[quote="Rushil"]Let $ f : [ \frac {1}{2} , \infty ) \rightarrow [ \frac{3}{4} , \infty ) $ where $ f(x) = x^2 -x + 1 $

Solve $ x^2 - x + 1 = \frac{1}{2} + \sqrt{ x - \frac{3}{4} } $[/quote]

[hide]Taking $ y=f(x) = x^2 -x + 1 $
we have $f^{ - 1} (x) = \frac{1}{2} + \sqrt {x - \frac{3}{4}}$

so, the given equation now asks for $f^{ - 1} (x) = f(x)$ which is possible only when $f(x)=x$

therefore,

$x^2 -x + 1 =x$

which gives $x=1$[/hide]

Let me also see.
[hide]
$x^2  - x + 1 = \frac{1}{2} + \sqrt {x - \frac{3}{4}}$

$\Rightarrow (x^2  - x + \frac{1}{2})^2  = x - \frac{3}{4}$

$ \Rightarrow x^4  - 2x^3  + 2x^2  - 2x + 1 = 0$

$\Rightarrow x^2  - 2x + 2 - \frac{2}{x} + \frac{1}{{x^2 }} = 0$

$(x + \frac{1}{x})^2  - 2(x + \frac{1}{x}) = 0$

which gives only $x=1$ as a real solution.[/hide]

CAn you plz hide your solution... This is indeed a standard question in India IIT MAths(dunno why). I want to see other's responses!!!!


abstract algebra
number theory unsolved
number theory

Prove that $\frac{(mn)!}{m!(n!)^m} \in \mathbb{N} \ \ \forall(m,n) \in \mathbb{N} \times \mathbb{N}$

Alexandros

Denote $f(m,n)=\frac{(mn)!}{m!(n!)^m}$. I'll use the short notation: $(k)_l: =k(k-1)(k-2)\dots(k-l+1)$

\begin{eqnarray*}f(m,n) &=& \frac{\frac{(n)_n}{n!}\cdot\frac{(2n)_n}{n!}\cdot\frac{(3n)_n}{n!}\dots\frac{(mn)_n}{n!}}{m!}\\ &=& \frac{\left(1\cdot\frac{(n-1)_{n-1}}{(n-1)!}\right)\left(2\cdot \frac{(2n-1)_{n-1}}{(n-1)!}\right)\left(3\cdot\frac{(3n-1)_{n-1}}{(n-1)!}\right)\dots\left(m\cdot\frac{(mn-1)_{n-1}}{(n-1)!}\right)}{m!}\end{eqnarray*}

(The identity used is $\frac{(an)_n}{n!}=\frac{an(an-1)_{n-1}}{n\cdot(n-1)!}=a\cdot\frac{(an-1)_{n-1}}{(n-1)!}$)

Obviously the leading numbers in each pair of parentheses multiply out to $m!$, which cancels out with the denominator, and the remaining fractions are integer (the product of $n-1$ consequent naturals is always divisible by $(n-1)!$)

Draw a rectangular grid with sides $m \times n$ points ($m$ rows of $n$ objects).
$(mn)!$ is the number of all permutations of these points, where $m! (n!)^m$ gives the number of permutations where we just permutate in the rows and then the rows itself.
Since both sets of permutations are closed under concatenating the permutations, by Lagrange's theorem we get what we want.


Hi

On behalf of the Indian Institute of Technology Madras, India, I am pleased to invite you to the [b]Online Math Contest of [url=http://www.shaastra.org/]Shaastra 2009[/url][/b] , an International Online Math Competition.

This years [b]Online Math Contest will consist of 5 problem sets with 6 problems in each set and about 5 days time will be given to submit the solutions[/b] to each problem set.

The topics covered in the problem sets with majorly be [b]Olympiad Mathematics[/b] along with [b] a few applications of College Mathematics[/b].

The [b]problem sets will be uploaded on the 5th, 10th, 15th, 20th and the 25th of September[/b]. For further details, please [url=http://www.shaastra.org/2009/site/events/online_events/online_math]visit the Online Math page of the Shaastra website.[/url]

The coordinators (the people who set the questions) of this event are [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=36236]chapplikka[/url] and myself.

Hope you will have fun in attempting the questions. We had fun in making them :)

Regards
Madness
[b]
PS: Please do post your suggestions and feedback in this thread.[/b]

[b]Note: The first problem set has been preponed to 2nd September. However the submission date will remain the same (10th September).[/b]

Now, let me take my turn in welcoming you'll to Online Math, Shaastra 2009. As many of you might already be knowing, Shaastra is the first ISO-certified student organised event in the world, and as students of IIT Madras, and as coordinators of Online Math, Shaastra 2009, madness and I take immense pleasure in welcoming you'll to participate in Online Math, and be a part of this mega fest.

The standards are of high school, and college levels, and that is the eligibility for participation as well. The problems are definitely in enjoyable taste, and I wish everyone a very pleasant experience with the problems in here. Have fun. All the best.. :)
Ever yours,
Chapplikka..

It will be nice if you don't include questions that would require any background beyond what IMO would require. That way students who haven't yet reached college will not be at a disadvantage. Obviously, with the IMO level subject coverage, one can set questions that can challenge even Masters students.  For that matter, just give some IMO papers to math profs and see :).

Well, most of the college topic problems we propose to give can be solved by IMO aspirants. Very few (<5 out of 30) require calculus and other similar topics, and those cannot be replaced because after all, Shaastra is a college festival and we have to have a few questions from calculus. Otherwise, most of the other questions are solvable by high school students.

And yeah, I agree with your opinion on the IMO :)

It is of if those calculus questions only require knowledge of single variable calculus.But if they require multivariable stuff then high school students will have a major disadvantage.

I tried registering and it doesn't let me  :roll:

There might be a few problems with the userportal. In which case, do mail us the solutions with your details (like name, school/college, class/year, country). After the problems are fixed, we shall inform you about the same.

Regards
Madness

They also have an Online Programming Contest:
[url]http://www.shaastra.org/2009/site/events/coding/online_programming_contest[/url]

The first problem set is up on the site http://www.shaastra.org/2009/site/events/online_events/online_math in the problem sets tab. It is a bit on the easier side because it's the first problem set! Hope you enjoy solving the questions! :)

[quote="arjunarul"]Could you please tell whether the online coding contests, especially the Online Programming Contest is open for school students?? I couldn't find any instruction for eligibility in the website.[/quote]
I think it is open to school students as well. As far as I know, Srivatsan participated in it last year.

With regards to online math, please post your feedback regarding the contest in this thread.

Can I participate if I'm not studying in College (therefore I don't know my College ID and the College stuff) and I'm not Indian?
P.S What is the summation of the first problem - cyclic or symmetric?

@madness
ur sig,u mean chaos theory?? :huh:

any international students participating in this?? (i mean students not from india)

Yeah  :)  Some of them are listed in this thread itself

[quote="arjunarul"]Could you please tell whether the online coding contests, especially the Online Programming Contest is open for school students?? I couldn't find any instruction for eligibility in the website.[/quote]

It is open for school students also. Sent an email to them.

[quote="madness"]You can participate if you are not an Indian and/or if you are a school student.
[u][b]
Details on how a school student can register on the userportal are as follows:[/b][/u]

Enter all details which apply to you as usual. Given below are directions to fill (the ambiguous) details asked for school students:

Enter your branch as your stream in school, for example: science, commerce, math or any other relevant stream.

If you do not have a mobile number, then enter your home phone number (with the international code).

You must be given a roll number in your school. If so, enter it in place of the college id. Otherwise, enter some arbitrary natural number.

Click on the can't find your college link and register your school name. After registering your school, type your school name in the college section.

[quote="broniran"]P.S What is the summation of the first problem - cyclic or symmetric?[/quote]

As for the above question, the summation is cyclic (as expected).[/quote] I can't register? Can I simply pick up the first state?
P. S I tried with one state but I got [quote]Invalid college name, or college does not exist. Please use add college form below to add your college if it does not already exist[/quote]. I study in Model Mathematical School "Academician Kiril Popov", Plovdiv, Bulgaria

I have the same problem. :( I've registered my school-name and it was written that it has been successfully registered but when I wrote it in the blank, it didn't work :( Madness, can you help us?

We are extremely sorry for the inconvenience caused by the userportal (which is being rectified). Until then, please send your solutions to onlinemath@shaastra.org along with your name, name of your school/college, grade and country.

Regards
Madness

Ok, I've done that way. :) When will the results be published?

I guess only after all the problem sets are completed. We wouldn't want to discourage people from attempting problem sets  :wink:

Problem Set 2 uploaded. Happy Problem solving!  :)

Problem Set 3 uploaded. Happy Problem Solving! :)

Problem Set 4 uploaded. Happy Problem Solving  :D


calculus
integration
logarithms
calculus computations

I need the integral of ln((1+x)^(1/2) + (1-x)^(1/2)) from 0 to 1. Any ideas?

hello, do you mean $ \int_{0}^1\ln(\sqrt{x\plus{}1}\plus{}\sqrt{1\minus{}x})\,dx$?
Sonnhard.

i do! exactly! i just do not know how to type it in like that.

Hello Dr. Thanks to make the question clear.
[hide="Solution"]
$ \int \ln(\sqrt {x \plus{} 1} \plus{} \sqrt {1 \minus{} x})\,dx$
use integration by parts .
$ \equal{} x \ln(\sqrt {x \plus{} 1} \plus{} \sqrt {1 \minus{} x}) \minus{} \int \frac {1}{\sqrt {x \plus{} 1} \plus{} \sqrt {1 \minus{} x}}( \minus{} \frac {1}{2\sqrt {1 \minus{} x}} \plus{} \frac {1}{2\sqrt {1 \plus{} x}}) x dx$
$ \equal{} x \ln(\sqrt {x \plus{} 1} \plus{} \sqrt {1 \minus{} x}) \minus{} \frac {1}{2} \int \frac {\sqrt {1 \minus{} x^2} \minus{} 1}{x} \frac {x}{\sqrt {1 \minus{} x^2}} dx$
$ \equal{} x \ln(\sqrt {x \plus{} 1} \plus{} \sqrt {1 \minus{} x}) \minus{} \frac {1}{2} \int \frac {\sqrt {1 \minus{} x^2} \minus{} 1}{\sqrt {1 \minus{} x^2}} dx$
$ \equal{} x \ln(\sqrt {x \plus{} 1} \plus{} \sqrt {1 \minus{} x}) \minus{} \frac {1}{2} \int [ \frac {\sqrt {1 \minus{} x^2}}{\sqrt {1 \minus{} x^2}} \minus{} \frac {1}{\sqrt {1 \minus{} x^2}}]dx$
$ \equal{} x \ln(\sqrt {x \plus{} 1} \plus{} \sqrt {1 \minus{} x}) \minus{} \frac {1}{2} (x \minus{} sin^{ \minus{} 1}(x)) \plus{} C$
after putting d values
$ \equal{} log(\sqrt {2}) \minus{} \frac {1}{2} \plus{} \frac {\pi}{4}$

  [/hide]
Thank u

I think we have the following formula:

$ \int\limits_{0}^{1}\ln(\sqrt[n]{1\plus{}x}\plus{}\sqrt[n]{1\minus{}x})dx\equal{}\frac{2\ln 2\minus{}1}{n}\minus{}\ln 2\plus{}2\int\limits_{0}^{1}\frac{dt}{(1\plus{}t^{n})(1\plus{}t)}$,

where $ n$ is a natural number, $ n\geq 2$.

why do you double post:

http://www.mathlinks.ro/viewtopic.php?t=267009


Let $a,b,c$ be real numbers.
(a) For $a+b+c=0$, find all values of the expression
$E=\frac{a^3b+b^3c+c^3a}{a^4+b^4+c^4}$;
(b)Prove that $3(a^4+b^4+c^4)+4(a^3b+b^3c+c^3a) \ge 0$.

[quote="arqady"][quote="Albanian Eagle"]The answer to (a) is:

$-\frac{1}{2}<E<1$

....(b) follows from(a)[/quote]
I think you be mistaken.
$\frac{a^3b+b^3c+c^3a}{a^4+b^4+c^4}=\frac{a^3b-b^3(a+b)-(a+b)^3a}{a^4+b^4+(a+b)^4}=\frac{-(a^2+ab+b^2)^2}{2(a^2+ab+b^2)^2}=-\frac{1}{2}.$ ;)[/quote]
sorry I wanted to say that its true without the codition $a+b+c=0$  :)

[quote="Albanian Eagle"]The answer to (a) is:

$-\frac{1}{2}<E<1$

....(b) follows from(a)[/quote]
[quote="Albanian Eagle"]
sorry I wanted to say that its true without the codition $a+b+c=0$  :)[/quote]
But this is not true ($-\frac{1}{2}<E<1$), since if $a+b+c=0$
and $a^2+b^2+c^2\neq0$  then $E=-\frac{1}{2}.$ ;)


On Thursday, February 26, we will host a free Math Jam to discuss problems from the 2009 AMC 10 B and 2009 AMC 12 B competitions.   The Math Jam will begin at 7 PM ET (4 PM PT), and last 1.5-2 hours.  We hope to see you there!  To attend, simply click Classroom up to 10 minutes before the Math Jam starts.

The Math Jam is over but the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=250]transcript[/url] is up on the Math Jams page.


calculus
integration
function
logarithms
algebra
rational function
complex analysis

Let $C$ be a piecewise smooth curve from $z=\alpha$ to $z=\alpha'$ and $R(z)$-rational function with finite poles $a_{1},...,a_{n}$ which don't lie on $C$. Prove that $ \int_{C}R(z)dz=-\sum_{s=1}^{n}res_{z=a_{s}}R(z) \ln \frac{z-\alpha}{\alpha'-z}-res_{z=\infty}R(z) \ln \frac{z-\alpha}{\alpha'-z}$

Here for $\ln \frac{z-\alpha}{\alpha'-z}$ we take any it's regular branch with the branch cut - $C$.

I managed to prove this result above. But it was just a step to prove that $\int_{C_{1}}R(z)dz = \int_{C_{2}}R(z)dz$ (*), where  $C_{1}=\{z=e^{i\phi}, 0\leq \phi \leq \pi \}, C_{2}=\{z=e^{i\phi}, \pi \leq \phi \leq 2\pi \}$  are the upper and lower unit semicircle.

$R(z)$ - here is a rational function which has no poles on the unit circle.

Actually i know how $R(z)$ looks like, but i don't post it for not to complicate things.
I think in order to prove (*) one could apply the result above  in this post and that was said there http://www.mathlinks.ro/Forum/viewtopic.php?t=84250.

But i do not know how exactly i can do it.  Maybe you have any ideas ? If it is really depends on how $R(z)$ looks like i'll post it.

[quote="mlok"]With $R(z)=1/z^{2}$, you get $\int_{\alpha}^{\alpha'}R(z)\,dz=1/\alpha-1/\alpha'$, which does not agree with your formula.

Try using the partial fraction decomposition and integrating each term separately.[/quote]

Thank you for you interest.
Tried, but failed.

Here is my $R(z) = \frac{z \frac{B(z)}{B(\alpha)}-\alpha \frac{B(\alpha)}{B(z)}}{z-\alpha}\frac{1}{z}$,

where $B(z) = \prod_{i=1}^{n}\frac{z-\alpha_{i}}{1-\bar{\alpha_{i}}z}$, $|\alpha_{i}|<1$ - Blashke product and $\alpha$ is a fixed complex number.

Does it help somehow?

[quote="Victor96"]I cannot see any hope for this formula to hold.
As you all may see, the right-hand side of the formula does not depend on the choice of the curve $C$, but obviously the left-hand side does!
And, can anyone tell me what does the term $res_{z=\infty}R(z)ln\frac{\alpha'-z}{z-\alpha}$ means?  $resR(\infty)\times ln(-1)$?[/quote]

1). My aim is not prove that statement in the first statement of this thread. In my second post i said i proved it.

2). The left side depends on the curve $C$ as well as the right one: you see, in the right side of this equality we have $\alpha, \alpha'$ , and in the problem statement it was said that $C$ is a curve joining points $\alpha$ and $\alpha'$, though in slighly different words.

3). $res_{z=\infty}R(z)ln\frac{\alpha'-z}{z-\alpha}$ means the residue of the function  $R(z)ln\frac{\alpha'-z}{z-\alpha}$ at infinity.

[b]eugene[/b], please be more precise in your posts. The formula from the  first post is false without additional assumptions on $R$. The equality of integrals over $C_{1}$ and $C_{2}$ depends on their orientation. And your formula for $R(z)$ looks strange, because $z-\alpha$ can be cancelled out.

:maybe:
[quote]
2). The left side depends on the curve C as well as the right one: you see, in the right side of this equality we have $\alpha$, $\alpha'$ , and in the problem statement it was said that C is a curve joining points $\alpha$ and $\alpha'$, though in slighly different words. [/quote]

I am sorry but I still don't get it, probably I have misunderstood the problem.  I am talking about the choice of $C$ with end points fixed.  This choice does not affect the right-hand side but does affect the left-hand side because once $C$ goes aross a pole the integral should change, say, $R(z)=\frac{1}{x}, \alpha=-1, \alpha'=1$, and the integral is different when you take $C_{1},C_{2}$ to be the upper and lower semi-circle, resp.


How do you simplify these?  I'm not sure where to begin.  Thanks a lot.

a) $\frac{a^{4}+b^{4}}{a^{2}+b^{2}}$

b) $\frac{3+2i}{3-2i}$

a) can't really be simplified further.  Are you sure you have the problem right?  Perhaps you meant

$\frac{a^{4}-b^{4}}{a^{2}+b^{2}}= \frac{(a^{2}+b^{2})(a^{2}-b^{2})}{a^{2}+b^{2}}= a^{2}-b^{2}$

As for b), this is a standard rationalization of the denominator.

$\frac{3+2i}{3-2i}= \frac{(3+2i)^{2}}{(3+2i)(3-2i)}= \frac{5+12i}{13}$

Thanks a lot.  I typed the problem right, but it must be a mistake in the worksheet.

This helps a lot.  Thanks.


With the sun directly overhead, a fly is moving outdoor above a tilted table defined by the plane x+y+10z=5, where z is the vertical axis and x,y are horizontal axes. If the fly's path is vector r=(1+t^2)i+(-4t+3)j+(1-t^2)k until it lands, what is the speed of the fly's shadow along the table at any time t before it lands?

[hide]I assume that $\vec{k}$ is the vertical vector.
We have that
- The sun is directly overhead (the projection of the shadow is in the direction of $\vec{k}$: in this easy case the shadow has the same $x$ and $y$ as the fly).
- The fly's position is $\vec{r}=x(t)\vec{\imath}+y(t)\vec{\jmath}+z(t)\vec{k}$.
- The table is defined by $z=ax+by+c$.
Then the shadow's position will be
$\vec{r}_{s}=x(t)\vec{\imath}+y(t)\vec{\jmath}+(ax(t)+by(t)+c)\vec{k}$,
the velocity of the shadow will be
$\vec{v}_{s}=x'(t)\vec{\imath}+y'(t)\vec{\jmath}+(ax'(t)+by'(t)+c)\vec{k}$,
and the speed of the shadow will be
$v_{s}=|\vec{v}_{s}|=\sqrt{(x'(t))^{2}+(y'(t))^{2}+(ax'(t)+by'(t)+c)^{2}}$.

If you need to calculate the landing time, it's when
$z(t)=ax(t)+by(t)+c$ (make sure $z'(t)<0$).[/hide]


Source: MAML 2006 States!!!

Find the number of intersection points in the interval $[0, 2\pi)$ of the graphs $y=\tan({x-\frac\pi6})$ and $y=4\cos({x+\frac\pi3})+\frac12$.

No calculators allowed

[quote="weatherbottle"]The correct answer is $\boxed 4$ as can be shown from a rough sketch. However, is there possibly a nonguessing method of finding the correct answer?[/quote]

I am assuming you can use your trig identities to simplify everything in terms of sine and cosine and then observing their graphs to get a rough idea. that would be the easiest way to do it in my opinion.

i don't know if i'd call drawing a rough sketch as guessing...for this problem, the graphs are easy enough to sketch, and then answer should be fairly clear. there are many problems in which a rough sketch is the best way to tackle the problem (e.g. How many real solutions are there for x such that $2x^4 + |x| = \pi$ - In this case, i would think giving a 'rough sketch' is actually the best method to tackle the problem.)

[hide="Makes it a little easier..."]
Let $x=a+\frac{ \pi}{6}$. Then we are solving $\tan{a}=4\cos(a+\frac{ \pi}{2}) + \frac{1}{2}$ for $a \in [\frac{-\pi}{6},\frac{11\pi}{6})$. Then:

$\tan(a)=4\sin(a)+\frac{1}{2}$
$2\sin(a)=8\sin(a)\cos(a)+\cos(a)$
$2\sin(a)=4\sin(2a)+\cos(a)$

I'm not sure if using the sum-product identities here would be of help or not...
[/hide]


prove that $\frac{a^2}{b-1} + \frac{b^2}{a-1} \geq 8$

if $a,b > 1, \in\Re$

this is my solution but I am sure it is not very good way:
[hide]
let $b-1=y$ and $a-1=x$

$\implies x, y>0$

now we need to prove that: $\frac{(x+1)^2}{y} + \frac{(y+1)^2}{x} \geq 8$       $\forall x, y >0\in\Re$

by Cauchy Scwarz:

$(\frac{(x+1)^2}{y} + \frac{(y+1)^2}{x})(y+x) \geq (x+y+2)^2$

$\implies \frac{(x+1)^2}{y} + \frac{(y+1)^2}{x} \geq \frac{(x+y+2)^2}{x+y}$

hence it is sufficient now to prove that $\frac{(x+y+2)^2}{x+y} \geq 8$

$\implies (x+y+2)^2\geq8x+8y$

$\implies (x+y-2)^2\geq0$  :D[/hide]

i think it can be done in an easier way:

 let $a \geq b$
 then $a^2 \geq b^2$......................
 also $a-1 \geq b-1$
 since both are positive,  we have :
 $1/(b-1) \geq 1/(a-1)$
 By the rearrangement inequality,
 the combination
$a^2/(b-1) + b^2/(a-1)$ is more than any other.
 therefore, we have:

$a^2/(b-1) + b^2/(a-1) \geq a^2/(a-1) + b^2/(b-1)$
but $a^2 \geq 4(a-1)$
for all $a$.
 the result follows.

[quote="sen"]prove that $\frac{a^2}{b-1} + \frac{b^2}{a-1} \geq 8$
f $a,b > 1, \in\Re$[/quote]
Very short solution:
\[ \frac{a^2}{b-1}+\frac{b^2}{a-1} \ge \frac{(a+b)^2}{a+b-2} \ge 8 \iff (a+b-4)^2 \ge 0  \]


factorial
LaTeX
calculus
geometric series

summation n = 0 to infinity

(2n)! / ( (n!)^2 x 5^n )

How do you solve this?

it turns out to be a rational number I think, but I am stuck on how to proceed

anybody trying out this problem, do you know if you need any special theorems for solving? Is basic knowledge on factorials, infinite series, geometric series enough?

In $ \text{\LaTeX}: \quad\sum_{n = 0}^{\infty}\frac {(2n)!}{(n!)^25^n}.$

Well, it's not a rational number. A simple closed form expression, but not quite rational.

One suggestion: I think this problem is in the wrong section; it makes more sense in the Calculus Computations and Tutorials section. Perhaps a moderator could move it.

Now the hint: the key here is Newton's binomial series: for $ |x| < 1$ and arbitrary $ \alpha,$

$ (1 + x)^{\alpha} = \sum_{n = 0}^{\infty}\binom{\alpha}{n}x^n,$ where

$ \binom{\alpha}{n}$ is defined as $ \frac {\alpha(\alpha - 1)(\alpha - 2)\cdots(\alpha - n + 1)}{n!}.$

That is,

$ (1 + x)^{\alpha} = 1 + \alpha x + \frac {\alpha(\alpha - 1)x^2}2 + \frac {\alpha(\alpha - 1)(\alpha - 2)x^3}{3!} + \cdots$

Another hint: the $ 5$ in the $ 5^n$ isn't particularly special, beyond the fact that it has to be bigger than $ 4$ for this series to converge.

i tried to set the equation to x, work with the summation until I can isolate x = some numbers.

I don't know much about the binomial series, maybe that is why I can't proceed on.

do you mean x^n in the equation instead of x^2

and so alpha = 2n so you get 2n choose n times 5^-1(n) summated from n = 0 to infinite

so thats just (1+1/5)^alpha

but in the series alpha is a number? can alpha be n which progresses and changes in the summation?

I fixed the typo.

$ \alpha$ must be a fixed number; it can't have anything to do with $ n.$

Here's an intermediate step:

$ \sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n(\minus{}1)(\minus{}3)(\minus{}5)\cdots(1\minus{}2n)}{n!}\left(\frac25\right)^n.$

Here's the [hide="answer."]

We can to one more rewrite of what I posted above and make it

$ \sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n(\minus{}\frac12)(\minus{}\frac32)(\minus{}\frac52)\cdots(\minus{}\frac12\minus{}n\plus{}1)}{n!}\left(\frac45\right)^n.$

That makes it $ \left(1\minus{}\frac45\right)^{\minus{}\frac12}\equal{}\sqrt{5}.$

Numerical estimation (that is, add up a bunch of terms and watch the pattern) does tend to confirm this answer.[/hide]


how would you write text in $\LaTeX$...if i put a $\$$ sign before random text it comes out wierd liek this:

$hello how is everyone today.$

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119174 for the answer.
[quote="7h3.D3m0n.117"]it comes out wierd[/quote]Normally characters in math mode (that is, between dollar signs) are regarded as mathematical symbols so are italicised as in $f(x,y)=x y$ and spaces ignored.

$\text{Like this}$

But the AoPS administration discourages it on AoPS because of server load (I guess).

[code]$\text{}$[/code]

or

[code]$\mbox{}$[/code]


trigonometry
ARML
AMC
AIME
modular arithmetic
algorithm
function

Find the smallest positive integer solution to $\tan 19x^\circ=\frac{\cos 96^\circ+\sin 96^\circ}{\cos 96^\circ-\sin 96^\circ}.$

This is the only progress I could make:

$\tan{19x}=\frac{\cos{96}+\sin{96}}{\cos{96}-\sin{96}}=\frac{\cos{192}}{1-\sin{192}}$

I'm interested to see how this is done.

I think it would be better to write it in terms of $\tan$, so far I've gotten it reduced to $\frac{1+\tan{96}}{1-\tan{96}}=\tan{(45+96)}$, so $\tan{141}=\tan{19x}$, so now we just need to find $141+360n=19x$ which I'm working on right now :?

EDIT: I give up...someone please solve this :stretcher:

[hide]Yep, do what Shobhit said (it should be 19x...not 18x though).  So then we have $19x=141+180n.$  Take this mod 19 and we get $9n+8\equiv 0 \mod 19$ or  $9n\equiv 11\equiv 144\mod 19.$  Divide both sides by 9 to get $n\equiv 16\mod 19.$  So we take $n=16.$  So then $19x=141+180\cdot 16\Rightarrow x=159.$[/hide]

[edit] Once you have $19x\equiv 8+9n$ you can solve for $n$, $n=2x+\frac{x-8}9$, and obviously the smallest positive integer $n$ occurs when $x=8$ so that $n=16.$

[quote="pkothari13"]Iversonfan, how did you get $\frac{1+\tan{96}}{1-\tan{96}}?$[/quote]

I multiplied the numerator and denomenator by $\frac{1}{\cos{96}}$

[hide="My Progress"]

$\frac{\cos{96^\circ} + \sin{96^\circ}}{\cos{96^\circ} - \sin{96^\circ}} = \frac{\cos{96^\circ} + \sin{96^\circ} \cdot (\cos{96^\circ} + \sin{96^\circ})}{\cos{96^\circ} - \sin{96^\circ} \cdot (\cos{96^\circ} + \sin{96^\circ})}$

= $\frac{(\cos{96^\circ})^2 + 2\sin{96^\circ}\cos{96^\circ} + (\sin{96^\circ})^2}{(\cos{96^\circ})^2 - (\sin{96^\circ})^2}$

= $\frac{1 + \sin{192^\circ}}{\cos{192^\circ}}$

So, $\tan{19x^\circ} = \sec{192^\circ} + \tan{192^\circ}$
[/hide]

[b]Edit[/b]

I just saw joml88's edited post and I probably have to refer to his solution now because I am utterly inept with mods.

[quote="amcavoy"]Could someone post another problem similar to this?  I would like to attempt something similar now knowing how this one was done.[/quote]

As a matter of fact, 1990 ARML team problem #10 looks [i]strikingly[/i] similar.  However, it's different enough to still be challenging after seeing the solution to the last one ;) Here ha go:

If \[ \tan(120^\circ-x)=\frac{\sin 120^\circ-\sin x}{\cos 120^\circ-\cos x},  \] where $0^\circ<x<180^\circ,$ compute $x.$

I'm going to have to take a hint on this one.  I have used the subtraction formula to get:

$\frac{\tan{120}-\tan{x}}{1+\tan{120}\tan{x}}=\frac{\sin{120}-\sin{x}}{\cos{120}-\cos{x}}$

...but I'm unsure of how to proceed.

For the first one, we can procede like this:

$\tan^2(19x)=\frac{1+2\cos96\sin96}{1-2\cos96\sin96}=\frac{1+\sin192}{1-\sin192}=\frac{1-\sin12}{1+\sin12}=\frac{1-\cos78}{1+\cos78}$

So:

$\tan(19x)=\sqrt{\frac{1-\cos78}{1+\cos78}}=\tan\frac{78}{2}=\tan36$

Therefore:

$19x = 39 + 180n$

and from here we can procede like [b]joml88[/b] did.

[quote="joml88"]If \[ \tan(120^\circ-x)=\frac{\sin 120^\circ-\sin x}{\cos 120^\circ-\cos x},  \] where $0^\circ<x<180^\circ,$ compute $x.$[/quote]

[hide="Solution"]
Use the difference to product formulas:

\begin{eqnarray*} \sin\alpha-\sin\beta &=& 2\sin\left(\frac{\alpha-\beta}2\right)\cos\left(\frac{\alpha+\beta}2\right)\\ \cos\alpha-\cos\beta &=& -2\sin\left(\frac{\alpha+\beta}2\right)\sin\left(\frac{\alpha-\beta}2\right) \end{eqnarray*}

This gives

\begin{eqnarray*} \tan(120-x) &=& \frac{\sin 120-\sin x}{\cos 120-\cos x}\\ &=& \frac{2\sin\left(\frac{120-x}2\right)\cos\left(\frac{120+x}2\right)}{-2\sin\left(\frac{120+x}2\right)\sin\left(\frac{120+x}2\right)}\\ &=& -\cot\left(60+\frac x2\right)\\ &=& -\tan\left(30-\frac x2\right)\\ &=& \tan\left(\frac x2-30\right) \end{eqnarray*}

So $120-x+180k=\frac x2-30\Rightarrow x=100+120k.$  So possible answers are $\ldots,-140, -20, 100, 220, 340,\ldots$ but $\boxed{100}$ is the only between 0 and 180.[/hide]

Wait, i got 131 instead of 141.  it reduces to -cot 51 or -tan 49 and thus tan 131 and tan 311.  Or did i mess up?>

grr, i have to learn those addition and subtraction formulas, they seem to make so many of these problems easy

Erm... since there seems to be no conclusive solution, I hope that I may be able to post one  :D

[hide]We have:

$\frac{\sin{19x}}{\cos{19x}} = \frac{\cos{96}+\sin{96}}{\cos{96}-\sin{96}}$

By cross-multiplying,

$\sin{19x}\cos{96} - \sin{19x}\sin{96} = \cos{19x}\cos{96} + \cos{19x}\sin{96}$

By moving stuff over, we have by the addition theorem,

$\sin{(19x - 96)} = \cos{(19x - 96)}$

$\sin{(19x - 96)} = \sin{(186 - 19x)}$

$19x - 96 = 186 - 19x$

$38x = 208$

$x = 141/19$

Plugging into $\tan{19x}$, we get $\tan{141}$. Since $\tan{(180 + x)} = \tan{x}$, we can say that:

$141 + 180m = 19x$

After HE** sticks with this Diophantine Equation (it took me about 6 minutes), we end up getting $m \equiv 16 \bmod{19}$. This means that $x \equiv 159 \bmod{180}$, which yields:

$x = 159^\circ$
[/hide]
Then again, my answer could definitely be incorrect. So yup.

[hide="A Different Approach"]
Square both sides to get

$ \tan^2 19x\equal{}\frac{\cos^2 96\plus{}\sin^2 96\plus{} 2\sin 96\cos 96}{\cos^2 96\plus{}\sin^2 96\minus{} 2\sin 96\cos 96}\equal{}\frac{1\plus{}\sin 192}{1\minus{}\sin 192}$.

Add $ 1$ to both sides, and use $ \tan^2\plus{}1\equal{}\sec^2$ on the left:

$ \frac{1}{\cos^2 19x}\equal{}\frac{2}{1\minus{}\sin 192}$.

Thus,

$ \minus{}\sin 192\equal{}2\cos^2 19x\minus{}2\equal{}\cos 38x$

or

$ \cos 38x\equal{}\cos 78$.

Since the original equation requires $ \tan 19x$ to be negative, this gives us the $ 19x\equal{}141\plus{}180n$ equation that has previously been solved.
[/hide]

This is an AIME question, isn't it?

Here's how I did it (everything's in degrees):

[hide="Solution"]$ \tan 19x\equal{}\frac{\sin 19x}{\cos 19x}\equal{}\frac{\cos 96\plus{}\sin 96}{\cos 96\minus{}\sin 96}$

$ \sin 19x\cos96\minus{}\sin19x\sin96\equal{}\cos19x\cos96\plus{}\sin96\cos19x$

$ \sin 19x\cos96\minus{}\sin96\cos19x\equal{}\cos 19x\cos 96\plus{}\sin 19x\sin 96$

$ \sin(19x\minus{}96)\equal{}\cos(19x\minus{}96)$.

$ \tan(19x\minus{}96)\equal{}1\implies 19x\minus{}96\equiv 45\pmod{180}\implies 19x\equiv141\pmod{180}$.

Applying the extended Euclidean algorithm, $ 19\cdot 19\equiv 1\pmod{180}$, so:

$ 19\cdot 19x\equiv x\equiv 141\cdot 19\equiv 159\pmod{180}$.

And therefore, the smallest solution is $ \boxed{159}$.[/hide]

Some of you guys seem to be overcomplicating the heck out of it  :wink:

(note: all numbers in the trig functions below are in degrees, not radians, so even if I don't write $45^o$ just know.)

basically you need to realize that the RHS is this in disguise:

$\frac{\sin{(45+96)}}{\cos{(96+45)}}$.

this can be done by noting that the lowest possible terminal point in which $x=y$ is $\sin{45}=\cos{45}=\frac{\sqrt{2}}{2}$ so this will cancel in the numerator with the denominator and you're left with:

$\tan{19x}=\frac{\sin{(45+96)}}{\cos{(96+45)}}=\frac{\sin{141}}{\cos{141}}$.

Since the period of $\tan{x}$ is $\pi$ it follows that we need $141+180k\equiv{0}\pmod{19}$ for some positive integer $k$.

The result follows after some simple mod arithmetic.

For storage...
[hide=Solution]
We can easy use the identity $sin(x)=cos(90-x)$ and then sum-to-product and difference-to-product identities to determine that the RHS is equal to $\frac{\cos 51^\circ}{\sin 51^\circ}= -cot 51^\circ= -tan 39^\circ= tan 141^\circ$. This yields that $19x=141+180k$ for some positiv e integer $k$. Dividing both sides by $19$ yields that $x= \frac{9k+18}{19}+9k+7$, so $9k+18=0 \textbf{mod 19}$. Guessing and checking easily gives that $k= 16 \textbf{mod 19}$, so $x=\boxed{159}$[/hide]

@above, \sin, \cos, \tan, \cot, etc. may be helpful to you. In addition, \pmod is a good way to express mods.


geometry
rectangle
analytic geometry
graphing lines
slope
perpendicular bisector

Hello everyone,

 Here is the math problem:

1. In the triangle ABC, angle C = 90°, AB = 20, AC = 12 and D is the midpoint of side AB. The perpendicular at D on AB intersects BC at E. Find the area of the quadrilateral ADEC.

--------

I am not sure if I have drawn the triangle correctly according to the question, and I am not sure which triangles I should compare.

I uploaded a rough diagram here:
http://img155.imageshack.us/my.php?image=problem4lo3.jpg

I know that triangle ABC is a 3:4:5 triangle so CB would be 16; but how should I proceed from here? I also added some lines to the triangle as shown in the uploaded picture.

Thank you very much in advance!

Patrick

[img]http://www.initdotd.com/~nikita/Pictures/Triangle.jpg[/img]

[hide]$ADEC$ can be split into a rectangle and a triangle. The area of the rectangle is $6\cdot8=48,$ and the area of the triangle is $\frac{6\cdot8}2=24.$ The area of $ADEC$ is $\boxed{72}.$[/hide]

Thanks for the reply, but I have the answers at the back of the sheet and it says for this question, the answer is 58.5.

I got 72 as well, but did we overlook something?

Thanks though; I appreciate your attempt.

Are you sure you are looking at the answer to this, and not some other, problem? I am $\infty\%$ sure the answer is 72.

$D$ is the midpoint of $AB$ and $\triangle ABC$ is right, so $D$ is the circumcenter. Therefore, the perpendicular bisectors of sides $BC$ and $AC$ intersect at $D.$ The area of $FDCE=48$ and the area of $\triangle ADF=24,$ so the area of $ADCE=72.$

Thanks for your responses.

I typed up the question correctly, but my diagram might be wrong; this is a word problem.

Yes, 58.5 is still the answer.

Thanks again!

[quote="vishalarul"]I think that you are drawing the diagram incorrectly.
I think that point E is the intersection of the perpendicular bisector of segment AB and segment CB.

In other words, you drew the perpendicular incorrectly. The line is perpendicular to segment AB and passes through D, intersecting at point E at segment BC.

I don't know if I am right, but maybe you misread the problem.[/quote]
Can you draw a diagram of your interpretation?

[quote="catalan"]1. In the triangle ABC, [b]angle C = 90°, AB = 20, AC = 12[/b] and [b]D is the midpoint of side AB[/b]. [b]The perpendicular[/b] at D on AB [b]intersects BC at E[/b]. Find the area of the quadrilateral ADEC.[/quote]
$D$ is the midpoint of $AB$ and $E$ is the midpoint of $BC.$ How can you draw any diagram different from mine that suits these facts?

How do you put a diagram in a message like you did?

I already made one; I just need to know how to put it here.

There are two ways; either host it on a website and use

[code][img]image url here[/img][/code]

or add an attachment, at the top of the post a reply box.

I think you're right.

[hide][img]http://www.initdotd.com/~nikita/Pictures/Triangle%202.jpg[/img]

Divide $ADEC$ into 3 parts: $\triangle ADF,\,\triangle DEG,$ and $FGEC.$

$\triangle ADF=24,\,\triangle DEG=13.5,\,FGEC=21$

The area of $ADEC=24+13.5+21=\boxed{58.5}.$[/hide]

How did you get all those side lengths?
Could you explain your method better?

$AB=20$ and $D$ is the midpoint, so $AD=10$

$AC=12$ and $F$ is the midpoint, so $AF=FC=GE=6$

$BC=16$ and $BC=2FD,$ so $FD=8$

We know the slope of $AB$ is $-\frac{3}{4}$ and $DE\perp AB,$ so the slope of $AB=\frac{4}{3}.$ If we put this on a coordinate system and do some algebra, we find $CE=FG=3.5$ and $EB=12.5.$

$FG+GD=FD=8$ and $FG=3.5,$ so $GD=4.5$

[quote="catalan"]Thanks for your replies, especially to i_like_pie and vishalarul.

Could you get the lengths without setting it up with a Cartesian graphic system, just by pythogorean and reasoning?

Thanks again![/quote]
Yes, you can.

We know $\triangle ABC\sim\triangle DEG,$ so $\triangle DEG$ is also a 3-4-5 triangle. Since we know $GE=6,$ we can find $GD$ and $DE.$

I got it!

Thank you very much, especially to i_like_pie!


[quote="Interval"]A man has a barn that is 20 ft by 10 ft. He tethers a cow to one corner of
the outside of the barn using a 50-ft rope. What is the total area that the
cow is capable of grazing?[/quote]
[hide="Hint"]If you draw it all out you see that the areas you need are a 3/4 circle with radius of 50, a quarter circle woth radius of 30 and another quarter circle with radius of 40. I'm sure know that you know what to do you can do the rest.[/hide]


The crepe store is having a sale- buy a first crepe for $ T, and get any additional crepe for$J less than the first crepe. How much would n crepes cost in total?

[hide="Solution"]We can prove by induction that the cost of $ n$ crepes is $ \boxed{n(T\minus{}J) \plus{} J}$.

Base case: $ n \equal{} 1$. The cost is $ T$, which is correct.

We assume the cost is correct for $ k$ crepes. The cost of $ k\plus{}1$ crepes would therefore be $ k(T\minus{}J) \plus{} J \plus{} (T\minus{}J) \equal{} (k\plus{}1)(T\minus{}J) \plus{} J$.

Therefore, by induction, the cost of $ n$ crepes is $ \boxed{n(T\minus{}J) \plus{} J}$.

[/hide]


inequalities
LaTeX
function
inequalities proposed

$ a, b, c$ are from the range $ (0, 1)$ and $ a\plus{}b\plus{}c\equal{}1$. Prove that:

\[ \frac{1\plus{}a^{2}}{1\minus{}a^{2}}\plus{}\frac{1\plus{}b^{2}}{1\minus{}b^{2}}\plus{}\frac{1\plus{}c^{2}}{1\minus{}c^{2}}\geq\frac{15}{4}.\]

It is a really easy problem I already have 5 solutions. The reason to be here is just to be popularized and someone may like it.

[color=red][Moderator edit: LaTeX (Arne)][/color]

very easy... for example one line with Jensen and function $ f(x)\equal{}\frac{x^{2}}{1\minus{}x^{2}}$ convex on $ [0,1[$

An easy one.
Denote $ S \equal{}\sum\frac{1\plus{}a^{2}}{1\minus{}a^{2}}\equal{}\sum\frac{1\plus{}a^{2}}{(1\plus{}a)(b\plus{}c)}\equal{}\sum\frac{a^{2}}{(1\plus{}a)(b\plus{}c)}\plus{}\sum\frac{1}{(a\plus{}1)(b\plus{}c)}$.
By CBS inequality:
$ \sum\frac{a^{2}}{(1\plus{}a)(b\plus{}c)}\geq\frac{(a\plus{}b\plus{}c)^{2}}{2(1\plus{}ab\plus{}ac\plus{}bc)}$
$ \sum\frac{1}{(1\plus{}a)(b\plus{}c)}\geq\frac{9}{2(1\plus{}ab\plus{}bc\plus{}ca)}$.
So we must prove that $ 10\geq\frac{15}{4}(2\plus{}2(ab\plus{}bc\plus{}ca))\Longleftrightarrow ab\plus{}bc\plus{}ca\leq\frac{1}{3}$.
Which is obviously true.

The original idea is to sum the following inequalities which are true for postive numbers different from 1 with sum 1.

(1+a)/(1-a)+(1+b)/(1-b)+(1+c)/(1-c)>=6 (easy to be proved by using 1=a+b+c and A/B+B/A>=2)

(1-a)/(1+a)+(1-b)/(1+b)+(1-c)/(1+c)>=3/2 (Bulgaria III round 1970) (may be proved by using 1=a+b+c and Nesbit inequality)

I saw many solutions - but no one uses my idea.

I think that this will be not much easier than my solution,isn't it?Or maybe i am wrong.Please use LaTeX.

It depends - your solutions is one of the simplest I saw - a man was published an article about the inequality you use - it is more general method than mine. But when I invented the inequality I didn't know this method. Other solutions I saw in the magazine I published it - use the proposal a<=b<=c.

P.S. I promised -  tomorrow I'll start learning Latex.

[quote="borislav_mirchev"]It depends - your solutions is one of the simplest I saw - a man was published an article about the inequality you use - it is more general method than mine. But when I invented the inequality I didn't know this method. Other solutions I saw in the magazine I published it - use the proposal a<=b<=c.[/quote]
Thank you.And what about the other solutions?what methods did they use?
[quote="borislav_mirchev"]
P.S. I promised -  tomorrow I'll start learning Latex.[/quote]
OK. :lol:

You can see other methods - and two generalizations...

1. If a1+a2+...+an=1; 1<>ai>0 (1<=i<=n) it follows that:
1/(1-a1^2)+1/(1-a2^2)+...+1/(1-an^2)>=n^3/(n^2-1)

2. If a,b,c,p are reals, a,b,c are in range (0;1), a+b+c=1, p>=1 prove that:
(1+a^p)/(1-a^p)+(1+b^p)/(1-b^p)+(1+c^p)/(1-c^p)>=3.((3^p+1)/(3^p-1)).

P.S. I'm sorry but I believe you can survive without using Latex from me for a day...

Yeah.It isn't problem.I understand many words in this book,because i know russian. :wink:
Thank you very much. :)

[quote="borislav_mirchev"]$ a, b, c$ are from the range $ (0, 1)$ and $ a\plus{}b\plus{}c \equal{} 1$. Prove that:
\[ \frac{1\plus{}a^{2}}{1\minus{}a^{2}}\plus{}\frac{1\plus{}b^{2}}{1\minus{}b^{2}}\plus{}\frac{1\plus{}c^{2}}{1\minus{}c^{2}}\geq\frac{15}{4}.\]
It is a really easy problem I already have 5 solutions. The reason to be here is just to be popularized and someone may like it.

[color=red][Moderator edit: LaTeX (Arne)][/color][/quote]

$ \sum\frac{1\plus{}a^{2}}{1\minus{}a^{2}}\ge\frac{15}{4}$

$ \iff\sum\left(\frac{1\plus{}a^{2}}{1\minus{}a^{2}}\minus{}1\right)\ge\frac{3}{4}$

$ \iff\sum\frac{a^{2}}{1\minus{}a^{2}}\ge\frac{3}{8}$

Now by Cauchy-Schwarz inequality we get

$ \sum\frac{a^{2}}{1\minus{}a^{2}}\ge\frac{(a\plus{}b\plus{}c)^{2}}{3\minus{}a^{2}\minus{}b^{2}\minus{}c^{2}}\ge\frac{1}{3\minus{}\frac{1}{3}}\equal{}\frac{3}{8}$

Where the last inequality follows from $ a^{2}\plus{}b^{2}\plus{}c^{2}\ge\frac{1}{3}(a\plus{}b\plus{}c)^{2}\equal{}\frac{1}{3}$.

Look at the generalizations. What do you think about them? I think they are very beautiful and are more difficult than the original inequality :) if you want you may try them.

P.S. It is not a book it is a small bulgarian magazine: "Mathematical Forum" in english. There are interesting inequalities. They will be easy for mathlinkers but there are some very good. I'll post them when I've more spare time.

[quote="borislav_mirchev"]
1. If $ a_{1}\plus{}a_{2}\plus{}\dots\plus{}a_{n}\equal{} 1.$ $ 1\leq a_{i}< 1,(1\leq i\leq n)$ it follows that:
$ S \equal{}\sum_{i \equal{} 1}^{n}\frac{1\plus{}a_{i}^{2}}{1\minus{}a_{i}^{2}}\geq\frac{n^{3}}{n^{2}\minus{}1}$
[/quote]
[quote="borislav_mirchev"]
Look at the generalizations. What do you think about them? I think they are very beautiful and are more difficult than the original inequality  if you want you may try them.
[/quote]
OK.If you talking about this problem.They are not very hard.The first one:
Solution:$ S\geq\frac{n^{3}}{n^{2}\minus{}1}\Longleftrightarrow\sum_{i \equal{} 1}^{n}\frac{2}{1\minus{}a_{i}^{2}}\geq\frac{n^{3}}{n^{2}\minus{}1}\plus{}n \equal{}\frac{2n^{3}}{n^{2}\minus{}1}$
Denote $ f(x) \equal{}\frac{2}{1\minus{}x^{2}}$.Then $ f$ is convex,so using Jensen inequality:
$ \sum f(a_{i})\geq nf(\frac{\sum a_{i}}{n}) \equal{}\frac{2n}{1\minus{}\frac{1}{n^{2}}}\equal{}\frac{2n^{3}}{n^{2}\minus{}1}$
BTW,the second one also can be solved by using Jensen ineq. :wink:

There are no difficult problems. There are problems invented by people with thinking different than yours, and different experience.

[quote="borislav_mirchev"]Kunny, you may take a look at the attachment to compare with your solution.[/quote]

I can't open your file for some reason. :P

Your problems is just related to a tangent line, so those are easy.

Kunny, do you have good download program? A tool that may open win zip. And jpeg viewer? If you  haven't some of them - you cannot open the file.

P.S. If someone have the same problem like Kunny, please, say me or say to moderators.


calculus
integration
trigonometry
complex numbers
real analysis
real analysis theorems

well here is some nice formula, always I see such these integrals most of the times when I'm solving some Differential Equation or a calculus problem, I was tired of using integrating by parts so here is a nice way to evaluate these two integrals where $\alpha$ and $\beta$ are real numbers:

$I_{1}=\int e^{\alpha x}\sin\beta xdx$
$I_{2}=\int e^{\alpha x}\cos\beta xdx$

we'll use complex numbers! :D  $i^{2}=-1$ and remember that $e^{ix}=\cos x+i\sin x$:

$I_{2}+iI_{1}=$

$\int e^{\alpha x}\left(\cos\beta x+i\sin\beta x\right)dx$

$I_{2}+iI_{1}=$

$\int e^{\alpha x}\cdot e^{i\beta x}dx=\int e^{(\alpha+i\beta)x}dx$

$I_{2}+iI_{1}=$

$\frac{e^{(\alpha+i\beta)x}}{\alpha+i\beta}$

$I_{2}+iI_{1}=$

$\frac{\alpha-i\beta}{\alpha^{2}+\beta^{2}}e^{\alpha x}\cdot e^{i\beta x}$

$I_{2}+iI_{1}=$

$\frac{\alpha-i\beta}{\alpha^{2}+\beta^{2}}e^{\alpha x}\left(\cos\beta x+i\sin\beta x\right)$

As a consequence:

$I_{2}+iI_{1}=$

$\frac{e^{\alpha x}}{\alpha^{2}+\beta^{2}}\left(\alpha\cos\beta x+\beta\sin\beta x\right)$

$+i\frac{e^{\alpha x}}{\alpha^{2}+\beta^{2}}\left(\alpha\sin\beta x-\beta\cos\beta x\right)$

and the integrals will be:

$\int e^{\alpha x}\sin\beta xdx=$

$\frac{e^{\alpha x}}{\alpha^{2}+\beta^{2}}\left(\alpha\sin\beta x-\beta\cos\beta x\right)$

$\int e^{\alpha x}\cos\beta xdx=$

$\frac{e^{\alpha x}}{\alpha^{2}+\beta^{2}}\left(\alpha\cos\beta x+\beta\sin\beta x\right)$

And Thats It! :lol:

Yes, I think this is well-known.  Similarly we can write them in terms of terms of the form $e^{ (a+b i) x}$ using the formulas

$\sin \beta x = \frac{e^{i \beta x}-e^{-i \beta x}}{2}$
$\cos \beta x = \frac{e^{i \beta x}+e^{-i \beta x}}{2}$

And the exercise I follow this with in my calculus class:

Assume $\alpha<0.$ (Making it a definite integral on $[0,\infty)$ makes it a little nicer, and this condition is to assure convergence.) Compute:

$\int_{0}^{\infty}xe^{\alpha x}\cos \beta x\,dx$

$\int_{0}^{\infty}xe^{\alpha x}\sin \beta x\,dx$

$J_{1}=\int^{\infty}_{0}xe^{\alpha x}\cos\beta xdx$
$J_{2}=\int^{\infty}_{0}xe^{\alpha x}\sin\beta xdx$

Wow that was nice and a little with much calculations :D I hope the answers are correct(and they seem to be :D ):

[hide="Answers"]$J_{1}=\frac{\alpha^{2}-\beta^{2}}{(\alpha^{2}+\beta^{2})^{2}}$
$J_{2}=\frac{2\alpha\beta}{(\alpha^{2}+\beta^{2})^{2}}$[/hide]

if:

$J_{1}=\int xe^{\alpha x}\sin\beta xdx$
$J_{2}=\int xe^{\alpha x}\sin\beta xdx$

[hide="Then"]$J_{1}=\frac{e^{\alpha x}}{(\alpha^{2}+\beta^{2})^{2}}\left[\left(\alpha^{3}x-\alpha^{2}-\beta^{2}\alpha x+\beta^{2}-2\alpha\beta^{2}x\right)\cos\beta x-\left(\alpha^{2}\beta x-\beta^{3}x+2\alpha^{2}\beta x-2\alpha\beta\right)\sin\beta x\right]$

$J_{2}=\frac{e^{\alpha x}}{(\alpha^{2}+\beta^{2})^{2}}\left[\left(\alpha^{2}\beta x-\beta^{3}x+2\alpha^{2}\beta x-2\alpha\beta\right)\cos\beta x+\left(\alpha^{3}x-\alpha^{2}-\beta^{2}\alpha x+\beta^{2}-2\alpha\beta^{2}x\right)\sin\beta x\right]$[/hide]

If we remember that $\alpha<0,$ and noting that $J_{2}$ should be positive, I believe we actually have $J_{2}=\frac{-2\alpha\beta}{(\alpha^{2}+\beta^{2})^{2}}.$

For my own sake, I would prefer not to attempt to write out the full antiderivative - I made it a definite integral problem precisely so I could avoid that step.

[hide="My version."]

$\int_{0}^{\infty}xe^{\alpha x}e^{i\beta x}\,dx=\int_{0}^{\infty}xe^{(\alpha+i\beta)x}\,dx$

$=\left.\frac1{\alpha+i\beta}xe^{(\alpha+i\beta)x}\right|_{0}^{\infty}-\frac1{\alpha+i\beta}\int_{0}^{\infty}e^{(\alpha+i\beta)x}\,dx$

$=0-\left.\frac1{(\alpha+i\beta)^{2}}e^{(\alpha+i\beta)x}\right|_{0}^{\infty}$

$=\frac{1}{(\alpha+i\beta)^{2}}= \frac{(\alpha-i\beta)^{2}}{(\alpha+i\beta)^{2}(\alpha-i\beta)^{2}}= \frac{(\alpha^{2}-\beta^{2})-2i\alpha\beta}{(\alpha^{2}+\beta^{2})^{2}}.$

Now we can read off the real and imaginary part of this as our answer for the definite integral.[/hide]


geometry
rectangle
probability
geometric transformation
rotation

Rectangle $ ABCD$ has center $ O$ and $ AB/AD\equal{}k$. A point is randomly chosen from the interior of rectangle $ ABCD$. What is the probability that it is closer to $ O$ than to any of the four vertices? Express your answer in terms of $ k$.
[asy]size(200);
draw((-250,100)--(250,100)--(250,-100)--(-250,-100)--cycle);
dot((0,0));
label("$O$",(0,0),N);
label("$A$",(-250,100),NW); label("$B$",(250,100),NE); label("$C$",(250,-100),SE); label("$D$",(-250,-100),SW);[/asy]

no, it can't.  even using your diagram...

if you call the lower left corner 0,0 then 2, 3.1 is closer to the center but not in your region....even 2,3.999 is closer to the center than to a corner.

You still ignored (2,3.9).  You must include 1,3 also, and abunch more..........

I solved this pretty quickly, but struggled to find a solution that involved k.  [hide]I considered using k/2k but decided that 1/2 had to be right... I would remove the "express your answer in terms of k" part of the problem.  Would k/2k have been accepted?[/hide]

I don't get it. Can someone help me out? Thanks!

[b]Nobody posted a solution I liked so...[/b]
[hide=The diagram is always not to scale]
We can assume any number for $k$, since it did not specify.
Assuming $k=1$, the resulting object is a square.
Now where is a random point closer to $O$ than any of the corners?
Another smaller square rotated $90$ degrees inscribed in it fits. Every place in that square is closer to $O$ than any other corner.
We know that the area of the small square is $\frac{1}{2}$ of the total area, so the
answer is equal to $\boxed{\frac{1}{2}}$
[/hide]
I hope everyone approves of my "solution"

[quote=bestzack66][quote name="ThriftyPiano" url="/community/p18118967"]
[b]Nobody posted a good solution so...[/b]

[/quote]

That's not very nice...[/quote]

Okay fine.

Cheese this with $k=1.$

@below oops lol :P

[quote=Lamboreghini]Cheese this with $k=1.$[/quote]

how would you know that the answer would be k/2 instead of (k+14213)/28428


geometry
IMO
circle
imo 2000
Sergei Berlov

Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP \equal{} EQ$.

let MN and AB intesect at K. Then $KA^2=KM*KN=KB^2.$ Hence KA=KB.
CD||AB. Hence PM=QM.
$\angle AEB=\angle ECD=\angle MAB,$ $\angle EBA=\angle EDC=\angle MBA.$
hence$\triangle AMB$ and$\triangle AEB$ congruity.
Hence $EM\perp AB.$ Hence $EM\perp PQ.$
Hence $EP=EQ.$

The radical axis MN of the circles $(G_1), (G_2)$ cuts the tangent length segment AB at its midpoint C. Since $PQ \parallel AB$, the triangles $\triangle ABN \sim \triangle PQN$ are centrally similar with similarity center N. Hence, M is the midpoint of the segment PQ. The radii $G_1A, G_2B$ of the circles $(G_1), (G_2)$ are perpendicular to the tangent AB, i.e, also to the chords AM, DM. Hence they cut these chords at their midpoints, which means that the triangles $\triangle ACM, \triangle BDM$ are isosceles and the angles $\angle ACM = \angle AMC, \angle BDM = \angle BMD$ are equal. Since the lines $AB \parallel CD$ are parallel and $M \in CD$, the angles $\angle MAB = \angle AMC = \angle ACM = \angle EAB$ are equal and similarly, the angles $\angle MBA = \angle BMD = \angle BDM = \angle EBA$ are also equal. Thus the diagonal AB of the quadrilateral AMBE bisects its opposite angles at the vertices A, B, which implies that this quadrilateral is a kite with AE = AM, BE = BM and and its diagonals $AB \perp EM$ are perpendicular to each other. Consequently, the lines $EM \perp CD \equiv PQ$ are also perpendicular to each other. Since M is the midpoint of the segment PQ, the line EM is the perpendicular bisector of this segment. It follows that the triangle $\triangle EPQ$ is isosceles with EP = EQ.

An "extra" question:

Prove that $EN$ bisects $\angle{CND}$.

Since $ AB$ is the common tangent of $ G_1$ and $ G_2$ and $ MN$ is the radical axis of $ G_1$ and $ G_2$, $ MN$ bisects $ AB$. A homothety centered at $ N$ that maps $ A$ to $ P$ and $ B$ to $ Q$ therefore maps the midpoint of $ AB$ to $ PQ$, so we have that $ M$ is the midpoint of $ PQ$.

It is sufficient to show that $ EM \perp CD$. Let $ R$ and $ S$ be the projections of $ A$ and $ B$ onto $ CD$, respectively; since $ AB \parallel{} CD$, $ AR \equal{} BS$. We first note that $ AC \equal{} AM$, since $ m \angle AMC \equal{} m \angle EAB \equal{} m \angle ACM$ (as $ AB$ is tangent to $ G_1$). We can similarly deduce that $ BM \equal{} BD$.

It also follows from this that $ CM \equal{} 2CR$ and $ DM \equal{} 2DS$. Let $ E'$ be the point such that $ E'M \perp CD$ and $ E'M \equal{} 2AR \equal{} 2BS$. A homothety centered at $ C$ that maps $ R$ to $ M$ must therefore map $ A$ to $ E'$, and a homothety centered at $ D$ must therefore map $ B$ to $ E'$. But this means that $ E'$ lies on both $ AC$ and $ BD$, implying that $ E' \equal{} E$. It follows that $ EM \perp CD$, so we are done.

[quote="Arne"]An "extra" question:

Prove that $ EN$ bisects $ \angle{CND}$.[/quote]

This extra question is also beautiful one.

First, from what we know  Let $ \angle{ANM}\equal{}\angle{BAM}\equal{}\angle{BAE}\equal{}x$ and $ \angle{BNM}\equal{}\angle{ABM}\equal{}\angle{ABE}\equal{}y$

Therefore, $ \angle{AEB}\plus{}\angle{ANB}\equal{} (\pi\minus{}x\minus{}y)\plus{}(x\plus{}y)\equal{}\pi$      $ \longrightarrow$      $ A,N,B,E$ is concyclic.

It follows that $ \angle{BNE}\equal{}\angle{BAE}\equal{}x$ and it is known that $ \angle{BND}\equal{}\angle{BNM}\equal{}y$
We get $ \angle{DNE}\equal{}x\plus{}y$

Similarly, we also have $ \angle{CNE}\equal{}x\plus{}y$  :lol:  :lol:  :lol:
This is what we want to prove.

$K$ is the midpint of $AB$. $NM$ is radical axis for $G_1,G_2$ circles. also we know that radical axis passes through the midpoints of comment tangents. Thus $N,M,K$ lies on the same line. $PQ||AB=>M$ is the midpoint of $PQ$.
Also it is easy to note this number of equations.
$\angle{PCA}=\angle{ANM}=\angle{BAE}=\angle{MAB}=\angle{PMA}=\angle{CNA}=\angle{ENB}$[*] the last equation follows from the fact that $AENB$ is cyclic.
Similarly we can get in the opposite side. Thus $\triangle {AMB}=\triangle {AEB}=>EM\perp PQ=>EP=EQ$
@Arne
from [*] it is easy to note it.

[quote="arshakus"]One more "extra" question!)
Prove that $2AB=CD$[/quote]
In your solution we prove that $ \triangle{AMB}=\triangle{AEB}=> EB=BM=BD$
and  $ CD\parallel AB =>  \frac{EB}{ED}=\frac{AB}{CD}=2 $
:)

Let $\angle ACM=\alpha$, then since $AB||CD$, we have $\angle EAB=\angle ACM=\alpha$ and since $AB$ is tangent to circle $G_1$, we have $\angle BAM=\angle ACM=\alpha$.  Thus $\angle EAB=\angle BAM=\alpha$.  Similarly, if we let $\angle BDM=\beta$, we find $\angle EBA=\angle ABM=\beta$.

Now let the intersection of $EM$ and $AB$ be $F$, then $\angle AFE=90^\circ$ and $\angle CME=\angle AFE=90^\circ$ and therefore proving $EP=EQ$ is equivalent to proving $PM=QM$ since $EM\perp PQ$.

Let the intersection of ray $NM$ and $AB$ be point $G$, then $GA=GB$ because $G$ is on the radical axis of the two circles and thus $GA^2=GB^2\implies GA=GB$.  Additionally, since $AB||CD$, we have $\triangle AGN~\triangle PMN$ and $\triangle BGN~\triangle QMN$ and therefore $PM=PQ$ as desired. $\blacksquare$

As the tangent is parallel to the chords So it follows that $ \triangle ACM$ and $ \triangle BDM$ are isosceles and $EA=AM=AC$ so $EM$ is perpendicular  to $PQ$.  Define $Z=MN \cap AB$.$Z$ lies on radical axis ,so $Z$ is the midpoint of $AB$. And hence $M$ is the midpoint of  $PQ$. So done.

OMG!! got a proof without bashing :O - if $MN$ meets $AB$ at $K$ then $AK^2=BK^2\implies AK=KB\implies PM=MQ\implies \angle{CMN}=\frac {\pi}{2}\implies EP=EQ$

I think $E,M,N$ are not at all  collinear.
I have drawn a ggb diagram. and checked it.

Yet additional requirement:

[b]the quadrilateral $ANBE$ is harmonic![/b]

Best regards,
sunken rock

[hide="Solution"]
Since $AB$ is tangent to $G_1$ at $A$, $\angle{ANM}=\angle{MAB}$. But $AB||CD$ and $AMNC$ is cyclic, so $\angle{EAB}=\angle{ECM}=\angle{ANM}=\angle{MAB}$. By similar reasoning, $\angle{EBA}=\angle{MBA}$. Also, $AB=AB$, so $\triangle{EAB}\cong\triangle{MAB}$ by ASA congruency. Then $AMBE$ is a kite, so $EM\perp AB$, implying that $EM\perp PQ$.

Let $MN$ intersect $AB$ at $X$. Since $MN$ is the radical axis of $G_1$ and $G_2$, the power of $X$ w.r.t. $G_1$ equals the power of $X$ w.r.t. $G_2$. Hence, $XA^2=XB^2$, or $XA=XB$. Since $PQ||AB$, $N,P,A$ are collinear, and $N,Q,B$ are collinear, a homothety centered at $N$ takes $PQ$ to $AB$. The homothety also takes the midpoint of $PQ$ to the midpoint of $AB$, which is $X$. It follows that $M$ is the midpoint of $PQ$. Combining this with $EM\perp PQ$ tells us that $\triangle{EPQ}$ is isosceles, and thus, $EP=EQ$. // [/hide]

Let $R=\overline{MN} \cap \overline{AB}$. Since $R$ lies on the radical axis of the two circles, we have that the power of $R$ with respect to both circles is equal. Thus, $\overline{MN}$ bisects $\overline{AB}$ and $\overline{PQ}$. We can now angle chase to get $\angle MAB = \angle PCA = \angle BAE$. From this, we know $MAEB$ is a kite, so $\overline{MN} \perp \overline{PQ}$, which finishes the problem.

Extending MN to intersect AB at X, we know by radical axis power of a point XA^2=XB^2, or XA=XB. Since PQ is parallel to this, sim triangles NPQ and NAB mean that MN must be the median of NPQ, as NK is of NAB. Now, we have <MAB=<MCA=<MCE=<BAE, and analogously with angles MBA and ABE. Then By ASA AEMB is a kite, or EM perp. AB \\ PQ gives EP=EQ by corresponding sides of congruent triangles EMP and EMQ (SAS).

[hide=storage]

Let $\rho(G_1,G_2)\cap \overline{AB} = K$
$$-1=(A,B;K,P_{\infty})\overset{N}{=}(P,Q;M,P_{\infty})$$
$\Longrightarrow \overline{PM} =\overline{QM}$\\
$$\triangle{EAB} \cong  \triangle{MAB} = \binom{\measuredangle{EAB} = \measuredangle{ACM} = \measuredangle{MAB}}{\measuredangle{EBA} = \measuredangle{BDM} = \measuredangle{ABM}}$$
$\Longrightarrow \overline{EM}\perp\overline{CD}$ hence $\overline{EP}=\overline{EQ}$

[/hide]

hmm
[hide = solution]
First remark that since $AB$ is tangent to $G_1$ we have $\angle MAB = \angle ACM = \angle ECM = \angle EAB$, where the last equality follows from $CD \parallel AB$. Thus $\angle MAB = \angle EAB$ and similarily $\angle MBA = \angle EBA$, implying that $E, M$ are reflections of each other about $AB$. In particular since $AB \parallel PQ$, it suffices to now show that $MP = MQ$. But this is clear since if the homothety at $N$ sending $PQ$ to $AB$ sends $M$ to $M'$, we have $M'A^2 = M'M \cdot M'N = M'B^2$ and we're done.
[/hide]
[hide = remarks]
I haven't explored a lot, but this configuration seems rather rich for now. We have for instance $EANB$ cyclic, moreover also harmonic. In fact, $EN$ is the angle bisector of $\angle CND$ due to $NM$ and $NE$ being isogonal conjugates in $\angle ANB$.
 [img]https://i.imgur.com/U0ZWCVF.png[/img]
[/hide]

Forgot the Parallel Postulate exists :flushed:

[hide]
Let $O=AB\cap MN$; PoP gives $OA^2=OM\cdot ON=OB^2$, so $O$ is the midpoint of $\overline{AB}$. Now, a homothety centered at $N$ sends $A\mapsto P$, $B\mapsto Q$, and $O\mapsto M$, meaning $M$ is the midpoint of $\overline{PQ}$. It suffices to show $\angle EMC=90^\circ$.

Bute note that $\angle EAB=\angle AMC=\angle MAB$ and $\angle EBA=\angle BMD=\angle MBA$, which implies by AAS that $\triangle EAB\cong\triangle MAB$. Specifically, $EAMB$ is a kite with $AB\perp EM$, so $CD\perp EM$ as desired. $\square$
[/hide]

It is easy to see that $\angle EAB = \angle ECD = \angle BAM$. Thus from a similar angle chase we find $\angle EBA = \angle MBA$. Then $\triangle EAB \cong \triangle MAB$. Thus $EM \perp CD$. It then suffices to show $PM = MQ$. However this is obvious from the homothety taking $\triangle ANB \mapsto PNQ$.

[color=#00f]Claim:[/color]
It suffices to show $EM\perp AB$
[i]Proof:[/i]

Note that $NM$ is the radical axis of $G_1$ and $G_2$. Therefore, $NM$ bisects $AB$. Let this point be $T$.\\

As $PQ\parallel AB$, a homothety centered at $N$ maps $PQ$ to $AB$. Therefore, $M$ must be the midpoint of $PQ$, so we have our desired claim $\blacksquare$\\

[color=#00f]Claim:[/color]
$EA:EC=1:2$

[i]Proof:[/i]

We now angle chase getting:
\[\angle EAB=\angle ECM=\angle BAM=\angle AMC.\]
From this, we see that $\triangle{AMC}$ is isosceles. Also, by the previous angle chase and SAS, $\triangle{AMB}\cong \triangle{AEB}$.\\

Therefore:
\[AC=AM=AE,\]
as desired $\blacksquare$\\

To finish, as $\angle MAB=\angle EAB$, and $\triangle{AEM}$ is isosceles, $AB\perp EM$, as desired $\square$

1) Phantom point to get $EAMB$ kite.
2) Note by PoP and homothety: $MN$ bisects $AB$ and therefore $PQ$.
3) Since $M$ is the midpoint of $PQ$ and $EM\perp CD$ we are done.

Let $NM \cap AB=X$ since $X$ lies on the radical axis of $G_1$ and $G_2$,  $XA^2=XB^2$ which means $XA=XB$.
By similarity $QM=PM$. Since $\angle BAM = \angle MCE= \angle BAE$ and $\angle MBA = \angle EDC= \angle EBA$ quadrilateral $AEBM$ is a kite. So $\angle EMC =90^{\circ}$. So $QEP$ is isosceles finishing the problem.

Note $MN$ bisects $AB$ by radical axis. Because $AB \parallel PQ$, there exists a homothety mapping $AB$ to $PQ$ centered at $N$ which also maps the midpoint of $AB$ to $M$. Hence $M$ is the midpoint of $PQ$.

The problem reduces to proving $EM \perp CD$, or $EM \perp AB$. At this point, points $M$, $N$, $P$, and $Q$ are irrelevant. We can angle chase to get
\[\angle BAM = \angle AMC = \angle ACM = \angle EAB,\]
\[\angle ABM = \angle BMD = \angle BDM = \angle EBA.\]

Consequently, $EAMB$ is a kite, giving our desired conclusion.

First, notice that we have
$$\angle CMA = \angle BAM = \angle ACM = \angle EAB$$
and similarly
$$\angle BMD = \angle ABM = \angle BDM = \angle EBA$$
Let $X$ be the intersection of $EM$ and $AB$. Now, notice that $$CA = AM = AE \text{ and } BM = BD = EB$$
And by congruency we have that
$$\angle AXE = \angle AXM = 90^\circ$$
Thus, $\angle CME = 90^\circ$ as well using Parallel Lines. Now, notice that $MN$ bisects $AB$ and thus using the parallel lines, $MN$ bisects $PQ$ as well. Thus, $PM=MQ$. Therefore, $\triangle EPM \cong \triangle EQM$ which gives us that $EP=EQ$ as required.

Wow I never posted soln for this problem.

$\angle EAB=\angle ECD=\angle ACM=\angle BAM$ and similarly $\angle EBA = \angle ABM \implies AEBM$ is kite $\implies EM\perp AB\implies EM \perp CD$. Now $M$ is midpoint of $PQ$ due to homothety at $N$ mapping $AB\mapsto PQ$. Thus $\triangle EPQ$ is isosceles $\implies EP=EQ$. :yoda:

Let $\overline{EN} \cap \overline{AB} = K$.
$\newline$
Notice that
\[\angle{ACM} = \angle{AMC} = \angle{MAB}\]
and similarly
\[\angle{MBA} = \angle{EBA}\]
so $\triangle EBA \cong \triangle MBA$ are directly similar $\implies EBMA$ is a kite.
This implies that $EN \perp AB$. Since $K$ lies on the radical axis of $G_1$ and $G_2$, $KA = KB$ so the homothety sending $AB$ to $PQ$ also sends $K$ to $M$.
$\newline$
Since the $E$-altitude and the $E$-median of $\triangle EPQ$ coincide(they intersect $PQ$ at $M$), $\triangle EPQ$ is isosceles, specifically $EP = EQ$, which finishes. $\blacksquare$

[b]My first problem after 1 month of INMO[/b]
It is well known that $M$ is altitude. Therefore $ EM\perp CD$. Also well known that $NM$ bisects $AB$. Therfore we get $MP=MQ.$ Now we get our desired result. :D  :D

[b]Claim 1:[/b] $EM \parallel BQ$
[i]Proof.[/i] Let $EM$ intersect $G_2$ at a second point $X \neq M$. First, we have $AB \parallel CD$, and angle chasing leads to $$\angle BMD = \angle ABM = \angle BDM$$ It follows that $$\angle MXD = m\overset{\Large\frown}{BD} = \angle BQD$$ and the claim follows. $\blacksquare$

Since we have $QB \perp AB$ and $AB \parallel CD$, we obtain $EM \perp CD$.

[b]Lemma (Trapezoid Lemma):[/b] In trapezoid $ABCD$ ($AB \parallel CD$), let $N$ be the midpoint of $CD$, and $X$ be the intersection of $AD$ and $BC$. The intersection $M$ of $XN$ and $AB$ is the midpoint of $AB$.
[i]Proof. [/i] By similar triangles, we obtain $$\dfrac{AM}{DN} = \dfrac{XM}{XN} = \dfrac{MB}{NC}$$ However, $N$ is the midpoint of $CD$, so it follows that $AM=MB$, or $M$ is the midpoint of $AB$. $\blacksquare$

Returning to the original problem, let $K = MN \cap AB$. Since $K$ lies on the radical axis of $G_1$ and $G_2$, we have $\text{Pow}_{G_1} (K) = \text{Pow}_{G_2} (K)$, which implies that $KA=KB$, or $K$ is the midpoint of $AB$. Applying the lemma above on trapezoid $ABQP$, we have $M$ is the midpoint $PQ$. Thus, $\Delta EPQ$ is isoceles with $EP=EQ$, as desired. $\blacksquare$


I tried to send my USAMTS solutions last night, but today when I still checked my e-mail, there was an error message saying that I had tried to send too big a message (I think it was the diagrams).  Can I resend my solutions with the error message as evidence, or am I toast?


[hide="1998#5"]Given that $ \displaystyle A_k=\frac{k(k-1)}{2}\cos\left(\frac{k(k-1)}{2}\pi\right)$, find $ |A_{19}+A_{20}+\ldots+A_{98}|$[hide="my solution"]
let $ G_k=\frac{k(k-1)}{2}$ then
$ \{G_k,G_{k+1}: 4\mid{k}\}\equiv0\mod{4}\Leftrightarrow A_k=G_k\Longrightarrow G_k+G_{k+1}=k^2$
$ \{G_{k-2},G_{k-1}: 4\mid{k}\}\equiv1\mod{4}\Leftrightarrow A_k=-G_k\Longrightarrow G_{k-2}+G_{k-1}=-(k-2)^2$

$ k^2-(k-2)^2=4(k-1)$
$ 4\mid{k}\Longrightarrow k=4n\Leftrightarrow 4(k-1)=16n-4$
$ \sum_{n=5}^{24}16n-4=16\sum_{n=5}^{24}n-\sum_{n=5}^{24}4=80\cdot57=4560$

Therefore,
$ S=|4560-A_{18}+A_{98}|$
$ S=|4560+153-4753|$
$ \boxed{S=40}$[/hide][/hide][hide="1998#7"]Let $ n$ be the number of ordered quadruples $ (x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $ \sum_{i=1}^4x_i=98$.  Find $ \frac{n}{100}$.
[hide="my work"]$ \sum_{i=1}^4x_i\equiv 2\mod{4}$
$ x_i\not\equiv0,2\mod{4}\Leftrightarrow x_i\equiv 1,3\mod{4}\Leftrightarrow$
$ (x_1,x_2,x_3,x_4)\equiv (1,3,3,3)$ or $ (1,1,1,3)\mod{4}$

That's as far as I've gotten.  I know that the numbers must have an average of $ 24.5$ and that the maximum possible value of $ x_i=95$.  I was hoping you guys could give me a bump in the right direction
[/hide]
[/hide][hide="1994#13"]The equation $ x^{10}+(13x-1)^{10}=0$ has ten complex roots $ r_1, \bar{r_1},r_2,\bar{r_2},r_3,\bar{r_3},r_4,\bar{r_4},r_5,\bar{r_5}$ where the bar denotes complex conjugation.  Find the value of $ \frac{1}{r_1\bar{r_1}}+\frac{1}{r_2\bar{r_2}}+\frac{1}{r_3,\bar{r_3}}+\frac{1}{r_4\bar{r_4}}+\frac{1}{r_5\bar{r_5}}$.

A solution for this is fine, but a hint would be much preferred.[/hide]
 :) Thanks!

[hide="#7"]Let $ x_1 \equal{} 2a_1 \plus{} 1,x_2 \equal{} 2a_2 \plus{} 1$, and so on. Then, $ 2(a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4) \equal{} 94$, and $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \equal{} 47$, which, by balls and urns has 50C3=19600 solutions. So 196.[/hide]

[hide="#13 hint"]
Just looking at the problem would suggest to to rewrite it as
$ \minus{} x^{10} \equal{} (13x \minus{} 1)^{10}$ and then taking the tenth root of both sides, resulting in $ e^{(\frac {\pi i}{10}\plus{}\frac{2k\pi}{10})}x \equal{} 13x \minus{} 1$.  Perhaps this makes the problem too easy, but whatever.  I think the answer is
[hide]850[/hide]
[/hide]


has anyone taken the amc12A yet?

i didn't. i took amc 10A. i got 150 (I guessed two) i was lucky...

what about you?

Wow, good job.

I took the 12A and totally bombed it... :wallbash:  :wallbash:

Wow, you're really smart!

I'm going to take the 12B and I'm probably going to fail it because I suck at math.


geometry
circumcircle
geometric transformation
rotation
angle bisector
geometry proposed

in triangle ABC, AD is angle    bisector of $ { \angle ABC}$  , $ {AQ}$ ⊥$ {PB}$,  $ {CP}$⊥ $ {PB}$ ;M is midpoint of segment AC;  $ {O}$ is circumcenter of triangle QMP . and circle $ {O}$ insect segment AC at point $ {H}$ , E is midpoint of segment $ {BC}$ . prove that points $ {O,H,M,E}$  are concyclic.

Dear Mathlinkers,
see for example: http://perso.orange.fr/jl.ayme/   vol. 4  Symetrique de OI p. 6-7
Sincerely
Jean-Louis

Take $ A'$ on $ (BC$ so that $ BA \equal{} BA'$, and $ C'$ on $ (BA$ so that $ BC' \equal{} BC$, then it's obvious that $ QM$ is midline in $ \triangle AA'C$, hence $ QM \equal{} \frac{A'C}{2}$, similarly for $ MP$. Consequently $ P, M$ and $ E$ are collinear. If $ F$ is midpoint of $ AB$, $ M, Q$ and $ F$ are collinear as well.
Since $ PE\equal{}\frac{BC'}{2}\equal{}\frac{BC}{2}\equal{}FM$, we get that a rotation of angle $ \measuredangle EMF \equal{} \measuredangle ABC$ about $ O$ will map $ P$ to $ M$ and $ E$ to $ F$, so $ \angle OEM \equal{} \angle OFM$ and $ O$ belongs to the [b]nine-point circle[/b] of $ \triangle ABC$.
From $ MP \parallel AB$ we get $ \angle MPQ \equal{} \angle QBA$ ( 1 ), but from $ PMQH$ cyclic $ \angle MPQ \equal{} \angle MHQ$ ( 2 ); from (1) and (2) $ AHQB$ is cyclic. As $ AQ \perp BD$ we get $ BH \perp AD$, hence $ H$ belongs to the same circle, done.

Best regards,
sunken rock


geometry
3D geometry
articles
complex numbers

[b]Can someone explain to me what nth roots of unity are? [/b] or, if you're feeling lazy, post links where tutorials are?

Because I tried looking on wikipedia (and wolfram) and I got really, really confused.
For example, on wikipedia...
[i] An nth root of unity, where n = 1,2,3,···, is a complex number, z, satisfying the equation

   $ z^n \equal{} 1 \, .$

Second roots are called square roots, and third roots are called cube roots.

An nth root of unity is primitive if

   $ z^k \ne 1 \qquad (k \equal{} 1, 2, 3, \dots, n \minus{} 1 ) \, .$

There are n different nth roots of unity:

$ z^k \qquad (k \equal{} 1, 2, 3, \dots, n ) \, ,$

where z is any primitive nth root of unity. These n roots are distributed evenly over the unit circle as can be seen in the plot on the right-hand side of the three 3rd roots of unity. The primitive nth roots of unity are those zk where k and n are coprime.

The notion of root of unity also applies to any algebraic ring with 1, namely a root of unity is any element of finite multiplicative order. [/i]

I am confused about how if $ z^k$ is a primitive nth root of unity, then according to wiki, there are n solutions to $ z^k \equal{} 1$ and the solutions are $ (1, 2, 3, ... n)$ But by the definition of a [i]primitive[/i] nth root of unity, $ (1, 2, 3, ..., n \minus{} 1)$ where k is an element of that set, $ z^k \ne 1$

So that is a contradiction, right?

CAN SOMEONE PLEASE HELP!!!!!

No. First, n-th root of unity is a number $ z$ such that $ z^n\equal{}1$. There are $ n$ such numbers.

It's primitive if $ z^k\neq,k\in\{1,2,\ldots,n\minus{}1\}$. Ok. That means that $ z$ isn't a k-th root of unity for given k.

And the solutions aren't 1,2,...,n. I think $ 2^n\neq 1$ for $ n>0$... Solutions are complex numbers...

[quote="theax"]Thanks :D

Anyone else?[/quote]

What do you need more? Ask now.

If you insist...

[quote]
No. First, n-th root of unity is a number z such that z^n=1. There are n such numbers.

It's primitive if z^k\neq 1,k\in\{1,2,\ldots,n-1\}. Ok. That means that z isn't a k-th root of unity for given k. [/quote]

Wiki says, "An nth root of unity is primitive if

    $ z^k \ne 1 \qquad (k \equal{} 1, 2, 3, \dots, n\minus{}1 ) \, .$"

However, z is still an nth root of unity, it's just primitive, right?

Then wiki says, "There are n different nth roots of unity:

  $ z^k \qquad (k \equal{} 1, 2, 3, \dots, n ) \, ,$

where z is any primitive nth root of unity."

Then we get that $ z^k \equal{} 1 \qquad (k \equal{} 1, 2, 3, \dots, n) \, ,$

But we were just given that   $ z^k \ne 1 \qquad (k \equal{} 1, 2, 3, \dots, n\minus{}1 ) \, .$"
by the definition of a primitive root.

Contradiction, right?

Can anyone tell me where exactly in this "proof" I went wrong?

Thanks!

Oh, and another thing...If you want to learn how to use it and apply it, you have to learn about complex numbers and their polar forms, and DeMoivre's Theorem.

In your line just after the quote: You have $ z^n \equal{} 1$, not $ z^k \equal{} 1$ for all $ k$ less than $ n$.

[quote="theax"]If you insist...
Wiki says, "An nth root of unity is primitive if

    $ z^k \ne 1 \qquad (k \equal{} 1, 2, 3, \dots, n \minus{} 1 ) \, .$"

However, z is still an nth root of unity, it's just primitive, right?
[/quote]
Right.
[quote="theax"]
Then wiki says, "There are n different nth roots of unity:

  $ z^k \qquad (k \equal{} 1, 2, 3, \dots, n ) \, ,$

where z is any primitive nth root of unity."

Then we get that $ z^k \equal{} 1 \qquad (k \equal{} 1, 2, 3, \dots, n) \, ,$

But we were just given that   $ z^k \ne 1 \qquad (k \equal{} 1, 2, 3, \dots, n \minus{} 1 ) \, .$"
by the definition of a primitive root.
[/quote]
Right.
[quote="theax"]

Contradiction, right?

Can anyone tell me where exactly in this "proof" I went wrong?

Thanks![/quote]
No. There is no contradiction.  Next time, can you be more explicit about what the contradiction is?  As someone who knows there is no contradiction, it is difficult for me to see where you might believe there is one and, therefore, difficult to help you out.  You can explain more if what follows doesn't address your question.

The first section is a definition that tells you what a primitive root of unity is.  For all values of $ k$ between $ 1$ and $ n\minus{}1$, if $ z^k\ne 1$, then we say $ z$ is a primitive root.
The second section gives a list of the $ n$th roots of unity.  For some primitive $ n$th root of unity $ z$, the $ n$ numbers $ z^1, z^2, z^3,\ldots ,z^n$ are the $ n$th roots of unity.  In other words, for all values of $ k$ between $ 1$ and $ n$, $ z^k$ is an $ n$th root of unity.  Nothing about $ z$'s definition as a primitive root contradicts that, for example, $ z^2$ is a solution of the equation $ x^n\equal{}1$.

Maybe I'm not getting something here...

$ z^k \ne 1$ AND $ z^n \equal{} 1$ ???

k is one of the numbers from 1 to n-1, not n!!!
So z is primitive n-th root if $ z^1\neq1,z^2\neq1,z^3\neq1,\ldots,z^{n\minus{}1}\neq1,z^n\equal{}1$.

Suppose we have a primitive $ n^{th}$ root of unity, $ z$ (that is, $ z$ such that the equation $ z^m \equal{} 1$ is only satisfied when $ m$ is a nonnegative multiple of $ n$, like $ 0$, $ n$, $ 2n$, and so on). What the article states is that the other roots of unity for the exponent $ n$ are necessarily the powers of $ z$ up to $ z^n$, for $ (z^k)^n \equal{} (z^n)^k \equal{} 1$, and they are all different complex numbers.


[color=red]Καλή αρχή και καλή επιτυχία στην όμορφη αυτή προσπάθεια που ξεκινάμε !!! [/color]

[b]α) [/b]Στο χώρο αυτό θα μαζέψουμε [b]15 ασκήσεις [/b]γεωμετρίας.
[b]β)[/b] Όλες θα αναφέρονται σε ύλη που λέει ο τίτλος : Μέχρι και εγγράψιμα.Όχι ομοιότητα ή μετρική γεωμετρία ή ειδικά δύσκολα θεωρήματα.Προτιμάμε αποδεικτικές ασκήσεις. Όχι τόπους ή κατασκευές , εκτός και αν έχουν κάτι το εξαιρετικό.Ανέχουμε χώρο , ίσως τις βάλουμε σε άλλη υποενότητα αυτές.
[b]γ)[/b] Αυτός που θα προτείνει μια άσκηση  , θα τη βάζει με τη λύση της ή με τις λύσεις της , αν γνωρίζει περισσότερες.
[b]δ) [/b]Γράψτε τις λύσεις με απλότητα , σαφήνεια και με τα απαραίτητα σχόλια , όπου νομίζετε.Ας μην πλατειάζουμε όμως υπερβολικά για προφανή πράγματα. Αν κάτι είναι λίγο σύντομα γραμμένο , ας το αλλάξει αυτός που θα δυσκολευτεί.
[b]ε) [/b]Θα έχετε τη λύση καθαρογραμμένη με [color=red]ΜΟΛΥΒΙ[/color]  . Μία άσκηση σε κάθε σελίδα.Μόλις τελειώσουν οι παρατηρήσεις , θα ολοκληρώνετε το γράψιμο και θα την κρατάτε στο αρχείο που θα φτιάξετε.
Θα δούμε στη συνέχεια πότε θα μου τις στείλλετε και πώς.
[b]στ) [/b]Τις βοηθητικές γραμμές του σχήματος θα τις βάζετε με κόκκινο.
[b]ζ)[/b] Δε χρειάζεται να πληκτρολογείτε στο LATEX , διότι δεν παίρνει ελληνικά, εκτός και αν κάποιος ξέρει να μας πει πώς βάζουμε ελληνικά.Τα κείμενα θα ξαναπληκτρολογηθούν έτσι και αλλιώς.
[b] η)[/b] Όποιος  μπορεί να φτιάχνει σχήματα σε κάποιο πρόγραμμα , είτε αυτός που προτείνει , είτε κάποιος που διαβάζει την άσκηση , παρακαλείται να το κάνει, για να έχουμε και εποπτεία.Ας είναι και λίγο πρόχειρο.
  [b]θ)[/b] Κάθε άσκηση είναι καλό να την περνάτε σε word ,  να την αποθηκεύετε όλοι σε σχετικό νέο φάκελο , να την τυπώνετε  , να μελετάτε τη λύση και να στέλνετε τις παρατηρήσεις σας . Ακόμα και το : '' την κοίταξα και είναι εντάξει '' είναι σημαντικό.
 [b] ι) [/b]Πρέπει να παρακολουθήσουμε όλοι μας όλες τις ασκήσεις και να διαβάσουμε προσεκτικά τη λύση τους. Δεν έχει νόημα να έχουμε γράψει ένα βιβλίο , να έχουμε μέσα δυο ασκήσεις μας και να μην ξέρουμε τι έχει το βιβλίο ! Όλοι είμαστε συνυπεύθυνοι για το περιεχόμενο.Και θα είμαστε χαρούμενοι να γνωρίζουμε τις ασκήσεις!
   Αυτά για αρχή !   θα μαζέψουμε και άλλες παρατηρήσεις και θα τις βάλουμε και αλλού. !

[i]Μπορείτε ακόμα να χρησιμοποιείτε για ευκολία ,αν θέλετε , και τα ψευδώνυμα. Τα κανονικά ονόματα θα τα φτιάξουμε έτσι και αλλιώς.Θα πρότεινα όμως σιγά σιγά να υπογράφετε με το πραγματικό σας όνομα και για λόγους αρχειακούς.[/i]

 [color=blue]ΚΑΛΗ ΔΥΝΑΜΗ ΚΑΙ ΚΑΛΕΣ ΕΜΠΝΕΥΣΕΙΣ  ! !! [/color]

Α! Να τονίσω ότι οι ασκήσεις θα αριθμούνται : [b]ΑΣΚΗΣΗ 1[/b] , κλπ

[color=blue][b]ΚΑΛΗ ΑΡΧΗ!!![/b][/color]

[color=red][size=150][b]ΑΣΚΗΣΗ 1[/b][/size][/color]

Βρίσκεται στο συνημμένο pdf . Έχει περαστεί σε $\LaTeX$.

 :D

[quote="stedes"]Καλή αρχή λοιπόν!  :lol:

[color=darkblue][b]ΑΣΚΗΣΗ 1[/b]-[i]διόρθωση[/i][/color]

Εγώ εντόπισα τρία τυπογραφικά λαθάκια στη λύση σου.
1)Στη δεύτερη σειρά (...της λύσης) το $CDZ$ μάλλον είναι $DCZ$,
2)στην 7η σειρά το $BIC=30$ μάλλον είναι $IBC=30$ και
3)στη 10η σειρά το $CI=AM$ μάλλον είναι $CI=MC$.
Αυτά! Λεπτομέρειες δηλαδή,αλλά θεώρησα καλό να αναφερθούν.  : )

Αν θες διόρθωσέ τα γιατί εγώ δεν μπορώ να επεξεργαστώ αυτή τη στιγμή pdf.

Μια σκέψη: Μήπως θα ήταν καλύτερα οι λύσεις να γράφονται κατευθείαν εδώ στο forum και όχι σε συνημμένο αρχείο;[/quote]

[b]Μπράβο ! [/b]

 [b]α) [/b]Αυτό είναι ! Να γίνεται ο έλεγχος αμέσως και να περνιούνται οι διορθώσεις από τον προτείνοντα.Αν βρεθούν και άλλες λύσεις, να σταλούν.
 Επίσης να στέλνετε και τις [color=red]τριγωνομετρικές λύσεις ή λύσεις με αναλυτική ή με διανύσματα ή μιγαδικούς[/color]. Όλα αυτά είναι γεωμετρία !!! Δε θα έχουμε ενότητες με αυτό το αντικείμενο , οπότε επιβάλλεται να παρουσιάσουμε μέσα στο κεφάλαιο της Γεωμετρίας και την αποδεικτική  δύναμη των  άλλων μεθόδων. Άλλωστε , αυτό που μετράει στους διαγωνισμούς είναι να βρεις λύση και όχι την ομορφότερη λύση. Φυσικά η καθαρά γεωμετρική λύση θα γράφεται πρώτη.[b]
 β)  [/b]Τώρα , τις λύσεις ή και τις εκφωνήσειες μπορείτε να τις στέλνετε και σε συνημμένα αρχεία , αν αυτό σας βολεύει. Φυσικά, όπως έγραψε κι ο stedes , βολεύει περισσότερο να φαίνεται απευθείας η άσκηση , αλλά είναι και κρίμα να πληκτρολογείτε δύο φορές. Πράξτε με τον καλύτερο για σας τρόπο.
 [b]
 γ) [/b]Υπενθυμίζω ότι ο καθένας μας πρέπει να φτιάχνει το [color=blue]δικό του αρχείο με όλες  τις ασκήσεις [/color], και αυτές που στέλνει και αυτές που στέλνουν οι άλλοι , ώστε να περνάει και τυχόν άλλες πρατηρήσεις ![b]
  δ) [/b]Αν βάζετε ασκήσεις που υπάρχουν ήδη στο  ελληνικό φορουμ , να βάζετε πάνω πάνω  και το [b]link [/b], ώστε να μπορούμε να ελέγχουμε  όλη τη σχετική αλληλογραφία.

      Μπάμπης

Thanks Stedes for pointing me out the errors  :D

Η άσκηση έχει διορθωθεί. Απλώς, για περισσότερους τρόπους λύσης δεν την έχω ψάξει.. έβαλα αυτόν που είχα στο αρχείο μου από παλαιότερα! Αν έχει κάποιος να προτείνει άλλον τρόπο λύσης, πολύ ευχαρίστως, όλες είναι ευπρόσδεκτες  :)

[size=150][color=red][b]ΑΣΚΗΣΗ 2[/b][/color][/size]

Είναι η άσκηση του Ευκλείδη που "έπεσε" φέτος, στη Β' Λύκειου ! Βρίσκεται στο συνημμένο αρχείο pdf . Έχει περαστεί σε $\LaTeX$ .

[u]Σημείωση[/u]: Δε θα δώσω άλλες ασκήσεις προς το παρόν, μεχρί να προταθούν και από άλλους μερικές !

gia ti 2 http://www.mathlinks.ro/Forum/viewtopic.php?t=129369

Καλή αρχή !!!!! Αν μπορώ θα βοηθήσω και εγώ..

[color=red]ΜΙΑ ΑΛΛΗ ΛΥΣΗ ΣΤΗ 2[/color] (δυστυχώς δεν μπορώ τώρα να βάλλω σχήμα  :( ) (Α περιέχει ομοιότητα και λίγο μετρικά αλλά αφού θα μπει ως λύση 2 δεν μας πειράζει έτσι δεν είναι ?)

Είναι $\angle{BDA}=135$ αφού βαίνει σε τόξο 270 μοιρών .
άρα $\angle{ADK}=45$ και $AK=45$ άρα $KD=KA$ (1) .
Τα τρίγωνα $CBD$ και $AB$ είναι όμοια άρα $\frac{CD}{CA}=\frac{BD}{AD}$  ή $\frac{CD}{2R\sqrt{2}}=\frac{BD}{AD}$ (2)

Από το θεώρμα της δύναμης σημείου ως προς κύκο έχουμε
$CD^{2}=CB\cdot CA=4R^{2}$  άρα $CD=2R$ και η (2) δίνει
$AD=BD\sqrt{2}$ (3)
Από τις (1) ,(3) έχουμε $DA=DK\sqrt{2}$ και $DA=DB\sqrt{2}$
άρα $DK=DB$  και πάλι με τη βοήθεια της (1) $KB=2KD=2KA$

Κοιτάξτε αν υπάρχουν λάθη και πείτε μου.

Βάζω μια από τισ πιο διδακτικές και πιο αγαπημένες μου ασκήσεις γεωμετρίας σε αυτή τη θεωρία .
[size=150][color=red][b]ΑΣΚΗΣΗ 3[/b][/color][/size]
Έστω $M$  το μέσον της πλευράς $AB$ ενός εγγεγραμμένου τετραπλεύρου $ABCD$ και $P$ το σημείο τομής της $MC$ με τη $BD$. Από το σημείο $C$ φέρουμε παράλληλη προς την $AP$ που τέμνει τη $BD$ στο $S$ .
Αν είναι $\angle{CAD}=\angle{PAB}=\frac{\angle{BMC}}{2}$ να αποδείξετε ότι $BP=SD$

TST for JBMO 2005 Greece
[color=red]ΛΥΣΗ[/color]
Έστω $\angle{PAB}=x$ τότε $\angle{BMC}=2x$ και άρα
$\angle{MPA}=x$ επίσης άρα $MP=MA=MB$ άρα ο κύκλος με διάμετρο την $AB$ περνάει από το $P$ άρα $\angle{APB}=90$(1)
Τώρα είναι $\angle{DAC}=\angle{DBC}=x$ και λόγω της (1)
$\angle{ABP}=90-x$ άρα $\angle{ABC}=90$ άρα $AC$ διάμετρος και άρα $\angle{ADC}=90$ επίσης .
Αφού από την υπόθεση $CS//AP$ θα είναι $\angle{CSD}=90$
Έστω $K$ το σημείο τομής της $CS$ με τον κύκλο .
Τότε $\angle{KCD}=90-\angle{SDC}=\angle{ADB}$
άρα τα τόξα $AB$ και $KD$ είναι ίσα άρα $AB=KD$ (2) και επιπλέον $AK//BD$ (οι ΑΚ και BD αποκόπτουν ίσα τόξα) άρα $KS=AP$ (3) ως ύψη του τραπεζίου $AKDB$ .

Από τις (2) και (3) έχουμε ότι τα τρίγωνα $KDS$ και $APB$ είναι ίσα αφού είναι και ορθογώνια .
Συνεπώς $DS=BP$ , το ζητούμενο

[color=red]ΔΙΟΡΘΩΣΕΙΣ:[/color]

[b]Άσκηση 2[/b]

[color=darkblue]Λύση mostel:[/color]
10η γραμμή από τέλος, αντί για γωνία KOD=90, [u]γωνία KOC=90[/u]

[color=darkblue]Λύση Nick Rapanos:[/color]
Στο πρώτο post αντί για γωνία ODB=90, [u]γωνία ΟDC=90[/u]  [size=75](έχει ήδη επισημανθεί)[/size]
Θα ήταν καλό νομίζω η λύση να γραφεί με κάποιες ακόμα λεπτομέρειες.  :oops:
[color=darkblue]
Λύση silouan:[/color]
Γραμμή 2: αντί για ΑΚ=45, [u]KAD=45[/u]
Γραμμή 3: αντί για ΑΒ μάλλον ήθελες να γράψεις CAD (από τις πλευρές που χρησιμοποιείς το συμπέρανα).
Δε θα ήταν καλό να αιτιολογηθεί και η ομοιότητα;

Από τις 3 αυτές λύσεις, μόνο του Nick Rapanos δεν περιέχει ομοιότητα.

Prokeitai gia mia poli omorfi askisi gewmetrias apo tin Diethni Olimpiada Mathimatikwn (IMO) to 1997. H lisi den iparxei sto IMO Compendium giati to provlima (opws tha deite stin Shortlist itan ligo diaforetiko kai to parapoiisan gia na ftiaksoun auto pou parathetw parakatw kai to opoio mpike stin IMO. H lisi pou dinetai sto IMO Compendium einai diaforetiki apo auti pou dinw sto episinaptomeno arxeio. Isws loipon na mporousame na valoume kai tis 2 liseis ('h an iparxei kai triti lisi apo esas).

[size=150][color=red]ASKHSH 4 [/color][/size]

H gwnia $\angle{A}$ einai h mikroterh sto trigwno $\stackrel{\triangle}{ABC}$. Ta shmeia $B$ kai $C$ xwrizoyn ton perigegrammeno kyklo se dyo toksa. Estw $U$ ena eswteriko shmeio twn tokswn metaksy $B$ kai $C$ poy den periexei to $A$. Oi mesokathetes twn $AB$ kai $AC$ temnoyn thn eytheia $AU$ sta $V$ kai $W$ antistoixa. Oi eytheies $BV$ kai $CW$ temnontai sto $T$. Na deiksete oti
\[AU=TB+TC. \]
Alexandros

Παραθέτω μια διδακτική, κατά τη γνώμη μου, άσκηση η οποία συνδιάζει βασικές γνώσεις στα ορθογώνια τρίγωνα και στον κύκλο. Επειδή όμως περιέχει προς στο τέλος της λύσης ομοιότητα τριγώνων, αν κριθεί απαραίτητο μπορούμε να την μεταφέρουμε στο topic Γεωμετρία 2 (...αν και δε νομίζω ότι χρειάζεται), εκτός βέβαια αν βρεθεί μια λύση χωρίς ομοιότητα.

[color=darkblue][b][size=150]Άσκηση 5[/size][/b][/color]
Έστω κύκλος με διάμετρο $AB$. Φέρνουμε την εφαπτομένη του κύκλου στο $B$ και έστω $G$ ένα τυχαίο σημείο πάνω σε αυτήν. Φέρνουμε την εφαπτομένη του κύκλου από το $G$ που δεν περνάει από το $B$ και έστω ότι αυτή εφάπτεται του κύκλου στο $C$. Φέρνουμε την κάθετη από το $C$ στην $AB$ και έστω ότι τέμνει την $AB$ στο $D$. Αν $E$ το σημείο τομής της $AG$ με την $CD$, να δειχθεί ότι $CE=ED$.

[color=darkblue][u]Λύση[/u][/color]
Έστω $F$ το σημείο τομής της $AC$ με την $BG$. Η γωνία $\angle{ACB}$ είναι ορθή (...επειδή βαίνει σε ημικύκλιο), συνεπώς το $BC$ είναι ύψος στο $\stackrel{\triangle}{ABF}$.
Ισχύει $GC=GB$ $(1)$ (...αφού είναι εφαπτομένες στον κύκλο από το ίδιο σημείο) συνεπώς το $\stackrel{\triangle}{BGC}$ είναι ισοσκελές άρα $\angle{CBG}=\angle{BCG}\Leftrightarrow 90-\angle{CFB}=90-\angle{FCG}\Leftrightarrow$

$\Leftrightarrow \angle{CFG}=\angle{FCG}\Leftrightarrow GC=GF$ και λόγω της $(1)$ θα είναι $GF=GB$ $(2)$.

H $GB$ είναι κάθετη στην  $AB$ άρα και παράλληλη στην $CD$, συνεπώς τα τρίγωνα $ADE$,$ABG$ είναι όμοια, άρα ισχύει $\frac{DE}{BG}=\frac{AE}{AG}$ $(3)$.
'Ομοια για τα τρίγωνα $AEC$ και $AGF$ ισχύει $\frac{EC}{GF}=\frac{AE}{AG}$ $(4)$.

Από $(3)$ και $(4)$ προκύπτει $\frac{DE}{BG}=\frac{EC}{GF}$ και λόγω της $(2)$, $DE=EC$.

Στέλιος

Vickil ευχαριστώ για τη διόρθωση αλλά δεν μπορώ να αλλάξω το ποστ γιατί μου λέει ότι θέλει άδεια moderator  :mad:  :(  .

εχω 2 εδω
ευκολα μεν σημαντικα κατα τη γνωμη μου δε..(δεν ειμαι σιγουρος για τις πηγες τους)

1/κυρτο τετρπλευρο ΑΒΓΔ μεσα Μ και Ν των ΑΒ και ΓΔ αντιστοιχα
Κ,Λ,Ζ και Ε μεσα των ΜΓ,ΜΔ,ΝΑ και ΝΒ αντιστοιχα να αποδειχθει οτι  ΚΛΖΕ παραλληλογραμμο

2/τυχαιο τριγωνο ΑΒΓ.Ε και Ζ οι προβολες της Α πανω στις διχοτομους των εξωτερικων γωνιων Β και Γ.να αποδειχθει οτι ΕΖ παραλληλη στην ΒΓ

ελπιζω να τις εκτιμισετε :)
μπορει σας παρακλω καποιος να δωσει τις λυσεις γιατι δεν εχω σκανερ...αν θελετε τις περιγραφω αλλα νομιζω οτι γι εσας θα ειναι παιχνιδακια

Δίνεται κύκλος κέντρου Ο και μια χορδή του ΑΒ. Θεωρούμε δύο χορδές ΔΕ και ΖΗ που διέρχονται από το μέσο Γ της ΑΒ με τα σημεία Ζ και Ε να βρίσκονται στο ίδιο τόξο σε σχέση με την ΑΒ. Αν οι χορδές ΔΖ και ΕΗ τέμνουν την ΑΒ (εκτός από το Γ) και στα Μ, Ν αντίστοιχα, να δειχθεί ότι ΑΜ=ΝΒ.

(Η άσκηση αυτή είναι η απόδειξη του θεωρήματος της πεταλούδας)

Λύση:

Θα δείξουμε ότι το Γ είναι το μέσο του ΜΝ, οπότε προφανώς προκύπτει και το ζητούμενο.

Τα τρίγωνα ΖΓΔ και ΕΓΗ είναι όμοια επειδή <ΔΖΗ=<ΗΕΔ (βαίνουν στο ίδιο τόξο) και <ΖΔΕ=<ΕΗΖ (ομοίως).

Άρα ΖΔ/ΖΓ = ΕΗ/ΕΓ. (1)

Θεωρούμε τα αποστήματα ΟΚ και ΟΛ προς τις ΖΔ και ΕΗ αντίστοιχα.

Η (1) γράφεται λοιπόν ΖΚ/ΖΓ = ΕΛ/ΕΓ.

Άρα τα τρίγωνα ΖΓΚ και ΕΓΛ είναι όμοια οπότε <ΖΚΓ=<ΕΛΓ. (2)

Τα τετράπλευρα ΚΜΓΟ και ΛΝΓΟ είναι εγγράψιμα επειδή έχουν δύο απέναντι γωνίες παραπληρωματικές.

Συνεπώς, <ΜΟΓ = <ΓΚΜ και <ΓΟΝ = <ΝΛΓ.

Όμως από τη (2) προκύπτει ότι τα δεύτερα μέλη είναι ίσα άρα και τα πρώτα, άρα <ΜΟΓ = <ΓΟΝ.

Τελικά ΟΓ ταυτόχρονα ύψος και διχοτόμος στο τρίγωνο ΜΟΝ οπότε είναι και διάμεσος και η απόδειξη ολοκληρώθηκε.

[quote="Protonios"]

[size=150][color=red]ΠΡΟΤΑΣΗ ( ΘΕΩΡΗΜΑ ΠΕΤΑΛΟΥΔΑΣ )[/color][/size]

[b]Δίνεται κύκλος κέντρου Ο και μια χορδή του ΑΒ. Θεωρούμε δύο χορδές ΔΕ και ΖΗ που διέρχονται από το μέσο Γ της ΑΒ με τα σημεία Ζ και Ε να βρίσκονται στο ίδιο τόξο σε σχέση με την ΑΒ. Αν οι χορδές ΔΖ και ΕΗ τέμνουν την ΑΒ (εκτός από το Γ) και στα Μ, Ν αντίστοιχα, να δειχθεί ότι ΑΜ=ΝΒ.[/b]

[color=red]Λύση:[/color]

Θα αποδείξουμε ότι το Γ είναι το μέσο του ΜΝ, οπότε προφανώς προκύπτει και το ζητούμενο.

Τα τρίγωνα ΖΓΔ και ΕΓΗ είναι όμοια επειδή <ΔΖΗ=<ΗΕΔ (βαίνουν στο ίδιο τόξο) και <ΖΔΕ=<ΕΗΖ (ομοίως).

Άρα ΖΔ/ΖΓ = ΕΗ/ΕΓ. (1)

Θεωρούμε τα αποστήματα ΟΚ και ΟΛ προς τις ΖΔ και ΕΗ αντίστοιχα.

Η (1) γράφεται λοιπόν ΖΚ/ΖΓ = ΕΛ/ΕΓ.

Άρα τα τρίγωνα ΖΓΚ και ΕΓΛ είναι όμοια ,  οπότε <ΖΚΓ=<ΕΛΓ. (2)

Τα τετράπλευρα ΚΜΓΟ και ΛΝΓΟ είναι εγγράψιμα επειδή έχουν δύο απέναντι γωνίες παραπληρωματικές.

Συνεπώς :  <ΜΟΓ = <ΓΚΜ και <ΓΟΝ = <ΝΛΓ.

Όμως από τη (2) προκύπτει ότι τα δεύτερα μέλη είναι ίσα άρα και τα πρώτα, οπότε   <ΜΟΓ = <ΓΟΝ.

Τελικά το ΟΓ είναι ταυτόχρονα ύψος και διχοτόμος στο τρίγωνο ΜΟΝ , οπότε είναι και διάμεσος και η απόδειξη ολοκληρώθηκε. c@Protonios[/quote]

Εϊναι [color=red]καλό συμπέρασμα και θα το βάλουμε[/color].Αξίζει να έχουμε και την απόδειξη. Καλά έκανες και την έστειλες.

[b]ΑΣΚΗΣΗ[/b]

Δίνεται ορθογώνιο τρίγωνο $ABC$, με $\hat A=90^\circ$ και έστω $A', B', C'$ τα μέσα των πλευρών $BC, CA, AB$ αντίστοιχα. Πάνω στην πλευρά $AB$ παίρνουμε σημείο $D$, τέτοιο ώστε $AD=\frac{AB}{4}$. Έστω $E$ και $Z$ οι προβολές του $D$ στην $BC$ και στην $AA'$. Να αποδειχθεί ότι $CC'\perp EZ$.

[b]ΛΥΣΗ[/b]

Έστω $P$ η τομή της $A'B'$ με την $CC'$. Το $P$ προφανώς είναι το μέσο της $A'B'$, οπότε, επειδή και το $D$ είναι το μέσο της $AC'$, γιατί $AD=\frac{AB}{4}$, η $PD$ θα είναι παράλληλη της $AC$ και $DP=\frac{AC}{2}$. Εξάλλου και $A'C'=\frac{AC}{2}$ και $A'C'\parallel AC$, άρα το $DPA'C'$ είναι οθρογώνιο παραλληλόγραμμο. Επομένως αυτό είναι εγγράψιμο σε κύκλο διαμέτρου $A'D$ και στον οποίο προφανώς βρίσκονται και τα σημεία $E$ και $Z$ γιατί $\hat E=\hat Z=90^\circ$. Εξάλλου είναι προφανές από το σχήμα ότι $\hat x_{1}=\hat C$ ως συμπληρώματα της $\hat B$ και ακόμη $\hat x_{2}=\hat x_{3}=\hat C\implies \hat x_{1}=\hat x_{2}$. Δηλαδή η $DP$ είναι η διχοτόμος της $\widehat{EDZ}$. Άρα $P$ το μέσο του τόξου $EPZ$ και επομένως η διάμετρος $PC'$, επειδή περνά από το μέσο $P$, θα είναι μεσοκάθετη της $EZ$.

[u]Σημείωση[/u]: Μερικές προφανείς αποδείξεις τις παρέλειψα λόγω έλλειψης χρόνου. Όποιος δεν καταλαβαίνει κάτι ας μου στείλει pm ή ας ρωτήσει εδώ.

[b]ΑΣΚΗΣΗ[/b]

Αν ένα πολύγωνο έχει για κορυφές τα σημεία επαφής των πλευρών ενός κανονικού πολυγώνου περιγεγραμμένου σε περιφέρεια και τυχόντος πλήθους πλευρών, το γινόμενο των αποστάσεων τυχόντος σημείου της περιφέρειας από τις πλευρές του εγγεγραμμένου πολυγώνου είναι ίσο με το γινόμενο των αποστάσεων του ίδιου σημείου από τις πλευρές του περιγεγραμμένου κύκλου.

[b]ΛΥΣΗ[/b]

Βρίσκεται μαζί με την εκφώνηση στο .pdf

[size=150][color=red]ΑΣΚΗΣΗ 21[/color][/size]

 Καλή και όχι τόσο γνωστή.

[quote="pontios"][size=150][color=red]ΑΣΚΗΣΗ 20[/color][/size]
Δίνεται ισοσκελές τρίγωνο $ABC$ με $\hat B = \hat C = 80^{o}$
Επί της $AB$ παίρνουμε σημείο $D$ έτσι ώστε $AD=BC.$
Να αποδειχθεί ότι $\angle{ACD}=10^\circ$

[size=150][color=red]ΛΥΣΗ[/color][/size]
Κατασκευάζουμε το ισόπλευρο τρίγωνο $AEC$, έτσι ώστε τα $E,D$ να βρίσκονται σε διαφορετικό ημιεπίπεδο της $AC$.

Είναι $\widehat{EAD}= \widehat{EAC}+\widehat{CAD}= 60^\circ+20^\circ = 80^\circ$

Τα τρίγωνα $EAD,ABC$ έχουν $EA=AB,AD=BC$ και $\widehat{EAD}=\widehat{ABC}=80^\circ$ άρα είναι ίσα [b](Π-Γ-Π)[/b]
Επομένως $EA=ED$ και $\widehat{AED}=20^\circ$.

Τα σημεία $A,D,C$ ανήκουν στον κύκλο $(E,EA)$, άρα $\widehat{ACD}=\frac{1}{2}\widehat{AED}=10^\circ$[/quote]

 Λουκά ,
 [color=red]έκανες πολύ καλά [/color]και έβαλες αυτό το κλασσικό και όμορφο θέμα.Ρίξε μια ματιά στις σημειώσεις σου και βρες και την τριγωνομετρική λύση. Θα το δω και εγώ κάποια στιγμή.Τώρα βλέπω ότι η λύση σου - πολύ καλή - είναι διαφορετική από αυτή που συνηθίζω και κάνω . Θα τις βάλουμε όλες τις λύσεις ,  γιατί η άσκηση αυτή κάνει πολλούς μαθητές να αγαπήσουν τη γεωμετρία  !!!
[color=blue] Καλή Ανάσταση ! [/color]

[color=blue]Σιλουανέ ,
 κάνε σε παρακαλώ ένα quote και γράψε τη λύση στα ελληνικά , ώστε να την έχουμε έτοιμη.[/color]

[size=150][color=red]ΑΣΚΗΣΗ 21[/color][/size]

[quote="stergiu"]In a  triangle $ABC$ segment $AD$ is  median, $E$ is a point on $AD$ and $EF$ is perpendicular to $BC$. Let $M$ be a point on $EF$ such that the line joining the feet of perpendiculars drawn from $M$  to sides $AC$  and  $AB$  passes through  point $E$.  Prove that $AM$ bisects angle $A$ .

  [/quote]

[b]SOLUTION[/b]

[quote="silouan"]I found a nice solution . Let the parallel through E to BC meet
AB at $B_{1}$ and AC at $C_{1}$ and let N,P the projections of M at AC , AB respectively . By Thales we have that E is the midpoint of $B_{1}C_{1}$ and since ME is perpendicular to $B_{1}C_{1}$ we have that $MC_{1}= MB_{1}$ . Now from the cyclic quardilaterals
$MPB_{1}E$ and $MENC_{1}$ we have that $MNP$ is isosceles so
$MP=MN$ and so $M$ is on the bisector of $A$[/quote]

[size=150][color=red]ΑΣΚΗΣΗ 21[/color][/size]

[quote="stergiu"] Δίνεται  τρίγωνο  $ABC$ , η διάμεσος   $AD$ ,    ένα σημείο $E$ στην $AD$ και $EF$ κάθετη στην $BC$. Έστω  $M$ σημείο στην  $EF$  τέτοιο ,  ώστε  ώστε η ευθεία που ενώνει τα ίχνη των καθέτων από το $M$  προς τις  $AC$  και $AB$ να περνάει από το $E$. Να αποδείξετε  ότι η $AM$  διχοτομεί τη γωνία $A$ του τριγώνου $ABC$ .
  [/quote]

[b]SOLUTION(by me)[/b]

[quote="silouan"]. Θεωρούμε την παράλληλη από το $E$ στη $BC$ που τέμνει την $AB$ στο  $B_{1}$ και την
   AC στο $C_{1}$.
  Έστω N,P οι προβολές του  M στις  AC , AB αντίστοιχα.Από το θεώρημα του Θαλή  συμπεραίνουμε  ότι το E είναι μέσο του $B_{1}C_{1}$ και αφού η  ME είναι κάθετη στη $B_{1}C_{1}$  προκύπτει  $MC_{1}= MB_{1}$ . Τώρα , από τα εγγράψιμα τετράπλευρα
  $MPB_{1}E$ και $MENC_{1}$  προκύπτει  ότι το $MNP$ είναι ισοσκελές. Άρα
$MP=MN$ και επομένως  το $M$ ανήκει στη διχοτόμο της γωνίας  $A$[/quote]

[size=150][color=red]ΑΣΚΗΣΗ 22[/color][/size]

Έστω  $A$ το ένα από τα σημεία τομής δύο κύκλων  $C_{1}$ και  $C_{2}$ με κέντρα  $O_{1}$ και  $O_{2}$

αντίστοιχα.Μια ευθεία που διέρχεται από το  $A$ τέμνει  τον κύκλο $C_{1}$ στο   $E$ και τον κύκλο  $C_{2}$ στο σημείο

$D$. Έστω  $M$ και $N$  τα μέσα των τμημάτων  $O_{1}O_{2}$ και  $ED$.Να αποδείξετε ότι  $MN=MA$.

[b] ΛΥΣΗ[/b]

  Αν $B,C$  είναι οι προβολές των σημείων  $O_{1}, O_{2}$ στην  $ED$ ,τότε $NE=ND$ ή  $NA+2AB=NC+CD$ ή

 $AC-CN+2AB= NC+CD$ ή    $2AB=2NC$ , δηλαδή     $AB=NC$.

    Αν  $P$  είναι η προβολή του  $M$ στην  $ED$ , τότε το   $P$  θα είναι και μέσο του  $BC$  και αφού

   $AB=NC$,  θα είναι επίσης και  $PA=PN$.

  Αυτό σημαίνει ότι το τρίγωνο  $MAN$ είναι ισοσκελές , αφού το ύψος του είναι και διάμεσος .

   Επομένως $MA=MN$.

   [u]Babis[/u]

[size=150][b][color=red]ΑΣΚΗΣΗ 23[/color][/b][/size]

 Μια βασική πρόταση στον εγγεγραμμένο κύκλο που αποτελεί το κλειδί για τη λύση αρκετών όμορφων ολυμπιακών θεμάτων.

 Επειδή η εικόνα δε φαίνεται καλά, στέλνω την άσκηση και σε pdf , ώστε να είναι ευανάγνωστη.

Μπάμπης

[color=red] EDIT[/color]

Τη συμπλήρωσα και με δύο καλές εφαρμογές.

Φοβερή πρόταση κύριε Στεργίου!! Είχα πολύ καιρό να τη δω! Σας ευχαριστώ πολύ.

Αλέξανδρος

Αλέξανδρε ,
 τυχαία τη συνάντησα σε κάποιο  θέμα . Και μένα μου άρεσε πολύ.
 Γενικά , στον εγγεγραμμένο κύκλο υπάρχουν πολύ καλά θέματα. ΄Κάθε φορά που συναντάω ένα , με συναρπάζει .Ελπίζω κάποια στιγμή να βρούμε χρόνο και να φτιάξουμε μαζί(θα μας βοηθήσοιυν και τα άλλα παιδιά) ένα αρχείο με όλα τα καλά θέματα και τις προτάσεις στον εγγεγραμμένο κύκλο.
 Θα είχε πολύ ενδιαφέρον. Νομίζω ότι σίγουρα θα βρούμε τουλάχιστον 20 καλά θέματα όλων των επιπέδων !

[b] Μπάμπης[/b]

[color=red] Α! Στέλνω και το παραπάνω αρχείο συμπληρωμένο με δυο πολύ καλές εφαρμογές.
 Τύπωσέ τες και ρίξτες μια ματιά.[/color]

[color=red][size=150][b]ΑΣΚΗΣΗ 24[/b][/size][/color]

 Στέλνω ένα ακόμα θέμα , σχετικά δύσκολο , με πολύ στοιχειώδεις γνώσεις. Το δίνω σε  word.Αν κάποιος θέλει να μαλακώσει λίγο τη λύση με περισσότερα σχόλια και εξηγήσεις , ας το κάνει και να μου ξαναστείλλει τη λύση να την τοποθετήσω.
 Μπάμπης

[color=red]edit : Ξαναστέλνω τη λύση με ένα καλό σχόλιο από το Θανάση Μπεληγιάννη[/color]

[size=150][color=red][b]ΑΣΚΗΣΗ 25[/b][/color][/size]

 Στέλνω ένα ακόμα θέμα στα εγγράψιμα. Το κείμενο είναι σε word. Τυπώστε το , συμληρώστε το ,αν χρειάζεται, και το ξαναστέλνετε.


What percent of 50 is $ 50\%$ of $ 50\%$ of 50?


Let $ \vartheta \left( n\right)$ is smallest natural number such that $ n|\vartheta \left( n\right) !$
Find all $ n\in\mathbb{N}$ such that $ \frac{\vartheta \left( n\right) }{n}\equal{}\frac{2}{3}$

Assume that $ n$ is a solution. Then $ 2n\equal{}3 \nu (n)$ so that $ n\equal{}3a$ and $ \nu(n) \equal{} 2a$ for some positive integer $ a$.
Now, $ 3a$ divides $ (2a)!$ but not $ (2a\minus{}1)!$.
But, if $ a \geq 4$ then $ 3 < a < 2a\minus{}1$ thus $ 3a$ divides $ (2a\minus{}1)!$. A contradiction.
It follows that $ a \leq 3$.

Direct checking shows that $ n\equal{}9$ is the only solution.

Pierre.


I took this test and got 130, 132, and 148 with SD 15, 16, and 32 respectively.EDIT: the test is slightly different from the one I took originally.  I got a 135 on this one the updated one I believe.  It looks reliable enough...  I'm guessing you guys will get higher.  The test takes 40 minutes, but I finished it with over 10 minutes left.  I thought I only missed 2 questions, the answers unfortunately are not posted.[/b]

oops yeah here

http://www.iqtest.dk/main.swf

How do you get other SDs? I got 130 with SD 15, too, but I'm sure I just clicked the wrong button on one of them or something. I had a bit more than 10 minutes left, too.

[quote="solafidefarms"]Eh, 108, but I spent but 10 minutes on it.[/quote]

Haha usually with online IQ tests I'll spend 10-15 minutes cruising through the first half and then say "Screw it" by the second half.

Had 21 minutes left, got bored, guessed on the last 5 or so (I don't remember exactly), and got 110.  Meh.  It's night.  My sleeping brain has an IQ of 110.  hehe.

ahahaha finished in 12 minutes, got 110 with st.dev 15

which is odd, because the last time i did an iq test, i got above mensa level

130... interesting test, I think I've taken it before. Some required quite a bit of thinking, and the last few were just too complicated.

Ooh, I remember that test. Screwed the last 2 or 3, I think, and my result was 130+ (I'm too lazy to take it again.)

I did it twice, 1st time 110, second time 126. Kinda annoying if you ask me. The test, I mean. Not the results.

Wow...I only did 11 questiosn and got a IQ of 90......I wonder how many I would get if I did all of them...

126 in sd 15. Used about 10 min, got lazy on last few problems. Not really satisfied with score, but oh well.

notice that if you don't answer any problems you get a 79  :huh:

i got a 133, sd15

hehe got 30 ques in 17 minutes, thought really hard on the next few, then screwed the last three or so with 12 minutes left

IQ 130..

Tickle tests, anyone? :D

i did the test before going to bed and i got around 120... i re-did it in the morning and got 135... still it's odd because several years ago i did an IQ test a psychologist gave me and had 155 so i'm rather skeptical about online tests

ya, I never do online tests.
A psychologist gave me a test and I got 158
but seriously online tests never really work.

I did the first 26 in about 5 minutes, then I heard my mom come and I put H the rest, so I got something between 100 and 110.


In $\triangle{ABC}$, $\angle{A}=60^o$, $AB > AC$, $O$ is the circumcenter. The altitude $BE$ and $CF$ meets at $H$. $M$ and $N$ are on $BH$ and $HF$ such that $BM=CN$.

Find the value of $\frac{MH+NH}{OH}$.

As <COB = 2 <CAB = 120° and <CHB = 180° - <CAB = 120°, CHOB is ciclyc. Then <HCO = <HBO therefore OMB and ONC are equal triangles from which <OMB = <ONC. This implies in turn that HNOM is ciclyc and by Ptolomeus MO*NH + NO*MH = NM*HO. Since MO = NO, it holds that (MH+NH)/OH = NM/OM = 3^.5.

[quote="wei_wenzhe9170"]This is 2002 China League[/quote]
It seems that I have messed up with the calendar......

Editted.