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circumcircle
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geometry proposed

Dear Mathlinkers,
let ABC be a triangle rectangular at A, (0) the circumcircle of ABC,
(1) the circle centered at B and cutting the segment BC at Q,
(2) the circle centered at C and cutting the segment BC at P,
D the second point of intersection of AQ with (0),
(3) the circle passing through A, P, D and
E the second point of intersection of BC with (3).
Prove that PQ = 2.CE.
Sincerely

Dear [b][size=100]Jean Louis[/size][/b], I guess that you mean the circle centered at $ B$ with radius $ BA$ and similarly, the circle centered at $ C,$ with radius $ CA.$

So, it is enough to to prove that $ (PQ)^{2} \equal{} 2(BP)(QC)$ $ ,(1)$ which is well known result, in the configuration of a right triangle $ \bigtriangleup ABC,$ with $ \angle A \equal{} 90^{o}$ and two points on the side-segment $ BC,$ such that $ BQ \equal{} BA$ and $ CP \equal{} CA.$

If I am not mistaken, I will post here later the proof of $ (1)$ I have in mind, based on the fact that the circles $ (B)$ ( centered at $ B$ with radius $ BA$ ) and $ (C)$ ( centered at $ C$ with radius $ CA$ ), are orthogonal each other and then, the points $ F,\ P,\ Q,\ Z,$ where $ F\equiv (B)\cap BC$ ( the other than $ Q$ ) and $ Z\equiv (C)\cap BC$ ( the other than $ P$ ), are in harmonic conjugation.

Kostas Vittas.

Dear Kostas,
sorry for my forgetfullness... You are right.
Is it possible to prove first the result of my problem, the to deduct the relation (of Ramanujan) you have mentionned?
Sincerely
Jean-Louis

Let me dear [b][size=100]Jean Louis[/size][/b] to restate the notations of your nice problem.

[b][size=100][color=DarkBlue]Let $ \bigtriangleup ABC$ be a right triangle with $ \angle A \equal{} 90^{o}$ and circumcircle $ (O).$ We denote the points $ P,\ Q$ on the side-segment $ BC,$ such that $ BQ \equal{} BA$ and $ CP \equal{} CA$ and let be the point $ D\equiv (O)\cap AQ.$ The circumcircle $ (O_{1})$ of the triangle $ \bigtriangleup APD,$ intersects the sideline $ BC$ of $ \bigtriangleup ABC,$ at point so be it $ E.$ Prove that $ PQ \equal{} 2(CE).$[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] - We draw the circles $ (B),\ (C),$ centered at verices $ B,\ C$ of the given triangle $ \bigtriangleup ABC,$ with radii $ BA \equal{} BQ$ and $ CA \equal{} CP$ respectively and we denote the points $ F\equiv (B)\cap BC$ ( the other than $ Q$ ) and $ Z\equiv (C)\cap BC$ ( the other than $ P$ ).

From the powers of the point $ Q,$ with respect to the circles $ (O),\ (O_{1}),$ we have $ (PQ)(QE) \equal{} (AQ)(QD) \equal{} (BQ)(QC)$ $ ,(1)$

From $ (1)$ $ \Longrightarrow$ $ (PQ)(QC \plus{} CE) \equal{} (BP \plus{} PQ)(QC)$ $ \Longrightarrow$ $ (PQ)(CE) \equal{} (BP)(QC)$ $ ,(2)$

From $ (2),$ in order to be $ PQ \equal{} 2(CE)$ it is enough to prove $ (PQ)^{2} \equal{} 2(BP)(QC)$ $ ,(3)$ ( I didn't know that this relation, is a result by the famous [b][size=100]Ramanujian[/size][/b] ).

From $ (3)$ $ \Longrightarrow$ $ (PQ)^{2} \equal{} 2(BQ \minus{} PQ)(PC \minus{} PQ)$ $ \Longrightarrow$ $ (PQ)^{2} \plus{} 2(BQ)(PC) \equal{} 2(PQ)(BQ \plus{} PC)$ $ ,(4)$

We will prove that the relation $ (4)$ is true.

$ \bullet$ Because of $ BA$ tangents to the circle $ (C)$ at point $ A,$ we have that $ (BP)(BZ) \equal{} (BA)^{2} \equal{} (BQ)^{2} \equal{} (BF)^{2}$ $ ,(5)$

From $ (5)$, based on the [b][size=100]Newton theorem[/size][/b], we conclude that the points $ F,\ P,\ Q,\ Z,$ are in harmonic conjugation as well.

So, we have that $ \frac {QP}{QZ} \equal{} \frac {FP}{FZ}$ $ \Longrightarrow$ $ (PQ)(FZ) \equal{} (FP)(QZ)$ $ ,(6)$

From $ (6)$ $ \Longrightarrow$ $ (PQ)[2(BQ) \plus{} 2(PC) \minus{} PQ] \equal{} [2(BQ) \minus{} PQ][2(PC) \minus{} PQ]$ $ \Longrightarrow$ $ 4(PQ)(BQ \plus{} PC) \equal{} 4(BQ)(PC) \plus{} 2(PQ)^{2}$ $ ,(7)$

From $ (7)$ $ \Longrightarrow$ $ (PQ)^{2} \plus{} 2(BQ)(PC) \equal{} 2(PQ)(BQ \plus{} PC)$ and the proof is completed.

[quote="jayme"]... Is it possible to prove first the result of my problem, the to deduct the relation of Ramanujan? [/quote]
Although I have the proof of some results, probably useful, however I don't know if I can prove first the relation $ PQ \equal{} 2(CE)$ of the proposed problem.

Best regards, Kostas Vittas.

Notice that $ BP \cdot CQ \equal{} (a \minus{} b)(a \minus{} c) \equal{} a^2 \plus{} bc \minus{} ac \minus{} ab$  and

$ PQ \equal{} a \minus{} (a \minus{} b) \minus{} (a \minus{} c) \equal{} b \plus{} c \minus{} a  $

$\Longrightarrow PQ^2 \equal{} b^2 \plus{} c^2 \plus{} a^2 \plus{} 2bc \minus{} 2ac \minus{} 2ab$

Since $ a^2 \equal{} b^2 \plus{} c^2  \Longrightarrow  PQ^2 \equal{} 2(a^2 \plus{} bc \minus{} ac \minus{} ab)$

But $ \ BQ \cdot QC \equal{} (a \minus{} b)(a \minus{} c) \equal{} a^2 \plus{} bc \minus{} ab \minus{} ac$  $\Longrightarrow  PQ^2 \equal{} 2BP \cdot QC \ (*)$

Inversion with center $ Q$ and power $ BQ \cdot QC$ maps $ C \mapsto B$ and $ E \mapsto P.$ Therefore

$ CE \equal{} BP \cdot \frac {BQ \cdot QC}{BQ \cdot PQ} \Longrightarrow  CE \cdot PQ \equal{} BP \cdot QC$

Then combinig with $ (*)$ yields $ CE \cdot PQ \equal{} \frac {_1}{^2}PQ^2  \Longrightarrow  PQ \equal{} 2CE.$

Dear Kostas, Luis, Virgil and Mathlinkers,
thank you for all your nice proofs.
The result involves by Kostas comes from Ramanujan (Bernd, B.C. (1994): Ramanujan's Noteboooks, Part IV. Springer -Verlag).
For my problem, I had in mind the Haruki's result : BP.CQ/PQ = CE... which comes from a more general case.
Sincerely
Jean-Louis

[quote="Virgil Nicula"][color=darkred] Let $ ABC$ be a triangle for which $ a > b$ , $ a > c$ . Consider the points $ \{P,Q\}\subset [BC]$ so that $ PC \equal{} b$ , $ BQ \equal{} c$ .

The line $ AQ$ cut again the circumcircle of $ \triangle ABC$ in $ D$ and the line $ BC$ cut again the circumcircle of $ \triangle APD$ in $ E$ .

Prove that $ \boxed {\ \frac {PQ}{CE} \equal{} 2 \plus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{(a \minus{} b)(a \minus{} c)}\ }$ and $ A \equal{} 90^{\circ}\ \Longleftrightarrow\ PQ \equal{} 2\cdot CE$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] Using the power of a point w.r.t. a circle obtain : $ QB\cdot QC \equal{} QA\cdot QD \equal{} QP\cdot QE$ . Since $ QB \equal{} c$ ,

$ QC \equal{} a \minus{} c$ , $ QP \equal{} b \plus{} c \minus{} a$ results $ QE \equal{} \frac {c(a \minus{} c)}{b \plus{} c \minus{} a}$ and $ CE \equal{} QE \minus{} QC$ $ \implies$ $ CE \equal{} \frac {(a \minus{} b)(a \minus{} c)}{b \plus{} c \minus{} a}$ .

Therefore,  $ \frac {PQ}{CE} \equal{} \frac {(b \plus{} c \minus{} a)^2}{(a \minus{} b)(a \minus{} c)} \equal{} 2 \plus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{(a \minus{} b)(a \minus{} c)}$ a.s.o.

[b][u]Remarks[/u].[/b]

$ 1\blacktriangleright$ If $ A \equal{} 120^{\circ}$ , then $ PQ < CE$ .

$ 2\blacktriangleright$ A very [u]interesting and nice algebraic identity[/u] ([b]own[/b] ?!) with many applications :

$ \boxed {\ \sum\frac {b^2 \plus{} c^2 \minus{} a^2}{(a \minus{} b)(a \minus{} c)} \equal{} \minus{} 2\ }$ , where $ \{a,b,c\}\subset\mathbb C$ and $ a\ne b\ne c\ne a$ .

$ 3\blacktriangleright$ Prove "easily" that in generally :[/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle for which $ a \ne b\ne c\ne a$ . Consider the points $ P\in (CB$ , $ Q\in (BC$ so that $ CP \equal{} b$ , $ BQ \equal{} c$ .

The line $ AQ$ cut again the circumcircle of $ \triangle ABC$ in $ D$ and the line $ BC$ cut again the circumcircle of $ \triangle APD$ in $ E$ . Prove that

 $ \boxed {\ \frac {\overline {PQ}}{\overline {CE}} \equal{} 2 \plus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{(a \minus{} b)(a \minus{} c)}\ }$ , where $ X(x)$ , $ \overline {XY} \equal{} y \minus{} x$ - an analytical geometry on the orientated line $ d \equal{} BC$ .[/color][/quote]
[color=darkblue]$ 4\blacktriangleright$ Prove similarly that [/color]:

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle . Consider the points $ \{P,Q\}\subset BC$ so that $ C\in (BP)$ , $ B\in (CQ)$ and  $ CP \equal{} b$ , $ BQ \equal{} c$ .

The line $ AQ$ cut again the circumcircle of $ \triangle ABC$ in $ D$ and the line $ BC$ cut again the circumcircle of $ \triangle APD$ in $ E$ .

Prove that $ \boxed {\ \frac {PQ}{CE} \equal{} 2 \plus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{(a \plus{} b)(a \plus{} c)}\ }\ .$ Is this  [u][b]Ramanujan[/b]'s problem[/u] ?![/color][/quote]

[color=darkblue]$ 5\blacktriangleright$ Prove similarly that :[/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle and consider the points $ \{P,Q\}\subset (BC)$ so that $ BQ^2 \plus{} CP^2 \equal{} BC^2$ . The line $ AQ$ cut again

the circumcircle of $ \triangle ABC$ in $ D$ and the line $ BC$ cut again the circumcircle of $ \triangle APD$ in $ E$ . Prove that $ PQ \equal{} 2\cdot CE$ .[/color][/quote]

[color=darkblue]$ 6\blacktriangleright$ Prove similarly that :[/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a $ A$-right triangle and consider the points $ \{X,Y,Z,T\}\subset BC$ (line !) so that

$ \left\|\begin{array}{cccccc} B\in (XY) & , & Y\in (BC) & , & BX \equal{} BY \equal{} AB \\
 \\
C\in (ZT) & , & Z\in (BC) & , & CZ \equal{} CT \equal{} AC\end{array}\right\|$ . Prove that $ \widehat {XAZ}\equiv\widehat {ZAY}\equiv\widehat {YAT}$ .[/color][/quote]

Dear Virgil and Mathlinkers,
your last message is very interesting.
The Ramanujan's problem after my source is:

let ABC be a triangulo rectangular at A. On the side BC take to points P and Q so that BQ=BA and CP=CA.
Prove: PQ²=2BP.QC.

Sincerely
Jean-Louis

Dear Mathlinkers,
my proof of Ramanujan's problem:
according to Haruki's theorem,  BP.CQ = PQ.CE
See email : http://www.mathlinks.ro/Forum/viewtopic.php?p=1555284#1555284
PQ = 2.CE
and we are done
Sincerely
Jean-Louis

It seems to me that Ramanujan came up with this identity PQ^2 = 2*BP * CQ

 immediately after reading  Propositions 5 thru 8

  of Book II of the Elements.  That one, after all,  is concerned with Geometric Algebra.

 It is hard to tell which of the above Propositions affected Srini the most.

 If it turns out that his is a new Proposition, then maybe we can squeeze it right between

 Props. 8 and 9 in the same Book II on the account of it's great importance.

 And it will be another confirmation of Fermat's notion that those bearded  Greeks

 did not know everything.

 It is also worth while investigating the "2r" connection of this problem with the Sangakus

 see  http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1150708027&t=264276

        http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1150708027&t=270926

Historical note:

 Young Srini failed math tests to make the Indian team to IMO  two years in a row.
 He managed to do Geometry problems, yet Number Theory tasks proved to be beyond
 his abilities.

 Also, his last name points towards Armenian origin  Ramanujian. It is interesting that Armenian
 mathematicians haven't yet claimed him to be their own.

 M.T.

Dear Mark,
thank you for your historical note and references.
You attached figure suggest an area approch...
Sincerely
Jean-Louis


Is there a way to do absolute value in latex?

Just use the | key on your keyboard.
[code]$|n|$[/code]
Outputs $ |n|$.

If the absolute value bars don't reach the whole part of the thing inside, use \left and \right before the bars.

Another thing I've been meaning to try is
[code]\newcommand{\abs}[1]{\left|#1\right|}[/code]
and then I could just stick in
[code]$|x| + \abs{ \frac{x}{2} } = 20$[/code]
and get
$ |x| \plus{} \left|\frac{x}{2}\right| \equal{} 20$.

personally, I like using

[code]
$\mid x \mid$
[/code]

spaces things out nicely - or maybe I'm just weird.


geometry
3D geometry
octahedron
ratio
pyramid

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is

$ \textbf{(A) } \frac{\sqrt3}{12} \qquad \textbf{(B) } \frac{\sqrt6}{16} \qquad \textbf{(C) } \frac16 \qquad \textbf{(D) } \frac{\sqrt2}{8} \qquad \textbf{(E) } \frac14$

First we cut the cube in half down the middle. Now we have a half-cube with a pyramid inscribed in it. The height of the pyramid is the same as the height of the half-cube. As for the base, the base of this pyramid is a square inscribed in the base of the half-cube, so it has half the area. Therefore, the area of the half-cube is $ bh$ and the area of the pyramid is $ \frac{1}{3}(\frac{b}{2})(h)\equal{}\frac{bh}{6}$ so our ratio is 1/6, corresponding to answer choice C.

[quote="ernie"]A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is

$ \textbf{(A) } \frac {\sqrt3}{12} \qquad \textbf{(B) } \frac {\sqrt6}{16} \qquad \textbf{(C) } \frac16 \qquad \textbf{(D) } \frac {\sqrt2}{8} \qquad \textbf{(E) } \frac14$[/quote]
[hide]Assume the length of the cube as $ s$.
The vertices of the octahedron is at the midpoint of each line, if you take a bird's eye view. So, $ \frac{s \sqrt{2}}{2}$ is an edge of the octahedron. Now, square it, to get the base of the pyramid. Now find the volume when the height is $ \frac{s}{2}$.
$ \frac{s^2}{2} \times \frac{s}{2} \times \frac{1}{3} \equal{} \frac{s^3}{12}$
So, $ 2$ times that value is $ \frac{s^3}{6}$, and dividing by $ s^3$, we have choice C, $ \frac{1}{6}$.[/hide]


function
algebra
polynomial
complex analysis
complex analysis unsolved

Is this possible to find a function $f$ analytic on the open disc with center $(0,0)$ and radius $1$, such that $lim |f(z)| \rightarrow +\infty$ as $|z| \rightarrow 1$ ?

It's been discussed before (I think I posted it myself, and you can find it on the first page, about 20 or so posts below your message).

Anyway, the answer is negative. Assume we can find such a function $f$. Then $f$ has a finite number of zeros, so if we divide it by a certain polynomial, the quotient will be a function $g$ analytic in the open unit disk with no zeros. Then we apply the maximum modulus principle to $\frac 1g$ and we're done.

Edit:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43428[/url]

Thanks Grobber. That's also what I did in the first place, but somehow I was not satisfy with the justification that $f$ can be written by the product of the "polynomial" and an analytic function with no zero. Actually I've just realized this was easy.


1) $5^{\log_5 {6} + \log_{5} {7}}$

2) $e^{\log_{e^2} {9}}$

[color=brown][size=75]mod edit: stop being lazy and learn $\LaTeX$!![/size][/color]

Wow!  Great job! I did not think anyone can solve those weird log questions.

these things ought to go in getting started, it would be helpful for that group to learn these things, but in this forum, those problems are automatic
[color=red][i]
mod edit: done[/i][/color]


[quote="1234567890"]How would you do this?

Find a polynomial with integer coefficients of lowest possible degree with a root at
a)$\cos10$
b)$\cos20$
c)$\cos40$
d)$\cos80$[/quote]

I'm not sure how you would minimize the total degree, but I'll show you what I've done for this problem so far.
$\cos{3x} = \cos{x}\cos{2x}-\sin{x}\sin{2x} = \cos{x}(2\cos{x}^2-1)-2\cos{x}\sin{x}^2 = \cos{x}(2\cos{x}^2-1)-2\cos{x}(1-\cos{x}^2) = 4\cos{x}^3-3\cos{x}$
This means $\cos{30} = 4x^3-3x \rightarrow 4(4x^3-3x)^2 - 3 = 0$, $8x^3-6x -1 = 0$, $8x^3-6x+1 = 0$, and $8x^3-6x+1 = 0$
So I guess the lowest degree polynomial would be the product, but you can exclude the last trinomial because $\cos{80}$ is a root of the third equation.  The lowest degree polynomial would then be 12.


What do you think was the most important play?  Who should have been MVP?

[quote="la mafioso"]OMFG what i can't believe was that they counted roethlessbergers jump as a td! That was soooo not.

And also, I can't believe that seahawks ran like 76 yards but had to redo play cuz of some penalty![/quote]

YES! A Seahawks supporter! It was completely evident that the roethelsberger (sp?) touchdown thingy was... urgh.he simply moved it over to make it look like a touchdown. and that one long catch which gained like 50 yards ... penalty?! and then that one touchdown seahawks made... "touchdown", where he supposedly went out of bounds, not sure if he did or not, but from the way i saw it... yeah. referees probably wanted to please the crowd. but i could always be wrong  :roll:

IMO, worst Super Bowl in history

^^^ If you think that's International Math Olympiad, you've been doing too much math. Go commune with your couch/office chair. :diablo:

:(

I think the most important play was probably the pass by the not-quarterback.

[quote="ch1n353ch3s54a1l"][quote="la mafioso"]OMFG what i can't believe was that they counted roethlessbergers jump as a td! That was soooo not.

And also, I can't believe that seahawks ran like 76 yards but had to redo play cuz of some penalty![/quote]

YES! A Seahawks supporter! It was completely evident that the roethelsberger (sp?) touchdown thingy was... urgh.he simply moved it over to make it look like a touchdown. and that one long catch which gained like 50 yards ... penalty?! and then that one touchdown seahawks made... "touchdown", where he supposedly went out of bounds, not sure if he did or not, but from the way i saw it... yeah. referees probably wanted to please the crowd. but i could always be wrong[/quote][size=200][color=orange]Super[/color] bowl [color=orange]rings[/color]:[/size]

[size=150][color=orange]Steelers[/color]: [/size]:winner_first: :winner_first: :winner_first: :winner_first: :winner_first:
[size=150]Seahawks:[/size]( :oops_sign: )

Stop making excuses for playing bad.  If you really were a championship-quality team you would not be blaming the refs for losing.  It was obvious that the Steelers played better.  So how many plays did you have the whole game? [size=200]1[/size].  How many did we have? [size=200]3[/size].  On Roethlisburgher's TOUCH DOWN (no quotes), do you really think the refs would have been fooled by him moving over the line about 3 seconds after the play was over? The Seahawks "touch down" where he was called out ... there was no question he was out.  Your other "touch down" was defiently a penalty.  It was the receiver's falt, not the refs'.  The only real touch down you got was when B.R. threw that interception right to you. The bottom line is Pittsburgh is a better team, and the trophey is in Pittsburgh where it belongs. [size=59]If you want I'll send you a Terrible Towel to wipe away your tears.[/size] :gathering:  :clap:  :clap2:  :wow:

[url=http://www.steelers.com]CLICK HERE[/url]

You hear what Roethlisberger siad after the superbowl?
http://www.seahawks.com/ArDisplay.aspx?ID=6506
http://www.mercurynews.com/mld/mercurynews/sports/13810016.htm
http://msn.foxsports.com/nfl/story/5313884
http://msn.foxsports.com/nfl/story/5310192
Oh well. It's over anyways. [size=200]Ahhh[/size][size=150]hhh[/size][size=100]hhh[/size][size=75]hhh[/size][size=59]hhh[/size]

[quote="ch1n353ch3s54a1l"]You hear what Roethlisberger siad after the superbowl? Go google that. Tell me what you find.[/quote]From the post-gazette:

"Ben Roethlisberger will not turn 24 years old until March 2 and thus can become the youngest quarterback to win a Super Bowl. But, he said, he's motivated to win the next game not for himself but for others. He also said he likely will have a ceremonial shaving of his beard if the Steelers beat Seattle.

"There is quite a bit that is my motivation," Roethlisberger said. "My grandfather passed this summer and never got to see a Steelers game. Getting Jerome back to Detroit and winning for coach Cowher. I also want to win for the offensive linemen.""

uhh...

try reading the WHOLE articles.  :)

I edited my response thingy, so yeah.

:huh: He didn't think he got in.  So?  He probably couldn't tell anymore than the refs could.  And anyway, the initial call was a touchdown so it would have to be 100% obvious he didn't get in for them to say it wansn't a TD.

"he couldn't tell any more than the refs could" what?! its like if you're playing a musical instrument. can you not tell if you play a wrong note than the audience? He's pro, he should be able to tell what mistakes he makes and what he does correctly.

HAHA. I'm watching the news while making the post and they mention the touchdown.  :)

Anyways, it doesn't really matter. Even if they rule the touchdown invalid, we can't change the champions.

The Seahawks didn't play badly. They had a few bad plays, but the refs should at least have called everything fairly. It's possible that the Steelers still might have won, but it would have been a much closer game, and the Seahawks would at least have the satisfaction of knowing that they lost fairly, not that they were cheated out of the Superbowl.

It's sad when people are such ardent fans that they're completely blind when something was actually unfair in favor of their team, and they think it was all perfectly fine.

I think the Steelers won the Super Bowl pretty cheaply.
First, the refs were HORRIBLE, and to the favor of the Steelers.
The one play where Matt Hasselback was called for low block? it cost the seahawks 15 yards and field position. HE WAS TACKLING THE STEELERS GUY WITH THE BALL. It was completely legal, and even the ref must have realized he was wrong.
Then the off. pass interference call. It cost the Seahawks 4 points, and the guy did NOT push off the guy enough to draw a penalty, however it was close. But close does not mean "automatically goes to the steelers".
Same thing with the Ben Roethlisberger touchdown.
It was close and could have gone either way, but AGAIN, it was in favor of the Steelers.
The holding call that cost them a touchdown?
Close, but it went to the Steelers, as usual.
Even besides the refs, the Seahawks  dominated the Steelers on both fronts.
There was ONE drive that the Seahawks failed to get a first down. 2 missed field goals, ne touchdown they were cheated, an the ONLY bad offensive play, the interception. The Steelers had only 3 good plays offensively, and all led to touchdowns. They also had no first downs until the 2nd quarter
the seahawks lost because of the refs, clock management, and luck.

Darn, I can't believe the Colts lost though.

actually i am a seahawks fan (for the superbowl only, if not, go bears and colts :) )and you could not see the balls tip go over the line so they could not overturn it (the original call had a huge effect)

and the td called bac cuz o a penalty- even steve young said it was a td- both jackson and the defender pushed off with their hands...

quite obviously, the seahawks are :first:

Stop pushing it. The game's over, and somehow I don't think they're going to overturn the results.


I need help with this one.

[url=http://g.imagehost.org/view/0971/24_p_181][img]http://g.imagehost.org/0971/24_p_181.jpg[/img][/url]

The coloured parts space is limited by three semicircle archs of which the smaller has the radius  r.

How should the radius MP be drawn, if you want it to divide the coloured space into two parts with
equally big areas?

Ok, this is might or might now work:

Let the area of the smaller shaded half circle be $ A_h \equal{} \dfrac{r^2\pi}{2}$

Let the area of the shaded region (not covering the smaller haded half circle) be $ A_r \equal{} \dfrac{(2r)^2\pi}{2} \minus{} \dfrac{r^2\pi}{2} \equal{} \dfrac{3r^2\pi}{2}$

Now when you divide the area the amount added to $ A_h$ is also subtracted from $ A_r$ so $ A_h \plus{} x \equal{} A_r \minus{} x \implies 2x \equal{} A_r \minus{} A_h$
\[ 2x \equal{} \dfrac{3r^2\pi}{2} \minus{} \dfrac{r^2\pi}{2} \equal{} r^2\pi\implies x \equal{} \dfrac{r^2\pi}{2}
\]
So the area in between the smaller shaded half circle and $ MP$ is $ \dfrac{r^2\pi}{2}$, which can be expressed as a fraction of the total area of the bigger circle $ \dfrac{4r^2\pi}{k} \equal{} \dfrac{r^2\pi}{2}\implies k \equal{} 8$, so the result of $ k$ means that the angle made from the smaller half circle to point P is $ \dfrac{360}{8} \equal{} 45^{\circ}$ in other words the arc $ AP \equal{} \boxed{45^{\circ}}$, and that's it.

What that means is that the angle $ \angle AMP \equal{} 45^{\circ}$.

I know this can be a little difficult to understand so if someone else has a better solution it would be best to post it.


if i understood your problem correctly, i got:

[hide]1/x?[/hide]

[quote]X+1___ DIVIDE __x2 +5x__
X2+3x+2 DIVIDE x2+7x+10

[/quote]

if you wont use LATEX, can you at least change X2 to x^2? that would really help.

all the terms cancel except for $x$
the answer is[hide]x^(-1)[/hide]


Find all pairs of positive integers satisfying the following equation:

$ a! \plus{} b! \equal{} a^b$

It's like, almost triv.  :rotfl:

This problem is a bit beyond the difficulty of HSB.  I guess this is fine in Intermediate, however.

I know $ a\equal{}2, b\equal{}3$ is a solution, but I don't have a rigorous answer.  Something about factorials containing way more factors than a power.

[quote="azn6021023"]I know $ a \equal{} 2, b \equal{} 3$ is a solution, but I don't have a rigorous answer.  Something about factorials containing way more factors than a power.[/quote]
What about $ a\equal{}2, b\equal{}2$?

There are three cases

[b]Case 1. [/b]: $ a \equal{} b$

Then we have  $ 2b! \equal{} b^b$

So $ (b \minus{} 1)|b \rightarrow b \minus{} 1 \equal{} 1 \rightarrow  b \equal{} 2$ and since $ 2*2\equal{}2^2$ we have a solution $ a\equal{}b\equal{}2$

[b]Case 2. [/b]: $ a < b$

We get  $ a!(1 \plus{} \frac {b!}{a!}) \equal{} a^b$

Again we have $ (a \minus{} 1)|a$ so $ a \equal{} 2$

If we plug in the value we get $ 2(1 \plus{} \frac {b!}{2}) \equal{} 2^b$

$ (1 \plus{} \frac {b!}{2}) \equal{} 2^{b \minus{} 1}$

Since $ 2^{b \minus{} 1}|1 \plus{} \frac {b!}{2}$ so $ b!$ must be divisible by $ 2$ but not by $ 4$. The only possible $ b$     that $ a < b$ is $ b \equal{} 3$

and since $ 2(1 \plus{} 3) \equal{} 2^3$ we have another solution $ a \equal{} 2$ and $ b \equal{} 3$

[b]Case 3.[/b] : $ a > b$

We have

$ b!(1 \plus{} \frac {a!}{b!}) \equal{} a^b$

Since $ b|a$ and $ (b \minus{} 1)|a$  so $ b \equal{} 2$

if we rewrite the term

$ 2(1 \plus{} \frac {a!}{2}) \equal{} a^2$

so $ 2|(1 \plus{} \frac {a!}{2})$ and again $ a!$ must be divisible by $ 2$ but not by $ 4$ and $ a > b$ so $ a \equal{} 3$

but $ a$ must also be divisible by $ 2$ and so there is no solution for this case.

The only two solutions $ (a,b)$ are $ (2,2)$ and $ (2,3)$

	$ \blacksquare$


If $ a > b > 0$, show that

$ \frac {(1 \plus{} a \plus{} a^{2} \plus{} ... \plus{} a^{n \plus{} 1})}{(1 \plus{} a \plus{} a^{2} \plus{} ...a^{n})} < \frac {(1 \plus{} b \plus{} b^{2} \plus{} ... \plus{} b^{n \plus{} 1})}{(1 \plus{} b \plus{} b^{2} \plus{} ... \plus{} b^{n})}$

and hence deduce that

$ \frac {(1 \plus{} x \plus{} x^{2} \plus{} ... \plus{} x^{n})}{(x \plus{} x^{2} \plus{} x^{3} \plus{} ... \plus{} x^{n \minus{} 1})} \ge \frac {(n \plus{} 1)}{(n \minus{} 1)}$.

[The message didn't state that $ x$ should be positive.]

[hide=hint]
I am confident that you will get far with the formula for geometric series:
$ 1\plus{}x\plus{}x^2\plus{}x^3\plus{} \ldots x^n\equal{}x\frac{x^{n\plus{}1}\minus{}1}{x\minus{}1}$

After all, the x-1 cancels

$ x$ has to be positive or it's false.
[hide="Solution"]Equality occurs when $ x \equal{} 1$. Otherwise, note that
\[ \frac {1 \plus{} \cdots \plus{} x^n}{x \plus{} \cdots \plus{} x^{n \minus{} 1}} \equal{} \frac {1 \plus{} \cdots \plus{} \left(\frac1x\right)^n}{\left(\frac1x\right) \plus{} \cdots \plus{} \left(\frac1x\right)^{n \minus{} 1}},\]
so without loss of generality we can assume that $ 0 < x < 1$. We cross multiply to get
\[ (n \minus{} 1)(1 \plus{} x^n) \plus{} (n \minus{} 1)\sum_{k \equal{} 1}^{n \minus{} 1}x^k\ge(n \plus{} 1)\sum_{k \equal{} 1}^{n \minus{} 1}x^k\Longleftrightarrow(n \minus{} 1)(1 \plus{} x^n)\ge2\sum_{k \equal{} 1}^{n \minus{} 1}x^k.\]
When $ n \equal{} 2$ we have
\[ 1 \plus{} x^2\ge2x\Longleftrightarrow(1 \minus{} x)^2\ge0.\]
[hide="Lemma"]We can prove that $ ix^{i \plus{} 1} \minus{} (i \plus{} 1)x^i \plus{} 1\ge0$ by induction on $ i$, noting that
\[ (jx^{j \plus{} 1} \minus{} (j \plus{} 1)x^j \plus{} 1) \minus{} x((j \minus{} 1)x^j \minus{} jx^{j \minus{} 1} \plus{} 1) \equal{} (1 \minus{} x)(1 \minus{} x^i)\ge0.\]
[/hide]
If $ (i \minus{} 1)(1 \plus{} x^i)\ge2\sum_{k \equal{} 1}^{i \minus{} 1}x^k$, then by the lemma,
\[ ix^{i \plus{} 1} \minus{} (i \plus{} 1)x^i \plus{} 1\ge0\implies ix^{i \plus{} 1} \plus{} 1 \minus{} ix^i \plus{} x^i\ge2x^i\]

\[ \implies i(1 \plus{} x^{i \plus{} 1}) \equal{} (i \minus{} 1)(1 \plus{} x^i) \plus{} ix^{i \plus{} 1} \plus{} 1 \minus{} ix^i \plus{} x^i\ge2\sum_{k \equal{} 1}^{i \minus{} 1}x^k \plus{} 2x^i \equal{} 2\sum_{k \equal{} 1}^{i}x^k,\]
so the desired inequality follows from induction.[/hide]


Given a $\triangle{ABC}$ with sides lengths $a,b,c$ such that $a+c=2b$ and $\angle{C}-\angle{A}=90^{\circ}$, find $a: b: c$.

[hide]
Law of sines, $\frac{a}{c}=\tan A \Rightarrow \boxed{a: b: c=1: \frac{\tan A+1}{2}: \tan A}$.[/hide]


Is it possible that the title of the topic you are currently browsing be the first part of the title, i.e. for this topic, instead of the title "Art of Problem Solving Forum", have it be "Topic Titles - Art of Problem Solving Forum"? When I'm browsing multiple topics, it's sometimes kind of hard to figure out what topic you are on.

That would be a really nice feature; it would make it much easier to be able to tell where you are on AoPS, besides the "Art of Problem Solvin..."

Alternative solution:
Make the font size really small so the entire title fits inside the tiny bar at the bottom of the screen  :rotfl:

[quote="JBL"]Motion seconded.  There's only one place on the entire page where I can find out what the title of the topic is.[/quote]Actually there are two places ;)

Will try to change that.

[quote="Valentin Vornicu"]Actually there are two places ;)

Will try to change that.[/quote]

Maybe in mathlinks skin -- in AoPS skin, there is only one.  And thanks!


The real numbers a,b satisfy
a^3−3a^2+5a −17 =0,
b^3−3b^2+5b +11 = 0.

Find a +b .


limit
logarithms
calculus
calculus computations

Which of the following limits is in the indeterminant form and why?

$ a)\lim_{x\rightarrow\infty}{(1/7)^x}$

$ b)\lim_{x\rightarrow\infty}{1^x}$

$ c)\lim_{x\rightarrow\infty}{5^x}$

$ d)\lim_{x\rightarrow\infty}{(1\plus{}1/x)^x}$

Only d) is an indeterminant form because you have two "competing forces" within the expression. The base, $ 1\plus{}1/x$, is trying to make the expression get closer to 1 as x goes to infinity; while the exponent, $ x$, is trying to make the expression go to infinity as x goes to infinity. This sort of "competing forces" is one of the signs of an indeterminant form.

For $ a > 0$, we have:  $ \boxed{a^{x} \equal{} e^{x \log a}}$


vector
geometry
3D geometry
geometry unsolved

hello;

I am writing a program ( flash actionscript3 ) that deals with 3d geometry;

I have a point in space, a direction vector from the point in space, and a plane with 3 known vertices and a known normal for the plane;

my goal is to find out:
1 - where the direction vector intersects the ( extended ) plane;
2 - is the intersection point contained within the boundaries of the plane;

any thoughts?

thanks,
Shannon

If the normal vector to the plane is $ {\bf n} \equal{} (n_1,n_2,n_3)$ and a point on the plane is $ {\bf a} \equal{} (x_0,y_0,z_0)$ then for all points $ {\bf x} \equal{} (x,y,z)$ on the plane we have
\[ {\bf n}\cdot ({\bf x} \minus{} {\bf a}) \equal{} n_1(x \minus{} x_0) \plus{} n_2(y \minus{} y_0) \plus{} n_3(z \minus{} z_0) \equal{} 0
\]
If the point in space is $ {\bf p} \equal{} (p_x,p_y,p_z)$ and the direction vector is $ {\bf v} \equal{} (v_x,v_y,v_z)$, then the points on the ray extending from $ {\bf p}$ along $ {\bf v}$ are of the form $ {\bf p} \plus{} t{\bf v}$ for real numbers $ t$.  Thus, to find the point of intersection we must find the real number $ t$ such that
\[ {\bf n}\cdot \left({\bf p} \plus{} t{\bf v} \minus{} {\bf a}\right) \equal{} n_1(p_x \plus{} tv_x \minus{} x_0) \plus{} n_2(p_y \plus{} tv_y \minus{} y_0) \plus{} n_3(p_z \plus{} tv_z \minus{} z_0) \equal{} 0
\]
Solving for $ t$ we find (assuming $ {\bf v}$ is not parallel to the plane so that $ {\bf n \cdot v}\neq 0$),
\[ t \equal{} \frac {{\bf n \cdot (a \minus{} p)}}{{\bf n\cdot v}} \equal{} \frac {n_1(x_0 \minus{} p_x) \plus{} n_2(y_0 \minus{} p_y) \plus{} n_3(z_0 \minus{} p_z)}{n_1v_x \plus{} n_2v_y \plus{} n_3v_z}
\]
The point of intersection is therefore
\[ \boxed{{\bf p \plus{} \left(\frac {{\bf n \cdot (a \minus{} p)}}{{\bf n\cdot v}}\right)v}}
\]
As for question 2, I guess it depends on how you're storing the 'boundary' of the plane.

JoeBlow, thanks for your response;

regarding point2:
my plane is actually a 3d [i]polygon[/i] ( specifically, a 3-vertex polygon ) as opposed to an infinite plane; so it is possible that when I find the intersection-point of the direction-vector that the intersection-point will not actually be contained within the polygon ( ie boundaries of the plane );

Shannon
Chicago

Let's say $ {\bf T_a,T_b,T_c}$ are the points of your 3-vertex polygon (also known as a triangle ;)), and let $ {\bf P}$ be the point that you're trying to determine whether or not it's in the triangle.

Define the vectors $ {\bf u_1 \equal{} T_b \minus{} T_a}$ and $ {\bf u_2 \equal{} T_c \minus{} T_a}$.  These vectors form a basis for the plane in which the triangle lies, and a vector of the form $ {\bf T_a}\plus{}\alpha {\bf u_1 } \plus{} \beta {\bf u_2}$ lies inside the triangle if and only if $ \alpha,\beta \geq 0$ and $ \alpha \plus{} \beta \leq 1$ lie inside the triangle (try drawing a picture to help see why this is so).  So, to determine if $ {\bf P}$ is inside the triangle you need to solve $ {\bf P} \equal{} {\bf T_a}\plus{}\alpha {\bf u_1} \plus{} \beta {\bf u_2}$ for $ \alpha$ and $ \beta$ and check whether or not the conditions hold.

ok; this seems do-able;

thanks for your time;

Shannon,
Chicago

Joe, are those dots representing dotproducts or multiplications?

such as in [b]n[/b] [u].[/u] ( [b]a[/b] - [b]p[/b] )

edit: NEVERMIND - they are indeed dotproducts;

thanks again;


I need help with #5, 10, 12, and 15. Thanks!

Well, one problem, at least.
[hide="#5"]$\text{Since }\tan\theta=-\frac{5}{12}, \text{and }\cos\theta<0, \text{then }\sin\theta \text{ must equal }\frac{5}{13}(\text{ because }5^{2}+12^{2}=13^{2}). \text{ Therefore}\csc\theta = 13/5.$[/hide]

[hide="#15"]Let $a=2^{x}-4,\ b=4^{x}-2,$ we can rewite your equation as $a^{3}+b^{3}=(a+b)^{3}\Longleftrightarrow 3ab(a+b)=0\Longleftrightarrow a=0\ or\ b=0\ or\ a+b=0.$[/hide]

[quote="t0rajir0u"][hide="#10"] $f \left( \frac{x}{1-x}\right) = \frac{1}{x}$

Now, we are going to find the inverse function of $y = \frac{x}{x-1}$.  This will be

$x = \frac{y}{y-1}\Leftrightarrow$
$xy-x = y \Leftrightarrow$
$y(x-1) = x \Leftrightarrow$
$y = \frac{x}{x-1}$

Now, substituting this into the original given we have

$f(x) = \frac{x-1}{x}$

Then

$f \left( \sec^{2}x \right) = \frac{ \sec^{2}x-1}{\sec^{2}x}= \frac{\tan^{2}x}{\sec^{2}x}= \boxed{ \sin^{2}x }$ [/hide][/quote]

I'm sorry, but could you explain why you're taking the inverse of $y=\frac{x}{x-1}$? I'm feeling rather stupid right now.

[hide="Essentially"] If $f \left( \frac{x}{1-x}\right) = \frac{1}{x}$ then

$f \left( \frac{y}{1-y}\right) = \frac{1}{y}$

We want to figure out what $f(x)$ is.  So let $x = \frac{y}{1-y}$ and then $y = \frac{x}{x-1}$, so

$f(x) = \frac{1}{y}= \frac{x-1}{x}$ [/hide]

[quote="Altheman"]where is this contest held/how could i take this test?

that would be some pwnage...lol[/quote]
Well for our school the ICTM this year was held at Parkland College.


Να βρεθούν όλες οι συναρτήσεις $ f: \mathbb{R} \mapsto \mathbb{R}$  τέτοιες ώστε $ \forall$ $ x,y \in \mathbb{R}$
\[ f(x - f(y)) = f(x + y) + f(y)
\]

[hide][hide][hide]
Υπάρχει και εδώ
http://www.mathlinks.ro/viewtopic.php?search_id=1550114382&t=249175 [/hide][/hide][/hide]

Ορίζω $ g(x) \equal{} f(x) \plus{} x$ και παίρνω

$ g(x \plus{} y \minus{} g(y)) \equal{} g(x\plus{}y) \minus{} y$ (1)

$ x \equal{} g(y) \minus{} y$ στην (1): $ g(g(y)) \equal{} y \plus{} g(0)$ (2)

$ y\equal{}0$ στην (2): $ g(g(0)) \equal{} 0$ άρα $ g(g(g(0))) \equal{} g(0)$ (3)

$ y \equal{} g(0)$ στην (2): $ g(g(g(0))) \equal{} 2g(0)$ (4)

(3) και (4): $ g(0) \equal{} 0$. (5)

(2) και (5): $ g(g(y)) \equal{} y$ (2')

$ x \equal{} \minus{}y$ στην (1): $ g(\minus{}(g(y)) \equal{} \minus{}y$ (6)

(6) και $ y \equal{} \minus{}g(z), x \equal{} w\plus{}g(z)$ στην (1): $ g(w\plus{}z) \equal{} g(w) \plus{} g(z)$ (7)

Άρα η g ικανοποιεί την (1) τότε $ g(g(x)0\equal{}x, g(x\plus{}y) \equal{} g(x)\plus{}g(y)$. Εύκολα ελέγχεται και το αντίστροφο.

Έχουμε συζητήσει αυτήν την συναρτησιακη εξίσωση  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=251937&start=20]εδώ[/url]

[quote="Demetres"]Ορίζω $ g(x) \equal{} f(x) \plus{} x$ και παίρνω

$ g(x \plus{} y \minus{} g(y)) \equal{} g(x \plus{} y) \minus{} y$ (1)

$ x \equal{} g(y) \minus{} y$ στην (1): $ g(g(y)) \equal{} y \plus{} g(0)$ (2)

$ y \equal{} 0$ στην (2): $ g(g(0)) \equal{} 0$ άρα $ g(g(g(0))) \equal{} g(0)$ (3)

$ y \equal{} g(0)$ στην (2): $ g(g(g(0))) \equal{} 2g(0)$ (4)

(3) και (4): $ g(0) \equal{} 0$. (5)

(2) και (5): $ g(g(y)) \equal{} y$ (2')

$ x \equal{} \minus{} y$ στην (1): $ g( \minus{} (g(y)) \equal{} \minus{} y$ (6)

(6) και $ y \equal{} \minus{} g(z), x \equal{} w \plus{} g(z)$ στην (1): $ g(w \plus{} z) \equal{} g(w) \plus{} g(z)$ (7)

Άρα η g ικανοποιεί την (1) τότε $ g(g(x)0 \equal{} x, g(x \plus{} y) \equal{} g(x) \plus{} g(y)$. Εύκολα ελέγχεται και το αντίστροφο.

Έχουμε συζητήσει αυτήν την συναρτησιακη εξίσωση  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=251937&start=20]εδώ[/url][/quote]

Αρα έχουμε απειρες λύσεις? :huh:

[quote="socrates"]
Αρα έχουμε απειρες λύσεις? :huh:[/quote]

Ναι.


integration
calculus
calculus computations

Hi. I have a question that I can't seem to get the right answer to, so I was wondering if someone could try it out.

"The base of a solid is the region enclosed by y=x^.5, the x-axis, and the line x=9. Each cross-section perpendicular to the x-axis is a semi-circle. Find the volume of the solid."

Thanks in advance.

[b]Solution[/b]
[hide]So we have $y=x^{\frac{1}{2}}$ from $x=0 \to 9$ where every cross section is a semicircle $\implies A=\frac{\pi r^{2}}{2}$.

$A(r)=\frac{1}{2}\cdot \pi r^{2}$ where $r=\frac{x^{\frac{1}{2}}}{2}$
$\iff A(x)=\frac{\pi x}{8}$

So volume of cross sections is:
$V=\int_{a}^{b}A(x) ~dx$
$\iff V=\int_{0}^{9}\frac{\pi x}{8}~dx$
$\iff V= \frac{\pi}{8}\cdot \int_{0}^{9}x ~dx$
$\iff V= \frac{\pi}{8}\cdot \left[ \frac{x^{2}}{2}\right]_{0}^{9}$
$\iff V= \frac{\pi}{8}\cdot \left[ \frac{9^{2}-0^{2}}{2}\right]$
$\iff V= \boxed{\frac{81 \pi}{16}}$[/hide]


No three diagonals of a convex 2000-gon meet at a point. The diagonals (but not the sides) are each colored with one of 999 colors. Show that there is a triangle whose sides are on three diagonals of the same color.

There are C(2000,2)=1999000 diagonals in the 2000-gon, the amount of the triangle which sides are the diagonals is C(2000,3)
Can anyone continue... GBU

It has been posted twice in the Combinatorics forum, http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=6008453&t=82444 and [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=6008453&t=6376[/url], and there is no solution posted in either thread.  It is harder than this forum is supposed to be.

I also note that Qruni has not read the problem very carefully.

yes this problem is very very hard,I remember lots of people spending lots of time on this and they didnt manage to solve this one...  :(


Alcumus
MATHCOUNTS
geometry
AMC 8
Support

Although I'm only in the diagnostic phase of alcumus, might I ask how hard the problems get. I understand that phase 1 complements introduction to counting and probability. Do these problems serve to prepare students for amc 12 problems, as in, are some of these problems at the amc 12 level? Or do these problems just help kids prepare for mathcounts, amc 8 and 10?

Most of the problems I see are MathCounts and easy AMC level.  However, some more experienced users might see harder problems.

There are some problems from the AMC 10 and 12.  In the next couple months, we'll probably add Mandelbrot questions, as well.

[quote="Bugi"]I think that there are too many C & P problems, and that there aren't many hard problems in other areas (only the easiest).[/quote]

Everything in Alcumus is C&P for now.  They're adding some problems from Intro to Algebra soon.

[quote="Bugi"]I think that there are too many C & P problems, and that there aren't many hard problems in other areas (only the easiest).[/quote]

As AIME15 notes, Alcumus currently only includes Introductory C&P problems.  If you find them too easy, then it's probably time for you to seek out [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=300]a[/url] [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]greater[/url] [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=102]challenge[/url].


$ x,y,z$ are positive numbers such that $ \frac{1}{x\plus{}4}\plus{}\frac{1}{y\plus{}4}\plus{}\frac{1}{z\plus{}4}\equal{}\frac{1}{3}$. Prove $ (x.y.z)_{min}\equal{}125$

What is wrong with this:

Wlog $ a\leq b \leq c$, $ a,b,c > 4$.

$ 3/c \leq 1/a \plus{} 1/b \plus{} 1/c \equal{} 1/3 \leq 3/a$, so we get $ a\leq 9$ and $ 9\leq c$. We now see that $ \min(a,b,c)\leq 9$ ? :huh:

[quote="scarface"]$ x,y,z$ are positive numbers such that $ \frac {1}{x \plus{} 4} \plus{} \frac {1}{y \plus{} 4} \plus{} \frac {1}{z \plus{} 4} \equal{} \frac {1}{3}$. Prove $ (x.y.z)_{min} \equal{} 125$[/quote]
That's wrong  :D .  $ (y \equal{}z\equal{}\frac{2x\plus{}20}{x\plus{}1},x \rightarrow 0) \Longrightarrow  (xyz \rightarrow 0)$
 :cool:


Let a, b, c be positive real numbers satisfying abc"<"or"="1. Prove that:
(a/c+b/a+c/b)">" or"=" (a+b+c)

[quote="TRAN THAI HUNG"]Let $a, b, c$ be positive real numbers satisfying $abc\leq1.$ Prove that:
$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\geq a+b+c$[/quote]
$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=\frac{a^2b+b^2c+c^2a}{abc}\geq\frac{a^2b+b^2c+c^2a}{\sqrt[3]{a^2b^2c^2}}=\sum_{cyc}\sqrt[3]{\frac{a^4b}{c^2}}=$
$=\frac{1}{3}\cdot\sum_{cyc}\left(\sqrt[3]{\frac{a^4b}{c^2}}+\sqrt[3]{\frac{a^4b}{c^2}}+\sqrt[3]{\frac{c^4a}{b^2}}\right)\geq\frac{1}{3}\cdot3\cdot\sum_{cyc}\sqrt[9]{\frac{a^4ba^4bc^4a}{c^2c^2b^2}}=\sum_{cyc}a.$ :)

Thank you for your answer.I have solved it for many hours.

[hide]
Let $abc=P$. If we can show that it works when $P=1$, then it can work for any $P\leq 1$ because
\[ \frac ac +\frac ba +\frac cb=\frac{\frac{a}{\sqrt[3]{P}}}{\frac{c}{\sqrt[3]{P}}}+\frac{\frac{b}{\sqrt[3]{P}}}{\frac{a}{\sqrt[3]{P}}}+\frac{\frac{c}{\sqrt[3]{P}}}{\frac{b}{\sqrt[3]{P}}}\geq \frac{a}{\sqrt[3]{P}} +\frac{b}{\sqrt[3]{P}}+\frac{c}{\sqrt[3]{P}}\geq a+b+c \]
So now we only consider the case where $P=1$. Then $\frac 1c =ab$ and the other variables are similar, so the left hand side is $a^2b+b^2c+c^2a$.

By AM-GM, $a^2b+2b^2c\geq 3\sqrt[3]{(a^2b)(b^2c)^2}=3b$. Similarly, $b^2c+2c^2a\geq 3c$ and $c^2a+2a^2b\geq 3a$, Adding up these inequalities and dividing by 3 produces the desired inequality.
[/hide]

[quote="TRAN THAI HUNG"]:? You are very good. Why don't you look this problem?
Let a trapezium ABCD (AB parallel with CD) satisfying  AngleC< Angle D.Prove that AC>BD
                           From:The examination question of Hungari 1955

Math is my life :rotfl:[/quote]

Um... this is very easy?

Let $h$ be the lenght of a perpendicular line between $AB$ and $CD$.
$sin{C}=\frac{h}{AC}$
$sin{D}=\frac{h}{BD}$
Since $\angle{C}< \angle{D}$, $sin{C}< sin{D}$
So $\frac{h}{AC}<\frac{h}{BD}$
thus $AC>BD$


inequalities
trigonometry
function
inequalities unsolved

For positive real numbers $a, b, c$ with $a+b+c = abc$, show that:

$\frac{1}{\sqrt{1+a^2}} + \frac{1}{\sqrt{1+b^2}} + \frac{1}{\sqrt{1+c^2}} \leq \frac{3}{2}$,

and determine when equality holds

[b]Note:[/b] Prove the inequality by dehomogenizing!

[quote="Armo"]Don't know about dehomogenuazing but $a=\tan{\alpha}, b=\tan{\beta}, c=\tan{\gamma} $works perfectly.[/quote]
Hmm...... I was going to post the same solution  :lol:
This is just Jenson inequality with $ \alpha+ \beta+ \gamma= \pi$  ;)

[quote="erdos"]
This is just Jenson inequality with $ \alpha+ \beta+ \gamma= \pi$  ;)[/quote]
Are you sure?

[quote="erdos"]Ok here si the complete proof :
put $ a=\tan{\alpha}, b=\tan{\beta}, c=\tan{\gamma} $
$\tan{\alpha}+\tan{\beta}+\tan{\gamma}= \tan{\alpha}\tan{\beta}\tan{\gamma}$
means that $\alpha,\beta,\gamma$ are the angles of a triangle, so $ \alpha+ \beta+ \gamma= \pi$
The inequality is equivalent to : $\cos{\alpha}+\cos{\beta}+\cos{\gamma} \leq \frac{3}{2}$
and this is Jenson inequality for the function $x \rightarrow \cos{x}$ which is concave  :lol:[/quote]

no, you need to prove the case that when one of them is negative. So it's not pure jensen.

Erdos, $f(x)=\cos{x}$ is convex on $x\in[\frac{\pi}{2},\pi)$ (that's why I asked and that's what siuhochung meant). And actually there is no any problem about proving that $\sum{\cos{A}}\le{\frac{3}{2}}$, it is very easy.

[quote="paladin8"]I think you can also do it using Cauchy-Schwarz and then Jensen pretty easily, too. But I don't know about dehomogenizing either...[/quote]

Yeah, this was actually the way I did it, but when I looked at the note with the problem, it said that if you dehomogenize, there is an excellent trig substitution, which would simplify the problem. This is probably what erdos and Armo showed.

By the way, what is dehomogenizing anyway?

Subsitute $1/x=a$, etc so $xy+yz+zx=1$, and by AM-GM,

\[\sum_{\text{cyc}} \frac{1}{\sqrt{1+a^2}} = \sum_{\text{cyc}} \frac{x}{\sqrt{x^2+xy+yz+zx}} = \sum_{\text{cyc}} \frac{x}{\sqrt{(x+y)(x+z)}} \]

\[\leq \frac{1}{2}\sum_{\text{cyc}} \left(\frac{x}{x+y}+\frac{x}{x+z}\right) = \frac{3}{2}.\]


limit
function
real analysis
real analysis unsolved

[quote="fedja"]Obviously does not exist. Is the function just the first combination of symbols that came into your head? :roll:[/quote]

No. My friend saw this in a book and asked me this sum.

[quote="Cyber-Shreyas"]Evaluate
$ \lim_{x \to \infty}|\cot x|^{lnx}$[/quote]

Excuse me , could you define the function $ f \left( x \right) \equal{} \cot x$

$ \lim_{x \to \infty}|\cot x|^{lnx} \equal{} \lim_{x \to \infty}| cotang x|^{lnx}$ ???  :huh:

$ \lim_{x \to \infty}| cotang x|^{lnx}$ is not exist.

Put :
$ x_{_k} \equal{} \frac{\pi}{4} \plus{} k \pi$
$ y_{_k} \equal{} \frac{\pi}{2} \plus{} k \pi$

We see $ x_{_k}$ , $ y_{_k}$ are two sup-sequence of x. We have :

$ \lim_{k \to \infty}| cotang x_{_k} |^{ln x_{_k}} \equal{} 1$
$ \lim_{k \to \infty}| cotang y_{_k} |^{ln y_{_k}} \equal{} 0$

$ \Rightarrow \not \exists \lim_{x \to \infty}| cotang x|^{lnx}$

[quote="brahman"][quote="Cyber-Shreyas"]Evaluate
$ \lim_{x \to \infty}|\cot x|^{lnx}$[/quote]

Excuse me , could you define the function $ f \left( x \right) \equal{} \cot x$

$ \lim_{x \to \infty}|\cot x|^{lnx} \equal{} \lim_{x \to \infty}| cotg x|^{lnx}$ ???  :huh:[/quote]

I'm sorry. I didn't get you???

[quote="Carcul"]Yes, $ \cot x$ is the cotangent function.[/quote]

It's not difficult to prove that : $ \lim_{x \to \infty}| \cot x|^{lnx}$ is not exist.

Put :
$ x_{_k} \equal{} \frac{\pi}{4} \plus{} k \pi$
$ y_{_k} \equal{} \frac{\pi}{2} \plus{} k \pi$

We see $ x_{_k}$ , $ y_{_k}$ are two sup-sequence of x. We have :

$ \lim_{k \to \infty}| \cot x_{_k} |^{ln x_{_k}} \equal{} 1$
$ \lim_{k \to \infty}| \cot y_{_k} |^{ln y_{_k}} \equal{} 0$

$ \Rightarrow \not \exists \lim_{x \to \infty}| \cot x|^{lnx}$ :D


advanced fields
advanced fields unsolved

I am trying to construct a simply connected covering space for $ \Sigma_2$, the compact genus 2 surface. A hint I got is to use the open disk model for the hyperbolic plane. But I don't have a clue what to do with it.

The hint does not strike me as very helpful, unless you already know the answer.

First think of how to construct the universal cover of a torus:  A torus is a square with opposite edges identified.  One can tile the Euclidean plane with squares in such a way that the edges match up "appropriately."  (Exercise: What the heck does "appropriately" mean?)

Meanwhile, the genus 2 surface is a regular octagon with certain edges identified.  It is not possible to tile the Euclidean plane with octagons in the same way that we did with squares, but it is IS possible to tile to hyperbolic plane with octagons (of the right size) in such a way that the edges match up "appropriately."  The disk model has very little to do with the actual solution, other than the fact that the tiling is perhaps easiest to *visualize* using this particular model of the hyperbolic plane.

Anyway, think about this, and if you still need to see more, try googling: hyperbolic plane octagon.


If $ n>1$Prove that $ (F_{n},n)\equal{}1$ when $ F_{n}$ is nth Fermat number.

Can anyone, please, explain to me how N.T.TUAN's hint can imply this problem.

If $ (F_n,n)>1$ then there exists a prime $ p$ such that $ p|n$ and $ p|F_n$. But then $ k\cdot 2^{n\plus{}2}\plus{}1\equal{}p\leq n$, contradiction!

[quote="N.T.TUAN"][hide="Hint"]Any prime divisor of the $ F_n$, where $ n\ge 2$, is of the form $ k\cdot 2^{n \plus{} 2} \plus{} 1$.[/hide][/quote]
how to prove it :?:

[quote="N.T.TUAN"][hide="Hint"]Any prime divisor of the $ F_n$, where $ n\ge 2$, is of the form $ k\cdot 2^{n \plus{} 2} \plus{} 1$.[/hide][/quote]
$ 5|F_{5k}$, therefore it is not true.
$ p|F_{p\plus{}r}$ were $ r\equal{}(5/p)$ ($ r\equal{}0, \  for \ p\equal{}5, \ r\equal{}1$ if $ p\equal{}\pm 1\mod 5$ and $ r\equal{}\minus{}1$ if $ p\equal{}\pm 2\mod 5$.
For example $ p|(n,F_n)$ if $ p(p\minus{}1)|n$ for $ p\equal{}\pm 1\mod 5$.

But $ F_n$ isn't  Fibonacci number.  :lol:

I am sorry, I think, that $ F_n$ is Fibonachi numbers.

[quote="keira_khtn"]Another beatiful property :$ F_n$ cannot be a part of amicable couple![/quote]
  Plz try it!It's very nice!


[hide="not quite the answer"]if you look at the first one and the last one, a=1, b=1, and c= -1. but that doesn't work with the second one. does it use complex numbers or anything?  would graphing it help?[/hide]

This is a nice question.

[hide="Hint"]From (1) and (2) you can get $ab+bc+ca$. From (2) and (3) you can get $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$. Now square $ab+bc+ca$.[/hide]


geometry
circumcircle
IMO
perpendicular bisector
isosceles
IMO 1997
Hi

It is known that $ \angle BAC$ is the smallest angle in the triangle $ ABC$. The points $ B$ and $ C$ divide the circumcircle of the triangle into two arcs. Let $ U$ be an interior point of the arc between $ B$ and $ C$ which does not contain $ A$. The perpendicular bisectors of $ AB$ and $ AC$ meet the line $ AU$ at $ V$ and $ W$, respectively. The lines $ BV$ and $ CW$ meet at $ T$.

Show that $ AU \equal{} TB \plus{} TC$.

[i]Alternative formulation:[/i]

Four different points $ A,B,C,D$ are chosen on a circle $ \Gamma$ such that the triangle $ BCD$ is not right-angled. Prove that:

(a) The perpendicular bisectors of $ AB$ and $ AC$ meet the line $ AD$ at certain points $ W$ and $ V,$ respectively, and that the lines $ CV$ and $ BW$ meet at a certain point $ T.$

(b) The length of one of the line segments $ AD, BT,$ and $ CT$ is the sum of the lengths of the other two.

Let $X$ be the second intersection of $BV$ with the circle. Since $AU,BX$ are symmetric chords in the circle wrt a line which passes through the center (the perpendicular bisector of $AB$), they have equal lengths, so in order to prove that $BT+TC=AU=BX=BT+TX$, it suffices to show that $TX=TC$. A simple angle chase will show that both $\angle TCX$ and $\angle TXC$ are equal to $\angle A$, so we're done.

P.S.

The condition on $\angle A$ ensured the fact that $BX=BT+TX$, and not $BX=TX-BT$, for example.

I note $x=m(\widehat {BAU}),y=m(\widehat {CAU})$. Thus, $m(\widehat {BTC})=2(x+y)=2A$

and $\frac{TB}{\sin (C-y)}=\frac{TC}{\sin (B-x)}=\frac{a}{\sin 2A}=\frac{R}{\cos A}\Longrightarrow$

$TB+TC=\frac{R}{\cos A}\cdot [\sin (B-x)+\sin (C-y)]=$

$=2R\cdot \cos \frac{B-C-x+y}{2}=2R\sin (C+x)=AU.$

We also discussed this in the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=32558 (in this topic, the problem was given in the weaker form "prove that $TB+TC\leq 2R$, where R is the circumradius of triangle ABC", but most of the solutions actually solved it in the stronger form).

  darij

Too easy for an IMO, isn't it! Just trigonometry.

I found really good solution:
Consider $X$ intersection $BT$ with cicrle around $ABC$. Now by easy angle chase we have $TX=TC$ . Now because $BVA$ is isosceles so is $VUX$ . Now we have $BX=AU$ and $BX=BT+TX=BT+TC$.

It's also necessary to argue that T is inside the circle.  If the angle BAC isn't the smallest in the triangle, T could be outside the circle, and then BT + TC  >  AU.  That's easy to see.

My solution ( essentially the same as [b]Wolowizard[/b] but still posting as it is an [i]IMO[/i] problem :P ).

[b]Proof.[/b] Firstly let us define a useful point. Let $K =^{Def} BT \cap (ABC)$. We proceed by the following claim.

[b]Claim:[/b] $TK = TC$.
$\rightarrow$ $TK=TC \iff \angle TCK = \angle TKC$. Since $ K \in (ABC) \implies \angle TKC = \angle BKC = \angle BAC$. Hence it suffices to show that $\angle TCK = \angle A$. Since $ \triangle WAC$ is [i]isosceles[/i] and so $\angle WCA= \angle CAW \implies \angle TCA = \angle CAW $. So it is now sufficient to prove that $\angle ACK = \angle BAV$. Which is again true as $\triangle VBA$ is [i]isosceles[/i].

Now we just need to show that $AU = BK$. $\triangle VAB \sim \triangle VKU \implies \triangle VKU$ is [i]isosceles[/i]. Therefore $\boxed { AU = AV + VU = VB + VK = BK = BT + TK = BT +TC} \blacksquare$
 [b][u]Note:[/u][/b] For those who are thinking that where was the first condition used, as [b]larkl[/b] correctly points out that if $\angle BAC$ is not the smallest then $T$ may lie outside $(ABC)$.

Let $CW$ meet $(ABC)$ again at $B'$. Simple angle chasing shows that $WB'=WU$. So, $AU=AW+WU=AW+WB'=CW+WB'=TC+TB'$. Extend $BT$ to meet $(ABC)$ again at $X$. Then, $$\angle B'BT=\angle B'BA+\angle ABT=\angle B'BA+\angle BAU=\angle B'CA+\angle BAU=\angle CAU+\angle BAU=\angle BAC=\angle BB'C=\angle BB'T$$
Hence, triangle $BB'T$ is isosceles. Thus, $AU=TC+TB'=TC+TB$ as wanted.

Almost feel bad for reviving it, but nobody mentioned that this is just follows from symmetry, no angle chase necessary. Just can't pass by.

Let $BV, CW$ meet $(ABC)$ at $X,Y$. Then $BX=CY$ like reflections of the same chord $AU$. So it's obvious that $BT+TC=BX=AU$.

Let $L,M,N$ be the midpoints of segments $\overline{AU},\overline{AB},\overline{AC}$, respectively.

[b][color=#f00]Claim: [/color][/b] $O$ is the $T \text{-excenter}$ of $\triangle TVW$.

[i]proof:[/i] An easy angle chase gives that $\overline{OV},\overline{OW}$ are the external angle bisectors of $\angle OVT,\angle OWT$, respectively. This proves the claim $\square$

Now as $L$ is the foot of perpendicular from $O$ onto line $\overline{VW}$, so it follows that $TV + VL = TW + WL = \text{semiperimeter of } \triangle TVW$. So we obtain that
\begin{align*}
BT + CT &= (BV-TV) + (CW + TW) = BV + CW + (TW - TV) = AV + AW + (VL - WL) \\
& = (AV + VL) + (AW - WL) = AL + AL = 2AL = AU
\end{align*}
This completes the proof of the problem. $\blacksquare$

[asy]
size(300);
pair C = dir(-30), B = dir(-140), A = dir(110), U = dir(-110) ;
dot("$A$",A,dir(90));
dot("$B$",B,dir(B));
dot("$C$",C);
draw(unitcircle);
pair O = (0,0), M = 0.5*A + 0.5*B, N = 0.5*A + 0.5*C ;
dot("$O$",O);
dot("$M$",M,dir(180));
dot("$N$",N);
draw(B--A--C);
dot("$U$",U,dir(-90));
draw(A--U);
pair V = extension(A,U,O,M), W = extension(A,U,O,N), T = extension(B,V,C,W);
dot("$V$",V,dir(30));
dot("$W$",W,dir(-140));
dot("$T$",T,dir(100));
draw(B--V);
draw(C--T);
draw(O--M);
draw(W--N);
pair L = 0.5*A + 0.5*U ;
dot("$L$",L,dir(-40));
draw(O--L);
draw(B--C);
[/asy]

[quote=avatarofakato]Let denote by $X$ a point of intersection of tangents in points $B,C$ to circumcircle of $\Delta ABC$ and by $Y$ an image of $C$ in symetry with respect to an angle bisector of $\angle BAU$ (obviosuly $Y$ lies on circumcircle of $\Delta ABC$). If $\alpha_1=\angle BAU, \alpha_2=\angle UAC$ then by simple angle chasing we get that $\angle WVT=2\alpha_1$ and $\angle VWT=2\alpha_2\Longrightarrow \angle BTC=2(\alpha_1+\alpha_2)=\angle BOC$. Now it is easy to see that $TBCX$ is cyclic so by Ptolemy's theorem we get: $BC\cdot TX = CT\cdot BX+BT\cdot CX=(BT+TC)BX$. But $\angle YAU=\angle BTX$, $\angle AUY=\angle BXT$ hence $\Delta AUY$ is similar to $\Delta TXB \Longrightarrow$ $\frac{TX}{BX}=\frac{AU}{BC}\iff \frac{BC\cdot TX}{BX}=AU \Longrightarrow AU=BT+TC$. qed[/quote]

can someone please tell what was the motivation behind constructing $Y$ ?

Let $D=BT\cap (ABC)$ and $E=CT\cap (ABC)$. Note that $\angle TCD=\angle ECD=\angle ECA+\angle ACD=\angle ABV+\angle WCA=\angle CAW+\angle WAB=\angle CAB=\angle CDT$, thus $TC=TD$, similarly $TB=TE$. Also as $CAEU$ is a isosceles trapezoid, hence $$AU=UW+AW=CW+WE=TC+TW+WE=TC+TE=TC+TB.$$
We are done.

Evan's video made me try this. I agree with him, this might be the easiest (kinda) recent IMO #2
Let $BT \cap \odot(ABC) = X$
Notice that $ABUX$ is an isosceles trapezoid because $$\angle BAU = \angle BAV = \angle ABV = \angle ABU$$
Now $\angle BTC = 2 \cdot \angle BAC$ which follows trivially from angle chasing. This implies that $$\angle BTC = 2\cdot \angle TXC \implies TB=TC$$ Now $$AU = BX = TB + TC \ \blacksquare$$

Let $D=\overline{BT}\cap (ABC).$ Since $$\angle BAU=\angle VBA=\angle DUA,$$ $ABUD$ is a cyclic isosceles trapezoid. Thus, $$\angle TDC=\angle BAU+\angle UAC=\angle ACD+\angle ACW=\angle DCT$$ and $TC=TD.$ Then, $$AU=AV+VU=TB+VT+VD=TB+VT+TC-VT=TB+TC.$$ $\square$

Let $BT$ meet $(ABC)$ again at $B_1$. Observe that $$\angle BTC = 180 - \angle TBC - \angle TCB = \angle BAC + \angle ABT + \angle ACT$$ $$= \angle BAC + \angle BAV + \angle CAW = 2 \angle BAC$$ and $$\angle TB_1C = \angle BB_1C = \angle BAC$$ so $$\angle TCB_1 = \angle BTC - \angle TB_1C = 2 \angle BAC - \angle BAC = \angle BAC = \angle TB_1C$$ which yields $TB_1 = TC$. Thus, we know $$TB + TC = TB + TB_1 = BB_1$$ so it suffices to show $ABUB_1$ is an isosceles trapezoid. Indeed, we have $$\angle ABB_1 = \angle ABV = \angle VAB = \angle UAB = \angle UB_1B$$ so $AB \parallel UB_1$, which finishes since $ABUB_1$ is cyclic. $\blacksquare$

[b]Remarks:[/b] This problem seems scarier than it actually is. Once you find $\angle BTC = 2 \angle A$, however, adding $B_1$ follows somewhat naturally, since we'd like to force $\angle BTC$ to be an exterior angle in order to form some isosceles triangle(s).

All you need to do is to construct a [s]parallelogram[/s] isosceles trapezoid!

Let $D = \overline{BT} \cap (ABC)$ and $E = \overline{CT} \cap (ABC)$. Observe that both $ABCD$ and $ABCE$ are isosceles trapezoids. Furthermore, by a simple arc chase, $DEBC$ is an isosceles trapezoid too. Ergo $$BT+CT=CE=AU.$$

Let $\angle BAU=\theta_b$ and $\angle CAU=\theta_c$. Of course, $$\angle ABT=\angle BCU=\theta_b$$ and $$\angle ACT=\angle CBU=\theta_c.$$ Note that $\theta_b+\theta_c=\alpha.$ We have $$\angle TBC=\beta-\theta_b$$ and $$\angle TCB=\gamma-\theta_c,$$ which means that $$\angle BTC=180-\beta-\gamma+\theta_b+\theta_c=2\alpha,$$ so $BTOC$ is cyclic.\\

WLOG the circumradius of $\triangle ABC$ is $\frac{1}{2}$. Then, $a=\sin\alpha$ etc. Thus, by law of sines, $$BT+CT$$ $$=a(\frac{\sin\beta-\theta_b+\sin\gamma-\theta_c}{\sin2\alpha})$$ and $$AU=\sin\angle ABU=\sin\beta+\theta_c.$$ Since $$\sin2\alpha=2\sin\alpha\cos\alpha=2a\cos\alpha$$, it suffices to show that $$\frac{\sin\beta-\theta_b+\sin\gamma-\theta_c}{2\cos\alpha}=\sin\beta+\theta_c.$$ By sum to product, we have $$\sin\beta-\theta_b+\sin\gamma-\theta_c$$ $$2\sin(\frac{\beta+\gamma-\alpha}{2})\cos(\frac{\beta-\theta_b-\gamma+\theta_c}{2})$$ $$=2\cos\alpha\cos(\frac{\beta-\theta_b-\gamma+\theta_c}{2}).$$ Therefore, the desired equation simplifies to just $$\cos(\frac{\beta-\theta_b-\gamma+\theta_c}{2})=\sin\beta+\theta_c.$$ However, we have $$(\beta+\theta_c)-(\frac{\beta-\theta_b-\gamma+\theta_c}{2})$$ $$=\frac{\beta+\gamma+\theta_b+\theta_c}{2}=90,$$ hence done.

[hide=sketch]
Let $BT$, and $CT$ intersect $(ABC)$, at $X$, and $Y$ respectively.
$$\angle UAC= \angle TCA= \angle YCA$$
So $UCAY$ is an Isosceles trapezoid, similarly, $BUXA$ is an Isosceles trapezoid. Since $\widehat{BUC}=\widehat{YAX}$, then $BCXY$ is an isosceles trapezoid.
$$TB+TC=TB+TC=XB$$
So it suffices to show $XB=AU$, but since $\widehat{XA}=\widehat{UB}$, we can "rotate" $XB$ to $AU$ on $(ABC)$. Which implies the problem $\square$

Sketch:

$\ell_1$ and $\ell_2$ denote the perpendicular bisectors of $AB$ and $AC$ respectively.

Reflect $U$ over $\ell_1$ to get $U'$ and over $\ell_2$ to get $U''$.

$UCAU''$ and $UBAU'$ are isosceles trapeziums.

We also get that $\overline{T-C-U''}$ and $\overline{T-B-U'}$ are collinear from a litlil bit of angel chessing.

Then chasing some more angels, we get $BCU'U''$ is also an isosceles trapezium $\implies AU = U''C = U''T + TC = TB + TC$. :yoda:


calculus
integration
logarithms
algebra unsolved
algebra

Hi,

I hope someone can please help me.  I am new here so if my question has been asked a thousand times before, Im sorry.

I need to integrate q^2 / ((1+q^2)^0.5) with respect to q.

I wouldnt know where to start.  Thats why I called my self just_a_hack    :oops:

Thanks very much for your help.

This is not the right place to post this question, but this is your first post, so welcome and:

recall the formula
\[
\int f^{\prime}g=fg-\int fg^{\prime},
\]
and observe that
\[
\frac{d}{dq}\left(  \sqrt{1+q^{2}}\right)  =\frac{q}{\sqrt{1+q^{2}}}.
\]
Therefore,
\[
A=\int\frac{q^{2}}{\sqrt{1+q^{2}}}dq=\int q\cdot\left(  \sqrt{1+q^{2}}\right)
^{\prime}dq=q\sqrt{1+q^{2}}-\int\sqrt{1+q^{2}}dq.
\]
On the other hand,
\[
\int\sqrt{1+q^{2}}dq=\int\frac{1+q^{2}}{\sqrt{1+q^{2}}}dq=\int\frac{dq}
{\sqrt{1+q^{2}}}+A=\ln\left|  q+\sqrt{1+q^{2}}\right|  +A.
\]
It follows that
\[
A=\frac{1}{2}\left(  q\sqrt{1+q^{2}}-\ln\left|  q+\sqrt{1+q^{2}}\right|
\right)  +\mathcal{C}.
\]
Next time, surf this site and find the appropriate place to post such questions :)
I have edited this to add: post your questions even if you are not sure where. You will be welcomed. This is one of the reasons this site exists (I hope :))


calculus
integration
LaTeX
algebra
polynomial
derivative
number theory

Is the set S={ {n :beta: } where  :beta: is any irrational no & n varies over positive integers}

I have tried to solve it but my solution became too long & confusing...  :(

Hmm.. Does it have so much to do with rationality? This holds, for instance, for all $\beta\in(0,1)$, because the sequence $a_n=n^{\beta}$ satisfies $a_n\to\infty$ and $a_{n+1}-a_n\to 0$. It's easy to see that for such a sequence, $\{a_n\}$ is dense. I think I remember a problem (its solution, actually, posted by Pierre, I believe :?) from the Number Theory section, which essentially showed that if $a_{n+1}-a_n\to\infty$, and $(a_{n+2}-a_{n+1})-(a_{n+1}-a_n)\to 0$, then $\{a_n\}$ is dense in $(0,1)$, but I'm really not sure about that. If it's true, this would show that it holds for $\beta\in(1,2)$ as well, and maybe it can be extended to all non-integral, positive $\beta$.

P.S.: I suppose you meant to pose the problem for positive $\beta$, right? Otherwise it's clearly untrue, since $n^{\beta}$ is positive and goes to $0$ as $n\to\infty$.

[quote="the game"]I am extremely sorry Grobber I havent completed the problem. The Problem is the following :Is the set S={ {n* :beta: } where  is any irrational no & n varies over positive integers} dense in [0,1].

By n* :beta: ,I meant n multiplied by  :beta:

I am very sorry for the error. :blush:[/quote]

Then this is [b]completely[/b] different, isn't it? :) Moreover, it's probably the fourth time I see it on the forum. I'll try to find the latest thread where it was discussed.

[quote="Myth"]Anyway, I want to know the complete proof for $n^\beta$.[/quote]

I, for one, do not have one :). I was just sort of "thinking out loud" :).

Here's that link I was talking about:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27475&postorder=asc&highlight=equidistributed[/url]

I ll definitely learn Latex because its horrible difficult for me to type the problems or the solutions.
By the way n :beta:  makes me recollect Beatty sequence..

O God I even posted in that problem..
But what Ravi said that the denseness can be proved by PHP has really interested me.I am trying to see if such an "elementary" solution exists or not..

What does PHP mean? Anyway, the denseness (not the equidistribution) is proven in a rather elementary way in the last post on that topic :?. Maybe you didn't see it because it was on the second page? :?

I sort of used that in the post I mentioned.

[quote="Myth"]Anyway, I want to know the complete proof for $n^\beta$.[/quote]
For non-integer $\beta>0$ this can be done using Weyl's lemma exactly in the same way as for fractional parts of polynomials whose leading coefficient is irrational. You start with the obvious situation of sequences $\{a n^\beta+o(1)\}$, $0<\beta<1$ and show that they are uniformly distributed on $[0,1]$. Then, for $\beta\in(1,2)$, you consider the sequences $x_n=a n^\beta+b n^{\beta-1}+o(1)$ and see that for each $k\geq 1$, $x_{n+k}-x_n=ak\beta n^{\beta-1}+o(1)$ has fractional part uniformly distributed over $[0,1]$. This allows you to apply Weyl's lemma and conclude that $\{x_n\}$ is also uniformly distributed on $[0,1]$ and so on. I remember that Myth dislikes using Weyl's lemma but I do not know whether any elementary proof exists. After all, the lemma was invented just to prove this kind of statements...

[quote="grobber"]What you're referring to as Weyl's Lemma is the same as [url=http://mathworld.wolfram.com/vanderCorputsTheorem.html]Van der Corput's Theorem[/url], right, fedja? :?[/quote]
Yes, grobber, your reference is more precise. I am just guilty of calling everything that easily follows from Weyl's criterion "Weyl's lemma". I'll try to be more accurate with names in the future though cannot guarantee perfection...

I think [b]too[/b] few people on this forum can say that even the case $\{an^{\beta}+o(1)\}$ is obvious.
Извините, Федор, но ?то звучит некорректно.

[quote="Myth"]I think [b]too[/b] few people on this forum can say that even the case $\{an^{\beta}+o(1)\}$ is obvious.[/quote]
I apologize, Myth. :blush: I have to figure out the level of the forum yet. Also I am guilty of writing an answer keeping in mind the level of the person who asked the question rather than that of the entire community. Sometimes it's inevoidable, but definitely this case isn't one of them. As to the case $an^\beta+o(1)$ with $\beta\in(0,1)$, it follows right from the definition of the uniform distribution that $o(1)$ doesn't matter. After that the key observation is that for large $n$, the derivative of $n^\beta$ doesn't change noticably whille the sequence $a n^\beta$ runs over an interval of length $1$, so essentially we deal with a sequence of arithmetic progressions running over $[0,1]$ with step of progressions tending to $0$. My main point was that this case is "elementary" in the sense that it doesn't require Weyl's criterion. The word "obvious" was used incorrectly, indeed... My apologies to anyone who might get offended by it. I'll try to avoid this word in the future...


geometry
3D geometry
probability
geometric transformation
rotation

Seven different colours are allowed to paint the faces of a cube.
Each face of the cube should be painted with exactly one of the colours .
Let N(k) denote the number of different ways of painting the cube with k colours.
( [b]two patterns are considered to be the same if one can be obtained from the other by a rigid body motion[/b].)
Then what is the maximum value of N(k)?

How is this question different from:

How many different (up to rigid motions) ways are there to paint the six faces of a cube, each with one of seven colors?

I think k can be 1,2,3,4,5,6,7.
Clearly, N(1) = 1*7=7 because if we only use 1 colour, then all faces have to be the same, and there is exactly one pattern for each colour.
Besides, N(7) = 0 because it is impossible to paint the 6 faces with 7 colours.
The other values of N(k) has to be considered too.
Then we have to find the maxiumum value of N(k) among k = 1,2,3,4,5,6,7;
or else you need to point out which k will get the maximum number of N(K) and also the corresponding value of N(k).

For you problem, you add up all the values of N(K), i.e. you will count N(1)+N(2)+N(3)+N(4)+N(5)+N(6)+N(7).

That is the difference between the two problems.

Aren't the two problems the same?? :huh: I think the answer is $N(7)$. This is how many different ways you can paint the cube with 6 of the 7 colors, plus number of ways with 5 of the 7 colors, 4 of the 7, etc.  :?

[quote="luimichael"]Probably I didn't translate the problem well.

N(k) = number of ways of painting the cube with [b]exactly[/b] k colours.

I think it will make the question clear now.[/quote]

Ok, using this interpretation.
[hide]I believe the highest $P(n)$ is when $n=6$. So we start with the top side. There are 6 possible color choices. Then we move to the 4 side faces. If we number the colors used, 1, 2, 3, and 4, there are 4 different configurations that are the same with rotations. For example, 1234, 2341, 3412, and 4123 are all the same. We know there are $4!$ configurations. So we have $\frac{4!}{4}=6$.
$6*6=36$
There is only one color left for the bottom, so the answer is still 36 ways. That seemed a bit too easy. I'm horrible at probability and stuff, so I'm positive this is wrong. I don't think my solution accounted for all rotations, just rotations around the vertical axis of the cube. :( [/hide]

[hide]The highest is definitely going to be at $k = 6$.  We have $7\cdot6\cdot5\cdot4\cdot3\cdot2$ possible colorings, but the cube has 24 rotations, so we have the desired number is 210.[/hide]


Let be given a function f which is continuous in [0, 1] and diffrentiable in (0, 1) satisfying g(0) = g(1) = 0. Show that there exist c from (0, 1) such that  g'(c) = g(c).

<i>Source: VUMC 2000 </i>

Consider the  function  f(x)=g(x)*e^(-x).
Its  easy  to  see  that  this  function  is  continous  at [0,1] , differenatble  at  (0,1)  and  f(0)=f(1)=0 . So  by  theorem  Rolle  there  exists  at  least  on  c  in (0,1)  such  f'(c)=0 <=> g'(c) = g(c).


Can someone post the sprint, target, team and solutions for this yr??

does anyone have it on their comp? one that they scanned from their packet would be fine

someone e-mailed mathcounts and it's a violation of copyright laws, to post the tests.

I'm glad that MathCounts gave out the answers, but it would be nice if we had the solutions like they have for chapter and state. Does anybody know some way to obtain them?

I have solutions for 2000, 2001, and 2002.  :|

Really, I just have seen some very rough solutions from the mid-90s, because our coach's mother used to be the head question writer for nationals.

No kidding! - I really like how they have the chapter and state ones online so we can see how to do those pesky problems. I guess I'll just have to make my own solutions using the problem and answer. :roll:

[hide][hide][hide]Then, make copies of your solutions and sell them!!!! ;)   :lol:  :lol:  :idea:    [/hide] [/hide][/hide]

[quote="mathgeek2006"]Then, make copies of your solutions and sell them!!!!1 ;)   :lol:  :lol:  :idea:[/quote]
good idea

[quote="09husbri"]Shhh! Nobody heard that but me, alright?! :?[/quote]
lol :D  :ninja:


I have a dell laptop with Windows Vista. The laptop works great. It's not the laptop. I just know to save my tex files very frequently because at unpredictable intervals Windows will tell me "TeXnicCenter has stopped working..." and close the program, and I'll have to re-open the program to continue working.

Any thoughts? How to fix this? Thanks!

Have a look at the Event Viewer in case that tells you anything. You can try reinstalling or installing the latest version at http://sourceforge.net/projects/texniccenter/. Also try Run as Administrator.

If that doesn't work see if you can find some common factor. Some people have reported crashes when using Bibtex and/or JabRef or the Navigator. Have a look at the [url=http://sourceforge.net/tracker/?group_id=36323&atid=416430]Bugs Tracker[/url], [url=http://sourceforge.net/tracker/?group_id=36323&atid=479426]User Reports[/url] and the [url=http://www.latex-community.org/forum/viewforum.php?f=22]Support Forum[/url].


Hello,
Could anyone post a solution to this problem:

If n > 9, then n(n+1)(n+2)(n+3) is divisible by a prime >= 13.

Cheers,
Thanos

very amusing indeed.  i did not think it was true, but I used a program to check over numbers up to ten million, and it is all true.

If our number is divisimble by a prime p can be found out by the formula:
[n+4/p]+[n+4/p^2]+ ........    -
[n/p]+[n/p^2]+......  <>0 or not.

Now using this formula for p=13,17,19,23...... and at some point, with a lot of patience, we will get that for any n, the expression will be correct.

There does exist sort of an exhaustive solution with many cases (and completely elementary at that), but it is rather cumbersome and, hmm, not terribly elegant! This is why I post the problem here; I'd like some brilliant olympiad argument.

Thanks anyway!

Cheers,
Thanos


[quote="paganinio"]on top left corner.'Olmypiad' is wrong.[/quote]Thanks I'll correct it.

[quote="Valiowk"]Did you pay any attention to what Pierre told you some time ago, Valentin? ;)[/quote]Yes I did change it then, but I reuploaded some old files when the board was down, so that's how this re-appeared :)


Is this acceptable?

certainly hope i'm not violating any copyrights by doing this >_<

thanks

13. How many positive integers n satisfy the following condition:

(130n) 50  > n 100  > 2 200

(A) 0  (B) 7  (C) 12  (D) 65  (E) 125

here's a hint. re-write everything in powers of 50, then take the 50th root of each part. the result is fairly easy to solve.

Oooh, I liked this one. Another hint: rewrite it as two separate inequalities--(130n)^50>n^100 and n^100>2^200. That makes it much easier to wrap your mind around.


geometry
perimeter
floor function
combinatorics unsolved
combinatorics

Let $ t(n)$ for $ n \equal{} 3, 4, 5, \ldots,$ represent the number of distinct, incongruent, integer-sided triangles whose perimeter is $ n;$ e.g., $ t(3) \equal{} 1.$ Prove that

\[ t(2n\minus{}1) \minus{} t(2n) \equal{} \left[ \frac{6}{n} \right] \text{ or } \left[ \frac{6}{n} \plus{} 1 \right].\]

Umm first of all actually it should be $[\frac{n}{6}]$ or $[\frac{n}{6}+1]$.I've found a long proof based on counting like number of solution to $x+y+z=n$ where each of them are less than $\frac {n}{2}$ and also there doesn't exist any triples such that $\{x,y,z\}=\{x',y',z'\}$ and that is same as,
$t(n)=\frac{n_s(x+y+z=n)-n_s(2x+y=n)}{6}+\frac {n_s(2x+y=n)-n_s(3x=n)}{3}+n_s(3x=n)$ where,
$n_s(a_1+a_2+....+a_k=n)$ denotes as number of solutions such that all are positive integers and less than $\frac {n}{2}$. If someone has direct argument please post.

For all $(x,y,z)$ such that $x,y,z$ are the sides of a triangle with perimeter $2n-1$ and $x \ge y \ge z$, the triple $(x+1,y,z)$ is almost always a triangle with perimeter $2n$. The only time it's not a triangle is if $x+1 = n$, so $x = n-1$ and so $y+z = n$; there are $\left\lfloor \dfrac{n}{2} \right\rfloor$ such triangles, corresponding to $z = 1, 2, \ldots, \left\lfloor \dfrac{n}{2} \right\rfloor$. We can show that every triple $(x,y,z)$ generates a distinct $(x+1,y,z)$ easily; just note that in $(x+1,y,z)$, $x+1$ is the unique largest element, and subtracting $1$ from it always give $(x,y,z)$.

However, we didn't count triangles that are in the form $(x,x,z)$ where $x+x+z = 2n$ and $x \ge z$. There are $\left\lfloor \dfrac{n}{3} \right\rfloor$ such triangles, corresponding to $z = 2, 4, \ldots, 2 \left\lfloor \dfrac{n}{3} \right\rfloor$.

So $t(2n-1) - t(2n)$ is equal to the number of triangles of perimeter $2n-1$ whose longest side is $n-1$ (so that it cannot be extended by the first paragraph), minus the number of triangles of perimeter $2n$ that has two longest sides (isosceles; cannot be constructed by the first paragraph). So $t(2n-1) - t(2n) = \left\lfloor \dfrac{n}{2} \right\rfloor - \left\lfloor \dfrac{n}{3} \right\rfloor$, which I suppose can be easily proven to be equal to either $\left\lfloor \dfrac{n}{6} \right\rfloor$ or $\left\lfloor \dfrac{n}{6} \right\rfloor + 1$ (just check all cases of $n \bmod 6$).


function
calculus
derivative
complex analysis
real analysis
real analysis unsolved

it was already posted here . Finaly I don't find simple solution , an by using information from wolfram an mathworld I astimate $ a_{n}$. you can try to find it .

Consider the equation $ e^{e^x} \equal{} \sum_{n \equal{} 0}^{\infty}a_nx^n$.  Now it is easy to see that this series has infinite radius of convergence and hence converges uniformly on any compact subset of the real line.  Moreover the power series represents an entire function so its derivative is a power series that also converges uniformly on any compact subset of $ \mathbb{R}$.

Hence, differentiating both sides of the equation we get:

$ e^xe^{e^x} \equal{} d/dx(\sum_{n \equal{} 0}^{\infty}a_nx^n) \equal{} \sum_{n \equal{} 0}^{\infty}(n \plus{} 1)a_{n \plus{} 1}x^n$

But $ e^x \equal{} \sum_{n \equal{} 0}^{\infty}\frac {x^n}{n!}$ so we get that $ (\sum_{n \equal{} 0}^{\infty}\frac {x^n}{n!})(\sum_{n \equal{} 0}^{\infty}a_nx^n) \equal{} \sum_{n \equal{} 0}^{\infty}(n \plus{} 1)a_{n \plus{} 1}x^n$

By multiplying out the power series on the left and equating coefficients we get a recurrence relationship among the coefficients $ a_n$ which may be used to estimate the $ a_n$'s.


number theory
greatest common divisor
LaTeX

[quote="Hidden Scofield"]find the $ gcd(\overbrace{33\cdots 3}^{30},\overbrace{22\cdots 2}^{30})$ ?

 :D[/quote]
$ gcd(3 * \overbrace{11\cdots 1}^{30},2 * \overbrace{11\cdots 1}^{30}) \equal{} \overbrace{11\cdots 1}^{30}$

that's really a nice question and anice sol. :rotfl:  :rotfl:

ahh i dont know this much LaTeX!

oh well 333333333333333333333333333333 (30) = 300000000030000000003 (3 9 0's 3 9 0's 3) * 1111111111(10)

and 22222222222222222222 (20) = 20000000002 (2 9 0's 2) * 1111111111 (10)
 then, using a TI n-spire :D you see that 20000000002 and 300000000030000000003 have no common factors. of course you could be semi intuitive about the fact that they have no common factors or you could see that 15000000000 * 20000000002 = 300000000030000000000 which is 3 less than 300000000030000000003 which means if 20000000002 is a factor of 300000000030000000003  it has to be divisible by 3, which it is not by the divisibility rule of 3

so if you got lost, the answer is 111111111 (10)

stevenmeow found the value of $ \gcd(\overbrace{33\dots3}^{30},\overbrace{22\dots2}^{20})$, which is a bit different from the original problem.

there was one of my post here ...where is it??? :huh:

anyway, good work  :D

isn't nice problem  :wink:

the one including only emoticons?yes i noticed too.Someone must have deleted it.


show that
$n=\sum_{i=0}^{\infty}\lfloor{\frac{n+2^{i}}{2^{i+1}}}\rfloor$

for all natural $n$.

[hide="Hint"]Notice that $\left\lfloor x+\frac12\right\rfloor= \left\lfloor 2x\right\rfloor-\left\lfloor x\right\rfloor$. [/hide]
[hide="Solution"]$\sum_{i=0}^{\infty}\left\lfloor \frac{n+2^{i}}{2^{i+1}}\right\rfloor= \sum_{i=0}^{\infty}\left\lfloor \frac{n}{2^{i+1}}+\frac12\right\rfloor= \sum_{i=0}^{\infty}\left\lfloor \frac{2n}{2^{i+1}}\right\rfloor-\left\lfloor \frac{n}{2^{i+1}}\right\rfloor=$ $\sum_{i=0}^{\infty}\left\lfloor \frac{n}{2^{i}}\right\rfloor-\sum_{i=0}^{\infty}\left\lfloor \frac{n}{2^{i+1}}\right\rfloor= \sum_{i=0}^{\infty}\left\lfloor \frac{n}{2^{i}}\right\rfloor-\sum_{i=1}^{\infty}\left\lfloor \frac{n}{2^{i}}\right\rfloor= \left\lfloor \frac{n}{2^{0}}\right\rfloor= n$. [/hide]

[quote="paladin8"][quote="Rzeszut"][hide="Hint"]Notice that $\left\lfloor x+\frac12\right\rfloor= \left\lfloor 2x\right\rfloor-\left\lfloor x\right\rfloor$. [/hide]
[/quote]

Wow... that is an excellent identity![/quote]

[hide="motivation for hint"]
well known:
$\lfloor x\rfloor+\lfloor x+\frac{1}{n}\rfloor+\lfloor x+\frac{2}{n}\rfloor+...+\lfloor x+\frac{n-1}{n}\rfloor=\lfloor nx \rfloor$[/hide]


function
geometry
3D geometry
complex analysis
number theory theorems
number theory

There exists the formula for [b]number[/b] of all divisord of some number. Whether exists something similar for the [b]sum[/b] of all divisors of the number?

Do you want $ \sigma(n) \equal{} \prod_{p|n} \frac{p^{v_p\plus{}1}\minus{}1}{p\minus{}1}$¿ (Where $ n\equal{} \prod p^{v_p}$ is it's prime factorisation)

[quote="ZetaX"]Do you want $ \sigma(n) \equal{} \prod_{p|n} \frac {p^{v_p \plus{} 1} \minus{} 1}{p \minus{} 1}$¿ (Where $ n \equal{} \prod p^{v_p}$ is it's prime factorisation)[/quote]
Yes! Thank you very much!
.... and I proved it!!! (after I seen the formula)  :D

The function $ \sigma_k(n) \equal{} \sum_{d | n} d^k$ has the property that $ \gcd(a, b) \equal{} 1 \implies \sigma_k(ab) \equal{} \sigma_k(a) \sigma_k(b)$.  This lets you calculate both $ \sigma_0$ (the number of divisors) and $ \sigma_1$ (the sum of the divisors) readily.  We can also calculate the sum of the squares, the cubes, the reciprocals...

I remember that when i still did math, i went to some math seminar and saw some formules on divisors of n represented as trigonometrical function of n. The special things is that those formules are proved using complex analysis and modular form!!!! ( I did not understand at all the seminar... :rotfl: , one of factors that leads to the choice of no-math career )

A cool result from modular forms is: $ \sigma_7(n) \equal{} \sigma_3(n) \plus{} 120 \cdot \sum_{k \equal{} 1}^{n \minus{} 1} \sigma_3(k) \cdot \sigma_3(n \minus{} k)$.
Or in short: $ \sigma_7 \equal{} \sigma_3 \plus{} 120 \cdot \sigma_3 * \sigma_3$.
There are a lot of these identities, all comming from the classification of modular forms.


1. A point is situated inside an angle of measure 60degrees, at a distance ‘a’ & ‘b’ from its sides. Find the distance of the point from the vertex of the given angle.

2. Four unit circles pass through the origin and have their centres on the co-ordinate  axes. What is the area of the quadrilateral whose vertices are the points of intersection(in pairs) of the circles ?

a.	1 sqr unit
b.	8^(1/2) sqr unit
c.	4 sqr unit
d.	Cannot be determined – insufficient information

Tell me HOW to solve these along with the solutons..

1
[hide]Let the point be $M$, $MA=a$, $MB=b$ and the vertex of the angle be $O$.
As $\angle MAO= \angle MBO = 90^{o}$ $OBMA$ is cyclic, the diameter of the circumscribed circle is $OM$ and $\angle AMB=180^{o}-\angle AOB$.
Now by cosine theoreme
$AB^{2}=AM^{2}+BM^{2}-2AM*BM* \cos AMB$
$=a^{2}+b^{2}+2ab* \cos AOB$.
By sine theoreme $OM=\frac{AB}{\sin AOB}$ so
$OM=\frac{\sqrt{a^{2}+b^{2}+2ab* \cos AOB}}{\sin AOB}$

Question: what will change if $M$ is outside the angle?[/hide]

The second is much simpler. [hide="Hint"] The quadrilateral is a square, and find the distance from the origin to a vertex. [/hide]

[quote="fizsle"]#2 - why is it a square and if so why is the distance from the vertex to the origin sufficient to solve? :huh:[/quote]
I just instinctively knew that it was a square with center at the origin.
[hide] Let's take the 2 circles with centers on the positive x-axis and positive y-axis. Let's call their centers $A$ and $B$ and their intersection point $P$. $AO=OB=BP=PA=1$, so it is a rhombus. Since $\angle AOB=90^\circ$, we can use the properties of rhombuses to show that $AOBP$ is a square. From there, we find the area to be $\boxed{c.\ 4}$. [/hide]


Let f: R -> R^2 be a smooth curve in the plane. Let K be the set of all r in R such that the circle of radius r about the origin is tngent to the curve f at some point. Show that K has empty interior in R.

I guess I have to define  a smooth map to R such that all r in K are critical values. Then apply Sard's Theorem (the set of critical values of a smooth map has measure zero).
What kind of map should I look for???

You asked for [quote]a smooth map to R such that all r in K are critical values[/quote] There you have it.


number theory
greatest common divisor
modular arithmetic
algebra
polynomial
number theory unsolved

Find the greatest common divisor of the numbers \[ 2^{561}\minus{}2, 3^{561}\minus{}3, \ldots, 561^{561}\minus{}561.\]

We shall prove that the answer is $d  =  \prod_{(p-1) \mid 560} p  =  2\cdot 3 \cdot 5 \cdot 11 \cdot 13 \cdot 17 \cdot  29 \cdot 41 \cdot 71 \cdot 281$.

First of all, since $d\mid a(a^{560}-1)$, for $a = 2,\ldots,561$, it is clear that if $p\mid d$, and $p<561$, then $d\neq 0 \pmod {p^2}$. So let's try to first determine the prime divisors of $d$ (the common gcd of all the numbers) smaller than 561. Let $p$ be such a prime divisor.

Let $a$ be a primitive root of $p$. Then $a^{560}\equiv 1 \pmod p$, so because $a$ is a primitive root, $p-1 \mid 560$. It is easy to check that if $p-1 \mid 560$, then $a^{560}\equiv 1$, so $p\mid d$. That explains the above result.

Now let's suppose $d$ has a prime divisor greater than 561. Then consider the polynomial $f(x) = x^{560}-1$ in the field $\mathbb Z/p \mathbb Z$. This polynomial has 560 distinct roots, therefore it must be equal to $(x-2)(x-3)\cdots (x-561)$. Identifying coefficients, we need \[ \dfrac {561 \cdot 562 } 2 -1 \equiv 0 \pmod p \ \Rightarrow 561 \cdot 561 \cdot 281 \equiv 1 \pmod p,\] or \[ p \mid 2^5 \cdot 5 \cdot 7 \cdot 563.\] Therefore $p = 563$. However in this case $2^{562} \equiv 1 \pmod {563}$, therefore $2^{561}\equiv 282 \neq 2\pmod{563}$, and we are done.

Note that while this might seem connected to [url=http://www.mathlinks.ro/viewtopic.php?t=150408]this PEN problem[/url], it is actually not that much. Also 561 being a Carmichael number doesn't help us here at all.

What's with this 561? Are we celebrating something I'm not aware of? :P (maybe this: http://en.wikipedia.org/wiki/Huniade_Castle ) 2008 or even n would probably be better choices than 561, at least less confusing, and involving less computations.
If p divides all those numbers, 0,1,...,n are roots of $ X^n \minus{} X$ mod p, so at least two of them have to be equal, which gives $ p\leq n$. It is then clear that no p can have multiplicity greater than one in the gcd, and it's also clear that p is a factor of the gcd iff $ X^p \minus{} X|X^n \minus{} X$ in $ \mathbb{Z}/p$ iff $ p \minus{} 1|n \minus{} 1$.

[quote="Valentin Vornicu"]
First of all, since $ d\mid a(a^{560} \minus{} 1)$, for $ a \equal{} 2,\ldots,561$, it is clear that if $ p\mid d$, and $ p < 561$, then $ d\neq 0 \pmod {p^2}$.

[/quote]

Why is this true?

Well take $ a\equal{}p$ since $ p<561$ :) $ d\mid p(p^{560}\minus{}1)$ and we're done.

Here's a topic slightly connected with this one for those interested: http://www.mathlinks.ro/viewtopic.php?t=181757

What's the mean of $ \mathbb Z/p \mathbb Z$?

Thanks

So, I'll generalize the problem: for $ n > 1$,
\[ D_n : \equal{} \gcd\left(1^n \minus{} 1,2^n \minus{} 2,\dots,n^n \minus{} n\right) \equal{} \prod\limits_{p \textrm{ prime} \atop p \minus{} 1 \mid n \minus{} 1} p \,.
\]
To show that $ p \leq n$ implies $ p \minus{} 1 \mid n \minus{} 1$ is just like what Valentin Vornicu did.  Also, the proof that $ p > n$ then $ p \nmid D_n$ is also similar to his proof, as follows:

Suppose $ p > n$ and $ p \mid D_n$.  Thus, the polynomial $ x^{n \minus{} 1} \minus{} 1$ has $ n$ roots in $ \mathbb{F}_p$, which are $ 1$, $ 2$, ..., $ n \minus{} 1$, and $ n$.  This cannot happen since the degree of the polynomial is $ n \minus{} 1$.  This proves the assertion.

One more question: What would $ D^\prime_n : \equal{} \gcd\left(1^n \minus{} 1,2^n \minus{} 2,\dots,(n \minus{} 1)^n \minus{} (n \minus{} 1)\right)$, $ n > 2$, be?

Answer: $ D^\prime_n \equal{} D_n$ if $ n > 2$.

To show that a prime $ p \leq n \minus{} 1$ divides $ D^\prime_n$ implies $ p \minus{} 1 \mid n \minus{} 1$ is identical to the earlier work.  Now, if $ n$ is a prime, then $ n$ divides $ D^\prime_n$ and $ n$ also occurs in the factorization of $ D_n$ (for $ n \minus{} 1 \mid n \minus{} 1$).  Consequently, $ D^\prime_n$ is divisible by $ D_n$.

Suppose that there exists a prime $ p > n$ such that $ p \mid D^\prime_n$.  Thus, the polynomial $ x^{n \minus{} 1} \minus{} 1$ has $ n$ roots in $ \mathbb{F}_p$, which are $ 1$, $ 2$, ..., $ n \minus{} 2$, and $ n \minus{} 1$.  Thus, $ x^{n \minus{} 1} \minus{} 1 \equal{} (x \minus{} 1)(x \minus{} 2)\cdots\left(x \minus{} (n \minus{} 1)\right)$.  Therefore, the coefficient of $ x^{n \minus{} 2}$ is $ 0 \equal{} \minus{}1 \minus{} 2 \minus{} \dots \minus{} (n \minus{} 1)$ in $ \mathbb{F}_p$.  Hence, $ \frac {n(n \minus{} 1)}{2} \equal{} 0$ in $ \mathbb{F}_p$.  This implies that $ p \leq n$, which is a contradiction.

Showing for all $n$ instead of $561$ .Let $p$ be a prime dividing all these numbers. If $p > n$, then $p|i^{n-1}-1$ for all $0<i<n+1$. Hence $1,2,....n$  are roots for the polynomial $x^{n-1}-1$ in $Z_p[x]$,
therefore $x^{n - 1} - 1$ has $n$ roots in $Z_p$, which should imply this is the null polynomial, absurd. If $p\leq n$, then $p|i^n-i$ for all $0<i<n+1$ it follows that $p|a^{n - 1} - 1,$ where $a$ in ${I, 2,...,p - 1}$, hence $1, 2,.....,p-1$ are roots for the polynomial $x^{n - 1} - 1$ in $Z_p[x]$. But they are also the roots of the polynomial $x^{p - 1} - 1$, from Fermat's Theorem. Therefore, $x^{p - 1} - 1|x^{n - 1} - 1$, whence $p - 1|n - 1,$ since for $n - 1 = q(p - 1) + r$, with $0 <r < p - 1$, we have $x^{n - l} - 1 = x^r(x^{q(p-1)} - 1) + (x^r - 1)$, so $x^ {p - l} - 1 |x^r-1 - 1,$ ,So finally $r=0$ Conversely, for $p - 1|n - 1,$ it follows from Fermat's Theorem that $p|a^n - a$ for all $a$, hence $p$ divides all the numbers in the statement. Finally, since $p^2$ does not divide $p(p^{n-l} - 1)$, so  $p^2$ cannot divide all numbers of the statement. Hence $gcd=\prod_{p-1|n-1} p$ as we got $gcd$ is square free.  :)

The reason $561 = 3\cdot 11\cdot 17$ has been chosen is because its factors are in the list of prime factors of that $\gcd$, meaning it divides all those terms; so solving this problem has as a byproduct the proof that $561$ [b]is [/b] a Carmichael number :)

My solution:
Lemma: if $n\mid m$ if and if only $x^n-1\mid x^m-1$(it's well known)
Let $d=\text{gcd}(2^{561}-2,3^{561}-3,\cdots ,561^{561}-561)$ consider two cases let $p$ be a prime number such that $p\mid d$ now consider two cases:
1)$p>561$ then $p\mid i^{560}-1$ for every $1\le i\le 561$ it mean's that polynomial $x^{560}-1$ has $561$ roots modulo $p$ which is absurd.
2)$p\le 561$ then p\mid $i^{560}-1$ for $1\le i\le p-1$ it means that $x^{560}-1$ has the roots $1,2,\cdots ,p-1$ modulo $p$ also they are the roots of $x^{p-1}-1$(remind ferma's theorem!!) so $x^{p-1}-1\mid x^{560}-1$ by converse of the lemma $p-1\mid 560$ since $p^2\nmid p^{561}-p$ number $d$ is equal to product of the prime numbers like $p$ such that $p-1\mid 560$ after checking we get that $d=2.3.5.11.17.29.41.71.113.281$
DONE

Let's generalize
We claim that $\[ D_n \stackrel{\text{def}}{=} \gcd\left(2^n - 2,\dots,n^n - n\right) = \prod\limits_{p \textrm{ prime} \atop p - 1 \mid n - 1} p \,.
\]$
Now [b][color=#f00]Claim:[/color][/b] $p<n$
[b][color=#f00]Proof:[/color][/b] Consider the congruence $X^n \equiv (X-1)(X-2)(X-3)(X-4)........(X-n)$ in $[F_\mathbf{p}]$
By Lagrange's theorem,it has pairwise distinct solutions $X=1,2,3,4,5,.......,n$ in $[F_\mathbf{p}]$ which isn't possible,since if we choose a prime $P$ which is $\lfloor{\frac{n}{2}}\rfloor<P<2 \lfloor{\frac{n}{2}}\rfloor$ which divides only two terms so it can't divide $D_n$ which is a contradiction to the first line.
Now since $p|a^n-a$ i.e $a^{n-1} \equiv 1 \pmod p$ for all $a$,so $p-1|n-1$(Corollary of Lagrange's theorem),proving the claim.

quick little generalization ;)
[b]Claim: [/b]For some positive integer $n$, we have $$\gcd(2^n-2,3^n-3,\cdots,n^n-n)=\prod_{p-1\mid d-1}p.$$
[b]Proof: [/b]Let $d=\gcd(2^n-2,3^n-3,\cdots,n^n-n)$ be our desired common factor, and let $p$ be some prime factor of $d$.

We first prove that $p\le n$; suppose by contradiction that $p>n.$ As $p\mid d=\gcd(2^n-2,3^n-3,\cdots,n^n-n)$, we have that $p\mid k^n-k=k(k^{n-1}-1)$ for all $k\in\{2,3,\cdots,n\}.$ Thus, either $p\mid k$ or $p\mid k^{n-1}-1$ since $p$ is prime. If $p\mid k$, then $p\le k\le n$, a contradiction. Hence, we must have $p\mid k^{n-1}-1$ for all such $k,$ implying that $k^{n-1}\equiv1\pmod{p}.$ It follows that the polynomial $P(X)=X^{n-1}-1$ has roots $2,3,\cdots,n$ modulo $p$; since $P$ is monic and of degree $k$, we deduce that $$X^{n-1}-1\equiv(X-2)(X-3)\cdots(X-n)\pmod{p}.$$ Applying Vieta's Formulas on the coefficient of the $n-2$th degree monomial gives
\begin{align*}
0&\equiv2+3+\cdots+n\pmod{p}\\
&\equiv(1+2+\cdots+n)-1\pmod{p}\\
&\equiv\frac{(n+2)(n-1)}2\pmod{p}.
\end{align*}
Thus $p\mid (n+2)(n-1)$, so either $p\mid n+2$ or $p\mid n-1.$ Since we assumed that $p>n$, we cannot have $p\le n-1$ or $p\le\frac{n+2}2,$ forcing the equality $p=n+2.$ As a result, it follows that $$2^{n-1}\equiv1\pmod{n+2},$$ or $$2^{n+1}\equiv 4\pmod{n+2}.$$ Using Fermat's Little Theorem on $n+2=p$, the above becomes $1\equiv4\pmod{n+2}$, so that $n+2\mid 4-1=3$, giving $n=3-2=1$, in which case $d=\gcd(2-2,3-3,\cdots,n-n)=\gcd(0,0,\cdots,0)$, which is clearly absurd. Thus we may assume that $p\le n$ for all nontrivial $n\ge2.$

Since $p$ is prime, let $a$ be a primitive root modulo $p$. Then $1<a<p\le n$, implying that $p\mid d\mid a^n-a,$ so that $$a^n\equiv a\pmod{p}.$$ Since $p$ is prime and $a<p$, we find that $a\nmid p$ and thus $$a^{n-1}\equiv1\pmod{p}.$$ Due to a well-known property of orders, this implies by primitive roots that $p-1=\text{ord}_p(a)\mid n-1.$ Thus any prime factor of $d$ must be $1$ modulo $n-1.$ We show that $\nu_p(d)=1$ for such $p.$

First note that $p-1\mid n-1$, implying that $m=\frac{n-1}{k-1}\in\mathbb{Z}.$ Then Fermat's Little Theorem gives
\begin{align*}
k^n-k&\equiv k(k^{n-1}-1)\pmod{p}\\
&\equiv k((k^m)^{k-1}-1)\pmod{p}\\
&\equiv k(1-1)\pmod{p}\\
&\equiv0\pmod{p},
\end{align*}
so our construction of $p$ works, giving $\nu_p(d)\ge1$ for all primes $p$ such that $p-1\mid n-1.$

It remains to prove that $\nu_p(d)<2.$ Note that $p\nmid p^{n-1}-1$ implies that $p^2\nmid p^n-p$. As a result, we conclude that $\nu_p(p^n-p)=1$; since $p\le n$, we have $p\in\{2,3,\cdots,n\}.$ Thus, using a well-known property of the $p$-adic numbers, we get
\begin{align*}
\nu_p(d)&=\min(\nu_p(2^n-2),\nu_p(3^n-3),\cdots,\nu_p(n^n-n))\\
&\le\nu_p(p^n-p)\\
&=1.
\end{align*}
Combining this with the fact that $\nu_p(d)\ge1$ for such $p$, it follows that $\nu_p(d)=1$.

We therefore conclude that $\nu_p(d)=1$ for all primes $p-1\mid n-1$ and $\nu_p(d)=0$ otherwise. Hence $$d=\prod_{p-1\mid d-1}p.$$ $\blacksquare$