$\log_{a}x = c$
then $a^c=x$.
so then $a^{\log_{a}x} = a^c = x$

I'm not good with proofs, so I'm not sure this is good. [hide]
$a^{\log_{a}x}=x$

$\log_{a}a^{\log_{a}x}=\log_{a}x$

$\log_{a}x*\log_{a}a=\log_{a}x$

$\log_{a}x=\log_{a}x$
[/hide]

looks good to me, although it wasnt the shortest.

[quote="AntonioMainenti"]yet another of my silly threads..

Find the fastest way to prove $a^{\log_a{x}}=x$, justifying each step.

[hide="my answer. give yours first!"]
$\log_a{x}=\log_a{x}\text{ (reflexive property) }\Longrightarrow{a}^{\log_a{x}}=x\text{ (definition of logarithm) }$  :D

I thought of this when I was tutoring a completely clueless girl from school in algebra 2 and I wanted to find the quickest way to explain it.
[/hide][/quote]

that wasnt a proof, really, especially if you want to show that to a "clueless" girl. however, i guess that is the fastest.

[quote="xxreddevilzxx"]that wasnt a proof, really, especially if you want to show that to a "clueless" girl. however, i guess that is the fastest.[/quote]Sure it was, I even justified both steps with fancy names.


inequalities
search
inequalities proposed

$ a,b,c > 0$ and $ c \equal{} max\{a,b,c\}$ then

$ (a \plus{} 3b)(b \plus{} 4c)(c \plus{} 2a) \ge 60abc$

[hide="Note:"]the original problem gives $ a \le b \le c$ but my solution uses only $ c \equal{} max\{a,b,c\}$[/hide]

[hide="solution"]
\[ (a \plus{} 3b)(b \plus{} 4c)(c \plus{} 2a) \geq 60abc
\]

\[ \Leftrightarrow 25 \plus{} 8\frac {a}{b} \plus{} 3\frac {b}{a} \plus{} 2\frac {a}{c} \plus{} 12\frac {c}{a} \plus{} 6\frac {b}{c} \plus{} 4\frac {c}{b}\geq 60
\]

\[ \Leftrightarrow 3\left( \frac {a}{b} \plus{} \frac {b}{a} \right) \plus{} 2\left(\frac {a}{c} \plus{} \frac {c}{a} \right) \plus{} 4\left(\frac {b}{c} \plus{} \frac {c}{b}\right) \plus{} 5\frac {a}{b} \plus{} 10\frac {c}{a} \plus{} 2\frac {b}{c}\geq 35
\]
Now since for positive real $ x$ we have $ \frac {1}{x} \plus{} x\geq 2$, the above is implied by
\[ \Leftarrow 5\frac {a}{b} \plus{} 10\frac {c}{a} \plus{} 2\frac {b}{c}\geq 17
\]
In the case $ b \geq a$ the inequality holds, since
\[ 5\frac {a}{b} \plus{} 10\frac {c}{a} \plus{} 2\frac {b}{c} \geq 5\frac {a}{c} \plus{} 10\frac {c}{a} \plus{} 2\frac {a}{c} \geq 14 \plus{} 3\frac {c}{a} \geq 17
\]
If on the other hand $ a\geq b$ it also holds, since
\[ 5\frac {a}{b} \plus{} 10\frac {c}{a} \plus{} 2\frac {b}{c} \geq 3\frac {a}{b} \plus{} \frac {a}{c}\left(2\frac {a}{b} \plus{} 2\frac {b}{a}\right) \plus{} 10\frac {c}{a} \geq 3 \plus{} 4\frac {a}{c} \plus{} 10\frac {c}{a} \geq 17
\]
[/hide]

[quote="JoeBlow"]
\[ \Leftrightarrow 3\left( \frac {a}{b} \plus{} \frac {b}{a} \right) \plus{} 2\left(\frac {a}{c} \plus{} \frac {c}{a} \right) \plus{} 4\left(\frac {b}{c} \plus{} \frac {c}{b}\right) \plus{} 5\frac {a}{b} \plus{} 10\frac {c}{a} \plus{} 2\frac {b}{c}\geq 35
\]
[/quote]

Nice solution. It can be streamlined as follows:

$ \frac{ 5\frac {a}{b} \plus{} 10\frac {c}{a} \plus{} 2\frac {b}{c} }{ 5\plus{}10\plus{}2 }\ge \sqrt[17]{ \left(\frac{a}{b}\right)^5\left(\frac{c}{a}\right)^{10}\left(\frac{b}{c}\right)^2}\equal{} \sqrt[17]{\frac{c^8}{a^{5}b^{3}}}\ge 1$ since $ c\ge a$, $ c\ge b$.

[quote="Altheman"]$ (a \plus{} 3b)(b \plus{} 4c)(c \plus{} 2a)\ge 4\sqrt [4]{ab^3}\cdot 5\sqrt [5]{bc^4}\cdot 3\sqrt [3]{ca^2} \equal{} 60\sqrt [60]{a^{55}b^{57}c^{68}}\ge 60\sqrt [60]{a^{60}b^{60}c^{60}}$ since $ c\ge a$, $ c\ge b$[/quote]

This is my first solution to it
[hide="The second"]expand and seperate all the terms (for instance $ 4a^2b \equal{} a^2b \plus{} .. \plus{} a^2b$, like Altheman did) and use AMGM to all of them, Its basically the same as the first.[/hide]

this problem is from Morocan finals of this year, and I doubt it's original, Some one know where is it originally?

[quote="FOURRIER"]

this problem is from Morocan finals of this year, and I doubt it's original, Some one know where is it originally?[/quote]
If i remember correctly,it is from Turkey National Olympiad of some year,maybe 1999.
After 45 minutes:
Here is the link:
http://www.mathlinks.ro/viewtopic.php?search_id=147490109&t=104592


x' = x(r-(2x^3)-(3x^2))*(r+x-1)

construct a bifurcation diagram in r,x plane.

find and classify each bifurcation.

describe the behavior fo the system between bifurcations.

This is the equation... its a little more clear... any help?

\[ x' \equal{} x(r\minus{} 2x^3 \minus{} 3x^2)(r \plus{} x \minus{} 1)
\]

[hide="bif points"]fixed pts are when $ 0\equal{}x(r\minus{}x^2(2x\plus{}3))(x\minus{}(1\minus{}r))$

so at $ (i) x\equal{}0, (ii) x\equal{} 1\minus{}r, (iii) r\equal{}x^2(2x\plus{}3)$

we know to check for bifurcations when:

$ (i)\equal{}(iii) \to r\equal{}0$
$ (i)\equal{}(ii) \to r\equal{}1$
$ (ii)\equal{}(iii) \to 2r^3\minus{}9r^2\plus{}13r\minus{}5\equal{}0$ which has one zero $ r_o \approx .6$
$ (iii)$ varies number of solutions :
note $ f(x)\equal{}x^2(2x\plus{}3) \to f'(x) \equal{} 6x(x\plus{}1) \to f' (x) < 0$ on $ (0,1)$
range of f is all reals, but number of solutions changes from one to three on $ [0,1]$
(note these two points r=0,r=1 are already considered)

so we can see that $ 0,r_o,1$ are the three bifurcation points.[/hide]

now you find out the types bifurcations..


Hello everyone. I began a few problems and I am getting a little confused. If anyone can help me I would appreciate it.

1. A 1.97 mF capacitor with an initial stored energy of 0.569 J is discharged through a 1.18 M resistor. (a) What is the initial charge (in C) on the capacitor? (b) What is the current through the resistor when the discharge starts? At time t = 602 s, find (c) the potential difference VC (in V) across the capacitor, [b](d) the potential difference VR (in V) across the resistor, and (e) the rate at which thermal energy is produced in the resistor.[/b]

[b]I answered parts (a) .0473 C (b) 2.034E-5 A and (c) 18.53 V. I am not sure how to answer questions (d) and (e). I would appreciate any feedback.[/b]

2. An underground cable 11 km long extends east to west and consists of two parallel wires, each of which has resistance 13 /km. An electrical short develops at distance x from the west end when a conducting path of resistance R connects the wires (Fig 2). The resistance of the wires and the short is 120  when measured from the east end and 200  when measured from the west end. What are (a)  x and (b)  R?

3. Two identical batteries of emf ε = 8.65 V and internal resistance r = 0.240 W are to be connected to an external resistance R, either in parallel (Fig. 3a) or in series (Fig. 3b). If R = 0.433 W, what is the current i in the external resistance in the (a) parallel and (b) in the series arrangements?

Thanks!

3a) The two batteries can be replaced with one of EMF $\mathcal{E}$ and internal resistance $r/2$ (resistors connected parallelly). The current passing through the circuit (and through resistor $R$) is $I =\frac{\mathcal{E}}{R+r/2}$.

3b) The batteries can be replaced with one battery of EMF $2\mathcal{E}$ and internal resistance $2r$ (resistors connected serially). The current through this circuit is $I = \frac{2\mathcal{E}}{R+2r}$.

1d) If you've calculated the current through the resistor (label it $I$), the voltage on the resistor is $U=IR$.

1e) The rate of change of thermal energy is merely the power of the resistor and is given by $P=UI=I^2R$ (expressed in Watts).

2) Um, what do you mean by 'measured' when you talk about resistance?

[quote="jeter123"]When they say measured I think they mean that the resistance read that particular amount. They didn't actually sit there with a ruler and measure.  :rotfl:[/quote]

Thanks for trying to help jeter123.

I have a general idea of what to do for question 2 but I'm not 100% sure.

2. An underground cable 11 km long extends east to west and consists of two parallel wires, each of which has resistance 13 /km. An electrical short develops at distance x from the west end when a conducting path of resistance R connects the wires (Fig 2). The resistance of the wires and the short is 120 when measured from the east end and 200 when measured from the west end. What are (a) x and (b) R?

My main question is does resistance vary linearly with the length of the cable? I believe that it does, and if this is the case I think my answer is correct, but [u]I would just like to know if someone could verify how to solve the problem for me. [/u]

Thank you!

Resistance $R_w$ of a conductor of a length l and constant cross-section area S (a wire) is equal to $R_w = \frac{\rho l}{S}$, where the resistivity $\rho\ [\Omega \cdot \text m]$ is a material property. What is given in the problem is $\frac{R_w}{l} = \frac{\rho}{S} = 13\ \Omega/\text m$.

Let $L = 11\ \text{km}$ be the cable length. Measuring from the east and west, respectively,

$R + 2(L - x) \frac{R_w}{l} = 120\ \Omega$

$R + 2x \frac{R_w}{l} = 200\ \Omega$

because in each case, we measure 3 resistors (2 of them equal to each other) connected in series. Substituting the values $L,\ \frac{R}{l}$ and solving the set of 2 equations for the unknowns R, x:

$R + 26(11 - x) = 120 \Longrightarrow\ R - 26x = -166$

$R + 26x = 200$

$2R = 200 - 166 = 34,\ R = 17\ \Omega$

$x = \frac{200 - R}{26} = \frac{183}{26} \approx 7.04\ \text{km}$

Thanks. That was very creative of you when setting up the problem, how you had ro / S = 13 ohm / m. That was the only thing that I got caught up with. Once you set that up, the rest of the problem works out nicely.

Thank you yetti. Have a good night.


linear algebra
matrix
integration
vector
function
superior algebra
superior algebra solved

Suppose $ a_{1},a_{2}, \ldots , a_{n} $ are natural numbers.
Let us define a matrix $ (a_{ij}), i,j = 1,2, \ldots ,n $, such that
$ a_{ij} = (a_{i} + a_{j})! $. Prove that the determinant of
$ (a_{ij}) $ is nonzero.

Sorry, I forgot to say that $ a_{1},a_{2}, \ldots , a_{n} $
are distinct.

Let $A$ the given matrix it is symetric real.

$(a_i+a_j)!=\Gamma (a_i+a_j +1)=\int_{0}^{+\infty}e^{-t}t^{a_i+a_j}dt$

$X$ vector column with entries

(x_1)
(x_2)
......
(x_n)

$X^{t}AX=\int_{0}^{+\infty}e^{-t}(\sum_{k=1}^{n}x_kt^{a_k})^2dt\geq 0$

If $X^{t}AX=0$ then $0 \leq \int_{[0,1]}e^{-t}(\sum_{k=1}^{n}x_kt^{a_k})^2dt \leq X^{t}AX =0$

the function with t variable $(\sum_{k=1}^{n}x_kt^{a_k})^2$ is $0$

so all $x_k=0$

$A$ is definite positive $det(A)>0$


AMC
AIME
USA(J)MO
USAMO
calculus
AMC 10
number theory

[quote="geometrictwilighter26"]How do I prepare? What classes should I take?[/quote]
This has seriously been the 15th or so thread asking this question since summer began. Just look through this forum, and you should find plenty of advice. The best advice, though, is
[hide="awesome advice"]DO PROBLEMS. You can find past AMCs, AIMEs, USAMOs, and tons more at the "Contests" section of this website (It's located in the top bar of the forums).[/hide].

[quote="geometrictwilighter26"]When do I sign up?[/quote]
Ask your school.

[quote="geometrictwilighter26"]Do I need teacher recommendation? Are there prerequisites?[/quote]
No, no. Anyone can take the AMCs, given that their school offers it.

[quote="geometrictwilighter26"]Also does anyone know a community college that offers classes for high school freshmen?[/quote]
This completely depends on where you live (unless if you mean online classes?), so you would be better off searching google or asking a school counselor.

I agree with anderson: Do problems. Also, if you can find a study buddy to help work through problems with you, that would help.

thank you very much. I will sure look around! But I bet I will find nothing better than awesome advice..  :wink: hee hee. :rotfl::huh: Sorry I make geeky jokes, no need to laugh if it is not funny to you!  :oops:

Oh, and about the community "coolege" classes, I searched on google 'till my fingers hurt, but they all required you to be junior or senior... school system needs a little help here in vegas..  :lol:  online classes will work fine if they are for credit.. Know any links? Anyhoo, I'll be sure to ask a counsler.. :)

thank you..

love,
 mia (the nerdy/ awesome/ so happy all the time chick)

Past pre-calculus, classes needed for AMC type problems are mostly those along the lines of discrete math (number theory, combinatoris, etc.) and are not in high school curriculums. So you won't get credit for any class preparing you for AMC.

[quote="geometrictwilighter26"]  :huh: :idea: :rotfl:

I am a high school freshman (yay.. finally) and I took Math Counts last year. I felt pretty happy and in my element so I think I might want to take the AMC 10 this year. How do I prepare? What classes should I take? When do I sign up? Do I need teacher recommendation? Are there prerequisites?

Also does anyone know a community college that offers classes for high school freshmen?

Girl down and freaking out!! Need help!  :weightlift:

Thanks!!![/quote]

1) The Introduction books give you most, if not all, knowledge for AMC 10
2) Maybe read [u]First steps for Math Olympians[/u] by Douglas Faires
3) Do practice problems
4) IF you want to push yourself more, get some AMC 12 practice too.


A farmer has a fixed amount of fence, say 1000 feet.  He wants to enclose an area of land next to a straight river.  He does not need to put a fence along the river.

  What is the maximum amount of land he can enclose with the fence (and the river)?

  Prove it!

 --hilbert
 :)

...oh, hilbert's your name...I thought this was one of the 23 problems

Anyways, [hide]I guess a half-circle is the best choice since circles tend to enclose the most area with the least perimeter. The radius of this half-circle is 2:pi:r/2=1000  r=1000/:pi: so the area is :pi:(1000/:pi:):^2:=1000000/:pi:[/hide]

That I can not do, I think you need to prove that the area increases as the number of angles increases, but that's just way too hard for me...

tare, are you an 8th grader?
What math are you taking?

yes, 10th Honors

P.S. It is customary here at AoPS to talk about private information and/or off-topic topics via Private Messaging, just to let you know. If you don't know how to use PM just ask at the Support Forum. Thanks.

1000 = 2w + l
l = length     w = width
w(1000 - 2w) = A(w)
w(1000 - 2w) = 0                w = 0, 500
Maximum is 500/ 2 = 250 = w.  A(w) = 125000
125000 ft:^2:
That's only if it's a rectangular area though, or else, a circle would give the maximum area.

Why is this in MATHCOUNTS? It's a variant on the isoperimetric problem, which is very difficult.

I think the concept is that circles and parts of circles enclose the most space in most cases, and to explore when htis stands and when this fails to stand. The prove it part was an excess though...

OK, here's an approach:

[hide]Suppose that you want to make two such, congruent fenced regions.  If you reflect the original across the river, it's easy to see that the result must be a circle for the maximum area.  Otherwise, you could increase the area of one or both of the halves.[/hide]

It's sort of informal, but it works.

-hilbert

p.s.  This extends easily to a rectangular region.  Play with it![/hide]

Gee... haven't we established that a semicircle will produce the greatest area and that a rectangle won't? Why play with it? Try this. What is the smallest area that can be enclosed? Let us assume the n sides of the chosen polygon are integers.

[list]If home is where the heart is and my heart is in this universe, what are you doing in my house?[/list][/quote]

[hide]

well, i suppose that the smallest would be a line right along the edge of the river

but if you actually wanted a polygon, then i would think it would be an equilateral triangle with length 500ft. then the area would be:

(250*sqrt(3)*500)/2 = 62,500*sqrt(3)

[/hide]

hmmm...interesting approach. i think you've got it the wrong way round though jeff. you think it'll be best if we have the least number of sides: 3. i think it'll be best if we have the largest number of sides, which is technically a semicircle. hmm...wonder how you could rigorize that. maybe take the limit of a polygon with n sides as n goes to infinity? nah...probably harder, judging by simon's post.

It is hard. It looks like a problem we wouldn't solve in my analysis class with Lebesgue integrals because it's too hard. Incidentally, I'm not convinced that the best we can do is a semicircle. It seems to me that maybe making the fence into an arc of a circle with less than pi radians is better. Or maybe half (or some other amount) of an ellipse is better. I really don't know.

I seem to remember that the circle gives you the best area-to-perrimeter ratio of any closed curve.  In that case, a semicircle should be the best by the argument of someone or other, which was to reflect it -- do the same thing on both sides of the river with twice as much fencing.

The question Jeff was answering with the equilateral triangle was the opposite question, Mystic -- what is the minimal area?  Of course, if you let me take any polygon I want, I can make the area as close to 0 as I choose.  If you force me to take a regular polygon, then I suppose that the equilateral triangle will be the worst.

yes, i was answering the least area question... i am also assuming you want a regular polygon because you can always make a triangle with sides 500, 500, and 1 so that would be pointless.

It seems that you get a better area:perimeter ratio as the number of sides in a regular polygon increases.
triangle area:perimeter = (s:^2::rt3:/4)/(3s) = s:rt3:/12 = .144s
square area:perimeter = (s:^2:)/(4s) = s/4 = .25s
hex area:perimeter = (3s:^2::rt3:/2)/(6s) = s:rt3:/2 = .866s
...
circle area:perimeter = (pi*r:^2:)/(2pi*r) = r/2 = umm...uh...oof...

The circle is effectively a polygon with an infinite number of sides, so if the trend increases, the area:perimeter ratio should increase, too (although it's hard to call right now b/c we have the area:perimeter ratio in terms of r.

[quote="mathfanatic"]The circle is effectively a polygon with an infinite number of sides[/quote]
yeah, that's what i was thinking (couple posts ago if you didn't see it), but that's not very mathematical. could we say that the limit as n goes to infinity of a regular polygon with n sides is a circle? well, that's still kinda intuitive. hrmph.

It's not as intutitive as you think -- both the perimeter and the area of regular polygons with large number of sides inscribed in or circumscribed about a circle approach the circumference of the circle.  I believe Archimedes (or someone in antiquity) actually did a pretty good calculation of pi using that fact.

Hmmmm I still don't understand it. How does that make sense @mathfanatic?


The results are out and i got 20 marks less than expected wth?  :bomb:   :cursing:   :censored:

Nothing for me to be proud of . 92.5 Percentile. WTH, I got 15 marks less than expected!!! :(:(

Esp in Physics I expected 50+32, I get 43+33??? God save the queen. My highest percentile is sadly in Maths 95.4, even after a bad performance.

I wonder why the cutoffs in Maths are this bad. The Ind Cutoff is 13-17.. :(

i got only 208
those bloody fools took my 10 marks away from me
a friend of mine has got i think the 2nd highest
hes got 266
the highest was 268
of course they r twins and they discuss each and every question
they make combined efforts
but it needs a applaud

IM VERY VERY VERY SORRY FOR U ANIT GRRRRRRRRRRR..........  :mad:  :mad:  :mad:

their rank prediction is top 20
my rank prediction is 101-200
neither of them r going to be true

Wth man they incr my mark in chem and reduced my mark soooooooooooooooo much in phy  :mad:  :mad: . And even maths marks was reduced  :mad:

It definitely isnt the worst, coz I got jus 20 marks below expected..
Got 98.1 percentile total 182..
Physics flopped this time 41 marks 62.9 percentile
Chem 52 marks 96 percentile
Maths 89 marks 100 percentile..(I jus don find this logically correct..)

i got 202 with 99.1 percentile. phys 96, chem 48, maths 58. hey anit_sahu how come you know your predicted rank???

just see the rank projection below your scorecard

Atleast after my FT-3 first paper not being evaluvated my fortunes have changed.
I got only a [b][i][color=red][size=200][u]PALTRY[/u][/size][/color][/i][/b] 176 :(  :(  :(
Got atleast 9 marks more but neways it was the worst.

[quote="hash_include"]all i can say is...
the subject shouldnt have been "wth..."
it should have been "WTF!!!!!!!!!!!!!????"  :mad:  :mad:

already their solns are PATHETIC and now they rob me of 16 marks!!!  :mad:[/quote]

Maybe u have a pt there  :wink: . Too pathetic too. They robbed me of 18 marks and 12 marks outta physics too the poor boy i am already in phy  :mad: . Even if they had changed the solns i shd have got atleast 20 marks more and wtbh i get 20 marks less  :mad:

I believe they have changed their solns. I see no other reason for some ppl losing marks and many others gaining.

Congrats Chappli for being Mathematics AIR 1. :DYes 100 percentile ironically is a mathematical fallacy. :D

Anit can you post your split up? And what do you mean they discuss everything. During the test or after the test?? :rotfl:

i m one of those people who got more than expected :)

but only 143 :(

but still thats more than expected!!

Got a 90.1 Percentile
Only thing to be proud of Chem-total 98 percentile
2nd paper 99.5 percentile


Supose you're leading a group of 8 Excursionists. You are in a Forest going to the camp, and there is still an hour of light. You find that there are 4 different paths and you don't know wich one you have to take. The only thing you konw is that the camp is 20 min away from where you are. There is a problem, there are two of the excursionists that are used to lie (they dont lie always)  and you don't know who they are. You have to go to the camp before it's dark. You have to divide the gruop into four  little groups such that after each groups takes a path and goes back(it takes 40 min, 20 to go 20 to go back) you can ask them, and be sure that you will find the camp before it's dark. How can you do it?

Hint: You always say the truth

[hide="solution?"]Divide into groups of 2,2,2, and 3. You are part of the group of 3.

Send each group down a path. Tell them to report back.

Note: We assume that your group goes down path "1" because you can name the paths any way you want.

[Scenario 1]: If: Your group finds the campsite, then you go back, tell everybody, and you arrive at the campsite safely. [Scenario 1 resolved]

[Scenario 2] If: Your group doesn't find the camp site, you know path 1 is clear. You then ask the people in your group whether you found the campsite. [Scenario 2.1] If: Both in your group say yes they found the campsite, they are both liars. You then ask all the other groups to figure out the campsite. Proceeding down the correct path, you arrive at the site safely. [Scenario 2.1 resolved].

[Scenario 2.2] If: One person in your group says that you found the campsite, then that person is the liar. You then ask all the other groups if they found the campsite. The group with both people agreeing that they found the campsite are the ones who went down the correct path. [Scenario 2.2 resolved]

[Scenario 2.3] If: Both people in your group say that they didn't find the campsite, then they are truth-tellers. You then ask all the other groups if they found the campsite.
[Scenario 2.3.1] If: In one group, both people claim to have found the campsite (and in the other groups, either no one or one person claims to have found it), go with the group in which both people agreed to have found the campsite. [Scenario 2.3.1 resolved]
[Scenario 2.3.2] If: Two groups claim to have found the campsite, then you know that the campsite is down one of those two paths. You also know that one of the groups contains the two liars (who went together). Because you don't know which group contains the liars (yet), you ask one person in one of the two groups a simple question (it could be anything, like whether you are a boy or a girl, or what color his hair is, or whether it is dark...etc.). [Scenario 2.3.2.1] If: The person lies, the group he comes from is the liar group. Follow the other group. [Scenario 2.3.2.1 resolved] [Scenario 2.3.2.2] If: The person tells the truth, the group he comes from is the group that found the campsite. Follow that group [Scenario 2.3.2.2 resolved]

I believe all the possibilities have been covered. Feel free to point out my mistakes or add suggestions/comments to my solution.[/hide]

It looks correct. :)
I can also ask Dr. Andy Liu for the intended solution(he usually has an ingenious solution). I see him often.

There is a problem, they dont lie always. I tought that it was clear but now i know it isn't.

So try to use the hint. It is useless that you go with someone, because you always say the truth. So try to solve it so.
And i think there's only one solution. :D

I talked to Dr. Liu yesterday. He told me that this is not his problem. The problem appears in "The Puzzling Adventures of Dr. Ecco" by Dennis Shasha, which I happen to own.

Below is Dr. Liu's solution:
[hide]First path: three excursionists
Second path:  three excursionists
Third path:  two excursionists
fourth path: me

Let's eliminate obvious cases.

case 1) I found the camp.
This case is too obvious.

case 2) The first group and the second group's opinions are split into 3:0.
We can trust both group.

case 3) The first group and the second group's opinions are split into 2:1.
We stil can trust both group.

The only non-trivial case is when
1) I couldn't find the camp.
2) (without loss of generality)The first group's opinions are split into 3:0, and the second group's opinions are split into 2:1.
3) The second group couldn't find the camp.

There are two cases:
case 1) The third group's opinions are split into 2:0.
Since there is a liar in the second group, the third group has at most one liar. Therefore, we can trust the third group.

case 2) The third group's opinions are split into 1:1.
Since there is a liar in the third group, the second group has at most one liar. Therefore, we can trust the second group.
ex)
First path: three excursionists   :noo:   :noo:   :noo:
Second path:  three excursionists  :noo:   :yup: :yup:
Third path:  two excursionists  :noo:  :yup:
fourth path: me  :noo:
Second path is the correct way.[/hide]
He asked me one more question: Is this still possible with 4 excurtionists and 100 minutes?


Songs of the Martian classical period had just two notes (let us call them $ x$ and $ y$) and were constructed according to rigorous rules:

1. the sequence consisting of no notes was deemed to be a song (perhaps the most pleasant);
2. a sequence starting with $ x$, followed by two repetitions of an existing song and ending with $ y$ was also a song;
3. the sequence of notes obtained by interchanging $ x$s and $ y$s in a song was also a song.

All songs were constructed using those rules.

(ii) Show that if there are $ k$ songs length $ m$ then there are $ 2k$ songs of length $ 2m \plus{} 2$. Deduce that for each natural number there are $ 2^n$ songs of length $ 2^{n\plus{}1}\minus{}2$.

Without using a linear recurrence relationship.


function
algebra unsolved
algebra
exponential equations

Solve the equation:
      $ 6^x+3^x=4^x+5^x$

Look at the monotonicity of the function $f_{S,a}:[\frac S2,S],\ f_{S,a}(x)=x^a+(S-x)^a$ by computing its derivative. It's strictly monotonic when $a\notin\{0,1\}$, and it's constant when $a\in\{0,1\}$.

Our problem translates to finding an $x$ for which $f(9,x)(5)=f_{9,x}(6)$. The monotonicity of functions of this type shows that this is only possible when $x\in\{0,1\}$.


inequalities
function
inequalities proposed

Dear kunny, all these inequalities follow directly from Jensen inequality for the power function or Power mean inequality.


search
MIT
college
Harvard
low quality discussion
2006 superior to 2021

[quote="shobber"]This problem has been solved by two Chinese Mathsmeticians. They are the first ones to get the Million Cheque of Clay.[/quote]

Really or joke????  Do you have any link???????

The poincare conjucture solved ????  :w00tb:

I seriously respect those chinese people who did it....:surrender:

A search in google doesnt seem to give any link to the solution though...... :(

It is the latest news. You can find the report here (in Chinese): http://news.sohu.com/20060604/n243550778.shtml

The goole search gives this link (in English): http://news.xinhuanet.com/english/2006-06/04/content_4644754.htm

Is there anything in English about their proof? When did they prove Poincare's conjecture? :?

Not to minimize the above work but most agree that quite a while ago, Grisha Perelman  had  (or knew ) the  proof (of Thurston's geometrization  conjecture which is an  extension of  Poincaré conjecture. ( Poincaré's conjecture is a special case of Thurston's conjecture)

Link:  http://mathworld.wolfram.com/news/2003-04-15/poincare/
(This was way back in 2002/2003 when he claimed that he had proof)

[quote="José"]I'm confused.

Who prove it, Perelman or the Chineese one?[/quote]

Well,  many believe that Perelman proved it, but he is  pretty reclusive, not too much interested in Clay Prize,   (not ready to follow their 'rules' and  publish everything etc...)   He was in USA a few years ago and gave  lectures at Stony Brook, MIT , Harvard etc, before 'disappearing'  but don't know if he ever published what was in his notes/preprints etc in a detailed manner.  (He may not  get Fields Medal , because one has to be under 40 years of age to get one, and it seems no one is sure how old he is  at present :) - He is  good but very  reclusive and strange !

The story, linked above sort of talks about it.

[quote="Buffalo"]The paper is on
ASIAN J. MATH. Vol. 10, No. 2, pp. 165–498

THE HAMILTON-PERELMAN THEORY OF RICCI FLOW
– THE POINCAR´E AND GEOMETRIZATION CONJECTURES

Perelman's proof has some mistakes and it's incomplete, there are still some gaps to the final proof. Mathematicians worked hard these years to fill these gaps. Zhu and his cooperators' work completes the proof, i.e. their work is the [b]final strike[/b] to the problem.
A group of mathematicians are checking the work now, if no mistake found, we can say that Poincare conjecture has been proved, after nearly 100 years, by mathematicians worldwide.[/quote]

God explanation!!! So, really, it is NOT officially proved.

[quote="José"]God explanation!!! So, really, it is NOT officially proved.[/quote]
Sorry, my mistake.
Zhu and Cao finished their work last summer, it passed the checking recently. It's hard to believe that all these mathematicians made mistake on this, so we are safe to say that Poincare conjecture has been proved.

Buffalo, do you have a link in english or spanish????????

It seems that Mr.Hamilton arrived in Beijing yesterday.

Here is some more information: http://blogs.guardian.co.uk/technology/archives/2006/06/06/has_poincares_conjecture_been_solved_the_conjecture_continues.html

what's up with the 14 year bump? Anyways yeah it's been solved

Yes... The poincare conjecture was solved by Grigori Perelman

[quote=pezzeepenguinstalker]Yes... The porncare conjecture was solved by Grigori Perelman[/quote]

iirc, you said that last time too :P

Great... I thought that was the spelling in the title was wrong.

@OP The statement is incorrect. The Poincare conjecture was first proven for all dimensions by Grigori Perelman(a Russian).


Prove that for $ a,b,c \in (0,1)$ we have:
$ (ab \plus{} bc \plus{} ca)(\sum \frac {1}{(\frac {c}{a} \plus{} \frac {c}{b}).(a \plus{} b)^2}) \geq \frac {9}{16}$


Prove $2\sqrt{x} < \frac{1}{\sqrt{x+1}-\sqrt{x}}$ for all $x > 0$.

Multiply denominator over to LHS, since $x>0$ we don't need to worry about sign changes

$2\sqrt{x^2 + x}<2x+1$

$4x^2+4x<4x^2+4x+1$

$0 < 1$

By rasionalising the denominator, RHS is equal to $\sqrt{x+1}+\sqrt{x}$ and $\sqrt{x}<\sqrt{x+1}$.
Result follows by adding $\sqrt{x}$ to both sides


Hi,

I have attached a picture of what I would like to create using LaTeX. I am namely interested in the diagram and don't really care about the annotations on the side.

Does anyone have any ideas on how I can go about creating the tree? (I am a complete amateur at using LaTeX and any help would be greatly appreciated!)

Many thanks for any ideas, beans.

If you search [url=http://tug.ctan.org/cgi-bin/search.py?firstNumber=0&metadataSearch=tree&numberResults=20]CTAN[/url] for [i]tree[/i] you will find 56 matches.
I haven't tried these but at least 3 of those matches look to do what you want:
[url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=qtree]qtree[/url]
[url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=synttree]synttree[/url]
[url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lingtrees]lingtrees[/url]
You'll have to choose the one you find easiest to use and install and which does what you want.

Hi Steve,

Thanks a lot. I'll have a proper look at them later tonight and let you know how I get on!

Hi Steve,

I had decided to use synttree but couldn't get it to work. (Upon clicking the .ins file it says my internet connection will be changed, after which nothing happens).

So then I decided to use lingtrees! I've saved the file on my desktop, but how do I get it to work? :oops: When I open one of the documents in the zip file, I can't build it as a pdf file because it says file can't be found! (Any ideas what I might be doing wrong this time round?)

EDIT: I have resolved the second problem--woohoo. (I came across a wonderful file called "... .log"! Any ideas why synttree might not be working? (It looked simpler to use!)

You didn't read the Readme instructions did you? Remember synttree is not a Windows program but can be used on any OS. It says:
[quote]To use the package, download synttree.dtx and synttree.ins. Then run

  tex synttree.ins

to generate synttree.sty. Place that file somewhere LaTeX will find it.[/quote]
BUT, don't do that. Just go to MiKTeX, Installed Packages, wait for it to find all the packages, right-click on synttree and then click Install.
OR Just put \usepackage{synttree} in your document, compile it and MiKTeX should offer to install it.

If neither of those work then follow the instructions above put synttree.sty in C:\<[i]your path to tex files[/i]>\tex\latex\synttree and then in MiKTeX, Settings press the Refresh FNDB button.

[quote="stevem"]You didn't read the Readme instructions did you? Remember synttree is not a Windows program but can be used on any OS. It says:

[quote]To use the package, download synttree.dtx and synttree.ins. Then run

  tex synttree.ins

to generate synttree.sty. Place that file somewhere LaTeX will find it.[/quote][/quote] I did read it, but didn't understand why the "tex" was written so just double clicked to install. :oops:

[quote]BUT, don't do that. Just go to MiKTeX, Installed Packages, wait for it to find all the packages, right-click on synttree and then click Install.
OR Just put \usepackage{synttree} in your document, compile it and MiKTeX should offer to install it.

If neither of those work then follow the instructions above put synttree.sty in C:\<[i]your path to tex files[/i]>\tex\latex\synttree and then in MiKTeX, Settings press the Refresh FNDB button.[/quote] Thanks a lot Steve!!  :first: I never thought of just writing \usepackage.

BTW, another dumb question but how do I find the "Installed Packages" thing, and the refresh button? (Maybe that will stop it asking me whether I want to install the same package whenever I compile). It's not in C:\program files\.... .

[quote="stevem"]Start, All Programs, MiKTeX 2.x then look for Browse Packages or for Settings[/quote] Cheers. I have just refreshed it now.

Another BTW, well actually I was wondering if you could "hold my hand" and help me install another package!

I am looking to install this package: http://tug.ctan.org/tex-archive/macros/latex209/contrib/trees/tree_tex/ and have downloaded the file onto my desktop.

Should I just write \usepackage{treetex} in my document, or do I have to do other things. (I mean in general too, would I just write usepackage and let it ask to be installed?)

[quote="beans"]I am looking to install this package: http://tug.ctan.org/tex-archive/macros/latex209/contrib/trees/tree_tex/ and have downloaded the file onto my desktop.[/quote]You don't need to do that. MiKTeX will download the latest versions for you.
[quote="beans"]Should I just write \usepackage{treetex} in my document, or do I have to do other things. (I mean in general too, would I just write usepackage and let it ask to be installed?)[/quote]Either do that or look for it in Browse Packages.

If you notice it comes from the latex209 directory. That is a very old version of latex so the chances of tree_tex working properly in a modern latex may be low. If you look at the README it's dated 1993.

[quote="stevem"]You don't need to do that. MiKTeX will download the latest versions for you.[/quote] What exactly do you mean by this? How will MiKTeX do it? I realise now, from the packages manager, that there are a million packages which haven't been installed by MiKTeX. Am I right in assuming that whenever I download a package that it picks it up?

[quote]Either do that or look for it in Browse Packages.[/quote] OK will do.

[quote]If you notice it comes from the latex209 directory. That is a very old version of latex so the chances of tree_tex working properly in a modern latex may be low. If you look at the README it's dated 1993.[/quote] I hadn't paid attention to the date! (In the documentation for synttree, the treetex package had been mentioned so I decided to see what it was about. Won't bother now--thanks.)

[quote="beans"]Am I right in assuming that whenever I download a package that it picks it up? [/quote] No. What happens is that MiKTeX uses its own files which it downloads from one of its own sites. It can't use what you've downloaded because it needs instructions on where to install everything and commands to run on the files.

An alternative, if you are happy with downloading 600MB, then you can install the Complete version of MiKTeX which means that everything is downloaded and installed ready to be used.

[quote="stevem"] No. What happens is that MiKTeX uses its own files which it downloads from one of its own sites. It can't use what you've downloaded because it needs instructions on where to install everything and commands to run on the files.

An alternative, if you are happy with downloading 600MB, then you can install the Complete version of MiKTeX which means that everything is downloaded and installed ready to be used.[/quote] That explains a lot! Thanks. I think I'll look into download the complete version. :thumbup:


geometry
calculus
function
integration
analytic geometry
LaTeX

More than half my MV calculus class is stumped at this question:

Find an approximate value, with four exact decimals, of the surface of the function e^(-X^2-Y^2) on a circle of radius 2.

For my part, I know that del Z/del X = -2Xe^(-X^2-Y^2) and del Z/del Y = -2Ye^(-X^2-Y^2). Square both (as required by the integral of a surface) and you have 4(X^2 + Y^2) e^2(-X^2-Y^2) + 1 under a square root.

Through the passage to polar coordinates, this becomes R sqrt (4R^2 x e^(-2R^2)) and I know that integrating D(theta) from 0 to 2 pi = 2 pi. Therefore, any answer to the integral of R sqrt (4R^2 x e^(-2R^2)) DR from 0 to 2 will be multiplied by 2 pi. However, I suspect R sqrt (4R^2 x e^(-2R^2)) DR has no simple primitive.

It asks for an approximate value. Express the integrand as Taylor series and integrate term by term. Then plug in your bounds while cutting off the series at say $ n = 5$ or something.

I didn't check if your set-up was correct. Please use $ \text{\LaTeX}$ to make things more readable (enclose expression in $ \$\$$ signs; see below for an example).

$ \int\limits_{0}^{1} \sum\limits_{n=1}^\infty \frac{1}{(n+x)^2} \, dx = 1$.

I'd call this question off because the teacher told us that there was no way for us to solve it by hand.

You can give arbitrarily close approximations using what I said, but yes, perhaps it's not that fun to go up to $ n \equal{} 20$ or so by hand.

You got to $ 2\pi\int_0^2\sqrt{1\plus{}4r^2e^{\minus{}2r^2}}\,r\,dr.$

At that point, I like making a modest substitution; I think I'll let $ u\equal{}2r^2,$ which gives me

$ \frac{\pi}2\int_0^8\sqrt{1\plus{}2ue^{\minus{}u}}\,du$

Yes, at that point, we see that it's not likely to have a closed-form antiderivative. And with that square root, power series are a rather unattractive option - the power series would be rather annoying to attempt to find more than two or three terms of. But it is a nice smooth function, for which all numerical integration methods should work about as well as they're supposed to. The simplest thing for you to do would probably be to use a calculator with a built-in numerical integrator.

I'm sitting at my desk with a spreadsheet program open. My more-or-less standard choice under that set of circumstances is Romberg integration. (Which is related to the trapezoidal rule and Simpson's rule but goes beyond them - it's kind of a neat technique.)

The whole thing should have taken me 5 minutes, but I made an entry error in the spreadsheet somewhere that needed to be tracked down - it just wasn't behaving right.

Using 33 function evaluations to set up the Romberg table, I get a value of $ 1.11235\cdot(4\pi)$ or $ 13.97825.$

I never even heard of Romberg integration anyway. The only thing I knew about is that the answer had multiples of pi in it because of the integral in theta.

The [url=http://en.wikipedia.org/wiki/Romberg%27s_method]Wickipedia article[/url] is on the brief an unenlightening side, but all the information is there. I personally find it very easy to remember and very easy to implement on a spreadsheet. I know it because I taught a beginning numerical analysis course a couple of times.

There are slicker numerical integration methods, notably Gaussian quadrature, and I think the numerical integrators in graphing calculators are based on Gaussian quadrature. But to implement Gaussian quadrature, you must already have computed a table of roots of Legendre polynomials and coefficients to use with those roots. It would be hard to do that from scratch.

I've picked it up from Kent as well; it's what I used for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273169]this integral[/url].

Actually, that error estimate in the Wikipedia article looks like a mistake. We should get two extra degrees per step, not a doubling.


geometry MIT college Harvard Princeton calculus percent

My GPA is 2.9 overall, not impressive, so I am ready to settle for a graduate school that is not so impressive, but most of all I must have a life.  I really didn't like Johns Hopkins and I am looking for a school that has a good social life and also in a moderate town that is not like as crappy and loud as New York City but has lots of friendly people.  Any advice?

I go to Hopkins so I take offense! :mad:  Tell me why you don't like it, and I will tell you why you're wrong.  Although, quite frankly, I don't think Hopkins would admit you with a 2.9 unless you have some truly incredible recs.

And I'm also from NY and it is the greatest city in the world.

[quote="sexynsmartjenny"]I really didn't like Johns Hopkins and I am looking for a school that has a good social life and also in a moderate town that is not like as crappy and loud as New York City but has lots of friendly people.[/quote]
So, do you imply that New York City (NYC) has unfriendly people?

[quote="JRav"]I go to Hopkins so I take offense! :mad:Although, quite frankly, I don't think Hopkins would admit you with a 2.9 unless you have some truly incredible recs. [/quote]

lol, Yeah, you're right! Along with the incredible recommendations, even to have a chance at a place like Hopkins or Stony Brook etc, I think she (I think sexynsmartjenny is "she" :P) needs to have obtained "perfect scores" in the Math subject GRE and the General-GRE. (I think it's highly improbable given the GPA, the correlation isn't strong at all)

In case it wasn't obvious, the O.P. already attended Hopkins (for undergraduate studies) -- she is not considering it for graduate school.

I notice that your list of desired properties doesn't say anything about academic strengths.  This is ... strange.  What areas of mathematics are you interested in working in?

I'll just add that Hopkins' faculty is mostly analysis/PDE so if you have no interest in that...

US News rankings should never be taken as fact, but for a general idea of who does what field, [url=http://grad-schools.usnews.rankingsandreviews.com/best-graduate-schools/top-science-schools]click here[/url].

[b]@OP:[/b] How about SUNY, Stony Brook? This particular school is located in the North shore of Long Island. Stony Brook is also a hamlet (and might not be as 'loud' as NYC -- which you think). You have been vague -- You haven't mentioned anything other than your GPA, you could have mentioned some things like:Math GRE, General GRE scores, extra-curricular activities, the quality of your Recos, etc. As [b]JBL[/b] had said you also hadn't mentioned anything about your interests in Math.

If you're opting for Differential Geometry (just in case) I would recommend SUNY, Stony Brook because I think it might conform to your prerequisites (Also, Stony Brook's is renowned for DG). I personally never attended SUNY, Stony Brook but I have a relative (cousin) who's currently pursuing his PhD in Math over there. (My cousin also said the campus atmosphere isn't 'loud' and is pleasant; He also had perfect scores in his Physics and Math subject GREs).

The way your post is formulated, it does send out connotations that you're frivolous about grad school.

[quote="Gen8"]Along with the incredible recommendations, even to have a chance at a place like Hopkins or Stony Brook etc, I think she (I think sexynsmartjenny is "she" :P) needs to have obtained "perfect scores" in the Math subject GRE and the General-GRE. (I think it's highly improbable given the GPA, the correlation isn't strong at all)[/quote]

I'd guess that less than 5% of graduate students at either of these schools have perfect scores on the Math subject GRE.  You really have no clue about graduate schools in math and as such should refrain from dispensing advice on this topic.

[b]@blahblahblah:[/b] Greetings!

Well, I was just suggesting (or rather 'forwarding' my cousin's suggestion) the [b]OP[/b] that [b] [i] just in case [/i] [/b] if she's interested in Differential Geometry, then Stony Brook is a very good place for it. The [b]OP[/b] had been very vague -- She hadn't mentioned anything about her interests in Math, the quality of her recommendations, etc.

I think the above 'suggestion' (though given in a hypothetical situation) is relevant and accurate -- My cousin (with whom I share a very close relationship) had told me that Stony Brook is a very good starting point for furthering one's interests in differential geometry. In his opinion, Stony Brook is certainly renowned for differential geometry. I had just thought of sharing his 'opinion' with the [b]OP[/b].

He is currently pursuing a PhD in Math at Stony Brook. He intends to further his interests in differential geometry. He's by no means a dolt -- He's highly informed.

Quite similarly as JRav had noted, I had observed that given the [b] OP's [/b] GPA, it would be quite difficult to secure admission in even a less decent/impressive Math grad school. In the present context, the word 'impressive' hasn't been clearly defined by the [b]OP[/b].

I felt the only way that a low GPA could be offset (to a certain extent) was by obtaining 'great' scores on the General and Subject GREs -- This is certainly quite difficult because there often exists a pretty strong correlation between GPA and scores on standardized tests -- Most times, people who had managed to secure "high scores" on these tests also tend to have a "high" GPA.

[quote="blahblahblah"] I'd guess that less than 5% of graduate students at either of these schools have perfect scores on the Math subject GRE. You really have no clue about graduate schools in math and as such should refrain from dispensing advice on this topic.[/quote]
I know many friends/acquaintances and relatives who're majoring in Math; I regularly converse with them.

Personally, From observing many of your 'non-technical' posts -- I think you would be better off engaging yourself in [b] constructive criticism [/b] rather than being [b] hypercritical [/b] most of the time.

I don't think that blahblahblah was objecting to your description of Stony Brook (which, as far as I can tell, is a perfectly respectable math department, and with an unusual amount of money washing around for geometry due to the generosity of Jim Simmons) but rather to your comment about the math subject GRE (on which very few people get a perfect score).

Anyhow, without some further input from the O.P., continuing this thread is not likely to increase its usefulness to anyone.

[quote="Gen8"]
Personally, From observing many of your 'non-technical' posts -- I think you would be better off engaging yourself in [b] constructive criticism [/b] rather than being [b] hypercritical [/b] most of the time.[/quote]

okay, how's this:

there are at least dozens of graduate schools that could fit the OPs description, depending on what he's interested in.  recommending a single graduate school on the basis that the OP may be interested in differential geometry is more or less useless, especially given that the closest major city is NYC, which the OP has professed a particular disdain for.

re: the GREs

first of all, it's doubtful that any graduate program places as high of a weight on the GRE subject exam as you think (i know several people who got into top graduate programs with scores in the 70th and 80th percentile).  second of all, it's an extremely difficult exam to get perfect.  i missed or omitted something like 10 questions and ran out of time, and still finished in the 98th percentile.  i'd guess that only a few people obtain a perfect score every year, but i guess it's possible that 'your cousin' did. (jbl: if it's not too personal, did you get a perfect score?  do you know of many people who did?)  a 99th percentile score would be another thing, I guess.  And competitive mathematics graduate programs (which SUNY Stony Brook and Johns Hopkins are) certainly don't give much weight to general GRE scores.

you may converse with a couple of math majors but you really don't know any mathematics, you haven't gone through the graduate admissions process or written the math subject GREs, and you actually don't know much about most of the universities you like to talk about (which was exemplified when you tried to console JBL for 'only' being a Ph.D student at MIT).  the advice you give is uniformly bad and in many cases misleading, and this subforum would be better off if you just refrained from giving advice on graduate studies in mathematics.

oh, and finally: stop backseat moderating.

[quote="blahblahblah"](jbl: if it's not too personal, did you get a perfect score?  do you know of many people who did?)[/quote]  I did not.  (I think I missed 2 or 3 questions?  I don't remember exactly, but it was a small enough number to impress a couple of friends :).)  It wouldn't shock me if a few people wandering around the MIT department had perfect scores (I mean, we have some of the very best test-takers in the country from my age cohort here, so I know who I would bet on), but I don't think I've discussed the GREs with anyone at MIT since my arrival.

For other readers: I'd like to concur with blahblahblah's assessment that, while a bad GRE score might have bad effects for applications to high-end grad schools, a good score is really quite meaningless in terms of applications.  Said another way, getting a handful of questions right or wrong isn't going to sway anyone's opinion of your suitability for mathematical research, but getting many many questions wrong might convince someone that you don't have a strong enough background to do math at a high level.

[quote="blahblahblah"]okay, how's this:
there are at least dozens of graduate schools that could fit the OPs description, depending on what he's interested in. [/quote]
If it isn't still obvious, the original poster is a girl.

[quote="blahblahblah"] recommending a single graduate school on the basis that the OP may be interested in differential geometry is more or less useless, especially given that the closest major city is NYC, which the OP has professed a particular disdain for. [/quote]
Well, According to my 'cousin' SUNY, Stony Brook has a pretty good social life-- It also has a pleasant ambiance. I think this conforms (to a certain extent) to the [b] OP's [/b] prerequisites for grad school. The only disadvantage might be that she might have to commute through NYC to reach SUNY, Stony Brook. Since I already went to Stony Brook (to meet my cousin) I can say it's pretty much a 'moderate' town. (Which the [b] OP [/b] is looking for)

Just as [b]JBL[/b] had noted before, Since the [b]OP[/b] had been very vague -- It's quite difficult to give an advice/suggestion that is highly relevant. I only recommended Stony Brook because I felt it did conform to some of the 'prerequisites' the [b] OP [/b] had in mind.

[quote="blahblahblah"]first of all, it's doubtful that any graduate program places as high of a weight on the GRE subject exam as you think[/quote]
Probably, But in the [b] OP's [/b]case-- I think given her GPA, the admissions committee might consider to accord more importance to her subject GRE score.

[quote="blahblahblah"]you may converse with a couple of math majors but you really don't know any mathematics, you haven't gone through the graduate admissions process or written the math subject GREs.[/quote]
Well, I think you should exhibit some caution while concluding that "[b] I [/b] absolutely don't know any mathematics."

I do readily acknowledge that I'm not mathematically proficient (proficient here means -- Having an advanced degree of competence,i.e. an expert) -- But it doesn't mean that I absolutely don't know "any" mathematics. One of the main reasons I couldn't contribute 'technically' (to some extent) to this great forum was due to a lack of great facility with LATEX. But I still do contribute to a certain extent in the 'general' and 'Puzzles and Brainteasers' sub-forums.

Like I have already reiterated I have conversed with many people (have friends/acquaintances/relatives) who have gone through the admissions process,etc. The suggestion that I had given to the [b] OP [/b] wasn't mine but my cousin's. I had gleaned enough information/knowledge about the admissions procedure,etc from interactions with my friends/relatives who're majoring in Math.

[quote="blahblahblah"] and you actually don't know much about most of the universities you like to talk about (which was exemplified when you tried to console JBL for 'only' being a Ph.D student at MIT). [/quote]
Now, that's an inept statement. I know pretty well about 'these' universities. First of all, I never tried to "console" [b] JBL[/b] or anything like that -- I had just expressed my astonishment of the fact that he was pursuing a PhD in Math from MIT after having completed his Bachelor's in Math from Harvard. I just found it to be slightly 'strange' because 'some' of my cousin's friends (whom I had interacted with) are pursuing their graduation in Math (from Harvard) right after their under-graduation from Harvard.

I had just expressed astonishment -- It wasn't meant to either "ridicule/insult" or "console" [b] JBL[/b] in any way; It's just your interpretation.

I know that [b] MIT, Harvard, University of Chicago, Princeton [/b] are consistently ranked among the Top Five programs in Math in the country. I also know MIT's Math program is really good.

[quote="blahblahblah"] the advice you give is uniformly bad and in many cases misleading, and this subforum would be better off if you just refrained from giving advice on graduate studies in mathematics.

oh, and finally: stop backseat moderating. [/quote]
In a similar vein, I feel (I'm sure some other members do too) that at most times your [b]non-technical views/opinions[/b] are consistently "hypercritical" in nature. Sometimes, I think you have a disposition to drawing invalid or absurd conclusions from a little piece of information.

Just because you're a "moderator" here doesn't entitle/guarantee you or your opinions/views to be completely immune to criticism. I think you should realize the fact that your views/opinions aren't always devoid of flaws --  Expressing a valid opinion/observation that's conflicting with your own or which disconcerts/discomfits you doesn't in any way amount to "back-seat moderating".  :roll:

[b] P.S.: [/b] Like JBL had earlier noted, Unless we receive sufficient information by the OP which dilutes the vagueness surrounding her interests,etc -- continuing this thread isn't going to be fruitful to anyone.

[quote="JBL"](I mean, we have some of the very best test-takers in the country from my age cohort here, so I know who I would bet on)[/quote]  Well, one of my bets would have been wrong: a friend (who it might be fair to describe as the top competitive math person in my academic year) got one question wrong.  (He actually got two questions wrong, but one of them was evidently not official.)

[quote="Gen8"][quote="blahblahblah"]okay, how's this:
there are at least dozens of graduate schools that could fit the OPs description, depending on what he's interested in. [/quote]
If it isn't still obvious, the original poster is a girl.
[/quote]

some day you will learn that not everyone who seems to be a girl on the internet is not, indeed, a girl.  (in fairness, my impression that the OP is male is based off of their old signature, which you probably were not around to see)

Going back to points made by blahblahblah and JBL, is it really that hard to earn a perfect score on the math subject test?  I will have to deal with it in a year and a half or so, so I would like to know what makes it particularly difficult.  Is it the time constraints or are the questions actually challenging?  I wouldn't there are proofs, right?

[quote="JRav"]is it really that hard to earn a perfect score on the math subject test?[/quote]
Just as [b] blahblahblah [/b] sir had said -- It seems like only very few people manage to secure a perfect score (i.e. 900) every year. IMO, it's very difficult to secure perfect scores on most of the subject GREs (say, for example-- Computer science, Physics,etc).

[quote="JRav"] I will have to deal with it in a year and a half or so, so I would like to know what makes it particularly difficult. Is it the time constraints or are the questions actually challenging? [/quote]
I had made these observations after attempting the "sample test". Actually, it's a combination of both. Difficult (sometimes very difficult) questions coupled with time constraints. In order to perform really well-- the test requires you to have broad knowledge across many disciplines in Math. I'm also contemplating of attempting the Math subject GRE (in 2-3 years). I think, Having a good calculus background is really useful for the test.

[quote="JRav"] I wouldn't there are proofs, right? [/quote]
This statement is highly incomprehensible. I interpreted it as such -- "Does the Math-subject GRE have questions that require proofs?" At present, The answer is an emphatic "No".

The Math subject GRE comprises 66 multiple choice questions; I think you haven't done much research regarding the Math-subject GRE. I suggest you to go to ETS' site and download the Math subject GRE practice booklet (it's a PDF file). There is a 'sample' test in the PDF file. The sample test consists of 66 multiple choice questions, and it is akin to the actual test. I'm pretty sure the actual test is going to be much harder than the "sample test".

First of all, a perfect score is 800, not 900.  Second, no I have not done much research or prep for the GRE's as it is not my primary focus at the moment.

You know, honestly, if you're going to correct someone you should be right: GRE subject tests are on a 200-990 scale.  (A score of around 900 is what you get with one or two wrong answers.)  The ETS website is not very link-friendly, but if you go to http://www.gre.org, click on "Test Takers," then on "GRE Subject Tests" then on "Scores" you can find this and other information.

The GRE Math Subject Test has a tough time limit combined with some genuinely nontrivial questions -- for a multiple-choice test that's mostly devoted to calculus exercises, there are a decent number of tough questions.  I don't remember precisely, but I think the example test was broadly comparable to the actual test, perhaps a little easier.

Back when I took the test (1982), I happened to get a 990.  I doubt that meant I got all the questions right: the associated percentile was 96%, which means 4% of folks got a 990.  Over the next decade or so, the percent of 990's increased to 18% in 1999.  As a result, the ETS rescaled the test in 2001, to allow grad schools to better distinguish kids at the top end.

Strangely, the sample test on the website has a compressed score range of 390 to 900.  That's probably where Gen8 got the number 900.

According to wikipedia (treat its accuracy however you want), when the exam was revised a few years ago, the exam was made much tougher and went from being one of the easier exams to one of the hardest.  [url=http://en.wikipedia.org/wiki/GRE_Mathematics_Test]This[/url] is the page.


Here are some hintz
Hint 1:
[hide]
try using cauchy schwarz to get a bound for x+y+z
[/hide]
Hint 2:
[hide]
try using hm gm inequality to find a  minimum value for xyz
[/hide]

Applying Cauchy and AM/GM/HM introduces irrationals? Are you sure?

 BTW I hope everyone noticed x,y,z are not necessarily positive.

Since everyone is math mood , I am bumping this.

[quote="Conjurer"]IF x^2 + y^2 + z^2 = 9 , x,y,z are reals.

Prove that 2(x+y+z) - xyz cannot exceed 10.[/quote]
Too tired and sleepy so I just found the limiting case :rotfl:
Will solve in the morning.
Take x=2, y=2, z=-1
2(x+y+z)-xyz
= 2(2+2-1) - 2X2X(-1)
=10


Find the next three terms in this sequece:

$0$, $1$, $5$, $14$, $30$, $55$ ...

[quote="math92"][hide]
the differnces are perfect squares so after 55 you would have
$\boxed{91, 140, 204}$
correct?[/hide][/quote]

Correct. Good Job!  :)


Find the number of polynomials of degree 5 with distinct
coefficients from the set ${1, 2, 3, 4, 5, 6, 7, 8}$ that are divisible
by $x^{2}-x+1$

Is there a nice way to do this without extreme amounts of casebashing?

If you consider an arbitrary fifth degree polynomial $ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f$ and take it $\pmod(x^{2}-x+1)$, noting that under that mod we have $x^{2}\equiv x-1$, then the polynomial has remainder $(e+d-b-a)x+(a+f-d-c)$, and we want $0$ remainder so we have the equations $d+e=a+b$ and $c+d=a+f$ that must be satisfied.

[quote=Elemennop]the equations $d+e=a+b$ and $c+d=a+f$ that must be satisfied.[/quote]

Now what?


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Find: $ \int^{\pi}_{\minus{}\pi} \frac{sin3x}{(1\plus{}2^{x})sinx} \,dx$

P.S. Do not use integration software :-)

[quote="borislav_mirchev"]Find: $ \int^{\pi}_{ \minus{} \pi} \frac {sin3x}{(1 \plus{} 2^{x})sinx} \,dx$

P.S. Do not use integration software :-)[/quote]

[hide=" hint"] By put $ x \equal{} \minus{} y$ we will get $ 2\int^{\pi}_{ \minus{} \pi} \frac {sin3x}{(1 \plus{} 2^{x})sinx} \,dx \equal{} \int_{ \minus{} \pi}^{\pi}(3 \minus{} 4\sin^2x)dx$ [/hide]

[color=darkblue]You can use $ \boxed {\ \int_{ - a}^af(x)g(x)\ \mathrm {dx} = \int_0^a[f(x) + f( - x)]g(x)\ \mathrm {dx}\ }$, where $ g$ is even.

Therefore, for the functions $ f(x) = \frac {1}{1 + 2^x}$ and $ g(x) = \frac {\sin 3x}{\sin x}$ obtain :

$ \{\begin{array}{c}
f(x) + f( - x) = 1\\\\
g( - x) = g(x)\end{array}$ and $ \int^{\pi}_{ - \pi} \frac {1}{1+2^x}\cdot\frac {\sin 3x}{\sin x}\ \mathrm {dx} = \int^{\pi}_0 \frac {\sin 3x}{\sin x}\ \mathrm {dx}$ a.s.o.[/color]

Your solution is very good. I believe you like this integral.

I'm not very good with integrals. Last time I active used them was before 5-6 years but I started to explore mathlinks integrals. I found this problem and two or three similar problems till now. I think your problem is very good.
http://www.mathlinks.ro/Forum/viewtopic.php?p=904452#904452. You can see what I found :-)

Not at all. I 'm glad I could help you.

Here is another problem from Japanese university entrance exam.

Keio University entrance exam/Medical 1974

Evaluate $ \int_{\minus{}\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sin ^ 2 x}{1\plus{}e^{\minus{}x}}\ dx$.

Cute; we can do it by setting $ u\equal{}\minus{}x$


Hi all,

  When I use the equation environment  (\begin{equation} ..... \end{equation}) in LaTeX, I have the equations numbered on the LEFT. Does anyone know how to change it to the RIGHT?

  Thanks,

  Trieu.

In the preamble [code]\usepackage[leqno]{amsmath}[/code]This will work with the eqnarray environment as well

Thanks, Steve. Actually me problem is as follows. I was using \documentclass{amsart} and so I had the equations numbered on the left. I wanted to chage it to the right so I posted the question. With your comments, I put the option "reqno" in \documentclass[12pt, reqno]{amsart} and it worked!

  Thanks again.


trigonometry
inequalities unsolved
inequalities

[quote="singapore1728"]This question is from 2006 BMO round 2, if I am not wrong... and since rust have the same solution as me, i shouldn't doubt that my simplistic answer is wrong[/quote]

You are right. I have posted mi solution, but I can't find my post.


A popcorn company wants to create a circular cylindrical container with diameter 10 inches and volume 1256 cubic inches. How many inches should the height of the container be? Express your answer to the nearest inch.

$ 1256\equal{}25\pi\cdot h\Rightarrow h\equal{}16$


group theory
abstract algebra
geometry
geometric transformation
superior algebra
superior algebra unsolved

Prove that A_n is the only proper subgroup of index <n of S_n for all n>= 5

Well, you may use the result (*) I proved in your previous post: http://www.mathlinks.ro/Forum/viewtopic.php?t=116164

Clearly, if we show that every proper subgroup $H$ of $S_{n}$ of index $<n$ must be normal, then we will be done by $(*)$. Now note that $S_{n}$ acts on the set $\Delta$ of cosets of $H$ by left translations, yielding an homomorphism $\phi: S_{n}\rightarrow S_{m}$, where $m<n$. Thus $Ker(\phi)$ is normal and nontrivial, therefore it is $A_{n}$ or $S_{n}$ (it actually cannot be $S_{n}$). This means $m=2$, which implies that $H$ is normal in $S_{n}$, and solves the problem.


$ x,y,z$are nonnegetive numbers such that $ xy\plus{}yz\plus{}zx\equal{}1$,prove that
\[ x^3\plus{}y^3\plus{}z^3\geq (x\minus{}y)(y\minus{}z)(z\minus{}x)\plus{}xyz(x\plus{}y\plus{}z)^2\]

I proved it with a very ugly method...Can someone prove it easily..

The inequality is equivalent to

\[ 3\sum_{cyc}x^3 \ge 3xyz\plus{}4\sum_{cyc}xy^2\minus{}2\sum_{cyc}x^2y\]

and we are done!


superior algebra
superior algebra unsolved

Let $G$ be a finite group. Show that if for each positive integer $m$ the number of solutions $x$ of the equation

$x^{m}= e$ in $G$ is at most $m$, then $G$ is cyclic


geometry
3D geometry
sphere
geometric transformation
rotation

I have a doubt in Rotational motion.
While a sphere is pushed up an incline plane at a velocity and also given some angular velocity to roll it upwards, $v>w r$,where $w$ is the angular velocity.Then in what direction should the frictional force be acting.
Here I was confused that whether the friction force acts to make the motion pure rolling or is it just against the motion.
Help is required urgently.

I'm not sure what exactly the situation you're talking about looks like but the friction force always acts in such direction that the object stops sliding.

if you think about it, the object has to stop rotating at its maximum height. The only way this would happen is if the friction force points up the incline.

I think that kinetic friction acts down the incline, decreasing translational velocity while increasing angular velocity until v = r*omega.  If you analyze the effect of gravity on the sphere, I think the translational acceleration is 2/5 of the rotational acceleration.  So the rotation of the ball slows down more than the translation up the incline.  Again, friction acts down the incline to balance the two.

Not too sure though.

I thought about it a bit more and I think I was wrong.  Once the kinetic friction and gravity act long enough (down the plane) to make v = r omega, the ball rolls without slipping.  Gravity, acting through the center of mass, only slows down the ball's translational velocity, so static friction needs to act up the plane to balance rotation and translation.


function
inequalities
analytic geometry
graphing lines
slope
limit
real analysis

Let $ f$ be a real-valued function on $ [0,1]$. Suppose that there is a constant $ C$ such that for any points $ a<c<b$ in $ [0,1]$ we have $ |f(c)\minus{}f(a)|\le C(|a\minus{}b|\plus{}|f(a)\minus{}f(b)|)$.

Prove or disprove: $ f$ has bounded variation.

Put $ C=1$. Note that every function of the kind $ f(x)=g(x)\pm \frac x3$ where $ g$ is monotone and continuous satisfies the condition with strict inequality for $ a\ne b$. Now start with $ f_0(x)=x$. Suppose $ f_k$ is a continuous piecewise linear function satisfying the condition with strict inequality for $ a\ne b$. Take an interval $ [t_i,t_{i+1}]$ of linearity of $ f_k$ and replace $ f_k$ by a sufficiently close "crocodile ridge" as shown on the picture:
[asy]size(200);
D((-3,1)--(0,0)--(2,5)--(3,1));
D((-3,-1)--(3,-1),black);
D((0,0)--D("t_i",(0,-1)),black);
D((2,5)--D("t_{i+1}",(2,-1)),black);

int n=7;
for(int k=2;k<2*n-2;++k)
D((k/n,5/2*k/n)--((k+0.1)/n,5/2*(k+1)/n+1/3*0.9/n)--((k+1)/n,5/2*(k+1)/n),red);[/asy]
(the negative slopes are $ -1/3$, the positive ones are much larger than the original slope)
If one the points $ a$ and $ b$ lied outside $ [t_i,t_{i+1}]$ and the change affected at least one point on $ [a,b]$, then $ |a-b|$ was bounded from below by some positive constant and, if the ridge was close enough to the original graph, the strict inequality was preserved (we choose the distance from the ridge ends to the endpoints of the interval of linearity first and the height of the ridge afterwards). If $ a,b\in [t_i,t_{i+1}]$, our new function of the form $ g(x)\pm x/3$ and we are fine again. On the other hand, the variation increased by almost $ \frac 23(t_{i+1}-t_i)$. Doing it on each interval of linearity of $ f_k$, we get $ f_{k+1}$ with $ |f_k-f_{k+1}|$ as small as we wish, $ f_{k+1}$ satisfying our condition, and $ V(f_{k+1})>V(f_k)+1/3$. Now just take $ f=\lim_{k\to\infty}f_k$ to get a counterexample.


Let $ A \equal{} (\minus{}4,0)$ and $ B \equal{} (4,0)$. Determine the number of points $ C \equal{} (x,y)$ on the circle $ x^2\plus{}y^2\equal{}16$ such that the area of the triangle whose vertices are A, B, and C is a positive integer.

Let AB be the base. We have $ A \equal{} \frac {bh}{2} \equal{} 4h$. There are four points on the circle where the height is .25, .5...3.75 and 2 points where the height is 4 so 62 total points.


A very attractive problem, but too straighforward for Pre-Olympiad problem.  The following is a method that will solve it for you.
[hide="Big Hint...Sketch of Solution"]Write each term in terms of $a_{1}$.  Then you'll see a patern in the denominators of the numbers.  You may prove by induction that the pattern holds for all n.  Then express the sum and see that the sum telescopes.  You'll then find $a_{n}$ in terms of $a_{1}=2001$.[/hide]
Masoud Zargar

[hide]Speedy... hope I didn't mess up

Write the given relation using $n+1$ in place of $n$, and subtract the original relationship from it, which gives you a formula for $a_{n+1}$ in terms of $n$ and $a_{n}$. From here, it can be shown that $a_{n}= \frac{2a_{1}}{(n+1)(n+2)}$, so $a_{2001}= \frac{2001}{2005003}$. Or something similar to that.

Edit: I can't count. Formula should say $a_{n+1}$ instead of $a_{n}$, so answer is $\frac{1}{1001}$[/hide]

[hide]

Using that, you get  if I didn't mess up which I PROBABLY DID.

[/hide]


Three ship captains are on the water (in the sea) and they want to meet. All ships have the same speed and they start in the same moment. Help them and find the best place for their meeting (best place is place where they meet so early as possible). Discuss all possibilities for their location (first ship can be in the Pacific, second in the Atlantic etc.  :D  :D  :D ).

For this problem, to consider every possiblity (in my opinioin is extravagant) because the coastlines of all the continents are wildly variant

ok, so let's say that we have three points A, B, C in the plane and we want to find the point for which is the longest width to one of A, B, C points minimal... ;)

Let's say they meet at point O.  If you want to minimize the longest of the 3 distances OA, OB and OC, then make all 3 distances the same by defining O to be the point of intersection of the perpendicular bisectors of AB and AC.  O is the center of the circumcircle through A, B and C.  This argument only applies if the speeds are the same.


Prove that for all $k\in{Z}$,there exists infinite $a,b,c\in{N}$ st :ab+k;bc+k and ca+k are perfect squares :D


Show that there exists a natural number $ n$ with the following property: If a regular triangle with side $ n$ is divided into $ n^2$ regular triangles with side $ 1$, then among the  ertices of these triangles one can choose $ 1993n$ points, no three of which are vertices of a regular triangle.


Hello

Indicate the name of each form of geometric shapes next?

THANKS
SALAHCOOL

Hi salahcool.

Not sure why you posted this.

[hide=" Click to reveal hidden content"]

square, octagon, pentagon, rectangle, triangle, circle, hexagon and heptagon.

[/hide]

But keep them coming: you have some very interesting problems.

Yes, they are a square, octagon, pentagon, rectangle, triangle, circle, hexagon and lastly, a heptagon.
But they are also all polygons! "[b]Poly[/b]" means many, and the ending "[b]gon[/b]" means "[b]sides[/b]." A list of polygons follows:

 :laugh: [hide="list of shapes"][b]3 sides:[/b] triangle
[b]4 sides:[/b] rectangle
[b]5 sides:[/b] pentagon
[b]6 sides:[/b] hexagon
[b]7 sides:[/b] heptagon
[b]8 sides:[/b] octagon
[b]9 sides:[/b] nonagon
[b]10 sides:[/b] decagon
[b]11 sides:[/b] hendecagon
[b]12 sides:[/b] dodecagon[/hide]

[hide="Wow"]
Square
Octagon
Pentagon
Rectangle
Triangle
Circle
Hexagon
Heptagon[/hide]


Find all function such that$ f: R\minus{}>R$
$ f(x^5\minus{}y^5)\equal{}x^2f(x^3)\minus{}y^2f(y^3)$


geometry
trigonometry
circumcircle
trig identities
Law of Sines
perpendicular bisector

Hi,

Okay, this might be first time I ever post on the pre-olympiad but I have this one question from very difficult geometry book I'm trying to at least, go through. --> Geometry Revisited.

On the first chapter of the Extended Law of Sines

Theorem 1.11 For a triangle ABC with circumradius R,

$\frac {a}{sin A} = \frac {b}{sin B} = \frac {c}{sin C} = 2R $

This is quite confusing becuase the law of sine is expressed on terms of radius, actually circumradius.

Is this too easy for pre-olympiad?

If it is, it's easy, remove it to intermediate.

Regards,

Silverfalcon

Okay, draw your triangle and your circumcircle.

We will prove R sin <A = a/2

Call the circumcenter O.  Draw OA, OB, OC.  Now draw D on BC such that OD is the perpendicular bisector of BC.  Now A is on the circle and O is the center of the circle so that 2<A = <BOC.  Thus <BOD = A.  Now apply basic trig to the right triangle BOD

We get that sin A = (r/2)/R --> 2R = r/sin A.  Proceed cyclically for the other sides.


Hi guys, I was a member of the Irish team and I'm just wondering whether anyone is eligible for Spain `08 next year. Oh and if you are make sure you look for the Irish team, we're always fun (after we all hopefully get on the team :thumbup: )

Hey, Galin :P
Laci (HUN4) and I (HUN3) are eligible. And sure you are :) e.g. I have a picture involving you and a bin :D

Hey Galin! Remember the Canadians? :moose:
For our team, everyone except CAN1 and CAN2 is eligible.
I hope to see you in Madrid!

Hey Galin =D

Everyone on the 07 Australian team except AUS1 and AUS6 is still eligible for the team. AMO is on the 5th and 6th- wish us luck ^^

(and best of luck to all of you too)

I didn't want to remember that bin - have limits now... or so i think  :roll: Good luck guys


Hi, :)  Math Geniuses. i'm having trouble solve these problems. so i need help...

solving the following without calculator.
Q1.cos²105°+cos²15°+cos105°cos15°
 X° the circle here represents zero, not degree...I think...

Q2.tan70°+tan15°-√3 tan70°tan15°

and I have answer so all I want to know is method.

good luck!

Looks like number one can easily be answered with cosine addition and subtraction formulae.

$ \cos{a\pm b}\equal{}\cos{a}\cos{b}\mp\sin{a}\sin{b}$.

For instance, $ \cos{105}\equal{}\cos{(45\plus{}60)}$ and since we know both of those cosine values, it is easy to expand it out with the formula and get an answer. I'll leave the rest to yourself.  :wink:

right, however my answer didn't match with the answer the text gave and I'm not 100% sure if it is the mistake of the text. eventhrough cos105=cos(45+60) or (90+15) ...it was bit awkward when I tried to solve it.
here's where I stucked.
original formula= [1/2+1/2cos2(45+60)]+[1/2+1/2cos2(45+60)cos105cos15
= 1+cos(90+120)+cos105cos15
=1+cos90cos120-sin90sin120+cos(90+15)cos15
=1-sin120-sin120cos15
=1-sin(pi-2pi/3)-sin(pi-2pi/3)cos15
=1-sin(-60)-sin(-60)cos15
=1+sin60+sin60cos15
=1+√3/2(1+cos15)
=1+√3/2(1+√[(1+cos30)/2]
=1+√3/2(1+-√{[1+√(3)/2]/2}
=1+√(9/2)/2
well... here are four choices: 1/4; 3/4; 5/4; √(6)/4 so obviously, my asnwer is wrong. :?:

$ [cos(60^{o}\plus{}45^{o})]^{2}$ + $ cos(45^{o}\minus{}30^{o})$ + $ cos(60^{o}\plus{}45^{o})cos(45^{o}\minus{}30^{o})$
Now just expand using trig iedntities
You can either solve this.

OR take common some terms and then proceed similarly

What year china maths is this?

------

number 2 looks difficult. How can you simplify $ tan70^{0}$

Well, for number 1, you don't have to handle a lot of expanding
Notice that $ 105^{o}\equal{}90^{o}\plus{}15^{o}$ so $ cos105^{o}\equal{}\minus{}sin15^{o}$
Now $ cos^{2}105^{o}\plus{}cos^{2}15^{o}\plus{}cos105^{o}cos15^{o}\equal{}sin^{2}15^{o}\plus{}cos^{2}15^{o}\minus{}\frac{sin30^{o}}{2}\equal{}1\minus{}\frac{1}{4}\equal{}\frac{3}{4}$

[quote="yui"]thanks for helping me out.
but, the text indicates the answer should be 5/4 instead of 3/4 for my first problem.

this problem is from year 2007 final exam.

and, I made a mistake on #2.

the equation here:

tan70°+tan15°-√3 tan70°tan15°

 should be

tan70°+tan50°-√3 tan70°tan50° I'm very sorry.  it's tan50. not tan15[/quote]

That makes a lot more sense.

[hide]Just start playing around with the sum formulas, we observe that:

$ \tan 120^{\circ}\equal{}\frac{\tan 70^{\circ}\plus{}\tan 50^{\circ}}{1\minus{}\tan 70^{\circ}\tan 50^{\circ}}$

(the motivation for this is that 70+50 = 120, which is an angle that we know the tangent of)

Remember that

$ \tan 120^{\circ}\equal{}\minus{}\sqrt{3}$,

so:

$ \minus{}\sqrt{3}\equal{}\frac{\tan 70^{\circ}\plus{}\tan 50^{\circ}}{1\minus{}\tan 70^{\circ}\tan 50^{\circ}}$

$ \minus{}\sqrt{3}\cdot (1\minus{}\tan 70^{\circ}\tan 50^{\circ}) \equal{}\tan 70^{\circ}\plus{}\tan 50^{\circ}$

$ \minus{}\sqrt{3}\equal{}\tan 70^{\circ}\plus{}\tan 50^{\circ}\minus{}\sqrt{3}\tan 70^{\circ}\tan 50^{\circ}$

So the answer is $ \boxed{\minus{}\sqrt{3}}$.[/hide]

Ok I have temporarily forgotten my trig math but I do remeber a few things.

Have you tried Tan(3(70)) the same way you would Tan(2(70))

[quote="yui"]thanks for helping me out.
but, the text indicates the answer should be 5/4 instead of 3/4 for my first problem.[/quote]

Well, I'm pretty sure that
$ \cos^{2}105^{o}\plus{}\cos^{2}15^{o}\plus{}\cos105^{o}\cos15^{o}\equal{}\frac{3}{4}$

Anything's wrong with your test answer? I tried it with my calculation also.  :ninja:

well, maybe they made mistake on answer.  thanks alot guys!

oh, for #2. I checked answer again and it was...well they say the answer is
(√3)/3 i wonder why..... if it's (√3)/3, then it should be tan30 or tan210 right?

Well, the answer for #2 is $ \minus{}\sqrt{3}$. Vishalarul solved it perfectly. What's wrong with your test answers?????  :o


Let ABCD be a cyclic quadrilateral. The perpendiculars to AD and BC
at A and C respectively meet at M, and the perpendiculars to AD and
BC at D and B meet at N. If the lines AD and BC meet at E, prove
that <DEN = <CEM.

<DEN=<DBN and <CAM=<CEM so it remains to prove <DBN=<CAM which is obvious
<DAM=<CBN=90° and <CAD=<DBC!!!
maybe i'm wrong !!!

I think you draw wrong picture...
perpendicular to BC at B and perpendicular to AD at D meet N
perpendicular to BC at C and perpendicular to AD at A meet M...

oh sorry my mistake..
good solution ,,thanks! :)

i think u wrote wrongly the prob, 'cause iran tst shouldn't have such an easy prob!!

$  \begin{array}{c} \angle{MAE} = \angle{MCE} = 90^0\Rightarrow AMCE - cyclic\Rightarrow \angle{AME} = \angle{ACB} \\
ABDD - cyclic\Rightarrow\angle{ACB} = \angle{ADB} \\
\angle{EDN} = \angle{EBN} = 90^0\Rightarrow EDBN - cyclic\Rightarrow\angle{ADB} = \angle{ENB} \end{array}\}\Rightarrow \\
\Rightarrow\angle{AME} = \angle{ENB}\Rightarrow \angle {MEA} = \angle{NEB}.$


[quote="djimenez"][quote="Bob75"]the sum of digits of (1111....1)² is 18 031, therefore what is the value of x, if the 1 in (111...1)² is repeating x times ?[/quote]

$x=2005$. A little trickly. Actually, it should be interesting to find the general formula for the sum of the digits of $(111\dots1)^2$ where there are $n$ ones in the number.

Best,[/quote]
Well, if $n=\mu+9d$ with $0\le\mu \le 9$ then $S((11\cdots 1)^2)=81d+(\mu+1)^2$ then as $18031=222\cdot 81+7^2$ then $n=222 \cdot 9+6=2004$.

[The trick is for every power of $9$ dissapear an $8$ digit and similar for an $1$.]

Best Regards.

[quote="Bob75"]If n = 2, S(11²) = 4, so the formula seems me false formula.
2005 is congrue at 4 modula 9 like 18 031, so the answer 2004 seems me false.
The good formula could be :
$\displaystyle S((11\cdots 1)^2)=81d+\mu^2$
then $\displaystyle 18031=222\cdot 81+7^2$
and
$\displaystyle n=222 \cdot 9+7=2005$.[/quote]
You need to make more calculatios of how you get the result, I am sure the correct result is $2004$ not $2005$, u forgot $+1$ in the $\mu$ coefficient, if you do some examples you can see sometimes u have an extra 8 and without a $1$.

Best Regards.

RobertuX

[quote="RobertuX"]You need to make more calculatios of how you get the result, I am sure the correct result is $2004$ not $2005$, u forgot $+1$ in the $\mu$ coefficient, if you do some examples you can see sometimes u have an extra 8 and without a $1$.
[/quote]

RobertuX, I would recomend YOU to review your calculations, because I even programed this (code at the end, in Scheme, because it handle integers of any length), and if you have 2004 1s, then the sum is 18018, but with 2005 1s, we actually have 18031.

[code]
(define (eneunos n)
  (do
    (
      (a n (sub1 a))
      (b 0 (add1 (* 10 b)))
      )
    ((zero? a) b)
    )
  )

(define (sumadigitos n)
  (do
    (
      (a n (quotient a 10))
      (b 0 (+ b (remainder a 10)))
      )
    ((zero? a)b)
    )
  )

(sumadigitos (* (eneunos 2004)(eneunos 2004)))

> 18018

(sumadigitos (* (eneunos 2005)(eneunos 2005)))

> 18031
[/code]

Best Regards

[quote="djimenez"]
RobertuX, I would recomend YOU to review your calculations, because I even programed this (code at the end, in Scheme, because it handle integers of any length), and if you have 2004 1s, then the sum is 18018, but with 2005 1s, we actually have 18031.
Best Regards[/quote]
Well, You r right!!, I forgot shift ONE in my program !!!   :?   :blush:  Now we have a nice formula for this, then I was thinking to do a similar with $S(2^n)$ maybe we can work to obtain this expression, you are interested to obtain this one?

Best regards.

Well, as we use to say "Errar es de humanos" (It is human to make mistakes)   :D

Best regards,

Could someone explain in more detail where $\displaystyle \displaystyle S((11\cdots 1)^2)=81d+\mu^2$ comes from?


inequalities
geometry
incenter
ratio
angle bisector
geometry proposed

Let ABC be a triangle ABC with incenter I and centroid G, and let $w_a$, $w_b$, $w_c$ be the lengths of the angle bisectors of triangle ABC. Prove the following inequalities:

[b](1)[/b]

$AG+BG+CG\geq AI+BI+CI$.

[This one was given with reversed sign by Fuzzylogic in the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16032]thread "Median, a nice topic"[/url], but no reply came.]

[b](2)[/b]

$\frac23 \left(w_a+w_b+w_c\right) \leq AI+BI+CI$.

  Darij

I just want to say that the solution of the problem in that thread has already posted by me in this following thread started by fuzzylogic:   http://www.mathlinks.ro/Forum/viewtopic.php?t=16365
  To your problem, it is easy to see that
    $\sum\frac{m_a}{h_a} \leq \frac{1}{3}(\sum m_a)(\sum\frac{1}{h_a}) = \frac{1}{3r}(\sum m_a)$ according to Chebychev 's inquality.
    Moreove from the thread, we have
   $\sum \frac{m_a}{h_a} \geq \sum{\frac{a}{b} + \frac{b}{a}}$
    Therefore it suffices to show that
  $\sum m_a \geq \frac{3}{2}r(\sum{\frac{a}{b} + \frac{b}{a}) \geq \frac{3}{2}(IA+IB+IC)}$
   From the inequality $\frac{1}{a} + \frac{1}{b} \geq \frac{4}{a+b}$ we have
$  r(\sum{\frac{a}{b} + \frac{b}{a}) \geq 4r(\sum \frac{sinA}{sinB+sinC} }$
   Hummm, I made a mistake. I did assume the rest was right. But maybe it is.
    I just didn't want to erase what I had tiredly typed. Ok, I had better sleep now and leave it to you.  :D It is a nice inequality though. I 'll think about it later. :maybe:

I think the following stronger inequality also holds though I have had no proof yet.
         $ (a^2+b^2+c^2+2\sqrt{3}S) \geq \frac{3}{2}(IA+IB+IC)^2$
      It is stronger as we apply the Finsler inequality to triangle with sides $m_a, m_b, m_c$.
     About the original inequality, can you post your solution, Darij ?

[b]Apologies!![/b]

Just now I have noted this thread. I should have posted [b](1)[/b] into Unsolved or Open Problems, since [b]I didn't have a solution and I still don't have it[/b]. Anyway, Treegoner, your [b]stronger version seems correct[/b] on my dynamic sketches, and I will try to prove it.

Here is, at least, a proof of [b](2)[/b]:

In order to prove $\frac23 \left(w_a+w_b+w_c\right) \leq AI+BI+CI$, it is enough to show that

  $\left(\frac23 w_a - AI\right) + \left(\frac23 w_b - BI\right) + \left(\frac23 w_c - CI\right) \leq 0$.

Well, if the lines AI, BI, CI intersect the sides BC, CA, AB of triangle ABC at the points A', B', C', respectively, then van Aubel's theorem yields $\frac{AI}{IA^{\prime}} = \frac{AB^{\prime}}{B^{\prime}C}+\frac{AC^{\prime}}{C^{\prime}B}$. But since an angle bisector in a triangle divides the opposite side in the ratio of the adjacent sides, we have $\frac{AB^{\prime}}{B^{\prime}C} = \frac{c}{a}$ and $\frac{AC^{\prime}}{C^{\prime}B}=\frac{b}{a}$, thus $\frac{AI}{IA^{\prime}} = \frac{c}{a}+\frac{b}{a}=\frac{b+c}{a}$. After a bit of algebra (using AA' = AI + IA'), this yields $\frac{AI}{AA^{\prime}} = \frac{b+c}{a+b+c}$. But $AA^{\prime}=w_a$; hence, $\frac{AI}{w_a} = \frac{b+c}{a+b+c}$ and $AI=\frac{b+c}{a+b+c}w_a$. Thus, $\frac23 w_a - AI = \frac23 w_a - \frac{b+c}{a+b+c}w_a = \frac{2a-b-c}{3\left(a+b+c\right)}w_a$. Denoting 3 (a + b + c) = D, we thus have

  $\frac23 w_a - AI = \frac{2a-b-c}{D}w_a = \frac{a-b}{D}w_a - \frac{c-a}{D}w_a$.

Similarly,

  $\frac23 w_b - BI = \frac{b-c}{D}w_b - \frac{a-b}{D}w_b$

and

  $\frac23 w_c - CI = \frac{c-a}{D}w_c - \frac{b-c}{D}w_c$.

Thus,

  $\left(\frac23 w_a - AI\right) + \left(\frac23 w_b - BI\right) + \left(\frac23 w_c - CI\right)$
  $= \left(\frac{a-b}{D}w_a - \frac{c-a}{D}w_a\right) + \left(\frac{b-c}{D}w_b - \frac{a-b}{D}w_b\right) + \left(\frac{c-a}{D}w_c - \frac{b-c}{D}w_c\right)$
  $= \frac{b-c}{D}\left(w_b-w_c\right) + \frac{c-a}{D}\left(w_c-w_a\right) + \frac{a-b}{D}\left(w_a-w_b\right)$
  $= \frac{\left(b-c\right)\left(w_b-w_c\right)+\left(c-a\right)\left(w_c-w_a\right)+\left(a-b\right)\left(w_a-w_b\right)}{D} \leq 0$,

since it is well-known that $\left(b-c\right)\left(w_b-w_c\right)\leq 0$, $\left(c-a\right)\left(w_c-w_a\right)\leq 0$ and $\left(a-b\right)\left(w_a-w_b\right)\leq 0$ (in fact, the number arrays (a; b; c) and $\left(w_a;\;w_b;\;w_c\right)$ are oppositely sorted - this is something one always shows when one proves the Steiner-Lehmus theorem). So the proof of [b](2)[/b] is complete.

[b](1) remains, to my knowledge, an open challenge to all friends of geometric inequalities on MathLinks.[/b]

  Darij

For [b](1)[/b] see [url=https://artofproblemsolving.com/community/c4h1174296][b]here[/b][/url], while for [b](2)[/b] we have a more general inequality which was proven [url=http://artofproblemsolving.com/community/c6h576972p6740851][b]here[/b][/url].


analytic geometry
geometry
similar triangles

[hide]Hint: Find the length of the internal tangent of the two circles, then use similar triangles to find the coordinates of the point where the tangent intersects the x-axis.[/hide]


calculus
integration
function
inequalities
absolute value
calculus computations

let f be a continus function wich have finite number of zero in the intervalle $ [0,1]$ ,and$ f(0)\equal{}f(1)\equal{}0$,
prove that $ lim_{\infty}$l$ \int_{0}^{1}e^{nt}f(t)$l$ \equal{}\plus{}\infty$

I suppose you need to use the triangle inequality and additivity of the integral on intervals for which f is 0 at the endpoints

The problem is unclear. Which variable is approaching infinity, and you are differentiating with respect to what variable.... Also, are you taking the absolute value of the function...

we differentiate with respect to the variable t ,and lim when n tend to infinity,

I think you could do integration by parts and then compute the limit.

Write the integral as a sum of integrals over the pieces between the zeros of $ f$. On each piece $ f$ has only one sign and has either a minimum or a maximum. It's then quite clear that the integral corresponding to the last piece (between the last zero in $ [0,1)$ and 1) will dominate all the other pieces, and this piece determines if the sequence of integrals goes to plus or minus infinity.


[quote="digitalboy1"]Let $n$ be a positive integer. Prove that there is no positive integer solution to the equation\[(x+2)^n-x^n=1+7^n\][/quote]

if $ n = 1 $, then obviously there's no solution.  so assume $ n \geq 2 $.

notice that if $x$ is odd, then one of $x, x+2$ is $1 (\mod 4)$ and the other is $-1 (\mod 4)$.  so then if $n$ is odd, the LHS is $\pm 2 (\mod 4)$ while the RHS is $ 0 (\mod 4)$, and if $n$ is even, the LHS is $0 (\mod 4)$ while the RHS is $ 2 (\mod 4) $.  in either case there's obviously no solution.

now, if $x$ is even, one of $x, x+2$ is $0 (\mod4)$ and the other is $2 (\mod 4)$, which imply that the LHS is $0 (\mod 2^n)$, and in particular that it is divisible by 4 (because we assumed $n \geq 2$).  now, for the RHS to also be divisible by 4, we need $n$ to be odd.  so $ n \geq 3 $.  for $n$ odd, however, the RHS is divisible by 8 but not by 16.  in view of the fact that the LHS is $0 (\mod 2^n)$, it follows that $n$ can only be 3.  but for n = 3, the equation becomes $(x+2)^3 - x^3 = 344$, and this can easily be seen to have no solutions...


The figure below shows a circuit made up of equilateral triangles. Inscribed triangles have length $ \frac {1}{2}$ times the outer triangle. The side of the outermost triangle is $ a$. There are infinite such triangles. The material has resistance per unit length $ \rho \; \frac {ohm}{m}$ [hide="."](what is the command for the "ohm" symbol??? There's one for "mho" $ \mho$ and none for "ohm" :cursing: )[/hide] Find equivalent resistance.

[hide="Doubt:"]
The problem is solved by assuming the equivalent resistance to be $ R_{effective}$ and the resistance of one side of outer most triangle to be $ R$ (ie: $ R \equal{} \rho a$) and then attaching $ R_{effective}$ across $ DE$ (see figure). The solution says no current enters the inner matrix through $ F$. The rest of the solution is well understood by me but I cannot appreciate as to why there will be no current through $ F$ into the matrix. Can anyone explain[hide="?"]
If you have an alternate method, I'd be grateful if you post it. :D [/hide][/hide]

[quote="hell_ever"]Use reversibility. The current at A splits equally. Now interchange the cell terminals. Because of the symmetric nature of the circuit, the current at B also splits equally. Put an equal and opposite cell into the picture. There is no current. This implies the current entering the branch at A along AF is equal to the current exiting along FB.[/quote]
How does the current "split equally"? The direct path from A to B has the least resistance out of all possible paths...if I'm interpreting the question correctly.


I've got a page like this:

[code]
\documentclass[a4paper,10pt]{article}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{fancyhdr}

\pagestyle{fancy}

%opening
\title{Something}
\author{Me}

\begin{document}

\maketitle

\begin{abstract}

\end{abstract}

\section{Asdfjkl}

\end{document}
[/code]

The only problem is, the sections are numbered 0.1, 0.2, 0.3... instead of 1,2,3 like I want them to be and as they look on this forum. How do I fix this?

Billy

Hi,
Are you sure you're using the article class? When I try the code you've posted, I get [b]1   Asdfjkl[/b] and not [b]0.1   Asdfjkl[/b] for the section number. You may have something else in your preamble that may be the problem. Did you mean in the document or the contents' table?
SSN.


linear algebra
matrix
induction
linear algebra unsolved

$ A\equal{}\begin{pmatrix}2006&1&\minus{}2006\\2005&2&\minus{}2006\\2005&1&\minus{}2005\end{pmatrix}$. identify all entries on the main diagon of the matrix $ S\equal{}I\plus{}A\plus{}...\plus{}A^{2006}$

[quote="sunlight"]$ A \equal{} \begin{pmatrix}2006 & 1 & \minus{} 2006 \\
2005 & 2 & \minus{} 2006 \\
2005 & 1 & \minus{} 2005\end{pmatrix}$. identify all entries on the main diagon of the matrix $ S \equal{} I \plus{} A \plus{} ... \plus{} A^{2006}$[/quote]
[hide]
$ A \equal{} I \plus{} M$ where I is the identity matrix and $ M \equal{} \begin{bmatrix}1& 1& 1\end{bmatrix}^t\begin{bmatrix}2005 &1& \minus{} 2006\end{bmatrix}$.
We can prove by induction that $ A^n \equal{} I \plus{} nM$
First note that $ M^2 \equal{} 0$
For $ n \equal{} 1$ we can obtain that $ A \equal{} I \plus{} M$ (True)
If $ A^k \equal{} I \plus{} kM$ for some positive integer $ k$, then for $ n \equal{} k \plus{} 1$, then $ {A}^{k \plus{} 1} \equal{} A^k.A \equal{} (I \plus{} kM)(I \plus{} M) \equal{} I \plus{} (k \plus{} 1)M)$ (True).
Then, $ I \plus{} A \plus{} A^2 \plus{} \cdots \plus{} {A}^{2006} \equal{} I \plus{} (I \plus{} M) \plus{} (I \plus{} 2M) \plus{} \cdots \plus{} (I \plus{} 2006M) \equal{} 2007I \plus{} (1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} 2006)M \equal{} 2007I \plus{} (1003.2007)M$
Then we can conclude that the entries of the main diagonal of the desired matrix is : $ {a}_{11} \equal{} 2007 \plus{} (1003.2007.2005)$,$ {a}_{22} \equal{} 2007 \plus{} 1003.2007.1$ and $ {a}_{33} \equal{} 2007 \plus{} 1003.2007.( \minus{} 2006)$.
Sorry if I am wrong.
[/hide]

thank you very much. you are true. i think it is a nice solution, but i wonder how can you find it?( why do you figure out the matrix M?) :D

The method Qruni suggested is great but intuitively using the identity $ (1 \minus{} x)(1 \plus{} x \plus{} .... \plus{} x^{n}) \equal{} (1 \minus{} x^{n \plus{} 1})$  then $ (1 \minus{} A)S \equal{} 1 \minus{} A^{2007}\Rightarrow S \equal{} (1 \minus{} A)^{ \minus{} 1}(1 \minus{} A^{2007})$. I guess  :lol: , $ A$ is diagonalizable  so we can find $ A^{2007}$


[hide]Let x be the number of humans, and y be the number of horses.

x + y = 74

Now the legs. There are 2 legs for each human and 4 for each horse.

2x + 4y = 196

x = 50 humans
y = 24 horses[/hide]

[quote="nonie"]One day, a person went to horse racing area, Instead of counting the number of human and horses, he instead counted 74 heads and 196 legs. Yet he knew the number of humans and horses there. How did he do it, and how many humans and horses are there?[/quote]

[hide]Say there are $x$ humans and $y$ horses.

$x+y=74$
$2x+4y=196$

$2x+2y=148$
$2x+4y=196$

$2y=48$
$\boxed{y=24}$

$x+24=74$
$\boxed{x=50}$[/hide]

[hide="I like this explanation"]If all the horses stood up, every animal would have two legs on the ground. So there would be 74*2=148 legs on the ground. So 196-148=48 legs would be in the air. Therefore there are 48/2=24 horses and 74-24=50 people.[/hide]

very creative, but it's a little hard for all the horses to stand up  :rotfl:

Then let's have them each put two feet on a stool, so only two feet are on the ground.

[i]<lets not spam>[/i]

[hide]i set up a sytem of equations:
$x+y=74\\\\2x+4y=196\\\\\text{solve and you get} 50 \text{humans and} 24 \text{horses}$
[/hide]

You're all right. It's a pretty simple question! :lol:

[hide]

The equation for the heads is
$x + y = 74$
The equation for the leg is
$2x + 4y = 196$
Solve and you get:
$\framebox{x = 50; y = 24}$[/hide]