Define the term $1:2:3:4:5:6:7:8:9:10.$  Put in parentheses in order to yield a required number. How many ways are there to obtain 7 ?

This basicly comes down to splitting them all up on a numerator and a denominator.

Factors 3 must divide eachother away: $\frac{9}{6.3}=\frac{1}{2}$
Factors 5 must divide eachother away: $\frac{5}{10}=\frac{1}{2}$
Factor 7 must be once in the numerator.
Factor 2 leaves us: 3 times 2, 1 time 4, 1 time 8. That leaves us 6 options.
Now the number 1 doesn't double it here, since that one starts in the numerator.
$7=$ $\displaystyle\frac{1.2.3.4.5.6.7}{8.9.10}=\frac{1.7.8.9.10}{2.3.4.5.6}=\frac{1.2.5.7.8.9}{3.4.6.10}=\frac{1.3.4.6.7.10}{2.5.8.9}=\frac{1.3.5.6.7.8}{2.4.9.10}=\frac{1.2.4.7.9.10}{3.5.6.8}$

This is an easy but very, very nice question! where did you find this one?

It is from "Hamburger Schlerzirkel Mathematik". So what is your answer to your questions ? How many possibilities are there ? Your solution does not seem to be complete at all.

Why can't I start a new topic in this thread. It tells me I have to be a moderator ?! Why that ? Shall I post in the intermediate and advanced sections and others move the problems if required ?  :?

yes ikap, you're right. there are only 3 instead of 6.

as for orl, it would really surprise me if there were more, since I generalized this for n and k, and I got that $f(k) = 3$ if $k=7$, $f(k)=1$ if $k=5$ and $f(k) = 0$ if $k \not= 7;5$

The real problematic generalization would be to change 10 by $n$. Now  f(k)=0, for any $k\neq p_2$, where p_2 is the largest prime less than n. As for $f(p_2)$, if $2p_1>n$, where p_1 is the next prime in decreasing order, it is also $0$. Now I think it is known that $p_{x+1}<\frac 43 p_x (1)$, for any $x\geq 4$, where $p_x$ are the prime numbers. So take $p_3<\frac 43p_2<(\frac 43)^2 p_1<n$, so $p_2$ is not the largest prime smaller than $n$, contradiction. So we must take n at most 10 etc.
Well I don't know if (1) is really correct. I only know something like $p_{x+1}\leq p_x\frac98$, for $p_x\geq 47$, so I suppose (1), also works.

[quote="ikap"]The real problematic generalization would be to change 10 by $n$.
...[/quote]

ikap, that's exactly what I did!

and the solution is much simpler than that: if you take n above 10, you got factor 11 to take care of, 7 and 11 means $n\geqq$ must be at least 77. But there we got 13, 17,19, etc. to disturb the party, so above it's trivial that there are no solutions for limited n.

For n=10 we got the solutions
for n=7 or 8 or 9 there are none because 5 and 7 require $n\geq35$
for n=6 we want it to be 5 of course, $\displaystyle 5=\frac{1.3.4.5}{2.6}$ only solution
for n=4 or 5 it's impossible because of 3
for n<4 it's also impossible of course :)

[quote="ikap"]Peter's solution is "more than complete". He forgot that 2 also is forced to stay somewhere, at the denominator. So this drastically reduces 6 to 3. I wonder if you wanted to trick him orl, with this "complete", or if I was tricked. 'cause there don't seem to be other solutions to me.[/quote]

I'm not seeing why 2 is forced to be in the denominator.
Is that something that comes from the instruction "add parenthesis" which doesn't make sense to me.

What would (1:2:3):(4:5:6:7:8):(9:10) be considered as representing?

$\frac {1.5.6.7.8.10}{2.3.4.9}$. I was talking about $1:2:3...$. You can't put parantesis here to put 2 in the numerator.
But Peter there is something that I really don't get.
[quote]7 and 11 means n must be at least 77[/quote]
I don't see any other way of managing these prime numbers but considering the biggest smaller than $n$.

what does "adding parenthesis" do?

i've never heard of using parentesis in ratios

yes, you're right.

maybe a better attempt is to find where 7,11,13... have even powers. you'll see the interval shrinks incredibly fast :) (just beware of multiples of $p^2$ here!)


hello, I am here with my first question:
   paralel light ray is falling to the wall by the angle $\alpha$ .if the wall is growing longer with velocity v=k.c (k is constant, c is light speed), what is the velocity of the shadow of the wall?


In an effort to improve the AoPS Mathcounts class, I am currently collecting past Masters Rounds to add to the curriculum.  Unfortunately I only have a few of them.  Is there anybody here who has copies of past Masters Rounds that they could email to me?  It would be very much appreciated.

Solutions are not necessary, but if anybody does any scanning, I wouldn't mind having them.

[quote="yif man12"]when do you need them by? and i've tried attaching stuff on a private message but it never works. how do you do it?[/quote]

I'm in no real hurry, but within the month would be nice.  You can email me at crawford@artofproblemsolving.com.  Thanks in advance.

i dont know if you already have it, but i still have copies of the masters round from last year if you want it.

If it's the one about using fractions and p/q used as decimals like .(p)(q), then I have it.  Otherwise I do not.  If you have a different one from last year (state or nationals) I would appreciate having copies.

Do you have the one from 2002 about harvesting? I might have that one somewhere.

I believe I have two masters rounds - one involves developing a method for finding a closed form for 1^k + 2^k + ... + n^k, and the other one I don't remember.  I'll post the exact dates when I get my old MATHCOUNTS binder out later.

I can TeX these up for you if you want them, but I don't want to if you already have them.

there is always this year's national masters round on dismal arithmetic.


If $ \frac{a}{b}$ and $ \frac{c}{d}$ are neighbor fractions, their difference has a numerator of plus-minus 1. Prove that $ \frac{a\plus{}b}{c\plus{}d}$ is between $ \frac{a}{b}$ and $ \frac{c}{d}$ and $ \frac{a\plus{}b}{c\plus{}d}$ is a neighbor fractions to both $ \frac{a}{b}$ and $ \frac{c}{d}$

thanks :)

Do we know that $ \frac{a}{b}$ and $ \frac{c}{d}$ are neighbor fractions, or is that first bit just the definition of neighbor fractions?

[b]@ Poincare:[/b] You double-posted, and wrote the problem wrong, I believe. Your combination fraction should be $ \frac {a \plus{} c}{b \plus{} d}$.

See your other version of the same thread for a (very nice) hint from [b]t0rajir0u[/b].


determine all real x  from the interval $[0,2\pi]$ for which
$\sqrt{\frac{4}{3}-\sqrt{\frac{4}{3}}}. \sin x .\cos x+(1+\sqrt{\frac{1}{3}}) \geq  1$

after moving stuff around, we get $\sin 2x\ge-\sqrt{3}-1$...which is obviously all of the values in that interval...perhaps there is a typo?


[color=darkblue][b]In the triangle $ ABC$ prove that the perpendiculars from the Gergonne's  point $ N$ to the interior bisectors are intersecting the sides of the triangle in $ 6$  points that are situated on the same circle, concentric with $ C(ABC)$.[/b][/color]

[color=red][b]If you don't know Gergonne's point see here[/b][/color]: [url]http://www.mathlinks.ro/viewtopic.php?t=33335[/url]

Anyone can solve this old problem? There seems hardly or yes?

Incircle $(I)$ of $\triangle ABC$ touches $BC, CA, AB$ at $D, E, F$. $N$ is symmedian point of $\triangle DEF$.
Perpendiculars to $AI$ through $N$ cut $AB, AC$ at $Z_a, Y_a$,
perpendiculars to $BI$ through $N$ cut $BC, BA$ at $X_b, Z_b$,
perpendiculars to $CI$ through $N$ cut $CA, CB$ at $Y_c, X_c$
$\Longrightarrow$ $EY_a = FZ_a,  FZ_b= DX_b, DX_c =  EY_c$.
$BC$ is antiparallel of $EF$ WRT $\angle FDE$ $\Longrightarrow$ $\triangle NX_bX_c \sim \triangle DEF$.
$DN$ is symmedian of $\triangle DEF$ $\Longrightarrow$ $ND$ is median of $\triangle NX_bX_c$.
Similarly, $NE, NF$ are medians of $\triangle NY_cY_a, \triangle NZ_aZ_b$
$\Longrightarrow$ $X_b,X_c,Y_c,Y_a,Z_a,Z_b$ are on circle concentric with $(I)$.


algebra
polynomial
number theory
relatively prime
algebra proposed

The polynomials $P,Q$ are such that $\deg P=n$,$\deg Q=m$, have the same leading coefficient, and $P^2(x)=(x^2-1)Q^2(x)+1$. Prove that $P'(x)=nQ(x)$

Classical problem in oral X: find all such couples of polynomials. It is nice problem. First, it is clear that $n=m+1$ and then by derivation of your given relation we have $ 2P(x)P'(X)=2XQ^2(X)+2(X^2-1)Q(X)Q'(X)$ and thus $Q$ divides $ P P'$ ; but your initial relation shows that $ Q$ and $P$ are relatively prime thus $ Q$ divides $P'$. SInce they have the same degree, they are associated and the condition that $P$ and $Q$ have the same dominant coefficient shows that your last relation must hold.


limit
trigonometry
calculus
inequalities
function
real analysis
real analysis unsolved

Evaluate $ \lim_{x\to0}\frac{\cfrac{\sin x}x\minus{}1}{1\minus{}\cos x},$ without power series or L'Hôpital's Rule.

Maybe assume already limits that are already known?  For example,
$ \lim_{x\rightarrow 0}\frac{\sin{x}}{x}\equal{}1$
can be proved by the squeeze theorem.

I don't know what are you talkin' about; the limit takes the form $ \frac00$ and I don't know a way to tackle it, that's why I posted it here.

hello, i have three times used the L'Hospital's rule to find the limit $ \minus{}\frac{1}{3}$.
Sonnhard.

One can show by a geometric argument (a triangle contained in and a triangle containing a wedge of a circle) that for first quadrant angles $ x,$ $ \frac12\cos x < \frac12 x < \frac12\tan x,$ which can be rearranged to $ \cos x < \frac {\sin x}x < 1$ What can you do with that?

(Note: the simple way of using that is not sufficient, by itself, to compute a limit.)

(Note 2: no Taylor series? I don't appreciate having my tool kit taken away from me.)

Yeah, no Taylor series or that Rule.

You know a way to solve this limit?

As I already pointed out, some of the limits in there, such as the $ \sin{x}/x$ limit is well known to converge to 1.  Are we allowed to already assume that true?

Once you know the individual limits, you can simply combine them since that is allowed by some theorems.

Why are you restricting the use of L'Hopital's Rule?  In practice, it is quite convenient.

[quote="JRav"]Are we allowed to already assume that true?[/quote]
Yes.

[quote="JRav"]Why are you restricting the use of L'Hopital's Rule?  In practice, it is quite convenient.[/quote]
I know, it's just that I wanna see a "clean solution."

Being able to calculate your limit is equivalent to being able to compute the limit $ \lim_{x \to 0} \frac{\sin x \minus{} x}{x^3}$.

Even if calculating this limit can be achieved without explicit application of Taylor theorem or something similar to this, your trial will reduce to a special case of the Taylor theorem. I think there would be no 'totally new' method to evaluate this limit which completely excludes Taylor theorem.

We should be able to do this using only $ \cos x < \frac {\sin x}{x} < 1$ and trigonometric identities (though calculus would make it a heck of a lot easier).  On Carcul's suggestion, we first evaluate the limit $ \lim_{x\to 0}\frac {\sin x - x}{x^3}$.  Suppose this limit exists and is equal to $ L$, then it must hold also if we make the substitution $ x\mapsto 3y$ so
\[ \lim_{y\to 0}{\frac {\sin 3y - 3y}{y^3} = 27L
}\]
by the triple angle formula $ \sin 3y = 3\sin y - 4\sin^3 y$ this implies
\[ \lim_{y\to 0}{\frac {3(\sin y - y) - 4\sin ^3 y}{y^3} = 27L
}\]
which is the sum of two limits that are assumed to converge, so we find $ L = - \frac {1}{6}$.  From this we find that the original limit must be $ - \frac {1}{3}$ as follows:
\[ \lim_{x\to 0}{\frac {\sin x - x}{x^3}} = \lim_{x\to 0}{\frac {\frac {\sin x}{x} - 1}{x^2}} = \lim_{x\to 0}{\frac {\frac {\sin x}{x} - 1}{\sin^2 x}} \\
= \lim_{x\to 0}{\frac {\frac {\sin x}{x} - 1}{(1 - \cos x)(1 + \cos x)}} = \frac {1}{2}\lim_{x\to 0}{\frac {\frac {\sin x}{x} - 1}{1 - \cos x}}
\]
Up to now we have only proven that [i]if[/i] the original limit exists, it must be $ - \frac {1}{3}$.  That's a big if though, and certainly requires justification (use the inequality mentioned earlier).

[hide="remark"]The system of functional equations
\[ f(x)^2 + g(x)^2 = 1 \\
f(x\pm y) = f(x)g(y)\pm f(y)g(x) \\
g(x\pm y) = g(x)g(y)\mp f(y)f(x) \\
0 < x g(x) < f(x) < x \ \ \ \text{for } 0 < x < 1
\]
uniquely defines $ f(x) = \sin x$ and $ g(x) = \cos x$.  For this reason the geometric identities alone (plus the inequality Kent Merryfield mentioned) are always enough, at least in principle, to prove any fact about the trigonometric functions that can be proven using Taylor series.[/hide]


How would you measure the number of decibels in a sound?  Would you need some kind of special (expensive?) device?

I think you can do it with a microphone attached to a computer, using any audio analysis software ...


geometry
trapezoid
ratio
vector
similar triangles
geometry proposed

Gods painted a regular hexagon with side $a$ on a plane. Also they gave you a ruler without any marks. You are to construct a segment of length $a/n$, $n\in\mathbb{N}$.

[i](3 points)[/i]

Label the regular hexagon $ABCDEF$, such that $a = AB$ is the side to be divided. The long diagonal $CF \parallel AB$ is parallel to this side. We could easily find its midpoint as the intersection of the long diagonals $AD, BE, CF$, but there is no need. Thus we have 2 parallel lines $AB \parallel CF$, which is all we need. Let the opposite sides $BC, FA$ of the trapezoid $ABCF$ intersect at a point $K_0$ and its diagonals $AC, BF$ at a point $X_0$. The line $K_0X_0$ cuts the segments $AB, CF$ in their midpoints $M_0, N_0$. Let the opposite sides $BC, N_0M_0$ of the trapezoid $BCN_0M_0$ intersect at a point $K_1$ and its diagonals $M_0C, BN_0$ at a point $X_1$. Furthermore, let the opposite sides $M_0N_0, FA$ of the trapezoid $M_0N_0FA$ intersect at a point $K_2$ and its diagonals $AN_0, M_0F$ at a point $X_2$. The line $K_1X_1$ cuts the segments $M_0B, CN_0$ in their midpoints $M_1, N_1$ and the line $K_2X_2$ cuts the segments $AM_0, N_0F$ in their midpoints $M_2, N_2$. Thus the segment $CF$ is divided in $4 = 2^2$ equal parts. By repeating this process a sufficient number of times, the segment $CF$ can be divided into $m = 2^k$ equal parts, such that $m \ge n$. Re-label the dividing points of the segment $CF$ as $C \equiv P_0, P_1, P_2, ..., P_m \equiv F$. Let the opposite sides $BC, P_nA$ of the trapezoid $BCP_nA$ intersect at a point $K$ and its diagonals $AC, BP_n$ at a point $X$. The lines $KP_1, XP_1$ cut the segment $AB$ at point $Q, R$, respectively, in the ratios

$\frac{QB}{AB} = \frac{P_1P_0}{P_nP_0} = \frac 1 n,\ \ \ \frac{AR}{AB} = \frac{P_1P_0}{P_nP_0} = \frac 1 n$

After some experimentation, it is easy to see that we can fill in the equilateral-triangular lattice of which the hexagon is a part. So if the hexagon is ABCDEF, where CF has midpoint G, define points along line CF by $Z_n=\vec{G}+n\vec{GC}$ (with vector notation). For $n\ge2$, if $BG$ and $AZ_n$ intersect at $X_n$, then by similar triangles $ABX_n$ and $Z_nGX_n$, $\frac{BX_n}{X_nG}=\frac{AB}{Z_nG}=\frac{1}{n-1}$, i.e. $BX_n=a/n$.


Find all $P(x)$ satisfying the condition:
\[P(2P(x))=2\cdot P(P(x)) + 2\cdot (P(x))^2\]

If it is a polynomial, then write $P(X) = a_nX^n+ a_{n-1}X^{n-1} + \cdots + a_1X+a_0$, where $a_n \neq 0$, and if $n^2>2n$, then the largest monom in the equality in the statement is
\[ 2a_n^{n+1} (2^{n-1}-1)X^{n^2} = 0 \] thus $n=1$ contradiction.

Therefore the degree of $P$ can only be $-\infty, 0, 1$ or $2$. Each case is simple to analise of the cases reveal the solutions $P(X)=0$, $P(X)= -\frac 12$, and another (possible) solution of degree 2, $P(X) = X^2+bX+c$, where $c$ can only be $0$ or $-\frac 12$, but I am too lazy to further compute this.

Yes, it was clear.
That is why I asked whether $P$ is a polynomial.


calculus
integration
trigonometry
logarithms
calculus computations

Here are some easy problems from Vietnamese High school Graduation Examination and University Entrance Examination:

[size=150][b]Evaluate:[/b][/size]

[b]Problem 1[/b]
\[ I_1 \equal{} \int_{0}^{1}{\left(1 \plus{} e^x\right)x dx}
\]
[b]Problem 2[/b]
\[ I_2 \equal{} \int_{ \minus{} 1}^{1}{x^2\left(1 \minus{} x^3\right)^4 dx}
\]
[b]Problem 3[/b]
\[ I_3 \equal{} \int_{0}^{\frac {\pi}{2}}{\left(2x \minus{} 1\right)\cos x dx}
\]
[b]Problem 4[/b]
\[ I_4 \equal{} \int_{0}^{\frac {\pi}{4}}{\cos x \sin x dx}
\]
[b]Problem 5[/b]
\[ I_5 \equal{} \int_{0}^{\frac {\pi}{6}}{\frac {\tan^4 x}{\cos 2x} dx}
\]
[b]Problem 6[/b]
\[ I_6 \equal{} \int_{1}^{2}{\frac {\ln x}{x^3} dx}
\]
[b]Problem 7[/b]
\[ I_7 \equal{} \int_{0}^{\frac {\pi}{4}}{\frac {\sin \left(x \minus{} \frac {\pi}{4}\right)}{\sin 2x \plus{} 2\left(1 \plus{} \sin x \plus{} \cos x\right)} dx}
\]

________

Sorry, I've edited !

Are problems 1 and 5 supposed to be identical?

[hide="problem 6"]
$ I_6 \equal{} \int_{1}^{2}{\frac {\ln x}{x^3} dx}$

suppose $ y \equal{} \ln{x}\rightarrow x \equal{} e^y \rightarrow dx \equal{} e^ydy$
the integral will be like this..
$ \int_{\ln 1}^{\ln 2} y e^{ \minus{} 3y} {e^yd(y)}$
$ \equal{} [ \minus{} \frac {1}{2}ye^{ \minus{} 2y} \minus{} \frac12\int e^{ \minus{} 2y}dy]_{\ln 1}^{\ln 2}$
$ \equal{} [\frac {1}{4} (\frac {2y \plus{} 1}{e^{2y}})]_{\ln 1}^{\ln 2}$
$ \equal{} \minus{} \frac {1}{16}(2\ln 2 \minus{} 3)$

[/hide]

sorry if I'm mistaken..
still confused with number 5..does it use $ \tanh^{\minus{}1}$?? if so,,I think Vietnam has great math level.. :maybe:
can anybody help?

[quote="mhar_teens"][hide="problem 6"]
$ I_6 \equal{} \int_{1}^{2}{\frac {\ln x}{x^3} dx}$

suppose $ y \equal{} \ln{x}\rightarrow x \equal{} e^y \rightarrow dx \equal{} e^ydy$
the integral will be like this..
$ \int_{\ln 1}^{\ln 2} y e^{ \minus{} 3y} {e^yd(y)}$
$ \equal{} [ \minus{} \frac {1}{2}ye^{ \minus{} 2y} \minus{} \frac12\int e^{ \minus{} 2y}dy]_{\ln 1}^{\ln 2}$
$ \equal{} [\frac {1}{4} (\frac {2y \plus{} 1}{e^{2y}})]_{\ln 1}^{\ln 2}$
$ \equal{} \minus{} \frac {1}{16}(2\ln 2 \minus{} 3)$

[/hide]

sorry if I'm mistaken..
still confused with number 5..does it use $ \tanh^{ \minus{} 1}$?? if so,,I think Vietnam has great math level.. :maybe:
can anybody help?[/quote]

My solution:

[img]http://i246.photobucket.com/albums/gg93/NguyenDungTN/VN.jpg[/img]

[quote="mhar_teens"]wah,,I just wanted to start my answer for number 5 (it's just same)..
sorry I made a wrong calculation before..
yup,,the solution is right..[img]http://l.yimg.com/us.yimg.com/i/mesg/emoticons7/113.gif[/img]
how about number 7??
I think I don't have enough time today..(goin' to church..)
happy Sunday[/quote]

Pro 7:

[img]http://i246.photobucket.com/albums/gg93/NguyenDungTN/Vn2.jpg[/img]

sorry for my errors.. :blush:

thanks for the correction..
@moderator::why can't users edit their previous post(s)?

[quote="undefined117"]You can edit them, just not after a certain amount of time.

You might also want to look over your solution for Question 3. :)

Your solution for Question 6 is awesome, though. :thumbup: I learned from it.[/quote]

sorry mistake again..(I might be in hurry in solving them..)

$ I_3 \equal{} \int_{0}^{\frac {\pi}{2}}{(2x \minus{} 1)\cos x dx} \equal{} \int_{0}^{\frac {\pi}{2}}(2x\cos{x} \minus{} \cos{x})dx$
$ \equal{} \boxed{[2x\sin{x} \plus{} 2\cos{x} \minus{} \sin{x}]_{0}^{\frac {\pi}{2}}} \equal{} \pi \minus{} 3$

I wrote the wrong one in the box..

thanks 4 the attention...

Wait oh no.. sorry it was my mistake :blush: - your first solution was correct. Sorry!


This is a question from one of the national targets, but i don't remember which.
There is a circle with radius 120. A bug walks halfway to one edge, stops and turns right 90 degrees. he then walks halfway to that edge, stops and turns right 90 degrees. He does this again twice more. What is the distance between him and his original starting point to the nearest whole number?
I know you can solve this with the pythagorean theorum a lot, but is there any easier ways?
Jeff, tell me if there's something wrong with the problem

[quote="anirudh"][quote="bpms"][quote="mgao"][hide="My Solution"]

The equation for a circle with radius r centered at the origin is

$x^{2}+y^{2}=r^{2}$.

so the equation in this case would be

$x^{2}+y^{2}=14400$.  Now, since the bug would walk to the point (0,60), we just plug in the value of 60 to this formula and get

$x^{2}+3600=14400\Longrightarrow x^{2}=10800$ and

$x=\sqrt{10800}$.

Then, the bug walks again, and the midpoint from where he walks is $\frac{\sqrt{10800}}{2}$, and we get $\frac{10800}{4}=2700$. So we have the coordinates $60, 30\sqrt{3}$.  We make him walk again,

$2700+y^{2}=14400$.  The general thing is to calculate this missing value and then find the midpoint to which he walks, and make him walk yet again, so we just find $y$:

$y^{2}=11700$ so $y=30\sqrt{13}$.  We find the midpoint and $y=15\sqrt{13}$.

Current coordinates:  $(60, 15\sqrt{13})$.

We make him walk again:

$x^{2}+2925=14400$, so subtracting,

$x^{2}=11475\Longrightarrow x=5\sqrt{459}\Longrightarrow 15\sqrt{51}$.

Divide by 2 again,

$\frac{15\sqrt{51}}{2}$.

Current coordinates:

$(\frac{15\sqrt{51}}{2}, 15\sqrt{13})$

Now we see how far he is from (0, 0):

$\sqrt{\frac{15\sqrt{51}}{2}^{2}+15\sqrt{13}^{2}}$

$=\sqrt{11475+2925}$

$=\sqrt{14400}$

$=120$

 :rotfl:  I'm probably wrong.[/hide][/quote]
Um yes. You are wrong. This problem isn't about a bug, it's about a guy named pedro :D. This problem is ridiculously hard.[/quote]
there is no one named pedro in the question  :P[/quote]
I meant in the original problem.

[quote="bpms"][quote="SplashD"]I don't remember the answer being an integer :maybe:[/quote]
It isn't. As yif man said to the nearest whole number but the answer [b]without[/b] rounding is some ugly radical.[/quote]

Yeah, I posted before Nationals I think.

[quote="lotrgreengrapes7926"]My exact answer is [hide]
$\sqrt{7143.75-675\sqrt{13}-15\sqrt{31725+2700\sqrt{13}}}$ :D [/hide][/quote]

No one posted a solution originally (how odd...)

I don't feel like writing out a solution, but the problem did involve Pedro, it wanted the answer rounded to the nearest whole number, and the answer was 41. I know the solution, and it involves setting up right triangles with radii as the hypotenuses.

By the way, I posted a solution a couple of years ago here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=77603#77603[/url].

The key to my solution was setting up a coordinate system.

By the way, this question is from the 1997 National Target, Question 6.


algebra
polynomial
quadratics
number theory unsolved
number theory

Is there a trinomial with integer coefficients f such that for all natural numbers n, all prime factors of f(n) are of the form 4k+3?

a trinomial is a polynomial of third degree, so $f(n)=an^3+bn^2+cn+d$?
the problem actually sounds beautiful! it there was such a polynomial then would could easily prove that there are infinitely many prime p such that p is 3 mod 4, but i guess there is no(actually that's the way one proves that there are infinitely many prime p such that p is 1 mod n for every integer n, just the n-th cyclotomic polynomial).

Peter

Actually, a trinomial is $an^2+bn+c$ (it has three monomials :)).

ah ok. i would call that a quadratic polynomial, but however, it makes the problem easier.

Peter

Can't we always find a prime of the form $4k+1$, which doesn't divide $a$, for which $\Delta=b^2-4ac$ is a quadratic residue? I think we can, and in this case, take $t$ s.t. $t^2=\Delta$ in $\mathbb Z_p$, then take $x=(-b+t)(2a)^{-1}$ in $\mathbb Z_p$, and we find $ax^2+bx+c=0$ in $\mathbb Z_p$, so $f(x)$ is divisible with $p$.

The nastiest case we can encounter is when $\Delta$ is a power of $2$, but that's still Ok, because it's either $2^{2\alpha}$, so it's a quadratic residue for all $p$, or $2^{2\alpha+1}$, so we must have $(\frac 2p)=1$, which can be achieved if $p=8k+1$ (we must keep the condition $p=M_4+1$).

I bet there's something wrong, but I can't see what :).

i actually think we can. just use quadratic reciprocity to make the condition of $\Delta$ being a square mod p into a condition of p being x mod n, where x and n have no common divisor(i don't want to explicitly write down the whole computation, but n=$\Delta$ or n=$4\Delta$ depending on whether $\Delta$ is divisible an even or odd time through 2 i think). by dirichlet there are infinitely many such p, so we're done. something like that should be the proof :D

Peter

[Edit: Why must i always do typos???]

Nice. Now, try to prove the following problem:
    For any non-constant polynomial f with integer coefficients and for any k, there always exist a number in the sequence f(1), f(2), f(3),... that has a prime factor of the form 1+nk.

Do you have a solution to this one, or were you wondering whether it's true or not?

As far as I know, it is true and moreover, extremeley difficult. I think I have an article with the proof, but I'm not sure. I will have to check when I come back in Romania. Anyway, the statement is true,; Hope...


linear algebra
matrix
quadratics
induction
inequalities
linear algebra unsolved

Let $ G$ be a skew-symmetric real square matrix, and let $ E$ be a symmetric positive semidefinite real square matrix. Prove that $ \det F\geq\det E$, where $ F\equal{}E\plus{}G$.

This is a known result, but how to show it.

[quote="miwalin"]Let $ G$ be a skew-symmetric real square matrix, and let $ E$ be a symmetric positive semidefinite real square matrix. Prove that $ \det F\geq\det E$, where $ F \equal{} E \plus{} G$.[/quote]

Here is how I would do it:

Since $ E$ is symmetric positive semidefinite, there exists a diagonal matrix $ D$ with all its entries nonnegative and an invertible matrix $ U$ such that $ E \equal{} U^TDU$. (This follows from the fact that every quadratic form has an orthogonal basis, which can be proven by Gram-Schmidt orthogonalization.) Let $ H \equal{} \left(U^{ \minus{} 1}\right)^TEU^{ \minus{} 1}$; then, $ H$ is skew-symmetric and $ E \equal{} U^THU$. Thus, $ F \equal{} E \plus{} G \equal{} U^TDU \plus{} U^THU \equal{} U^T\left(D \plus{} H\right)U$. Hence, $ \det F\geq \det E$ rewrites as $ \det\left(D \plus{} H\right)\geq\det D$.

Now, we use the following fact: For any two $ n\times n$ matrices $ A$ and $ B$, we have

[color=green][b](1)[/b][/color] $ \det\left(A \plus{} B\right)$
$ \equal{} \sum_{\text{two subsets }I\text{ and }J\text{ of }\left\{1,2,...,n\right\}\text{ with equal cardinality}}\pm\left|\text{the }\left(I,J\right)\text{ \minus{} minor of the matrix }A\right|$
$ \cdot\left|\text{the }\left(\left\{1,2,...,n\right\}\setminus I,\left\{1,2,...,n\right\}\setminus J\right)\text{ \minus{} minor of the matrix }B\right|$,

where for any two subsets $ U$ and $ V$ of $ \left\{1,2,...,n\right\}$, the $ \left(U,V\right)$-minor of a $ n\times n$ matrix $ Q$ is defined as the $ \left|U\right|\times\left|V\right|$ matrix which is obtained if we take the matrix $ Q$ and remove the $ i$-th line for every $ i\notin U$ and remove the $ j$-th column for every $ j\notin V$. Note that the $ \pm$ sign in the sum depends on $ I$ and $ J$, but it is always a $ \plus{}$ sign if $ I \equal{} J$.

Applying the above formula [color=green][b](1)[/b][/color] to $ A \equal{} D$ and $ B \equal{} H$, we see that the sum simplifies, because all terms with $ I\neq J$ are zero (since the matrix $ D$ is diagonal, and thus only its diagonal minors, i. e. its $ \left(I,J\right)$-minors for $ I \equal{} J$, can have nonzero determinant). Hence, only the terms with $ I \equal{} J$ remain, and all of these terms are nonnegative (since the $ \pm$ sign is a $ \plus{}$ for $ I \equal{} J$, besides every diagonal minor of $ D$ is a diagonal matrix with no negative entries and thus has nonnegative determinant, and besides every diagonal minor of a skew-symmetric matrix is a skew-symmetric matrix and thus has a nonnegative determinant as well). Thus, $ \det\left(D \plus{} H\right)$ is a sum of nonnegative terms; one of these terms is $ \det D$ (namely, the term for $ I \equal{} J \equal{} \left\{1,2,...,n\right\}$), and another one is $ \det H$ (the one for $ I \equal{} J \equal{} \emptyset$), so we obtain $ \det\left(D \plus{} H\right)\geq\det D \plus{} \det H$, which translates to $ \det F\geq\det E \plus{} \det G$, which is even more than we wanted.

I hope this was of some help to you (sorry, I can't go into details now, since this proof would take 10 pages if detailed).

  darij

Let's try proving the theorem by induction on $ n.$

The theorem is vacuously true for $ n \equal{} 1,$ and we can prove it for $ n \equal{} 2$ as darij has done by assuming WLOG that $ E$ is diagonal which makes $ E \equal{} \begin{bmatrix}a & 0 \\
0 & b\end{bmatrix}$ and $ G \equal{} \begin{bmatrix}0 & \minus{} c \\
c & 0\end{bmatrix}$ in which case $ \det(E \plus{} G) \equal{} ab \plus{} c^2\ge ab \equal{} \det(E).$

Now suppose the theorem has been proved for $ n \minus{} 1$ and we have $ n\times n$ matrices $ E$ and $ G$. Proceed to matrices $ D$ and $ H$ as in darij's post.

$ D \plus{} H \equal{} \begin{pmatrix}d_1 & \minus{} v^T \\
v & D' \plus{} H'\end{pmatrix}$ where $ D'$ is an $ (n \minus{} 1)\times(n \minus{} 1)$ positive diagonal matrix and $ H'$ is that size of skew-symmetric matrix.

Perform row elimination on the first column to get

$ \det(D \plus{} H) \equal{} \det\begin{pmatrix}d_1 & \minus{} v^T \\
0 & D' \plus{} H' \plus{} d_1^{ \minus{} 1}vv^T\end{pmatrix}.$

From that, $ \det(D \plus{} H) \equal{} d_1\det((D' \plus{}d_1^{\minus{}1} vv^T) \plus{} H')\ge d_1\det(D' \plus{} d_1^{ \minus{} 1}vv^T)$

where the last inequality represents our inductive hypothesis. That inductive hypothesis applies because $ d_1^{ \minus{} 1}vv^T$ is a positive semidefinite symmetric matrix, so $ D' \plus{} d_1^{ \minus{} 1}vv^T$ is positive definite symmetric.

All we need to finish the proof is that $ \det(D' \plus{} d_1^{ \minus{} 1}vv^T)\ge\det(D'),$ or more generically, if $ A$ is symmetric positive definite and $ B$ symmetric positive semidefinite, then $ \det(A \plus{} B)\ge\det(A).$

Yes, the solutions are good. Under the same condition is it true that $ det F\ge det E \plus{} det G$?

[quote="miwalin"]Yes, the solutions are good. Under the same condition is it true that $ det F\le det E \plus{} det G$?[/quote]

Yes, as I said in my solution.

  darij


inequalities
function
analytic geometry
graphing lines
slope
inequalities proposed

Let $ a,b,c,d>0$ that satisfy $ a\plus{}b\plus{}c\equal{}1$. Prove:
$ a^3\plus{}b^3\plus{}c^3\plus{}abcd \ge min${$ \frac{d\plus{}3}{27}, \frac{1}{4}$}

[quote="bambaman"]Let $ a,b,c,d > 0$ that satisfy $ a \plus{} b \plus{} c \equal{} 1$. Prove:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} abcd \ge min${$ \frac {d \plus{} 3}{27}, \frac {1}{4}$}[/quote]
I have seen the problem $ a^3 \plus{} b^3 \plus{} c^3 \plus{} abcd \ge min${$ \frac {d }{27},\frac{1}{9}, \frac {1}{4}$
The both proof is used pqr or schur three agree with  2 case $ d\ge 3.d\le 3$

[quote="bambaman"]Let $ a,b,c,d > 0$ that satisfy $ a \plus{} b \plus{} c \equal{} 1$. Prove:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} abcd \ge \min\{\frac {d \plus{} 3}{27}, \frac {1}{4}\}$[/quote]
Let $ \min\{\frac {d \plus{} 3}{27}, \frac {1}{4}\} \equal{} k$ then
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} abcd \ge k\Leftrightarrow a^3 \plus{} b^3 \plus{} c^3 \plus{} abcd \ge k(a \plus{} b \plus{} c)^3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left((1 \minus{} k)a^3 \minus{} 3ka^2b \minus{} 3ka^2c \plus{} \left(\frac {d}{3} \minus{} 2k\right)abc\right)\geq0.$
If $ c\rightarrow0$ and $ a \equal{} b \equal{} \frac {1}{2}$ then $ 2(1 \minus{} k) \minus{} 6k\geq0,$ which gives $ k\leq\frac {1}{4}.$
But for $ k \equal{} \frac {1}{4}$ your inequality is equivalent to
$ \sum_{cyc}\left(\frac {3}{4}a^3 \minus{} \frac {3}{4}(a^2b \plus{} a^2c) \plus{} \frac {3}{4}abc \plus{} \left(\frac {d}{3} \minus{} \frac {1}{2} \minus{} \frac {3}{4}\right)abc\right)\geq0,$
 which true for $ \frac {d}{3} \minus{} \frac {1}{2} \minus{} \frac {3}{4}\geq0$ ( and not true for  $ \frac {d}{3} \minus{} \frac {1}{2} \minus{} \frac {3}{4}<0$ ), which gives $ d\geq\frac {15}{4}.$
Id est. it remains to check that $ \frac {\frac {15}{4} \plus{} 3}{27} \equal{} \frac {1}{4},$ which indeed true.

Hello arqady, why do you take $ a\equal{}b\equal{}\frac{1}{2}$ while $ c\rightarrow{0}$? Doesn't it refer to a particular case?? :huh:
Sorry if I sound stupid....

[quote="Sunkern_sunflora"]Hello arqady, why do you take $ a \equal{} b \equal{} \frac {1}{2}$ while $ c\rightarrow{0}$? Doesn't it refer to a particular case?? :huh:
[/quote]
Because it proves the inequality. :wink:

[quote="Sunkern_sunflora"][quote="arqady"]Because it proves the inequality. :wink: [/quote]
I know, I know! Still, isn't there any general procedure so that we won't have to assume something for our proof? :maybe:[/quote]
Of cause there is a general procedure!
We need to look when the equality holds. Usually it happens when $ a \equal{} b \equal{} c$ or $ a \equal{} b,$  $ c \equal{} 0.$
Indeed we see that for $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$ the expression $ a^3 \plus{} b^3 \plus{} c^3 \plus{} abcd$ equal to $ \frac {d \plus{} 3}{27}$ and
for $ a \equal{} b \equal{} \frac {1}{2},$ $ c \equal{} 0$ the expression equal to $ \frac {1}{4}.$
It makes a proof starting. :wink:

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} abcd\geq min\left\{\frac {d \plus{} 3}{27},\ \frac {1}{4}\right\}\ \cdots [*]$.

$ a \plus{} b \plus{} c\geq 3\sqrt [3]{abc}\Longleftrightarrow 1\geq 3\sqrt [3]{abc}\ \therefore 0 < abc\leq \frac {1}{27}$.

Lemma: $ \boxed{a^3 \plus{} b^3 \plus{} c^3 \plus{} \frac {15}{4}abc\geq \frac {1}{4}}\ with\ a,\ b,\ c > 0,\ a \plus{} b \plus{} c \equal{} 1$

$ 1^\circ\ \frac {d \plus{} 3}{27}\geq \frac {1}{4}\Longleftrightarrow d\geq \frac {15}{4}$

$ [*]\Longleftrightarrow a^3 \plus{} b^3 \plus{} c^3 \plus{} abcd\geq \frac {1}{4}$

$ \Longleftrightarrow (abc)d \plus{} a^3 \plus{} b^3 \plus{} c^3\geq \frac {1}{4}$.

Proof: Let $ f(d) \equal{} (abc)d \plus{} a^3 \plus{} b^3 \plus{} c^3$, Use the linear function for $ d$, since the slope is  positive,

$ f(d) \geq f\left(\frac {15}{4}\right) \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} \frac {15}{4}abc\geq \frac {1}{4}$.

$ 2^\circ \ \frac {d \plus{} 3}{27}\leq \frac {1}{4}\Longleftrightarrow 0 < d\leq \frac {15}{4}$

$ [*]\Longleftrightarrow \left(abc \minus{} \frac {1}{27}\right)d \plus{} a^3 \plus{} b^3 \plus{} c^3\geq \frac {1}{9}$.

Proof: Let $ g(d) \equal{} \left(abc \minus{} \frac {1}{27}\right)d \plus{} a^3 \plus{} b^3 \plus{} c^3$, Use the linear function for $ d$, since the slope is  negative or 0,

$ g(d)\geq g\left(\frac {15}{4}\right) \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} \frac {15}{4}abc \minus{} \frac {5}{36}\geq \frac {1}{9}$. $ Q.E.D$.


calculus
derivative
function
vector
linear algebra
matrix
probability

Suppose $ g(x)$ is a twice differentiable function in the open neighborhood of the origin, and let $ \mathbf{x}$ and $ \mathbf{y}$ be two unit vectors in $ \mathbb{R}^n$, not necessarily orthogonal to each other. Show that in an open neighborhood of the origin, $ \partial_{\mathbf{x}} \partial_{\mathbf{y}} g(\mathbf{0}) \equal{} \partial_{\mathbf{y}} \partial_{\mathbf{x}} g(\mathbf{0})$. Why is it to be expected that if $ \mathbf{0}$ is a local maximum of the function $ g$, then usually we will have $ \partial_{\mathbf{x}} \partial_{\mathbf{x}} g(\mathbf{0}) < 0$? Demonstrate an instance where this is not so.

uhhhh sry maybe this should be moved to the unsolved probs section.

sry.

For a sufficiently smooth $ f,$ we have

$ f(a\plus{}\mathbf{u})\equal{}f(a)\plus{}Df\cdot\mathbf{u}\plus{}\frac12\mathbf{u}^TH\mathbf{u}\plus{}o(\|\mathbf{u}\|^2)$

where $ H$ is the Hessian matrix consisting of all of the second partial derivatives of $ f.$ Note that $ Df$ and $ H$ are evaluated at $ a$ in the statement above. Note also that $ H$ is a symmetric matrix.

One directional derivative works out to

$ \partial_{\mathbf{x}}f(a\plus{}\mathbf{u})\equal{}Df\cdot\mathbf{x}\plus{}\frac12(\mathbf{x}^TH\mathbf{u}\plus{}\mathbf{u}^TH\mathbf{x})\plus{}o(\|\mathbf{u}\|)$

$ \equal{}Df\cdot\mathbf{x}\plus{}\mathbf{x}^TH\mathbf{u}\plus{}o(\|\mathbf{u}\|)$

where we just used that $ H$ is a symmetric matrix.

A second directional derivative gives

$ \partial_{\mathbf{y}}\partial_{\mathbf{x}}f\equal{}\mathbf{x}^TH\mathbf{y}$

Since $ \mathbf{x}^TH\mathbf{y}\equal{}\mathbf{y}^TH\mathbf{x},$ it's not going to matter which order we take this in. That answers your first question.

As for the second question: there is no good reason at all to suppose that we "usually" have $ \partial_{\mathbf{y}}\partial_{\mathbf{x}}f(0)<0$ at a maximum. In fact, $ \mathbf{x}^TH\mathbf{y}$ is an inner product of $ \mathbf{x}$ and $ \mathbf{y}.$ It's bilinear. Merely taking $ \minus{}1$ times one of those vectors will reverse the sign of that inner product. Generically, if we choose $ \mathbf{x}$ and $ \mathbf{y}$ at random, the probability that $ \partial_{\mathbf{y}}\partial_{\mathbf{x}}f(0)<0$ should be $ \frac12,$ and that's the same whether the point is a maximum, a minimum, a saddle point, or not even a critical point at all.


Does anyone know exactly how much mathematical knowledge you need to compete at IMO level? Like for example how much of number theory or calculus do you need to know? Also is it better to just buy problem books and constantly solve problems  or to study a subject for e.g. residue classes and solve problems related to it?
Thanks for your reply. :)

[quote="mathematica"]Does anyone know exactly how much mathematical knowledge you need to compete at IMO level?[/quote]Does anyone know how many Fermat primes are out there? No, and this is exactly the same answer to your question :lol:

Anyway, I can tell you this for sure: you don't need to know any calculus. Stick around this forum (and site) and you will gradually get more and more prepared for the IMO ;)

So, you basically mean the imo comittee has no regulations for the subject of the questions only if it is in elementary mathematics?

I may not be the right guy to answer your questions, but one thing is for sure, and this one can't be denied by no one: [b]practice makes perfect[/b]! I had very very good results when i used this as a motto :)

In conclusion, the only thing you have to do is... practice and solve as many problems as you can.

I agree with icx. What you have to do is practising a lot, and learning some theory gradually, not everything together, just practise, practise, practise, and add gradually some important theory (teorems, formulas)


abstract algebra
linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra unsolved

Prove that the sum of two algebraic integers is an algebraic integer ?

Use the criterion that an element $ a$ in a number field $ K$ is an algebraic integer iff there is a finitely generated $ \mathbb Z$-module $ M$ in $ K$ with $ aM\subset M$.

I am not able to get it if you don't mind send the complete proof :(javascript:emoticon(':(')
Sad
    I don't know what is Z module

If $ \alpha, \beta$ are algebraic integers, then the matrix of multiplication by $ \alpha \plus{} \beta$ in $ \mathbb{Q}[\alpha, \beta]$ has integer coefficients, so its characteristic polynomial is monic.


geometry
geometric transformation
rotation
conics
parabola
analytic geometry
trigonometry

[quote="Calculus"]Is there any way to apply a rotation to a graph? For example, if you had a parabola, y = x^2, how would you manipulate the algebraic equation to tilt the graph 45 degrees clockwise?[/quote]

I completely doubt this is correct, but for a graph like $y=x^2$, you could take the line of symettry and simply rotate it to the graph $y=x$ and draw the graph from there. If not, you migth be able to use parametrics to somehow rotate the graph with different angle measurements....  :?  :surrender:

[quote="scorpius119"]for general rotations, polar coordinates can be used. If the angle of rotation is $\alpha$ clockwise, then $\theta \rightarrow \theta+\alpha$ if $x=r\cos \theta$ and $y=r\sin \theta$.

Using trig addition, $x\rightarrow x\cos \alpha -y\sin \alpha$ and $y\rightarrow y\cos \alpha +x\sin \alpha$.[/quote]

wait sorry. I don't see what you mean by $\theta \rightarrow \theta+\alpha$.

[quote="boxedexe"]The original angle is $\theta$ and the added angle is $\alpha$, making angle of $\theta +\alpha$.  As $\theta$ acquires the value of $\theta +\alpha$...

Masoud Zargar[/quote]

Oh, right. Hehe, thanks :dry:

[quote="jackb1115"]In a similar vein, does anyone know how to easily do reflections across a line.

For instance, AIME 1988 #14:

The graph on xy = 1, when reflected across the line y = 2x, leaves the graph

12x^2 + bxy + cy^2 + d = 0

Find b*c

JB[/quote]

Since $xy=1$ is its own reflection across $y=x$ and the composition of two line reflections is a rotation by twice the angle between the lines, you can treat the reflection as a rotation by an angle of $2\theta-\frac{\pi}{2}$, $\theta=\tan^{-1}2$.
[hide="If you want the answer"]84[/hide]

I'm not sure what to do with an arbitrary function, though.


Suppose I have two sets, $ A$ and $ B$, and that $ |A| \equal{} m$ and $ |B| \equal{} n$. Suppose that $ |A \cap B| \equal{} k$. Now I wanna choose $ x$ elements from $ A$ and $ y$ elements from $ B$ (I can only choose the same elements once). In how many ways can I do that?

Let $ S_a\subset A$ and $ S_b\subset B$ such that $ |S_a| \equal{} x$ and $ |S_b| \equal{} y$. And let $ N$ be the number that we are supposed to find.

Verify that if $ |S_a\cap S_b| \equal{} i$ then $ N_i \equal{} \binom{m}{x \minus{} i}\binom{n}{y \minus{} i}\binom{k}{i}$.

Therefore $ N \equal{} \sum_{i \equal{} 0}^{\min\{x,y,k\}}\binom{m}{x \minus{} i}\binom{n}{y \minus{} i}\binom{k}{i}$.

That should be $ \binom{m \minus{} k}{x \minus{} i} \binom{n \minus{} k}{y \minus{} i} \binom{k}{i}$ ways to choose your elements such that $ i$ of them lie in the intersection, and then you sum over $ i$.


Let $m$ be a positive integer and let the lines $13x+11y=700$ and $y=mx-1$ intersect in a point whose coordinates are integers. Then $m$ is:

$\text{(A)} \ 4 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 6 \qquad \text{(D)} \ 7 \qquad \text{(E)} \ \text{one of the integers}~ 4,5,6,7~\text{and one other positive integer}$

[hide]$mx-y=1 \text{ and } 13x+11y=700$. Then by multiplying the first equation by 11, and adding the two we get
$(13+11m)x=711$. Then 711 must be divisible by $13+11m$ for x to be an integer. Since the factorization 0f 711 is $3^2*79$,
$13m+11=79$, $m=6$. Then $x=9 \text{ and } y=53$[/hide]


hey...I recently tok the open exam, and currently studying for the semifinal exam. Can anyone give me tips on how to study for the semifinal exam, because i know it's way harder than the open exam. Thanks~

:(  I gave my teacher the USABO site and stuff but he didn't register.

I'd go over evolution, and organisms (protists, plants, eubacteria, etc.).

where can I check? I cant seem to find the loation


linear algebra
matrix
algebra unsolved
algebra

Note that in $ P$, there is no constraint on $ a,b,c$. Hence, in order to find the minimum, we wish to find the region of $ (x,y,z)$ so that $ a,b,c$ exists.

A little bit transformation gives:

$ \begin{pmatrix} 1 & \minus{}z & \minus{}y \\ \minus{}z & 1 & \minus{}x \\ \minus{}y & \minus{}x & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} \equal{} \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

We then come up with the condition for the above linear system: $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$.

Now, it is easy (at least it is easier).

[quote="mathangel"]Note that in $ P$, there is no constraint on $ a,b,c$. Hence, in order to find the minimum, we wish to find the region of $ (x,y,z)$ so that $ a,b,c$ exists.

A little bit transformation gives:

$ \begin{pmatrix} 1 & \minus{} z & \minus{} y \\
\minus{} z & 1 & \minus{} x \\
\minus{} y & \minus{} x & 1 \end{pmatrix} \begin{pmatrix} a \\
b \\
c \end{pmatrix} \equal{} \begin{pmatrix} 0 \\
0 \\
0 \end{pmatrix}$

We then come up with the condition for the above linear system: $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$.

Now, it is easy (at least it is easier).[/quote]
How did you come up with the condition? (Also remember that $ a,b,c > 0$)

[quote="pro_math"]The positive real numbers $ a$, $ b$, $ c$, $ x$, $ y$, and $ z$ are such that $ \begin{cases}cy \plus{} bz \equal{} a \\
az \plus{} cx \equal{} b \\
bx \plus{} ay \equal{} c\end{cases}$
Find the smallest possible value of the expression
$ P \equal{} \frac {x^2}{1 \plus{} x} \plus{} \frac {y^2}{1 \plus{} y} \plus{} \frac {z^2}{1 \plus{} z}$[/quote]
We want to solve this in $ (a,b,c)$:
$ \left( \begin{array}{ccc|c}
\minus{}1 & z & y & 0  \\
z & \minus{}1 & x & 0  \\
y & x & \minus{}1 & 0  \end{array} \right)$

But it is equivalent to (By $ zP_1 \plus{} P_2, yP_1 \plus{} P_3$):
$ \left( \begin{array}{ccc|c}
\minus{}1 & z & y & 0  \\
0 & z^2\minus{}1 & x\plus{}yz & 0  \\
0 & x\plus{}yz & y^2\minus{}1 & 0  \end{array} \right)$

Thus the system is equal to(By $ \minus{}\frac{z^2\minus{}1}{zy\plus{}x}P_3 \plus{} P_2$):
$ \left( \begin{array}{ccc|c}
\minus{}1 & z & y & 0  \\
0 & 0 & x\plus{}yz \minus{} \frac{(y^2\minus{}1)(z^2\minus{}1)}{x\plus{}yz} & 0  \\
0 & x\plus{}yz & y^2\minus{}1 & 0  \end{array} \right)$

Since $ a,b,c > 0$ and $ c(x\plus{}yz \minus{} \frac{(y^2\minus{}1)(z^2\minus{}1)}{x\plus{}yz}) \equal{} 0$ we know:
$ x\plus{}yz \minus{} \frac{(y^2\minus{}1)(z^2\minus{}1)}{x\plus{}yz} \equal{} 0$ $ \iff$

$ (x\plus{}yz)^2 \equal{} (y^2\minus{}1)(z^2\minus{}1)$ $ \iff$

$ x^2\plus{}y^2z^2 \plus{} 2xyz \equal{} y^2z^2\minus{}y^2\minus{}z^2 \plus{} 1$ $ \iff$

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$.

Including $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$, we just have to solve:
$ \left( \begin{array}{ccc|c}
\minus{}1 & z & y & 0  \\
0 & x\plus{}yz & y^2\minus{}1 & 0  \end{array} \right)$

But it's easy to see that it has positive solutions when $ 0 < x,y,z < 1$. (But none can be zero!!!)

So the problem is equivalent to:
Find min of $ P \equal{} \sum_{cyc} \frac{x^2}{x\plus{}1}$ when $ x,y,z > 0$ and $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$.


[quote="#H34N1"]What is the maximum number of distinct lines that can be drawn through ten distinct points if no three of the points are collinear?[/quote]

[hide=" :maybe: "]wouldnt it just be ${10\choose 2}=45$?[/hide]

Oh. I shouldn't have been that naive. Why didn't my brain function :wallbash:  :wallbash:

[quote="indianamath"]10C2=45

The message is too small. Please make the message longer before submitting.[/quote]

uhh...what message?  are you talking about his second post?  i dont think it was to small.  without your complaint, your post would have been 7 characters long :wink:

[quote="10000th User"]What is the maximum number of regions created by these 10 lines? :) Now that's something to do.[/quote]

it would be 56

i know its 0 lines-1, 1 line-2, 3 lines-4, so the pattern is 1,2,4,7,11,16,22,29,37,46,56... so 56 is 10 lines (you just add the number of lines to the previous number) but can someone tell me the generic formula for caclulation regiions-thx

It's just one more than $\binom{n+1}{2}$.

yeah. I drew out 10 points and drew 45 lines in 2 minutes :rotfl: Thank goodness #30 was simple.

Extension:
10 lines and 3 circles make how many intersection points?

[quote="machack"]10 lines and 3 circles can produce 111 intersections at most
and
i derived (by myself yay :D ) another fomula for finding the number of regions  :arrow: (x^2+x+2)/2
now lets prove why that equals 1+(x+1C2)  (:P just proved it as i was typing)

x+1C2 = ((x+1)*x)/2=
(x^2+x)/2[/quote]
Dude, how did you get 111?

[quote="jhredsox"][quote="Fanatic"]Editing extension:

10 lines and 3 circles make [b]at most[/b] how many intersection points?[/quote]

[hide="fshizzle (by bpms)"]

line, line= up to 1 intersection
line, circle= up to 2 intersections
circle, circle= up to 2 intersections

45 intersections at most w/ 2 lines
60 intersections at most w/ 1 line, 1 circle
6 intersections at most w/ 2 circles

adds up to 111 intersections at most[/hide]
jorian[/quote]
Dude, I'm not understanding your solution.
111 is correct.
I think you are misinterpreting the question, but maybe I'm just being stupid.

i don't THINK i am. 2 lines can only intersect once at most. a line and a circle as well as 2 circles can only intersect twice at most, right?

jorian

ask Mr. Rusczyk as I got this from the geo book

Yep, I get the same answer using the same method as jorian.


number theory
least common multiple
greatest common divisor
LaTeX
prime factorization

What is the least common multiple of 51, 68 and 85?

This is my answer $ \boxed {1020} $ :D :)

[hide="Legit Solution"]
So, the GCD (Greatest Common Denominator) of $51$, $68$, and $85$ is $17$. And, like james4l said, it would be
$17 \times{3} = 51$
$17 \times{4} = 68$
$17 \times{5} = 85$

This is where james4l got it wrong.
The LCM (Least Common Multiple) of $3$, $4$, and $5$ is $\bf{60}$, not $120$. So, the correct answer for the LCM of $51$, $68$, and $85$ would be $17 \times{60} = \boxed{1020}$ [/hide]

[hide="weird but works"] So we can take 2 numbers first. Let's take 51 and 68. If I simplify 51/68 I get 3/4..... Then I multiply 51/68(lazy to do LaTex) by reciprocal of 3/4. I get 204/204 meaning the LCM of those 2 numbers is 204. Now we have 204 and 85. 85/204=5/12....so i do 85/204 * reciprocal of 5/12...I get 1020/1020 so the answer is $\boxed{1020}$  [/hide]

[quote="james4l"]51 = 3*17
68 = 4*17
85 = 5*17
lcm: 3*4*5*17=120*17=2040[/quote]

I believe 3*4*5=60 and not 120[quote="markoboy"][hide="this is my answer"]you will 1st want to get the prime factorization of both all the #'s

51=3*17

68=4*17

85=5*17

so the lcm is 4*3*5*17 =2040 or $\boxed{2040}$[/hide]

[b][i][u]rate my post thanks !![=[/u][/i][/b]

please be fair though [= thanks!

 :play_ball:  :play_ball:  :play_ball:  :play_ball:[/quote]

How did you get 4*3*5*17=2040 because it seems like 4*3*5*17=1020


A square piece of paper is folded in half to form a rectangle. If the resulting rectangle has a perimeter of 18 inches, what is the area of the original square, in square inches?

Call the side length of the square $ x$. The rectangle has dimensions $ 0.5x$ by $ x$. So we have:

$ 0.5x \plus{} 0.5x \plus{} x \plus{} x \equal{} 18$

$ 3x \equal{} 18$

$ x \equal{} 6$

Thus our answer is $ 6^2$ or $ \boxed{36}$.


function
real analysis
real analysis unsolved

find all continuous functions $f: [0,1]\rightarrow[0,1]$ such that $f(f(f(x)))=x$ for all $x$ in $[0,1]$

[hide] my answer- $f(x)=x$[/hide]

$f$ has an inverse, namely $f \circ f$, so it's bijective.

Since it's continuous, it must be strictly monotonic.

(a) $f$ is strictly increasing;
If $f(x) > x$ for some $x$, then $f(f(x)) > f(x)$ and $x = f(f(f(x))) > f(f(x))$, from where we get that $x > x$, contradiction.
If $f(x) < x$ for some $x$, then we get a contradiction in a similar manner.
Hence, $f(x) = x$, for all $x \in [0,1]$, which indeed works.

(b) $f$ is strictly decreasing.
We have that $f(0) = 1, \, f(1) = 0$. But $f(f(f(0))) = f(f(1)) = f(0) = 1$, contradiction.


MATHCOUNTS
AMC
AMC 8
trigonometry
trig identities
Law of Cosines
bump

what is the first time between 4:00 and 5:00 that the hour hand and the minute hand are exactly 10 degrees apart

i think this might be a coincidence but all the mathcounts problems i've seen that involved the hands of a clock being certain degrees apart is always from 4 to 5.

yeah, they had something about it on this years amc8 also

I believe I also saw something like this on the MAML Olympiad this year...oh wait not really: "28.) The hands of a clock are 4 inches and 5 inches long, respectively. At some time between 1:35 and 2:00, the tips of the hands are 8 inches apart. The time is closest to which of the following?
A.) 1:40
B.) 1:55
C.) 1:43
D.) 1:48
E.) 1:41"

Use law of cosines and then when the angle is found,

:theta: = l 30 - 5.5m l where m is the number of minutes.  Spoiler: [hide]B[/hide]

Oooooh, law of cosine, didn't see that...I just guessed :P

I guess it's just like this problem after you get the angle.

[quote=HamstPan38825]It's quite simple when you think of it as a speed problem, with the differences of the speeds between the hour and minute hand being 5.5 degrees/minute.[/quote]

how do u possibly even find a 2003 thread... no reason to bump
don't know law of cosines so can't say if its trivial

Actually, there is an entire topic called "clock problems" in Chinese olympiad mathematics.


trigonometry
calculus
calculus computations

Can someone show me the steps to solve this differential equation?

$ x^2\sin(\frac{\pi}{4} \plus{} \ln(x))y'' \plus{} 2x\cos(\frac{\pi}{4} \plus{} \ln(x))y' \plus{} [\cos(\frac{\pi}{4} \plus{} \ln(x)) \minus{} \sin(\frac{\pi}{4} \plus{} \ln(x))]y \equal{} 0$

Would it help to simplify the above differential equation to:

$ x^i[y''(1 \minus{} i) \minus{} 2y'(1 \plus{} i) \plus{} 2yi] \plus{} x^{ \minus{} i}[y''(1 \plus{} i) \minus{} 2y'(1 \minus{} i) \minus{} 2yi] \equal{} 0$

It seems like there's still something I can do with this, any ideas?

Well I noticed that if we define $ A \equal{} y''(1 \minus{} i) \minus{} 2y'(1 \plus{} i) \plus{} 2yi$, we can rewrite the differential equation in terms of just $ A$:

$ Ax^i \plus{} \overline{A}x^{ \minus{} i} \equal{} 0$

I'm not sure if this helps though.


function
inequalities proposed
inequalities

Prove or Disprove that
$ 2\sqrt{ab\plus{}bc\plus{}ca} \ge (a\minus{}b)^2 \plus{} (b\minus{}c)^2 \plus{} (c\minus{}a)^2$
for $ a,b,c$ positive reals greater than 1.

[quote="geniusbliss"]Prove or Disprove that
$ 2\sqrt {ab \plus{} bc \plus{} ca} \ge (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2$
for $ a,b,c$ positive reals greater than 1.[/quote]

not true, try $ a\equal{}2,b\equal{}c\equal{}6$.

for what condition  of $ a,b,c$ is it true ? or is it not true for any $ a,b,c$ ?

You can possibly check the degree of the function first and after that, suggest the condition of the problem. :)


function
geometry
geometric transformation
reflection
angle bisector
algebra unsolved
algebra

What does the graph of [tex] y=x^y [/tex] look like and what are some properties of it?

We got that it's just a point (1,1) because you can extend it to [tex]y={x^{x^{x^{{...}^y}}}}[/tex] but what do you guys think?

[quote="vinny919"]What does the graph of [tex] y=x^y [/tex] look like and what are some properties of it?[/quote]

If you want to know how it will "look like", then you can just plot the function [tex]y=\sqrt[x]{x}[/tex] and then reflect this plot in the first angle bisector (i. e. in the line with the equation y = x). What you get is the plot of the equation [tex]x=\sqrt[y]{y}[/tex], or, equivalently, [tex]y=x^y[/tex].

  Darij

[quote="vinny919"]because you can extend it to [tex]y={x^{x^{x^{{...}^y}}}}[/tex] but what do you guys think?[/quote]

can you explain this please? I can't follow :?

Let [tex]y=x^y[/tex].

We then replace the [tex]y[/tex] on the right side with [tex]x^y[/tex] giving [tex]y=x^{x^y}[/tex].

Repearing this process gives [tex]y=x^{x^{x^y}}[/tex], and so on producing the statement [tex]y=x^{x^{x^{...^y}}}[/tex] above.

So then, the function changes and $ y\equal{}x^{1/x}$ doesn't work since it will be something different.  So again, would that mean that there is only one point on this graph?  Is this function diffeerentiable?


topology
calculus
derivative
integration
analytic geometry
function
LaTeX

After processing the document it said "oerrors, 0 warnings, 2 bad boxes, overfull" .
They both are related to some big equations arrays. How can I solve the problem?

Thanks
Jean

Maybe the equations need to be broken up but you will need to post the code to be sure of what needs to be done.

It says: overful boxes in paragraph at lines 256-256

I copied line 256: %section{The fractional jet bundle of order $ k$
on a differentiable manifold; geometrical
objects}"

Actually line 256 is only objects}

and the same lines 182-196: 182 is 283756 line and 196 is label {16} line , I pasted more.

 %begin{equation} \label{ZEqnNum283756} x^{i} (0)=x^{i},\quad  y^{i(\alpha a)} =\frac{1}{\Gamma (1+\alpha a)} D_{t}^{\alpha a} x^{i} (t)\left|_{t=0} \right. , \end{equation}
for $ i \equal{}\overline{1, n}$ and $ a \equal{}\overline{1, k}$.\\
By definition, the fractional osculator bundle of order
$ r$ is the fibre bundle
$ (Osc^{\alpha k}(M), M)$ where
$ Osc^{\alpha k}(M) \equal{}\bigcup_{x_{0}\in M}Osc_{x_{0}}^{\alpha k}(M)$
and
$ \pi_{0}^{\alpha k}: Osc^{\alpha k}(M)\to M$
is defined by
$ \pi_{0}^{\alpha k}([\rho ]_{x_{0}}^{\alpha k}) \equal{} x_{0}$,
$ (\forall )[\rho ]_{x_{0}}^{\alpha k}\in Osc^{\alpha k}(M)$. \\
\indent For $ f\in{\mathcal F}(U)$, the fractional derivative
of order $ \alpha$, $ \alpha\in (0,\; 1)$, with respect to
the variable $ x^{i}$, is defined by
%begin{equation} \label{16)} \begin{array}{l} {(D_{x^{i} }^{\alpha } f)(x)=} \\ {\frac{1}{\Gamma (1-\alpha )} \frac{\partial }{\partial x^{i} } \int _{a^{i} }^{x^{i} }\frac{f(x^{1} ,...,x^{i-1} ,s,x^{i+1} ,...,x^{n} )-f(x^{1} ,...,x^{i-1} ,a^{i} ,x^{i+1} ,...,x^{n} )}{(x^{i} -s)^{\alpha } }  ds,} \end{array} \end{equation}
where $ x^{i}$ are the coordinate functions on $ U$,
$ \frac{\partial }{\partial x^{i}}$,

If you put the code like this [code]$Osc^{\alpha k}(M) = \bigcup_{x_{0}\in M}Osc_{x_{0}}^{\alpha k}(M)$[/code] between code tags it won't be converted by AoPS and show the dollar signs.

The problem is that LaTeX is having difficulty breaking the lines as there are long words or equations where it wants to put a new line. So here are some options:
1. do nothing if the output looks OK to you and is easy to read;
2. put in \\ line breaks to make the lines shorter so LaTeX doesn't have to choose what to do;
3. use wider margins - but I expect that's not a good idea;
4. use displayed equations rather than inline which breaks things up;
5. use \raggedright which stops LaTeX trying to right justify each line and allows it to break lines at other places.

Be guided by published papers in your subject area which you think look good.

You will find that this code is much easier to read with none of the formulae touching each other. Displaying them makes everything eaiser on the eye. I also suspect Osc should be upright not italics. The code doesn't have any bad boxes
[code]\begin{raggedright}
\section{The fractional jet bundle of order $k$ on a differentiable manifold; geometrical objects}
\end{raggedright}
\newcommand{\Osc}{\text{Osc}}

\begin{equation}
	\label{ZEqnNum283756}
	x^{i} (0)=x^{i},\quad y^{i(\alpha a)} =\frac{1}{\Gamma (1+\alpha a)} D_{t}^{\alpha a} x^{i} (t)\left|_{t=0} \right. \end{equation}
for $i = \overline{1, n}$ and $a = \overline{1, k}$.

By definition, the fractional osculator bundle of order $r$ is the fibre bundle $(\Osc^{\alpha k}(M), M)$ where
\[\Osc^{\alpha k}(M) = \bigcup_{x_{0}\in M}\Osc_{x_{0}}^{\alpha k}(M)\]
and
\[\pi_{0}^{\alpha k}: \Osc^{\alpha k}(M)\to M\]
is defined by
\[\pi_{0}^{\alpha k}([\rho ]_{x_{0}}^{\alpha k}) = x_{0},
(\forall )[\rho ]_{x_{0}}^{\alpha k}\in \Osc^{\alpha k}(M)\].

For $f\in \mathcal{F}(U)$, the fractional derivative of order $\alpha, \ \alpha\in (0,1)$, with respect to
the variable $x^{i}$, is defined by
\begin{multline} \label{16)}
	(D_{x^{i}}^{\alpha} f)(x)= \\
	\frac{1}{\Gamma (1-\alpha )} \frac{\partial}{\partial x^{i}} \int _{a^{i}}^{x^{i}}\frac{f(x^{1} ,...,x^{i-1} ,s,x^{i+1} ,\dots,x^{n})-f(x^{1} ,\dots,x^{i-1} ,a^{i} ,x^{i+1} ,\dots,x^{n})}{(x^{i} -s)^{\alpha}}ds
\end{multline}
where $x^{i}$ are the coordinate functions on $U$, $\dfrac {\partial }{\partial x^{i}}$, [/code]


The third term of an arithmetic sequence is $ 5$ and the sixth term is $ \minus{}1$.  Find the twelfth term of this sequence.

First we want to find the common difference. We know the sixth term is $ \minus{}1$ and the third term is $ 5$.

$ \minus{}1 \minus{} 5 \equal{} \minus{}6$

$ \minus{}6 \div 3 \equal{} \minus{}2$

So the common difference is $ \minus{}2$. We want to find the 12th term, so we can use the formula:

$ a_n \equal{} a_1 \plus{} d(n\minus{}1)$

to find the $ n$th term, where $ a_1$ is the first term of the sequence, $ a_n$ is the $ n$th term, and $ d$ is the common difference.

Before we can substitute numbers in, we need to find the first term. We know the third term is 5, and the common difference is -2, so when we work backwards, we find the first term is 9. Now we can substitute our numbers.

$ a_{12} \equal{} 9 \plus{} (\minus{}2)(12 \minus{} 1)$

$ a_{12} \equal{} 9 \minus{}2(11)$

$ a _{12 }\equal{} 9 \minus{} 22$

$ a_{12 }\equal{} \minus{}13$

So our 12th term is $ \boxed{\minus{}13}$


Don't we need to know the radius of the circle?

well the first circle and CE I have to figure out the length of CE
really bad ways of explaining and expressing my answer... :(
[hide]
each section made by two letters is 30 degrees and we know a triangle is made by 180 degrees so this is a 60-60-60 and CE=1
the area of the segment of the circle made by CE is 1/6(pi) and since the area of an ellipse is major*minor*pi we can say since it is  1/2 of that or 1/2*1/2*sqrt3/2*pi
1/6-sqrt3/8(pi)
[/hide]

[quote="13375P34K43V312"][hide="funcia read"]If the radius is 1, the length of the inscribed equilateral triangle is not 1. Think of drawing in DE and CD and dropping a perpendicular from D. This gives two 30-60-90 triangles, then you can find the sidelength from there.

And this problem does not involve ellipses, only circles and equilateral triangles are neccessary.[/hide][/quote]
[hide="i see that but what is the problem  :( "]
wait a minute by length do you mean the one that is sqrt3/2
or the one that is 1/2 and if so i think i knew that please show me which part of my thing i went wrong because i am sure that i could try to fix it
[/hide]

[quote="13375P34K43V312"]Six points $A,B,C,D,E,F$ are placed going clockwise equally spaced around a circle. The circle with center $A$ and radius $AC$ is constructed. Find the area of the smallest region enclosed by the first circle and arc $CE$.

btw how do you draw an arc over two letters in LaTeX?[/quote]
okay then maybe this works
[hide]
5/6(pi) is the area of 5/6 of circle
sqrt3/4 is area of that equilateral triangle add them and subtract them from pi to get
pi/6-sqrt3/4[/hide]
and could you clarify? I am saying the first circle has radius 1

I think a diagram would be helpful to solve the problem...

[quote="sonny"][hide="solution?"]Is the radius of the original circle 1 or the new circle 1? If AC=1 then i believe the area of the region shared by the two circles is $\frac{\pi-\sqrt{3}}{2}$ and the area of the smallest is $\frac{3\sqrt{3}-\pi}{6}$.... [/hide][/quote]
why is $\frac{\pi-\sqrt{3}}{2}$ the area of the region shared by the 2 circles?

[quote="funcia"]i can draw one for you[/quote]

Sure......

As for ProtestanT's question It is the area because the area of the share region is the equilateral triangle plus three small sector things. The area of the sectors are all the same. Now the area of one equilateral triange and one sector thing is $\frac{\pi}{6}$ Subtract the area of the triangle from that to get the area of the small sector thing. Multiply that by three since there are three of those and add the area of the equilateral triange to the product......

it turned out in vain because i am so not skilled with protractor and compass

oh **** i hurt myself with this compass but anyone who does draw it please shade in the region we are finding because i think ...oh well...

i made one the big circle doesn't matter much does it?  i mean maybe i am wrong though because i hurt myself making it i'll scan it in right now :lol:
EDIT: uploaded!-tell me what is wrong

The correct answer has been given, but if you are struggling with what's going on this is the picture for this problem.  Hope that helps.

Just to make sure no one is confused, draw an equilateral triangle and find the areas of the 3 sectors.

[hide="the correct answer is"]$\frac{3\sqrt{3}-\pi}{6}$ or $\frac{\sqrt{3}}{2}-\frac{\pi}{6}$[/hide]

[quote="funcia"][quote="13375P34K43V312"]Six points $A,B,C,D,E,F$ are placed going clockwise equally spaced around a circle. The circle with center $A$ and radius $AC$ is constructed. Find the area of the smallest region enclosed by the first circle and arc $CE$.

btw how do you draw an arc over two letters in LaTeX?[/quote]
i believe that this should be worded differently if that diagram was correct....[/quote]

Yes I agree, it is a little confusing to understand but read it carefully.


I have 15 indistinguishable cherry flavored lollipops that I want to distribute among the 5 members of my school's math team. In how many ways can I do that?

So if you do balls an urns, would it be $ 15^5$ or $ 5^{15}$?

Assuming each member recieves atleast 1 lollipop, you can use stars and bars-

14 spaces between objects, put in 4 dividers

I believe this would be 14 choose 4-

$ (14*13*12*11)/(4*3*2)$

$ 7*13*11$

$ 1001$
maybe...

What if each member did not have to have at least one lollipop?

Well, according to AoPSWiki, this would be the indistinguishable to distinguishable situation.  We are distributing $ n$ indistinguishable items to $ k$ distinguishable items.  The equation for that is $ \binom{n \plus{} k \minus{} 1}{k \minus{} 1}$.  We get out of that $ \binom{19}{4}$, which is 3876 with a calc  :P .

This is nothing but Hockey-stick. 4 dividers here.


algebra
polynomial
function
real analysis
real analysis unsolved

The problem of finding $ f(x)$ such that f(R+) -> R+ and $ f(f(x)) \plus{} f(x) \equal{} 2x$ was proposed in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1115382#1115382[/url].  I showed that the only possible polynomials $ f$ could be are $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{} 2x \plus{} b$ (with certain restrictions).  I was wondering if this said anything about possible Taylor series representations of $ f$, and thus about what other functions that are differentiable (or maybe even continuous) would work for $ f$.  Any ideas?


geometry
geometric transformation
reflection
search
Support
function
invariant

Does God only value those who have a determined belief in Him, and send those who are not strong enough to believe in him to hell? If God existed, then why did he make me too weak to be faithful in Him?

Of course, god [b]probably[/b] does not exist.

Now, I'm not religious, so the following isn't a fact. I'm just stating what I know. (In fact, you can follow my lack of knowledge with the question marks noted)

Christianity was offered as a religion of an understanding god, and not of one where a god mercifully punishes with little rain for a harvest season or eternal suffering in hell. If you believed, and had faith, then He was with you, through troubled times. If you didn't have faith, then you are left to your own accord.

I recently read (today in fact) [i]Things Fall Apart[/i], a novel written based on a man in Nigeria who experiences the "white men invasion" and the spread of Christianity throughout his clan. The missionaries didn't force Christianity into the hearts and minds of the Nigerians, but they did set up a clan, and preach to all those who listened. Those that didn't simply didn't matter. This implies that God himself has no resentment or condemnment to those that didn't believe. (?)

Any fans of Stargate SG-1 (the latter seasons) will note that a religion can be spread many ways, but one of the more forceful is under threat of death and suffering unless you give in. This series is purely fiction of course, but it drives home the point that missionaries often reflect their gods - and if the Christian missionaries were not forceful or violent, then their God isn't either.

I think I've given my refutation of Pascal's Wager before. If I choose to believe in God X, either I'm wrong and nothing happens, I'm right and I win (very improbable), or I'm wrong and I piss off God Y who happens to not like believers of other gods but not particularly care about atheists (far more probable).

God loves atheists.  :wink:
Well it says so in the bible, he loves everybody. Why would a so loving person send someone to be tormented in flames for eternity? I think Hitler, Stalin and Mao would be jealous. If we combine all their victims, the suffer would not, at all, be near the suffer of just one of Gods victims.
Btw, why would you want to go to heaven? Sounds like a boring place. I don't want to spend an eternity just praising the lord and singing him songs (I'm not a good singer).

There are too many assumptions in Pascals wager. Maybe God loves atheists or maybe he doesn't. But then the question is: which one of all Gods is the existing one? The flying spaghetti monster?

[quote="7h3.D3m0n.117"]and if the Christian missionaries were not forceful or violent, then their God isn't either.[/quote]This depends entirely on which missionaries you're talking about. It is not a coincidence that the main cathedral in Mexico City is built on the site of a major Aztec temple- the Spanish were trying very hard to eliminate and replace the native religion.
On the frontiers, some of the larger missions also served as forts, offering military defense for the native converts.

There are a lot of missionary efforts throughout history that followed the conquerors and relied on state/military support. The conquering soldiers themselves weren't effective missionaries, but the noncombatant priests they brought along were.

Another question: if a supernatural deity existed that was obsessed with recognition of his existence by humans (as the Abrahamic god is), why does he provide [i]so much[/i] overwhelming evidence against this? He seems much more like a poltergeist than a god.

[quote="MysticTerminator"][quote="Altheman"]If God existed, then why did He make me too weak to be faithful in Him?[/quote]

I do not think he made you too weak to be faithful to him. Implicit in all people, even the very wicked, is the possibility that they can be saved through faith in the Lord Jesus Christ. It is up to each person and their command of their own free will, however, to accept the salvation that Jesus offers.[/quote]

Right but the point is that some people do not have "command of their own free will" to do that.

@worthawholebean: This doesn't have a whole lot to do with pascal's wager.

[quote="daboa"]Btw, why would you want to go to heaven? Sounds like a boring place. I don't want to spend an eternity just praising the lord and singing him songs (I'm not a good singer).[/quote]
Certainly sounds better than hell, though.

The point of my question was not really so much wanting to go to heaven or something like that, it is more why God would want to d a m n somebody in the circumstances that I described.

What if He doesn't? What if there is no hell? Who said there's a hell? The existence of God does not imply that God is an [color=red][b]-------[/b][/color].

Belief in God comes before belief in any particular religion, not as a result of.

[b]Proposition:[/b]

Agreement or disagreement with "filled-in" details of a religion (ex, having to sing songs in Heaven, hanging from your tongue in hell for lying, caste system, etc) should not impede one's belief in the concept of God, nor should any religion's attempt to personify the nature or will of God.

Start constructively, axiomatically.

Who said there is a hell? Jesus said there was in the New Testament.

What do you mean by "Start Constructively; axiomatically." It feels like an afterthought... Also just because this is the accepted method in mathematics doesn't mean it is the best method for approaching other subjects. Axioms are chosen because they are simple statements that produce systems with nice properties. I don't think this is the best way to approach questions of existence, purpose, morality, etc.

and

What do you mean by filled in details? In your opinion, where does one look to find answers about religion? Must a belief come from one's self? In your words, I don't accept the axiom that I know best. Do you consider the New Testament to be 'filled in'?

I partly believe that God is real, but I don't really know. Like 7h3.D3m0n.117, I'm stating from what I know.

God created us. He loves his creations. He wouldn't want to send his creations to a horrible place.

However, his relative Kronos (yes, I'm basing this off the Greek Gods and Titans as well) kept eating the Gods Zeus, Poseidon, and Hades (I think). Obviously, God gave Zeus, Poseidon and Hades powers to live inside Kronos's stomach, and he loves his mini-Gods. Soon, the power began to get manipulated by birth and soon there demi-Gods. Then, they kept going into the line of regular humans.

We're all very distant relatives of God. But don't push it. I think He would be willing to send someone to Hell if they did horrible things (like Kronos, when he ate his children, and Zeus threw him into that big pit). So Zeus would be a lesser God compared to God, but they could be almost the same thing because they were both willing to allow Kronos to be thrown in that deep dark pit.

Either my assumptions are incorrect or just inaccurate, I don't know.

[b]@ Altheman:[/b] My organization and articulation was a bit weird, so I'll try to clarify

[b]Background and adopted definition of religion - [/b]I am inspired by teachings of Ramakrishna and Vivekananda. Ramakrishna believed that religion is just a path to God, and that one can take different paths to God. Though he was a Hindu preist, he also achieved enlightenment through practice of Christianity, and through practice of Islam.

So now, what path do you take? While you can take suggestions from others, its your path.

[b]Why constructively/axiomatically?[/b]

Since you are a mathematician, you're more likely to develop a strong conviction in some "system [of belief] with nice properties" based on "simple statements," rather than some text of myths and lores*. If there's any truth to the "religions being a path to God" thinking, then you are best off taking a path you can actually follow and believe in. For someone else who is less logical, perhaps a blind-faith approach would have more potential. It depends on the person.

There are some less person-specific advantages, as well, I think. The resulting "religion" will be clean and self-consistent, and it is easier to capture the big picture. These properties are very helpful, especially when handling ethical/moral dilemas.

[b]What are the "axioms"?[/b]

Up to you! Concepts of God, kartavya/duty, existence, the soul, and such might go here. The concept that, with meditation/contact with one's conscience, one can distinguish between right and wrong, will go here. Heaven goes here, Hell goes here, if you want them. etc. And then, you test your system. And it will probably be refined as you experience life.

Notice that the God concept is here in the axioms. I don't think many people would want to spend the time trying to prove that God exists (FLT took long enough). However, notable others who have successfully taken "logical" approaches say that the most "perfect" and harmonious system comes from assuming the existence of God; that a God concept ties the picture together. It is for this reason that Vivekananda uses the phrase "realize God".

[b]Re: "Filled-in details":[/b]

By filled-in details, I vaguely meant anything that is too specific to be a part of your core philosophy. For example rituals, religious history, details about life in Heaven/Hell, scare tactics, and anything else that does not enrich the core philosophy. Which, in turn, depends on what you define to be your core philosophy/axioms. Someone might say, hey, what's wrong with gay marriage? Why can't two men love eachother? That's a trivial and useless restriction in the grand scheme of things. Another person might assume the man/woman definition of marriage from the start, and their thought process would be quite different. To me, much of the New Testament is "filled-in". Some others, who start off by assuming the whole Bible is the ultimate truth, would say otherwise.

[b]Re: answers, and beliefs:[/b]

Answers about religion can come from anywhere. Books, texts, other people, etc. They can also come from yourself, through meditation. But while a statement obviously doesnt have to originate from the self, ones belief in that statement does. It's your belief, its something very personal. One approach is the open-minded approach: find as many answers as possible, from external and internal sources, but then stringently test them through deep thought.

--------------------

I hope that made at least a little more sense...  :maybe:

*- I don't mean to discredit any particular work here

ok, so I'm fully aware there's a seeming contradiction here. yes, god is all-powerful. yes, humans still have the ability to make choices for themselves by exercising their free will. so what? (I also outlined a few pages ago how this may not be a contradiction)

furthermore, I didn't say everyone carries out god's will. some people reject god. they go to hell.

and you were 100% sure because it was essentially a question that had already been asked many times and to which I had given the answer many times.

Hmmm...I reread the thread, and it seems as if you started to give a nice full explanation, but it never came out explicitly. Are you referring to the quantum thing?

If so, it seems as if you are saying that God sets up the wavefunctions, but how they collapse at the end is up to us. So he knows all the information (omniscient) but doesn't know the outcome of everyone's lives. Is this correct?

I didn't mean to imply that there was a direct parallel so that we can directly apply quantum mechanics to the principles of god and the universe. I was merely explaining how something that's determined at one level of information may still be undetermined at another level of information, and thus how god may be omnipotent while allowing humans to retain free-will. Indeed, I actually doubt that you can directly apply the quantum mechanical example; I merely wanted to point out subtleties that people often overlooked when making their otherwise rigorous arguments.

Hello, I guess the topic has gotten pretty far away from what I started with. Anyway, just in case it isn't clear, I wasn't seriously debating that question.  I think arnav has argued the idea that there is a lot more to belief than the threats of hell that came about in the middle ages and such. However, this response isn't really satisfying because it doesn't answer what somebody who believes in god punishing people by sending them to hell and such. It is more avoiding the question than anything. It merely says that it is impossible for somebody to not accept god if they try.

I appreciated the response from Protestant. The idea of loving god being a choice is one of the fundamental questions that is related to my question . Furthermore, the fundamental way that god loves us is that he gives us free will. It was spot on! This is a core idea of the passage that inspired my question here.

Really what the question is driving at is this:  [i]The way that god loves us is that he gives us free will. When god sends someone to hell, it shows that god has created a person that he cannot extend free will (and love) to. In sending someone to hell, he is not giving them the choice of free will and thus is not loving his creation.[/i]

Anyway, the book that inspired this question offered a much better answer than anything that anybody gave in this discussion. It is given by a fictional elder (a leader of a monastery) who is dying.

The jist of his argument is that there are two groups of people who go to hell. In life it is possible for somebody to not accept god. Their nature made it such that they chose to defy god until the end. However, when they are facing god and the entrance to heaven, they hear of the people who prayed for them and cared for them.

At this, they feel a sense that they wish to embrace their fellow man. But in the depreciated state that they are in, they are unable to (for the acts of love are things that can only be truly exercised in life; in death, you cannot make such connections in the depreciated state; you are a soul possibly). Anyway, these people recognize that they have done wrong and they feel spiritual torment. They choose to suffer because they wronged others. And in this agony, they feel some solace even though they cannot ever truly love. It is a sort of resignation to humility.

Then there is the second class of people that are not moved by love. They feed on arrogance and pride and they are unreachable to the love of god. The sight of god angers them and they feed on their pride. By seeing god, they realize that they were wrong. Since they are motivated by pride, they would rather burn in hell than to live in the presence of god and his love. They surrender themselves to Satan and burn in anger, wishing that they would just not exist.

So anyway, I think that this is a sensible way to answer the question if you actually believe in these things. God is not choosing to send people to hell, they are, by their free will, choosing to go. That is, if one wants to go to heaven, they can. Hell is for the ones who don't want to go to heaven. Free will is preserved and those who are in hell have a reason for being there. God continues to give his gift of free will and thus still loves them. The reason that god cannot be reached is because the people are bewitched by the devil.

It seems that the first type of person would be inclined to choose the "wrong" path and go to heaven, since can they seriously feel so bad that they want to suffer forever? Apparently this sort of "cheating the system" is allowed? The second type choosing hell is more plausible.

Finally, something I can relate to. This isn't "cheating the system," because God is loving. He would never want someone to suffer forever. Hell is not a punishment that God sends people to.

This isn't my belief (I don't believe in hell at all), but if one does believe in hell, I think this perfectly gets around the contradictions we've been discussing.

I don't really see how that's anything other than a longer souped-up version of the simplicity that "if you do not accept god, it doesn't make any sense to spend the whole eternity accepting him"

first sentence of #139

YO DUDES CHECK IT OUT

http://www.smbc-comics.com/index.php?db=comics&id=752#comic

ahahahaha nice...here is another one that is related to what we've been saying lol

http://www.smbc-comics.com/index.php?db=comics&id=668#comic

i totally believe in heaven, but sometimes people say that's a bad thing. i dont get that...

if Satan was a "bad person" he would probably have a whole stash of drugs and other 'bad things"
In a way this would actually be a "bad persons" "heaven"

So god created a whole race of beings for the sole purpose of worshipping him, and then he condemns the ones who don't to an eternity in hell.  Really, it seems to me that god's a bit insecure... :maybe:

naw, man, god's pretty cool

also hey it turns out you're wrong

if you want to see why try reading the thread

all this sort of "god" stuff is really too cotradictory for me. why not just live off the facts? here is an interesting little argument a classmate made: if all the questions about god come with so much contradictory, the fact that god exists cannot be proved. this makes god a THEORY. if god created everything, this makes our entire world a THEORY. therefore, there exists NOTHING in the world that can be proved, and all matter technically defys god's "perfection". if god is not perfect, there is no god. in conclusion, GOD DOES NOT EXIST.

this is what i think: i believe that ther IS a god, not a god that can do anything or send people to hell and stuff. this god is only a sort of good luck charm, and is there watching you always. but he/she/it does not have a direct impact on your life. i dont care if people say he/she/it doesnt exist. i mean, maybe he/she/it really doesnt exist. i have no way of knowing, and will probably never know. if he/she/it sudenly left me forever, i would still be able to get on with my ordinary life and i would NOT GO TO HELL. mainly because i do not believe in a hell. my perception of death is that you do NOT go to heaven or hell, but instead get reincarnated, but you dont know it. kinda like buddism.

also, i agree with cyberspace.

dang we should probably keep this to the other thread in the polls forum. this specific topic's kinda played itself unless you're directly responding to AL THE MAN or something

so uh yeah post it in the other thread and then I will give you my totally AWESOME REBUTTAL which you TOTALLY WANT TO SEE cause IT IS AWESOME OKAY


trigonometry
geometry
trapezoid
circumcircle
radical axis
cyclic quadrilateral
power of a point

Let $C$ and $D$ be two points on the semicricle with diameter $AB$ such that $B$ and $C$ are on distinct sides of the line $AD$. Denote by $M$, $N$ and $P$ the midpoints of $AC$, $BD$ and $CD$ respectively. Let $O_A$ and $O_B$ the circumcentres of the triangles $ACP$ and $BDP$. Show that the lines $O_AO_B$ and $MN$ are parallel.

Let O be the midpoint of AB, the center of the semicircle (O) with the points C, D forming a convex quadrilateral ABDC. $OM = OA\ \cos \widehat{AOM}$ and $ON = OB\ \cos \widehat{BON}$. $ON \parallel AD$ is a midline of the triangle $\triangle ABD$ and $OM \parallel BC$ is a midline of the triangle $\triangle ABC$. Let $E, F$ be the midpoints of PC, PD. Denote $\alpha = \angle (OO_A, PE) = \angle (OM, PC)$ and $\beta = \angle (OO_B, PF) = \angle (ON, PD)$. From the trapezoid $OO_AEP$, $OO_A = \frac{PE}{\cos \alpha}$ and from the trapezoid $OO_BFP$, $OO_B = \frac{PF}{\cos \beta}$. Since $OA = OB = \frac{AB}{2}$ and $PE = \frac{PC}{2} = \frac{CD}{4} = \frac{PD}{2} = PF$, we get

$\frac{OM}{ON} = \frac{\cos \widehat{AOM}}{\cos \widehat{BON}},\ \ \ \frac{OO_A}{OO_B} = \frac{\cos \beta}{\cos \alpha}$

But from the circle (O) and form $AD \parallel ON, BC \parallel OM$,

$\alpha = \angle (OM, PC) = \angle (BC, DC) = \angle BCD = \angle BAD = \angle BON$

$\beta = \angle (ON, PD) = \angle (AD, CD) = \angle ADC = \angle ABC = \angle AOM$

Consequently, $\frac{OM}{ON} = \frac{OO_A}{OO_B}$, which means $O_AO_B \parallel MN$.

Let $ L$ be the midpoint of $ AB$ and $ Q \equiv \odot(ACP) \cap \odot(BDP),$ different from $ P.$ Clearly  $ O_AO_B \perp PQ$ and lines $ CA,DB$ and $ PQ$ concur at the radical center $ E$ of $ \odot(ACP),$ $ \odot(BDP)$ and the circumference with diameter $ AB.$ Since $ CD$ is antiparalell to $ AB$ with respect to $ EA,$ $EB,$ then $EP$ and $ EL$ are  isogonals with respect to $ \angle AEB,$ i.e. $ EP$ is the E-symmedian of $ \triangle EAB.$ Since  $ EMLN$ is a cyclic quadrilateral on account of $ \angle EML \equal{} \angle ENL \equal{} 90^{\circ},$ we have $ \angle MEL \equal{} \angle MNL \equal{} \angle PEN$ $ \Longrightarrow$ $ MN \perp PE.$ Hence $ O_AO_B \parallel MN.$

Let be the point $ Q\equiv (O_{a})\cap (O_{b}),$ the other than $ P,$ where $ (O_{a}),\ (O_{b})$ are the circumcircles of the triangles $ \bigtriangleup ACP,\ \bigtriangleup BDP,$ respectively.

The line segments $ AC,\ PQ,\ BD,$ as the radical axis of the circles $ (O),\ (O_{a}),\ (O_{b})$ taken per two of them, are concurrent at one point so be it $ R,$ as their radical center, as well.

$ \bullet$ Because of now, $ PM\parallel AD$ and $ PN\parallel BC$ and $ AD\perp BD$ and $ BC\perp AC,$ we conclude that $ P$ is the orthocenter of the triangle $ \bigtriangleup RMN$ and so, we have that $ RP\perp MN$ $ \Longrightarrow$ $ PQ\perp MN$ $ ,(1)$

From $ (1)$ and beause of $ O_{a}O_{b}\perp PQ$ we conclude that $ O_{a}O_{b}\parallel MN$ and the proof of the proposed problem is completed.

Kostas Vittas.

Take $\{T\}= BD \cap AC$, then easily $P$ is the orthocenter of $\triangle TNM$, $PT \perp MN$, but $P$ and $T$ belong to the radical axis of the 2 circles, hence $TP \perp O_AO_B$.

Best regards,
sunken rock

Project all 7 points $A$,$M$, $O_A$, $O$, $O_B$, $N$, and $B$ onto $CD$, then it's trivial!

Let $h_{X}$ denote the projection of $X$ onto $CD$. Let $O$ be the midpoint of $AB$. Then $O$,$O_A$, and $M$ are collinear, similarly $O$,$O_B$, and $N$ are collinear.
Then since $AO=OB$, $h_{A}P=h_{B}P$. Also, $CP=PD$, so $h_{A}C=h_{B}D$. Since $M$ and $N$ are the midpoint of $AC$ and $BD$, $h_{M}P=h_{N}P$. Also, since $O_A$ and $O_B$ are on the perp bisectors of $CP$ and $PD$, $h_{O_A}P=h_{O_B}P$. Thus,
$\frac{MO_A}{O_AO}=\frac{h_{M}h_{O_A}}{h_{O_A}P}= \frac{h_{N}h_{O_B}}{h_{O_B}P}= \frac{NO_B}{O_BO}$
So $MN \parallel O_AO_B$.

let AC,BD ntersect at E.circle $O_A,O_B$ intersect at P and T
it suffices to prove that PT is perpendicular to MN
by equal power theorem,E,P,T collinear
then $MP\bot EN,NP\bot EM$
so P is the orthocemter of triangle EMN
hence $EP\bot MN$


Can someone explain how to use them and what they are?

You really need to be more specific.  You can start [url=http://en.wikipedia.org/wiki/Sine#Sine]here[/url].


A runner hopes to complete the 10,000 meter run in less than 30.0 minutes. After exactly 27 minutes, there are still 1100 meters to go. The runner must then accelerate at 0.20 $ m/s^2$ for how many seconds in order to achieve the desired time?

So far the runner has run 8900 metres in 27 × 60 = 1620 seconds; that’s an average speed of $ \frac {8900}{1620}\approx5.49\ \textrm{m}\,\textrm{s}^{ \minus{} 1}$. However, if he is to run the last 1100 metres in 180 seconds, he will need to do it at an average speed of no less than $ \frac {1100}{180}\approx6.11\ \textrm{m}\,\textrm{s}^{ \minus{} 1}$. So if he accelerates for $ t$ seconds at $ 0.20\ \textrm{m}\,\textrm{s}^{ \minus{} 2}$, we have

$ 5.49 \plus{} 0.20\times t\ \geq \ 6.11$

Now solve for $ t$.

Jane_Bennet, the runner can run the last 1100 meters in 179.999 seconds if he wishes...

Also, the minimum value of $ t$ in your inequality doesn't yield an average speed of 6.11

[hide]
continuing at the rate the runner had been going,
he/she would have finished 8900*10/9 meters, leaving
1000/9 meters

so now, you need to find what's the minimum t that
$ 0.1t^2\plus{}(180\minus{}t)(0.2t) \ge 1000/9$

by quadratic, you get
$ t \ge 360 \minus{} 80\sqrt{11}$

hence, the runner needs to accelerate for at least
$ 360 \minus{} 80\sqrt{11} \equal{} 94.67001 sec$ at a rate of 0.2m/s^2

[/hide]

I see what I did wrong. I ignored the fact that after the runner has accelerated for a time, the distance left to cover and the time left to do it in both change and so the average speed needed also changes. I’ll try again.

Suppose he accelerates for a time $ t$. His speed goes up from $ 5.49$ to $ 5.49 \plus{} 0.20t$. The distance he runs in this time is $ 5.49t \plus{} 0.10t^2$. Thus the distance left to go is $ 1100 \minus{} 5.49t \minus{} 0.10t^2$. If he stops accelerating and runs this remaining distance at constant speed, his total time for the last 1100 metres will be $ t \plus{} \frac {1100 \minus{} 5.49t \minus{} 0.10t^2}{5.49 \plus{} 0.20t}$. Hence

$ t \plus{} \frac {1100 \minus{} 5.49t \minus{} 0.10t^2}{5.49 \plus{} 0.20t}\ \leq\ 180$

That’s what needs to be solved to find $ t$. Am I correct now? :oops:

NB: $ 5.49$ is an approximation of $ \frac{8900}{1620}$. The latter should be used to get a more accurate calculation.

The answer is 3.41 seconds, so I'm not sure. :maybe:

All right, let’s solve the quadratic inequality I got:

$ t \plus{} \frac {1100 \minus{} 5.49t \minus{} 0.10t^2}{5.49 \plus{} 0.20t}\ \leq\ 180$

To be more accurate, let’s replace the 5.49 by $ \frac{8900}{1620}\equal{}\frac{445}{81}$:

$ t \plus{} \frac {1100 \minus{} \frac{445}{81}t \minus{} 0.10t^2}{\frac{445}{81} \plus{} 0.20t}\ \leq\ 180$

Thus

$ \frac{445}{81}t \plus{} 0.20t^2 \plus{} 1100 \minus{} \frac{445}{81}t \minus{} 0.10t^2\ \leq\ 180\left(\frac{445}{81}\plus{}0.20t\right)$

It looks horrible, but it can actually be simplified to

$ 9t^2\minus{}3240t\plus{}10000\ \leq\ 0$

Hence

$ \frac{3240\minus{}\sqrt{3240^2\minus{}360000}}{18}\ \leq\ t\ \leq\ \frac{3240\plus{}\sqrt{3240^2\minus{}360000}}{18}$

that is to say

$ t\ \geq\ 3.11334$ (we can ignore the upper limit)

So my answer (to 2 d.p.) is 3.11 seconds (or 3.13 seconds if using 5.49 for the initial speed). I would say it’s pretty close. :wink:


Find all natural number whose first digit is 1, such that if we move the first digit to the last, the resulting number is three times greater than the original number.

[hide] Let the number be $ 10^n\plus{}c$ where $ c<10^n$.  Then we have

$ 3(10^n\plus{}c)\equal{}10c\plus{}1$

$ \Rightarrow \frac{3\cdot10^n\minus{}1}{7}\equal{}c$

It is obvious that $ \frac{3\cdot10^n\minus{}1}{7}<10^n$ for all non negative $ n$.  Thus we are only concerned with the fact that $ c$ is an integer.  If $ 3\cdot10^n\minus{}1\equiv0\pmod{7}$ then $ 3^{n\plus{}1}\equiv1\pmod{7}$.  Thus, $ n\equiv\minus{}1\pmod{6}$.  So we have all solutions take the form

$ 10^{6t\minus{}1}\plus{}\frac{3\cdot10^{6t\minus{}1}\minus{}1}{7}$[/hide]


probability
function
integration
Functional Analysis
complex analysis
complex analysis unsolved

Any idea of how to solve this problem?
Let $t>0$ and let the probability measure $dA_{t}$ given by: $dA_{t}(z)=\frac{t}{\pi}e^{-t|z|^{2}}dA(z)$, where $dA$ is the ordinary Lebesque measure on C.

Let $A^{1}=L^{1}(C,dA_{t}) \cap H(C)$, where $H(C)$ is the set of analytic functions on $C$, i.e., $A^{1}$ is nothing else but the set of analitic functions on $C$ which are integrable with respect to $dA_{t}$.

Let $f$ be an analytic function, let $a\in C$ fixed and let $g(z)=f(z+a)$.

Prove or disprove:

If $f\in A^{1}$ then $g\in A^{1}$.

In other words the implication is equivalent to:

$f\in A^{1}$ if and only if $g\in A^{1}$.

[quote="didilica"]Any idea of how to solve this problem?
[/quote]
Look at the functions $f(z)=e^{nz}$ with large $n$. They are all in your space. How does their norm change when you shift them?

Thanks for the interest in the problem.

Let`s see if I understood your argument:

$\int_{C}|e^{nz}|e^{-t|z|^{2}}dA(z)\leq \int_{C}e^{n|z|}e^{-t|z|^{2}}dA(z)<\infty$.

Now, let $g(z)=e^{n(z+a)}$. Then,

$\int_{C}|e^{n(z+a)}|e^{-t|z|^{2}}dA(z)\leq |e^{na}|\int_{C}e^{n|z|}e^{-t|z|^{2}}dA(z)<\infty$.
I do not think that this is a counterexample. Am I right?

I meant a slightly different thing: assume $a>0$. Then, if $f(z)=e^{nz}$, we have $g(z)=e^{na}f(z)$, so the norm of $g(z)$ is much ($e^{na}$ times) greater than the norm of $f(z)$. From here there is just one step to the counterexample.

We get that,

$\parallel g\parallel _{1}=e^{na}\parallel f\parallel _{1}$,

I still do not see the step to the counterexample.
The equality states, since $f\in A^{1}$, that $g\in A^{1}$.

I'm not perfectly sure I understand, but if I understand correctly, dA_t is essentially integrable because of exponential decay (or some level of decay).  Now, if the norm of g is large enough to offset that decay, perhaps, you have a counterexample.

I am probably wrong, because I don't really know what's going on, but that's a guess.

If you know some functional analysis, the end is easy: assume that $g(z)=f(z+a)\in A^{1}$ for all $f\in A^{1}$. Then the mapping $f\mapsto g$ is a linear mapping with closed graph, so it should be bounded (it is not too difficult to check that $A^{1}$ is a Banach space). But we saw that it cannot be bounded; the end.

If you don't know functional analysis, you can explicitly construct a counterexample as a sum $f=\sum_{k}c_{k}e^{n_{k}z}$ where $n_{k}$ is a fast increasing sequence of positive numbers and $c_{k}$ are chosen appropriately.


geometry
3D geometry
sphere
calculus
integration
special factorizations
calculus computations

Hello, i have stumbled upon these problems. please help me if you have the time :)

1. Sketch and describe in words the region in R3 represented by the equation
x^2 + y^2 + z > 2x.
I tried plottings points, but eventually got confused.. >.<

2. Find the volume of the solid that lies inside both of the spheres
x^2 + y^2 + z^2 = 4
and
x^2 + y^2 + z^2 + 4x − 2y + 4z + 5 = 0.

after completing the square for the 2nd equation, we get
(x+2)^2+(y-1)^2+(z+2)^2=0, which has center at (-2,1,-2)

now im stuck.. i dont know how to find the intersecting point, let alone the volume afterwards.

please help~~!!

thanks.

For #1, first complete the square:

$ x^2\minus{}2x\plus{}y^2\plus{}z>0.$

$ x^2\minus{}2x\plus{}1\plus{}y^2\plus{}z>1$

$ (x\plus{}1)^2\plus{}y^2\plus{}z>1$

$ z>1\minus{}[(x\plus{}1)^2\plus{}y^2]$

Does putting it in that form help you see?

#2: You got the constants wrong- that second sphere does have positive radius. Once you fix that, the only parts that matter are the two radii $ r,s$ and the distance $ d$ between the two centers. Choose a new axis along the line between the centers, which we put at zero and $ d$, then slice perpendicular to that axis. The area of a slice of the intersection is $ \pi\left(\min\left(\sqrt{r^2\minus{}x^2},\sqrt{s^2\minus{}(x\minus{}d)^2}\right)\right)^2$, wherever at least both of those square roots are nonnegative (from $ d\minus{}s$ to $ r$, unless $ |r\minus{}s|>d$ and one sphere is entirely contained in the other). Find where those two expressions are equal, so you can split the integral into two parts.

Kent, that does make it alot more simpler. I guess the only thing left is for me to plug and test. Any idea how to graph 3D's by hand?

jmerry, yeah you're right. the radius should be 2, instead of the 0 i put. I'll look into what you posted. Thanks.


hi,
prove that there exists an infinite natural numbers n s.t.
\[ \varphi(n) \equal{} \frac{n}{3}.\]

$ n\equal{}2^a3^b$ works (if $ a,b\geq 1$), since $ \phi(n)\equal{}\phi(2^a)\phi(3^b)\equal{}2^{a\minus{}1}2\times 3^{b\minus{}1}\equal{}2^a3^{b\minus{}1}\equal{}\frac{n}{3}$


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14405[/img]

 :(

You can download from the site http://ilib.mccme.ru the Manuel de construction des triangles, written by the brazilian amateur Luis Lopes. the item is the number 40 of the list. Most of the items of the list are in russian, but this is written in french. :D

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14423[/img]

 :clap: