geometry
conics
parabola
incenter
analytic geometry
geometry solved

Point D on side BC of triangle ABC is connected to A and a circle is inscribed

   in ABD.  Point X is the tangent point of this incircle with AD.

   What is the locus of X when D moves on BC ?

  Thank you.

   M.T.

[quote="armpist"]Point D on side BC of triangle ABC is connected to A and a circle is inscribed

   in ABD.  Point X is the tangent point of this incircle with AD.

   What is the locus of X when D moves on BC ?[/quote]

Interesting question, but I fear it's nothing that simple. I have checked that it is not a line and not a circle. It looks like a parabola, but I am not the kind of person to start speculating without having any reasonable data, so I leave this question for later...

[b]EDIT:[/b] I have checked using dynamic geometry software that the locus is [b]not[/b] a conic, so, in particular, not a parabola (six points on the locus don't satisfy the Pascal condition). It looks like a singular cubic, but may be something much more complicated.

  darij

I have a polar equation for this locus. let $I$ be the incenter of $ABD$ .If $\angle{BAI}=\theta$ then $AI=\frac{c sin(B/2)}{sin(B/2+\theta)}$ then $AX=AI cos(\theta)= \frac{csin(B/2)cos(\theta)}{sin(B/2+\theta)} =c sin(B/2)\frac{cos(\theta+B/2)cos(B/2)+sin(\theta+B/2)sin(B/2)}{sin(\theta+B/2)} =\alpha cot(\theta+B/2)+\beta$.
If we choose a suitable coordinate system, we have the polar equation $r=\alpha cot(\frac{\theta}2)+\beta$


Sequence {a_n} is defined by a_1 = 2 and

a_(n+1) = (2 + a_(n)) / (1 - 2a_(n))

Prove that all terms of the sequence are not zero.

Alternatively, $ a_n \equal{} \tan n\tan ^{\minus{}1}2$ and that's never zero.


i would really appreciate any help for this guys: given two triangles, equal lengths of their angle bisectors (the three of them) does NOT imply that the triangles are the same. In other words, a triangle is not uniquely determined by the lengths of its angle bisectors, i've been searching for a counterexample, i.e. different triples (a,b,c) and (d,e,f) of triangle sides having the same bisectors, but unsuccessfully so far...

You will not find counter example  because two triangles with respectively
    equal angle bisectors  are congruent.

     read this

http://www.hti.umich.edu/cgi/t/text/text-idx?c=umhistmath;idno=AAN3028.0001.001

    Maj.  Pestich

thanks a lot !!!!
yet the source you mentioned seems quite complicated and long, hopefully it gives me the desired answer!
once again, thanks pestich!


For the non zero numbers a, b, and c, define
$\clubsuit (a,b,c) = \frac{abc}{a+b+c}$
Find $\clubsuit (2,4,6)$
(A) 1
(B) 2
(C) 4
(D) 6
(E) 24

[hide="And the answer is..."]Substitute a=2, b=4, c=6:
(2)(4)(6)/2+4+6 = 4
THE ANSWER IS C.
[/hide]

[quote="vishalarul"][hide="And the answer is..."]Substitute a=2, b=4, c=6:
(2)(4)(6)/2+4+6 = 4
THE ANSWER IS D.
[/hide][/quote]
[hide="actually"]$\frac{2\cdot 4\cdot 6}{2+4+6}=\frac{48}{12}=4$. The answer is NOT D, It is C[/hide]


礼物1：求一切f(x):R→R满足

(i)f(1)=1

(ii)f(2x)=f(x)

(iii)对任意数列\[ u_n  \]若\[ \lim_{n\to oo}u_n  \]存在，则\[ \lim_{n\to oo}f(u_n)=f(\lim_{n\to oo}u_n)  \]
(oo 即“无穷大”）
礼物2：在一个单位立方体中是否可放2个棱长为1的正四面体？

此二题均为2006清华大学保送生暨自主招生冬令营考试试题

第2题不是很难嘛。两个四面体各一条棱为正方体的相距为$\sqrt{2}$的棱，注意到正四面体的一组对棱间的距离为$\frac{\sqrt{2}}{2}$，中间共用一条棱。

关于这个第二题，如果我没记错，可以放入三个。


[quote="makmib"]Find all functions  f : R -> R  such that for every real x,y we have :
      f(xf(x) + f(y)) = (f(x))^2 + y[/quote]
Already posted (without response) at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=112328[/url]. You were 15 minutes late.  :D


Use your knowledge of math and ability to recognize key numbers to do the following questions.  Do NOT use a calculator for any of them, however, you may use pen(cil) and paper to multiply/divide/add/subtract complicated decimals.  The problems at first may not seem connected, but they are.

$a)$  Show that $\frac{3.14}{1.41}\approx \frac{7}{3.14}$.
$b)$  Show that $\sqrt[4]{2}\approx 1.19$.
$c)$  Show that $\sqrt{7}\approx 2.65$.

For a, do not simply divide it out; show how it might be found.
For b, you may wish to use Newton's approximation formula.
For c, yes, you can use Newton's approximation formula, but the purpose of the question is to use (a) and (b) to find (c).  There is a very sneaky way.

[hide="a"]$\frac{3.14}{1.41}$
$\Downarrow$
$\frac{< \sqrt{10}}{\approx \sqrt{2}}$
$\Downarrow$
$< \sqrt{5}$
$\Downarrow$$< \frac{\sqrt{50}}{\sqrt{10}}$
$\Downarrow$
$\frac{\ll \sqrt{50}}{< \sqrt{10}}$
$\Downarrow$
$\frac{7}{3.14}$[/hide][hide="b"]Zuton's $b$ method is better than mine.[/hide][hide="c"]Just $\LaTeX$ing Elemennop's method.
According to (a),
$\frac{7}{3.14}\approx \frac{3.14}{1.41}$
$\Downarrow$
$7 \approx \frac{(3.14)^{2}}{1.41}$
$\Downarrow$
$\sqrt{7}\approx \frac{3.14}{\sqrt{\sqrt{2}}}$
According to (b), $\sqrt[4]{2}\approx 1.19$, so
$\sqrt{7}\approx \frac{3.14}{1.19}\approx \boxed{2.65}$[/hide]


[quote="Interval"]Find the center (h, k) and radius r of the circle below.

Then, graph each circle.

x^2 + (y - 1)^2 = 1[/quote][hide]The graph can also be written like this: $(x-0)^{2}+(y-1)^{2}=1^{2}$ so then the center is (0,1) and the radius is 1. (the regular formula is $(x-h)^{2}+(y-k)^{2}=r^{2}$ where $(h,k)$ is the center and $r$ is the radius)
I have attached the graph[/hide]

Absolutely beautiful picture and great reply.

This is an easy question.

Thanks!


Hello

Let $(a,b) \in \mathbb{R}^{2}$ where $1\le a^{2}-ab+b^{2}\le2$
1. Show that $\frac{2}{9}\le a^{4}+b^{4}\le8$
2. Show that $a^{2n}+b^{2n}\ge\frac{2}{3^{2n}}$
for every natural number $n$ where $n\ge3$


HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:

good job doing chinese... how did you do that?

 MAI GUH ZUHN SHING NIEHN KWAI LUH!!!!


Tare-5
Scrambled-3
mathfiend-2

1. If n is a positive integer, an integer which is always divisible by 3 is
a.) n(n +1)(n + 4)
b.) n(n + 2)(n + 4)
c.) n(n+2)(n +3)
d.) n(n + 3)(n - 3)
e.) (n + 2)(n + 3)(n + 5)

2. If a + b = 1, and a^2 + b^2 = 2, then what is a^4 + b^4?

3. A bluebird arrives at a birdfeeder at 8:00am and immediately begins to eat at a constant rate.  From past experience it is know that this bird can eat all the feed in 6 hours.  However at exactly 9:00am the bluebird leaves and a finch arrives at the feeder and by itself eats all the remaining feed in 7 hours.  How much time would it take for both birds to eat all the feed in the birdfeeder if they both started eating at the same time?

4. What is the last digit of 3^126?

5. A computer is programmed to scan the digits of the counting numbers.  For example if it scans 1 2 3 4 5 6 7 8 9 10 11 12 then it has scanned 15 digits.  The computer begins its task and scans the first 1788 digits.  What is the last counting number scanned?

cou7ld you inform us when the next contest will be?

Scrambled incorrect for #1.
mathfiend incorrect for #4.

Tare incorrect for #2.
BHorseMath incorrect for #5.

Tare gets #3.

#'s1,2,& 5 still out there.  Only one guess for multiple choice questions allowed.

Tare gets #2.

Remember, only 1 guess for mult. choice.

I've just realized all other contestants left, so shall we continue this tomorrow?

Oh yeah for when it starts, how's 7:30 EST for everyone?

I have play practice until 7:00pm CST & then I have to babysit my 5 younger siblings until my parents get home from stuff.  Depending on how much homework I have, I'll see if people are online.  Plus, it's AMC 10 tomorrow & maybe people will be studying or something.  We'll just see how it goes.

I just like to post the problems when there are a good number of people online.  It gives people some new problems to blow through if they feel like it.  The keeping score thing basically means nothing, so I don't worry too much about everyone being able to answer.  I just really think the more problems being posted, the better.


[hide]
$\angle BOG = \frac{1}{2}\angle AOH = \frac{1}{2}(\angle AOE+\angle EOH)\\ \angle BOF = \frac{1}{4}\angle AOH\\ \angle BOD = \frac{1}{8}\angle AOH\\ \angle BOC = \frac{1}{4}\angle AOE\\ \angle COD = \angle BOD - \angle BOC\\ \angle COD = \frac{1}{8}\angle AOH-\frac{1}{4}\angle AOE\\ \angle COD = \frac{\angle EOH-\angle AOE}{8}$

[/hide]


calculus integration limit algebra function domain logarithms

Let $ f(x)\equal{}\left(1\plus{}\frac{1}{x}\right)^{x}\ (x>0)$.

Find $ \lim_{n\to\infty}\left\{f\left(\frac{1}{n}\right)f\left(\frac{2}{n}\right)f\left(\frac{3}{n}\right)\cdots\cdots f\left(\frac{n}{n}\right)\right\}^{\frac{1}{n}}$.

[hide="A possible reasoning"]We can extend the domain of $ f$, by defining $ f(0) \equal{} 1$, so that $ f$ is continuous at $ [0, 1]$.

Then this limit equals to $ \exp\left[\int_{0}^{1}\ln f(x)\; dx\right]\; \equal{}\;\sqrt{e}$ by the fundamental theorem of calculus.[/hide]

[quote="kunny"][color=darkred]Let $ f(x) =(1+\frac{1}{x})^{x}\ (x > 0)$. Find $ \lim_{n\to\infty}\{f(\frac{1}{n})f(\frac{2}{n})f(\frac{3}{n})\cdots\cdots f(\frac{n}{n})\}^{\frac{1}{n}}$.[/color][/quote]

[color=darkblue][b]Proof.[/b] The function $ g: [0,1]\rightarrow [0,\infty)$, where $ g(x) =\{\begin{array}{ccc}0 & , & x = 0\\ \\ x\cdot\ln(1+\frac{1}{x}) & , & 0 < x\le 1\end{array}$ is continue because $ \lim_{x\searrow 0}g(x) =\lim_{x\searrow 0}[x\ln(1+\frac{1}{x})]= 0 = g(0)$.

$ a_{n}\equiv[\prod_{k = 1}^{n}f(\frac{k}{n})]^{\frac{1}{n}}=[\prod_{k = 1}^{n}(1+\frac{n}{k})]^{\frac{1}{n}}$ $ \implies$ $ \ln a_{n}=\frac{1}{n}\cdot\sum_{k = 1}^{n}\ln(1+\frac{1}{\frac{k}{n}})\rightarrow\int_{0}^{1}x\cdot\ln(1+\frac{1}{x})\ dx =\int_{0}^{1}g(x)\ dx =\ln\sqrt{2e}$ $ \implies$  $ \lim_{n\to\infty}a_{n}=\sqrt{2e}$.[/color]

[quote="Virgil Nicula"][quote="kunny"][color=darkred]Let $ f(x) = (1 + \frac {1}{x})^{x}\ (x > 0)$. Find $ \lim_{n\to\infty}\{f(\frac {1}{n})f(\frac {2}{n})f(\frac {3}{n})\cdots\cdots f(\frac {n}{n})\}^{\frac {1}{n}}$.[/color][/quote]

[color=darkblue][b]Proof.[/b] The function $ g: [0,1]\rightarrow [0,\infty)$, where $ g(x) = \{\begin{array}{ccc}0 & , & x = 0 \\
 \\
x\cdot\ln(1 + \frac {1}{x}) & , & 0 < x\le 1\end{array}$ is continue because $ \lim_{x\searrow 0}g(x) = \lim_{x\searrow 0}[x\ln(1 + \frac {1}{x})] = 0 = g(0)$.

$ a_{n}\equiv[\prod_{k = 1}^{n}f(\frac {k}{n})]^{\frac {1}{n}} = [\prod_{k = 1}^{n}(1 + \frac {n}{k})]^{\frac {1}{n}}$ $ \implies$ $ \ln a_{n} = \frac {1}{n}\cdot\sum_{k = 1}^{n}\ln(1 + \frac {1}{\frac {k}{n}})\rightarrow\int_{0}^{1}x\cdot\ln(1 + \frac {1}{x})\ dx = \int_{0}^{1}g(x)\ dx = \ln\sqrt {2e}$ $ \implies$  $ \lim_{n\to\infty}a_{n} = \sqrt {2e}$.[/color][/quote]
[color=darkblue]Thank! I like this problem.[/color]

[quote="kunny"]Let $ f(x) \equal{} \left(1 \plus{} \frac {1}{x}\right)^{x}\ (x > 0)$.[/quote]

[quote="Virgil Nicula"]$ \left[\prod_{k \equal{} 1}^{n}f\left(\frac {k}{n}\right)\right]^{\frac {1}{n}} \equal{} \left[\prod_{k \equal{} 1}^{n}\left(1 \plus{} \frac {n}{k}\right)\right]^{\frac {1}{n}}$[/quote]

Wait... we defined $ f$ as above, so $ f\left(\frac {k}{n}\right)$  should be equal to $ \left(1 \plus{} \frac {n}{k}\right)^\frac {k}{n}$.
Or maybe I've simply misunderstood  :huh:

Goodbye
GM

I get a value of 1/2 as the value of the indefinite integral.

so the answer should be
\[ \sqrt {e}
\]

can someone verify?

[b]Johnny[/b], yes, this is right answer.

$ \ln a_n \equal{} \frac {1}{n}\ln\left\{\prod_{k \equal{} 1}^{n}f\left(\frac {k}{n}\right)\right\} \equal{} \frac {1}{n}\sum_{k \equal{} 1}^{n}\frac {k}{n}\cdot \ln\left(1 \plus{} \frac {1}{\frac {k}{n}}\right) \equal{} \int_{0}^{1}x\cdot\ln\left(1 \plus{} \frac {1}{x}\right)\;dx \equal{} \frac {1}{2}$ when $ n\to\infty$.

$ \int x\cdot\ln\left(1 \plus{} \frac {1}{x}\right)\;dx \equal{} \frac {x}{2} \minus{} \frac {1}{2}\ln\left(1\plus{}x\right) \plus{} \frac {x^2}{2}\ln\left(1 \plus{} \frac {1}{x}\right) \plus{} \mathbb{C}$


[hide]Subtracting the first equation from the second gives
$y=2$
Substituting back into the first gives $x=12$[/hide]

[quote="trgtft"]x-y=10
x-2.y=8
x=?[/quote]

[hide]
All we want is $x$, so let's eliminate $y$ (sorry, you'll get your chance another day).  Multiplying the first equation by 2 gives us $2x-2y=20$.  Subtracting the first from the second gives us $\boxed{x=12}$.
[/hide]

Not to spam, but http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

Read up on LaTeX. For simpler equations like these, it's so easy to make them into

$x-y=10$
$x-2\times y=8$
$x=?$

solving using the elimination method:

$x-y=10$
$x-2y=8$

$-1(x-y=10)$

$-x+y=-10$

add this equation to x-2y=8

and you get -y=-2
which means y=2 and x=12

[hide="Time for Something Different"]
I will be using cramer's rule for no particular reason.

$d_{1}=Determinant\begin{bmatrix}1&1\\1&2\end{bmatrix}=1$
$x_{1}=Determinant\begin{bmatrix}10&1\\8&2\end{bmatrix}=12$
$x=\frac{12}{1}=\boxed{12}$
[/hide]

[quote="trgtft"]x-y=10
x-2.y=8
x=?[/quote]
                                                                     Solution(by galois01)
                     x-y=10 (1)
                     x-2y=8 (2)
                     from (1) we have: x=10+y (3)
                     so (2) gives 10+y-2y=8 => y=2
                     finally (3) gives x=12

[quote="hunter34"][hide="Time for Something Different"]
I will be using cramer's rule for no particular reason.

$ d_{1}=Determinant\begin{bmatrix}1&1\\1&2\end{bmatrix}=1$
$ x_{1}=Determinant\begin{bmatrix}10&1\\8&2\end{bmatrix}=12$
$ x=\frac{12}{1}=\boxed{12}$
[/hide][/quote]

I don't think that most middle school kids learn cramer's rule or matrices or determinants

[quote="trgtft"]x-y=10
x-2.y=8
x=?[/quote]
if thats a decimal point, then the second equation could also b writen as $ x-2+\frac{1}{10}y=8$. solving gets us $ y=-10x+100$
using substitution we get $ x-(-10x+100)=10$ so $ 11x=110$ and $ x=10$.

EDIT:is that $ (x-2).y or x-(2.y)$?

[quote="Temperal"]It's supposed to be a multiplication sign. e.g. $ \times$[/quote] then it could just have been $ 2y$. it looked like a decimal point to me

Some people just don't use standard notation. But trust me, no problem would use a variable after a decimal point.

yeah  there was a problem like this on 1 of my tests. you just say $ 2.y=2+\frac{1}{10}y$. remeber. expanded notation. ive had to do this since like 2nd grade.

Okay, fine. But most people don't use it, so just assume.


The circumference of a circle increases from $ 20\pi$ cm to
$ 30\pi$ cm. By how many centimeters will the radius increase?

20pi = 2pi*r
r=10 before

30pi = 2pi*r
r=15 now

15-10= $ \boxed{5}$ cm

Also, a solution is that you know that the circumference of a circle is $2\pi{r}$. The difference between the two circumferences are 10, but since the radius is multiplied by 2 each time, than the actual difference between the radius is two times less than the circumference, which is $ \boxed{5}$.


How many different convex hexagons can be formed using only sides and/or diagonals of regular decagon ABCDEFGHIJ? Two such hexagons to be included are ABCDEF and BCDEFG.

In decagon ABCDEFGHIJ, we must choose $ 6$ vertices to make a hexagon.

There are $ \dbinom{10}{6}\equal{}\boxed{210}$ ways to do this.

Hey guys, i forgot how to do 10 choose 6 like Akhil did above, could someone write it out for me? Thanks!

$ \binom{10}{6} \implies \frac{10!}{6!4!} \implies \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6  \cdot 5}{6!} \implies \frac{10 \cdot 9 \cdot 8 \cdot 7}{4!} \implies \frac{5 \cdot 3 \cdot 7 \cdot 2}{1} \equal{} 210$


There is fraction in which denominator>numerator by3.If 1 is added to both,the differnce betwen new and old frctions is 1/24.find fraction.

x=original numerator

$\frac{x+1}{x+4}-\frac{x}{x+3}=\frac{1}{24}$
$\frac{3}{x^2+7x+12}=\frac{1}{24}$
$x^2+7x-60=0$
$x=5$ or $x=-12$

Leaving out -ves (since they'll cancel out in the frac., and you'll get improper answers)
the fraction is 5/8

but why can't we take 12/9?If we add one to both numerator and denom.,it will be diff from what we obtain by doing so in 4/3 :huh:

IS my above post right?Are there 2 valid answers?

[quote="JesusFreak197"]$\frac{12}{9}$ doesn't work... $\frac{13}{10}-\frac{12}{9}=-\frac{1}{30}$. :huh:

Not only that, but it says that the denominator is greater than the numerator, not the other way around. It has to be for the difference to be positive.[/quote]

First of all [b][size=150][color=orange]HAPPY NEW YEAR!![/color][/size][/b]

Secondly,jesusfrk,take the fraction as -12/-9 not 12/9...then carry on with the calculations [b][u]without[/u][/b] cancelling any signs annnyywheree except when you multiply the denominators while taking L.C.M.It is coming as $1/24$

The problem with that is that $\frac{-12}{-9}=\frac{12}{9}$, and it doesn't work for $\frac{12}{9}$. It probably should have specified that both the numerator and denominator were positive to avoid confusion.

So there are 2 answers taking $-12/-9$ as one.

Yes there are.
By JesusFreak197's reasoning, $\frac 58$ wouldn't work because it is equivalent to $\frac{10}{16}$, which doesn't work.

[quote="JesusFreak197"]No, because the simplest form of $\frac{10}{16}$ is $\frac{5}{8}$. Same thing with $\frac{-12}{-9}$ - it becomes $\frac{12}{9}$, which doesn't even meet the conditions. That's where the problem lies.[/quote]
So the only answer is 5/8 :?:
And shouldn't they have specified that it had to be positive and in simplest form, to avoid  misunderstandings?

[quote="JesusFreak197"]Well, on MC tests, they say that all answers must be in simplest form... and if you put any of those other attempts into simplest form, they're either the correct answer or don't work. ;)[/quote]
Which other attempts, exactly?

But when you solve it all like ptotype did,you directly get 5/8 and not 10/16 and we directly get -12/-9.Our sir said that -12/-9 is extranoeous for some reason :huh:

Confusing......

Yes, read my post... $\frac{-12}{-9}$ is not simplest form. All answers on MC must be in simplest form, so that cannot possibly be an answer. The one they're looking for is $\frac{5}{8}$, which is in simplest form.

[quote="JesusFreak197"]Actually, I guess the only other suggestion was $\frac{-12}{-9}$, but in simplest form that would be $\frac{12}{9}$.[/quote]

Ummm actually $\frac{-12}{-9}$ is $\frac43$ in simplist form.

I think that the two answers are $\frac 43$ and $\frac 58$. $\frac 43$ is an answer because $\frac{-12}{-9}$ is an answer and in simplest form is $\frac 43$. However, I would out $\frac 58$ in a contest.

in mc, if a question has 2 answers, they'll ask for both answers.
and i wouldn't put 4/3, i'd put -12/-9 as it is an answer to the problem.

JUST BECAUSE IT'S NOT IN SIMPLEST FORM DOESN'T MEAN IT'S NOT A FRACTION...

Yes, but it states in the MC rules that the fractions MUST be in simplest form unless the problem otherwise specifies. In this case it says nothing about leaving the fraction in its original form, and therefore you should put $\frac 43$.

well i would put the best possible answer...in this case $\frac{5}{8}$.

Wow, seriously, people... $\frac{-12}{-9}$ is not in simplest form, and is therefore not an answer. $\frac{4}{3}$ obviously doesn't work, so that can't be the answer, even though it's the simplest form of $\frac{-12}{-9}$. They would not have an answer to a problem that was not in simplest form and then tell you to put it in simplest form. :huh: It just doesn't make sense.

$\frac{5}{8}$ works in simplest form, and you're only supposed to put one answer unless specified otherwise, which this problem didn't do. Therefore, the intended answer is obviously $\frac{5}{8}$.

Who is talking about mathcount,this is just a general question i posted here bcos i think its mathcount level.I understand that this forum has to follow the rules of mathcount,so if i posted this in Getting startted the answers would be $-12/-9,5/8$. :D

For mathcount,5/8 :)


function
analytic geometry
graphing lines
slope

The speed v (in feet per second) of a car is a linear function of time t (in seconds) for 10 is less than or equal to t, which is less than or equal to 30.  If after each second the speed of the car has increased by 5 feet per second and if after 20 seconds the speed is 80 feet per second, how fast is the car going after 30 seconds?  Also, find the function v = v(t).

[hide]

On a graph, this function would have a slope of 5, since the speed increases 5 feet per second. After 20 seconds, the speed is 80, so a point on this graph is (20,80). Plugging the slope and this point into the slope-intercept formula:

$v-v_{1}=m(t-t_{1})$
$v-80=5(t-20)$
$v-80=5t-100$
$v=5t-20$

Putting 30 seconds in for $t$

$v=5(30)-20$
$v=150-20$
$v=130$

[/hide]

You guys are amazing tutors!  I would have never done this alone.

Thanks!


analytic geometry
graphing lines
slope
vector
geometry
3D geometry
sphere

(For those of you who lost your packets, the AHSME problems from activity 13 are [url=http://www.math.uncc.edu/~hbreiter/AHSME/1996st.pdf]here[/url].)

This is a transplant from the old forum.  I think we came to a consensus on these:

15. b
16. d
17. e
18. d
19. d
20. c
21. d
22. b
23. b
24. b (or d?)
25. b
26. b
27. d
28. c
29. c
30. e

Solutions:
24 (from Sara S.)
For #24, I said that the sum of the first n blocks is n^2:
n=1, 1, 1 term
n=2, 1+(2+1)=4, 3 terms
n=3, 1+(2+1)+(2+2+1)=9, 6 terms
...
set # of terms = n(n+1)/2=1234, n~49, 49*50/2=1225

so the sum is 49^2+2(1234-1225)=2401+18=2419, B

but there are probably other ways of doing it

[b]25[/b] (from Sara C.)
for 25, you can rearrange the equation:
(x-7)^2 + (y-3)^2 = 8^2
and you want the largest value of 3x + 4y... let that equal a... so y = (-3/4)x + a/4. so you want the largest value of a so you want the highest line with slope (-3/4) that still passes through that circle. and i guess there are lots of ways to go from there.. work some vector magic, etc...

[b]26[/b] (from Kevin)
the way i did 26 was to write out what the probabilities of each was, in terms of r,w,b,g (the number of marbles of each color), and n, the total number of marbles (which ends up canceling out). out of that you get a bunch of relationships such as r-3=4w, from which you can find the minimum number of each color, and then the total number.

[b]27[/b] (from Joe)
For 27, there's only one value for z that's an integer (5). Plug in 5 for z in one of the generic sphere equations for one of them, and you get an equation for a circle. Find lattice points...
For 30, um... I think I have a method but I sort of calculated things wrong and just skipped it in the end... I'll explain over email if you want.

(from Kevin)
for #27, they're balls, ie not just the surface. i think it's easier just to look at how far they extend along the z axis--one reaches to 11/2, and the other reaches back to 9/2, and 5 is the only integer between the two, so all the latice points have a z-coordinate of 5. from theres its just a 2D problem.

[b]30[/b]

Let the hexagon be ABCDEF with AB = BC = CD = 3, DE = EF = FA = 5. Let O be the center of the circumscribed circle.

From a bit of angle chasing, we find that angles BCD and EFA are both equal to 120 degrees. So from the law of cosines, BD = EA = 7. So from the formula for circumradius of a triangle,

R = (abc)/(4[ABC])

which follows from the law of sines and the familiar formula [ABC] = 1/2*ab*sin C

we know that the circumradius of BCD is
R = (3*5*7)/(15*sqrt(3)) = 7/sqrt(3)

But the circumcircle of BCD is the same as that of the entire hexagon, so we now know the radius of the circle.

Let angle AOB equal x. Then applying the law of cosines in reverse to triangle AOB yields cos x = 71/98.

Almost done! We want to find AD, and triangle AOD has two sides of lenght R and an included angle of size 3x. So we basically want to find cos 3x in order to apply the law of cosines.

Using the cosine addition formula (or DeMoivre's Theorem, perhaps), we have cos 3x = 4 cos^3 x - 3 cos x = -76751/117649 (good thing calcs are allowed).

Finally, from the law of cosines, AD = \sqrt{49/3 + 49/3 + 2*49/3*cos 3x} = 360/49.

Thus the answer is 409, E.

For 28, find the volume of ABCD two different ways, and viola! your answer

For 30, I think this is probably an easier way:
[hide]Say we have AB=BC=CD=3, and DE=EF=FA=5. Note that angle CDE is 120 degrees. From the law of cosines, we can find that CE=7. Now let BE=x. From Ptolmeys theory on BCEF, we find that x=BE=8. Ptolemy's on BCDE, we can find BD, and since BD=AC, Ptolmey's on ABCD should do the trick.[/hide]

for 24, you can solve it by realizing that there are 49 1's in the series by solving
                            n 2 +n=1234
                            n=49.13....
then just multiplying 1234 by 2 and subtracting 49 for the 49 ones.
                            2468-49=2419=B


analytic geometry
graphing lines
slope
conics
ellipse

Line $y=1-x$ meets the ellipse: $mx^2+ny^2=1$ at $M$ and $N$. $P$ is the midpoint of $MN$. If the slope of $OP$ is $\frac{\sqrt{3}}{3}$, then what is the value of $\frac{m}{n}$.

$O$ is the origin ..

[hide="solution"]

both the line and ellipse intersects at $mx^2+n(1-x)^2=1$ or $(m+n)x^2-2nx+(n-1)=0$

Letting $M(x_1,1-x_1)$ , $N(x_2,1-x_2)$ so $x_1+x_2=\frac{2n}{m+n}$

Hence $P(\frac{n}{m+n},\frac{m}{m+n})$

So $OP^2=\frac{n^2}{(m+n)^2}+\frac{m^2}{(m+n)^2}$ $\implies m^2-mn+n^2=0$

Hence $m=\frac{1+\sqrt{5}}{2}n$ (positive because for ellipse , $m,n>0$)

So $\frac{m}{n}=\frac{1+\sqrt{5}}{2}$[/hide]


calculus
limit
integration
calculus computations

1. If $ 0<a<b$, find $ \lim_{t\to 0} \left\{\int_0^1 [bx+a(1-x)]^tdx\right\}^{\frac{1}{t}}$.

2. Evaluate $ \int_0^1\left(\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}\right)dx$. I'm pretty sure the fact the two things are inverses is to be used.

For 2, I think $ u \equal{} x^{7} \iff du \equal{} 7(x^{3})^{2} dx$ is the important substitution step.

[quote="aspiringchemlete"]2. Evaluate $ \int_0^1\left(\sqrt [3]{1 \minus{} x^7} \minus{} \sqrt [7]{1 \minus{} x^3}\right)dx$. I'm pretty sure the fact the two things are inverses is to be used.[/quote]You are right, their being inverses gives the answer immediately.  Draw a picture and you will see why.

The problem has been posted a couple of times already, but is fun to think about.


[hide="Dont think this is right.."]Using the sum of infinite series:
$\text{S}=\frac{a}{1-r}$
$\alpha=\frac{1}{1-z}$
$z=\frac{1}{1-y}$
$y=\frac{1}{1-x}$

$\Rightarrow z=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$
$\Rightarrow \alpha=\frac{1}{1-\frac{x-1}{x}}=x\implies \alpha^{2}=x^{2}$

$x^{2}=\left(\frac{1}{1-\frac{\sqrt{3}}{2}}\right)^{2}=\left(\frac{2}{2-\sqrt{3}}\right)^{2}=\boxed{4\left(7+4\sqrt{3}\right)}$[/hide]

Well the second sum doesn't converge... :huh:

ya i saw that when i was doing the solution...hence I knew it was likely wrong.

i got 4+2*sqrt(3)

i would show my work... but i dunno how to type it up correctly...

This is not right. The first one yields, by the sum of geometric series formula:
$x = \frac{cos\ 30\textdegree}{1-cos\ 30\textdegree}= \frac{\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}= \frac{\sqrt{3}}{2-\sqrt{3}}= 2\sqrt{3}+3$

This is greater than 1, so the second one will not converge.

oops....   I didn't check to see if it WOULD converge.....

Sorry, folks. :oops_sign:

edit: hey! The smilies counter says there was four smilies when there was only two! What's going on?!


[quote="thachpbc"]Find all continuous functions $f: R \to R$ so that for all real x, we have  $f(f(x))= 2 f(x)+1$.[/quote]
$f(2f(x)+1)=f(f(f(x)))=2f(f(x))+1=4f(x)+3$. Therefore if $y=2f(x)+1$ then $f(y)=2y+1$
1. One solution $f(y)=-1 \ \forall y\in R$.
2. Let exist $y_{0}$, suth that $f(y_{0})\not =-1$, then exist $y_{1}=2(f(y_{0})+1)-1, f(y_{1})=2y_{1}+1, \ y_{2}=2^{2}(f(y_{0})+1)-1,f(y_{2})=2y_{2}+1,...$
If $f(y_{0})>-1$ then for all $z>f(y_{0})$ exist x, suth that $f(x)=z$ (by continiosly), therefore for all $x>2f(y_{0})+1$ $f(x)=2x+1$.
Therefore all continiosly function given by:
1. $f(x)=-1 \forall x\in R$.
2. Exist $x_{0}\ge-1$, suth that $f(x)=2x_{0}+1$ if $x\le x_{0}$ and $f(x)=2x+1$ if $x>x_{0}$.
3. Exist $x_{0}<-1$, suth that $f(x)=2x_{0}+1$ if $x>x_{0}$ and $f(x)=2x+1$ if $x<x_{0}$.
4. $f(x)=2x+1$.


I'm stuck with these problems... could you pls explain it.
if it is too much any of them will be appreciated.. thx

1) When (a+b+c+d)^10 is expanded and like terms combined, how many terms are in the result?

2)Suppose that 7 boys and 13 girls line up in a row. Let S be the number of places in the row where a boy and a girl are standing next to each other. for example, for the row GBBGGGBGBGGGBGBGGBGG we have S=12. If all possible orders of these 20 people are considered what is the average value of S? Can you generalize this result to a group of m boys and n girls?

3) Let n be a fixed positive integer. Find the sum of all positive integers with the following property: In base 2, it has exactly 2n digits consisting of n 1's and n 0's. (the first digit canno be 0)

all from AoPS Vol 2....  i don't understand the solution.

Is the answer for the first one 322 terms?
If so I can explain how I got it for you.

the solution says "hence, as discussed in the text, there are
( 13 )
(  3  )   =   286 terms


find all $ (a,b,c) \in N^3$ with $ a,b,c \neq 0$ such that: $ a^2\plus{}2^{b\plus{}1}\equal{}3^c$

[hide="outline"]$ \mod 4$ we see $ c \equal{} 2k$.
Hence $ 2^{b \plus{} 1} \equal{} (3^k \minus{} a)(3^k \plus{} a)$, and so $ 3^k \minus{} a \equal{} 2^m$ and $ 3^k \plus{} a \equal{} 2^m$ with $ m, n \geq 0$.
Since $ 3$ doesn't divide $ a$ then $ (3^k \minus{} a, 3^k \plus{} a) \equal{} 2$.
Hence we have $ 3^k \minus{} a \equal{} 2$ and $ 3^k \plus{} a \equal{} 2^n$, from which $ 3^k \equal{} 2^{n \minus{} 1} \plus{} 1$. It's well known (Engel, for example) that the only solutions are for $ n \equal{} 2$ and $ n \equal{} 4$.
Finally the solutions are $ (1, 2, 2)$ and $ (7, 4, 4)$.[/hide]

Taking $ mod 4$ then $ c \equal{} 2k$. Taking $ mod 3$ then $ b \equal{} 2k$. So

$ (3^k \minus{} a)(3^k \plus{} a) \equal{} 2^{2l \plus{} 1} \ \ (1)$. But $ (3^k \minus{} a,3^k \plus{} a) \equal{} 2$ so $ \left(\frac {3^k \minus{} a}{2},\frac {3^k \plus{} a}{2}\right) \equal{} 1$ and then the equation $ (1)$ becomes $ \frac {3^k \minus{} a}{2}\cdot\frac {3^k \plus{} a}{2} \equal{} 2^{2l \minus{} 1}$.

So $ \frac {3^k \minus{} a}{2} \equal{} 1$ and $ \frac {3^k \plus{} a}{2} \equal{} 2^{2l \minus{} 1}$ and then $ 3^k \equal{} 2^{2l \minus{} 1} \plus{} 1 \ \ (2)$

i) For $ l\geq 2$, take $ mod4$ and we get $ k \equal{} 2k'$ so the equation $ (2)$ becomes $ (3^{k'} \minus{} 1)(3^{k'} \plus{} 1) \equal{} 2^{2l \minus{} 1}$. But

$ \left(\frac {3^{k'} \minus{} 1}{2},\frac {3^{k'} \plus{} 1}{2}\right) \equal{} 1$ so $ \frac {3^{k'} \minus{} 1}{2} \equal{} 1$ and $ \frac {3^{k'} \plus{} 1}{2} \equal{} 2^{2l \minus{} 3}$.

Then $ l \equal{} 2$ so $ b \equal{} 4$ and  $ k' \equal{} 1$ so $ c \equal{} 4$ and $ a \equal{} 7$. We have the solution $ (7,4,4)$.

ii) If $ l \equal{} 1$ we get the solution $ (1,2,2)$.

Alexandros

[quote="bboypa"]$ (7,4,4)$? :roll:[/quote]
You're right! Nella fretta ho dimenticato l'altra.  :blush:


function
integration
algebra
polynomial
real analysis
real analysis unsolved

Is it possible for a continuous function $ f$ on $ [0,1]$ to be such that $ \int_0^1 xf(x)\equal{}1$ and $ \int_0^1 x^n f(x)\equal{}0$ for $ n\equal{}0,2,3,4,...$?

Thanx in advance.

No.  The polynomials (even with no linear term) are dense in the continuous functions on $ [0, 1]$, so the second condition implies that $ \int_{0}^{1} f(x)^2 \equal{} 0 \Rightarrow f(x) \equal{} 0$.

Let $ A \subseteq \Bbb{N}$ be a set of positive integers.  Then the following are equivalent:

(i) If $ f$ is continuous on $ [0,1]$ and $ \int_0^1 f(x) x^s\,dx \equal{} 0$ for all $ s \in S$, then $ f \equal{} 0$

(ii) $ \sum_{s \in S} \frac{1}{s} \equal{} \infty$

I've just got to understand t0rajir0u's solution. As for transseries's suggestion, could you provide some hint? Thanx.

It seems that if $ \sum_{s\in S}\frac{1}{s}\equal{}\infty$, then the subspace spanned by $ \{x^s\}_{s\in S}$ is dense in the space of continuous functions. I do not see the point why. In t0rajir0u's suggestion, I can apply Stone-Weierstrass theorem. But this does not work in the above case...

What transseries wrote is a direct consequence of Muntz's theorem.


Let s={1, 2, 3, ... , n}. For what vaule of n can s be split into two disjoint subsets A and B whose union is s and the sum of the numbers in A is the same as the sum of the numbers in B.

Like $ 1\plus{}2\plus{}...\plus{}n\equal{}\frac{n(n\plus{}1)}{2}$
Then $ n\equiv1(4)$ or $ n\equiv3(4)$
In the both cases we can do an rearrangement that prove the problem
For $ n\equiv1(4)$:
See that $ n\equal{}4k$
$ (4k)\plus{}(1)\equal{}(4k\minus{}1)\plus{}2\equal{}...\equal{}(2k\plus{}1)\plus{}(2k)$
For $ n\equiv3(4)$:
See that $ n\equal{}4k\minus{}1$
$ (4k\minus{}1)\equal{}(4k\minus{}2)\plus{}1\equal{}...\equal{}(2k)\plus{}(2k\minus{}1)$


$\sum_{k=0}^n \frac{{n \choose k}}{{m \choose k}} = ?$

We have
\[S=\sum_{k=0}^n \frac{{n \choose k}}{{m \choose k}}=\frac{n!(m-n)!}{m!}\sum_{k=0}^n {{m-k \choose m-n}}=\frac{n!(m-n)!}{m!}\sum_{k=0}^{n} {(m-n)+k \choose m-n}.\]
Using induction on $n$, it ieasy to show that
\[\sum_{k=0}^{n}{s+k \choose s}={s+n+1 \choose n}.\]
Therefore
\[S=\frac{n!(m-n)!}{m!}\cdot {(m-n)+n+1 \choose n}=\frac{m+1}{m-n+1}.\]

 :D  :D  :D

Thank's.
What about this sum:
$\sum_{k=0}^n k {m-k-1 \choose m-n-1}$
It seems to similar but i get:
$\sum_{k=0}^n k { s+k \choose s}$
Where $s = m-n-1$


How many solutions are there for the following equation?
2arcsin2x + arccos2x = 4

I got 1 but the answer is 0. Could someone explain me?

hello, your equation $ 2\arcsin(2x)\plus{}\arccos(2x)\minus{}4\equal{}0$ is equivalent to
$ \minus{}4\plus{}\pi\minus{}\arccos(2x)\equal{}0$, this equation has no real solution.
Sonnhard.


You have two identical crystal orbs. You need to figure out how high an orb can fall from a 100 story building before it breaks. You know nothing about the toughness of the orbs: they may be very fragile and break when dropped from the first floor, or they may be so tough that dropping them from the 100th floor doesn't even harm them.
What is the largest number of orb-drops you would ever have to do in order to find the right floor? (i.e. What's the most efficient way you could drop the orbs to find your answer?)
You are allowed to break both orbs, provided that in doing so you uniquely identify the correct floor.


geometry
geometric transformation
reflection
circumcircle
geometry proposed

Let $ B,C$ be fixed points on circle $ (O)$. $ A$ is a mobile point on $ (O)$. $ D,E$ are the reflections of $ B,C$ through $ AC,AB$. Prove that the line passes $ A$ and perpendicular to $ DE$ passes a fixed point when $ A$ moves on $ (O)$

[hide="Hint"]The line passes $ A$ and perpendicular to $ DE$ always passes through the circumcenter of triangle $ OBC$[/hide]

We have to prove that:
\[ AE^{2}-AD^{2}=EU^{2}-DU^{2}\]
where $ U$ is the center of $ \triangle{OBC}$ (for this implies $ AU\perp{DE}$). Equivalently,
\[ b^{2}-c^{2}=-2a\cdot{BU}(\sin(2C)-\sin(2B)). \]
Since $ BU=\frac{R}{2\cos(A)}$, the relation becomes:
\[ \frac{1}{2}(\cos(2C)-\cos(2B))=\tan(A)\cdot{\cos(B+C)\cdot{\sin(B-C)}}\]
whic is, of course, true.

It is also interesting to notice that $ AD$ (and respectively $ AE$) passes through a fixed point $ M$ which is found on the circumcircle of $ \triangle{ABC}$ and $ MC\perp{UC}\Longleftrightarrow MC=BC$.


[quote="madness"]See the sum below . It reminds me of the FTSE sum we had and all argued. Nostalgia
Chappli? Shreyas?
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=182460 .[/quote]

Probably both from the same source. FIITJEE isn't framing their one sums incase you didn't know that. I got that wrong like a fool of a took/


Find all functions $f: \mathbb R\to \mathbb R$ satisfying the equation $f(x^{2}+y)+f(f(x)-y)=2f(f(x))+2y^{2}$ for all $x,y\in\mathbb R$.

[hide]
First plug in $y=0$ to get $f(x^{2})+f(f(x))=2f(f(x))$, or $f(x^{2})=f(f(x))$.

Then plug in $y=f(x)-x^{2}$ to get $f(f(x))+f(x^{2})=2f(f(x))+2y^{2}$, which simplifies to $0=2y^{2}$. Therefore $y=0$ and $\boxed{f(x)=x^{2}}$.
[/hide]


An 11x11x11 wooden cube is formed by gluing together $11^3$ unit cubes. What is the greatest number of unit cubes that can be seen from a single point?

$\text{(A)} \ 328 \qquad \text{(B)} \ 329 \qquad \text{(C)} \ 330 \qquad \text{(D)} \ 331 \qquad \text{(E)} \ 332$

[hide]You can see a maximum of 3 faces from a single point. This is $11^2\cdot3-11\cdot3+1=331$ cubes. $\boxed{D}$[/hide]

[hide="Answer"]If we look at the cube with one vertex as the closest point, then we see all but a 10x10x10, so we see $331$ cubes. $\boxed{D}$[/hide]

[hide]We can see 3 faces. 11*11*3 = 363 - 10*3 (we counted these twice..) - 2 (this point was counted 3 times instead of once) = 331
$\boxed{D}$[/hide]

FWIW -  Another easy way to see this is:
[hide]$11^3-10^3$  (You  can seen all on  the layer outside the 10x10x10 cube ) [/hide]


find all $ x,y \in R$ that :
\[ sin^3x\plus{}sin^3y\equal{}(cos^3x\plus{}cos^3y)tan^3(x\plus{}y)\]
\[ cos^2x\plus{}cos^2y\equal{}cos(x\minus{}y)\]


1. What exactly does getting a commended solution mean?

2. Are the problems in round 1 easier than the ones in round 4? Do they get progressively harder as the year goes on, or do they stay about the same?

Commended = the grader thought your solution was elegant and well-written.

We do expect the difficulty of the problems to increase over the year.


geometry
search
combinatorics unsolved
combinatorics

Find the least $ m$ for which any five equlilateral triangles of the total area $ m$ can cover an equilateral triangle of area 1.

Was just posted very recently, and no response there yet either:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=152414865&t=307289
Perhaps you should try a search to see if there are other past instances?

(I found that one very quickly by searching for the words "five" and "equilateral".)


Suppose that $ \tau$ is a metrizable topology on a set $ X$ of cardinality less than or equal to continuum. Prove that there exists a separable and metrizable topology on $ X$ that is coarser that $ \tau$.

[i]L. Juhasz[/i]

The aim is to prove that we can find countably many real-valued continuous functions on $ (X,\tau)$ which separate the points of $ X$ (that is, for every two $ x\ne y\in X$ there is a function $ f$ in this countable family with $ f(x)\ne f(y)$). Such functions give a one-to-one continuous map of $ (X,\tau)$ into a countable product of real lines, and the topology on $ X$ generated by this map has the required properties (it identifies $ X$ to a subspace of a countable product of real lines, and this latter space is a separable metric space).

Now to find our countable family of functions on $ X$. For each $ n\ge 1$, let $ X_n$ be a maximal family of points in $ X$, each two at a distance $ \ge\frac 1n$ from each other. Then, for each $ m > n$, let $ X_n^m$ be the union of the closed balls of radius $ \frac 1n \minus{} \frac 1m$ centered at the points of $ X_n$. The set $ X_n$ has at most continuum cardinality, by the hypothesis, and the set $ X_n^m$ is closed for every pair $ m > n$ of positive integers. Moreover, each of the balls of radius $ \frac 1n \minus{} \frac 1m$ mentioned above is an open subset of $ X_n^m$. It follows that we can find a real-valued continuous function $ f_{n,m}$ defined on $ X_n^m$ which is constant on each ball, but which assigns different values to different balls. Extend $ f_{n,m}$ to a continuous real-valued function on $ X$ (Tietze Extension Theorem) denoted by the same symbol. It should be clear now that the functions $ f_{n,m}$ separate the points of $ X$ (two distinct points will be members of different balls in some set $ X_n^m$), and this finishes the proof, according to the previous paragraph.


calculus
Euler
function
integration
analytic geometry
derivative

I came across the problem in multi-variable calculus in  a study called "Jerky Mechanics".We have to find the Euler-Lagrange equation if the Lagrangian is a function of three variables.
$ L(q,q',q'')$
We have to argue that the quantity
$ \delta S \equal{} \int_{t_1}^{t_2}{L(q,q',q'')}dt \equal{} 0$ and then find the Euler-Lagrange equations.Please show each and every steps.

First your equation doesnt make sense you wrote $ \delta S \equal{} ...$ but what you wrote is the action itself not the variation of the action. You should have written $ \delta S \equal{} \delta ... \equal{} 0$

So I guess you want to vary the $ q$ coordinate right?

I assume that $ q$ is just $ q$ i.e. one coordinate not a set of coordinates $ q_1,q_2\dots$

Well just vary it, take:

\[ S[q\plus{}\delta q] \equal{} \int_{t_1}^{t_2}L(q\plus{}\delta q, q'\plus{}\delta q', q'' \plus{} \delta q'') dt \equal{} \int_{t_1}^{t_2}\left(L(q,q',q'')\plus{}\frac{\partial L}{\partial q}\delta q \plus{}\frac{\partial L}{\partial q'}\delta q' \plus{}\frac{\partial L}{\partial q''}\delta q''\right)dt \equal{} \int_{t_1}^{t_2}L(q,q',q'')dt \plus{} \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\delta q \plus{}\frac{\partial L}{\partial q'}\delta q' \plus{}\frac{\partial L}{\partial q''}\delta q''\right)dt \equal{} S[q] \plus{}\delta S\]

Hence we only have to evaluate:

\[ \delta S \equal{} \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\delta q \plus{}\frac{\partial L}{\partial q'}\delta q' \plus{}\frac{\partial L}{\partial q''}\delta q''\right)dt\]

First of all we can write:

\[ \frac{\partial L}{\partial q'}\delta q' \equal{} \frac{\partial L}{\partial q'}\frac{d\delta q}{dt}\]

and:
$ \frac{\partial L}{\partial q''}\delta q'' \equal{} \frac{\partial L}{\partial q''}\frac{d\delta q'}{dt}$

Now we have to use partial integration:

\[ \int_{t_1}^{t_2}\frac{\partial L}{\partial q'}\delta q' dt \equal{} \int_{t_1}^{t_2}\frac{\partial L}{\partial q'}\frac{d\delta q}{dt} dt \equal{} \left[\frac{\partial L}{\partial q'} \delta q\right]_{t_1}^{t_2} \minus{} \int_{t_1}^{t_2}\frac{d}{dt}\left(\frac{\partial L}{\partial q'}\right)\delta q dt\]

We dont vary on the boundary points, since we fixed them this means that the first term in the brackets should be zero since it has to be evaluated at the endpoints and contains the variations.

Lets do the same with the other term:

\[ \int_{t_1}^{t_2}\frac{\partial L}{\partial q''}\delta q'' dt\equal{} \int_{t_1}^{t_2}\frac{\partial L}{\partial q''}\frac{d\delta q'}{dt}\equal{}\left[\frac{\partial L}{\partial q''} \delta q'\right]_{t_1}^{t_2} \minus{} \int_{t_1}^{t_2}\frac{d}{dt}\left(\frac{\partial L}{\partial q''}\right)\delta q' dt \equal{} \minus{}\int_{t_1}^{t_2}\frac{d}{dt}\left(\frac{\partial L}{\partial q''}\right)\frac{d\delta q}{dt} dt \equal{} \minus{}\left[\frac{d}{dt}\left(\frac{\partial L}{\partial q''}\right)\delta q\right]_{t_1}^{t_2} \plus{} \int_{t_1}^{t_2}\frac{d^2}{dt^2}\left(\frac{\partial L}{\partial q''}\right)\delta q dt\]

Again all of the evaluated terms at the boundaries are zero.

Combining our results:

\[ \delta S \equal{}  \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\minus{}\frac{d}{dt}\left(\frac{\partial L }{\partial q'}\right) \plus{} \frac{d^2}{dt^2}\left(\frac{\partial L }{\partial q''}\right)\right)\delta q dt\]

We want $ \delta S \equal{}0$, which means that the above expression has to be zero for any variation $ \delta q$.

With this we obtained the Euler lagrange equations:

\[ \frac{\partial L}{\partial q}\minus{}\frac{d}{dt}\left(\frac{\partial L }{\partial q'}\right) \plus{} \frac{d^2}{dt^2}\left(\frac{\partial L }{\partial q''}\right) \equal{} 0\]

Thanks yeahnigga.Nice proof.But I have 1 doubt.Why should we take
$ \delta q'\equal{}0$ at the boundary points?Taking $ \delta q\equal{}0$ is understandable as the particle begins and ends at the same point.But I don't understand how $ \delta q'\equal{}0$. Please elaborate.

we have:
\[ \delta q' \equal{} \frac {d}{dt}\left(\delta q\right)\]
Basically we want to calculate the rate of change of $ \delta q$ at the boundary, but we know that at the boundary $ \delta q$ doesnt change plus, we know that it is zero. This means that it doesnt vary wrt time implying that the above derivative is zero at the boundary.


inequalities
function
integration
calculus
trigonometry
real analysis
real analysis solved

If $g:[0,\pi]\to\mathbb R$ is defined by $g(x)=f\left(\frac x\pi\right)$, then we must prove $\int^\pi_0g^2dt\le\int^\pi_0g'^2dt$, given the assumption that $g(0)=g(\pi)=0$.

We can regard both $g$ and $g'$ as $\pi$-periodic $L^2$ functions defined on $[-\pi,\pi]$ (simply extend them to the entire segment $[-\pi,\pi]$ so as to preserve periodicity, and forget about the continuity and all that; all we need is the fact that they are in $L^2$). We now compute the Fourier coefficients $\hat g(n)$ and $\hat g'(n)$ and reach the conclusion that due to the hypothesis $g(0)=g(\pi)=0$, we have $\hat g'(n)=n\cdot i\cdot \hat g(n),\ \forall n$ (it's just integration by parts).

We will eventually get $\|g\|_2=\sum|\hat g(n)|^2\le\sum n^2|\hat g(n)|^2=\sum|\hat g'(n)|^2=\|g'\|_2$, as desired.

The one thing grobber didn't mention about his extension is that we should extend $f$ to be an odd function, thereby causing the constant term of the Fourier series to vanish.

What follows is the same as grobber's argument; I'm just putting it here to give readers some alternate terminology.

Extend the original function $f$ to be an odd function on $[-1,1]$ and then as periodic of period 2. This extension is continous and $C^1.$ The Fourier series of such a function converges everywhere and, as an odd function, we may write

$f(x)=\sum_{n=1}^{\infty}b_n\sin(n\pi x).$

Then, $f'(x)=\pi\sum_{n=1}^{\infty}nb_n\cos(n\pi x).$

By Parseval's identity:

$\int_0^1(f(x))^2dx=\frac12\sum_{n=1}^{\infty}b_n^2$ and

$\int_0^1(f'(x))^2dx=\frac12\sum_{n=1}^{\infty}\pi^2n^2b_n^2$

Thus, we are asked to show that

$\pi^2\sum_{n=1}^{\infty}b_n^2\le\pi^2\sum_{n=1}^{\infty}n^2b_n^2.$

This is clearly true. Just as clearly, the extremal case would be given by a function for which $b_n=0$ for $n\ge2.$ That is, we have equality in this inequality if and only if $f(x)=b\sin\pi x$ for some $b\ne0.$

Opps! Yes, I forgot about that free term. In fact, when I extended $f$, I took it to be odd, but then forgot, and when I wrote the reply I just took it to be $\pi$-periodic :).


A geometric sequence of positive integers is formed for which the first term is 2 and the fifth term is 162. What is the sixth term of the sequence?

Divide the fifth term (162) by the first term to get 81. Thus the common ratio is 3. Our desired answer is then $ 162\cdot3\equal{}\boxed{486}$.


Given: $a,b,c \in \mathbb{C}$
$\left|a\right| = \left|b\right| = \left|c\right| \text{ (1)}$
$a + b + c = a^4 + b^4 + c^4 \text{ (2)}$
$a^2 + b^2 + c^2 = a^8 + b^8 + c^8 \text{ (3)}$

Prove one of the three:
$\text{a.)} \sum a = 0$
$\text{b.)} \sum a^3 = 3abc$
$\text{c.)} \sum a^2 = \sum ab$

[quote="Pork_Chop8"][quote="mathmanman"]So, if we have these conditions, $a, b$ and $c$ are the affixes of an equilateral triangle. :)[/quote]
In other words yes, but that still offers no concrete proof that (1) (2) and (3) imply that.[/quote]
Yes, I know, that was just a comment for those who wouldn't see the interest of this stuff. ;)

Nudge, Nudge, Nudge, No takers?
Any thoughts at all are appreciated by me. A full solution is not fully necessary. A hint, or more of a hunch, I think that a non-symmetric change to the givens is what will solve this.

Let $\{a,b,c\}\subset C\ ,\ abc\ne 0\ ,\ |a|=|b|=|c|=r>0$
$(a\cdot \overline a=r^2,\ldots)\ ,\ \sum a=\sum a^4\ ,\ \sum a^2=\sum a^8\ .$
=========================================
[b]Lemma.[/b] $ab+bc+ca=0\Longleftrightarrow a+b+c=0\ .$ Indeed,
$\overline {ab+bc+ca}=\overline a\overline b+\overline b\overline c+\overline c\overline a=r^4\cdot \left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{r^4}{abc}\cdot (a+b+c)\ .$
=========================================================
$2(ab+bc+ca)=(a+b+c)^2-\left(a^2+b^2+c^2\right)=\left(a^4+b^4+c^4\right)^2-\left(a^8+b^8+c^8\right)=$
$2\left(a^4b^4+b^4c^4+c^4a^4\right)\Longrightarrow \boxed {\ ab+bc+ca=a^4b^4+b^4c^4+c^4a^4\ }\ \ (1)\ .$
$\overline {a+b+c}=\overline {a^4+b^4+c^4}\Longleftrightarrow r^2\left(\frac 1a+\frac 1b+\frac 1c\right)=r^8\left(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right)\Longleftrightarrow$
$(abc)^3\cdot (ab+bc+ca)=r^6\cdot \left(a^4b^4+b^4c^4+c^4a^4\right)$ and the relation $(1)\ \Longleftrightarrow$
$\left[(abc)^3-r^6\right]\cdot (ab+bc+ca)=0\Longleftrightarrow \boxed {\ ab+bc+ca=0\ \ \vee\ \ abc=r^2\omega,\ \omega^3=1\ }\ .$
[b]Remark.[/b]
$1.\blacktriangleright$ If $ab+bc+ca=0$ (see the above lemma), then $a+b+c=0$ [b]a.s.o.[/b]
$2.\blacktriangleright$ If $abc=r^2\omega$, then $r=1\ (a=\overline a\ ,\ \ldots)$ and $abc=\omega$ [b]a.s.o.[/b]

Virgil, Eureka, you have found it! I have been searching for this solution for months and have never been able to formulate one and now you have done so, so easily and succinctly. This is pretty much verbatim from a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43976]post[/url] by JHaussmann5.
Given (1), (2), and (3) JH asks to prove that $a$, $b$, and $c$ are vertices of an equilateral triangle in the complex plane. Once we come to $a^2 + b^2 + c^2 = ab + bc + ca$ we have finished the proof. I'll [hide="hide"]
$a^2 + b^2 + c^2 = ab + bc + ca$
$\Rightarrow a^2 + b^2 + b^2 + c^2 + c^2 + a^2 = 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca$
$(a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ca + a^2) = 0$
$(a - b)^2 + 2(a - b)(b - c) + (b - c)^2 + (c - a)^2 = 2(a - b)(b - c)$
$2(a - b)(b - c) = (a - b + b - c)^2 + (c - a)^2 = (a - c)^2 + (c - a)^2 = 2(c - a)^2$
$(a - b)(b - c) = (c - a)^2$ Since our numbers are complex, by equating magnitudes, we can say $|a - b||b - c| = |c - a||c - a|$. If we say $x = |a - b|$, $y = |b - c|$, and $z = |c - a|$ (where $x,y,z$ are positive reals since they are magnitudes) we then have $z^2 = xy$. When we added $2(a - b)(b - c)$ to each side we could have just as easily added $2(b - c)(c - a)$ or $2(c - a)(a - b)$, so by symmetry we also have $x^2 = yz$ and $y^2 = xz$. Summing, we get $x^2 + y^2 + z^2 = xy + yz + zx$. By the same manipulation we did on $a,b,c$ we easily arrive at
$(x - y)^2 + (y - z)^2 + (z - x)^2 = 0$. Though, since $x,y,z$ are real numbers, the squared quanities must all be greater than or equal to zero, thus all must be 0 which gives us $x = y = z$. Coincidentally, this tells us $|a - b| = |b - c| = |c - a|$, in other words, each side of the triangle has the same length.
[/hide] the reason why, but know that we must almost always reach that to say they are in fact vertices. The fact $a + b + c = ab + bc + ab = 0$ gives us
$0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$= a^2 + b^2 + c^2 + 2(0) = a^2 + b^2 + c^2$
$\Rightarrow a^2 + b^2 + c^2 = 0 = ab + bc + ca$
That follows simply from $ab + bc + ab = 0$ and Virgil's Lemma, but the other fact: $abc = \omega$, is not as simple to prove with. I leave it as a challenge to show that they are indeed vertices.

Next Proposal:
Use $|a| = |b| = |c| = r > 0$, $(abc)^3 = r^6$, (2), and (3) to prove $a$, $b$, and $c$ are vertices of an equilateral triangle in the complex plane.

[quote="luimichael"]Since $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$,
we have the following inter-relationship:
(b) $\Leftrightarrow$ (a) or (c).[/quote]
Actually I just put in (b) and (c) for the heck of it because (a) implies both of them and everything else I wished was implied.
$a + b + c = 0 \Rightarrow (a + b + c)(a^2+b^2+c^2-ab-bc-ca)$
$= 0(a^2+b^2+c^2-ab-bc-ca) = 0$
$\Rightarrow a^3 + b^3 + c^3 - 3abc = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc \Rightarrow \boxed{\text{b.)}}$
Obviously, with Virgil's Lemma $\text{a.)} \Rightarrow \boxed{\text{c.)}}$. Without it, we can still do it:
$\sum a = 0 \Rightarrow 0 = (\sum a)^2 = \sum a^2 + 2\sum ab$
$\sum a^4 = \sum a = 0$
$(\sum a^2)^2 = \sum a^4 + 2\sum a^2 b^2 = 2\sum a^2 b^2$
$2(\sum ab)^2 = 2(\sum a^2 b^2 + 2abc \sum a) = 2(\sum a^2 b^2 + 0) = 2\sum a^2 b^2$
$0 = \sum a^2 + 2\sum ab \Rightarrow \sum a^2 = -2\sum ab$
$\Rightarrow 2(\sum ab)^2 = 2\sum a^2 b^2 = (\sum a^2)^2 = (-2\sum ab)^2 = 4(\sum ab)^2$
$2(\sum ab)^2 = 4(\sum ab)^2 \Rightarrow 2(\sum ab)^2 = 0 \Rightarrow 0 = 2(\sum ab)^2 = (\sum a^2)^2$
$\Rightarrow \sum ab = 0 = \sum a^2$
P.S. Sorry for all of these posts, but I have done a ton of work and I want to show a little bit of it otherwise it is just feels like I am worthless since I gave up anyways and made someone else solve it.

\[{ \underline {\underline {\mathrm {Two\ aplications\ of}\ \boxed {\ \vee\ }\mathrm {\ ,\ i.e.\ }\boxed {\ or\ }\ .}}}  \][b]Lemma[/b] (trivial, yet very useful). $x=0\ \ \vee\ \ y=0\Longleftrightarrow xy=0\ .$

$1.\blacktriangleright$ [b]Example.[/b] Ascertain $m\in R$ so that between the roots $x_1,\ x_2$ of the equation $x^2+(m-4)x+(m+1)=0$ there is the relation $x_1=x^2_2+1\ .$

[u]Demonstration.[/u] Denote $S=x_1+x_2=4-m$, $P=x_1x_2=m+1$, $S_2=x^2_1+x^2_2=S^2-2P=m^2-10m+14$, $S_3=x^3_1+x^3_2=S^3-3PS=-m^3-45m^2+3m+52\ .$
But the roots $x_1,x_2$ are [color=red]twins[/color] (the index is useless), i.e.
 $ax^2+bx+c=0\Longleftrightarrow x\in \{x_1,x_2\}$. Thus,
$\boxed {\ x_1=x^2_2+1\Longleftrightarrow x_1=x^2_2+1\ \vee\ x_2=x^2_1+1\ }\Longleftrightarrow$
$x^2_1-x_2+1=0\ \vee\ x^2_2-x_1+1=0$ (and lemma) $\Longleftrightarrow$
$(x^2_1-x_2+1)(x^2_2-x_1+1)=0\Longleftrightarrow P^2-S_3+S_2+P-S+1=0\Longleftrightarrow$
$P^2-S^3+3PS+S^2-P-S+1=0\Longleftrightarrow$
$(m+1)^2-(4-m)^3+3(m+1)(4-m)+(4-m)^2-(m+1)-(4-m)+1=0\Longleftrightarrow$
$m^3-13m^2+51m-39=0\Longleftrightarrow (m-1)(m^2-12m+39)=0\ .$
But $m\in R$. Therefore, $m=1\ .$ [color=darkred][i]Verification :[/i] indeed, for $m=1$ the equation becomes
$x^2-3x+2=0$ with the roots $\{1,2\}$ and $2=1^2+1\ !\ .$[/color]

$2.\blacktriangleright$ [b]Example.[/b] Denote $X(x)$ the point $X$ with the affix $x$. Prove that
the triangle $ABC$ is equilateral if and only if $a^2+b^2+c^2=ab+bc+ca\ .$

[u]Proof.[/u] Denote $\omega=\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}$, i.e. $\omega^3=-1$, $\omega^2-\omega+1=0\ (\omega^2=\omega -1)$
and $\overline {\omega}=\frac{1}{\omega}=-\omega^2\ .$ Then the triangle $ABC$ [b]is equilateral if and only if[/b]
$a-b=\omega (c-b)\ \ \vee\ \ (c-b)=\omega (a-b)\Longleftrightarrow$
$a-b=\omega (c-b)\ \ \vee\ \ a-b=-\omega^2 (c-b)\Longleftrightarrow$
$a+(\omega -1)b-\omega c=0\ \ \vee\ \ a-(1+\omega^2)b+\omega^2 c=0\Longleftrightarrow$
$a+\omega^2b-\omega c=0\ \ \vee\ \ a-\omega b+\omega^2c=0\Longleftrightarrow$
$(a+\omega^2b-\omega c)(a-\omega b+\omega^2c)=0\Longleftrightarrow$
$a^2+b^2+c^2-\omega (ab+bc+ca)+\omega^2 (ab+bc+ca)=0\Longleftrightarrow$
$a^2+b^2+c^2=(\omega -\omega^2)(ab+bc+ca)\Longleftrightarrow$
$\boxed {\ a^2+b^2+c^2=ab+bc+ca\ }\ .$

I assume Virgil just wanted an example but there is a easyer way to find $m$ such that $x^2+ (m-4)x+m+1=0$ has real roots that satisfy $x_1=x^2_2+1$.

As $x_2$ is a root of the equation $\Rightarrow x^2_2=(4-m)x_2-m-1$ so we want $x_1=(4-m)x_2-m$ or $x_1+x_2=4-m=(5-m)x_2-m$ so $x_2=\frac{4}{5-m}$. Substitute this in the original equation to get  $m^3-13m^2+51m-39$

[hide="Only for Delta."][color=red]My dear Delta , [/color][color=blue]and your solution is nicely and correctly. But if the relation between the roots of the equation is, for example $\boxed {\ x^4_1+x_1=8x^3_2+6x^2_2+3x_2+1\ }$, then how work you it ? In my opinion, the my method has a general symmetrical character,  I must ascertain the sums $S_k,\ k\in \overline {2,4}$. In the your method you must ascertain the all needful powers of the $x_i,\ i\in \overline {1,2}$ linear depending on the $x_i$, i.e. $x^k_i=A_kx_i+B_k$, where $k\in \overline {2,4}$. I offered some examples for the use of an above trivial and useful Lemma and only so much ![/color][/hide]

So, what happens to this case?
$2.\blacktriangleright$ If $abc=r^2\omega$, then $r=1\ (a=\overline a\ ,\ \ldots)$ and $abc=\omega$ a.s.o.


Suppose you are in a conversation with a person whom you intensely dislike, but who doesn't realize how you feel. You want to end the conversation and leave. Basically, there are two courses of action open to you:

(1) Be honest with the person. Express in candid terms your dislike, regardless of how this makes him/her feel, and then leave.

(2) Be tactful (deceitful!) and hide your true feelings while trying to worm your way out of the situation without arousing any ill feelings.

Which of these do you think is the morally proper course of action? Why? Is it necessarily true that you would actually choose the option that you think (while studying this abstract case) is the morally proper one? If not, why not?

It kind of depends on why you dislike them. If they are rude/mean, then I think that choice number one is fine. If it is an irrational dislike, or because of something that isn't their fault, or something like that, then you should think about why you dislike that person.

i think the first choice is the better one. being honest doesnt neccessarily mean being rude. i think its the best choice under the given conditions. you can worm out of the situation once or twice but what if it keeps happening? i'm pretty sure a person with a decent amount of intelligence can make out if you like him/her just by looking at the way you interact with them. its better you tell before they find out. and still if it gets you a bad name, well you dont like him anyway... :D

When I am in that situation, I would opt for the latter course, since I would find it hard to bluntly tell someone I dislike them, particularly if they don't realise it.
However, if I put myself into the other person's position, I would prefer to be told by someone that they didn't like me, rather than to be left guessing. I especially hate it when I feel as if someone might be being tactful/deceitful (like when ending a conversation).
Of course it is only fair for me to be told when someone doesn't like me if I am prepared to do the same. I'm convinced everything would be better if everyone said exactly who they do and don't like: but it's sometimes hard to be blunt.

Would you prefer someone who didn't like you to say so, instead of just keeping quiet about? If so, then you should probably do the same.

i would go with 2nd choice but i am not sure why ?

i go with 1 but that's why i have a lot of people who want me dead

It's not an easy question, is it?  Xevarion, is this hypothetical? :wink:

Anyway, I think I'd come up with a "there's my train" or "I gotta go ask that guy something" excuse. :|
But what do you mean by disliking? Sometimes there are people who tell you lots of stuff while you don't really like them, but you don't dislike them either you wanna spare them? What kind of "disliking" are we talking about?

There was some saint whose name I can't quite remember. He wrote a lot about friendships. He said that deceitful friendships are the worst kind and the only kind that can actually be bad to have. This is basically a deceitful friendship (sort of). I would say it is best to express yourself in as unhurtful a manner as possible while still being honest.

As a side note, I originally disagreed, at least partially, with this saint person. Then after hearing a talk about their life and stuff and this belief they held about friendships, after I thought about it I changed my mind and agreed with them, so I've been on both sides of the fence.

Hm I think the saint was like Saint Francis of Assisi.

I'll go with the "do unto others..." principle: a significant number of people may, and probably do, dislike me, and I don't really want to have exchanges of the first kind on a regular basis.

I would be tactful but drop occasional hints that I dislike the person. Not obvious ones, but enough to make him figure out that I disliked him if he thought about them a little bit

that way, he'll probably avoid talking to me later on and no confrontation

If there is real reason why I dislike someone, I usually say it openly. Problem is when I don't like someones company without specific reason. In that case I try to avoid communication and be polite because I don't want to offend a person who didn't do anything wrong.

I think that it depents and on the person that you dislike. For exable if being honest will cause damage to his/her mental health then you should try to be a bit more tactful

I'll do neither. I have often been at the receiving end of both, and it is never nice. And by the way, I am yet to meet somebody I have an intense dislike for.

Go with #1 option definitely. When I don't like someone, they usually know it pretty well from all the insults they receive.

I would go with number 2. I did number 1 before, tried it, hated it. I felt sooo guilty, and the person I told started spreading lies about me. This was like 2 years ago.

Never make anyone mad on purpose in a bad way unless he is being violent.

I suggest that you look to the Golden Rule.  You need to be motivated by what is best for the other person and not what is easy for yourself.

Why would you consider telling them you disliked them?

Will it help them or just help you feel better then them?

If you are going to tell them, in what setting, and with what words?

One shouldn't lie, but it doesn't help to go out of your way to hurt someone.

Is you dislike such that you can/they can change it?  Can they become a friend at a distance, rather than someone you hurt?

Again, I would carefully consider why you choose to say something about this and how to carry out whatever you decide to say.

If you are only motivated by your own short term interests, consider if later they become attractive to you -- they can take only one person to a dance or something you want to attend -- it would be sad if you burned the bridge that prevented the two of you enjoying something like this later.

Also consider, that moral questions usually are only resolved correctly by you doing something unpleasant, that hurts yourself, costs you money, causes you embarassement -- otherwise it isn't a moral question.

Be careful with this and act like a carpenter, measure (think) twice, cut (act) once.

It really depends on the person, I'd assess what his/her possible reactions would be to both of those options and choose the better course. But I probably would do neither and do something like pull them aside and tell them why I dislike them and how they could fix it if they're not emotional and overreactive, and if they are, I'd just attempt sidestepping them and avoiding them.

Very appropriate topic for a certain individual who is somehow my "friend."  He butts into conversations and immediately tries to become the center of attention, while displaying a false sense of "expertise" in any discussion.  I know I'm not the only one who feels this way about this person; many of my peers agree 100%.  The problem is, he is VERY ignorant.  The only way to get him to understand anything that he doesn't want to believe is to hit him over the head with it REPEATEDLY, because he is so arrogant and sure of himself.  It also doesn't help that he asks stupid questions and doesn't understand why others don't find him charming.  IS THERE ANY WAY TO DEAL WITH THIS???

I'm sure any response I get I will have already tried multiple times, but let's just say I'm desperate to reformat him or move to Antarctica.

in response to the wizard:
it's my personal belief that maintaining a painful and false relationship with a person to have a shot at going to the dance is indicative of a 1st grader/total FONEE

in response to bad answers:
vagueness doesn't help  (ie be nice AND be honest..where's the realism?)

Since this thread has been revived anyway, I'll answer this one:

[quote="fredbel6"]It's not an easy question, is it?  Xevarion, is this hypothetical? :wink: [/quote]
Yes and no. I don't worry about this particular situation, but I have other similar ones -- there is an issue that's difficult to bring up with someone, but failing to bring it up causes (or fails to prevent) some other problem, and one must decide between the two unpleasant options.

But I am also taking a philosophy course right now, and this has gotten me interested in some questions of how to define harm and how to decide between two undesirable situations.

I have also encountered this double standard, that people sometimes hold themselves to different moral standards than those to which they hold others (or that people want different things done to them than they think they should do to others). I am very interested in this phenomenon (I am particularly guilty of it), and I want to try to understand how it fits into an overall moral view of the world, so to speak.

Anyway yeah. There you go.

[quote="jhcreinhardt"]Very appropriate topic for a certain individual who is somehow my "friend."  He butts into conversations and immediately tries to become the center of attention, while displaying a false sense of "expertise" in any discussion.  I know I'm not the only one who feels this way about this person; many of my peers agree 100%.  The problem is, he is VERY ignorant.  The only way to get him to understand anything that he doesn't want to believe is to hit him over the head with it REPEATEDLY, because he is so arrogant and sure of himself.  It also doesn't help that he asks stupid questions and doesn't understand why others don't find him charming.  IS THERE ANY WAY TO DEAL WITH THIS???[/quote]

You know, I was faced with the same kind of problem a few weeks ago. There's this guy I see at work with exactly the characteristics you describe. He's not exactly very ignorant but compared to what he says he knows the guy is close to having the intelligence of a coral. It also happens that I'm a year older, which at our age means that either he is the genious of the season or I will know way more than he does. The day I got tired I dealt with this by telling him kind of crudely to stop claiming to know about things he doesn't and explaining him with much detail why he didn't knew a word about this and that.  After two or three times of me pointing him when he was doing it again out relation has gotten a bit colder but I assure you, he has stopped.
Don't know if it is the best way but there is a change of attitude. Also notice that different cultures carry this things differently. I surely wouldn't have done it if I were japanese or something. How do you feel your environment handles this stuff?

Ursula

Well in America, I imagine it wouldn't be as frowned upon as some other cultures, but it's more of a question of the individual.

What most people view as moral questions, I don't, in general, and visa versa.

This isn't a moral question so much in my opinion, thus generalized.  I'm going to say that, in general, it would depend on what the results are of each option.

To try and sort of answer the question, what you do will determine what kind of relationship you have with the person (this is true of most questions like this, ie: should on elie to person X in situation Y)  If you take the first option, it's all good, since it pretty much brings your thoughts out into the open, which is ALWAYS acceptable, IMO.  However, depending on your relation to a person (actually, the relationship depends more on how honest, open and essentially trusting one is), it is acceptable to be more or less truthful, open and trustful.  Since you assumingly do not have or want to create a meaningful relationship with this person, lying to them is not counterproductive and therefor the second option is also acceptable.


In the showdown of mathematics we will have four teams. First two sets (each set containing two teams) will compete in a showdown! In a showdown, one team has to make a problem and the other has to solve. So, how do you win or lose? First you start off with 10 points. every problem solved gets 10 points to the team. Every problem that is made and doesn't have a solution loses the maker of the problem's team 15 points and gives the other team 5 points. there are two teams "chicken pox" and "small pox" if you want to be with someone specific please just post here. I will post further rules after more people sign up. Btw, i am going to play too :P

Are we supposed to sign up by PM or post? And what are you going to do if not enough people sign up?

[quote="Poincare"][b]In the showdown of mathematics we will have four teams.[/b] First two sets (each set containing two teams) will compete in a showdown! In a showdown, one team has to make a problem and the other has to solve. So, how do you win or lose? First you start off with 10 points. every problem solved gets 10 points to the team. Every problem that is made and doesn't have a solution loses the maker of the problem's team 15 points and gives the other team 5 points. [b]there are two teams "chicken pox" and "small pox"[/b] if you want to be with someone specific please just post here. I will post further rules after more people sign up. Btw, i am going to play too :P[/quote]
How many teams are there?
Guessing there's 2...
Meh, I'll join.

[quote="pythag011"]I join.

Hmm...I'm gong to guess we will get some really hard problems.

Hmm...Who's good at writing problems?
Hm.....[/quote]
You?
I'll join your team if you want.
Or we could have a problem solving death match. I mean showdown.

Sure.

Hmm...I can't think of any good problems...

If MysticTerminator decided to join, that would not be a good thing for his opposing team.

I will JOIN! And make the opposing team suffer with "uber" problems!

[quote="pythag011"]Hmm....does the Riemann Hypothesis count as a legitamate problem?[/quote]
[quote="Poincare"]Every problem that is made and doesn't have a solution loses the maker of the problem's team 15 points and gives the other team 5 points.[/quote]
Sure...

The Riemann Hypothesis has a solution.

It's just that nobody has figured it out yet.

I believe it's implied that you have to be able to come up with the solution.

[quote="pythag011"]Hmm....does the Riemann Hypothesis count as a legitamate problem?[/quote]

ooooo yeah i forgot to mention NO OPEN PROBLEMS and the Riemann hypothesis is a conjecture its not a problem :P

EDIT: Also, if the opposing team cannot answer the problem you have to show a solution :)

[quote="budi713"]I guess i will join.

If stevenmeow or ra wants to join i would like to be on their teams.[/quote]
can i be on ur team? it needs some balance  :D

I'd like to join pythag's team if possible.

Er, when can we start. I'm somewhat eager to see the setup.

we shall start as soon as about 5-10 more people join :D


A) Να δειχθεί ότι για κάθε φυσικό $ n\ne 0$ υπάρχει μοναδικός φυσικός $ k_n$ ώστε $ 2^{k_n}<n\le 2^{1\plus{}k_n}$
B) Να υπολογιστεί συναρτήσει του $ n$ το άθροισμα $ \displaystyle\sum_{i\equal{}1}^n \frac {1}{2^{k_i}}$

[quote="Athinaios"]Καλησπερα. για το πρωτο ερωτημα πιστευω οτι η ισοτητα πρεπει να βρισκεται στο LHS στην ανισοτικη σχεση,διοτι ετσι για n=2 δεν ειναι αληθης.
(επισης και στο αλλο μελος να παει πιστευω οτι στην θεση του φυσικος θα ηταν καλο να μπει μη αρνητικος ακεραιος,διοτι εχουμε και n=1. Βεβαια για το αν περιεχεται στους φυσικους το 0 ή οχι ειναι μια αλλη κουβεντα..) πειτε μου αν δεν κανω λαθος στα παραπανω για να βαλω μια λυση για το (α)[/quote]
[b]έχεις δικιο Συγνώμη για την αβλεψία[/b]

Μια προχειρη προσεγγιση (λαμβανοντας υποψιν και τη διορθωση του Αθηναιου):

[hide]Για το Α προφανος ισχυει $ k_n \equal{} \max{\{k \in N / 2^k \leq n\}}$ το οποιο ειναι μοναδικο (αν αυτο δεν αρκει, πειτε μου να γραψω μια αλγεβρικη αποδειξη)

Για το Β, αν δεν μου εχει ξεφυγει κατι, ειναι: $ S_n \equal{} \frac {n \plus{} 1}{2^{k_n}} \plus{} k_n \minus{} 1$[/hide]

[quote="NickNafplio"]Μια προχειρη προσεγγιση (λαμβανοντας υποψιν και τη διορθωση του Αθηναιου):

[hide]Για το Α προφανος ισχυει $ k_n \equal{} \max{\{k \in N / 2^k \leq n\}}$ το οποιο ειναι μοναδικο (αν αυτο δεν αρκει, πειτε μου να γραψω μια αλγεβρικη αποδειξη)

Για το Β, αν δεν μου εχει ξεφυγει κατι, ειναι: $ S_n \equal{} \frac {n}{2^{k_n}} \plus{} k_n \minus{} 1$[/hide][/quote]

Για το Β νομίζω πως δεν πρόσθεσες τον τελευταίο όρο. (Ίσως όμως να κάνω εγώ λάθος.)

Εχεις δικιο, ειχα κανει ενα μικρο λαθακι στον υπολογισμο λογο βιασινης. Edited.

καλησπερα.
αυριο γραφω φυσικη και εκλεψα λιγο χρονο για να στειλω αυτο το μηνυμα :)

για το (α) νικο δεν ειμαι σιγουρος κατα ποσο εισαι ok απλα λεγοντας οτι οριζουμε το max του συνολου αυτο.
Η λυση που ειχα φανταστει εγω ειναι να ορισουμε το ${ S = \{s \in N / 2^s > n\}}$
Να πουμε οτι δεν ειναι φραγμενο γιατι ευκολα με επαγωγη
$ (1+1)^s \geq 1+s > s$
οτι παρουσιαζει ελαχιστο στοιχειo εστω $ r$
ομως $ r=k+1$ για καποιο $ {k \in N}$
αρα ισχυει
$ 2^{k+1} > n \geq 2^{k}$ , διοτι $ k<k+1$
η μοναδικοτητα ειναι ευκολη.
Τωρα για το (β)...

χρησιμοποιεις στο αθροισμα σου το k_n . Πιστευω οτι ειναι απαραιτητο να δειξεις πως το προσδιοριζεις(δηλαδη πως εχεις την δυνατοτητα να το χρησιμοποιησεις) .
Λεγοντας συναρτησει του n εγω θεωρω οτι το μονο γνωστο ,σαν δεδομενο , ειναι το n, και οχι το k_n.
Μια ιδεα που μου ρθε τωρα ενω γραφω το μηνυμα(ισως υπαρχει πιο ευκολος τροπος ή δεν αποκλειω κατι να μην σκεφτηκα και να ειναι λαθος) ειναι να το εκφρασεις ως το πληθος των πολλαπλασιων του 2 και μονο του 2 μεχρι το n ( με το ακεραιο μερος και μια αρχη εγκλεισμου-αποκλεισμου)
Δηλαδη, οπως το βλεπω αν παρεις την διαφορα του n απο των πολ/σιων ολων των περιττων αριθμων απο 3 εως n(και αρα ολων των αλλων πρωτων πιθανων διαρετων) ισως εισαι οκ.

so, δεν δεσμευομαι για το τελευταιο γιατι τωρα το σκεφτηκα.
για τα αλλα θα ηθελα να μου πεις την γνωμη-τις παρατηρησεις σου και εσυ αλλα και οποιος αλλος επιθυμει διοτι γενικα κατι τετοια θεματα σε διαφορες περιπτωσεις με ταλαιπωρουν(ευχαριστα παντα) στην αιτιολογηση.

Καλο σας βραδυ.

Ηλιας

[quote="Athinaios"]καλησπερα.
αυριο γραφω φυσικη και εκλεψα λιγο χρονο για να στειλω αυτο το μηνυμα :)

για το (α) νικο δεν ειμαι σιγουρος κατα ποσο εισαι ok απλα λεγοντας οτι οριζουμε το max του συνολου αυτο.
Η λυση που ειχα φανταστει εγω ειναι να ορισουμε το ${ S = \{s \in N / 2^s > n\}}$
Να πουμε οτι δεν ειναι φραγμενο γιατι ευκολα με επαγωγη
$ (1 + 1)^s \geq 1 + s > s$
οτι παρουσιαζει ελαχιστο στοιχειo εστω $ r$
ομως $ r = k + 1$ για καποιο $ {k \in N}$
αρα ισχυει
$ 2^{k + 1} > n \geq 2^{k}$ , διοτι $ k < k + 1$
η μοναδικοτητα ειναι ευκολη.
Τωρα για το (β)...

χρησιμοποιεις στο αθροισμα σου το k_n . Πιστευω οτι ειναι απαραιτητο να δειξεις πως το προσδιοριζεις(δηλαδη πως εχεις την δυνατοτητα να το χρησιμοποιησεις) .
Λεγοντας συναρτησει του n εγω θεωρω οτι το μονο γνωστο ,σαν δεδομενο , ειναι το n, και οχι το k_n.
Μια ιδεα που μου ρθε τωρα ενω γραφω το μηνυμα(ισως υπαρχει πιο ευκολος τροπος ή δεν αποκλειω κατι να μην σκεφτηκα και να ειναι λαθος) ειναι να το εκφρασεις ως το πληθος των πολλαπλασιων του 2 και μονο του 2 μεχρι το n ( με το ακεραιο μερος και μια αρχη εγκλεισμου-αποκλεισμου)
Δηλαδη, οπως το βλεπω αν παρεις την διαφορα του n απο των πολ/σιων ολων των περιττων αριθμων απο 3 εως n(και αρα ολων των αλλων πρωτων πιθανων διαρετων) ισως εισαι οκ.

so, δεν δεσμευομαι για το τελευταιο γιατι τωρα το σκεφτηκα.
για τα αλλα θα ηθελα να μου πεις την γνωμη-τις παρατηρησεις σου και εσυ αλλα και οποιος αλλος επιθυμει διοτι γενικα κατι τετοια θεματα σε διαφορες περιπτωσεις με ταλαιπωρουν(ευχαριστα παντα) στην αιτιολογηση.

Καλο σας βραδυ.

Ηλιας[/quote]
$ k_n=[log_2(n)]$ οπου [χ] ἆκἐραιο μέρος του χ

Κάτι που μου φαίνεται ενδιαφέρον είναι ότι το α ερώτημα απλοποιείται σημαντικά στο δυαδικό σύστημα αρίθμησης και μπορούμε να το δούμε δια γυμνού οφθαλμού.

Κάθε αριθμός της μορφής $ 2^k$ αποτελείται από έναν άσσο ακολουθούμενο από μηδενικά. Άρα αυτό που πρέπει να δειχθεί είναι ότι κάθε αριθμός στο δυαδικό σύστημα είναι ανάμεσα σε 2 αριθμούς της μορφής 100…00 όπου ο πρώτος έχει ένα ψηφίο λιγότερο από τον δεύτερο… (πράγμα προφανές αφού κάθε αριθμός είναι μεγαλύτερος ή ίσος από τον αριθμό που έχει ίδιο πλήθος ψηφίων και αποτελείται από άσσο και μετά μηδενικά, ενώ είναι μικρότερος από τον αριθμό της ίδιας μορφής που έχει ένα ψηφίο παραπάνω).
Η ισότητα είναι για τις δυνάμεις του 2.

Στο πιο οικείο μας δεκαδικό σύστημα, οι δυνάμεις του 10 παίρνουν τη θέση του 2 και πχ οι τριψήφιοι είναι ανάμεσα στο 100 και το 1000. Κάθε αριθμός φράσσεται από 2 διαδοχικές δυνάμεις του 10.

Πιστεύω δηλαδή ότι η άσκηση γενικεύεται για κάθε σύστημα αρίθμησης πειράζοντας κατάλληλα τη βάση της δύναμης που υπάρχει στην ανισότητα. Το $ k_n \minus{} 1$ είναι απλά το πλήθος των ψηφίων στο συγκεκριμένο σύστημα αρίθμησης (εκτός από τις δυνάμεις της βάσης του συστήματος αρίθμησης που είναι το $ k_n$ ).

Ελπίζω να μην κατάλαβα κάτι λάθος.

G

[quote="r_boris"]
$ k_n \equal{} [log_2(n)]$ οπου [χ] ἆκἐραιο μέρος του χ[/quote]

cool!!

[quote="r_boris"]A) Να δειχθεί ότι για κάθε φυσικό $ n\ne 0$ υπάρχει μοναδικός φυσικός $ k_n$ ώστε $ 2^{k_n} < n\le 2^{1 \plus{} k_n}$
B) Να υπολογιστεί συναρτήσει του $ n$ το άθροισμα $ \displaystyle\sum_{i \equal{} 1}^n \frac {1}{2^{k_i}}$[/quote]
Για τον υπολογισμό του αθροίσματος θα δώσω μια υπόδειξη
Δείξτε ότι για κάθε φυσικό $ n\ne 0$ υπάρχει μοναδικοί φυσικοί $ x,y$ ώστε $ n\equal{}2^y\plus{}x$ με $ 0\le x<2^y$ και είναι $ y\equal{}k_n , \displaystyle\sum_{i \equal{} 1}^n \frac {1}{2^{k_i}}\equal{}k_n\plus{}\frac{x\plus{}1}{2^{k_n}}$

[quote="NickNafplio"]Βασικα εγω το αθροισμα το βρηκα $ S_n \equal{} \frac {n \plus{} 1}{2^{k_n}} \plus{} k_n \minus{} 1$ ως εξης:

[hide]Μεταξυ των αριθμων $ 2^{k_n}$ και $ n$ υπαρχουν ακριβως $ n \minus{} 2^{k_n} \plus{} 1$ αριθμοι $ m$ με $ k_m \equal{} k_n$ (συμπεριλαμβανονται και οι:$ 2^{k_n}, n$). Πριν απο αυτους εχουμε $ k_n$ δυναμεις του 2 $ (1, 2, ..., 2^{k_n \minus{} 1})$, ενω σε καθε διαστημα $ [2^a, 2^{a \plus{} 1}), a \in N$ εχουμε ακριβως $ 2^a$ φυσικους $ m$ με $ k_m \equal{} a \Rightarrow \frac {1}{2^{k_m}} \equal{} \frac {1}{2^a}$. Αρα $ S_n \equal{} 1 \plus{} 1 \plus{} 1 \plus{} ... \plus{} \frac {n \minus{} 2^{k_n} \plus{} 1}{2^{k_n}} \equal{} \frac {n \plus{} 1}{2^{k_n}} \plus{} k_n \minus{} 1$.[/hide][/quote]
ΣΩΣΤΑ ($ k_n\equal{}[log_2(n)],x\equal{}n\minus{}2^{k_n}$)


hello

please help me understand this equation
please solve step by step
thanx

Solve the recurrence relation
an = 6an−1 − 12an−2 + 8an−3
with a0 = −5, a1 = 4, and a2 = 88.

[quote="hyderman"]i looked at google but still if someone can show me the right steps to solve this recurrence

thanx[/quote]

He is doing it the right way.

The characteristic polynomial is obtained by guessing that $a_{n}$ is represented in exponential form. This guess can be made using inequalities to look at consecutive terms of the sequence.

[quote="cincodemayo5590"][quote="worthawholebean"][hide="Uh oh..."]
Look up characteristic polynomials of recurrence relations on google.

The characteristic polynomial of this is $x^{3}-6x^{2}+12x-8$, which factors to $(x-2)^{3}$, so the general formula for the relation is $a_{n}=A(2)^{n}+B(2)^{n}+C(2)^{n}$, where A, B, and C are constants, or $a_{n}= (A+B+C)2^{n}= K(2^{n})$, where K is also a constant.  Plugging in $a_{0}$, we get... wait... What am I doing wrong?[/hide][/quote]

[hide="When we have multiple roots like this, it takes the form: "]
\[A\cdot 2^{n}+Bn \cdot 2^{n}+Cn^{2}\cdot 2^{n}\]
etc, i.e. it goes up to a power of $n^{k-1}$ for a root with multiplicity $k$. [/hide][/quote]

I read the thing you put up on the other thread about characteristic polynomials and it was very interesting. Why do you need the $n^{k-1}$ though?

this is confusing what is the final solution..... please help by showing the steps i need this because my next exam will be a bout this subject .....
please help me

thanx


Find $n$ max such that:
   $6x$ $+$ $10y$ $+$ $15z$ $+$ $7t$ $=$ $n$
has no solution in $N*$

Hmm... a higher version of the Frobenius Coin Problem.  Well in two variables, you have \[\max_{n\neq ax+by}\left\{n\right\}=ab-a-b.\] I'm not sure if this may be extended to four variables in the same manner, but [b]assuming[/b] it can, we would have \[n=6\cdot 10\cdot 15\cdot 7-6-10-15-7=\boxed{6262}\] but this may be completely wrong.

Nice solution Rust.  Clearly that formula doesn't extend to cases with more than $2$ variables.   :roll:


Find all real solutions to the equation:
$ x\left(\frac{6\minus{}x}{x\plus{}1}\right)\left( \frac{6\minus{}x}{x\plus{}1} \plus{}x \right) \equal{} 8$

Substituting $ t \equal{} x \plus{} 1$ we get

$ (t \minus{} 1)\left({7\over t} \minus{} 1\right)\left({7\over t} \plus{} t \minus{} 2\right) \equal{} 8$

$ \left(7 \minus{} t \minus{} {7\over t} \plus{} 1\right)\left({7\over t} \plus{} t \minus{} 2\right) \equal{} 8$

Now $ y : \equal{} {7\over t} \plus{} t$ gives

$ (8 \minus{} y)(y \minus{} 2) \equal{} 8\iff y^2 \minus{} 10y \plus{} 24 \equal{} 0\iff y\in\{4,6\}$

1. $ t \plus{} {7\over t} \equal{} 4\iff t^2 \minus{} 4t \plus{} 7 \equal{} 0$ which has no real solutions.

2. $ t \plus{} {7\over t} \equal{} 6\iff t^2 \minus{} 6t \plus{} 7 \equal{} 0\iff t_{1,2} \equal{} 3\pm\sqrt {2}$

Now $ x \equal{} t \minus{} 1\iff x_{1,2} \equal{} 2\pm\sqrt {2}$

Let $ f(x)=\frac{6-x}{x+1}$ and $ x\neq{-1}$ we have: $ xf(x)^2+x^2f(x)=0\rightarrow{ f(x)=-x}$
Suppose $ f(x)=-x+c$ with c constant $ \rightarrow{x=\frac{c+\sqrt{c^2+4c-24}}{2}}$ or $ x=\frac{c-\sqrt{c^2+4c-24}}{2}$
Substituting in the equation:
$ (\frac{c+\sqrt{...}}{2})(\frac{-c-\sqrt{...}}{2}+c)c=8 \rightarrow{c(c^2-c^2-4c+24)=32}\rightarrow {c=4}$ or $ c=2$ $ \rightarrow{x=2+\sqrt{2}}$
$ (\frac{c-\sqrt{...}}{2})(\frac{-c+\sqrt{...}}{2}+c)c=8 \rightarrow{{c=2}}$ or $ {c=4} \rightarrow{x=2-\sqrt{2}}$

Just to see what was the idea for this equation:

I-st method:
If we substitute: $ y \equal{} \frac {6 \minus{} x}{x \plus{} 1}$
We get the system:
$ xy(x \plus{} y) \equal{} 8$;
$ xy \plus{} (x \plus{} y) \equal{} 6$

II-nd method:
If we substitute: $ y \equal{} x.\frac {6 \minus{} x}{x \plus{} 1}$
We get the system:
$ yz \equal{} 8$;
$ y \plus{} z \equal{} 6$

It is extracted from my textbook. I think it is beautiful equation. Do you think so?


Let $ A$ and $ B$ be positive integers satisfying $ B\equal{}3A$. Assume that the digits of $ B$ are a permutation of the digits of $ A$. Prove that $ B\equal{}3A$.

It is not so hard !
If B=3A then B=3A. Really very nice :D

Well, if you assume that $B=3A$ there is nothing to prove so that I don't consider it as reaal hard, and neither really beautiful  :D
In another hand, from the statement, it seems that you have solution so is it an open problem or not?

Pierre.

K09, maybe you had mistakely typewritten the conditions or conclusions, would you please check it out?

I am sorry, I change ........Prove that A devisible by 27

[Moderator edit: Topic locked, problem to be discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=23737 .]


Let $ y \equal{} f(x), f: R \to R$ is an odd differentiable function such that $ f'''(x) > 0$ and $ g(\alpha, \beta) \equal{} \sin^8 \alpha \plus{} \cos^8\beta \minus{} 4 \sin^2\alpha \cos ^2\beta$.

If $ f''(g(\alpha, \beta)) \equal{} 0$, then $ \sin^2 \alpha \plus{} \sin 2 \beta \equal{} ?$

[quote="Conjurer"]
EDIT. stewpid mistake

[/quote]
Sorry! Should've been in the Calculus section. This is what happens when you've got too many tabs open in one window!

[quote="aadil"]this should be in the jee thread not here.
[/quote]
Not necessarily!


number theory
least common multiple
LaTeX
prime factorization
prime numbers

what is the smallest number u can find that has 1 2 3 4 5 6 7 8 and 9 as factors

edit:

2: 1-20 as factors
3: 1-100 as factors

[quote="jackdillon"][hide]is the answer 362880?[/hide][/quote]

i got sumthing lower i wanna see lowest u guys can find

[quote="Ihatepie"][hide]i now have 2020 is there anything lower?
lol first i only got 1/2 of ur answer but now I realized u were way off[/hide][/quote]
There is nothing lower than your answer ihatepie

[hide]I got 2520 by 2x2x2x3x3x5x7[/hide] :huh:

[quote="mgao"][hide]

You don't have to use guess and check...

We know that the max number of 2s in the prime fac of the LCM is 3.   Similarly, for 3 it is 2.  Then put 5 and 7 in the mix and you have

8 x 9 x 35= 2520,

Ihatepie your answer is wrong, 2020 is not divisible by 3 or 9...[/hide][/quote]

yeah i got 504 and then imultiplied it by 4 not 5
messed up...

[quote="jackdillon"][quote="humblemastermind"][hide="2"]
$2^{4}\cdot3^{2}\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19 = 232792560$ [/hide][/quote]
$Thats\;even\;bigger\;then\;my\;answer!$[/quote]

thats number 2 or 1-20

Sorry, but I don't have Latex...

[hide]
1. If you do prime factorization, you get 2^3*3^2*5*7
and you get 2520 which is divisible by all integers 1-9 (inclusive)

2. If you do the same thing, you get
2^4*3^2*5*7*11*13*17*19=232792560

3. Again, with prime factorization, you get 2^6*3^4*5^2*7^2*11*13*17*19*23*29*31*37*41*43*47*53*59*61*67*71*73*79*83*89*97=
69720375229712500000000000000000000000000
[/hide]

 :P  Not sure if I'm right, but I'm pretty sure

[quote="Quevvy"]Sorry, but I don't have Latex...[/quote]
You don't need to "have" it. It's installed on the site. Just put whatever you want to be in $\LaTeX$ between dollar signs.

[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]

his equation is also provided.
it should be that big of a number
because
$1~100$ = a lot of numbers.

just get that number and try dividing it by random numbers from 1~100

do that number / randint(0,100,1) if you have a ti-calculator
and u should see if it's correct or not!

Here's an easy way to figure this out,
First, say you're starting with 1-9, you figure out the power of two where the result is less than 9 (2^x<9 (or equal to)) and that is 2^3. Then you repeat the process with all the other prime numbers and then you multiply those numbers together to get your final answer!

Here's #1
[hide]2 - 2^x[u]<[/u]9 x=3   2^3
3 - 3^x[u]<[/u]9 x=2   3^2
5 - 5^x[u]<[/u]9 x=1   5^1
7 - 7^x[u]<[/u]9 x=1   7^1
Multiply together get 2520!
That's why #3 has all the primes up to 100![/hide]

[quote="bpms"][quote="jackdillon"][quote="humblemastermind"][hide="2"]
$2^{4}\cdot3^{2}\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19 = 232792560$ [/hide][/quote]
$Thats\;even\;bigger\;then\;my\;answer!$[/quote]
1. Humblemastermind is a genius and is always right.
2. Don't post text in latex, for the people with dial up.
3. [hide="1-100"]$2^{6}\cdot 3^{4}\cdot 5^{2}\cdot 7^{2}\cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47 \cdot 53 \cdot 59 \cdot 61 \cdot 67 \cdot 71 \cdot 73 \cdot 79 \cdot 83 \cdot 89 \cdot 97$ [/hide] That took me much less time then I thought it would :|[/quote]

how do u do that? I get how u r doing every prime number but how do u determine the powers?

[quote="Ihatepie"][/quote][quote="bpms"][quote="jackdillon"][quote="humblemastermind"][hide="2"]
$2^{4}\cdot3^{2}\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19 = 232792560$ [/hide][/quote]
$Thats\;even\;bigger\;then\;my\;answer!$[/quote]
1. Humblemastermind is a genius and is always right.
2. Don't post text in latex, for the people with dial up.
3. [hide="1-100"]$2^{6}\cdot 3^{4}\cdot 5^{2}\cdot 7^{2}\cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47 \cdot 53 \cdot 59 \cdot 61 \cdot 67 \cdot 71 \cdot 73 \cdot 79 \cdot 83 \cdot 89 \cdot 97$ [/hide] That took me much less time then I thought it would :|[/quote][quote="Ihatepie"]

how do u do that? I get how u r doing every prime number but how do u determine the powers?[/quote]
Prime factorize every number 1-100 and combine the greatest powers.

[quote="i_like_pie"][/quote][quote="Ihatepie"][/quote][quote="bpms"][quote="jackdillon"][quote="humblemastermind"][hide="2"]
$2^{4}\cdot3^{2}\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19 = 232792560$ [/hide][/quote]
$Thats\;even\;bigger\;then\;my\;answer!$[/quote]
1. Humblemastermind is a genius and is always right.
2. Don't post text in latex, for the people with dial up.
3. [hide="1-100"]$2^{6}\cdot 3^{4}\cdot 5^{2}\cdot 7^{2}\cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47 \cdot 53 \cdot 59 \cdot 61 \cdot 67 \cdot 71 \cdot 73 \cdot 79 \cdot 83 \cdot 89 \cdot 97$ [/hide] That took me much less time then I thought it would :|[/quote][quote="Ihatepie"]

how do u do that? I get how u r doing every prime number but how do u determine the powers?[/quote][quote="i_like_pie"]
Prime factorize every number 1-100 and combine the greatest powers.[/quote]
Look at my explanation and it helps to go faster! Because factoring every number takes a long time...


Hi, just wondering if someone could assist me with this question. It has just completed got me stuck. The question is

How many 6-digit numbers are there that don’t have a 3 followed immediately by a 7?

Thanks for the help.

In problems of this sort it is often useful to count what you do not want rather than what you do want.

So, consider:  How many 6-digit numbers [b]do[/b] have a $3$ followed by a $7$?

:)

Sorry i don't quite follow - please explain.

$\text{Number of 6-digit numbers}= (\text{Number of 6-digit numbers that don't have a 3 followed by a 7})+(\text{Number of 6-digit numbers that do have a 3 followed by a 7})$

Rather than solving for the second quantity directly, we calculate the first (which is easy) and the third (which is not quite as easy, but easier than the second) and subtract.  :)

Nope sorry - could someone explain this to me with some proof? Still lost with it :(

All $6$ digit numbers either have a $37$ somewhere in them, or they don't have a $37$. So when you combine the set of all the $6$ digit numbers that have a $37$ in them with the set of all $6$ digit numbers that don't have it, you get the set of all $6$ digit numbers.

[quote="emilgouliev"]There are 100000 6-digit numbers.

There are five cases for 3 to be immediately followed by 7.

37####, #37###, ##37##, ###37#, ####37 (# represents a number)

The latter four are equal number.

There are 10000 (10*10*10*10) possibilities for the first one. The others have 9000 (9*10*10*10) possibilities. So there are 10000+9000*4 possibilities for 3 to be immediately followed by 7, which equals 46000. 1000000 - 46000 = 954000 is our answer.[/quote]
That's incorrect. The number $373737$, for example, is counted 3 times, instead of once. Also, there are $900000$ 6-digit numbers.


conics
hyperbola
geometry
Euler
circumcircle
analytic geometry
geometric transformation

Let $M(a,b)$ and $N(-a,-b)$ be two points with $b\neq 0$ that lie on the hyperbola $H : x^2-y^2=2$. The circle with radius $MN$ and center $M$ intersects the hyperbola at $A,B,C$ different from $N$. Prove that there is one segment from $NA,NB,NC$ such that its length is equal to the sum of the lenghts of the other $2$.

This problem is from the Moldavian Republican Olympiad 2005 first day. No one fully solved this problem. Good luck!!!

[color=red][Moderator edit: Made a few stylistic changes. For instance, the sign "is not equal to" is $\neq$ (\neq) in LaTeX.][/color]

That's an absolute insider problem! How could it be given on a competition? Anyway, here is a solution.

We begin with a lemma due to Jean Victor Poncelet:

[color=blue][b]Lemma 1.[/b] If A, B, C are three points on a rectangular hyperbola h, then the center of this hyperbola h lies on the nine-point circle of triangle ABC.[/color]

(Poncelet also showed that the orthocenter of triangle ABC lies on the hyperbola h, but we won't need this in the following.)

Please don't ask me how to prove this all; there is a neat PDF document by Jean-Pierre Ehrmann (JPE) containing the proof, but I am not sure whether it should be distributed. Some literature should also contain the proof. Just wanted to say that the above Lemma 1 was actually the way Poncelet discovered the nine-point circle of a triangle - as the locus of the centers of all rectangular hyperbolas passing through the vertices of the triangle.

Now, in [1], I defined the [i]Euler point[/i] of a cyclic quadrilateral. We will show now:

[color=blue][b]Lemma 2.[/b] If ABCD is a cyclic quadrilateral and h is a rectangular hyperbola passing through its vertices A, B, C, D, then the center of this hyperbola h is the Euler point of the cyclic quadrilateral ABCD.[/color]

[i]Proof of Lemma 2.[/i] Applying Lemma 1 to the points A, B, C on the rectangular hyperbola h, we see that the center of the hyperbola h lies on the nine-point circle of triangle ABC. Similarly, the center of the hyperbola h lies on the nine-point circles of triangles BCD, CDA and DAB. In other words, the center of the hyperbola h is a common point of the nine-point circles of triangles BCD, CDA, DAB and ABC.

On the other hand, after [1], Theorem 3, the nine-point circles of triangles BCD, CDA, DAB and ABC all pass through the Euler point of the cyclic quadrilateral ABCD. Hence, the Euler point of the cyclic quadrilateral ABCD is also a common point of the nine-point circles of triangles BCD, CDA, DAB and ABC.

Since the nine-point circles of triangles BCD, CDA, DAB and ABC cannot have more than one point in common (in fact, any two of these nine-point circles have in common a midpoint of one of the segments AB, BC, CD, DA, AC, BD, and at least three of these points must be distinct), it follows that the center of the hyperbola h and the Euler point of the cyclic quadrilateral ABCD, both being common points of the nine-point circles of triangles BCD, CDA, DAB and ABC, must coincide. This proves Lemma 2.

Finally a really well-known lemma:

[color=blue][b]Lemma 3.[/b] If ABC is an equilateral triangle, and P is a point on its circumcircle, then one of the distances PA, PB, PC equals the sum of the two others.[/color]

[i]Proof of Lemma 3.[/i] The point P lies on the circumcircle of triangle ABC; hence, it lies on one of the arcs BC, CA, AB. We can WLOG assume that the point P lies on the arc BC. Then, the quadrilateral ABPC is a convex cyclic quadrilateral. Thus, by the Ptolemy theorem, $PA\cdot BC=PB\cdot CA+PC\cdot AB$. But since the triangle ABC is equilateral, its sidelengths BC, CA, AB are all equal, so we obtain PA = PB + PC. And this shows that one of the distances PA, PB, PC equals the sum of the two others; thus, Lemma 3 is proven.

Now to the solution of the problem:

The circle with radius MN and center M passes through the points A, B, C and N. Thus, the quadrilateral ABCN is a cyclic quadrilateral with circumcenter M.

The hyperbola H, given by the equation $x^2-y^2=2$ is, obviously, a rectangular hyperbola, and its center is the origin O of the coordinate system.

Now, since ABCN is a cyclic quadrilateral, and the rectangular hyperbola H passes through its vertices A, B, C, N, Lemma 2 yields that the center O of this hyperbola H is the Euler point of the cyclic quadrilateral ABCN.

But after the definition of the Euler point of a cyclic quadrilateral ([1], Theorem 2), if H is the orthocenter of the triangle ABC, then the Euler point O of the cyclic quadrilateral ABCN lies on the segment NH and bisects this segment. In other words, the point H is the reflection of the point N in the point O.

On the other hand, since O is the origin of the coordinate system, while the points M and N have the coordinates M (a, b) and N (-a, -b), respectively, the reflection of the point N in the point O is the point M.

Hence, the point H coincides with the point M.

But the point H is the orthocenter of the triangle ABC, while the point M is the circumcenter of triangle ABC (since the circle with radius MN and center M passes through the points A, B, C). It follows that the orthocenter of triangle ABC coincides with the circumcenter of triangle ABC. But this immediately yields that the triangle ABC is equilateral.

The point N lies on the circumcircle of this equilateral triangle ABC (since we know that the points A, B, C and N lie on one circle). Thus, by Lemma 3, one of the distances NA, NB, NC equals the sum of the two others. $\blacksquare$

[color=green][b]References[/b][/color]

[b][1][/b] Darij Grinberg, [i]The Euler point of a cyclic quadrilateral[/i], http://de.geocities.com/darij_grinberg/ .

  Darij