????? ($ x\approx{12.624}$).
2) That of Timlaiminh with 3 root-signs in LHS; then $ P$, the $ x$-minimal polynomial, is of degree 8 ($ x\approx{12.54989}$).
Moreover the $ P$-Galois group is $ S_{8}$; thus it isn't solvable by radicals and ??????
For example $ \forall{n}$ $ \mathbb{Q}(x)\not=\mathbb{Q}(u_{n})$ with $ u_{n}=cos(\frac{2\pi}{n})$ because the $ u_{n}$-minimal polynomial is of degree $ \frac{\Phi(n)}{2}$ and thus $ \Phi(n)=16$; there don't exist $ n$ s.t. $ \Phi(n)=16$ and s.t. the $ u_{n}$-minimal polynomial admits $ S_{8}$ as Galois group.


Let $  (O) $ be a circle with diameter $ PQ $ . $ A, B $ belong to the same plane with the edge $  PQ $ . The tangent at $ B $ of $ (O) $ intersects the tangent at $ B $ of $ (O) $ at $ C $ . The tangent at $ Q $ of $ (O) $ intersects $ PA , PB , PC $ at $ A' , B' , C' $  respectively . Prove that : $  C' $ is the midpoint of the segment $ A'B' $

Did you mean this one: http://www.mathlinks.ro/Forum/viewtopic.php?t=15088 .

  darij

Yes , Darij and we will have six solutions  ;)
 Let $ PC $ intersect $ (O) $ at $ T $
      $ OC $ intersect $ AB $ at $ M $ $ \to M $ is the midpoint of $ AB $
 We have : $ \triangle PA'B' $ is similar to $ \triangle PBA $
   So $ C' $ is the midpoint of $ A'B' \iff \triangle PMA $ is similar to $ \triangle PA'B' \iff \triangle PMA $ is similar to $ \triangle PTB $
  $ ATBP $ is a harmonic quadrilateral . This means $ TA.BP = TB.AP $
 By Ptoleme's theorem : $ TA.BP + TB.AP = TP.AB  \iff 2.TB.AP = TP.AB \iff TB.AP = TP.MA' $
 The last one infers $ \triangle PMA $ is similar to $ \triangle PTB $  ;)

  Is it OK ?


I joined about a couple months ago, and I've been trying to send PM's out to people.... but they're stuck in the outbox? Can someone please help? Is this a glitch or does this take time?

The Outbox represents messages the person you sent to have not read yet. Once they are read, they are moved to the sentbox. This system allows you to know which of your messages have been read.

Adding onto that, you can also retract messages / edit messages if they are in the outbox.


advanced fields
advanced fields unsolved

The nth statement in the list of 100 statement is exactly n
of the statements in this list are false.
(a) what conclusion can you draw from these statements.
(b) answer part (a) if the nth statement is "Atleast n of
     the statements in this list are false".
(c) answer part (b) assuming that the list contains 99
     statements.

Please tell the right way to approach?

Well, for (a), observe that all of the statements conflict. So either 1 or 0 is true. Which is possible?

For (b), none of the statements conflict immediately. But "not $n$" conflicts with $100-n$ (and all higher ones). What can you conclude from that?

For (c) it'll be the same kinda ideas, but w/ 99 instead of 100 statements..


[quote="Dr. No"][quote="riddler"]A spring is compressed and kept like that and dissolved completely in acid. What happens to the elastic potential energy of the spring. with reasons

I could not get it :(  :?[/quote]

It is released as heat into the acid as the spring disolves.[/quote]

assuming that all rxns with metal and acid are exothermic. i agree..

are there any that arent? (i guess this is more of a chem question)

if the rxn was endothermic. it wouldnt require as much heat .. i think.

i never said it was a chemistry question.. i just said it mattered on whether or not the rxn was endothermic or exothermic.. i think you are supposed to assume that ALL such rxns are exothermic..

sry if that was confusing..

as for reasons:

my guess would be that somehow compressing a spring compresses the metal atoms, which release more kenetic energy when they are allowed to seperate..

but thats just a guess.

That can be an appropriate solution but.....  I am
 :?
ISNT THERE A PROPER WAY TO SOLVE SUCH QUESTIONS?

i dont think here is a general method for getting answers to this type of question..

actually its a really interesting question... the underlying question, is how is the energy stored in a spring? why does it exert a restoring force... ???

i dont know.

its pretty much obvious that the spring reacts with the acid and completely dissolves. But the question is where does the E.P.E go?

[quote="riddler"]skylarth, this cant be a chemistry question because i got it from a physics book
and Dr.no why does it release more heat?????????????????? I want the solution also, pls.
 :([/quote]

According to the 1st Law of Thermodynamics (in differential form)

The change in internal energy of the system (spring) is equal to the heat added to the system minus the work done by the system:

$\mathrm{d}E=\delta Q - \delta W$

$-\delta W = -kx\math{d}x$

Compressing the spring adds internal energy to the system.

$\mathrm{d}E = \delta Q - kx\mathrm{d}x$

$\delta Q = 0$ before the spring enters the acid.

Dissolving the spring while still in its compressed state implies $\mathrm{d}x = 0$

This leaves $\mathrm{d}E = \delta Q$; the only way for the additional energy to be released is as heat.


Try to solve this problem. I have only a pretty ugly solution.

An $8\times8$ chessboard  is completely tiled by $2\times1$ dominoes. Prove that there exist a king's tour
of that chessboard such that every cell of the board is visited exactly once and such that king goes domino by
domino, i.e. if king moves to the first cell of a domino, it must move to another cell in the next move. (King
doesn't have to come back to the initial cell. King is an usual chess piece.)

What happens if king must come back to the initial cell? I think that closed king's tour exists too, but I can't prove it. And you?

The closed from the open is simple:

a graph can be drawn in 1 line without holding up your pencil if and only if there's 0 or 2 points where an odd amount of lines meet. 0 means closed, 2 means open.

Since the chessboard is 4-sided symmetrically (and each construction of a size in king's moves reach is thus at least 4 times possible).
So if there was one point with an odd amount of ways, there would a domino placement possible so that the open tour cannot be made either, contradiction.

So if an open tour is possible, then a closed tour is possible as well.

The solution for the open tour can easily be loong and ugly, but if you say you have it then you now know it's ok for the closed tour. :)

I didn't quite get that, but I have some questions: basicly, you guys say that, no matter how we place the dominoes, there will always be some closed path guided by the dominoes? Wouldn't your reasoning work for $4\times 4$ tables as well, Peter? However, I don't think there's a closed path on the following board (two positions with the same number denote the two squares of a domino):

$\begin{array}{cccc}1&1&3&4\\2&2&3&4\\8&7&6&5\\8&7&6&5\end{array}$. That's because if a closed path is possible then such a path exists starting from any square (the graph is both Euclidean and Hamiltonian). However, you can easily convince yourself that a path starting from the lower right corner (a $5$ in the diagram) cannot be closed.

Am I misunderstanding something?

[quote]Am I misunderstanding something?[/quote]

Yes. That pattern does have a closed path starting at the lower right 5. We trace the array
1a 1b 3a 4a
2a 2b 3b 4b
8a 7a 6a 5a
8b 7b 6b 5b
in the order 5b,5a,4b,4a,3a,3b,1b,1a,2a,2b,8a,8b,7b,7a,6a,6b,(5b).
Remember that a king can move diagonally.

The last line you wrote is the key to all my unsuccessful struggling: I had for gotten that the king cn move diagonally :) (I do know how to play chess, and was once quite good, so that's not the reason :D).

You see, the thing is that I think there's always an open path if we use a "restricted" king, i.e. one that can move only horizontally and vertically. $8\times 8$ boards are too messy, so while playing around with the problem I've only used $4\times 4$ boards, and I don't even know if it's true in this case, but I've tried lots of examples and they all seem to work.


Maybe it was asked, but what's the point of locking posts after a certain time? It's a bad feature.

It's a good feature, and no, it will not be changed.


It takes four painters working at the same rate $ 1.25$ work-days to finish a job. If only three painters are available, how many work-days will it take them to finish the job, working at the same rate? Express your answer as a mixed number.

it takes $ 1.25$ work-days for $ 4$ people to finish the job, so $ 1$ person doing the job would take $ 5$ days
if we only have $ 3$ people, it would take $ \frac {5} {3}$ days to finish the job

ans is $ \boxed{\frac{5} {3}}$

I got a different answer.

Say each painter can finish a job in $ a$ days. So we have:

$ \frac{1}{a} \times 4 \equal{} 1.25$

$ \frac{1}{a} \equal{} \frac{5}{16}$

$ 1 \equal{} \frac{5a}{16}$

$ 16 \equal{} 5a$

$ 3.2 \equal{} a$

So each painter can finish a job working alone in 3.2 days. So 3 painters can finish a job in:

$ \frac{1}{3.2} \times 3 \equal{} \boxed{\frac{15}{16}}$

Vallon's method is correct.

Isabella's cannot be correct because there's no way 3 people will finish a job in less time than 4 people, given that they all work at the same rate.  You basically multiplied by 3/4, instead of 4/3.

That said, the actual answer is $ \boxed{1\frac23}$, since the problem asks for a mixed number.


calculus
integration
geometry
3D geometry
analytic geometry
calculus computations

Let $ S\equal{}\{(x,y,z): a\le x\le y\le z\le b\}$. Show that:
$ \int_S f(x)f(y)f(z)dxdydz\equal{}\frac{1}{6}(\int_a^b f)^3$

The cube $ Q\equal{}\{(x,y,z)\colon x,y,z\in [a,b]\}$ consists of $ 6$ essentially disjoint parts (they have common boundary points): $ S$ and its images under permutations of coordinates, e.g. $ \{(x,y,z): a\le z\le x\le y\le b\}$. Since the integrand $ f(x)f(y)f(z)$ is unchanged under permutations of variables, it follows that $ \iiint_S f(x)f(y)f(z)\,dx\,dy\,dz\equal{}\frac16\iiint_Q f(x)f(y)f(z)\,dx\,dy\,dz$. The latter integral is equal to $ \frac16\left(\int_a^b f(t)\,dt\right)^2$ by Fubini's theorem.


Let $\triangle ABC$ be a right triangle such that $\angle B$ is the right angle.  Let $D$ be a point on the hypotenuse $\overline{AC}$.  Show that \[\min(AD, CD) \le BD \le \max(AD, CD) \, .\]

[hide="Solution"]Without loss of generality let $AD \geq CD$. The condition is equivalent to proving that $CD \leq BD \leq AD$. Draw the circumcircle of $\triangle ABC$, and extend $BD$ to meet the circle at point $E$. Since $AC$ is a diameter of the circle, $DE=BD$. By the Power of a Point Theorem, $BD^{2}=BD \times DE=AD \times CD$. We must then have $CD \leq BD \leq AD$ as desired.[/hide]

[quote="Ravi B"]Why is $DE = BD$?[/quote]
Oh, sorry, in the diagram I drew, it looked like $\overline{BD}$ was perpendicular to $\overline{AC}$, which is why I made that mistake.  :blush:

[hide="My next attempt at a solution"] WLOG assume that $AD \geq CD$ again. Let the midpoint of $\overline{AC}$ be $M$. Then since $D$ is on $\overline{CM}$, $\angle CBD \leq \angle MBC=\angle MCB=\angle DCB$ (since $BM=CM=AM$ so $\triangle MBC$ and $\triangle MBA$ are isosceles). This means that $BD \geq CD$. We can prove that $AD \geq BD$ similarly. Thus $CD \leq BD \leq AD$, and we are done.[/hide]

Ravi B, is my solution correct, or is there something I need to fix?

Sorry, I didn't notice your second solution before.  It is an excellent proof!

Without Loss Of Generality.  It's used when a problem-solver handles one case to show that the other cases are equivalent.

[quote="John Liao"]What does WLOG stand for? :blush:[/quote]
It stands for "without loss of generality," as you can see if you look at my first "proof."  :P
It basically means if it doesn't matter if something is one way or another way, you can just assume it is one of them to make what you are trying to prove more specific.

For example, if you wanted to show that there are no distinct positive integers $x$, $y$, and $z$ such that $x+y+z \leq 5$, then you would assume without loss of generality that $x<y<z$, so that you have $x \geq 1$, $y \geq x+1 \geq 2$, and $z \geq y+1 \geq 3$, so that $x+y+z \geq 6$, which proves the statement. This only works if the objects ($x$, $y$, and $z$ in this case) are under equivalent conditions. For example, you could not assume WLOG that $x>y$ in the equation $2x+y=6$.

[quote="John Liao"]PS what does the brakets stand for?[/quote]
Um... which brackets are you talking about?  :wink:

[quote="John Liao"][quote]min [b]([/b]AD, CD[b])[/b][/quote]

The brakets that contains AD, CD[/quote]
Min is just which one of the two lengths is the smaller one, and max(AD, CD) is just which one of the two lengths is the larger one. For example, min(3, 5) is 3, and max(3, 5) is 5.


I am trying to construct a counterexample for a problem and here is the way I am trying to construct the counterexample and it seems to me like this will work but I have the feeling that I am missing something...

Let's say we have a measure spaces $ ([0,1], A, \mu)$ and $ ([1,2],A',\vu)$.

If we define a subset E of [0,2] to be measurable if and only if its intersection with [0,1] and with [1,2] is in A and A' respectively and then define the measure of that subset to be:

$ \mu (E \cap [0,1]) + \nu (E \cap [1,2] )$

do we get a measure space?

Yes. $ \tilde \mu(E): \equal{}\mu(E\cap [0,1])$ is a measure on $ [0,2]$ (or on any other superset of $ [0,1]$). And the sum of two measures is a measure.


Sorry if I'm spamming or anything, but does anyone know when the AMC 8 is being held?

See complete dates and details at the AMC website, http://www.unl.edu/amc.
The Teacher's Manual contains all regulations for the administration of the AMC 8.

Yes, the official date is Tuesday, November 13.

Steve Dunbar
MAA Director, American Mathematics Competitions

just asking for clarification...calculators will not be allowed this year on any AMC competitions correct?..i read that, but then today my teacher said that they were going to be allowed

See FAQ AMC8 #3, on
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

See also the Teacher's Manual, Part III, item 7, on page 4
and the instruction 5 on the reproduction of the contest cover page on page 23.

Steve Dunbar
MAA Director, American Mathematics Competitions


AMC AIME inequalities USA(J)MO USAMO geometry ratio

So now they've pretty much explained what they can explain about the courses/tracks...

Which track looks the best for you?

Personally, I'm liking course 2 a lot, though I think I could maybe go for course 3 if I REALLY had to pick a different one.
I need the Euclidian Geometry course, because my geometry is horribly weak... and I think I'll benefit more from a nt course than inequalities. At first I was thinking about doing combinatorics, because I'm weak in that subject too, but the fact that it's not paired with geometry, and because I think a lot of it will be pretty much the exact same thing as mop has me thinking that I'd be better off not taking the combo course this year.

I'm leaning towards track 3, because geometry is just awesome, but I don't know too much about it, inequalities are also just awesome, but I suck at it. I would also like do to some combinatorics, but meh, combo isn't as awesome as geo or ineqs

I really want to have 3. I am bad in both areas.

Course three sounds interesting, but I'm worried that it might be a little too hard (maybe I'm just still reeling from this year's AIME, though).  Otherwise, I would probably take course 5, because Algebraic techniques sounds a little too much like things I already know (and I need help with geometry).  Oh, and I scored 7.5 on the admissions test.  Any thoughts?

I'm thinking 3 as well. 7 AIME + 4 USAMO (dang...) puts me somewhere near there, and I'm pretty weak in both subjects. I can do AIME algebra all right (provided I'm not caffeine-high), but once I get to the USAMO, that, combined with inequalities, terrorizes me.

Actually, I don't have much experience with more olympiad problems. My issue with Geometry is that most of the time it's like, #6, which is pretty impossible anyways. Unless I can somehow improve to that level with AMSP ;)

I'm going to go for 2 or 3, preferably 3 because my geometry SUCKS

Humm...before I saw the tracks, I was thinking either

i) I want to really challenge myself; lets do my two worst subjects; combo+NT
ii) Anything with geometry...
iii) Not to be arrogant, but I expect to be doing the "harder half" of the tracks.

--

Now I can

i) get owned by track 1 like i was originally thinking
ii) maybe track 2; geo+ nt, i ono though...i feel like i know the stuff in the course description
iii) blargh i suck at inequalities...i just feel like i memorize them...i don't really like them

I wonder if the organizers will mind that I got a 4 on the AIME and a "not taken officially but i probably had like 19-21 based on WOOT ppl's feedback" on the USAMO... :P

[quote="PenguinIntegral"]Haha, I love this.

There is high level NT paired with everything but 3 and 6.

So as to not appear to be a wimp, we all pick 3.

This is hilarious  :rotfl:[/quote]

All the girls are in three

Gosh

[quote="mustafa"]
All the girls are in three

Gosh[/quote]

"Oh look a girl"

"What's a girl?"

 :rotfl:

lol, he'll probably give you a math problem for the answer.

Or he could just say $ i$ and be done with it.

Maybe we should start a movement to get all the AoPSers to pick track 3 :winner_third:  :jump: .

dude track 3 will be stacked
and because of the girls
i must pick track 3

but also i want to do algebra and geometry

but the girls is the primary reason

I don't know what to choose.  In terms of contest scores, I got a 6 on this year's AIME, including a few stupid mistakes (such as #2 with the person off the walkway).  Recently, I looked through a Putnam exam and was able to knock off 3-4 questions, although I haven't worked through a USAMO (I'll try one tomorrow and the day after).

In terms of each course, I basically did a math fair project on the stuff within "Foundations of Number Theory", although I doubt preparing for a 15 minute presentation is anywhere close to an AwesomeMath course.

In terms of Algebra/Geometry/Combinatorics, I am at similar levels... decent, but not quite up to the levels I would want.

In all probability, no choice is a bad choice, but I still don't know what track I'll take.

[quote="jli"]i have to pick track 3 because im not good enough for 1 or 2 and i cant do 4-6 because ive already been there last year and itll make me look like a noob.  (i also like the classes)

i wonder how many are going to track 6[/quote]
dude wtf is with this "omfg im not good enough for level 1 or 2"
1-3 should be like the same difficulty, just different topics
same with 4-6

To some extent, the organizers read this message board. You need to have a convincing argument to place yourself in track 3. The whole: "OMG I AM NEVER AMONG GIRLS SO LETS DO THE ONE WITH GIRLS" is not a convincing reason...

I think I'd better take an AIME for practice some time next week, because I don't think that my score on this year's AIME A is an accurate measure of my skill level.  I certainly think that track three has more interesting classes than tracks four through six, and I want to have some way to demonstrate that I'm actually ready for it (I worked out this year's USAMO question 1, for what that's worth...).

dude yeah the AIME 1 sucked so bad

it was like "HELLO I AM THE AIME 1 AND I SUCK"

LOL...That [b]4[/b] on the [i]AIME[/i] and that [b]NOT TAKEN[/b] on the [i]USAMO[/i] don't exactly reflect my ability either...

I should get together some stuff...like I have been working on some IMO SL geometry recently [with moderate success].

do you guys think that almost everyone will get their first choice for tracks?
also if too many people want a track would they just keep it that way, split the track in half, or just divert them into other tracks?

The track selection system is REALLY BAD, in my opinion. I wouldn't be surprised if it was changed. It seems like they pair one really applicable subject (algebra (inequalities), geometry) with something stupid (NT or combo). And by stupid, I mean subjects that aren't going to improve my AIME/USAMO score versus studying on my own, whereas I really need some mental foundation and intuitiveness help on the former two.

(And really, isn't like a third of the AIME geometry, compared to like one or two combo problems? I think a similar ratio holds for algebra to NT.)

It's just everyone has a few subjects that they really suck in, and this track thing makes it so you pick one subject you need help with and one that you don't, most of the time. Maybe that's what Titu said to do in the email, but the greatest improvement in my math ability is going to come from fixing all the gaping holes in my knowledge first.

Though I must say this is pretty stupid criticism if I can get the source .pdfs to all the handouts (assuming the handouts contain easier problems to practice with and some instructional material. If it starts out with doing like AIME problems already with no foundation I'm going to be a little more precise with my track choice).

Assuming that, I might go for track 5, because trig and complex numbers hurt my brain, or more likely 2, because I don't want to wimp out and I need geo help.

But really, 3 looks like the best choice for me.

Not to sound ungrateful or anything  :roll: .

Mkay...I think they were very deliberate with the track system. I don't think that it is our place to complain for we are not the experts; they are. But if by some chance, they happen to consider changes, I would like

[b]Geometry + Combo[/b].

I feel like olympiad combinatorics is one of the hardest subjects. I believe that this is the least technical of the topics. I think it relies the most on ingenuity and I think that it would be a good pair with geometry, which is one of the most technical topics. Admittedly, line addition in geometry is important, but theorems play a large role in that subject. Consequently, this grouping would represent a fusion of technical and non-technical thinking that would further our olympiad problem solving abilities.

@PenguinIntegral: I don't see how inequalities is going to improve your AIME or USAMO score. Might I mention that among the 36 problems from the 2007 AIME I, AIME II, and USAMO, there was a whopping total of half an inequality problem [which really had no inequality in it...it was just geometry since it reduced to $ R\ge 2r$].

I guess I'm the first one to say I was thinking about taking track 4? I kinda want to do a geometry course though...

Earlier today I took the 2003 AIME 1 just for fun.  I solved the first seven problems in either $ 1\frac14$ or $ 2\frac14$ hours, I'm not sure which (the timer on my watch simply rolls over to zero after an hour, and I wasn't paying quite enough attention to the time when I started), and anyway, I answered five of the problems correctly (made silly mistakes on 5 and 6), and I feel that this, although hardly formal, gives a better idea of my skill level than does this year's AIME.  So basically, I scored a five without as much effort as I would put into the actual test.  I guess this can be considered a "why I feel I'm ready for track three" post.  Besides, it sounds more interesting than tracks 4-6 :wink: .

[quote="PenguinIntegral"][quote="Altheman"]
@PenguinIntegral: I don't see how inequalities is going to improve your AIME or USAMO score. Might I mention that among the 36 problems from the 2007 AIME I, AIME II, and USAMO, there was a whopping total of half an inequality problem [which really had no inequality in it...it was just geometry since it reduced to $ R\ge 2r$].[/quote]
There's pretty consistently inequalities in the USAMO and IMO, but mainly it is a very large gap in my knowledge I would like to get fixed. I do not grasp geometry and inequalities in the same way I do other things.[/quote]

dude yeah I think technically 3 of the questions on the last IMO were inequalities

IT WAS PRETTY CRAZY

I'm kinda stuck looking at this. I see many of you are looking at track 3, which is probably exactly what I need because I can't do geom for my life and all these inequalities in the induction packet are killing me. if it comes down to who has high scores on contests, I may drop to 4, because a 5 on AIME (4 arithmetic/reading mistakes) and a not taken USAMO don't look too good for me (from what I worked of the USAMO, I got estimations on landing in the 6-11 range). Mod arithmetic just doesnt excite me too much, but I really don't know what to expect.

If you can get a 6 - 11 range on the USAMO, then you should look at the lower tracks.

If you don't know what to expect out of mod arithmetic, you might want to consider it. I told my college friend exactly the same, and he said that mod arithmetic was tough for him to get into, but it's deceivingly powerful.

okay. I was looking at 3-4 because the email said 10/10 AIME/USAMO should look at 1-3 and 7/3-4 AIME/USAMO should look at 3-4. So if I get into 3 I get into 3, and if I don't, hopefully I'll get into 4.


[quote="marko avila"]sorry i posted it two times wont happen again  :blush:  :blush:[/quote]
Hello newmember :D  :D  :D


I can't answer the poll because I didn't like it. :P

ya i am not too sure what the point of this thread is...............................


Let $ a,b,c >0$.prove that:
$ 3(\sum {\sqrt{\frac{b\plus{}c}{2a}}})^2\geq\frac{8(a\plus{}b\plus{}c)^3}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

[quote="toanIneq"]Let $ a,b,c > 0$.prove that:
$ 3(\sum {\sqrt {\frac {b \plus{} c}{2a}}})^2\geq\frac {8(a \plus{} b \plus{} c)^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$[/quote]
We have
$ K\equal{}(\sum 2a(b\plus{}c)^2)(\sum \sqrt{\frac{b\plus{}c}{2a}})(\sum \sqrt{\frac{b\plus{}c}{2a}})\ge 8(a\plus{}b\plus{}c)^3$
We will prove
$ 3(a\plus{}b)(b\plus{}c)(c\plus{}a)\ge 2(a(b\plus{}c)^2\plus{}b(c\plus{}a)^2\plus{}c(a\plus{}b)^2)$
<=> $ \sum c(a\minus{}b)^2\ge 0$
so $ 8(a\plus{}b\plus{}c)^3\le K\le (a\plus{}b)(b\plus{}c)(c\plus{}a).LHS$
<=> $ RHS\equal{}\frac{8(a\plus{}b\plus{}c)^3}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\le LHS$
so the ineq was proved  :D


We see that $ FD\equal{}\sqrt{18}$, $ FC\equal{}\sqrt{18\plus{}9}$, $ FB\equal{}\sqrt{27\plus{}9}$, and $ FA\equal{}\sqrt{36\plus{}9}\equal{}\sqrt45\equal{}3\sqrt5$.

Notice that the hypotenuse of each smaller triangle is a leg of the next. Therefore, we can nest the lengths inside each other in the formula.

$ AF\equal{} \sqrt{(AB)^2\plus{}(BF)^2}$
$ AF\equal{} \sqrt{(AB)^2\plus{}(BC)^2\plus{}(CF)^2}$
$ AF\equal{} \sqrt{(AB)^2\plus{}(BC)^2\plus{}(CD)^2\plus{}(DF)^2}$
$ AF\equal{} \sqrt{(AB)^2\plus{}(BC)^2\plus{}(CD)^2\plus{}(DE)^2\plus{}(EF)^2}$
$ AF\equal{} \sqrt{3^2\plus{}3^2\plus{}3^2\plus{}3^2\plus{}3^2}$
$ AF\equal{} \sqrt{5(3^2)}$
$ AF\equal{} \sqrt{5}(\sqrt{3^2})$
$ AF\equal{} \boxed{3\sqrt{5}}$


conics
ellipse
rotation
analytic geometry
graphing lines
slope
geometry

If the symmetrical curve of $C$ with respect to $x+y=1$ is  $\frac{(x-2)^2}{25}+\frac{(y+4)^2}{9}=1$, find $C$.

[hide]

Because the scenario is only an ellipse and a line, we can merely find the symmetric center of the ellipse, and then rotate in the old figure into the place of the new image.

As the line only has slope $-1$, we turn the axes of the ellipse by $\pi/2$ radians.

The center of the ellipse is $(2,-4)$, so the center of the image is $(5,-1)$

Thus $C$ is $\frac{(x-5)^2}{9} + \frac{(y+1)^2}{25} = 1$

[/hide]

[quote="Altheman"]do you mean reflect the ellipse about the line?[/quote]
Yes.


LaTeX trigonometry function algebra polynomial algorithm logarithms

The attached file is a handout I'm going to give to a class of mine this week. It's a junior-level "real analysis" class, but there's nothing in here that isn't calculus.

Feel free to suggest changes or additional problems.

Edit: old Word file replaced with $\LaTeX$ version. Added some more examples.

[quote="liyi"]Does the infinitesimal $\sin\sqrt{x+1}-\sin\sqrt{x} \ (x\to\infty)$ has an order?[/quote]
That's $O\left(\frac1{\sqrt{x}}\right).$ We can see this because that is the order of $\sqrt{x+1}-\sqrt{x}$ and $\sin$ is a Lipschitz function.

Remember that big-O doesn't mean "always the same size as", it merely means "no larger than (a constant multiple of)."

I meant to find $o(\frac{1}{x^\alpha})$... It is quite easy for big-Oh

something must be wrong here, but some stuff I cannot read.

some things get replaced by "Fout!" , which is dutch for "Error!"

do I need to install something specific on word? I've always wondered what those O's were...

I downloaded the .pdf and it looks good, but it looks like you did the document in Microsoft Word and didn't use $\LaTeX$. It turns out fine though.

[quote="Kent Merryfield"]liyi, one can see that

$\limsup_{x\to\infty}\frac{|\sin(\sqrt{x+1})- \sin(\sqrt{x})|}{\frac1{\sqrt{x}}}=\frac12.$

We can do this by choosing $x$ close to $n^2\pi^2.$

So you can't do any better than $O\left(\frac1{\sqrt{x}}\right).$[/quote]

Well, it may be my fault... Perhaps I didn't describe it clearly.
Yours is right. But I just asked small-Oh... In this sense, the order does not exist...

Woah this notation is very neat.  It's basically making a function that you don't care about as much, except for it's largest polynomial degree depending on whether you're getting big or small.  But what is a lim sup?

lim sup = limes superior. Some functions do not have a real 'limit', like f(x)=sin(x). So we cal lim sup the highest value it takes in the limit. e.g. for sin(x) that's simply 1. :)

[quote="FMako"]Woah this notation is very neat.  It's basically making a function that you don't care about as much, except for it's largest polynomial degree depending on whether you're getting big or small.[/quote]
And not limited to polynomials: just listen to a bunch of computer scientists discussing whether such-and-such algorithm has complexity that is $O(N\log N)$ or $O(\log\log N)$ or whatever. (Of course, "as $N\to\infty$" is understood there.)

I invite comparison of two forms of Taylor's theorem: for a sufficiently smooth $f,$ we can say that

$f(x+h)=\sum_{k=0}^n\frac{f^{[k]}(x)h^k}{k!}+\frac{f^{[n+1]}(\xi)h^{n+1}}{(n+1)!}$ for some $\xi$ between $x$ and $x+h$.

Or we could say that

$f(x+h)=\sum_{k=0}^n\frac{f^{[k]}(x)h^k}{k!}+O(h^{n+1})$ as $h\to 0.$

Now the first statement contains more information than the second statement, with its relatively precise estimate of the remainder term. But a very large fraction of the time, you don't really need that additional information, and the second, slightly less precise, statement invites us to perform rapid and effective manipulations involving those terms for which FMako said "you don't care about as much."

In your exact definitions of the "o" functions, you used a backwards E among other characters, what does the definition mean in english or could you define some of your notation or link me to a site with that notation defined?

It's a logical quantifier. There are two logical quantifiers:

$\exists,$ which means "there exists." ($\text{\TeX}$ code \exists), and

$\forall,$ which means "for all" or "for every." ($\text{\TeX}$ code \forall.)

If everyone understood what you said, did you really misspeak?  :ninja:

[quote="Carcul"][quote="Bos1234"]What does iff mean?[/quote]

"Iff" means "if and only if".[/quote]

 :) ，it was invented by P.R.Halmos

I have always used the notation

$ f(x)\ll{g(x)}$

to denote that $ f(x)$ is "much less" than $ g(x)$

Where what I think you guys are using it for is what I define with

$ f(x)\prec{g(x)}$

Which means that $ \lim_{x\to\infty}\frac{f(x)}{g(x)}\equal{}0$

Or in a more holistic sense, as x gets arbitrarily large that $ g(x)$ "overpowers" $ f(x)$.

What about this on page 2 ... $ O(g_1(x))\plus{}O(g_2(x)) \equal{} O(\max(g_1(x),g_2(x)))$ ???
You have allowed $ g_1(x)$ and/or $ g_2(x)$ to be negative, after all.

I was implicitly assuming that the functions in question were nonnegative.

We just don't say things like $ O(\minus{}x^2)$ or $ O(\sin x)$ (as $ x\to\infty).$ The functions used inside a $ O$ or $ o$ symbol should be nonnegative and (most likely) monotone.

[quote="Kent Merryfield"]I was implicitly assuming that the functions in question were nonnegative.

We just don't say things like $ O( \minus{} x^2)$ or $ O(\sin x)$ (as $ x\to\infty).$ The functions used inside a $ O$ or $ o$ symbol should be nonnegative and (most likely) monotone.[/quote]

Indeed, that is a standard thing to assume.  But you explicitly DID NOT assume it at the beginning.  Just "eventually monotone and nonzero" or something.

?????? No idea of what the heck your saying.

Actually the first person who came up with the big-O, little-o notations was Bachmann in a 1894 textbook. Landau used Bachmann's notations, mentioning his name in 1909.

I replaced the old Word file by a newer $\LaTeX$ file. It's mostly the same, but I added an example or two. It's back up in the first post on this topic.

iff= if and only if, a short what of stating a bi-conditional statement

i heard about the BIG-O for the first time today, can anyone please explain how do I use it, and what does it means?

[quote=fedja]The formally correct approach to the big and little O notation, in his opinion, should be the following. Given a positive $ g$ defined in a punctured neighborhood of $ x_0$, denote by $ O_{x_0}(g)$ the class of all functions $ f$ such that the ratio $ f/g$ is bounded in some punctured neighbourhood of $ x_0$.[/quote]

For example Corman & al. define the notation as $O(g(n))=\{f(n):$ there exist a positive constants $c$ and $n_0$ such that $0\leq f(n)\leq cg(n)$ for all $n\geq n_0\}$. Now for example if we take $g(n)=0$ then $g(n)\in O(0)$. But in the new notation we can't say that $g(n)\in O_{x_0}(0)$. So although we can find situation where the notation $O_{x_0}(g)$ is more rigorous, we can't simply modify all old text that uses notation $O(g)$ simply to use the notation $O_{x_0}(g)$ if $g$ has zeros nearer and nearer to $x_0$.

There are also computer scientists who use $O$-notations as inequalities. For example http://www.cs.yale.edu/homes/spielman/365/asymptotics.pdf


I've sent it to the Pre-Olympiad, but have recieved no reasonable answer:
[b]Angles A, B, C and diagonals e, f are given, $A+C\neq \pi$.
Draw the quadrilateral.[/b]
It's not as easy as it seems... I want you to find some synthetic solution.
Thanx in advance.

Some 214 people have seen this and noone has solved it... i REALLY wonder wheather there is a solution...


function
inequalities
topology
real analysis
real analysis unsolved

[quote="kalazealot"]Consider ODE of $y(x)$, real-valued function on $R$, $y''=f(x,y')$. Where $f$ is $C^{r}$, $r>2$,
and satisfy: $f(s,ct)=c^{2}f(s,t)$ for any constant $c$, (thus $f(s,0)=0$).
How to show that: for every the local solution $y(x)$ defined on an open set
$U$ of $x$, either $y(x)$ is a constant in $U$ or $y'(x)$ is never zero in $U$.

(the constant part is easy, but I think $y'(x)=0$ implies $y''(x)=0$, and can't get anything more than that...). Can someone help? thanks.[/quote]
Set $z(x) = y'(x)$ (as [i]mlok[/i] suggested). Then $z'(x) = f(x,z(x)\cdot1) = f(x,1) z(x)^{2}$. Now set $w(x) = \frac{1}{z(x)}= \frac{1}{y'(x)}$ in this Riccati equation and write down the differential equation for $w$ to check that $w(x)$ is defined everywhere.

[quote="mlok"]Since $y$ does not appear in the equation, you can treat $y'$ as an unknown function $z(x)$, so that $z'=f(x,z)$. [url=http://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality]Gronwall's lemma[/url] can help.[/quote]

I can understand HPE's argument, didn't use Gronwall as mentioned above. Can someone
show how to use Gronwall here? Thanks.

Yes, you don't actually need Gronwall's inequality here. But if you want to use it, notice that $z'(x)=f(x,1)z(x)^{2}\le f(x,1)z(x)$ near $x=0$ (where $0\le z(x)\le 1$). Or just use the fact that the solution $z\equiv 0$ has the same initial values ($z(0)=0=z'(0)$) and the uniqueness theorem applies.


probability
logarithms
integration
probability and stats

$ (X_n)$ random variables such that
$ \log{n}*E\left(\frac{|X_n\plus{}5|}{1\plus{}|X_n\plus{}5|}\right)\minus{}>1$ as $ n\minus{}>\plus{}\infty$.
Study the convergence in probability of $ X_n$.

It is well known that convergence in probability $ X_n\to X$ is equivalent to convergence in metrics $ \rho(X_n,X) \equal{} \int \frac {|X_n \minus{} X|}{1 \plus{} |X_n \minus{} X|}dP$.


I started my indepedent studying for physics and so far, it seems fun because it's actually "math" science. By that, I mean that it's bit harder than chemistry and I have to use my brain (no offense to chemistry.. I was only in general chemistry so I don't know all chemistry but so far yeah). So, it seems like a good choice to major physics but I'm planning to become a doctor. Is this major an OK choice?

Physics majors can follow lots of interesting careers. Yes, they can become medical doctors if they want to. People who follow any college major can become doctors if they meet medical school admission requirements. There are English majors, math majors, chemistry majors, music majors, and history majors who are doctors today.

Math and Physics majors tend to do very well as doctors; this is mostly a result of those two majors being very challenging and actually requiring very good problem solving skills, which is very useful in med school.

Indeed, physics and math majors do fare well in medicine since these students are science majors by nature rather than of necessity.

"Pre-med" students with a primary interest in medicine rather than one in science often major in biological sciences.  In general, physical science, math and humanities majors outscore biological science majors on the MCAT.   See the MCAT summary statistics at the AAMC site:  [url]http://www.aamc.org/students/mcat/examineedata/pubs.htm[/url]

The downside to majoring in physics (or anything other than biology, chemistry, or variations thereof) is that the classes you are expected to take to get into med school (pre-med classes) won't overlap with your major, so you will have fewer pure elective courses.  In the grand scheme of things, this is not that terrible (and it's less of a problem for physics than it is for, say, English).

Another thing worth noting is that "what will I major in?" is really not one of those questions that merits a lot of worrying in advance.  At most colleges, you don't decide on your major until after 2, 3 or 4 semesters and it tends to be easy to switch between majors (so long as you are switching into a major in which you will be able to complete the coursework to graduate on time).


Let $ABC$ be a triangle and $k$ be a positive real number. Take $D,E,F$ s.t $AD=kBC, BE=kCA, CF=kAB$ and $AD\bot BC, BE\bot CA, CF\bot AB$, and $D,E,F$ are in the $\angle CAB,\angle ABC,\angle BCA$.Show that the centroid of $ABC$ is the centroid of $DEF$.

we have
$R^{+\frac{\pi}{2}}(a\frac{1}{AD}\vec{AD}+b\frac{1}{BE}\vec{BE}+c\frac{1}{CF}\vec{CF})=\vec{BC}+\vec{CA}+\vec{AB}=\vec{0}\Rightarrow a\frac{1}{AD}\vec{AD}+b\frac{1}{BE}\vec{BE}+c\frac{1}{CF}\vec{CF}=\vec{0}\Leftrightarrow \frac{1}{k}(\vec{AD}+\vec{BE}+\vec{CF})=\vec{0}\Leftrightarrow \vec{AD}+\vec{BE}+\vec{CF}=\vec{0}\Leftrightarrow$ they have the same centroid. :D


function
trigonometry
real analysis
real analysis unsolved

Prove that there aren't any differentiable functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the relation:

\[ \boxed{f(x)\plus{}f'(x)\equal{} \begin{cases} \sin x,x \le 0\\\cos x,x>0\end{cases},x \in \mathbb{R}}\]

[quote="moldovan"]Prove that there aren't any differentiable functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the relation:
\[ \boxed{f(x) \plus{} f'(x) \equal{} \begin{cases} \sin x,x \le 0 \\
\cos x,x > 0\end{cases},x \in \mathbb{R}}\]
[/quote]

Hint: $ f'$ cannot have a jump discontinuity.


Yeah, I dont get it. Also called Factor labeling...

It's kind of complicated.
Say you have 1337m. Convert it to miles.

$ \left(\frac {1337m}{1}\right)$ You are given this. Converting to km:

$ \left(\frac {1337m}{1}\right)\left(\frac {1km}{1000m}\right)$ Now what is conversion factor. It's like 1.6

$ \left(\frac {1337m}{1}\right)\left(\frac {1km}{1000m}\right)\left(\frac {1mile}{1.6km}\right)$

Now see how the units cancel out leaving you with miles?

Which simplifies to 1337/1600 milez.

EDIT:  you can find out how many meters to one mile so you don't have to do tis for every prob.

Wasn't this in the other thread, too? Because I answered it there... :maybe:  :maybe:  :maybe:


I was looking through the grand winners of each year's ISEF, and I was wondering why some people were 1st, 2nd, or 3rd place but did not do the same in Siemens or STS, which are limited to the US. On the other hand, there were people who won Siemens or STS and were not on the ISEF list. I don't think that ISEF is less presigious, so why is there this discrepency (or maybe I haven't been searching closely enough)?

1) People might not apply to all of the contests.
2) They are judged according to different standards.
3) Subjectivity abounds.

I could have done that in paragraph form but I figured I'd spare you and just make it short.


f(X)= X^10 + ?X^9 + ?X^8 + ... + 1
A and B write a number in ? continuously. If f(X) doesn't have roots, A win. If f(x) has roots, B win.
A is the first player.
Who can win? Find the strategy.

Have a look here
http://mathcentral.uregina.ca/mp/previous2005/apr06sol.php

Pierre.


geometry
circumcircle
geometric transformation
homothety
ratio
geometry proposed

Let a triangle ABC. Find the position of a ppoint M inside triangle ABC such that circumcenters of triangles BMC,CMA,AMB form a triangle which has the same area as triangle ABC.

Call X,Y,Z,O,O' the circumcenter of BCM,CAM,BAM,ABC,XYZ. With and homothety with center M and ratio 2, XYZ become the antipedal triangle of M wrt ABC. Then the thesis become: find the locus of points M s.t. the area of the antipedal triangle of M wrt ABC is 4 times the ara of ABC.

Lemma 1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=214577[/url]
Lemma 2: let 3 circles with the same radius that pass all from a point P, then the circle that pass from the 3 point of intersection of those 3 circle ($ \neq P$) have the same radius of the 3 cirles.

by the lemma 1 the thesis is true iff $ M \equiv O'$. Then if $ M \equiv O'$ the triangles BCM,CAM,BAM have the same circumradius that for the lemma 2 is equal to the circumradius of ABC. Then X,Y,Z are the symmetric of O wrt BC,CA,AB, so XYZ and ABC are homothetic and then M is the orthocenter of ABC. (also if M is the orthocenter of ABC clearly [XYZ]=[ABC])


geometry
circumcircle
3D geometry
prism
search
geometry unsolved

Let $ (O)$ and $ (I)$ respectively be the circumcircle and the incircle of $ \triangle ABC$ and P is an animating point on $ (O)$. The tangents of $ (I)$ from $ (O)$ intersects the line $ BC$ at $ E,F$. Prove that the circumcircle of $ \triangle PEF$ passes through a fixed point.

(I have found out that fixed point is the point of tangent between $ (O)$ and the mixtilinear incircle of angle $ A$ of $ \triangle ABC$).

Dear Mathlinkers,
this problem has been proposed by
Pohoata C., Mixtilinear incircles and somehow Poncelet's porism, Mathlinks (01/02/2008) ;
	http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1968592250&t=186117.
Sincerely
Jean-Louis


I used to play it, when it had just came out.
It was addicting, definitely (I rock with the sword).
But now I'm kinda tired of it.

I'm on there too now, same name as here

IT NEEDS MORE WEAPONS
other than that, rockin game :D

isn't it so addicting and awesome!!!!????!!!
i love it. :D

look at the files to see what rank 7 and 8 look like :)

how do u get higher rank? by kills or winning rounds?

its by kills
Rank 1- 10 kills
Rank2- 20 kills
Rank 3- 100 kills
Rank 4- 500 kills
Rank 5- 1000 kills
Rank 6-2000 kills
Rank 7- 5000 kills
Rank 8- 10,000 kills
Rank 9- 20,000 kills
Rank 10- 50, 000 kills
I currently have 7,718 kills which is Rank 7.
Check [url=http://www.xgenstudios.com/play/stickarena/]here[/url] for stick arena high scores. I am number 35 i think.

That sir, is insane.  :o

Being an amateur player it's really annoying when someone from the top score list comes in and schools all of us...

Ya like I went to a place called "newb arena" cuz it was my first time.

I looked up the player that won, he had 12,000 kill or so,like 3rd or 4th place!

[quote="math92"]i'm number 25 on the high score table now!!!
I have 8489 kills!!!
 :D  :D[/quote]

how much time do u spend on that game?

there are people more obsessed than me on that game. :o

[quote="math92"][quote="Hexaditidom"]What is your name on there?[/quote]
math92 is my name on there

[quote="juicybooty911"]how do u understand the rank icon thing?[/quote]
One stripe is Rank 1
Two stripes is Rank 2
Three stripes is rank 3
Three Sripes with a star on top is rank 4
three stripes with a star and a circle is rank 5
a shield thing is rank 6
a right parenthesis turned so that its looks like a frown with two stars in it is Rank 7
a big star with stripes under it is rank 8
a shield with four stars around it is rank 9

thats all i know up to. I am currently rank 7

Edit:
I found two cheats for Stick arena that one should never use
These kinda ruin the game.
http://www.hot-shareware.com/games/speed-gear
http://www.artmoney.ru/

the first site just speeds up the game, so you go really fast and everyone else is at normal speed
the second site just adds kills to your kill count so you can get a higher rank[/quote]

HOW do u use artmoney in stick arena i dont get it!!!!!!!!!!!!!!!!!!!! TELL ME!! ________________________
I HAVE RETURNED.

[quote="math92"]go [url=http://www.xgenstudios.com/play/stickarena/]here[/url]
its a game called stick arena.  Its a game that is really addicting.[/quote]

its not i dont like it, too many ppl cheat..

[quote="Dan."]No bumping? But this is a very good thread. Math92 was one of the best at this game. Does he still come around?[/quote]

Yes, he frequents the High School Basics forum. I don't think he posts here that much anymore.

[quote="1=2"]@Dan: Please read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11598]this.[/url]

Yeah, There is this guy that goes really fast, he is a beast with the sledgehammer.[/quote]
I think sometimes it is fine to bump a discussion rather than start a new thread, and this is such a situation.

There's a stick Arena Trainer, I'll use it to own you all.

This game caused my keyboard to not be able to type the letter "a"! Now I have to use my Dad's computer for all eternity!

Seriously, don't play the game.

The bizarre typing problem stopped. I don't think it's Stick Arena anymore.

[quote="math92"]Wow, major thread revival.   :)

Yeah, I'm still here, although I haven't played Stick Arena in over a year and a half.  I probably still have some skizzles though.   :D[/quote]

I was doing some research, because I'm making the SA hall of fame continuation, but in a freewebs site. Noticed this forum, and the topic from the magical world of google. So this is where you been hiding out? I should have guessed. You really are a math fiend.

[quote="Dan."]You really are a math fiend.[/quote]

Most of the people here are. :wink:


group theory
abstract algebra
modular arithmetic
number theory
superior algebra
superior algebra unsolved

Let $ G$ be a finite group of order $ n$. Prove that for any $ a\in G$, $ a^n\equal{}e$ where $ e$ is the identity element of $ G$.

I know a proof using Lagrange theorem, but I would like to know is there any other method without quoting Lagrange Theorem? Thanks

You can reproduce the proof of Lagrange's theorem by taking $ H \equal{} \left< a \right>$, but what would the point of that be?  Proving special cases of a theorem when the general case isn't any harder is missing the forest for the trees.

Actually, the original problem from a book is

Let $ G$ be a finite cyclic group of order $ n$ generated by $ a\in G$. If $ a^s,a^t\in G$ for some $ s,t\in \mathbb{Z}$, prove that
$ \left<a^s\right>\equal{}\left<a^t\right>$ if and only if $ gcd(n,s)\equal{}gcd(n,t)$.

This question is before Lagrange theorem is introduced, but there is a part in my proving that we need the statement of the question I post. That is why I am asking whether there is a proof for that question without quoting Lagrange

Since you are only dealing with cyclic subgroups of cyclic groups (actually, all subgroups of cyclic groups are cyclic, but we don't need that here) you can't possibly need your theorem except in the case where $ G$ is cyclic, in which case it is immediately obvious.

In fact i use it when i try to prove $ gcd(n,s)\equal{}gcd(t,s)\implies \left<a^s\right>\equal{} \left<a^t\right>$

I first show that both $ \left<a^s\right>$ and $ \left<a^t\right>$ has same order then try to prove that $ \left<a^s\right>\subset \left<a^t\right>$.

Where i notice that if $ gcd(n,s)\equal{}d$ then there exist $ a_1,b_1\in\mathbb{Z}$ such that

$ a_1n\plus{}b_1s\equal{}d$

Since $ d|t$ so we can write $ t\equal{}dk$ for some integer $ k$.

hence $ t\equal{}a_1kn\plus{}b_1sk$ which in turn gives us $ a^t\equal{}\left(a^s\right)^{b_1k}$ if we can show that $ a^n\equal{}e$.

Of course for this particular question we only need to consider cyclic case ,but im just curious about the general case  :lol:

[quote="shyong"]Actually, the original problem from a book is

Let $ G$ be a finite cyclic group of order $ n$ generated by $ a\in G$. If $ a^s,a^t\in G$ for some $ s,t\in \mathbb{Z}$, prove that
$ \left < a^s\right > \equal{} \left < a^t\right >$ if and only if $ gcd(n,s) \equal{} gcd(n,t)$.
[/quote]

One can reduce the group theory to number theory by modeling the cyclic group $ C(n)$ as fractions $ \mathbb{Z}/n \pmod{ \mathbb{Z}}.$ Here the order of $ s/n$ is the least k such that $ ks/n \in \mathbb{Z}.$ But $ n|sk \iff n/d|k$ where $ d \equal{} gcd(n,s)$, as follows by factoring out d then using Euler's lemma. Hence $ s/n$ has order $ n/gcd(n,s)$ and the result follows immediately. It may aid the intuition to view the fractional arithmetic $ \pmod 1$ circularly, e.g. as "$ n$-clock arithmetic"


show how to cut any triangle by two straight cuts into symmetric parts.

which triangle can be cut in two symmetric parts by one straight cut?

its  too easy...

i wonderd ,nobody answerd it :!: .

I guess no one knows what you mean by symmetric.
Symmetric with respect to what? Perhaps you mean congruent? Similar? Something else?


[quote="Vasc"]It is problem 18 in Algebraic Inequalities, page 401.
I hope Mircea Lascu sent you my book, where you appear as author of 3 problems.  :([/quote]

Dear Vasc,

Thanks for the info - if I know that I certainly will not post the problem. On the other hand, the very fact that I posted it implies ... well, you know ;-)

[quote="ductrung"]For any positive real numbers $a,b$ and $c$,
\[(a+b+c)^{2}(ab^{2}+bc^{2}+ca^{2}) \ge 9abc(a^{2}+b^{2}+c^{2}) \]
[/quote]
It's stronger than
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33612

In Vasc's book it is nicely proved that if $a+b+c=3$ we have that $(a+b+c)^{6}\geq 27(ab+bc+ca)^{2}(a^{2}+b^{2}+c^{2})$

[quote="arqady"][quote="ductrung"]For any positive real numbers $a,b$ and $c$,
\[(a+b+c)^{2}(ab^{2}+bc^{2}+ca^{2}) \ge 9abc(a^{2}+b^{2}+c^{2}) \]
[/quote]
Yes, I think it's true. But my proof is very ugly.[/quote]

Surely it's true because I have a very short proof ;-).
I do not have Vasc's book so I don't know the proof there - may somebody post that proof - as I want to know.
For me this inequality arises very [b]naturally[/b]: I want to "compare" the right-hand sides of two inequalities of mine, namely
http://www.mathlinks.ro/Forum/viewtopic.php?t=84160
and
http://www.mathlinks.ro/Forum/viewtopic.php?t=87192

[quote="Vasc"]It is problem 18 in Algebraic Inequalities, page 401.
I hope Mircea Lascu sent you my book, where you appear as author of 3 problems.  :([/quote]
Sorry, Ductrung. The problem 18 is not your inequality, it is a weaker inequality:
http://www.mathlinks.ro/Forum/viewtopic.php?t=33612
I found your inequality is equivalent to
$\sum S_{a}(b-c)^{2}\geq 0$, where $S_{a}=\frac 2{c}+\frac a{bc}-\frac 6{a+b+c}$.

Ductrung, can you post your very short proof?

[quote="Vasc"]Your initial inequality (for which no solution was posted) is very nice: $\frac a{b}+\frac b{c}+\frac c{a}\geq a^{2}+b^{2}+c^{2}$.[/quote]
Awesome!

[quote="Vasc"]A stronger inequality is the following:
  If $a,b,c$ are positive numbers such that $a+b+c=3$, then

$6(\frac a{b}+\frac b{c}+\frac c{a})+3 \geq 7(a^{2}+b^{2}+c^{2})$.

Ductrung, have you also a "very short proof" for this ?[/quote]
No. This is very sharp. Reusing the result from my initial inequality, I can quickly transform this one to an SOS form. But I am unable yet to make the last evaluation without the help of the computer. So it is not nice. I will try again.

Lovasc, be patient  :wink: (many problems here are without proof in many months). However, a hint can be found the above.

[quote="ductrung"]For any positive real numbers $a,b$ and $c$,
\[(a+b+c)^{2}(ab^{2}+bc^{2}+ca^{2}) \ge 9abc(a^{2}+b^{2}+c^{2}) \]
[/quote]

Ductrung, it is also the problem in my note. I solve it by SOS. How about you?

Yes, I solved by SOS and Vasc already gave a hint above.

yes, but i have a different hint with VasC.

[quote="Vasc"]A stronger inequality is the following:
  If $a,b,c$ are positive numbers such that $a+b+c=3$, then

$6(\frac a{b}+\frac b{c}+\frac c{a})+3 \geq 7(a^{2}+b^{2}+c^{2})$.

Ductrung, have you also a "very short proof" for this ?
Your initial inequality (for which no solution was posted) is very nice: $\frac a{b}+\frac b{c}+\frac c{a}\geq a^{2}+b^{2}+c^{2}$.[/quote]
hint:
\[\sum S_{a}(b-c)^{2}\ge 0, S_{a}=\frac{b}{c}+\frac{a}{b}+\frac{2a}{c}-3 \]
In the case $a \le b \le c$, it is easy. And in the case $a \ge b \ge c$, you can easily check that
\[S_{a}\ge 0, S_{b}+S_{c}\ge 0,S_{a}+2S_{b}\ge 0,S_{c}+2S_{b}\ge 0 \]
:)

[quote="Vasc"]It is problem 18 in Algebraic Inequalities, page 401.
I hope Mircea Lascu sent you my book, where you appear as author of 3 problems.  :([/quote]

When can i find this book?


I think there is a bug in the resources section.When you click on the Download the PDF version, it takes you back the the forum index.
For example:
http://www.mathlinks.ro/Forum/resources.php?c=80&cid=90&year=2006
Click on the Downalod the PDF version and it will take you back to http://www.mathlinks.ro/Forum/

I think it has been like that for months, (though it might have been fixed and recurred in the meantime)

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=resources+pdf&t=101982

Now there are shortcuts that link to download.php?id=xxx not found. :maybe:

We don't have the PDF generator working right now...

Is there any way you can fix it ? :maybe:


Edit:  I should note that this is problem 11, part (d) of Chapter 3, Principles of Mathematical Analysis, Rudin.

$ \sum{a_{n}}$ diverges, $ a_{n} > 0$ for all $ n$.  Describe the behavior (convergence or divergence) of $ \sum{\frac {a_{n}}{1 \plus{} na_{n}}}$

Is there, by some chance, a missing condition? If $ a_n$ is decreasing, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=25574]here[/url].

If it's not decreasing, there can be cases in which the new series converges.

There's no missing condition.  I've shown that the series diverges when it's decreasing, and when {$ a_{n}$} does not converge to $ 0$, but I need to describe its behavior completely.  This has given me an idea, though...

To buttress jmerry's claim, let's consider the sequence $ (a_n)$ defined by

$ a_n \equal{} \begin{cases} 1 & \text{if } n \equal{} 2^{k} \text{ for some } k \in \mathbb{N} \\ 2^{\minus{}n} & \text{otherwise} \end{cases}$.

Then $ \sum_{n\equal{}1}^{\infty} a_n$ diverges while $ \sum_{n\equal{}1}^{\infty} \frac{a_n}{1\plus{}na_n} \leq \sum_{n\equal{}1}^{\infty} \frac{2^{\minus{}n}}{1\plus{}n2^{\minus{}n}} \plus{} \sum_{k\equal{}1}^{\infty} \frac{1}{1\plus{}2^{k}} < \infty$.

This example obviates the trial to search for a single simple answer, since the convergence of the new series depends on yet unidentified property of $ (a_n)$.

Your example also contradicts my claim in the above post that the series diverges when {$ a_{n}$} does not converge to $ 0$.  I went back and found my proof, and there was indeed a mistake.  I revised it to say that the series diverges when {$ a_{n}$} has no subsequence which converges to $ 0$.  It's not necessary that there be a single, simple answer; but I need to identify the properties of {$ a_{n}$} that determine whether or not the series converges.


well, heres the question

A school play ran for two nights, with audiences totalling 1390 adults and students. They paid $\$$4285 for admission. One adult ticket cost $\$$4.00 and one student ticket cost $\$$2.50. The ratio of adults to students was 3 : 5 on the first night and 2 : 3 on the second night. How many students attended each night?

thanks for the help if can give any

[hide="Half-solution"]If there were $a_{1},a_{2}$ adults and $s_{1},s_{2}$ students on the first and second nights respectively, then we have following equations:

${a_{1}\over s_{1}}={3\over 5}$

${a_{2}\over s_{2}}={2\over 3}$

$a_{1}+s_{1}+a_{2}+s_{2}=1390$

$4(a_{1}+a_{2})+2.5(s_{1}+s_{2})=4285$

Now express $a_{1}$ and $a_{2}$ from the first two equations, plug that into the other two and solve the system in $s_{1}, s_{2}$.[/hide]


A circle is inscribed in a triangle with sides $ 8$, $ 15$, and $ 17$.  The radius of the circle is:

$ \textbf{(A)}\ 6 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 5 \qquad
\textbf{(D)}\ 3 \qquad
\textbf{(E)}\ 7$

[hide="Solution"]
Note that a triangle with side lengths 8, 15, 17 is right triangle.

Thus, its area is $ 8*15\div2\equal{}60$

$ \text{Area}\equal{}rs$ where $ r$ is incircle's radius and $ s$ is semiperimeter of the triangle.

$ 60\equal{}r*20 \implies r\equal{}3$

Answer: D[/hide]

Another way to solve this is to note that
[hide="alternative solution"]$ r\equal{}\frac{a\plus{}b\minus{}c}2$ for any right triangle.[/hide]


induction
LaTeX
analytic geometry
combinatorics unsolved
combinatorics

A trajectory from the point (0,0) to the point  (x,y) is called  the broken line that connects the points (0,0,) , (1,s[1]) , (2,s[2]),......, (k,s[K]), where:

    s[x]=y

                        and    s[i]-s[i-1]= {    0 or 1 or -1

 x is integer,x>=0
 y is integer.

Let  T[x][y] is the number of trajectory from (0,0) to (x,y).

Prove that

   T[x][y] =

[(x-|y|)/2]
=    Σ     ____________________ x!______________________
 k=0   (  (y+|y|)/2 +k   )! (    (|y|-y)/2 +k   )! (  x- |y| - 2k)!

where [ (x-|y|)/2]  is the biggest integer that does not exceed  (x-|y|)/2 .

I tried to prove this with induction on x , but it became very ugly and i gave up.

that looks pretty messy, and the lack of LaTeX isn't helping... can someone rewrite that in LaTeX lol :)

I am sorry that i have written this problem in this way.... i  can't use programs like Latex or Matematika.

 I will try to explain the problem again.

 A trajectory is  broken line that looks for example in this way ( the '*' is for nothing)

****** _
***** / *\_
(0,0)  /*** *\
********** \......        ....(x,y)

The end of every little line ( i mean / or _  or \)   has coordiantes (m,s[m]) ,  where
m is integer and it is >=0 and s[m] is only integer (it can be <0 or >=0).

The general equation of  a trajectory is  s[i]-s[i-1]  can be equal to 0 or to 1 or to -1.
For example, for the trajectory above is true that:
s[0]=0.
s[1]=1.
s[2]=2.
s[3]=2.
s[4]=1.
s[5]=1.
s[6]=0.
s[7]=-1.
.... etc.
s[x]=y.
On the drawing it seems that s[3] is not 2 but it is because it appears in this way....
The third little line has to appear axactly between the second little line and the fourth little line.I don know why it does not appear in this way.

Let's T(x,y) is the number of trajectory that connects the point  (0,0) and the point (x,y)
x,y are integer ,x>=0.
Prove that T(x,y) is equal to the sum that is given in my post above.
k  is integer  changing from  0 to [ (x-|y|)/2)] , where [p] is the biggest interger, that
does not exceed  p.

I will give a little hint that may be will be useful.
If we have to solve a  problem where the trajectory is defined  as in the problem above, but without horizontal moves( without horizontal little lines), the general equation  of
such kind of trajectory will be:
s[i]-s[i-1]   can be equal to 1 or -1. (  it can not be equal to 0 this time).
Such kind of trajectory is more simple than the above. If we look at specific trajectory(not all that connects the points (0,0) and (x,y) but  look at only one that connects them) it will be determined only by its  little lines, that are draw in upper direction, i mean  that it will be determined only by such points that has second coordinate s[i] such as:
s[i]>s[i-1].
In this way if M is the number of this points, the total number of trajectory that connects (0,0) and (x,y) is  x*(x-1)*....*(x-M+1) / (x-M)!

And M is easy to found if it is given the point (x,y) , but i will not write how now.

This second problem can be used to be solved a problem like this:
We have m+n people that are on queue to buy  tickets.n of these people has a  banknote  of 5 dollars(every person has only one banknote) and the other m has  one
banknote of 10 dollars. The price of one ticket is 5 dollars and in the paybox  in the begining has 0 dollars..Find all the ways that can be located the customers  in the queue, so that noone has to wait  to  receive  a change( 5 dollars actually).
This problem can be solved quite easy finding all  trajectories that does not cross(or touch) the line y=-1.

But in the first problem it is given that  the trajectory can have and  sero jumps, not only positive and negative ( it can have and horizontal little lines) and this is difficult for me....
If someone can help, i will be thankful

first lets work only with y positive, if y is negative is exactly the same.

you just need to see that you have 3 kinds of moves, up, front and down.

you need to do y moves up and x-y moves front to get to (x,y).
each move you do down you need to do one more up and two less front.
and at most you can do [(x-y)/2] moves down.

lets call k to the moves down. you do y+k up moves and x-y-2k moves front, now you just need to chose the order
of the moves and you can do that in
$\sum_{k=0}^{[(x-y)/2]} \frac{((y+k)+k+(x-y-2k))!}{(y+k)!(k)!(x-y-2k)!}= \sum_{k=0}^{[(x-y)/2]} \frac{x!}{(y+k)!(k)!(x-y-2k)!}$

I think i  understood the solution.  Thank you.


function
real analysis
real analysis unsolved

Let $ (X,d)$ be a metric space. Let $ f: X\to \mathbb{R}$. The [b]level sets[/b] of $ f$ are of the form $ \{x: f(x)\le k\}$, $ k\in \mathbb{R}$. For what kind of functions are the level sets compact?
For lower semi continuous functions what can we say about the level sets?

In fact for my problem $ f: X\to [0,\infty)$. So all I am asking is whether $ f^{\minus{}1}([0,k])$ is compact when $ f$ is lower semi continuous?


Suppose $p(z)=z^2-2$. Suppose: \[H_{\infty}=\{z \in C| \lim_{n \rightarrow \infty} ||p^n(z)||= \infty \}\] Find everything you can about $H_{\infty}$(I proved $C-H_{\infty}= [-2,2] \subset R$)

Well known trick?
Write $z=y+\frac{1}{y}$. Then we have $p^n(z)=y^{2^n}+\frac{1}{y^{2^n}}$. And we know that $|y|=1$ only for $z\in [-2,2]$. So Omid's answer is correct.

Right?

Bravo!! It was exactly my solution  ;)  ;)


If $ x$ and $ y$ are negative, $ x^{2}y\equal{}1$ won't hold, as $ x^{2}y<0$. We see that neither of them are equal to $ 0$. If only $ y$ is negative, $ x^{2}y<0$ again. If $ x<0$, than $ y>0$, but $ x^{x\plus{}y}\equal{}y^{xy}$ won't hold. So, $ x$ and $ y$ are both positive.
$ y\equal{}\frac{1}{x^{2}}$ gives: $ x^{x\plus{}\frac{1}{x^{2}}\plus{}\frac{2}{x}}\equal{}1$. So, $ x\plus{}\frac{1}{x^{2}}\plus{}\frac{2}{x}\equal{}0$ or $ x\equal{}1$.So, $ x\equal{}1$ and $ y\equal{}1$ or $ x\equal{}\minus{}1$ and $ y\equal{}1$.


algebra
polynomial
real analysis
real analysis unsolved

Let $ f\in C[0,1]$, show that you can approximate $ f$ arbitrarily closely  by polynomials of the form $ P(x) \equal{} a_{0} \plus{} a_{3}x^{3} \plus{} ... \plus{} a_{3k}x^{3k}$

Edit:

I think i've got it. Using this  [url=http://books.google.com/books?id=QALoZC64ea0C&pg=PA136&dq=weierstrass+approximation+bernstein&lr=&ei=8M8JSNSqFYPUzATQ9syhCA&hl=pl&sig=8ZYpj-zDSPNzvTw5yDQ4xgcH-0Y#PPA137,M1]proof
[/url] of weierstrass approximation theorem and treating bernstein polynomials followingly:
\[ B_{n}(f) \equal{} \sum_{k \equal{} 0}^{n}x^{3k}(1 \minus{} x^{3})^{n \minus{} k}f(\frac {k}{n})\binom{n}{k}
\]
won't change that proof dramatically, will it ?

mlok's way is the simplest, but you probably can use [i]stone[/i]-weierstrass as well

Natural question: For which sets $ A\subset \mathbb N$ is the linear span of $ \{x^n\colon n\in A\cup \{0\}\}$ dense in $ C[0,1]$?

all $ A\subset\mathbb{N}$ such that $ \sum_{k\in A} k^{\minus{}1} \equal{} \infty$ by muntz-szasz?