inequalities
trigonometry
function
inequalities unsolved

Hi! Can you please help me to prove inequalities in this form:

Given that, $ A$, $ B$, $ C$ are angles of triangle, and $ k$ is some positive integer. Find minimum of

$ \frac {1}{k \minus{} cos{A}} \plus{} \frac {1}{k \minus{} cos{B}} \plus{} \frac {1}{k \minus{} cos{C}}$

By the way , it doesn't have minimum for all $ k$

This would be kind of generalization of inequality that Vasile Cirtoaje gave in Crux Mathematicorum:

$ \frac {1}{2 \minus{} cos{A}} \plus{} \frac {1}{2 \minus{} cos{B}} \plus{} \frac {1}{2 \minus{} cos{C}} \geq 2$

I think we can use this fact : $ \cos{A} \plus{} \cos{B} \plus{} \cos{C} \leq \frac {3}{2}$  and Cauchy Schwarz.
But i am not sure.

Your Help is much appreciated!  :roll:

I think the minimum must be  $ \frac{6}{2k\minus{}1}$ may be Jensen inequality works for the function $ f(x)\equal{}\frac{1}{k\minus{}cosx}$ :blush:

[quote="othman"]I think the minimum must be  $ \frac {6}{2k \minus{} 1}$ may be Jensen inequality works for the function $ f(x) \equal{} \frac {1}{k \minus{} cosx}$ :blush:[/quote]
othman, for you: $ f''(x) \equal{} \minus{}\frac {\cos^2x \plus{} k\cos x \minus{} 2}{(k \minus{} \cos x)^3}.$ :wink:


Dear friends,

Let $ ABC$ be a triangle. The incircle $ C\left(I\right)$ touch the sides $ BC$, $ CA$, $ AB$ at the points $ K$, $ E$, $ F$ respectively and $ M$ is the midpoint of the altitude $ AD$. If $ \left\{N,K\right\}\equal{}KM\cap C\left(I\right)$, $ \left\{X,N\right\}\equal{}BN\cap C\left(I\right)$, $ \left\{Y,N\right\}\equal{}CN\cap C\left(I\right)$ then prove that the lines $ MK$, $ XE$, $ YF$ are concurrent.  :)

Dear Petry, the problem you have posted was mentioned by [b]mr.danh[/b] many months ago.
You can see at [url]http://www.mathlinks.ro/viewtopic.php?search_id=1142805182&t=206246[/url]


inequalities
quadratics
inequalities unsolved

Suppose real numbers $A,B,C$ such that for all real numbers $x,y,z$ the following inequality holds:

\[A(x-y)(x-z) + B(y-z)(y-x) + C(z-x)(z-y) \geq 0.\]

Find the necessary and sufficient condition $A,B,C$ must satisfy (expressed by means of an equality or an inequality).

very not sure about my "proof" but thought i would post anyway

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

$\displaystyle A(x-y)(x-z) + B(y-z)(y-x) + C(z-x)(z-y) \geq 0.$ (*)

Suppose (*).  Without loss of generality, $x \geq y \geq z$. [We can repeat the same argument if $x \geq z \geq y$, etc.]

Let $a = x-y$ and $b = y-z$ and understand that $a,b$ are positive reals.

Then:

$(*) \Leftrightarrow A(x-y)(x-y + y-z) +  C(x-y + y-z)(y-z) \geq B(x-y)(y-z)$

$\Leftrightarrow Aa(a+b) + C(a+b)b \geq Bab$

$\Leftrightarrow A(\frac{a}{b})^2 + (A+C-B)(\frac{a}{b}) + C \geq 0$  (If $b = 0$, the condition would be just $A \geq 0$).

Notice that $\frac{a}{b} \geq 0$.

If $A \leq 0$, for extreme values of $a/b$, we have it being negative.  So certainly $A > 0$.

If there is a positive root, then some segment of the graph of the quadratic is in the 4th quadrant.

So we want $- (A+C-B) + \sqrt{A^2 + B^2 + C^2 - 2AB - 2BC - 2CA} < 0$,

or $A+C-B > \sqrt{(A+C-B)^2 - 4AC}$, which is true if $AC > 0$.

From here, using AMGM it is clear that the condition is $4\sqrt{AC} \geq B$.  Because our steps are reversible, it is clear it is sufficent.

I think your proof is wrong!
There you can't suppose x>y>z cause it surely loses generality. Remember the inequality is a family of inequalities about A,B,C ,you can't just change the condition of the inequities.
My proof is to suppose z=0,the use discriminant of a quadratic equation to get that the square root of A,B,C forms a triangle.

[quote=programjames1]I think that you can suppose $x>y>z$. If you do this, depending on if it's true or not, you can just change $A$, $B$, and $C$ to fit the inequality.
Also, It would be nice if you could elaborate on how
$\Leftrightarrow Aa(a+b) + C(a+b)b \geq Bab$
$\Leftrightarrow A(\frac{a}{b})^2 + (A+C-B)(\frac{a}{b}) + C \geq 0$[/quote]
But if we don't suppose x>y>z,from this inequality we can get A^2+B^2+C^2-2AB-2AC-2BC<=0,and A,B,C >=0, it's much stronger than what Singular get.


[url=http://6-six-6.myminicity.com/]Proof of Riemann Hypothesis[/url]

minicity?  :D

Show this to pythag011, he'll click it a million times.

DONT CLICK ITS TO HIS MINICITY AND HE'LL JUST GET MINICITY MONEY!!!!!!!!!!!!!!

[url=http://ladue-ms.myminicity.com]my first[/url]

[url=http://ernieinre.myminicity.com]my second[/url]

The first beast the second infinitely, in my opinion. :P

MO is 6th in total posts....we're almost past North Carolina which will make us 5th... :o

Eventually all of this silly stuff would boost the post count to 1st. :D


A store is selling an item for $ 1000$ dollars.But you also have the option  to pay it in $ 20$ parts of $ 80$ dollars each, one part per month.What is the interest rate (per month) ?

I don't know what PG stands for, but it seems to me that you pay $ 1600$ instead of $ 1000$ so the interest rate is $ 60$ per cent. Or, each month, instead of $ 50$, you pay $ 80$, again $ 60$ per cent.

How do you get a percentage sign and a dollar sign in Latex?

[quote="AndrewTom"]
How do you get a percentage sign and a dollar sign in Latex?[/quote]

$ \$$ ,$ \%$

[code]$\$$ ,$\%$[/code]

Thanks. Of course! The edited title (which should have been obvious from GP) makes it clear what was being asked.


Your cow is tied to a 50 foot length of rope attached 5 feet in from the corner of one of the long sides of a 30 ft by 40 ft barn. How many square feet can the cow graze?

A) 5871 ft2 B) 7854 ft2 C) 5694 ft2 D) 6654 ft2

[quote="sri340"]... attached 5 feet in from the corner ...[/quote]What do you mean by this?

If the barn is $ ABCD$ where the long sides are $ AB\equal{}DC\equal{}40$ and the short sides are $ AD\equal{}BC\equal{}30$, could it be that the rope is attached to a point $ E$ where $ AE$ or $ DE \equal{} 5$?

he just asks questions, doesn't reply or say thanks so don't expect an answer to your question.


I am not sure whether or not I should take this course next year. The prerequisites only include Algebra II and Geometry. I havent taken any courses in Basic though. Since AP Comp Sci A at our school teaches JAVA, I was wondering if you think one could get an A in the class and pass the AP without taking basic

Does not require any prior computer programming skills, although they can be useful...

It also depends on the teacher though.

Take me for example... I only know a bit (really little!!) JavaScript and HTML and some TI-83 programming before I took this class. I have an A for now... And I am in AB!

You don't need BASIC before you start AP computer science.


geometry
analytic geometry
floor function
logarithms

[quote="fizsle"]1) The number of lines that are parallel to 2x + 6y +7 = 0 and have an intercept of length 10 between the coordinate axes is? => how to solve this?

2) if log 2 =.301 find the number of digits in $ \ 125^{25}$[/quote]

What does "intercept of length 10" mean?
[hide="2"]
The number of digits is equal to $ \lfloor\log_{10}{125^25} \plus{} 1\rfloor$
$ \equal{} 25(\log_{10}{125}) \plus{} 1$ (I know the floors are gone, I'll add them back later)
$ \equal{} 25(\log_{10}{\frac {1}{8}\cdot 1000}) \plus{} 1$
$ \equal{} 25(3 \minus{} .903) \plus{} 1$
$ \equal{} \lfloor85.25\rfloor$ (There they are)
$ 85$
[/hide]

subtle answer dude :D
check latex.  :wink:

Whooaaa, major system error.  Thanks for pointing that out,


If we snail mail our solutions on Monday night, will it count?  If there is something as postmark time (I don't know how mail works very well) and it says that it was postmarked on Tuesday at 5 AM, then will we get the benefit of the doubt? The round runs for over one month, so I'm hoping 5 hours "late" is okay?  I will probably mail my solutions around 9 or 10 PM Monday Arizona Time, so will that be okay?

Just wondering, wouldn't it be much eaiser to e-mail the solutions first, and then, in addition to that (in case something happens) send a  copy of your solutions via snail mail? Of course, I am assuming that you TeXed your solutions...

Of course I will be emailing them, but I'm saying if the email doesn't arrive safely, and I am relying on my snail mail copy, would this be ok?

In that case, I would PM either Mr. Crawford or Mr. Ruscyzk

PM them now and ask what to do or PM them right when I mail the solutions?

You email client will have on record when you submitted your paper (not webmail stuff like hotmail, but gmail does).  In the unlikely event that your email doesn't reach them, you could always use that as proof that you submitted it on time.  In all honesty though, snail mail is more likely to get lost.

If you email and snail mail, please make it clear on both that you are sending your solutions via two methods.


What is the largest number of prime addends with a sum of $ 125$?


calculus
integration
function
limit
real analysis
real analysis theorems

I quoted this theorem in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216615]this topic[/url].

------

Suppose $ g$ is a locally integrable function on $ [0,\infty)$ and suppose that if $ G(x)\equal{}\int_0^xg(t)\,dt,$ then $ \lim_{x\to\infty}G(x)\equal{}L.$ (Note that this forces $ G,$ which is continuous, to be bounded.)

(We're saying that $ \int_0^{\infty}g(x)\,dx\equal{}L$ as an improper integral.)

Prove that $ \lim_{r\to0^\plus{}}\int_0^{\infty}e^{\minus{}rx}g(x)\,dx\equal{}L.$

[quote="Kent Merryfield"]

Prove that $ \lim_{r\to0^ \plus{} }\int_0^{\infty}e^{ \minus{} rx}g(x)\,dx \equal{} L.$[/quote]

I am not sure if the last "Prove..." was rhetorical, but could you make the argument that since

$ e^{ \minus{} rx}g(x)\in\mathcal{C}\left(\mathbb{R}^\plus{}\right)$ We know this due to its integrability

And that

$ e^{ \minus{} rx}g(x)\leq{max_{r\in(0,\infty)}\left(e^{ \minus{} rx}\right)g(x)} \equal{} g(x)$

And since we know that

$ \int_0^{\infty}g(x)\,dx \equal{} L < \infty$

That this integral is convergent

Therefore by the Weierstrass M-test we know that our integral is uniformly convergent for $ r\in\mathbb{R}^ \plus{}$

Now seeing that if we define $ D \equal{} \left\{r: r\to{0^ \plus{} }\right\}$ that $ D\subset\mathbb{R}^ \plus{}$

Therefore

$ \lim_{r\to{0}}\int_0^{\infty}e^{ \minus{} rx}g(x)\,dx \equal{} \int_0^{\infty}\lim_{r\to{0}}e^{ \minus{} rx}g(x)\,dx \equal{} \int_0^{\infty}g(x)\,dx \equal{} L$

Is that correct at all?

All criticism welcomed.

The "prove" was not rhetorical; I was proposing it to the community as a problem.

Your proof is incorrect. The only way something even remotely resembling this could be a proof would be if $ \int_0^{\infty}g(x)\,dx$ were absolutely convergent - that is, $ \int_0^{\infty}|g(x)|\,dx < \infty.$

But I very carefully did not make that a hypothesis, and all of the interesting applications of this theorem involve cases in which that integral is merely conditionally convergent.

The same can be said for the original Abel's Theorem:

If $ \sum_{n \equal{} 0}^{\infty}a_n \equal{} L,$ then $ \lim_{r\to1^ \minus{} }\sum_{n \equal{} 0}^{\infty}a_nr^n \equal{} L$

If $ \sum_{n \equal{} 0}^{\infty}a_n$ is absolutely convergent, then we can invoke the Weierstrass M-test to obtain uniform convergence and then use that the uniform limit of continuous functions is continuous. But that's not Abel's theorem! The theorem is really only used in cases in which the series is merely conditionally convergent. For instance, we need Abel's theorem to claim that $ \sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^k}{2k \plus{} 1} \equal{} \frac {\pi}4.$ That's a conditionally convergent series.

[quote="Kent Merryfield"]

Your proof is incorrect. [/quote]

Sorry Dr. Merryfield, I should not have attempted it. Apparently it is well beyond my skill level as of yet.

Alex


If $ ABC$ is an equilateral triangle with an area of $ \sqrt3$ square units, how many units long is each side?

The area of an equilateral triangle, where $ s$ is the side length, is $ \frac{s^2\sqrt{3}}{4}$. So, we have

$ \frac{s^2\sqrt{3}}{4}\equal{}\sqrt{3}$

$ \implies s^2\equal{}4 \implies s\equal{}\boxed{2}$.


integration
trigonometry
calculus
calculus computations

Evaluate

$ \int_{x\equal{}(1\minus{}\sqrt{5})/2}^0 \sqrt{x\plus{}\sqrt{x\plus{}1}} \, dx$.

$ x \plus{} 1 \equal{} t\Longrightarrow dx \equal{} 2tdt$,

$ \int_{\frac {1 \minus{} \sqrt {5}}{2}}^{0} \sqrt {x \plus{} \sqrt {x \plus{} 1}}\ dx$

$ \equal{} \int_{ \sin \frac {\pi}{5}}^ 1 \{\sqrt {t^2 \plus{} t \minus{} 1}(t^2 \plus{} t \minus{} 1)' \sqrt {t^2 \plus{} t \minus{} 1}\}dt$

$ \equal{} \left[\frac 23(t^2 \plus{} t \minus{} 1)^{\frac {3}{2}}\right]_{ \sin \frac {\pi}{5}}^1 \minus{} \int_{ \sin \frac {\pi}{5}}^ 1 \sqrt {t^2 \plus{} t \minus{} 1}dt$


irrational number
number theory unsolved
number theory

I post it in the unsolved section, because I would like to see a purely number-theoretic approach to this (the solution I have, which is of course that of Dani, uses some extremely difficult results on flows on homogeneous spaces):
  Theorem (Dani)  Prove that for any irrational number $x$, the quantity
\[\frac{1}{T}\sum_{0<m\leq T\{mx\}, \gcd(m,[mx])=1}{\frac{1}{\{mx\}}}\] has limit $\frac{6}{\pi^{2}}$ as $T$ tends to $\infty$.

To be sure I'm understood, the sumation is take after those positive $m$ which satisfy $\gcd(m,[mx])=1$ and $m\leq T\{mx\}$. Here, of course, $[]$ and $\{\}$ are the integer and fractional parts.

Obviously, as stated it is wrong! I forgot to mention "tends to " after the ugly expression (here $T$ becomes larger and larger and tends to infinity).


Solve system of equations
$ y^2\plus{}(4x\minus{}1)^2\equal{}\sqrt[3]{4x(8x\plus{}1)}$
 $ 40x^2\plus{}x\equal{}y \sqrt{14x\minus{}1}$

[quote="pi3.14"]Solve system of equations
$ y^2 \plus{} (4x \minus{} 1)^2 \equal{} \sqrt [3]{4x(8x \plus{} 1)}$
 $ 40x^2 \plus{} x \equal{} y \sqrt {14x \minus{} 1}$[/quote]
We have
$ 2(y^2 \plus{} (4x \minus{} 1)^2) \plus{} 4(40x^2 \plus{} x) \equal{} \sqrt [3]{2.16x(8x \plus{} 1)} \plus{} 2.2\sqrt {y^2(14x \minus{} 1)}$
$ \le \frac {2 \plus{} 16x \plus{} 8x \plus{} 1}{3} \plus{} 2(y^2 \plus{} 14x \minus{} 1) \equal{} 36x \minus{} 1 \plus{} 2y^2$
or $ 192x^2 \minus{} 48x \plus{} 3\le 0 \leftrightarrow 3(8x \minus{} 1)^2 \le 0 \leftrightarrow x \equal{} \frac {1}{8}$.
So $ y \equal{} \frac {\sqrt {3}}{2}$


articles
geometry
algorithm
analytic geometry

Recently I’ve come across an article of some unknown author. Has anyone heard about Andrey Shvets? Who’s he?  :blush:

“Modern Leonardo da Vinchi” or a Belorussian Dreamer?

I’ve got acquainted with Andrey Shvets' works after a friend of mine called him seriously or for fun modern Leonardo da Vinci. The breadth of scientific interests of this Belarusian author is really amazing. He offers new original ideas almost in every field of knowledge. However, nobody knows whether his ideas are proved or caused by the desire of the author to look for paradoxes even if they are not present.
For example, it is considered to be that the “thermal death” of the system comes in case of its maximal entropy value, and in this case all processes stop in it. Therefore the system can be only in one static condition. Shvets approves that if the system has only one possible condition, its entropy is equal to zero. Hence, the recourse of the system and its approaching to the point of the “thermal death” is accompanied by the reduction of the entropy, and the progress – by the increase of the entropy. So the author makes the conclusions in the field of sociology and the economic theory. The divergence with the accepted position explains the mess between the information entropy and the thermodynamic entropy. I’m not an expert in this area, but I’m inclined to trust much more to the opinion of several generations of scientists-physicists than to Shvets.
Let’s take another example. The paradox of Zenone about Ahille and the turtle has not been a riddle for a long time. Nobody doubts that the runner would overtake the turtle. And the paradox is based on the ignorance of the ancient people of the basics of the differential calculus. If people in Zenone’s time knew that the sum of the infinite quantity of the decreasing pieces can be equaled to any final number, there would have been no paradox at all. But Shvets writes that the opportunity of division of the final piece to the infinite number of infinitesimal pieces had been known earlier. In the very statement of this paradox the certain algorithm of the runner’s movement is incorporated. While Ahille is running to the point where the turtle was, he does not know where it is moving at this time. Only having reached this point the runner can see the following point to which the reptile gets. That is, if Ahille chases the turtle not knowing how it moves and in what point it appears next, he could never catch up with it. It’s impossible to catch up with the thing that moves absolutely vaguely, appearing here and there. However, the photons of light move like this. So Shvets makes the assumption that because of the uncertainty of the occurrences of photons of light, the more predicted particles could never "catch up" with them at the microlevel.
Shvets has also formulated “the most general principle of relativity” which says that any observer has the right to consider himself based. The principle of relativity of Galilee gives such a right to the rectilinearly moving observer at the regular intervals, the general theory of relativity of Einstein - to the observer accelerating in the gravitational field. And still more “general principle of relativity” says that it’s possible to explain any physical phenomenon operating physical laws by having accepted any system of coordinates for based.
Andrey Shvets goes beyond physics. It mentions economy, psychology, mathematics, philosophy and religion. But he does not try to join together various religions and philosophical doctrines as many people are trying to make it now. He has created his own direction, but at the same time all religious and mystical currents have found their place in this construction. This doctrine looks strange enough, and some religious dogmas have got absolutely other sense. It is supposed that we represent some comprehensions which are having dreams. These dreams are synchronized among themselves as well as the programs of the players in the computer network. As a result, the uniform world which we consider to be real is formed. This world-dream carries out a role of a simulator for those comprehensions which have not got stronger yet for the perception of the real world. Gradually the sleeping man changes his dreams and sees himself as different animals and different people passing on higher levels. It’s called the resettlement of souls. In the process of strengthening the comprehension starts to feel the illusoriness of this world-dream, and it becomes less interested in it, and passions and desires of this illusive world dominate less over it. Finally, the prepared comprehension ceases to require these dreams-simulators and wakes up at long last breaking off the circuit of regenerations. The sin is all that prevents to perceive the real world. And if the sin is still present, the comprehension is born in this world-dream again, and the idea of the “fall of the man” is concluded in it.
Just as the communication between the computers can be broken, the communication between the sleeping and seeing one dream comprehensions can be broken, and then all of them would wake up. Our world-dream would disappear. There would come a doomsday. And then the comprehension prepared for the perception would apprehend the true reality as pleasure or the paradise. For those who has not had time to get stronger or for sinners, the perception of this reality would cause intolerable tortures like listening too loud music with too bright flashes of light. Indeed Shvets doesn’t insist on this model of the world.
Beside the listed areas of human activity Andrey Shvets is interested in the Chinese philosophy and interprets it in a new fashion. In spite of the fact that it is exciting to read his works, probably it is necessary to recognize that the time of the talented and "omnivorous" self-educated persons is over. The future is for the narrow experts who have given to the field of knowledge a significant part of their life.
William Barret

Look, I’ve found a couple of references: :D

http://www.quodlibet.net/shvets-model.shtml
http://www.createspace.com/3345050


x=2 is the only solution.

Case 1. x>2
The left hand side increases by a factor of at most 12^k, while the right hand side changes by at least 13^k as x increases, causing the two sides to diverge apart.

Case 2. x<2
The left hand side decreases by a factor of at most 12^k, while the right hand side decreases by a factor of at least 13^k, so the 2 values diverge from each other as well.

Please forgive, if I mixed up some of the facts, but the general statement should be easy to understand. :lol:


It is not a hard problem. AC or UMV theorem kill it enought. But have anything else without that theorem. Let think at least one time  :)
Let $ a,b,c\ge 0$. Prove that:
\[ a^3\plus{}b^3\plus{}c^3\plus{}9abc\plus{}4(a\plus{}b\plus{}c)\ge 8(ab\plus{}bc\plus{}ca)\]

Can we prove $ 9abc \plus{} 7(a\plus{} b\plus{}c)\geq 8(ab \plus{} bc \plus{} ca) \plus{} 6$?, If so, I can prove your inequality.

But it is wrong. Try $ a\equal{}b\equal{}c\equal{}1/2$ then the inequality is equivalent to: $ \minus{}3/8\ge 0$  :D


Inside a room of area 5, you can place 9 rugs, each of area 1 and arbitrary shape. Prove that there are two rugs which overlap by at least 1/9.

You have $ \binom92$ ways to choose two carpets, and you have 4 overlap space. $ \frac{4}{\binom92}=\frac19$, so we are done.


real analysis
function
inequalities
integration
calculus
real analysis unsolved

$ \phi: R\to R$ such that for every measurable $ f : R \to R$ ,the inequality holds
$ \int_0^1 \phi(f) \geq \phi(\int_0^1 f)$
prove that $ \phi$ is a convex function

shouldn't it be convex?? or switch the inequality sign?

yup,im sorry  :blush:  .Prove $ \phi$ is convex

It's really easy. this seems to be a converse to the integral form of Jensen's inequality.

Consider only functions of this type:
$ a,b\in \mathbb{R}$ and $ 0\le t\le 1$
$ f(x)\equal{}a$ if $ x\le t$ and $ f(x)\equal{}b$ otherwise.
What information do they give you?

thank u  , i didn't think it was simple like that  :blush:


[b][size=100][color=DarkBlue]The semicircles $ (O_{1}),\ (O_{2}),\ (O_{3}),$ taken as diameters the side-segments $ BC,\ AC,\ AB,$ of a given triangle $ \bigtriangleup ABC,$ intersect its altitudes $ AD,\ BE,\ CF,$ at points $ Z,\ M,\ N,$ respectively ( the point $ Z$ is inwardly to $ \bigtriangleup ABC$ and the points $ M,\ N,$ are outwardly to it ). We denote the points $ M'\equiv AB\cap ZM,\ N'\equiv AC\cap ZN$ and $ K\equiv MN'\cap M'N.$ Prove that $ KM \equal{} KN.$[/color][/size][/b]

Kostas Vittas.

[quote="kaka__2004"]nice!I have no idea about it, could you give me a clue,dear vittasko?[/quote]
[hide]Must prove $ AM \equal{} AN$ and then, $ \angle ASM \equal{} \angle ASN,$ where $ S\equiv BN\cap CM.$ So, if $ S,\ A,\ K$ are collinear, we are done.[/hide]
Kostas Vittas.

MA^2=NA^2=AF*AD=AE*AC   =>  MA=NA
Denote MC∩NB=S  SA∩MN’=K  we have SM=SN
Our task is to prove K,M’,N are collinear, in other words, we have to show that
KA/KS * NS/BN *BM’/M’A=1   (#)
Notice that we have
KA/KS * SM/MC * CN’/AN’=1
NH/NC * CN’/AN’* AZ/ZH=1
MH/MB * BM’/AM’* AZ/ZH=1
=> NH/NC * CN’/AN’= MH/MB * BM’/AM’
If (*) is true , then SM/MC * CN’/AN’= NS/BN *BM’/M’A
 〈=〉SM/MC * NH/NC　＝　NS/BN * MH/MB
 〈=〉MH/NH * NB/MB * NC/MC =1
〈=〉sin∠HNM/ sin∠HMN * sin∠NMB/ sin∠MNB * sin∠NMC/ sin∠MNC=1  (##)
Because (##) is obviously true , (#) is true
We have S,A,K are collinear  =>  MK=NK
Thus we are done!

[quote="thaithuan_GC"]There are many transformations in plane geometry's proof.
Dear vistasko, do you have a synthetic proof?[/quote]
In the solution (synthetic, but rather long) I have in mind, the collinearity of the point $ S,\ A,\ K,$ is derived from the perspectivity of the triangles $ \bigtriangleup MN'C$ and $ \bigtriangleup NM'B.$

That is, we can prove that the line segments $ MN,\ M'N',\ BC,$ are concurrent at one point and I will post here later, more details about this result.

Kostas Vittas.

For Vittasko's idea, which guided to prove $ MN,M'N'$ and $ BC$ are concurrent, I think we can apply Desagues's theorem:

Consider $ \triangle ZMN$ and $ \triangle ABC$, $ (ZM,AB),(ZN,AC),(MN,BC)$ respectively intersect each other and $ M',N',P'$, since $ AZ,BM,CN$ are concurrent as they are three altitudes of $ \triangle ABC$, $ M',N',P'$ are collinear, which we have to prove.

Thank you dear [b][size=100]mashimaru[/size][/b], for the pointing out the only one needed argument, for the concurrency of the line segments $ MN,\ M'N',\ BC.$

It was exactly the reason I posted this problem here, because of in the solution I have in mind, this result has been arisen by a complicated way and I was sure that there is a simpler approach to prove it ( but not easy to see the perspectivity of the triangles $ \bigtriangleup ZMN,\ \bigtriangleup ABC$ ).

When I have solved a problem, I am after very lazy to try for a simplification of the proof.

Thank you again, Kostas Vittas.

PS. Could you please tell me your name and where do you come from ( Japan )? I would like to preserve it, in my notes about this problem.

Could anybody prove that NM' and MN' are the bisectors of the angles ANB and AMC respectively?

Thanks & best regards,
sunken rock

Still nobody's able to prove my remark??

Best regards,
sunken rock

Let $X$、$Y$、$L$ be the second intersection of $BH$、$CH$、$AH$ with $\odot (AC)$、$\odot (AB)$、$\odot (BC)$ . Let $M' \equiv MZ \cap LX$，now we will prove $M'$ is on $AB$ . Easy found $CL^2=CX^2=CZ^2=CM^2=CH \cdot CF$，so $LXZM$ are concyclic，then $\angle XM'Z=\angle CXH+\angle CZH=\angle XFZ$，so $XZFM'$ are concyclic，so $\angle CFM'=180-(\angle CFZ+\angle M'XZ)=180-(\angle CZL+\angle M'XZ)=90$，so $M'F \perp CF$，i.e. $M'$ is on $AB$ .

Acording to $CX^2=CZ^2=CH \cdot CF$ we have $\frac{FX}{FZ}=\frac{XH}{ZH} \Longrightarrow \frac{\sin{\angle FXZ}}{\sin{\angle FZX}}=\frac{\sin\angle HLM}{\sin\angle HML}$

$\Longrightarrow \frac{\sin{\angle AM'M}}{\sin{\angle BM'L}}=\frac{\sin\angle AMM'}{\sin\angle BLM'}$ $($Note $AM$、$BL$ are the tangent$)$ $\Longrightarrow \frac{AM'}{BM'}=\frac{AM}{BL}=\frac{AN}{BN}$  .

So $NM'$ is the bisector of $\angle ANB$，similary，we have $MN'$ is the bisector of $\angle AMC$，so $\angle KMA=\angle KNA=45$，thus we get $KM=KN$ . Done .

Many thanks, my friend!

Best regards,
sunken rock


geometry
rectangle
geometric transformation

If ABCD is a rectangle and P is inside ABCD and PA=9, PB=7, PC=2, and PD=6 then find the maximum area of ABCD.

and also minimum area.

Do a parallel translation that maps one parallel side to another. Then you get a closed quadrilateral that is half of the area or something. Then you can note that a cyclic quad has the largest area for fixed sides.


geometry
trigonometry
circumcircle
parallelogram
rectangle
geometric transformation
homothety

Let ABCD be a cyclic quadrilateral in which the diagonals AC and BD intersect at right angles at E.  Let O be the center of its circumscribing circle.  Let P be the point of intersection of the tangent lines at A and B.  Let Q, R, S be similarly defined for the pairs B and C, C and D, D and A, respectively.  It is known that PQRS is a cyclic quadrilateral.  Let T,U,V,W be the orthocenters of triangles AOB, BOC, COD, DOA respectively.  Let F,G,H,K be the orthocenters of triangles POQ, QOR, ROS, SOP respecitively.
Prove that TUVW and FGHK are straight lines intersecting at right angles at E.

I will denote the sides of the quadrilateral $ABCD$ as $a = AB, b = BC, c = CD, d = DA$ and the orthocenters of the triangles $\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$ as $H_1, H_2, H_3, H_4$. WLOG, assume that lengths of the sides $a > b > d > c$ follow in this order. This is OK, because the shortest side of a cyclic quadrilateral with perpendicular diagonals is necessarily against the longest side. Consider the triangles $\triangle AEH_1, \triangle CEH_2$.

$\measuredangle EAH_1 = \measuredangle BAH_1 - \measuredangle BAE = \measuredangle ACB - \measuredangle BAC$

$\measuredangle ECH_2 = \measuredangle BCA - \measuredangle BCH_2 = \measuredangle ACB - \measuredangle BAC$

$\frac{AH_1}{CH_2} = \frac{AB}{2 \cos{\widehat{ACB}}} \cdot \frac{2 \cos{\widehat{BAC}}}{BC} =
\frac a b \cdot \frac{\cos{\widehat{BAC}}}{\cos{\widehat{ACB}}}$

Since the diagonals $AC \perp BD$ are perpendicular,

$\frac{AE}{CE} = \frac{AB}{\cos{\widehat{BAC}}} \cdot \frac{\cos{\widehat{ACB}}}{BC} = \frac a b \cdot \frac{\cos{\widehat{BAC}}}{\cos{\widehat{ACB}}}$

Thus the triangles $\triangle AEH_1 \sim \triangle CEH_2$ are similar by SAS. This implies that the diagonal intersection $E$ and the orthocenters $H_1, H_2$ are collinear. Note that these are orthocenters of the adjacent triangles $\triangle AOB, \triangle BOC$. This implies that by walking around the cyclic quadrilateral $ABCD$ (i.e., by a cyclic exchange of the vertices $A, B, C, D$ and the orthocenters $H_1, H_2, H_3, H_4$), we can demonstrate in an entirely similar way that the orthocenter pairs $H_2, H_3;\ H_3, H_4;\ H_4, H_1$ are also collinear with the diagonal intersection $E$. Hence, all 4 orthocenters of the 1st group of triangles are collinear with the diagonal intersection and the points $H_1, H_4, E, H_3, H_2$ follow on the line of collinearity in this order.

Since the diagonals $AC \perp BD$ of the cyclic quadrilateral $ABCD$ are perpendicular, the tangents to the circle $(O)$ at the vertices $A, B, C, D$ form a bicentric quadrilateral $PQRS$ with the incircle $(O)$ and a circumcircle $(Z)$. There are 2 simple proofs of this fact:

[b](1)[/b] Invert the tangential quadrilateral $PQRS$ in its incircle $(O)$. The vertices $P, Q, R, S$ are carried into the intersections $P', Q', R', S'$ of the angle bisectors $PZ, QZ, RZ, SZ$ with the sides $AB, BC, CD, DA$, forming a parallelogram $P'Q'R'S'$ with the opposite sides parallel to the diagonals $AC, BD$. Since this parallelogram is inscribed in the inverted circumcircle $(Z')$, it is a rectangle and consequently, the diagonals $AC \perp BD$ are perpendicular. Conversely, if they are perpendicular, the midpoints of the sides $AB, BC, CD, DA$ form a rectangle, which can be circumscribed by a circle $(Z')$ and inverting this rectangle in the circle $(O)$ yields a bicentric quadrilateral $PQRS$.

[b](2)[/b] Let $O_1, O_2, O_3, O_4$ be the circumcenters of the triangles $\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$. The circumcircle pairs $O_1, O_2;\ O_2, O_3;\ O_3, O_4;\ O_4, O_1$ lie on the perpendicular bisectors of the segments $BO, CO, DO, AO$, respectively. Hence

$\measuredangle O_4O_1O_2 = 180^o - \measuredangle AOB = 180^o - 2\measuredangle ACB$

$\measuredangle O_2O_3O_4 = 180^o - \measuredangle COD = 180^o - 2\measuredangle CBD$

$\measuredangle O_4O_1O_2 + \measuredangle O_2O_3O4 = 360^o - 2(\measuredangle ACB + \measuredangle CBD) =$

$= 360^o - 2(\measuredangle ECB + \measuredangle CBE) = 360^o - 2 \cdot 90^o = 180^o$

Thus the quadrilateral $O_1, O_2, O_3, O_4$ is cyclic. But the tangential quadrilateral $PQRS$ is centrally similar to the cyclic quadrilateral $O_1O_2O_3O_4$ with homothety center $O$ and homothety coefficient 2, i.e., it is also cyclic and consequently, it is bicentric.

Let $G_1, G_2, G_3, G_4$ be the orthocenters of the triangles $\triangle POQ, \triangle QOR, \triangle ROS, \triangle SOP$. Since $(O)$ is the incircle of the bicentric quadrilateral $PQRS$, the sides $PO, QO, RO, SO$ of the triangles $\triangle POQ, \triangle QOR, \triangle ROS, \triangle SOP$ are the angle bisectors of the angles $\measuredangle P = \measuredangle SPQ$, $\measuredangle Q = \measuredangle PQR$, $\measuredangle R = \measuredangle QRS$, $\measuredangle S = \measuredangle RSP$ and thus the perpendicular bisectors of the sides $AB, BC, CD, DA$ of the isosceles triangles $\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$, i.e., also the altitudes of the later triangles. Hence, the orthocenters $H_1, H_2, H_3, H_4$ lie on the lines $PO, QO, RO, SO$. Likewise, the orthocenters $G_1, G_2, G_3, G_4$ lie on the lines $BO, CO, DO, AO$, because the radii $BO, CO, DO, AO$ are perpendicular to the tangents $s = SP, p = PQ, q = QR, r = RS$ at the vertices $A, B, C, D$, i.e., they are also altitudes of the triangles $\triangle POQ, \triangle QOR, \triangle ROS, \triangle SOP$. Again, we have to show that these 4 orthocenters $G_1, G_2, G_3, G_4$ are all collinear with the common diagonal intersection $E$ of the quadrilaterals $ABCD, PQRS$ (It is a well known fact that a tangential quadrilateral and its contact cyclic quadrilateral have a common diagonal intersection. Usually, this is shown by a repeated application of Pascal's theorem for the contact quadrilateral. For a convex tangential quadrilateral, this can also by demonstrated by projecting the incircle into a circle and the line $KL$ to infinity, where $K \equiv AB \cap CD$, $L \equiv BC \cap DA$ are the intersection of the opposite sides of the contact quadrilateral.)

The triangles $\triangle PG_1Q \sim \triangle SOR$ are similar, because

$\measuredangle ORS = 90^o - \measuredangle OSD = 90^o - \frac{\measuredangle COD}{2} = 90^o - \measuredangle CAD =$

$= \measuredangle ADB = \frac{\measuredangle AOB}{2} =  \measuredangle POB = \measuredangle QPG_1$

and similarly $\measuredangle OSR = \measuredangle PQG_1$. Since $G_1$ is the orthocenter of the triangle $\triangle POQ$, $O$ is the orthocenter of the triangle $\triangle PG_1Q$. From similarity of these triangles,

$\frac{r_O}{G_1B} = \frac{OD}{G_1B} = \frac{OG_3}{G_1O} = \frac{G_3D}{OB} = \frac{G_3D}{r_O}$

where $r_O$ is radius of the circle $(O)$. The triangles $\triangle EBG_1 \sim \triangle EDO$ are also similar. The angles $\measuredangle EBG_1 = \measuredangle EDO$ are equal, because they are the equal angles of the isosceles triangle $\triangle OBD$. Similarly to an earlier result,

${\frac{AE}{CE} = \frac a b \cdot \frac{\cos{\widehat{BAC}}}{\cos\widehat{ACB}}}= \frac a b \cdot \frac{\cos{\frac{\widehat S}{2}}}{\cos{\frac{\widehat R}{2}}}$

${\frac{BE}{DE} = \frac b c \cdot \frac{\cos{\widehat{CBD}}}{\cos\widehat{BDC}}}= \frac b c \cdot \frac{\cos{\frac{\widehat P}{2}}}{\cos{\frac{\widehat S}{2}}}$

In addition,

$\frac{G_1B}{OD} = \frac{BQ}{\cos{\frac{\widehat R}{2}}} \cdot \frac{2 \cos{\frac{\widehat R}{2}}}{CD} = \frac{BC}{2} \cdot \frac{\cos{\frac{\widehat P}{2}}}{\cos{\frac{\widehat S}{2}}} \cdot \frac{2}{CD} = \frac b c \cdot \frac{\cos{\frac{\widehat P}{2}}}{\cos{\frac{\widehat S}{2}}}$

Hence, the triangles $\triangle EBG_1 \sim \triangle EDO$ are similar by SAS and the angles $\measuredangle OG_1E \equiv \measuredangle BG_1E = \measuredangle DOE \equiv G_3OE$ are equal. Since the orthocenters $O, G_1$ of the similar triangles $\triangle PG_1Q \sim \triangle SOR$ cut their altitudes $G_1B, OD$ in the same ratio, the angles $\measuredangle OEG_1 = \measuredangle OEG_3$ are also equal and the orthocenters $G_1, G_3$ are collinear with the diagonal intersection $E$. Similarly, it can be shown that the orthocenters $G_2, G_4$ are also collinear with the diagonal intersection $E$. However, since these are orthocenters of the opposite triangle pairs $\triangle POQ, \triangle ROS$ and $\triangle QOR, \triangle SOP$, this does not yet imply that all 4 orthocenters $G_1, G_2, G_3, G_4$ are collinear with the diagonal intersection.

Collinearity of the points $E, G_1, G_3$ was demonstrated with the help of similar triangles $\triangle EBG_1 \sim \triangle EDO$ and $\triangle OEG_1 \sim \triangle G_3EO$. Collinearity of the points $E, G_2, G_4$ could be demonstrated in an entirely similar way with the help of similar triangles $\triangle EAG_4 \sim \triangle ECO$ and $\triangle OEG_4 \sim \triangle G_2EO$. It follows that

$\measuredangle OEG_1 + \measuredangle OEG_4 = 180^o - (\measuredangle EG_1O + \measuredangle EOG_1) + 180^o - (\measuredangle EG_4O + \measuredangle EOG_4) =$

$= 360^o - (\measuredangle G_1OG_4 + \measuredangle G_3OG_2) = 360^o - (\measuredangle AOB + \measuredangle COD) =$

$= 360^o - 2(\measuredangle ACB + \measuredangle CBD) = 360^o - 2 \cdot 90^o = 180^o$

This means that the orthocenters $G_1, G_4$ are collinear with the diagonal intersection $E$. Combining with the previous results, all 4 orthocenters of thre 2nd group of triangles are collinear with the diagonal intersection and the points $G_1, G_3, E, G_2, G_4$ follow on the line of collinearity in this order.

[quote="pestich"]Dear Yetti,

    It seems to me that there is no big need to dance around the quads and
    compare angles.
    The vertices P,Q,R,S are the reflections of orthocenters of AOB,....
    and they are concyclic, which is the same as being collinear. All you need is properly
    invert the circle PQRS.

    Maj.  Pestich[/quote]
Hello,

I already proved the collinearity of both the orthocenters of the triangles $\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$ and the orthocenters of the triangles $\triangle POQ, \triangle QOR, \triangle ROS, \triangle SOP$ with the diagonal intersection $E$, there is no question about that. What remains is to show that the 2 lines of collinearity are perpendicular. I even found the 8 right angle triangles, each formed by the diagonal intersection $E$ and 2 vertices from one orthocenter group each, which are similar to the right angle triangles, into which the perpendicular diagonals $AC, BD$ divide the contact cyclic quadrilateral $ABCD$. Now, similarity of at least one pair of the triangles has to be demonstrated.

The fact that the the reflections of 4 points in 4 different lines are concyclic does not imply that the original 4 points are collinear or even concyclic. Certainly, if the first group of orthocenters is collinear (which was one item to prove), there is an inversion carrying the circumcircle of the bicentric quadrilateral $PQRS$ into the collinearity line of the first group of orthocenters. The inversion center must be on the circumcircle of $PQRS$. But if the points $P, Q, R, S$ should go to the orthocenters of the triangles $\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$, the inversion center must be the concurrency point of the lines connecting $P, Q, R, S$ with the corresponding orthocenters. This point is the center $O$ of the incircle of the bicentric quadrilateral $PQRS$, which clearly must be inside of its circumcircle. Thus the inversion carrying the concyclic points $P, Q, R, S$ into the first group of collinear orthocenters does not exist.

Yetti.

It remains to be shown that the lines $h \equiv H_1H_3H_2H_4$ and $g \equiv G_1G_3G_2G_4$ intersecting at the diagonal intersection $E$ are perpendicular to each other. The 2 pairs of right angle triangles $\triangle AEB \sim \triangle DEC$, $\triangle BEC \sim \triangle AED$, into which the diagonals $AC, BD$ divide the contact quadrilateral $ABCD$, are similar for a number or reasons. For example, the power of the point $E$ to the circle $(O)$ is $-EA \cdot ED = -EB \cdot ED$ and consequently, $\frac{EA}{EB} = \frac{ED}{EC}$. This fact has been used on a number of occassions. The last step of the proof has to show that the triangles $\triangle H_1EG_1, \triangle H_3EG_2, \triangle G_4EH_1, \triangle G_3EH_3 \sim \triangle AEB$ and $\triangle H_2EG_2, \triangle H_4EG_3, \triangle G_1EH_2, \triangle G_4EH_4 \sim \triangle BEC$.

Collinearity of the orthocenters $H_1, H_2$ has been demonstrated using the similar triangles $\triangle AEH_1 \sim \triangle CEH_2$ with homothety coefficient $\eta_1 = \frac{AE}{CE}$. Collinearity of the orthocenter pairs $H_2, H_3;\ H_3, H_4;\ H_4, H_1$ could have been demonstrated in the same way from similarity of the triangle pairs $\triangle BEH_2 \sim \triangle DEH_3$, $\triangle CEH_3 \sim \triangle AEH_4$, $\triangle DEH_4 \sim \triangle BEH_1$ with homothety coefficients $\eta_2 = \frac{BE}{DE}$, $\frac{1}{\eta_1} = \frac{CE}{AE}$, $\frac{1}{\eta_2} = \frac{DE}{BE}$, respectively. Collinearity of the orthocenter pairs $G_1, G_3;\ G_2, G_4$ has been demonstrated from similarity of the triangle pairs $\triangle EBG_1 \sim \triangle EDO$, $\triangle EOG_1 \sim \triangle EG_3O$ and $\triangle EAG_4 \sim \triangle ECO$, $\triangle EOG_4 \sim \triangle EG_2O$ with similarity coefficients $\eta_2 = \frac{BE}{DE}$, $\eta_1 = \frac{AE}{CE}$, respectively. This could have been done equally well from similarity of the triangles $\triangle ECG_2 \sim \triangle EAO$, $\triangle EOG_2 \sim \triangle EG_4O$ and $\triangle EDG_3 \sim \triangle EBO$, $\triangle EOG_3 \sim \triangle EG_1O$ with similarity coefficients $\frac{1}{\eta_1} = \frac{CE}{AE}$, $\frac{1}{\eta_2} = \frac{DE}{BE}$, respectively. Because of the assumption $a > b > d > c$, $\eta_1 < \eta_2$, which fully determines the order of the orthocenters on the lines of collinearity $h, g$.

Using the [color=blue]cosine theorem[/color] for the triangles $\triangle AEH_1, \triangle EAG_4$,

$EH_1^2 = AE^2 + AH_1^2 - 2 AE \cdot AH_1 \cos{\widehat{EAH_1}} =$

$= a^2 \left(\cos^2{\widehat{BAC} + \frac{1}{4 \cos^2{\widehat{BOP}}} - \frac{\cos{\widehat{BAC}}}{\cos{\widehat{BOP}}} \cos{(\widehat{ACB} - \widehat{BAC})}\right) = }$

$= 4r_O^2 \sin^2{\frac{\widehat R}{2}} \left(\cos^2{\frac{\widehat S}{2}} + \frac{1}{4 \cos^2{\frac{\widehat R}{2}}} - \frac{\cos{\frac{\widehat S}{2}}}{\cos{\frac{\widehat R}{2}}} \cos{(\widehat C - 90^o)}\right) = $

$= 4r_O^2 \sin^2{\frac{\widehat R}{2}} \cos^2{\frac{\widehat S}{2}} \left[1 + \frac{1}{4 \cos^2{\frac{\widehat R}{2}} \cos^2{\frac{\widehat S}{2}}} - \frac{1}{\cos{\frac{\widehat R}{2}} \cos{\frac{\widehat S}{2}}} \cos{\left(\frac{\widehat Q}{2} + \frac{\widehat R}{2} - 90^o\right)}\right] = $

$= 4r_O^2 \sin^2{\frac{\widehat R}{2}} \cos^2{\frac{\widehat S}{2}} \left[1 + \frac{1}{4 \cos^2{\frac{\widehat R}{2}} \cos^2{\frac{\widehat S}{2}}} - \frac{1}{\cos{\frac{\widehat R}{2}} \cos{\frac{\widehat S}{2}}} \cos{\left(\frac{\widehat R}{2} - \frac{\widehat S}{2}\right)}\right] = $

$= r_O^2 \tan^2{\frac{\widehat R}{2}} \left(1 - 4 \sin{\frac{\widehat S}{2}} \cos{\frac{\widehat S}{2}} \sin{\frac{\widehat R}{2}} \cos{\frac{\widehat R}{2}}\right) = r_O^2 \tan^2{\frac{\widehat R}{2}}\ (1 - \sin{\widehat S} \sin{\widehat R})$

$EH_1 = r_O\ \frac{BE}{CE}\ \sqrt{1 - \sin{\widehat S} \sin{\widehat R}}$

Earlier, we obtained

$\frac{G_1B}{r_O} = \frac{G_1B}{OD} = \frac b c \cdot \frac{\cos{\frac{\widehat P}{2}}}{\cos{\frac{\widehat S}{2}}}$

By cyclic exchange of the vertices $A, B, C, D$ and $P, Q, R, S$,

$\frac{G_4A}{r_O} = \frac{G_4A}{OC} = \frac a b \cdot \frac{\cos{\frac{\widehat S}{2}}}{\cos{\frac{\widehat R}{2}}}$

$EG_4^2 = AE^2 + AG_4^2 - 2 AE \cdot AG_4 \cos{\widehat{EAG_4}} =$

$= a^2 \left(\cos^2{\widehat{BAC} + \frac{r_O^2}{b^2} \cdot \frac{\cos^2{\frac{\widehat S}{2}}}{\cos^2{\frac{\widehat R}{2}}} - \frac{2r_O}{b} \cos{\widehat{BAC}} \cdot \frac{\cos{\frac{\widehat S}{2}}}{\cos{\frac{\widehat R}{2}}} \cdot \cos{(90^o - \widehat B)}\right)}$

$= 4r_O^2 \sin^2{\frac{\widehat R}{2}} \left[\cos^2{\frac{\widehat S}{2}} + \frac{1}{4 \sin^2{\frac{\widehat S}{2}}} \cdot \frac{\cos^2{\frac{\widehat S}{2}}}{\cos^2{\frac{\widehat R}{2}}} - \frac{\cos{\frac{\widehat S}{2}}}{\sin{\frac{\widehat S}{2}}} \cdot \frac{\cos{\frac{\widehat S}{2}}}{\cos{\frac{\widehat R}{2}}} \cdot \cos{\left(90^o - \frac{\widehat P}{2} - \frac{\widehat Q}{2}\right)}\right]$

$= 4r_O^2 \sin^2{\frac{\widehat R}{2}} \cos^2{\frac{\widehat S}{2}} \left[1 + \frac{1}{4 \sin^2{\frac{\widehat S}{2}} \cos^2{\frac{\widehat R}{2}}} - \frac{1}{\sin{\frac{\widehat S}{2}} \cos{\frac{\widehat R}{2}}} \cdot \sin{\left(\frac{\widehat R}{2} + \frac{\widehat S}{2}\right)}\right]$

$= r_O^2\ \frac{\tan^2{\frac{\widehat R}{2}}}{\tan^2{\frac{\widehat S}{2}}} \left(1 - 4 \sin{\frac{\widehat R}{2}} \cos{\frac{\widehat R}{2}} \sin{\frac{\widehat S}{2}} \cos{\frac{\widehat S}{2}}\right) =  r_O^2\ \frac{\tan^2{\frac{\widehat R}{2}}}{\tan^2{\frac{\widehat S}{2}}}\ (1 - \sin{\widehat R} \sin{\widehat S})$

$EG_4 = r_O\ \frac{AE}{CE}\ \sqrt{1 - \sin{\widehat S} \sin{\widehat R}}$

From similarity of the triangles $\triangle ACH_1 \sim \triangle ECH_2$ with homothety coefficient $\eta_1 = \frac{AE}{CE}$, we immediately obtain

$EH_2 = \frac{CE}{AE} \cdot EH_1 = r_O\ \frac{BE}{AE}\ \sqrt{1 - \sin{\widehat S} \sin{\widehat R}}$

Using the cosine theorem for some of the remaining triangles $\triangle BEH_2 \sim \triangle DEH_3$, $\triangle CEH_3 \sim \triangle AEH_4$, $\triangle DEH_4 \sim \triangle BEH_1$ and $\triangle EBG_1, \triangle ECG_2, \triangle EDG_3$, we can find in exactly the same way that

$EH_3 = r_O\ \frac{CE}{BE}\ \sqrt{1 - \sin{\widehat S} \sin{\widehat R}}$

$EH_4 = r_O\ \frac{AE}{BE}\ \sqrt{1 - \sin{\widehat S} \sin{\widehat R}}$

$EG_1 = r_O\ \frac{BE}{DE}\ \sqrt{1 - \sin{\widehat S} \sin{\widehat R}}$

$EG_2 = r_O\ \frac{CE}{AE}\ \sqrt{1 - \sin{\widehat S} \sin{\widehat R}}$

$EG_3 = r_O\ \frac{DE}{BE}\ \sqrt{1 - \sin{\widehat S} \sin{\widehat R}}$

As a result we obtain the following relations for the triangles $\triangle H_1EG_1, \triangle H_3EG_2, \triangle G_4EH_1, \triangle G_3EH_3$ and $\triangle H_2EG_2, \triangle H_4EG_3, \triangle G_1EH_2, \triangle G_4EH_4$:

$\frac{EG_1}{EH_1} = \frac{EG_2}{EH_3} = \frac{EH_1}{EG_4} = \frac{EH_3}{EG_3} = \frac{BE}{AE} = \tan{\frac{\widehat S}{2}}$

$\frac{EH_2}{EG_2} = \frac{EH_4}{EG_3} = \frac{EG_1}{EH_2} = \frac{EG_4}{EH_4} = \frac{BE}{CE} = \tan{\frac{\widehat R}{2}}$

In addition,

$\frac{EH_1}{EH_2} = \frac{EH_4}{EH_3} = \frac{AE}{CE} = \frac{\tan{\frac{\widehat R}{2}}}{\tan{\frac{\widehat S}{2}}}$

$\frac{EH_2}{EH_3} = \frac{EH_1}{EH_4} = \frac{BE}{DE} = \tan{\frac{\widehat R}{2}} \tan{\frac{\widehat S}{2}}$

$\frac{EG_1}{EG_2} = \frac{EG_4}{EG_3} = \left(\frac{BE}{CE}\right)^2 = \tan^2{\frac{\widehat R}{2}}$

$\frac{EG_1}{EG_4} = \frac{EG_2}{EG_3} = \left(\frac{BE}{AE}\right)^2 = \tan^2{\frac{\widehat S}{2}}$

From similarity of the triangles $\triangle ECG_2 \sim \triangle EAO$, the line $g \equiv EG_2 \equiv EG_3$ is a reflection of the line $EO$ in the diagonal $BD$ and from similarity of the triangles $\triangle EDG_3 \sim \triangle EBO$, the line $g \equiv EG_3$ is a reflection of the line $EO$ in the diagonal $AC$. Denote $\phi = \measuredangle{CEG_2}$. Using the [color=blue]sine theorem[/color] for the triangle $\triangle ECG_2$,

$\frac{G_2C}{EG_2} = \frac{\sin{\widehat{CEG_2}}}{\sin{\widehat{ECG_2}}} = \frac{\sin \phi}{\sin{(90^o - \widehat B)}} = \frac{\sin \phi}{\cos{\widehat B}}$

By cyclic exchange of the vertices $A, B, C, D$ and $P, Q, R, S$,

$\frac{G_2C}{r_O} = \frac{G_2C}{OA} = \frac c d \cdot \frac{\cos{\frac{\widehat Q}{2}}}{\cos{\frac{\widehat P}{2}}} =  \frac b a \cdot \frac{\cos{\frac{\widehat R}{2}}}{\cos{\frac{\widehat S}{2}}}$

$\sin \phi = \frac{G_2C}{EG_2} \cos{\widehat B} = \frac b a \cdot \frac{\cos{\frac{\widehat R}{2}}}{\cos{\frac{\widehat S}{2}}} \cdot \frac{AE}{CE} \cdot \frac{\cos{\widehat B}}{\sqrt{1 - \sin{\widehat S} \sin{\widehat R}}} = \frac{\cos{\widehat B}}{\sqrt{1 - \sin{\widehat S} \sin{\widehat R}}}$

Denote $\psi = \measuredangle{BEH_2}$. Using the sine theorem for the triangle $\triangle BEH_2$,

$\frac{BH_2}{EH_2} = \frac{\sin{\widehat{BEH_2}}}{\sin{\widehat{EBH_2}}} = \frac{\sin \psi}{\sin{(90^o - \widehat B)}} = \frac{\sin \psi}{\cos{\widehat B}}$

By cyclic exchange of the vertices $A, B, C, D$ and $P, Q, R, S$,

$AH_1 = r_O \tan{\frac{\widehat R}{2}},\ \ BH_2 = r_O \tan{\frac{\widehat S}{2}}$

$\sin \psi = \frac{BH_2}{EH_2} \cos{\widehat B} = \tan{\frac{\widehat S}{2}} \cdot \frac{BE}{AE} \cdot \frac{\cos{\widehat B}}{\sqrt{1 - \sin{\widehat S} \sin{\widehat R}}} = \frac{\cos{\widehat B}}{\sqrt{1 - \sin{\widehat S} \sin{\widehat R}}}$

Thus the angle $\phi = \measuredangle{CEG_2}$ between the line $g \equiv G_1G_2G_3G_4$ and the diagonal $AC$ is equal to the angle $\psi = \measuredangle{BEH_2}$ between the line $h \equiv H_1H_2H_3H_4$ and the diagonals $BD$. Since the diagonals $AC \perp BD$ are perpendicular, the lines of collinearity $h \perp g$ are also perpendicular. Since the line $g$ is a reflection of the line $EO$ in either of the perpendicular diagonals $AC, BD$ of the contact quadrilateral $ABCD$ and since $AC \perp BD$ and $h \perp g$, the line $h$ is an isogonal of the line $EO$ with respect to the diagonals $AC, BD$.  :cool:

Yetti

There are additinal properties of this configuration:

    The third diagonals of the 3 complete quads ABCD, PQRS and
    XYZW - the tangential to PQRS  is a common line thru common focus of
    2 parabolas - the Miquel point.

    Tangents to parabolas at points of intersection with the third diagonal  pass
     thru point E.

    The common tangent to both parabolas is the radical axis of circles ABCD
    and PQRS.

    The quad XYZW  is not cyclic and ABCD is not inscriptible - so the process of
    constructing bicentric quads stops after 2 circles.

    If this is not a complete proof ( without words! ) of the
    Last Fermat Theorem, then I do not
    know what possibly could be (well,  Darij does not share my belief and places
    me firmly into that happy, although overpopulated, camp of the confused and
    accomplished Fermatists).

    Sgt.   Pepper

I did not really believe what I posted before about the 2 lines of collinear orthocenters being perpendicular, but it has been fixed by editing the previous post. Using the cosine theorem (of all things !), it is surprisingly easy to calculate the distances of all 8 orthocenters from the diagonal intersections and to get not only equal ratios for some of these distances, but the actual values of these ratios. Using these distances and the sine theorem (what else ?), it easily follows that the angle of the line $g = G_1G_2G_3G_4$ with the diagonal $AC$ is the same as the angle of the line $h = H_1H_2H_3H_4$ with the diagonal $BD$. Since the diagonals $AC \perp BD$ of the contact quadrilateral are perpendicular, the lines $h \perp g$ are also perpendicular.

Yetti.


linear algebra
matrix
superior algebra
superior algebra solved

Can we show, without too much advanced theory, that $M_{n}(\mathbb{R})$ and $M_{m}(\mathbb{R})$ are not isomorphic as rings? What if we replace the real numbers with another field?

Uhm, both centres are isomorphic to $\mathbb{R}$?

The dimensions over the centers are not the same.

Look at nilpotents. In $M_{n}(K)$, there is a nilpotent matrix $A$ with $A^{n-1}\neq0$, and $N^{n}=0$ whenever $N$ is nilpotent.


Try it:
Solve in reals:
$2x^2+6y^2+5z^2+2y+1=2x+4xz+8yz$


My hint:

If $0<x\le y\le z$ and $0<a\le b \le c$, then
$\frac{1}{a+y}+\frac{1}{b+z}+\frac{1}{c+x} \le \frac{1}{a+x}+\frac{1}{b+y}+\frac{1}{c+z}$.


trigonometry
calculus
calculus computations

can you solve  :lol:

$ y \cdot (y'' \plus{} y) \equal{} x\cdot \sin\; 2x$

I think it has [hide="4 solutions"]
$ f(x)\equal{}xsinx$
$ f(x)\equal{}\minus{}xsinx$
$ f(x)\equal{}ixcosx$
$ f(x)\equal{}\minus{}ixcosx$[/hide]

[quote="shape_shift2005"]I think it has [hide="4 solutions"]
$ f(x) \equal{} xsinx$
$ f(x) \equal{} \minus{} xsinx$
$ f(x) \equal{} ixcosx$
$ f(x) \equal{} \minus{} ixcosx$[/hide][/quote]

How did you come up with these solutions?


geometry
analytic geometry
geometry unsolved

Let $ ABC$ be a triangle,$ D$ is a point in $ [AC]$ such that $ BD \equal{} DC$.a parallel to $ (BD)$ meets $ [BC]$ in $ E$ and $ [AB]$ in $ F$.and $ (AE) \cap (BD) \equal{} G$.Prove that: $ \angle BCG \equal{} \angle BCF$
[u][i]Figure:[/i][/u]
[img]http://img194.imageshack.us/img194/1793/mmlikn.png[/img]

[hide="hint"] use $ (x,y)$ coordinates,sorry if it isn't nice...[/hide]

Let $ J \equal{} BC\cap GF$.
$ \frac {GD}{DC} \equal{} \frac {GD}{BD} \equal{} \frac {EH}{FH} \equal{} \frac {CH}{FH}$
Therefore $ \Delta GDC\sim \Delta CHF$
$ \Rightarrow \frac {GC}{FC} \equal{} \frac {GD}{CH} \equal{} \frac {GD}{EH} \equal{} \frac {AB}{AF} (1)$
On the other side, applying Menelaus's theorem for triangle $ AFG$ with line $ (BJE)$ we have:
$ \frac {JG}{JF}.\frac {BF}{BA}.\frac {EA}{EG} \equal{} 1\Rightarrow \frac {JG}{JF} \equal{} \frac {BA}{BF}.\frac {BF}{AF} \equal{} \frac {BA}{AF} (2)$
From $ (1)$ and $ (2)$ we are done.


Consider all the two element subsets $ (m,n)$ of $ (1,2,3,4,5,6,7)$ such that $ m < n$. If you choose two different sets $ (m_1,n_1)$ and $ (m_2,n_2)$ without replacement, what is the probability that $ m_1 \minus{} m_2 < n_1 \minus{} n_2$?

[hide="No case work I think"]Equivalent to the probability that $ n_1 \minus{} m_1 < n_2 \minus{} m_2$. We observe that the number of ways for $ n_1 \minus{} m_1$ to be less than $ n_2 \minus{} m_2$ is equal to the number of ways for $ n_1 \minus{} m_1$ to be greater than $ n_2 \minus{} m_2$. Thus, we can find the number of ways they can be equal, subtract that from the total possibilities, and divide by 2 to get the answer.

Suppose $ n_1 \minus{} m_1 \equal{} x$ (so $ 1 \leq x \leq 6$). We can find that there are $ 7 \minus{} x$ ways for this to happen by basic counting. So the number of choices for $ (m_1,n_1)$ so that they have difference $ x$ is $ 7 \minus{} x$, and the number of choices for $ (m_2,n_2)$ to have the same difference is $ 7 \minus{} x \minus{} 1 \equal{} 6 \minus{} x$ (minus $ 1$ since without replacement). So we find $ \sum_{i\equal{}1}^{6}(7\minus{}x)(6\minus{}x)$. We compute this to be $ 70$.

Now for the number of total ways to pick to sets. $ m$ can be any of seven elements, and $ n$ can be any of the $ 6$ others, so $ 7 \cdot 6$. But since $ m < n$, exactly one of $ (m,n)$ and $ (n,m)$ works. So we divide by $ 2$: thus there are $ 7 \cdot 6 / 2 \equal{} 21$ total ways to pick a two element subset. Then in picking the second one, we just have one less choice. So the total number of ways to pick $ (m_1,n_1)$ and $ (m_2,n_2)$ is $ 21 \cdot (21 \minus{} 1) \equal{} 420$.

$ 420 \minus{} 70 \equal{} 350$ is the total number of ways to pick two sets where $ n_1 \minus{} m_1 \neq n_2 \minus{} m_2$. Exactly half of them satisfy the condition, so $ 350/2 \equal{} 175$ is the total number of ways. The probability is then $ 175/420 \equal{} \boxed{5/12}$.[/hide]


Let $ a,b,c,k$ be nonnegative real numbers. Prove that
\[ \frac{a^2\plus{}bc}{b\plus{}kc}\plus{}\frac{b^2\plus{}ca}{c\plus{}ka}\plus{}\frac{c^2\plus{}ab}{a\plus{}kb} \ge \frac{2}{k\plus{}1}(a\plus{}b\plus{}c)\]

Remark. For $ k\equal{}1$, we get the known inequality if Hojoo Lee
\[ \frac{a^2\plus{}bc}{b\plus{}c}\plus{}\frac{b^2\plus{}ca}{c\plus{}a}\plus{}\frac{c^2\plus{}ab}{a\plus{}b} \ge a\plus{}b\plus{}c\]
:)

I don't if my approach is correct , but please correct me if I am mistaken.
If we plug $ k\to \frac{1}{k}$ in the inequality above :\[ \frac{a^2\plus{}bc}{b\plus{}kc}\plus{}\frac{b^2\plus{}ca}{c\plus{}ka}\plus{}\frac{c^2\plus{}ab}{a\plus{}kb} \ge \frac{2}{k\plus{}1}(a\plus{}b\plus{}c)\] we get \[ \frac{a^2\plus{}bc}{kb\plus{}c}\plus{}\frac{b^2\plus{}ca}{kc\plus{}a}\plus{}\frac{c^2\plus{}ab}{ka\plus{}b} \ge \frac{2}{k\plus{}1}(a\plus{}b\plus{}c)\]
So if we prove the inequality for $ k\geq 1$  , I think that in a similar way we can also prove for $ 0 \leq k\leq 1$.
Now w.l.o.g. we take $ k \geq 1$.From Cauchy we obtain:\[ \left( \frac{a^2\plus{}bc}{b\plus{}kc}\plus{}\frac{b^2\plus{}ca}{c\plus{}ka}\plus{}\frac{c^2\plus{}ab}{a\plus{}kb}  \right)\left( \sum (a^{2}\plus{}bc)(b\plus{}kc) \right) \geq \left( \sum a^{2}\plus{} \sum bc \right)^{2}\]
It is enough to prove that
\[ \left( \sum a^{2}\plus{} \sum bc \right)^{2} \geq \frac{2}{k\plus{}1}\left( \sum (a^{2}\plus{}bc)(b\plus{}kc) \right) (a\plus{}b\plus{}c)\]
The left side is equal to \[ \sum a^{4} \plus{}3\sum b^{2}c^{2} \plus{} 2\sum a^{3}(b\plus{}c) \plus{} 4 \sum a^{2}bc\]
The right side is equal to \[ \frac{4}{k\plus{}1} \left(\sum_{cyc}a^{3}b\plus{}k\sum_{cyc} ab^{3}\plus{}(k\plus{}1)\sum b^{2}c^{2}\plus{}(k\plus{}1)\sum a^{2}bc\right)\]
After some simplifications the inequality can be writen in the form
\[ (k\plus{}1) \left( \sum a^{4}\minus{}\sum b^{2}c^{2} \plus{} 2\sum ab(a^{2}\minus{}b^{2}) \right) \geq 4\sum ab(a^{2}\minus{}b^{2})\]
Because $ k\geq 1$ the rest depends on proving that \[ \sum a^{4}\minus{}\sum b^{2}c^{2} \geq 0 \Longleftrightarrow \sum (a^{2}\minus{}b^{2})^{2} \geq 0\]

If my first statement is incorrect , at least I can say I managed to prove the inequality for $ k\geq 1$.


Alright, so let's get going on this important issue. Based on the problems in Day 1, I'm thinking at this point a 35-27-18 is a good cutoff.

I would lower that... Based on D1:

33-24-18

34-22-15
I believe last year's was like easy as this years
so cutoffs will be similar.

34-25-17. Indeed, D1 was surprisingly easy, but I don't think that it'll take the cut-offs sky high.

okay, I am ready to take my prediction back - the second day turned out to be slightly harder than the first...

here is my new prediction -

35 - 25 - 18

I think:

[size=150]Bronze 16
Silver 24
Gold 33[/size]

After again viewing and analysing problems, I must agree with Valentin on higher... Problem 6 wasn't hard at all...

35-26-20

Well... Sometimes it isn't how easy the problem is, it's always how well-covered the problem is, and of course, how most people feel about it.

Hope that the silver cutting will be not higher than 24... I want a silver this year... deep in my heart!

I propose an idea to determine who's prediction is more accurate, it is a formula we used to gamble cutoffs in IbMO:
Let $ x, y$ and $ z$ your predictions for gold silver and bronze, and let $ x_0 , y_0$ and $ z_0$ be the real cutoff, who minimizes the following expresions wins
$ 4(x\minus{}x_0)^2\plus{}3(y\minus{}y_0)^2\plus{}2(z\minus{}z_0)^2$

33-24-15 <- Silver editted to allow me get Silver  :D

my prediction 17/23/33
and
my friend's prediction 16/24/33

[quote="chichi"]I propose an idea to determine who's prediction is more accurate, it is a formula we used to gamble cutoffs in IbMO:
Let $ x, y$ and $ z$ your predictions for gold silver and bronze, and let $ x_0 , y_0$ and $ z_0$ be the real cutoff, who minimizes the following expresions wins
$ 4(x \minus{} x_0)^2 \plus{} 3(y \minus{} y_0)^2 \plus{} 2(z \minus{} z_0)^2$[/quote]

Nice. So you do the calculations :D

My prediction
             BRONZE 15,
             SILVER 23,
             GOLD    31

15 - 23 - 30

seems reasonable cutoffs for me

Well, after pbornzstein's post in Problem 6 discussion topic, it seems that the gold edge might even be 30,31,32

[quote="Teppic"][quote]Gold cut-off must be greater than 35![/quote]
With less than 10 people with a nonzero score in problem 6?[/quote]

However, most powerful (enough) students can successfully achieve 35 by finishing T1~T5.

[quote="ZHANGWENZHONGKK"]
However, most powerful (enough) students can successfully achieve 35 by finishing T1~T5.[/quote]

Then it's no longer "greater" :) But anyway.. I seriously doubt there'll be 30-40 people getting 35 points..

[quote="Teppic"][quote]Gold cut-off must be greater than 35![/quote]
With less than 10 people with a nonzero score in problem 6?[/quote]Is this "official"? Wow :)

No, I think pbornzstein said it was a bit more... But less than 20
http://www.mathlinks.ro/viewtopic.php?p=1565600#1565600

[quote="jiahan"]Do they award silver medals to the minimum number of contestants that > 1/4 of the total, or the maximum number that < 1/4?[/quote]

Well.. the answer is no..
They give medals to at most half of the contestants, and they try to do so, so that the ratio of gold:silver:bronze is as close to 1:2:3 as possible.

Official results [url=http://www.imo-official.org/year_info.aspx?year=2009]are published[/url].

  darij

Getting such low index isn't the greatest, the greatest thing of all is that, I got 24, and 24 is exactly the silver cutting XD.

Er, a lower index meant a better prediction. :P

EDIT: Oh, I see.

He meant it was great, but hitting the cut off was greater :D

[quote="MellowMelon"]So the actual cutoffs were 32, 24, 14. Using a slightly modified version of chichi's metric, the index is $ \sqrt {4(g_p \minus{} g_a)^2 \plus{} 3(s_p \minus{} s_a)^2 \plus{} 2(b_p \minus{} b_a)^2}$ where $ g,s,b$ denote gold silver bronze and $ p,a$ subscripts denote predicted, actual. The top 5 predictions in this thread:

$ \begin{tabular}{|rc|c|c|c|c|} \hline & & GOLD & SILVER & BRONZE & INDEX \\
\hline \hline 1st & mszew & 32 & 24 & 15 & 1.4 \\
2nd & KenHungKK & 33 & 24 & 15 & 2.4 \\
3rd & cefer & 32 & 24 & 16 & 2.8 \\
4th & Kiki & 31 & 23 & 15 & 3.0 \\
5th & TomciO & 32 & 23 & 16 & 3.3 \\
5th & ggbg90 & 32 & 25 & 16 & 3.3 \\
\hline \end{tabular}$

Hopefully didn't make any typos for them or anyone else.[/quote]

 :D Maybe I should have learned these skills (prediction ones) when I was participating on the Imo,
or at least have this amazing forum to have access to more olympiad level problems to improve my performance with the practice... However I will try to contribute to the next Imo 2012 in Argentina, As I did during the 1997 one.

IMO in Argentina is 2012, 2011 is in Netherlands.


calculus
integration
trigonometry
function
calculus computations

Please help me with this problem. Thank you.
Calculate:
\[ \int\limits_{\minus{}1}^{1}\frac{\sqrt{1\minus{}x^2}}{x\plus{}a}\ \ dx\]
a is a constant

Let $ x = \sin \theta \left( - \frac {\pi}{2}\leq \theta \leq \frac {\pi}{2}\right)$,

$ \int_{ - 1}^1 \frac {\sqrt {1 - x^2}}{x + a}\ dx$

$ = \int_{ - \frac {\pi}{2}}^{\frac {\pi}{2}} \frac {|\cos \theta|}{\sin \theta + a}\cos \theta \ d\theta$

${ = \int_{ - \frac {\pi}{2}}^{\frac {\pi}{2}}} \frac {\cos ^ 2 \theta}{\sin \theta + a}\ d\theta \ \because - \frac {\pi}{2}\leq \theta \leq \frac {\pi}{2}\Longrightarrow \cos \theta \geq 0.$

$ = \int_{ - \frac {\pi}{2}}^{\frac {\pi}{2}} \frac {1 - \sin ^ 2 \theta}{\sin \theta + a}\ d\theta$

$ = \int_{ - \frac {\pi}{2}}^{\frac {\pi}{2}} \left(a - \sin \theta + \frac {1 - a^2}{\sin \theta + a}\right)\ d\theta$

$ = \pi a + \int_{ - \frac {\pi}{2}}^{\frac {\pi}{2}}\frac {1 - a^2}{\sin \theta + a}\ d\theta \ \because \sin \theta \ \text{is an odd function}$

Now set $ \tan \frac {\theta}{2} = t\ \left(-\frac {\pi}{2} < \phi < \frac {\pi}{2}\right)\Longrightarrow \frac {1}{2\cos ^ 2 \frac {\theta}{2}} d\theta = dt\Longleftrightarrow \frac 12(1 + t^2)d\theta = dt$, yielding $ d\theta = \frac {2}{1 + t^2}dt$,

$ \int_{ - \frac {\pi}{2}}^{\frac {\pi}{2}} \frac {1 - a^2}{\sin \theta + a}\ d\theta$

$ = (1 - a^2)\int_{ - 1}^1 \frac {1}{\frac {2t}{1 + t^2} + a}\cdot \frac {2}{1 + t^2}\ dt$

$ = 2(1 - a^2)\int_{ - 1}^1 \frac {1}{at^2 + 2t + a}\ dt$

Afterwards, we are to classify several cases: $ a = 0$ or $ a\neq 0$ or $ |a|\leq 1$ or $ |a| > 1.$

[quote="kunny"]Afterwards, we are to classify several cases: $ a \equal{} 0$ or $ a\neq 0$ or $ |a|\leq 1$ or $ |a| > 1.$[/quote]
If $ |a| < 1,$ then the integral diverges, since the integrand looks like $ \frac {C}{x \plus{} a}$ for some nonzero $ C$ in a neighborhood of $ \minus{} a.$


Τι θα λεγατε να βαζουμε σε ενα τοπικ καποια θεματα απο την ΕΜΕ ,
κατα προτιμηση  απο διαγωνισμους πριν το 98,δυστηχως δεν εχω παρα ελαχιστα και θα ηθελα πολυ να τα βρω αφου ειναι πολυ ωραια ,πιστευετε οτι  ειναι δυνατο?

exo merika an 8elis kosta...epidi ta exo idi saromena ston ipologisti.....mipos kses pos mporo  na ta stilo san sinimeno arxio...??

 :D

ευχαριστω Ραλλη,οχι δυστηχως δεν ξερω πως να τα στειλεις .αν θες προτεινε κανα δυο εδω στο φορουμ  :)

Δοκίμασε να ανατρέξεις σε θέματα προηγούμενων χρόνων από διαγωνισμούς άλλων χωρών. Όλο και σε κάποιο παρόμοιο (αν όχι πανομοιότυπο) θέμα θα πέσεις

 :D

πιθανον να ισχυει , αλλα δεν θα το κανω  :D
δυστηχως δεν ειναι ευκολο να τα βρεις οπως πχ τα θεματα της Αμερικανικης ολυμπιαδας

[quote="silouan"]Μπορείτε να τα βρείτε στο cd  με όλα τα τεύχη του περιοδικού Ευκλείδης.... :wink:[/quote]
εχει τευχη απο το 90 και μετα αλλα και παλι δεν γινεται στα windows xp ,

[b]telikos esoterikos diagonismos gia tin 37 imo 1996[/b]

1) esto s_{n} to a8risma ton n proton,proton ari8mon.na apodiksete oti metaksi ton s_{n} kai s_{n+1} iparxi ena tulaxiston telio tetragono fisiku.

2)esto oksigonio trigono abc egegrameno se kiklo k kai ta iyi tu aa1,bb1,cc1 pu temnonte sto H.esto akoma ε i efaptomeni tu K sto a kai i ka8etes bb2,cc2 stin ef8ia ε.( ta b2,c2 simia tis ε).na apodiksete oti i ef8ies b1b2,c1c2,aa1 sintrexun.

3)a)na prosdiorisete 4 omades fisikon ari8mon pu ka8e omada periexi 4 ari8mus oste ka8e fisikos ari8mos n (0<=n<=225)
 na ine a8risma 4 ari8mon,enos apo ka8e omada.
b)ka8e atomo apo mia omada n atomon ,simpa8i k atoma apo tin omada.pia ine i megaliteri timi tu n gia dosmeno k,oste na iparxun tulaxiston 2 atoma stin oomada pu simpa8unte amivea?

4)[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=114094[/url]  i askisi 2....

 :D

[quote="RAKAR"]

1) esto s_{n} to a8risma ton n proton,proton ari8mon.na apodiksete oti metaksi ton s_{n} kai s_{n+1} iparxi ena tulaxiston telio tetragono fisiku.

 :D[/quote]

καμια ιδέα για αυτη την πανεμορφη ασκηση ?

Για την άσκηση 1 μπορείτε να δείτε μια λύση εδώ:
http://www.mathlinks.ro/Forum/viewtopic.php?t=66311

Και για μια γενίκευσή της μπορείτε να δείτε εδώ:
http://www.mathlinks.ro/Forum/viewtopic.php?p=320842

[quote="RAKAR"][b]telikos esoterikos diagonismos gia tin 37 imo 1996[/b]

...

2)esto oksigonio trigono abc egegrameno se kiklo k kai ta iyi tu aa1,bb1,cc1 pu temnonte sto H.esto akoma ε i efaptomeni tu K sto a kai i ka8etes bb2,cc2 stin ef8ia ε.( ta b2,c2 simia tis ε).na apodiksete oti i ef8ies b1b2,c1c2,aa1 sintrexun.

...

[/quote]

Για ευκολία, έχει αλλαχτεί η ονοματολογία των διαφόρων σημείων στο σχήμα.

Είναι γνωστό ότι οι ευθείες $B'C',$ $B''C'',$ είναι παράλληλες μεταξύ τους, ως αντιπαράλληλες της πλευράς $BC,$ ως προς τις ευθείες $AB,$ $AC,$ του $\bigtriangleup ABC.$

Επειδή η ευθεία $B''C'',$ είναι εφαπτομένη του περιγεγραμμένου κύκλου $(O),$ του $\bigtriangleup ABC$ στην κορυφή $A,$ προκύπτει ότι $\angle ACB = \angle BAB''$ $\Longrightarrow$ $\angle CAA' = \angle ABB''$ $,(1)$

Από το εγγεγραμμένο τετράπλευρο $AA'BB''$ $\Longrightarrow$ $\angle ABB'' = \angle AA'B''$ $,(2)$

Από $(1),$ $(2)$ $\Longrightarrow$ $\angle CAA' = \angle AA'B''$ $,(3)$

Από $(3)$ $\Longrightarrow$ $A'B''\parallel AC$ και με παρόμοιο τρόπο $A'C''\parallel AB.$

Άρα τα τρίγωνα $\bigtriangleup A'B''C''$ and $\bigtriangleup AB'C',$ έχουν τις πλευρές τους παράλληλες μία προς μία και επομένων είναι προοπτικά.

Σύμφωνα με το θεώρημα του Desarques, συμπεραίνουμε ότι οι ευθείες $AA',$ $B'B'',$ $C'C'',$ τέμνονται στο ίδιο σημείο και η πρόταση έχει αποδειχθεί.

Κώστας  Βήττας.


real analysis
combinatorics proposed
combinatorics

Consider words of length $ n$ in four letters, $ a,b,c$ and $ d$. Define a convolution in a word as a contiguous repetition of the same block, for instance $ aa$ is a convolution and so is $ abab$. Call a word convolution free if it does not contain any convolutions.
Prove that the number of convolution free words of length $ n$ is atleast $ 2^n$.

It is Romania TST 2003, and can be found in the Resources.
Also at [url]http://www.mathlinks.ro/viewtopic.php?p=118745#118745[/url]

The technical term is [color=red]squarefree words[/color].

The bound seems far from sharp, since [url]http://mathworld.wolfram.com/SquarefreeWord.html[/url] says
[quote]The number of squarefree quaternary words of length $ n \equal{} 1, 2, ...$ are $ 4, 12, 36, 96, 264, 696, ...$
(Sloane's [url=http://www.research.att.com/~njas/sequences/A051041]A051041[/url]).[/quote]


1.An estate was divided among three heirs. After Andrea received 20% of the estate, her mother received 50% of what remained, leaving 20,000 for Andrea's uncle. What was the total value of the estate?

2. ABCD is a square and AE=AF=CG=CH. Given AB=5 and the shaded region is five-ninths of the are of ABCD, find AF Express your answer as a common fraction

3. Stacey can trim the shrubbery in six hours working alone. Her father can do it in five hours. They worked together until dinner but trimmed 11/15 of the shrubbery. How many hours did they work together?

4. In a certain sequence of numbers, the sum of any three consecutive terms is 3. If the second term is 4 and the fourth term is -2, what is the 18th term?

5.If 3 is equal to or less than Y and Y is less than or equal to less than 22, for how many ordered pairs of positive integers (x,y) does Y exceed X by at least 2 ?

here is pic for #2

1. Andrea got 20%, her mother got $ 0.8 \times 0.5 \equal{} 40$ percent, so her uncle got 40 percent. The problem says that he got 20000 dollars, so the total value of the estate is $ 20000 \div 0.4 \equal{} 50000$.

#2 is obviously impossible w/o a pic.

3. They can do 1/6+1/5=11/30 per hour. (11/15)/(11/30)=2 hours.

#4- The third number is obviously $ 1$.

$ 4$, $ 1$, $ \minus{}2$

$ 1$, $ \minus{}2$, $ 4$

$ \minus{}2$, $ 4$, $ 1$

$ 4$, $ 1$, $ \minus{}2$

It's a pattern!!!

Thus, find $ 16 \mod{3}$, which is $ 2$.  Thus, the answer is $ \boxed{4}$.

$ 3\le Y \le 22$ and $ Y\ge X\plus{}2$
For $ Y\equal{}3$ $ X$ has one value, for $ Y\equal{}4$ it has 2 values, . . . , for $ Y\equal{}22$ it has 20 values.
$ 1\plus{}2\plus{}...\plus{}20\equal{}\frac{20\cdot 21}{2}\equal{}210$

Please don't now try to explain some parts of my solution to get more posts.

mewto, isn't there a 'download' thing of the pic for #2 at the bottom?

I won't explain Bugi's post, just this part:
$ 1\plus{}2\plus{}3...\plus{}n\equal{}\frac{n(n\plus{}1)}{2}$

This is also expressed as $ \displaystyle\sum_{k\equal{}1}^{n}\equal{}\frac{n(n\plus{}1)}{2}$

As for #2, let $ AE\equal{}AF\equal{}CG\equal{}CH\equal{}x$
Then, $ 5^2\minus{}2\left(\frac{1}{2}(5\minus{}x)(5\minus{}x)\right)\equal{}\frac{5}{9}5^2$ (The area of the shaded region is equal to the area of the square - the area of the triangles.)

$ 25\minus{}x^2\plus{}10x\minus{}25\equal{}\frac{125}{9}$

$ 0\equal{}x^2\minus{}10x\plus{}\frac{125}{9}$

$ 0\equal{}9x^2\minus{}90x\plus{}125$

$ x\equal{}\frac{\minus{}(\minus{}90)\pm\sqrt{(\minus{}90)^2\minus{}4(9)(125)}}{2(9)}$

$ x\equal{}\frac{90\pm\sqrt{8100\minus{}4500}}{18}$

$ x\equal{}\frac{90\pm\sqrt{3600}}{18}$

$ x\equal{}\frac{90\pm60}{18}$

$ x\equal{}\frac{90\plus{}60}{18}\equal{}\frac{150}{18}\equal{}8\frac{1}{3}$
OR
$ x\equal{}\frac{90\minus{}60}{18}\minus{}\frac{30}{18}\equal{}1\frac{2}{3}$

Clearly, the answer must be less than 5, so the latter is the answer.

Let $ AE\equal{}AF\equal{}CG\equal{}CH\equal{}5\minus{}x$ such that we have $ 2[BEH]\equal{}5^2\minus{}\frac595^2\implies x^2\equal{}\frac{100}9$. Thus, $ x\equal{}\frac{10}3$ and our answer is $ 5\minus{}x\equal{}\boxed{\frac53}$.

The other case is similar too (if the triangles are shaded). Then, we would just have $ x^2\equal{}\frac{125}9$, so $ x\equal{}\frac{5\sqrt5}3$ and $ 5\minus{}x\equal{}\boxed{\frac{15\minus{}5\sqrt5}3}$.

[quote="shortstack28"]
2. ABCD is a square and AE=AF=CG=CH. Given AB=5 and the shaded region is five-ninths of the are of ABCD, find AF Express your answer as a common fraction
[/quote]

What region is shaded?

[quote="ernie"]#4- The third number is obviously $ 1$.

$ 4$, $ 1$, $ \minus{} 2$

$ 1$, $ \minus{} 2$, $ 4$

$ \minus{} 2$, $ 4$, $ 1$

$ 4$, $ 1$, $ \minus{} 2$

It's a pattern!!!

Thus, find $ 16 \mod{3}$, which is $ 2$.  Thus, the answer is $ \boxed{4}$.[/quote]

Isn't it the $ 18^{th}$ number?  Then it would be $ 18 \mod {3}$, which is $ \boxed{1}$

On #5, where does the $ x$ come in.  Why does it say "less than or equal to less than"?  (typo?)  Is the second Y supposed to be $ x$?

The answer is $ 1$. And for #5 "if $ 3\leq Y\leq22$, for how many ordered pairs of positive integers $ (X,Y)$ does $ Y$ exceed $ X$ by at least $ 2$?", it simply means that find the number of pairs such that $ X$ is at least $ 2$ less than $ Y$. As for your previous post, refer above.

[quote="ernie"]#4- The third number is obviously $ 1$.

$ 4$, $ 1$, $ \minus{} 2$

$ 1$, $ \minus{} 2$, $ 4$

$ \minus{} 2$, $ 4$, $ 1$

$ 4$, $ 1$, $ \minus{} 2$

It's a pattern!!!

Thus, find $ 16 \mod{3}$, which is $ 2$.  Thus, the answer is $ \boxed{4}$.[/quote]

The first number is $ \minus{} 2$, so ajain is correct, the answer is $ \boxed{\boxed{\boxed{1}}}$.


This problem is taken from the Italian selection for the National Competition of the Math Olympiad...

Find all the pairs (m, n) of positive integers such that

\[ \frac{3^{m} + 3}{2^{n} + 2^{n-1}} \]
Is an integer number.

We have
${3^{m-1}+1 \over 2^{n-1}}$

but
$3^{2k+1} =3 \pmod 8$
$3^{2k}=1 \pmod 8$

so we have $n\leq 3$
for
$n=3$ it is easy to verify that  $m=2k$ is good
for
$n=2$ it is easy to verify that all $m$ is good
for
$n=1$ it is easy to verify that all $m$ is good

I understood nothing lol

where did the (3^m-1   +  1) come from?

and where did mod8 come from??
I know nothing...


Convert the following Riccati equation into a Bernoulli equation:

dy/dx = xy^2 + y + (1/x^2) where y = -1/x is a known solution to the equation.

Here is what I did so far. I used the substitution y = w - (1/x) and thus came up with the following equation:

dw/dx = x(w^2 - (2w/x) + (1/x^2)) + (w - (1/x)) + (1/x^2). So after all the Algebra which I have checked several times (although dumb ol me could still have made a mistake), I end up getting

dw/dx - w = xw^2 - 2w + (1/x^2).

This WOULD be a Bernoulli equation if it wasn't for that darned (1/x^2) term. I figure this must cancel out somewhere but so far I don't find any flaws in my Algebra. Where did I go wrong?

hello, we want to solve the differential equation
$ \frac{dy}{dx}=xy^2+y+\frac{1}{x^2}$. By substituting $ y=\frac{1}{z}-\frac{1}{x}$ we get by differentiating $ y^'=-\frac{z^'}{z^2}+\frac{1}{x^2}$. Now has our equation the following form:
$ -\frac{z^'}{z^2}+\frac{1}{x^2}=x\left(\frac{1}{z}-\frac{1}{x}\right)^2+\frac{1}{z}-\frac{1}{x}+\frac{1}{x^2}$, simplifying this we have:
$ 0=z^'-z+x$. This equation is linear and has the solution
$ z(x)=Ce^x+x+1$. From here follows our solution of the Riccati-equation from above:
$ y(x)=-\frac{Ce^x+1}{x(Ce^x+x+1)}$.
Sonnhard.

[quote="ROLEX4life"]

$ \frac {dw}{dx} \equal{} x\left(w^2 \minus{} 2\frac {w}{x} \plus{} \frac {1}{x^2}\right) \plus{} w \minus{} \frac {1}{x} \plus{} \frac {1}{x^2}$

[/quote]

If I'm not mistaken, the error is here, because it should be

$ \frac {dw}{dx} \plus{} \frac {1}{x^2} \equal{} x\left(w^2 \minus{} 2\frac {w}{x} \plus{} \frac {1}{x^2}\right) \plus{} w \minus{} \frac {1}{x} \plus{} \frac {1}{x^2}$

$ \left(\frac {d(w \minus{} x^{ \minus{} 1})}{dx} \equal{} \frac {dw}{dx} \minus{} \frac {d(x^{ \minus{} 1})}{x} \equal{} \frac {dw}{dx} \plus{} \frac {1}{x^2}\right)$

i.e.

$ \frac {dw}{dx} \equal{} x\left(w^2 \minus{} 2\frac {w}{x} \plus{} \frac {1}{x^2}\right) \plus{} w \minus{} \frac {1}{x}$


Dose anyone have this book or source to download,please tell me
Thank for your help :help:

[quote="mathangel"]It 's not for free. If you are from an IHL (Institute of Higher Learning) then search its digital library. Otherwise, approach the National/City Library. In worst case, buy it online.[/quote]
Did you mean the book by Sierpinski?
As far as I understand, you're from Russia, so it seems plausible that there are plenty of different russian web-sites with opportunity to download this book...for instance:
http://www.math.ru/lib/book/djvu/serp-250-tch.djvu


That's a good question.

Technically, saying that a person can exert a maximum force of $ F$ implies that if that person were holding a rope and another person person pulled the other end of the rope with a force greater than $ F$, the first person wouldn't be able to hold on.  Therefore, I would say that the tension in the rope increases until it becomes $ F_B$, at which point person B can't hold on anymore and lets go, reducing the tension in the rope to zero.

Each person would experience an acceleration towards the other, and the rope would immediately go limp. The source of the acceleration would be the force the rope exerts on the person, and a limp rope exerts no such force.

If the rope is slightly elastic, a small but positive time elapses between the beginning of the force and the rope going limp. Each person experiences a change in momentum given by the impulse, or force times time, exerted by the rope. By conservation of momentum, the two impulses must be the same; therefore the only consequence of the one person exerting a greater force is that the time until the rope goes limp on his end is less, making the impulse the same. The two will collide with each other at their combined center of mass.

If we make that a rigid pole of positive mass rather than a rope, then the we could have one person exert a stronger force than the other for a significant time; the result would be that the pole accelerates towards the person who pulls more strongly. If the pole is light compared to the people, its acceleration is quite rapid compared to the accelerations of the people. The person who pulls less strongly sees the pole move away from him, until at the full extension of his arms, it is pulled out of his hands. (If it isn't pulled out of his hands, then he is exerting a greater force on it - or more to the point, the pole is exerting that greater force on him.) The two people collide at a point that is lies on the weaker person's side of their original center of mass, and with a net momentum after the collision (if they remain together) that drives the stronger person forward and the weaker person backwards - all compensated by the pole flying rapidly off into space behind the stronger person.

Yes, yes, that's right.

Edit: Oh yeah, I was imagining that each person would be pulling him/herself along the rope as they moved toward their center of mass, hence continually exerting a force over time rather than an instantaneous one - in that case, the tension in the rope would increase as I described, albeit very quickly.

Edit again: Actually, since the rope is a massless one, "very quickly" would mean "instantaneously," so the weaker person would immediately have to let go.

[quote="Kent Merryfield"]By conservation of momentum, the two impulses must be the same; therefore the only consequence of the [b]one person exerting a greater force[/b] is that the time until the rope goes limp on his end is less, making the impulse the same.[/quote]
However, I thought the two people actually exert the same force since we have $ F_A\equal{}F_B$?

Well I mean, person B cannot exert $ F_A$ since the max he can pull is $ F_B<F_A$. So I was thinking that person A actually does not pull as hard as he can, even though he tries (he is only pulling with force $ F_B$).

[quote="aspiringchemlete"]
However, I thought the two people actually exert the same force since we have $ F_A \equal{} F_B$?

Well I mean, person B cannot exert $ F_A$ since the max he can pull is $ F_B < F_A$. So I was thinking that person A actually does not pull as hard as he can, even though he tries (he is only pulling with force $ F_B$).[/quote]

That is exactly right - each person is acted on by the same impulse for the same amount of time due to the tension in the rope, which never exceeds $ F_b$.  The impulses are the same not because the stronger person pulls the rope for a shorter time (since they pull for the same time) but because he's never able to exert as much force as he would be able to.

I think it helps to think of each person gradually pulling the rope harder rather than suddenly exerting their maximum forces to start with, and that would be more plausible anyway.

[quote="aspiringchemlete"]Say you had two people, pulling on opposite sides of a massless rope in outer space. Person A can exert a maximum force of $ F_A$ and person B can exert a max force of $ F_B$ where $ F_A > F_B$. They both pull as hard as they can.

If you consider the rope, since it is massless $ F_B \minus{} F_A \equal{} 0$ and so $ F_A \equal{} F_B$. What exactly is the interpretation of this, since we assumed person A can pull harder than person B?[/quote]

[hide="Consider"]

If there is no net torque on the tug o' warriors then

\[ \sum \tau\colon m_{A}(x_{A}\minus{}x_{CM})a_{n} \plus{} m_{B}(x_{B}\minus{}x_{CM})a_{n} \equal{} 0\]

The acceleration $ a_{n}$ is the acceleration of gravity due to the nearest massive object. In space $ a_{n}$ would be minute, but not identically zero. Therefore we can cancel it, leaving,
\[ m_{A}(x_{A}\minus{}x_{CM}) \plus{} m_{B}(x_{B}\minus{}x_{CM}) \equal{} 0 \\ \\ \Rightarrow m_{A}x_{A} \plus{}m_{B}x_{B} \equal{} (m_{A} \plus{} m_{B})x_{CM}\]

$ x_{CM}$ is the coordinate of the center of mass.

We can differentiate with respect to time (because of the conservation of mass, the change of mass with respect to time is zero) twice:
\[ m_{A}\frac{\mathrm{d}^2 x_{A}}{\mathrm{d}t^2} \plus{} m_{B}\frac{\mathrm{d}^2 x_{B}}{\mathrm{d}t^2} \equal{} (m_{A} \plus{} m_{B})\frac{\mathrm{d}^2 x_{CM}}{\mathrm{d}t^2}\]

If we now let,
\[ a_{A} \equal{}\frac{\mathrm{d}^2 x_{A}}{\mathrm{d}t^2} \\ \\ a_{B} \equal{} \minus{} \frac{\mathrm{d}^2 x_{B}}{\mathrm{d}t^2}\\ \\ \&\quad a_{CM} \equal{} \frac{\mathrm{d}^2 x_{CM}}{\mathrm{d}t^2}\]

then

\[ m_{A}a_{A} \minus{} m_{B}a_{B} \equal{} (m_{A} \plus{} m_{B})a_{CM} \\ \\ \therefore a_{CM} \equal{} \frac{F_{A} \minus{} F_{B}}{m_{A} \plus{} m_{B}}\]
[/hide]