Alright guys,
So how bout we bring some stuff to do when we need physical exercise (I know...exercise?) or fun mid breaks (though problem solving is generally my fun mind break, so I guess fun mind break of a fun mind break).

Physical Exercise  :juggle: :
I'm guessing Basketball and Ultimate Frisbee are cool with most of the people. These games probably won't require much skill, considering this is math camp not the NBA or The Ultimate Frisbee International Summer League (U-FISL). Maybe we should have some sort of league. You know, have like 3 to 8 teams drafted on the first day and round robin afterwords.

Mind Break:
Chess is fun.
We could also play scrabble, monopoly, or something along those lines.
Definitely bring cards!!!
I can bring chips, but someone else bring some in case. (and everyone bring your bet...jk :P)

Any other ideas?

And don't worry Cali people....I'll copy and paste one for those of us going there.
Luckily I'm going to both  :D

haha there WILL be frisbee, I'll make sure of that :D

Last year, mafia was pretty epic so I hope that happens again.

There is also a pool which is nice to hang out in.

Darn we need starcraft on this list.
I do play chess (well I used to, beat a few 1700s before, now I don't exactly "play"), poker, and I can hold very long conversations about certain Russian composers, and about piano in general, although I don't really have the technique and record of competition participation and winning to match. Maybe if I stop failing at NT sometime during camp, hopefully there will be some ISL problems that I can find.

Last year volleyball was kinda popular...altho the sand can get hot...and swimming was probably the best...and the game room (pool...the game). Last year a bunch of people (me included) played king of the hill on the pool deck. oh, and i dont remeber a basketball court...was there one???hm..... the scavenger hunt was kinda fun(especially emily's veggetable- monty hall thing...we threw the celery and tomatos at birds) o and there was running every now and then in the morning.

at utd, there's an outdoor basketball court close to the soccer fields where ppl play frisbee

By the way is soccer a lost sport there?

I can bring risk :D best board game ever ftw.

Temperal says he's going to play 4' 33'' for the talent show.

AMP costs too much (for me).

4'33" is the best song ever. It requires very little practice!

people should bring some board games,

settlers of catan

if somebody has it, it would be nice to bring :D

I'm probably going to bring a card game called Munchkin. And a deck of cards, of course.

you ought to bring an ethernet cable and get hamachi or something because it is unlikely that you'll get the internet for starcraft

hmmm since I'm not from Texas I won't be bringing a computer, so I'm counting on everyone else!

tHE only sports I play COMPETITIVELY are tennis, and then swimming - swim team.. I do play little b-ball (or used to..  :D ) and like most of you say, Ultimate (or frisbee) is like for everyone, so. Again, does anyone play tennis? It seems like I am the only person who plays tennis..  :( .. and you can't play TENNIS by YOURSELF.. That's just horrible.

Um, "indoors" entertainment... UNLIKE most talented people @ AMSP, I tend to suck @ chess.. I also suck @ most card games except for ERS.. (does anyone know ERS or am I just being an alien..) ..

And.. as Altheman says... Mafia. Some party game..  :D .. You can't really be "good" at mafia, can you? hmm...

Looking forward... this is a local speaking btw (20 minute drive from UTD) and very looking forward to meeting everyone tomorrow!! (It's 7:52 AM local time here.)

yeah... we played ERS a lot at MC Nats... it was fun... we couldn't beat the NJ people though. :mad:

I haven't played since MC Nats, but my USCF rating for chess is 1564, so I don't think I'll be the best person there... hopefully there will be someone who will play w/ me... I'm not bringing a chess set though, but i'm bringing a laptop! :D

yeah, maybe i will bring my chess set... idk

neways, I suck at all athletic activities (although, in swimming i have state cuts for 50 breast, 100 breast, 200 breast, and 200 IM :D ). I'll be missing the state meet at AMSP, though... :(

so here is what I may do at AMSP:
chess
read
math

i'm not bringing my violin b/c i don't think i'll have time to practice... :(

Yeah, mafia is really fun (especially with Lehigh Valley students on the bus to ARML, I don't think you'll find 15 player games that take 2-3 hours in many places :P). I do think that some people are "better" at mafia than others, though, because when you play with the same people for some time, you eventually figure out how they act differently when they are mafia. And keeping track of voting patterns and connections, etc.

Oh, I'm probably also going to bring a set of scrabble tiles, because speed scrabble is really amazing.

Other fun party games include the watermelon game and thumper, both of which don't require any equipment.

Frisbee is about the only sport I don't epic fail at. :D

They have chess sets there(one floor down from where we eat)...will anyone bring SET???i learned it at camp last year and it was FUN!!!there was pool there and a whole lot of video games...^^

[quote="gatorlover124"]will anyone bring SET???i learned it at camp last year and it was FUN!!![/quote]

I <3 Set too... I play daily set online almost everyday! Although I don't have the SET cards... D: D:

I do hope someone brings set, also it is a brainy game? LOL... (In fact, so is chess and most card games.. D: lol)

All of you be prepared for the heat... It's not going to be majorly hot when you get here - it's forecasted to rain.. mid 90's.. But it's going to shoot up to mid 100's often, so yep.

Does it cool down much at night? Texas is in the desert right...

i didnt think it was that hot last year (then again my roomate put the temp @ 50 degrees last year COUGH COUGH TOM) and yea...also im from florida so...woot?

Unfortunately, from being a local, I know the weather well, and cooling down during the summer is just not what Texas is made for, even if it's, well, "desert."

It's not like a "desert" but it's still flat and all, and very explosed to the sun..

I have heard of many room people turning down the a/c to the 50's, thus requiring the need of heavy blanketing or people catching a cold and stuff, as my friend came out of AMSP with a cold. a 50 degrees temp difference is MAYBE SOMEWHAT HARD TO ADJUST TO.

If I were you, I'd bring not a "light blanket" but a moderately heavy one, or try to bring at least 2. Just in case, you know?

Guys guys guys I am right outside the UTD campus right now.  :o It's really kind of a pity that registration doesn't start for another few hours.

Remember to bring piano music, serial. I'm only bringing a bunch of fail pieces I play, and I need some stuff to try to sight read.

umm seriosly
if you decide to put the AC to 50 degrees during the day...
MAKE SURE TO TURN IT BACK UP.
UNLESS YOU WANNA GET SICK THAT IS
:]]]

ehem...my roommmmatteeeee [...cough MICHELLE cough...]
fajkhdsfahfslajfd um yeah.
and bring a moderate blanket. i freezed my butt off with just a fleece blanket UGHHH


trigonometry
geometry
trig identities
Law of Sines
geometry proposed

Please solve the following problem.

Let triangle $ABC$ be an  isosceles triangle such that $AB=AC$ and $BC=1$. We take a point $D$ on $AB$ such that $AD=1$.

If $\angle DAC=\frac{\pi}{7}$ , then find the length of $DC$.

Best regards!

kunny

Put x=cos(Pi/7); by sinus law AC = sin(3Pi/7)/sin(Pi/7)=4x^2-1
By cosinus law, CD^2 = 1+(4x^2-1)^2-2x(4x^2-1)
We can easily check that u = 8x^3-4x^2-4x+1 = 0;
as CD^2 = 2+2ux, it follows that CD = root(2)

Thank you for your reply, jpe.

You are right. To tell the truth, this problem was  given by junior high school student, so I would like to know the solution without using sine and cosine rule. Can anyone post such a solution?

kunny

I have just written a little note on that problem, "An adventitious angle problem concerning sqrt(2) and pi/7", avaliable from [url=http://de.geocities.com/darij_grinberg]my geometry website[/url]. I provide two synthetic solutions (one by Stefan V., which is a pseudonym, and one by me) to the problem together with its natural converse ($DC=\sqrt2\Longrightarrow\measuredangle DAC=\frac{\pi}{7}$). Hope you will like the proofs (sorry that the note was written quite in a hurry).

[b]EDIT:[/b] Thanks, Sprmnt21, for the notice - the problem has been discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=1439 .

  darij

Fantastic! Thank you, darij grinberg.

kunny

I saw same problem in japanese magazine.
"koukou heno suugaku" you see?

I prepare this problem for you, kunny.

line segment AB = line segment AC
Point P and Q are espectively in AC and AB
such that BC=BP=PQ=QA.
solve for AB. [AB=?×PQ]

from 00section shin kazama

[quote="shobber"]My calculation shows that $AB=2.24698 \cdot PQ$...[/quote]
Yeah, I calculated $\frac{1}{2}\csc \frac{\pi}{14}$ and got the same answer.

BTW, shin kazama, nice to meet you :)
[hide="If you are interested"]
Please post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45479]here[/url] ;) [/hide]

EDIT- I typed shin kazama's name wrong :oops:  Edited.

Nice to meet you, shin kazama! :)

[quote="shin kazama"]I saw same problem in japanese magazine.
"koukou heno suugaku" you see?[/quote]

Oh,really? Can you tell me the number of the issue?

Best regards.

kunny

Let, angle ACB=x, angle PBC=y.

 BC=BP=PQ=AQ
 angle BPC=angle BCP
 angle PBC=angle PQB
 angle QPA=angle QAP
 angle ABC=angle ACB

 Since the base angles of an isosceles triangle are congruent.

 angle QBP+angle PBC=angle ABC=angle ACB=x
 angle QBP+y=x
 angle QBP=x-y
 angle A=180-angle ABC-angle ACB
 {Sum of the angles of a triangle is 180}
 angle A=180-x-x=180-2x
 anglePQA+anglePQB=180{Supplementary angles}
 angle PQA=180-(x-y)

 I hope that explains why all the angles are labeled the way they  are.

 OK, now...

 add the angle of triangle PQA
 180-2x+180-2x+180-(x-y)=180
 540-5x+y=180
 360=5x-y
 also for triangle BPC
 2x+y=180
 so
 5x-y=360
 2x+y=180
_________add these two equations

 7x=540
 x=77 1/7
 y=180-2x=25 5/7
 x-y= 51 3/7
 and angle BPQ=180-2(51 3/7)=180-(102 6/7)=77 1/7=x
 so by the law of sines
 BQ/sin x=PQ/sin (x-y)
 BQ=PQ(sin x)/sin(x-y)
 AB=BQ+AQ=PQ(sin x)/sin(x-y)+AQ
 AB=PQ(sin x)/sin(x-y)+PQ
 AB=PQ[(sin x)/sin(x-y)+1]
 AB=PQ[(sin 77 1/7)/(sin51 3/7) +1]
 AB=2.25PQ

 so the side is 2.25 times the base.

 I'm very exhausted.

 To kunny
 I'm verry sorry. I can't remember the number.
 Probably, last year.

 00section shin kazama

Nice solution :)

You can actually just solve for one variable like this:

Let $a=\angle BAC$(=$\angle QAP$).
Since $QA=QP$, $\angle QAP=\angle QPA=a$.

$\angle BQP=\angle QAP+\angle QPA=a+a=2a$
$\angle QBP=\angle BQP=2a$

$\angle QPA+\angle BPC=\angle BQP+\angle QBP$, so $\angle BPC=2a+2a-a=3a$

Then $\angle BPC=\angle BCP=3a$ ($\because BP=BC$)

Also $\angle PBC=3a-2a=a$

Then, we see that $7a=180^\circ$ or $a=\frac{180^\circ}{7}$ ($=25\frac{5}{7}^\circ$)

From here, I simply used $\frac{AB}{\frac{1}{2}BC}=\frac{AB}{\frac{1}{2}PQ}=\csc\left(\frac{1}{2}\cdot \frac{180^\circ}{7}\right)$.

I also found a solution to the problem above:
You can find a point $E$ on $AC$ so that $AD=DE=EB=BC$ (that can easily be prooved). Now draw a circle so that the points $B,C,E$ lie on it. The point which lies on this circle and on $DC$ ist called $F$.  As $BC=EB$, $\angle{BEC} = \angle{ECB}$. And as $B,C,E,F$ lie on a circle, also $\angle{BFC}=\angle{BEC}$.
And together with $\angle{CBA}=\angle{ACB}$ follows $BCF\sim DCB$ and therefore $CF\cdot CD = 1$ and $\angle{BDC} = \angle{CBF}$. And as $\angle{CBF}=\angle{AEF}$(because $B,C,E,F$ lie on a circle), $\angle{BDC} = \angle{AEF}$ and it follows that $D,F,E,A$ lie on a circle, too.
$\angle{DAE}=\angle{AED}$ because $AD=DE$. And as $D,F,E,A$ lie on a cirle, $\angle{AED}=\angle{AFD}$ and it follows that $FAD\sim ACD$, because $\angle{AFD} = \angle{DAE}$. Therefore  $FD\cdot CD=1^{2}$. And together with $CF\cdot CD = 1^{2}$  it follows that $FD=CF=\frac{1}{2}CD$. $\Rightarrow \frac{1}{2}CD\cdot CD=1^{2} \Rightarrow CD=\sqrt{2}$


geometry
circumcircle
cyclic quadrilateral
projective geometry
geometry proposed

Let $ABCD$ is a circumscribed quadrilateral of $(O)$. $AC$ meets $BD$ at $I$, $AB$ meets $CD$ at $M$, $AD$ meets $BC$ at $N$. Prove that $OI$ is perpendicular with MN

Circle (O) tagent AB,BC,CD,DA at L,K,J,P
We have M is the polar of JL wrt (O),  N is the polar of PK wrt (O)
So the intersection of JL,PK is the polar of MN wrt (O)
Otherwise JL,PK,AC,BD concurrent at I so I is the polar of MN wrt (O)
Thus OI,MN are perpendicular q.e.d

The segment lines $JL,$ $PK,$ ( Huyen Vu's wording ) pass through the point $I\equiv AC\cap BD,$ by [b][size=109]Newton's theorem,[/size][/b] as well.

So, because of $I\equiv BD\cap JL\cap PK,$ the triangles $\bigtriangleup BKL,$ $\bigtriangleup DPJ$ are perspective and hence, by [b][size=109]Desarques’s theorem,[/size][/b]

we have that the points $Q\equiv KL\cap PJ,$ $M\equiv AB\cap CD$ and $N\equiv BC\cap AD,$ are collinear.

That is, the point $Q$ lies on the segment line $MN$ and similarly for the point $R\equiv KJ\cap PL.$

Hence, as well known, we conclude that $OI\perp QR\equiv MN$ and the proof is completed.

$($ In every cyclic quadrilateral (convex or non-convex), the line connecting the circumcenter with the intersection point of it’s diagonals, is perpendicular to the line connecting the intersection points of it’s opposite sidelines $).$

For an elementary proof see at: http://www.mathlinks.ro/Forum/viewtopic.php?t=110887

Kostas Vittas.

As an elementary approach to the polar theory, let us to present a solution, explanatory to the Huyen Vu’s one.

[b][size=109]LEMMA.[/size][/b] - Through an arbitrary point $I$ inwardly to a given circle $(O),$ we draw a line which intersects $(O)$ at points, so be it, $P$ and $K.$
We denote as $N,$ the intersection points of the tangents of $(O)$ at points $P,$ $K$ and as $S,$ the orthogonal projection of $N,$ on the segment line $OI.$
Prove that the point $S,$ is the harmonic conjugate of $I,$ with respect to $X,$ $Y,$ as the intersection points of $(O),$ from the segment line $OI.$

[b][size=109]PROOF OF THE LEMMA.[/size][/b] - We draw a line through $I$ and perpendicular to $OI\equiv XY,$ which intersects $(O),$ at points, so be it, $E,$ $F.$

The tangents of $(O),$ at points $E,$ $F,$ are intersected at point, so be it $S,$ lies on the segment line $OI,$ as well.

It is easy to prove that the point $S,$ is the harmonic conjugate of $I,$ with respect to the points $X,$ $Y$ and it is enough to prove that $NS\perp SO.$

We denote as $U,$ $V,$ the intersection points of the segment line $SF,$ from the tangents of $(O),$ at points $X,$ $Y.$

From $XU\parallel EF\parallel YV$ $\Longrightarrow$ $\frac{SX}{SY}= \frac{XU}{YV}= \frac{FU}{FV}= \frac{IX}{IY}$ $,(1)$

From $(1),$ we conclude that the points $S,$ $I,$ $X,$ $Y,$ are in harmonic conjugation and so, by [b][size=109]Newton’s theorem,[/size][/b] we have $(OX)^{2}= (OI)\cdot (OS)$ $,(2)$

$\bullet$ Let $X',$ $Y'$ be, the intersection points of $(O),$ from the segment line $NO$ and also let be the point $T\equiv PK\cap NO.$

By the same way as before, we have that the points $N,$ $I,$ $X',$ $Y',$ are in harmonic conjugation and so, we take $(OX')^{2}= (OT)\cdot (ON)$ $,(3)$

From $(2),$ $(3)$ and because of $OX = OX',$ $\Longrightarrow$ $(OI)\cdot (OS) = (OT)\cdot (ON)$ $,(4)$ and hence, the quadrilateral $ISNT,$ is cyclic, taken as diameter the segment $NI$ $($ because of $NT\perp IT$ $).$

We conclude now, that $NS\perp IS$ and the proof of the lemma is completed.

$\bullet$ Return to the proposed problem, we see that the wanted conclusion, is a direct application of the above Lemma, because of the orthogonal projections of $M,$ $N,$ on the segment line $OI,$ are coincided with the point $S,$ as the harmonic conjugate of $I,$ with respect to the points $X,$ $Y.$

Kostas Vittas.

[quote="Yosh..."]Which one is "newton theorem"?
I just know Newton line..

Can anyone give the source or give more explanation for that theorem?
Thx..[/quote]
This is another well known [b][size=100]Newton's theorem[/size][/b], which states that in every circumscribed quadrilateral, the lines connecting the tangent points of the incircle to its opposite sides, pass through the intersection point of its diagonals.

Unfortunately, I have not in mind any link about it and if someone knows a proof somewhere in the Forum, kindly please to inform us.

I will post later (next week ) a well known proof, if will be not any reference.

Kostas vittas.


function
limit
number theory
relatively prime
number theory open

For each n is positive integer  let f(n) is greatest number such that $2^{f(n)}|\phi(n)$.find all n satisfys f(n)=2006

[quote="ZetaX"]It would be nice to define $\phi$ ;)[/quote]

In number theory, $\phi(n)$ is used to denote the Eulerian $\phi$-function:
The number of integers in 1..n that are relatively prime to n.

I know, but that makes no sense here (just use the product expansion to see that you need some primes with $p-1$ having given $2$-adic valuations).

Does there exist $\lim_{n\to\infty}{f(n)}$

No, e.g. it gets the value $1$ infinetely often, but also the values $>1$: for the frist, any prime $\equiv 3 \mod 4$ works, for the second those $\equiv 1 \mod 4$.

If $gcd(m,n)=1$ then $f(mn)=f(m)+f(n)$ $\Rightarrow$ no exist $lim f(n)$

All solution's equation: $f(n)=m$ are $n=2^{l_0}p_1^{l_1}...p_k^{l_k}$, were $p_i$ are odd primes with condition:
$\sum_i ord_2(p_i-1) \le m.$ It give $k\le m,$ and if $l=m-\sum_i ord_2(p_i-1)$, then $l_0=l+1$ if l>0, else $l_0=0 \ or \ 1$.


Let $p_1, p_2, ..., p_n$ be distinct primes each larger than 3. Prove that number $2^{p_1p_2...p_n}+1$ has at least $4^n$ divisors.

this  problem is from shortlist 2002.you can find the solution from there.


geometry
parallelogram
circumcircle
cyclic quadrilateral
perpendicular bisector
geometry unsolved

Denote by $O$ the circumcentre of a cyclic quadrilateral $ABCD$. Let the lines $AC$ and $BD$ intersect each other at $P$. The circumcentres of triangles $\triangle{ABP}$, $\triangle{BCP}$, $\triangle{CDP}$ and $\triangle{DAP}$ are $O_{1}$, $O_{2}$, $O_{3}$ and $O_{4}$.

Show that the lines $OP$, $O_{1}O_{3}$ and $O_{2}O_{4}$ are concurrent.

$O_1O_2O_3O_4$ is a parallelogram, because $O_1O_2 \perp BD$ is the perpendicular bisector of BP, $O_3O_4 \perp BD$ is the perpendicular bisector of DP. Hence $O_1O_2 \parallel O_3O_4$ and similarly, $O_2O_3 \parallel O_4O_1$. The diagonals of this parallelogram cut each other in half at a point M. The quadrilateral $OO_1PO_3$ is also a parallelogram, because $OO_1 \perp AB$ is the perpendicular bisector of AB and $PO_3 \perp AB$ is the isogonal conjugate of of $PO_1$ in the triangle $\triangle ABP \sim \triangle DPC$ passing through the conjugate $H_1$ of its circumcenter $O_1$, i.e., through its orthocenter. Hence, $OO_1 \parallel PO_3$ and simiarly, $OO_3 \parallel PO_1$. The diagonals of this parallelogram also cut each other in half at the midpoint M of $O_1O_3$. Hence, OP passes through $M \equiv O_1O_3 \cap O_2O_4$.


How fast does a sound wave travel in air at a temperature of -6 degrres Celcius

The temperature dependence of the speed of sound in air is given by

$ v = 331.3 \sqrt{1+\frac{T}{273.15}}$

where v is the velocity and T the temperature in degrees Celsius. So for T = -6ºC we get v = 327.6 m/s.


calculus
analytic geometry
graphing lines
slope
derivative
algebra
linear equation

x^2 + y^2 = 169 is an equation for a circle.
how can i find the x and y intercepts? how can i find an equation of the line that is tangent to the circle at point (-12,5)?

For the x-intercepts set $ y\equal{}0$ and solve for $ x$. For the y-intercepts set $ x\equal{}0$ and solve for $ y$.

One way the tangent line can be found is with Calculus, just differentiate the equation and find the slope for the point.  Use the slope and point to find the y-intercept of the line to get the linear equation for the tangent.

[quote="linzk3"]how do i set x^2 + y^2 = 169  to the form of y=   ?[/quote]

Ever heard of algebra?

$ x^{2}\plus{}y^{2}\equal{} 169$
$ y^{2}\equal{} 169\minus{}x^{2}$
$ y \equal{}\pm\sqrt{169\minus{}x^{2}}$

Also, since all circles have an equation of $ (x\minus{}h)^{2}\plus{}(y\minus{}k)^{2}\equal{} r^{2}$ where $ (h,k)$ is the center and radius $ r$. If you are just graphing on graph paper, just draw a circle with radius 13 centered at the center..

And for your original question, note that this circle has radius 13 and that 5-12-13 triangles are always right triangles, and you can go from there. Differentiation isn't really needed here...


Let $x,y,z \geq  0$ and $x+y+z=1$, then prove the following ineq:

$\frac{x+1}{x+yz}+\frac{y+1}{y+xz}+\frac{z+1}{z+xy}\geq  9$

Interesting, I get it to form of \[\sum_{cyc}\frac{2x+y+z}{2x^{2}+y^{2}+z^{2}}\geq 9\] which looks aesthetic and plausibly has an easy solution


One  pro similar to pro http://www.mathlinks.ro/Forum/viewtopic.php?t=128397
Let triangle $ ABC$ inscribed circle $ (O)$. Let line $ \delta$ not congruent to $ BC$ and not through $ A$, $ \delta$ intersect $ AB, AC$ at $ M, N,$ respectively. $ J$ be intersection of circles $ (ABN)$ and $ (ACM)$. Find the locus of $ \delta$ such $ AJ$ through intersection of tangets from $ B, C$ of circle $ (O)$.

no one solve above problem
I thinks it not hard

Consider the inversion has $ A$ to pole and $ AB.AC$ to constant.
We have $ J' \in B'N' \cap C'M'$
$ AJ$ is the symmedian of $ \Delta ABC$, and not that $ \Delta AB'C'$ is symmetry with $ \Delta ACB$ wrt the angel bisector of $ \angle BAC$, so $ AJ'$ is the median of $ \Delta AB'C'$
It means $ M'N' \parallel B'C'$
So $ MN \parallel BC$


Let $ a$ and $ b$ be distinct real numbers. Prove that for each positive real number
$ x$,
\[ \frac {8x^2}{|a \minus{} b|} \plus{} \frac {a^2 \plus{} b^2}{x}\ge 6x
\]

Inequality is equivalent to $ \frac{\frac{2x}{|a\minus{}b|}\plus{}\frac{2x}{|a\minus{}b|}\plus{}\frac{a^2\plus{}b^2}{2x^2}}{3}\ge 1$
By AM-GM, we have $ \frac{\frac{2x}{|a\minus{}b|}\plus{}\frac{2x}{|a\minus{}b|}\plus{}\frac{a^2\plus{}b^2}{2x^2}}{3}\ge \sqrt[3]{\frac{2(a^2\plus{}b^2)}{(a\minus{}b)^2}}$.
We have $ \frac{2(a^2\plus{}b^2)}{(a\minus{}b)^2}\ge 1 \Longleftrightarrow (a\plus{}b)^2\ge 0$.


[hide="hint"]consider a pt. mass at $B$ since it is in equilibrium consider horizontal components of forces[/hide]


inequalities
trigonometry
real analysis
real analysis unsolved

$0<t<1$, how to show:

$|e^{it}-1|<t$?

(I tried Typlor, but is not straightforward as I thought.
There must be a simple proof...)

${t \geq 0}$

${\left|e^{i t}-1\right|\leq t \Rightarrow |\cos (t)-1+i \sin (t)|\leq t}$

${(\cos (t)-1)^{2}+\sin^{2}(t)\leq t^{2}\Rightarrow 2-2 \cos (t)\leq t^{2}}$

${t^{2}+2 \cos (t)\geq 2}$

Since the minimum is 2 and ${\frac{\partial^{2}\left(t^{2}+2 \cos (t)\right)}{\partial t^{2}}\geq 0}$


Given a sequence: $1$, $0$, $1$, $0$, $1$, $0 \cdots$ From the 7th term, each term equals to the [b]last digit [/b]of the sum of the 6 numbers before it. Show that there cannot be another series of $1$, $0$, $1$, $0$, $1$, $0$ in that sequence.

Editted to correct a mistake.

Um... are you sure that you didn't mistype this problem?

I'm a bit confused.

[hide]
As written, the sequence always increases from the 7th term onward, so the sequence will never reach 1 or 0 again, so the sequence never happen again.
[/hide]

Maybe I just don't understand something?

Sorry I made really a huge mistake here.
It should be: The[b] last digit [/b]of the sum of the first six numbers.

Interesting Problem:
[hide]
Let the series described be series $A$.
Let's assume that a series of $1,0,1,0,1,0$ can be repeated.
So looking back at the 6 previous entries, we have:
$a,b,c,d,e,f,1,0,1,0,1,0$.
Let $x$ and $y$ be nonegative arbitrary integers.
Let $y$ be the number that is excluded from consecutive series.
Basically $a+b+c+d+e+f=10x+1$ (Because first term of series that's present is 1).
$b+c+d+e+f+1=10x+2-y=10z$.  $\Rightarrow y=2$.
$c+d+e+f+1+0=10x-y=10z+1$.  $\Rightarrow y=9$.
$10x-9+1-y=10z\Rightarrow y=2$.
$10x-10-y=10z+1\Rightarrow y=9$.
$10x-19+1-y=10z\Rightarrow y=2$.
$10x-20-y=10z+1$.

So $2,9,2,9,2,9,1,0,1,0,1,0$.

But $3(2)+3(9)=33$. Contradition.

[b]QED[/b]
[/hide]


vector
geometry
geometric transformation
rotation
linear algebra
matrix
linear algebra unsolved

this one is easy but seems interesting:
let $M \in S_n^{+}(R)$ be a symetric positive definite matrix.
$(v_1,..,v_n)$ a family of vector such that:
1/$(v_i|v_j)=0$ for $i \neq j$
2/$(v_i|M(v_i))=1$ for all $i$.
(it is of course possible since $(v_i|M(v_i))=1$ means that $v_i$ is on an ellipsoid of equation $(v|M(v))=1$)

prove that $\sum_{i} \frac{1}{|v_i|^2}$ is a constant that
depends only of $M$.

Looks like if we are given two distinct systems of vectors $(v_i)_i,(v'_i)_i$ satisfying the hypotheses, we can turn one of them into the other one through rotations which fix a certain direction of the $n$ mutually orthogonal directions: assuming that $v_1\ne v'_1$, we keep $v_1$ fixed, and rotate the other $n-1$ directions in the hyperplane they generate until $v_2$ enters the $n-2$-plane orthogonal to the plane generated by $v_1,v'_1$. After this, we keep $v_2$ fixed and rotate the others until $v_1$ comes on top of $v'_1$. From now on, we only work in the hyperplane generated by $v_2,\ldots,v_n$.

In this manner, we have reduced the problem to $n-1$ dimensions, and we can keep doing this until we reduce it to the case of $2$ dimensions, and we only need to prove the assertion for an ellipse. This should be easy enough :).

I think I have easier proof. First, a quick look shows that it suffices to prove with $M$ diagonal with entries $ m_1,...,m_n$. Let $e_i=\frac{1}{||v_i||}{ v_i}$. They form an orthonormal basis. We have $\frac{1}{||v_i||^2}=<e_i,De_i>$. When we make the sum, unsing that $e_i$ are orthonormal basis we find $m_1+...+m_n$ as sum. Is it correct or I made again stupid mistakes?

Let $Q$ be the matrix with $i$-th column $\frac{v_i}{\|v_i\|}$. Then $Q$ is orthogonal and $N=Q^{-1}MQ=Q^TMQ=\begin{bmatrix}\frac{v_1^TMv_1}{\|v_1\|^2}&n_{12}& \cdots\\n_{21}&\frac{v_2^TMv_2}{\|v_2\|^2}&\cdots\\ \vdots&\vdots&\ddots\end{bmatrix}$
Since $v_i^TMv_i=1$ for all $i$, $\sum\frac1{\|v_i\|^2}=\mathrm{tr }Q^{-1}MQ=\mathrm{tr }M$.


Αν a(i)συμβολισω τον a με δεικτη i και  ισχύει a(i+1)-2a(i)+a(i-1)>=0 για κάθε δείκτη 1<=k<=n-1 για κάποιους δωσμένους αριθμούς a(0),a(1),...,a(n) όπου a(0)=a(n)=0 ΝΔΟ a(i)<=0 για κάθε 1<=i<=n-1.Θα ήθελα 2 τρόπους!Αλήθεια το συμβολίσω γράφεται με ι ή η?

έχουμε ότι  $a_{r+1}-a_r \geq a_r-a_{r-1}$. άρα επαγωικά δείχνουμε ότι αν  $a_s - a_{s-1}> 0$ ,τότε $a_n > a_s$.Υποθέτουμε ότι ο $a_s$ είναι ο μεγαλύτερος  θετικός, τότε   $a_s > a_{s-1}$, άρα  $a_n > 0$.  άτοπο.

Τη γραφική λύση του την εχεις δει?Είναι πολύ κάλη

Προτείνω και εγώ μια λύση.

Από τη δοθείσα σχέση έχουμε α(ν-2)>=2α(ν-1)   (*)
Αποδεικνύουμε επαγωγικά ότι α(κ+1)/(κ+1)>=α(κ)/κ για κάθε κ με 1=<κ<=ν-2 οπότε αρκεί να δείξουμε ότι α(ν-1)<=0
Όμως  α(ν-1)/(ν-1)>=α(ν-2)/(ν-2)>=2α(ν-1)/(ν-2)  και άρα [ν/(ν-2)]α(ν-1)<=0 οπότε το ζητούμενο αποδείχτηκε.

Όχι Κώστα δεν έχω δει τη γραφική της λύση αλλά το γεγονός ότι   $a_{r+1}-a_r \geq a_r-a_{r-1}$  μας δείχνει ότι οι τεθλασμένη γραμμή με άκρα τα σημεία

$(r,a_r)$  είναι κυρτή σωστά ? ;)

Σωστά αυτό εννοώ!

ΚΑΛΩΣΗΡΘΕΣ ΑΓΑΠΗΤΕ ΚΩΣΤΑ ΚΙ ΚΑΛΗ ΠΟΡΕΙΑ
ΧΑΙΡΕΤΩ ΚΙ ΤΑ ΥΠΟΛΟΙΠΑ ΠΑΙΔΙΑ


OK! Now 5.1 just one tough problem (not so much). but u know my exam is running at the same time IMO and IOI is knocking at the door. so i don't have any way than seeking help from webbie :D

Anyhow, i am talking about the latin square prime. backtraking? if so then i need to walk through special way like the Prime problem at 4.5 right? what's that way webbie?

And another the star problem it is ez. but i think i need some time to solve it. I hate these kinda probs which is straight but difficult to tough (time wasting) :mad:

So webbie reply to this post ok?

Section 5 is harder than the sections before (obviously)... Mainly because some of the task doesn't have any "standard" solution.

Fortunely, this task isn't all THAT hard. ;)

With a simple BFS algorithm (just keep track of which numbers in rows/columns that hasn't been used yet), you can make N <= 6 or so, but you'll fail on the hardest test case (N = 7).

What we need is a few optimisations...

One thing that might be useful is that number of latin squares is N! * (N-1)! * K where K is number of latin squares having "1, 2, 3, ... N" as both row and column! (for usaco, the answer would be only (N-1)! * K, because first row must start with "1, 2, 3, ... N")

For example:

If you want to calculate number of latin squares for N = 5, calculate how many that have this pattern:

1 2 3 4 5
2 ? ? ? ?
3 ? ? ? ?
4 ? ? ? ?
5 ? ? ? ?

And multiply with 5! * 4! (or, only 4! = 4 * 3 * 2 *1 to get the answer for USACO).

You'll probably need other optimisations too... (can't remember any right now :P).

Quick question, did both of you guys make the camp?

well i am not an USA citizen. again i don't have such a financial position that i can fly to usa for the camp. so ...

i hope u understood!

Nope... Didn't compete in the "proctored" Gold Division...

And, didn't have enough points either. ;) A small flaw on first one and couldn't think of any nice DP solution to the last one... But, the last task turned out to have a greedy solution... &%%¤#&%#&%#!!


geometry
circumcircle
modular arithmetic
geometry proposed

An acute triangle $ ABC$ inscribed in circumcircle $ (O)$. $ A_1,B_1,C_1$, respectively, are the feet of the perpendiculars from $ A,B,C$ to $ BC,CA,AB$. Let $ AO$ intersects $ (O)$ at the second point $ D$. Denote $ M,N,P$ the projections of $ D$, respectively onto $ BB_1,BC,CC_1$. $ AP$ intersects $ (O_c)$, with diameter $ AB$, by the point $ P'$, $ AM$ intersects $ (O_b)$, with diameter $ AC$, by the point $ M'$, $ AN$ intersects the $ (C)$ with diameter $ AH$ by the point $ N'$ (of course, $ N',M',P'$ are all distinct and different from $ A$). The symmedian from $ A$ wrt triangle $ AB_1C_1$ intersects $ (C)$ by the the point $ X$.
Prove that: $ X,M',N',P'$ are concyclic. :D

Still no solution  :lol: . Here is mine (A little ugly):
Consider the inversion pole $ A$, power $ \overline {AC_1}.\overline {AB} \equal{} \overline {AB_1}.\overline {AC} \equal{} k^2$, which turns: $ C_1\mapsto B$, $ B_1\mapsto C$. Therefore, $ I(A,k): BC\mapsto (C)$, $ CC_1\mapsto (O_c)$, $ BB_1\mapsto (O_b)$. Hence, through the inversion pole $ I$, power $ k^2$, we have, $ I(A,k): N\mapsto N'$, $ P\mapsto P'$, $ M\mapsto M'$, $ X\mapsto M_a$. So in order to prove $ X,M',N',P'$ concyclic, we need to prove that $ M_a,M,N,P$ are concyclic.
Indeed, let $ H$ be the orthocenter of triangle $ ABC$. Consider the symmentry center $ M_a$, which turns $ H\mapsto D$, $ N\mapsto A_1$, $ P\mapsto P_1$, $ M\mapsto M_1$. Therefore, we get, $ HP_1\bot BD$, $ HM_1\bot DC$. Hence, $ S(M_a): (MNP)\mapsto (M_1N_1P_1)$. Now, according to a very well- known result, we get, $ \overline {HA}. \overline {HA_1} \equal{} \overline {HB}. \overline {HB_1} \equal{} \overline {HC}. \overline {HC_1} \equal{} p^2$. Consider the inversion through pole $ H$, power $ p^2$,  which turns  $ A_1\mapsto A$, $ B_1\mapsto B$, $ C_1\mapsto C$, $ P_1\mapsto P'_1$, $ M_1\mapsto M'_1$, $ M_a\mapsto Q$. But it is easy to notice that $ (B,C_1,H,A_1,P_1)$ are concyclic, hence $ P'_1\in AC$.  With the same argument,  we get, $ M'_1\in AB$, $ Q\in (AC_1HB_1)$, further, $ Q\in (O)$.
Because the fact that $ HP_1\bot BD$, $ BD//CC_1$, we get, $ HP_1\bot CC_1$ by the point $ H$. Therefore, $ P_1HPD$ is a rectangle. Hence, $ (HP_1,HM_a)\equiv (P_1H,P_1M_a) \pmod \pi$. In the other hand, because $ \overline {HP_1}. \overline {HP'_1} \equal{} \overline {HM_a}. \overline {HQ}$, which leads to the fact that $ P'_1, Q, P_1, M_a$ are concyclic, hence $ (P_1H, P_1M_a)\equiv (P_1P'_1,P_1M_a)\equiv (QP'_1,QM_a) \pmod \pi$. Then $ (HP_1,HM_a)\equiv (HP'_1, HQ) \equiv (QP'_1,QM_a)\equiv (QP'_1,QH)\pmod \pi$, $ (1)$.  With the same argument,  $ (HM'_1,HQ)\equiv (QM'_1, QH) \pmod \pi$, $ (2)$. From $ (1)$ and $ (2)$, we get, $ (HP'_1, HM'_1)\equiv (QM'_1,QP'_1) \pmod \pi$. But it is easy to notice that $ (HP'_1,HM'_1)\equiv (AM'_1,AP'_1) \pmod \pi$. Hence, $ (AM'_1,AP'_1)\equiv (QM'_1,QP'_1) \pmod \pi$.  Therefore,  $ A,Q,M'_1,P'_1$ are concyclic, which leads to the result of the problem.
Our proof is completed. :lol:

[b]Remark[/b]: From this problem, I think there is an important notice about concyclic quadrilateral:
" Let triangle $ ABC$ be acute and inscribed in circumcircle $ (O)$. Let $ H$ be its orthocenter. Through $ H$, let $ d_1,d_2$ be the line respectively parallel to $ AB,AC$, intersects $ AC, AB$, respectively, by the point $ X,Y$. The circle with $ AH$ its diameter intersects $ (O)$ again at $ Q$. Then $ A,Q,X,Y$ is concyclic, further, $ XY$ is also parallel to $ AQ$." :D

No one is interested in this problem?? :blush:  :lol:

Still nobody is interested in this one?? :blush:

Yes, I am. But I think the second part of your solution has something lengthy and unnecessary. After the inversion to show that we must prove $ M,N,P,M_a$ are concyclic, we just need a little angle chasing:

$ \widehat{PNC} \equal{} \widehat{PDC} \equal{} 90^o \minus{} \widehat{MDP} \equal{} 90^o \minus{} \widehat{HDM} \minus{} \widehat{HDP} \equal{} \widehat{M_aMH} \minus{} \widehat{HMP} \equal{} \widehat{M_aMP}$

Thus, $ M,N,P,M_a$ are concyclic and we are done.

[b]Comment:[/b] This second part is quite simple, the key of the problem is that you use an inversion to create a harder concyclic set of point, isn't it? :P

@[b]mathVNpro[/b]: Poor you, "autofuner" :lol:


[quote="LeFromage"]Given real numbers $ a$, $ b$, and $ c$, such that $ a,b,c>0$, compute the minimum value of $ \frac{a^{3}}{4b}+\frac{b}{8c^{2}}+\frac{1+c}{2a}$. Find all ordered triplets $ (a,b,c)$ that yield this minimal value.[/quote]

$ \frac{a^{3}}{4b}+\frac{b}{8c^{2}}+\frac{1+c}{2a}$
$ =\frac{a^{3}}{4b}+\frac{b}{8c^{2}}+\frac{1}{2a}+\frac{c}{4a}+\frac{c}{4a}$
$ \geq5(\frac{a^{3}}{4b}\cdotp\frac{b}{8c^{2}}\cdotp\frac{1}{2a}\cdotp\frac{c}{4a}\cdotp\frac{c}{4a})^\frac{1}{5}$
$ =\frac{5}{4}$
Min for $ a=c=2$ and $ b=8$


Alerian can mow the same lawn twice as fast as Bryceson. Together tget can mow the lawn in 2 hours and 40 minutes. How long would it take Alerian to mow the lawn by himself?

How would you do this problem?

Hmm... I'm probably wrong but I remember reading somewhere that work = 1/T{_1} + 1/T{_2}... but then the answer is different... does anyone remember the formula?

Let Alerian's rate be $A$ and Bryceson's rate be $B$. Then, we have the two equations $2A=B$ and $160\left( \frac{1}{A}+\frac{1}{B}\right) = 1$.

Solving for $A$ yields $A=240\,\, minutes$ and this is $\bf{4\,\, hours}$.

There's no need to memorize a formula.  Everything stems from a very basic concept:

$\text{Amount}= \text{Rate}\times \text{Time}$

In this case, let's say that the entire lawn constitutes an $\text{Amount}$ of $1$ (that is, one lawn).  Alerian can mow twice as fast as Bryceson; therefore,

$\text{Rate}_{A}= 2 \times \text{Rate}_{B}$

And finally, if Alerian and Bryceson are working together, their rates add; in other words,

$\text{Rate}_{AB}= \text{Rate}_{A}+\text{Rate}_{B}= \frac{3}{2}\times \text{Rate}_{A}$

Inverting, we know that

$\text{Time}_{AB}= \frac{2}{3}\times \text{Time}_{A}$

From which the solution is easily deducible.  In general,

$\frac{1}{ \text{Time}_{AB}}= \frac{1}{ \text{Time}_{A}}+\frac{1}{ \text{Time}_{B}}$


geometry
geometric transformation
reflection
rectangle
conics
hyperbola
parallelogram

Given rectangle $ABCD \ (AB=a,BC=b)$  find locus of points $M$ , so that reflections of $M$  in the sides are concyclic.

It is, of course, equivalent to the fact that the projections are concyclic.

It appears to be a hyperbola passing through the four vertices of the rectangle and having the following two points as vertices: among the two perpendicular bisectors of the sides, take the longest one. The circle having it as diameter cuts the long sides of the rectangle in four points (or two, if the rectangle is a square). These four points are projected in two points (or one, if the rectangle is a square) on the mentioned perpendicular bisector. These two points are the vertices of the hyperbola, and in the degenerate case (rectangle=square), the locus seems to consist of the two diagonals.

I'll try to prove it (I'm not even sure it's true).

[quote="grobber"]It appears to be a hyperbola passing through the four vertices of the rectangle and having the following two points as vertices: among the two perpendicular bisectors of the sides, take the longest one. The circle having it as diameter cuts the long sides of the rectangle in four points (or two, if the rectangle is a square). These four points are projected in two points (or one, if the rectangle is a square) on the mentioned perpendicular bisector. These two points are the vertices of the hyperbola, and in the degenerate case (rectangle=square), the locus seems to consist of the two diagonals.[/quote]

This is all correct. Actually, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15103]for a general quadrilateral ABCD, the locus of all points whose reflections in the sides of the quadrilateral are concyclic is a cubic, the so-called [b]isoptic cubic[/b] of the complete quadrilateral formed by the lines AB, BC, CD, DA[/url]. But if the quadrilateral ABCD is a rectangle - but also, more generally, if the quadrilateral ABCD is an arbitrary parallelogram! -, then this cubic degenerates into the union of a rectangular hyperbola and the line at infinity. Since we will consider only finite points here, we ignore the line at infinity, and thus we will just show that the locus is a rectangular hyperbola.

In fact, I begin with some preliminaries:

We will use directed angles modulo 180° throughout the following.

If ABC is a triangle, and g is a line which doesn't pass through any of the triangle's vertices, then the isogonal conjugate of the line g with respect to triangle ABC (i. e., the locus of the isogonal conjugates of all points on g with respect to triangle ABC) is a conic passing through the vertices A, B, C of triangle ABC. If the line g passes through the circumcenter of triangle ABC, then this conic is a rectangular hyperbola. One example of a line g passing through the circumcenter of triangle ABC is, of course, the perpendicular bisector of the side CA. Thus, the isogonal conjugate of the perpendicular bisector of the side CA with respect to triangle ABC is a rectangular hyperbola passing through the vertices A, B, C of triangle ABC. This hyperbola will be called the [b]B-medial hyperbola of triangle ABC[/b]. Its center is the midpoint of the side CA. Of course, we can similarly define a C-medial and an A-medial hyperbola of triangle ABC (by considering the perpendicular bisectors of the sides AB and BC rather than of the side CA).

Now, the B-medial hyperbola is an important hyperbola and has lots of equivalent definitions:

[b]Theorem 1.[/b] Let M be a point in the plane of a triangle ABC. Consider the following possible properties of the point M:

[b](a)[/b] The isogonal conjugate of the point M with respect to triangle ABC lies on the perpendicular bisector of the segment CA.
[b](b)[/b] We have < MCB = < BAM.
[b](c)[/b] If Z and X are the orthogonal projections of the point M on the sides AB and BC of triangle ABC, respectively, then the midpoint of the segment CA lies on the perpendicular bisector of the segment ZX.
[b](d)[/b] If the point D is such that the quadrilateral ABCD is a parallelogram (so, the point D is the point where the parallel to the line AB through the point C meets the parallel to the line BC through the point A), then < BMC = < AMD.
[b](e)[/b] If Q is the point satisfying the vector equation $\overrightarrow{MQ} = \overrightarrow{BA}$ (in other words, the point Q is the image of the point M under the translation along the translation vector $\overrightarrow{BA}$), then the points A, M, D, Q are concyclic.
[b](f)[/b] The orthogonal projections of the point M on the lines AB, BC, CD, DA are concyclic.
[b](g)[/b] The reflections of the point M in the lines AB, BC, CD, DA are concyclic.

Then, each of these properties characterizes the B-medial hyperbola of triangle ABC. In other words, each of these properties holds if and only if the point M lies on the B-medial hyperbola of triangle ABC.

I will give a [i]proof[/i] of Theorem 1 without the property [b](c)[/b]; actually, I had given property [b](c)[/b] just for the sake of completeness (we won't need it anywhere further; you can prove [b](c)[/b] as a nice exercise). Since property [b](a)[/b] is clearly equivalent to M lying on the B-medial hyperbola of triangle ABC (since the B-medial hyperbola of triangle ABC is the isogonal conjugate of the perpendicular bisector of the segment CA), it is enough to show that the properties [b](b)[/b], [b](d)[/b], [b](e)[/b], [b](f)[/b], [b](g)[/b] are all equivalent to property [b](a)[/b]. We will show this by subsequently proving the following equivalences:

[b](a) $\Longleftrightarrow$ (b)[/b];
[b](b) $\Longleftrightarrow$ (e)[/b];
[b](e) $\Longleftrightarrow$ (d)[/b];
[b](d) $\Longleftrightarrow$ (f)[/b];
[b](f) $\Longleftrightarrow$ (g)[/b].

The [i]proof of [b](a) $\Longleftrightarrow$ (b)[/b][/i] is very easy: If M' is the isogonal conjugate of the point M with respect to triangle ABC, then the lines CM and CM' are symmetric to each other with respect to the angle bisector of the angle BCA. Thus, < MCB = < ACM'. Also, the lines AM and AM' are symmetric to each other with respect to the angle bisector of the angle CAB. Thus, < BAM = < M'AC. Hence, the condition < MCB = < BAM is equivalent to < ACM' = < M'AC. But we have < ACM' = < M'AC if and only if the triangle CM'A is isosceles, i. e. if and only if CM' = AM', what is clearly equivalent to the assertion that the point M' lies on the perpendicular bisector of the segment CA. Hence, we have < MCB = < BAM if and only if the isogonal conjugate M' of the point M lies on the perpendicular bisector of the segment CA. Hence, the equivalence [b](a) $\Longleftrightarrow$ (b)[/b] is proven.

Now we are going to [i]prove [b](b) $\Longleftrightarrow$ (e)[/b][/i]. In fact, since $\overrightarrow{MQ} = \overrightarrow{BA}$, the lines MQ and BA are parallel, and thus < QMA = < (MQ; AM) = < (BA; AM) = < BAM. On the other hand, since the quadrilateral ABCD is a parallelogram, we have $\overrightarrow{CD} = \overrightarrow{BA}$; together with $\overrightarrow{MQ} = \overrightarrow{BA}$, this yields $\overrightarrow{MQ} = \overrightarrow{CD}$. Therefore, the quadrilateral CMQD is a parallelogram, so that DQ || CM. On the other hand, DA || CB since the quadrilateral ABCD is a parallelogram. Thus, < QDA = < (DQ; DA) = < (CM; CB) = < MCB. Altogether we have proved < QMA = < BAM and < QDA = < MCB. Thus, we have < MCB = < BAM if and only if < QDA = < QMA, what is clearly equivalent to the assertion that the points A, M, D, Q are concyclic. This proves the equivalence [b](b) $\Longleftrightarrow$ (e)[/b].

Now we will [i]establish [b](e) $\Longleftrightarrow$ (d)[/b][/i]. In fact, using $\overrightarrow{MQ} = \overrightarrow{CD}$, we have shown DQ || CM, and similarly, from $\overrightarrow{MQ} = \overrightarrow{BA}$, we can derive AQ || BM. Thus, < AQD = < (AQ; DQ) = < (BM; CM) = < BMC. Hence, < AQD = < AMD holds if and only if < BMC = < AMD holds. But < AQD = < AMD clearly means that the points A, M, D, Q are concyclic. Hence, the points A, M, Q, D are concyclic if and only if we have < BMC = < AMD. Thus, the equivalence [b](e) $\Longleftrightarrow$ (d)[/b] is proven.

Let's now [i]settle [b](d) $\Longleftrightarrow$ (f)[/b][/i]. In fact, let R, S, T, U be the orthogonal projections of the point M on the lines AB, BC, CD, DA. We must show that < BMC = < AMD if and only if the points R, S, T, U are concyclic. Well, the points R, S, T, U are concyclic if and only if < STU = < SRU. Thus, it remains to show that < BMC = < AMD if and only if < STU = < SRU.

Well, since < MSC = 90° and < MTC = 90°, the points S and T lie on the circle with diameter MC, and thus < STM = < SCM. In other words, < STM = < BCM. Similarly, the points T and U lie on the circle with diameter MD, and thus < MTU = < MDU, so that < MTU = < MDA. Hence, < STU = < STM + < MTU = < BCM + < MDA. Analogously, < SRU = < CBM + < MAD. Thus,

  < STU - < SRU = (< BCM + < MDA) - (< CBM + < MAD)
                = (< BCM - < CBM) - (< MAD - < MDA)
                = (< (BC; CM) - < (BC; BM)) - (< (AM; DA) - < (DM; DA))
                = < (BM; CM) - < (AM; DM) = < BMC - < AMD.

This immediately shows that < BMC = < AMD if and only if < STU = < SRU. And thus, the equivalence [b](d) $\Longleftrightarrow$ (f)[/b] is proven.

Finally we have to [i]verify [b](f) $\Longleftrightarrow$ (g)[/b][/i]. But this is almost trivial: The reflections of the point M in the lines AB, BC, CD, DA are nothing but the orthogonal projections of the point M on the lines AB, BC, CD, DA, streched with center M and stretch factor 2. Hence, these reflections are concyclic if and only if the projections are concyclic. This proves the equivalence [b](f) $\Longleftrightarrow$ (g)[/b].

Altogether, Theorem 1 (without part [b](c)[/b]) is established.

Now, of course, for an arbitrary parallelogram ABCD, Theorem 1 part [b](g)[/b] answers the question on the locus of all points M such that the reflections of the point M in the lines AB, BC, CD, DA are concyclic: This locus is the B-medial hyperbola of triangle ABC. Completely analogously, we can see that this locus is also the C-medial hyperbola of triangle BCD, the D-medial hyperbola of triangle CDA, and the A-medial hyperbola of triangle DAB. Thus, as a bonus, we have obtained the following fact: For any parallelogram ABCD, the B-medial hyperbola of triangle ABC, the C-medial hyperbola of triangle BCD, the D-medial hyperbola of triangle CDA, and the A-medial hyperbola of triangle DAB are actually one and the same hyperbola. I propose calling this hyperbola the [b]medial hyperbola[/b] of the parallelogram ABCD. The center of this hyperbola is the center of the parallelogram ABCD (because of the symmetry!), and it is a rectangular hyperbola.

We can thus summarize what we have established: For every parallelogram ABCD, the locus of all points M such that the reflections of the point M in the lines AB, BC, CD, DA are concyclic is the medial hyperbola of the parallelogram ABCD.

Now, the original problem of Pestich asked only for the case when the parallelogram ABCD is a rectangle. Actually, not much changes in this special case (except for the very special case when the rectangle ABCD is a square - in this case, the locus actually degenerates to the union of the two diagonals of the square, as you can easily see). For a generic rectangle ABCD, the locus is indeed a rectangular hyperbola through the vertices of the rectangle, and Grobber's [hide="construction of the vertices of this hyperbola"][quote="grobber"]among the two perpendicular bisectors of the sides, take the longest one. The circle having it as diameter cuts the long sides of the rectangle in four points (or two, if the rectangle is a square). These four points are projected in two points (or one, if the rectangle is a square) on the mentioned perpendicular bisector. These two points are the vertices of the hyperbola,[/quote][/hide] directly follows from the fact that the hyperbola is the locus of all points whose reflections in the lines AB, BC, CD, DA are concyclic. Thus, everything is proven.

  Darij

Very, very nice and informative post !!!

   A lot of members and guests alike  will read it with a great benefit to
  themselves.

  (I noticed one typo in (c) :  the mid-perpendicular to the segment
  connecting  the feet of orthogonals is passing thru middle of AC, not BC )

   Also, for completeness, it would be nice to have location of foci
   and , if possible,  the location of the center of concyclicity  wrt the
   orthogonal hyperbola.

  Pestich.

Thanks, Pestich!

[quote="pestich"](I noticed one typo in (c) : the mid-perpendicular to the segment
connecting the feet of orthogonals is passing thru middle of AC, not BC )[/quote]

Thanks. Now I have corrected this typo.

[quote="pestich"]Also, for completeness, it would be nice to have location of foci
and , if possible, the location of the center of concyclicity wrt the
orthogonal hyperbola.[/quote]

About the foci I can't say anything special, but the locus of the centers of concyclicity is really interesting - it is the medial hyperbola of the parallelogram ABCD again!

In fact, we will show the following:

[b]Theorem 2.[/b] Let N be the center of the parallelogram ABCD (i. e., the common midpoint of the diagonals AC and BD). Also, let M be a point on the medial hyperbola of the parallelogram ABCD. Then,

[b](a)[/b] The center of the circle passing through the orthogonal projections of the point M on the lines AB, BC, CD, DA is the point N.
[b](b)[/b] The center of the circle passing through the reflections of the point M in the lines AB, BC, CD, DA is the reflection of the point M in the point N.

As soon as this theorem will be proven, we will be able to conclude from [b](b)[/b] that the center of the circle passing through the reflections of the point M in the lines AB, BC, CD, DA lies on the medial hyperbola of the parallelogram ABCD; in fact, this center is the reflection of the point M in the point N, but the point M lies on the medial hyperbola of the parallelogram ABCD, and this medial hyperbola is symmetric with respect to the point N (in fact, any symmetrically defined object of the parallelogram ABCD is symmetric with respect to the center N of this parallelogram), so we can conclude that the reflection of the point M in the point N also lies on the medial hyperbola. Hence, the locus of the centers of the circles passing through the reflections of the point M in the lines AB, BC, CD, DA will be the medial hyperbola of the parallelogram ABCD.

Remains to prove Theorem 2. This is very easy:

For [b](a)[/b], denote the orthogonal projections of the point M on the lines AB, BC, CD, DA by R, S, T, U again. Then, we have to show that the center of the circle through the points R, S, T, U is the point N. Indeed, the center of the circle through the points R, S, T, U must lie on the perpendicular bisectors of the segments RT and SU (since these segments are chords in this circle). But remembering once again that our quadrilateral ABCD is a parallelogram, we see that the perpendicular bisectors of the segments RT and SU are the "midparallels" between the parallel sides of this parallelogram and intersect each other at the center N of the parallelogram. Since the center of the circle through the points R, S, T, U lies on both of these perpendicular bisectors, it therefore coincides with their point of intersection, i. e. with the point N. This proves Theorem 2 [b](a)[/b].

In order to prove [b](b)[/b], we note again that the reflections of the point M in the lines AB, BC, CD, DA are the images of the orthogonal projections of the point M on the lines AB, BC, CD, DA under the stretch with center M and factor 2. Since the center of the circle through the orthogonal projections is the point N, it thus follows that the center of the circle through the reflections is the image of the point N under the stretch with center M and factor 2; in other words, this center is the reflection of the point M in the point N. This completes the proof of Theorem 2 [b](b)[/b].

Thank you again for your beautiful problem!

  Darij

Hello Darij,

   As they say these days in Siberia 'pas mal, pas mal at all'.

   I hope you guessed
   about the origin of this problem,  and the reason it was a rectangle:
   yes,you are right about the concyclic incenters in circumscribed
   quadrilateral.   What's left now for you to do is to navigate the 'bridge'
   between  the two, show that diagonals intersect on the medial
   hyperbola,   the relation of medials and original hyperbolas, etc.

 Pestich.

  BTW,
  The circle center is also on the line thru vertex B,  M' (iso. con. of M),
  and mid-point of segment RS (it is on mid-perpendicular to RS).

Hello Darij,

   The fact that circumcenter (O)  is on perpendicular bisector of AC and
   orthocenter (H)  is iso. con.  of it  means that H is on medial hyperbola
   and mid-perpendicular to the segment of it's projections on AB and BC
   passes thru mid-point of AC.   This explains why the internal tangent
   to 2 incircles of triangles AHB and BHC (other than the altitude from B)
   also passes thru mid AC.

   Pestich.

Hello Darij,

   I found you a parallelogram.  If you draw the external angle bisectors
   of circumscribed quadrilateral then the 4 triangles formed by the
   externals and the neaby sides  have their orthocenters in vertices of
   a parallelogram with sides parallel to the diagonals of the original
   quadrilateral.

   Happy holidays.

   Pestich.

Hello Pestich,

I have been trying for a rather long time to understand your remarks about the Bradley conjecture (both here and in email), but I must say I didn't understand them. Could you be not so mystic? However, your remark on the "common tangents" problem really seems to hit the point:

[quote="pestich"]   The fact that circumcenter (O)  is on perpendicular bisector of AC and
   orthocenter (H)  is iso. con.  of it  means that H is on medial hyperbola
   and mid-perpendicular to the segment of it's projections on AB and BC
   passes thru mid-point of AC.   This explains why the internal tangent
   to 2 incircles of triangles AHB and BHC (other than the altitude from B)
   also passes thru mid AC.[/quote]

The fact that the "mid-perpendicular to the segment of it's projections on AB and BC passes thru mid-point of AC" is actually trivial without any application of the medial hyperbolas: The projections of C and A on AB and BC lie on the circle with diameter CA, and hence, the perpendicular bisector of the segment joining these projections passes through the center of this circle, i. e. through the midpoint of the segment CA. But you are right that there is a nice relationship between such perpendicular bisectors and the common tangents of the incircles. I even suspect now that:

[b]Conjecture 3.[/b] Let M be a point on the B-medial hyperbola of the triangle ABC. Then, the internal common tangent of the incircles of triangles AMB and BMC different from the line BM passes through the midpoint of the side CA.

Perhaps the converse is also true. I'll try now to find a proof of this conjecture. I don't see why it is so clear as you claim but probably it is not too hard.

Thank you!

  Darij


geometry
geometric transformation
rotation

In a plane, there are $n \geq 3$ circular beer mats $B_{1}, B_{2}, ..., B_{n}$ of equal size. $B_{k}$ touches $B_{k+1}$ ($k=1,2,...,n$); $B_{n+1}=B_{1}$. The beer mats are placed such that another beer mat $B$ of equal size touches all of them in the given order if rolling along the outside of the chain of beer mats.
How many rotations $B$ makes untill it returns to it's starting position¿

[hide]The total arc length can be determined by considering that for each beer mat, $B$ is in contact with every point on the outside arc of that mat, excepting a 60 degree arc stemming from each tangency point. (a picture would help, i guess).  Then, using that the sum of the arc lengths inside is $180(n-2)$ degrees, we have a total arc length not covered of $300n-360$, maeaning $B$ makes $1+\frac{n}{6}$ revolutions.[/hide]I'm not sure if this makes any sense right now, I'll clear it up and post a picture when I return from school.

By the way, why beer mats?  What exactly is a beer mat?

[quote="K81o7"]By the way, why beer mats?  What exactly is a beer mat?[/quote]

Maybe it means coaster?  (The thing you place a glass of beer on to protect the table?)

Yes. My translator said beer mat and google agreed to know the word. Seems to be British English.


vector
superior algebra
superior algebra theorems

Hi! What does the notation in the following problem in higher algebra mean?

Determine $ [\mathbb{Q}(\sqrt{2},i): \mathbb{Q}]$.

Thanks!

Los

$ \mathbb{Q}(\sqrt {2},i)$ is a field of minimal size containing the rationals, a square root of $ 2$, and a square root of $ \minus{} 1$. This has a natural vector space structure over $ \mathbb{Q}$, and the index $ [\mathbb{Q}(\sqrt {2},i): \mathbb{Q}]$ is the dimension of that space - in this case, $ 4$.


The four-digit numeral $ 3AA1$ is divisible by 9. What digit does $ A$ represent?

3AA1 is divisible by 9, so the 3+A+A+1 must be a even multiple of nine.

The smallest even multiple is 18. 18-3-1=14=2A

Thus, A=7.


Two water wave sources, separated by a distance $L$, create a standing wave. If the frequency of the standing wave is the minimum possible, how many nodes and antinodes will there be (including at the source itself, if there is any)?

[quote="Kamil Witek"]2 nodes and 1 antinode[/quote]

Reasoning?


floor function
geometry
cyclic quadrilateral
national olympiad

[size=134][b]Day 1[/b] (FST1)[/size]

[b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83358]Question 1[/url][/b]
Let $E$ be the intersection of the diagonals of the cyclic quadrilateral $ABCD$. Let $F$ and $G$ denote the respective midpoints of the sides $AB$ and $CD$. Prove that the line through $G$ which is perpendicular to $AC$, the line through $F$ which is perpendicular to $BD$ and the line through $E$ which is perpendicular to $AD$ are concurrent.

[b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83363]Question 2[/url][/b]
Let $S = \{1, 2, 3, \ldots, 2006\}$. Prove that there are 14 subsets of $S$ which have the following property: if $s \in S$, then there are 7 of these 14 sets which have $s$ as their only common element.

[b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83362]Question 3[/url][/b]
Does there exist an infinite subset $T$ of the positive integers such that whenever $a, b$ are in $T$, then $a^{2}-ab+b^{2}$ divides $a^{2}b^{2}$.

[size=134][b]Day 2[/b] (FST2)[/size]

[b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83364]Question 1[/url][/b]
Each integer is coloured either red, blue, green or white. Let $x$ and $y$ be odd integers such that $|x| \neq |y|$. Show that there exists two natural numbers of the same colour whose difference is one of $x, y, x+y, x-y$.

[b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=49710]Question 2[/url][/b]
A point $P$ is in the interior of the cyclic quadrilateral $ABCD$ and has the property $\angle BPC = \angle BAP+\angle PDC$. The feet of the perpendiculars from $P$ to $AB$, $AD$ and $DC$ are respectively denoted $E$, $F$ and $G$. Show that $\triangle FEG$ and $\triangle PBC$ are similar.

[b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83361]Question 3[/url][/b]
Find all real numbers $x$ such that there is a positive integer $n$ with $15\lfloor x \rfloor^{2}+10\lfloor x \rfloor x+18\lfloor x \rfloor+2x+6-4^{n}= 0$. ($\lfloor x \rfloor$ denotes the greatest integer which is not more than $x$.)

Could someone please post the questions in their appropriate subforums and get the links back here? Thanks :)

FST2 was definitely much harder than FST1. I believe some of these questions were old Slovenian Selection Test questions.

The squad of 8 (team plus 2 reserves in some not yet determined order):

Tom Eccles
Peter Ford
Jos Gibbons
Saul Glasman (myself)
Jonathan Lee
Freddie Manners
Jack Shotton (cadge_nottosh)
Lee Zhao (centy)

I recognize Question 1 from FST 2 as a IMO 1999 Shortlist problem, and I recognize at least one Slovenian question indeed :)

FST2, Question 2 appeared during last year's Oundle camp. Still didn't stop the entire squad scoring 0 on it.  :oops:

Could someone put the UK FST questions into the resources page? I have more FST questions from previous years that could also be added.

Cool, I'll do that, Lee.

(Edit: seems like someone did that already. Anyways, just post all FST questions you have ;))

Could you tell me how many marks you need on this FST to make the IMO squad?

The marks are never disclosed so I'm not sure. However, depending on squad strength it varies from year to year. Some years it can be high as 20+ but I think low teens are usual.

(By the way, its been called the Final Selection Test in the resources page and as the subject of all my questions rather than the First selection test. Given as there's another selection test, the NST (Next Selection Test) after the FST, that could be confusing. :blush: )

[quote="Centy"](By the way, its been called the Final Selection Test in the resources page and as the subject of all my questions rather than the First selection test. Given as there's another selection test, the NST (Next Selection Test) after the FST, that could be confusing. :blush: )[/quote]

I fixed it :)

:D

Time to dig out all my old FST exams...

And will there be LST (Last Selection Test)?  :D

Oh we're far to optimistic for that.

There's a FST (First) and a NST (Next).


trigonometry
linear algebra
matrix
geometry
geometric transformation
rotation
LaTeX

Consider the matrix  $A=\left(\begin{array}{cc}t\cos{\varphi}&-t\sin{\varphi}\\ t\sin{\varphi}& t\cos{\varphi}\\ \end{array}\right)\; .$
 [b]
a) Find $A^{k}\; ,\; k\in \{2,3,...\}$
b) Using the fact that if $I-A$ is nonsingular, then $\sum_{k=0}^{n}A^{k}= \left(I-A^{n+1}\right)(I-A)^{-1}$ find some trigonometric summation formulas.
[/b] (Note that $A^{0}: =I \; .$ )

a) since for t=1 $A$ is the matrix of a counterclockwise rotation by angle $\varphi$, it's obvious that $A^{k}$ is rotation by $k\varphi$ (it's A superposed k times); so $A^{k}= t^{k}\left(\begin{array}{cc}t\cos k{\varphi}&-t\sin k{\varphi}\\ t\sin k{\varphi}& t\cos k{\varphi}\\ \end{array}\right)\;$

Take care ! something is wrong ([b] [u](only)[/u]  for first question a)  [/b] !

neworder's post had a $\LaTeX$ error which prevented it from rendering properly. I've fixed that for him. It's still not quite correct, but it's closer. And even the problem it still has is probably a typo. I'm guessing he meant to say

$A^{k}= t^{k}\left(\begin{array}{cc}\cos k{\varphi}&-\sin k{\varphi}\\ \sin k{\varphi}& \cos k{\varphi}\\ \end{array}\right)$

[quote="Kent Merryfield"].... [b]And even the problem it still has is probably a typo. [/b] ......I'm guessing he meant to say

$A^{k}= t^{k}\left(\begin{array}{cc}\cos k{\varphi}&-\sin k{\varphi}\\ \sin k{\varphi}& \cos k{\varphi}\\ \end{array}\right)$
[/quote]
Hi Kent,
of course your form of $A^{k}$ is corect !
By the way, if you find an error (typo) in my proposed question please inform me ! Thanking in advance,Alex

One comment: $2\times 2$ matrices of this particular form can be identified with complex numbers. The trigonometric summmation identities that can be found in this setup would involve objects familiar from Fourier analysis such as the Dirichlet kernel and Poisson kernel. I'm used to deriving those identities from geometric sums of complex numbers - which is what these matrices essentially are.


Previously, were a state to have no ordinary USAMO qualifiers, their highest index individual was invited to take USAMO. We now have two termination tests. How will this work?

I can see three ways:

They take the USAJMO if below 11th, else USAMO

The top 11th or 12th grader is invited to USAMO, and the top underclassman is invited to USAJMO

They take the USAJMO in all cases, so that a not-so-good upperclassman isn't dominated by the test.

How about they take USAMO if their AMC 12 index was closer to the cutoff and USAJMO if their AMC 10 index was closer?

This prompts me to think of an modification of the current selection procedure to provide upperclassmen more opportunities:

For 10th graders or below, create a spot on the AIME form to indicate if they'd rather take JMO if they qualified for USAMO. After AIME, select the top 270 AMC 12 indices, then move those who chose the said option into the JMO qualifiers. Fill the USAMO qualifiers from the remaining AMC 12 indices and JMO qualifiers from the remaining AMC 10 indices. (This also applies for the top in state rule)

Page 3 of the PDF attached to the first post in the announcement lists the steps that are taken to select the USAMO and USAJMO participants. The last two steps are
[quote="Announcement PDF"]6) Select the top index from any state not already represented in the USAMO.
7) Approximately the top 220-230 students with AMC10 based USAMO indices and not already
selected to the USAMO via an AMC12 based index will be invited to the JMO.[/quote]
So it sounds like the state qualifiers rule will only apply to the USAMO, which makes sense.

[quote="timwu"]For 10th graders or below, create a spot on the AIME form to indicate if they'd rather take JMO if they qualified for USAMO.[/quote]
One of AMCDirector's first replies in the announcement thread says that USAMO qualifiers won't get the option to take the USAJMO; they want students "working at the best of their abilities."


function
real analysis
real analysis unsolved

Let the continuous functions $ f_n(x), \; n\equal{}1,2,3,...,$ be defined on the interval $ [a,b]$ such that every point of $ [a,b]$ is a root of $ f_n(x)\equal{}f_m(x)$ for some $ n \not\equal{} m$. Prove that there exists a subinterval of $ [a,b]$ on which two of the functions are equal.

The sets $ \{ X: m,n \in \mathbb{N} \mbox{ and } f_m \equal{} f_n \equal{} 0 \mbox{ on } X\}$ form a countable set of closed subsets of $ [a,b]$.
So, according to Baire, one of these sets contains an (open) interval, because $ [a,b]$ is complete.


Let $ a,b,c$ positive reals such that $ \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\equal{}1$. Prove that

$ \frac{a^2\minus{}b^2\plus{}c^2}{a\plus{}b}\plus{}\frac{a^2\plus{}b^2\minus{}c^2}{b\plus{}c}\plus{}\frac{\minus{}a^2\plus{}b^2\plus{}c^2}{c\plus{}a}\geq\frac{1}{2}$

[quote="Ligouras"]Let $ a,b,c$ positive reals such that $ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \equal{} 1$. Prove that

$ \frac {a^2 \minus{} b^2 \plus{} c^2}{a \plus{} b} \plus{} \frac {a^2 \plus{} b^2 \minus{} c^2}{b \plus{} c} \plus{} \frac { \minus{} a^2 \plus{} b^2 \plus{} c^2}{c \plus{} a}\geq\frac {1}{2}$[/quote]
[hide="Solution"]The given inequality is equivalent to $ a\minus{}b\plus{}\frac{c^2}{a\plus{}b}\plus{}b\minus{}c\plus{}\frac{a^2}{b\plus{}c}\plus{}c\minus{}a\plus{}\frac{b^2}{c\plus{}a}\ge \frac{1}{2}$, or $ \frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{c\plus{}a}\plus{}\frac{c^2}{a\plus{}b}\ge \frac{1}{2}$.

By Titu's Lemma, we have $ \frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{c\plus{}a}\plus{}\frac{c^2}{a\plus{}b}\ge \frac{(a\plus{}b\plus{}c)^2}{2(a\plus{}b\plus{}c)}.$

Hence, it suffices to prove that $ \frac{(a\plus{}b\plus{}c)^2}{2(a\plus{}b\plus{}c)}\ge \frac{1}{2}$, or $ a\plus{}b\plus{}c\ge 1$.

However, this follows from the condition $ \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\equal{}1$, since we have $ a\plus{}b\plus{}c\equal{}\sqrt{a^2}\plus{}\sqrt{b^2}\plus{}\sqrt{c^2}\ge \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\equal{}1$.

We are done.[/hide]


[quote="kunny"]Let  $ x,\ y,\ z$ be real numbers such that $ x \plus{} y \plus{} z \equal{} 0,$ and $ x,\ y,\ z\in[p,\ q]$ for the constant numbers $ p,\ q\ (p < q)$.
Find the maximum value of $ x^2 \plus{} y^2 \plus{} z^2$.[/quote]

I assume you mean $ p < 0 < q$

Since $ f(u)\equal{}u^2$ is convex, then

$ max\equal{}f(\minus{}2q)\plus{}2f(q)  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  for \ \ \ \ \ p\le \minus{}2q$;
$ max\equal{}f(p)\plus{}f(\minus{}p\minus{}q)\plus{}f(q) \ \ \  for \ \ \minus{}2q\le p \le \minus{}\frac q{2}$;
$ max\equal{}2f(p)\plus{}f(\minus{}2p) \ \ \ \ \ \ \ \ \ \ \ \ \ \   for \ \ \minus{}\frac q{2}\le p<0$.

$ p\leq x\leq q\Longrightarrow (p\minus{}x)(x\minus{}q)\geq 0$ e.t.c.

$ \therefore (p\minus{}x)(x\minus{}q)\plus{}(p\minus{}y)(y\minus{}q)\plus{}(p\minus{}z)(z\minus{}q)\geq 0$.

$ \Longleftrightarrow \minus{}(x^2\plus{}y^2\plus{}z^2)\plus{}(p\plus{}q)(x\plus{}y\plus{}z)\minus{}3pq\geq 0$

$ \Longleftrightarrow x^2\plus{}y^2\plus{}z^2\leq \minus{}3pq$


geometry
parallelogram
geometry proposed
U sviti matematyky

The circle $ \omega$ passing through the vertices $ B,C$ of triangle $ ABC$ with $ AB\neq AC$ intersects the sides $ AB,AC$ at $ R,S$ respectively. Let $ M$ be the midpoint of $ BC$. The line perpendicular to $ MA$ at $ A$ intersects $ BS,CR$ at $ K,T$ respectively. Suppose that $ TA\equal{}AK$.

Prove that $ MS\equal{}MR$.

[quote="mr.danh"]Apply the butterfly theorem to the circle (w),[/quote]

Apply how? I can't see how related to butterfly...

[hide][quote="mr.danh"][b]The Butterfly theorem:[/b] Two chords AB,CD of the circle (O) meet at I. The perpendicular line at I to OI cuts AD,BC at E,F. Then, IE=IF. [/quote][/hide]

But which two chords are you using, and how

[quote="mr.danh"]Let O is the center of (w). The perpendicular line to OA at A intersects BS,CR at K',T'. Apply the butterfly theorem to the circle (w), we have that AT'=AK'. Suppose O is other than M, then TT'KK' is a parallelogram, tht's wrong. It implies that O=M, hence result.
Note: This problem is a consequen of the inversive [b]Butterfly[/b] theorem.[/quote]

[quote="mr.danh"]In our problem, the circle (O) has two chords BR,CS meet at A. The perp line to OA at A cuts BS,CR at K',T'. Then AK'=AT' sice Butterfy theorem.[/quote]


Rationalize the denominator: $ \sqrt{\frac23}$.

\begin{align*}
\sqrt{\frac{2}{3}}
&= \frac{\sqrt{2}}{\sqrt{3}} \\
&= \frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
&= \boxed{\frac{\sqrt{6}}{3}}
\end{align*}


1. Show that if p is prime then p^2  -  1 is divisible by 24

2.  for any a greater than 1, show that  a^13  - a is divisible by 546

3. find 1000 non-prime  consecutive naturel numbers .

4. Show that if p and 8p-1 are primes then 8p+1 is not a prime.

Thanks.

Well for 1, an odd squre is equivilent to 1 (mod 8) And $p^2$ is equivelent to 1 (mod 3) so if we fill in the gaps we are done. This does not include the prime 3, of course

2)

It is enough to apply Fermat's little theorem for $p=2,3,7,13$ (I wrote last by mistake, lol)

4)

Clearly, because $8p-1$ is prime, we have $p+1\not\equiv 0\bmod 3$.  So $p=3k$ or $p=3k+1$.  We may eliminate the first case, so it's enough to say $p=3k+1$.  But then $8p+1=8(3k+1)+1=3(8k+3)$.  Test the small cases and we're done.

4. we have that if p=3 then 8p-1=23 but 8p+1=25 and so it's not prime. Now if [tex]p \neq 3[/tex] then or 3|8p-1 or 3|8p+1. And because of the primality of 8p-1 we finally get that 3|8p+1 and so it isn't prime.

2. Some application of fermat's little theorem and we'll get 2730|a^13-a

d'oh!!! Too late

1. P^2-1=(p-1)(p+1).  Since p is prime, both of these are even, and one has to be a multiple of 4 (alternating multiples of 2 are multiples of 4).  Then, in every 3 consecutive integers, one has to be divisble by 3.  p-1, p and p+1 are consecutive.  p is prime and is not the multiple of 3, so one of the other two has to be.  The only case where this falls is when p=3.

2. 546=2*3*7*13.  a^13-a=a(a^12-1)=a(a^6+1)(a^3-1)(a^3+1)=a(a^6+1)(a-1)(a^2+a+1)(a+1)(a^2-a+1).  For all these factors, one clearly ahs to be divisible by 2 (either a is even, or if it's odd, then all the others are even).  Then, out of a, a-1 and a+1, one must be divisible by 3.  And then I don't know how to proceed.  I know almost nothing about mods.

3) The 1000 consecutive integers between  $1001! + 2$ and $1001! + 1001$ are surely composite.

Any idea about finding the least series of 1000 consecutive non-primes?

[url=http://mathworld.wolfram.com/PrimeGaps.html]"No general method more sophisticated than an exhaustive search is known for the determination of first occurrences and maximal prime gaps (Nicely 1999)."[/url]

The first run of 1000 composite numbers begins after the prime $p = 1,693,182,318,746,371$ and in fact contains 1,131 composite numbers.  For comparison, the first term in that sequence of composites, 1,693,182,318,746,372, has 16 digits while the first term in the sequence given by hammond has 2568 digits.

4.  If 8p-1 and 8p+1 are prime, then 8p must be a multiple of 6, and thus 3|p.  p is prime, so this is a contradiction. (Note: if p=3, 8p+1 is composite)

The sequence of composites beginning with 1000! extends only to 1000! + 1283, a prime number, in case anyone cares.

[quote]1. Show that if p is prime then p^2 - 1 is divisible by 24 [/quote]
hmm.. if you can show that for ALL primes,  I can  prove that  2 and 3 are not prime! :D
(The problem must state " if p is prime >3" ..)

[quote="JBL"][url=http://mathworld.wolfram.com/PrimeGaps.html]"No general method more sophisticated than an exhaustive search is known for the determination of first occurrences and maximal prime gaps (Nicely 1999)."[/url]
[/quote]
Thanks for the link JBL !


function
ARML
MATHCOUNTS
Gauss
number theory
prime factorization
relatively prime

Find the remainder when $4^{1996} + 5^{1997}$ is divided by 9.

I know this involves Fermat's Little Theorem, and I went as far as to look it up, but it still doesn't help me.  I also saw a solution for this specific problem, but it doesn't help either.  Can someone post a solution that is very easy to understand?  I know this is an easy concept, but it has not been explained to me well.

To understand Fermat's Little Theorem, lets start out by looking at the pattern in the powers of 4 mod 9.

$4 \equiv 4 \bmod 9$
$4^2 \equiv 7\bmod 9$
$4^3 \equiv 1\bmod 9$
$4^4 \equiv 4\bmod 9$
.
.
Therefore the powers of 4 repeat in cycles of 3, and any time 4 is raised to a multiple of 3 it will be equal to 1 in mod 9.

Lets try it for powers of 5 mod 9.

$5 \equiv 5 \bmod 9$
$5^2 \equiv 7 \bmod 9$
$5^3 \equiv 8 \bmod 9$
$5^4 \equiv 4 \bmod 9$
$5^5 \equiv 2 \bmod 9$
$5^6 \equiv 1 \bmod 9$
$5^7 \equiv 5 \bmod 9$
.
.
Whew, that was a long one! But we see that powers of 5 repeat in cycles of 6 in mod 9, and any time 5 is raised to a multiple of 6 it will be equivalent to 1 in mod 9.

So different numbers repeat in different cycles. But is there a pattern between them?

Well yes. It turns out that any number that does not have a factor in common with 9 (such as 4 and 5) when raised to powers will repeat in blocks of 6 in mod 9!! [Notice that if something repeats in a cycles of 3 it also repeats in cylcles of 6 - sort of like how the number .351351351... repeats in blocks of 3, but also in blocks of 6.]

This is no accident! The reason for this is explained by Fermat's Little Theorem.

[u]Fermat's Little Theorem[/u] says that:

$a^{\phi(n)} \equiv 1 \bmod n$, for any integer $a$ that does not have any factors in common with n, and where $\phi(n)$ (pronounced "phi of n") is the number of natural numbers less than n having no factors in common with n. (More about this "phi" function later).

The proof of this theorem takes a little bit of work, but is not impossible to understand. A little experimenting however should convince you that it is true, and also convince you why its important for $a$ and $n$ to have no common factors. Try it!!

So now applying Fermat's Litttle Theorem to your problem tells us that:

$4^{1996} \equiv 4 \bmod 9$ (since 1996 is one more than a multiple of 3), and $5^{1997} \equiv 2 \bmod 9$ (since 1997 is one less than a multiple of 6) $\Longrightarrow$ $4^{1996} + 5^{1997} \equiv 4 + 2 \equiv 6 \bmod 9$. Therefore the given number leaves a remainder of $6$ when divided by 9.

There are other ways you could do this also. You could replace 5 with -4 (since they're the same mod 9), and look at it as $(-4)^{1997} \equiv -7 \bmod 9$, and so $4^{1996} + 5^{1997} \equiv 4 - 7 \equiv -3 \equiv 6 \bmod 9$. This could be helpful if for instance the exponents had been 1997 and 1997 respectively, because a number raised to an odd power plus its negative raised to the same odd power always equals zero!

[quote="Iversonfan2005"]hmmm...did you get this problem from an old ARML Sean? looks falmiliar.....[/quote]

Yeah, it's from old ARML Team practice.  I dunno where I found it though.  But I want to learn how to do this one because I see it on MathCounts a lot.

[quote="gauss202"]

Well yes. It turns out that any number that does not have a factor in common with 9 (such as 4 and 5) when raised to powers will repeat in blocks of 6 in mod 9!! [Notice that if something repeats in a cycles of 3 it also repeats in cylcles of 6 - sort of like how the number .351351351... repeats in blocks of 3, but also in blocks of 6.]
[/quote]

Cool!  Does this work for and mod, or only mod 9?  If it was mod 7, would we still have blocks of six?

You would have blocks of $\phi(n)$.  Coincidentally, $\phi(7)=6$.

By the way:

\[ \phi(n) = n\left(1-\frac 1{p_1}\right)\left(1-\frac 1{p_2}\right)\cdots\left(1-\frac 1{p_n}\right) \]

where \[ n=p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n} \]

and the $p_i$ are primes.

[quote="mysmartmouth"]
Cool!  Does this work for any mod, or only mod 9?  If it was mod 7, would we still have blocks of six?[/quote]

Yes, it works for any mod. In mod 7 it would repeat in blocks of $\phi(7)$; in mod 8 it would repeat in blocks of $\phi(8)$; in mod 10 it would repeat in blocks of $\phi(10)$; and so on.

The way to compute $\phi(n)$ for any number n is as follows:

Find its prime factorization, $n = \prod_{i = 1}^{n} p_i^{a_i}$. Then phi of n is given by, $\phi(n) = \prod_{i = 1}^{n} p_i^{a_i - 1}(p_i - 1)$. Or in other words take off ONE of each of the primes divisors of n and decrease it by one.

As examples, $\phi(8) = (2)(2)(2-1) = 4$, $\phi(10) = (2-1)(5-1) = 4$,
$\phi(2^45^6) = (2^3)(1)(5^4)(4) = 2^55^4$, and
$\phi(p) = p-1$ for any prime p.

See if you can understand why this method works for giving you the number of natural numbers less than n having no common factors with n. [A mathematical formula is only as good as how well you understand why it works!!]

A great book to learn this kind of thing (aside from the MEGSSS number theory book), is a little book called [u]Number Theory[/u] by G.E. Andrews. I've seen it at a lot of Barnes and Nobles for about 8 bucks. Some of the later parts of the book may be too advanced for young high school students, but the early chapters on number theory tools like mods and $\phi(n)$ and so forth are very good.

yeah so for example, since $36$ has prime factors of just $2$ and $3$, $\phi(36)=\frac12 \cdot \frac23 \cdot 36 = 12$.

Just in terms of speed (and avoiding fractions), I like my method just a liiiittle better. $\phi(36) =(2)(1)(3)(2) = 12$.

Notice that if m and n are relatively prime (i.e. have no factors in common), then $\phi(mn) = \phi(m)\phi(n)$. This can sometimes be very helpful.

yeah, just break it down to factors and subtract and stuff like gauss did, that's the way i learned it

[quote="joml88"]Additional problems:

(Furman '06) What is the remainder when $4^{87}+6^{87}$ is divided by 25?

Find the remainder when $20^{1802}$ is divided by $99$.[/quote]

[hide="hintforsecondone"]Binomial Theorem[/hide]

well, actually, i guess we should be doing it through Fermat's and stuff

I'm confused on
[hide="Number 1"]Well, we have numbers raised to the 87th power, so $\phi{87} = 28/29 * 2/3 * 87=56$.  This means it repeats in blocks of 56?  Now what do I do?  I'm confused.  I couldn't write it out 56 times.

[/hide]

Sean: \[ a^{\phi(m)}\equiv 1\mod m. \]

Look at it again and see what you did wrong.

\[ \text{remainder when divided by 25} \not\Leftrightarrow \mod 87  \]

Wait, does it repeat in blocks of $\phi{(25)}$, whick would be $20$?

The powers will repeat in cycles of $\phi(25) = (5)(4) = 20$. Use the base of the mod, not the exponents to determine the length of the cycle.

But Shobit has a good point. When you CAN use the binomial theorem it is sometimes better to do so. Especially when the cycle is as large as it is here.

I don't think the binomial theorem is something that Mathcounts expects students to know. They will usually keep the bases of the mods small. But ARML on the other hand will expect you to know the Binomial Theorem. So if you have time you may want to try and brush up on it.

[hide="Solution using the Binomial Theorem"]

$4^{87} + 6^{87} = (5-1)^{87} + (5+1)^{87}$
$= \Big[5^{87} - \binom{87}{1}5^{86} + \ldots - \binom{87}{84}5^2 + \binom{87}{1}5 - 1\Big] + \Big[5^{87} + \binom{87}{1}5^{86} + \ldots + \binom{87}{84}5^2 + \binom{87}{1}5 + 1\Big]$
$= 25m + 2(87)(5)$, because every term preceeding the last two in each bracket is a multiple of 25.

Therefore, $4^{87} + 5^{87} \equiv 2(87)(5) \equiv 870 \equiv 20 \bmod 25$[/hide]

oh no gauss...i was talking about $20^{1802}$...i didn't even think about using BT for the other one

$(19+1)^{1802} \Longrightarrow .......+\binom{1801}{1}19+\binom{1802}{0}1$ and take that $\mod{99}$

Actually this second one you can do with either Fermat's Little Theorem or the Binomial Theorem! Although Fermat's L.T. is probably just a liiiiittle bit faster!

Only experience will tell you which one to use in which situation.

So that's one way to do it.  But I would really like to learn how to do it with Fermat's.  So let me review:

Q. What is the remainder when $4^{87}+6^{87}$ is divided by $25$?

[hide]We know it repeats in blocks of $\phi{(25)}$, which is $20$.  So $4^{87} = 4^7 \mod 25$
Also, $6^{87} = 6^7 \mod 25$.  This is equal to $(11 \mod 25) + (9 \mod 25)$, giving us a final answer of $\boxed{20}$.
[/hide]

Cool.  Let me try number 2.

Q. Find the remainder when $20^{1802}$ is divided by $99$.

[hide="Answer"]
Well, first we know that $\phi{(99)}=60$. So $20^{1802} = 20^2 \mod 99$  That is $400$ in $\mod 99$, which is equal to $\boxed{4}$.[/hide]

[quote="gauss202"]Again Correct! :thumbup:[/quote]

 :ddr:  :ddr:


$x, y, z \geq 0$
$m, n \geq 0$ , $n+1 \geq m$

${x^{m}}{y^{n}}{(y-z)}+{y^{m}}{z^{n}}{(z-x)}+{z^{m}}{x^{n}}{(x-y)}\geq 0$  (Hoping that this is what you wanted to mean)

wlog(I hope I can do it) $x \geq y \geq z$

Then $z-x \leq 0$

Let's rewrite the inequality:

${x^{m}}{y^{n}}{(y-z)}+{z^{m}}{x^{n}}{(x-y)}\geq{y^{m}}{z^{n}}{(x-z)}$

${y^{m}}{z^{n}}{(x-z)}={y^{m}}{z^{n}}{(x-y)}+{y^{m}}{z^{n}}{(y-z)}$

Then making subtitutions we get

${x^{m}}{y^{n}}(y-z)-{y^{m}}{z^{n}}(y-z)+{z^{m}}{x^{n}}(x-y)-{y^{m}}{z^{n}}(x-y) \geq 0$

${x^{m}}{y^{n}}(y-z)-{y^{m}}{z^{n}}(y-z)=(y-z)({x^{m}}{y^{n}}-{y^{m}}{z^{n}})$

Since $x \geq y \geq z$ then $(y-z)({x^{m}}{y^{n}}-{y^{m}}{z^{n}}) \geq 0$

and ${z^{m}}{x^{n}}(x-y)-{y^{m}}{z^{n}}(x-y)= (x-y)({z^{m}}{x^{n}}-{y^{m}}{z^{n}}) \geq 0$

and this means that

${x^{m}}{y^{n}}{(y-z)}+{y^{m}}{z^{n}}{(z-x)}+{z^{m}}{x^{n}}{(x-y)}\geq 0$

And now go to the test forum and learn how to write in LaTeX. If someone rated you as spam because of non-using LaTeX he is totally right.

Basically, the ineuqlaity is cyclic not symmetric... :oops:

Then I can only suppose that $x \geq y \geq z$ if the inequality is cyclic?

(If yes then sorry :blush: )

I wanted to say simmetric, anyway now I've understood why I was wrong. Thanks   :)

We had already  proved  the inequality hords for m>=0 ,n>=1,m<=n+1,or n>=m>=0.But  if  0<=n<m<1, then the inequality does not  hold .

Generally we have:

      Let x,y,z>0, k>=0,n>=m>=0,or 0<=m<=n+k<=2*n ,then

         x^m*y^n*(y^k-z^k)+y^m*z^n*(z^k-x^k)+z^m*x^n*(x^k-y^k)>=0.

  See:

        http://www.irgoc.org/bbs/dispbbs.asp?boardid=19&id=2483&star=1#2483

it seems that the link doesn't work. :maybe: ..could you please fix it?


We've got two charged semispheres(with same center) one with charge $q$ and radius $r$, the second with charge $Q$ and radius $R$. What is the force of attraction(detraction)?

doesn't the orientation matter? is one inside the other? or are they opposite each other? I would assume the force between them would be different.

intersting.. the standard approach for a sphere (splitting into 2 hrmispheres) probably won't work here!!! There will be extraneous effects!!!

Can you give some hints!!!

OK I'll tell you that $R>r$ this one I forgott to mention.
You want hint ok use the fact that the electrical force inside the sphere is 0.

This is a tough problem, I'll post my solution for the case r=R.
I'm assuming that the hemispheres are hollow and are oriented to face each other.

If we placed a hemisphere of charge +Q with radius R+x and a hemisphere of charge +Q with radius R-x, x << R, inside and outside of the hemisphere with charge +q, there would be no net force on it. Therefore, let's add these hemisphere's into the configuration and find the charge on the hemisphere with charge +q. It is clear that the force in this configuration is equivalent to the sum of the forces in the configurations with the hemisphere with charge +q inside and outside, respectively, a complete sphere of charge +Q. We can discard the former case, as the field inside a hollow sphere is zero. Now we can shrink down the sphere of charge +Q to a point, so we are trying to figure out the force between a hemisphere of charge +q, and a point with charge Q at its center. (This is equivalent to finding the height of the center of mass of a hemisphere.) The answer is surprisingly simple, at in the end we get
$\frac{-1}{2}\frac{kqQ}{R^2}$

[quote="schulmannerism"]This is a tough problem, I'll post my solution for the case r=R.
I'm assuming that the hemispheres are hollow, and they are facing each other, i.e. one is not inside the other.
[/quote]
There is no reason to assume any of this based on the problems! "hollow" implies that they are conducting shells. Nowhere is this stated. I believe the problem wants solid unfirm charge distributions, because otherwise it should have been stated as such.

Furthermore, I asked about the orientation, which I am still convinced makes a difference. If the one is contained within the other, the properties are completely different than if they are completely separate, or slightly overlapping. e.g. is one the north east semisphere and one the south west semisphere?... Nothing in the problem says they are facing each other.

The equal case is quite simple by some theoerms...

It's the other case which is difficult!!!

Some hints plz!

OK people it is a problem which needs beutiful solution otherwise you're gonna solve it very very long. The orientation doesn't matter you'll see from my solution. Let's $F$ be that force which we are searching for if we add the second hemisphere with radius $R$ which complete the first hemisphere with radius $R$ to sphere with the same radius. The force between the  second hemisphere and  little hemisphere is $F$ because the resulting force of two hemispheres is 0(according to my hint). Let's move out the second hemisphere and complete the little hemiphere to sphere with radius $r$. The force between this little sphere and big hemisphere is $2F$ (try to understand it yourself, it's simple). The sphere with charge $2q$ creates the same electrical field as the point with the same charge in it's center(it can be easily proved by Gause's theorem), so it's left only to count down the force between the hemisphere wit radius $R$ and charge $Q$ and a point-charge in its center with charge $q$, it will be the force we are searching for. The last part I'm leaving for you :)

*edit* Actually, come to think of it. I can't find the definition of a "semishpere" anywhere. I guess I assumed it was something less than a hemisphere :). I don't think the term semisphere exists.

What's the difference? On mathworld.com I haven't even found what a semisphere is...

Listen people I'm not very good in English esp. in scietific words so consider semisphere as half of the sphere, ok?


Consider the harmonic series so modified

$ 1 \minus{} \underbrace{({1\over 2} \plus{} {1\over3})}_{two\ terms} \plus{} \underbrace{({1\over 4} \plus{} {1\over5} \plus{} {1\over6})}_{three\ terms} \minus{} \underbrace{({1\over 7} \plus{} {1\over8} \plus{} {1\over9} \plus{} {1\over10})}_{four\ terms}\ldots$

Prove the series converges.

Consider the harmonic series by grouping $ 2^k,$ $ k \equal{} 1,2,\ldots$ terms each time.
Prove its divergence  [u] (sorry, the k-th group has the opposite sign of the (k+1)-th group).  [/u]

Let the series be $ c_{1} \minus{} c_{2} \plus{} c_{3} \minus{} c_{4} \plus{} ...$

$ c_{n} \equal{} \frac {2}{(n \minus{} 1)n \plus{} 1} \plus{} ... \plus{} \frac {2}{n(n \plus{} 1)} < \frac {2n}{n^{2} \minus{} n \plus{} 1}$.

So we have a sequence of positive numbers $ {c_{n}}$ that decreases to zero as $ n \to \infty$ ans so by Leibniz's test (or the alternating series test), the series $ \sum_{n \equal{} 1}^{\infty} ( \minus{} 1)^{n \minus{} 1}c_{n}$ is convergent.

For the harmonic series, let $ N\equal{}2^{n}$.

Then, $ s_{N} \equal{} 1 \plus{} \frac{1}{2} \plus{} (\frac{1}{3} \plus{} \frac{1}{4}) \plus{}(\frac{1}{5} \plus{} ... \plus{} \frac{1}{8}) \plus{} ... \plus{}(\frac{1}{2^{n\minus{}1}\plus{}1} \plus{} ...\plus{} \frac{1}{2^{n}})$

$ >\frac{1}{2}\plus{}\frac{1}{2}\plus{} 2\cdot\frac{1}{4}\plus{}4\cdot\frac{1}{8}\plus{} ...\plus{}2^{n\minus{}1}\cdot\frac{1}{2^{n}}$.

That is, $ s_{N} > \frac{n\plus{}1}{2}$ and so is unbounded.

Therefore $ \sum_{n\equal{}1}^{\infty} \frac{1}{n}$ diverges.


Here's a very interesting problem from a textbook:

For a pole erected on the equator, how fast is the length of the shadow moving at time t?

(Just assume the earth is flat for this, please)

We need more information, including the height of the pole, the time of the year, and the time of day.

Let's take a simple case: assume that the date is the spring or fall equinox. Then the sun will be directly overhead at noon (giving zero shadow) and otherwise the shadow will fall along the equator itself. Then the angular inclination of the sun changes at a rate $ \frac{d\theta}{dt}=\frac{2\pi}{24\text{ hr}}.$ The rate of change of the shadow will then be a related rates problem involving a triangle and the definition of a trigonometric function.

For a date other than the equinox, it gets a little more complicated.


trigonometry
inequalities proposed
inequalities

Let  is a triangle, $ w_a,w_b,w_c$ be bisectors of angles,prove that:

$ \frac {\cos\frac {A}{2} \plus{} \cos\frac {B}{2} \plus{} \cos\frac {C}{2}}{\sin\frac {A}{2}} \geq \frac {4w_a \plus{} w_b \plus{} w_c}{a}$

or:
$ (\cos\frac {A}{2} \plus{} \cos\frac {B}{2} \plus{} \cos\frac {C}{2})\cos\frac{A}{2} \geq \frac{4w_a\plus{}w_b\plus{}w_c}{4R}$

BQ

[size=150]In most cases, people Found  is a very casual,
But Often There is necessity.
BQ[/size]

$ (\cos\frac {A}{2} \plus{} \cos\frac {B}{2} \plus{} \cos\frac {C}{2})\cos\frac {A}{2} \le \frac {4w_a \plus{} w_b \plus{} w_c}{8r}$


When Sarah turns 36 years old, her father will be 62. Currently, Sarah is $ \frac13$ of her father's age. How many years old is Sarah now?

So this means she is $ 62\minus{}36\equal{}26$ years younger. Now if you are $ \frac13$ of your dad's age, it means he is 2 times your age older than you, which you already know is $ 26$   so now divide that by 2, you get your answer, $ 13$


calculus
integration
limit
trigonometry
calculus computations

Evaluate

\[\lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \frac{[n\sin x]}{n}\ dx\]

where $ [x] $ is the integer equal to $ x $ or less than $ x $.

[quote="kunny"]Evaluate

\[ \lim\limits_{n\to\infty} \int\limits_0^{\dfrac{\pi}{2}} \dfrac{[n\sin x]}{n}\ dx \]

where $[x]$ is the integer equal to $x$ or less than $x$.[/quote]

I will post the solution if my answer '1' is correct.

[quote="vidyamanohar"]
I will post the solution if my answer '1' is correct.[/quote]

That's right  :lol:
 :cool:

$[n\sin x] = k \Longleftrightarrow \arcsin\dfrac kn \leq x < \arcsin\dfrac {k+1}n$ , so
$\int\limits_0^{\pi /2} \dfrac{[n\sin x]}{n}\ dx =\dfrac 1n \sum\limits_{k=0}^{n-1} k(\arcsin\dfrac {k+1}n-\arcsin\dfrac kn) = \dfrac{(n-1)\pi}{2n} - \dfrac 1n \sum\limits_{k=0}^{n-1} \arcsin\dfrac kn$ , and finally
$\lim\limits_{n\to\infty} \int\limits_0^{\pi /2} \dfrac{[n\sin x]}{n}\ dx = \dfrac{\pi}2 - \int\limits_0^1 \arcsin x\ dx = 1$
:cool:


when i put three smilies in my post, it says that i have put too many smilies!! :mad:

Can someone help me?  :help:

Thanks!!

Try putting more content in your posts and only use smilies sparingly.

Listen: there is a search function.  Put it to use.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=91613&search_id=1290836374


could you render assistant?
how many permutations A1A2...An of {1,2,3...n} have exactly k indices j such that :
a  : Ai<Aj for all i<j?(such j are called "left to right maxima")
b  : Aj>j?(such j are called "excedance"

For a). See D. Knuth "The art of computer programming", vol.1, 1.2.10. Answer is related with Stirling's numbers.

See also "Concrete Mathematics" by R.L.Graham, D.E.Knuth and O. Patashnik.
Chapter 6 - Special Numbers BOUNS PROBLEMS 63 (p.316)
and the answer can be found at the page of 557.

Sam-n! Did you tackle this problem?
Please, answer.

Happy New Year!

You are realy quick Myth it seems that you wont rest until you get the solution of all problem in Unsolved section  :D
I just find recursion for a) and b).a): assume that the number of permutation is $a_{n,k}$,then if $A_n=n$ we have $a_{n-1,k-1}$ permutation which satisfy the conditions,and if $A_n<n$ then we have $(n-1)a_{n-1,k}$ permutation (because we have n-1 possible number for A_n ) and for b):like before assume that the  number of permutation is $a_{n,k}$ .then assume that we have $B=A_1A_2\dots A_{n-1}$ if we place n in j th place of permutation and place $A_j$ at the end of permutation  if B has k excedances then we have k+1 possible value for j such that the permutation that was construct above has k excedances and $n-k-1$ remaining value for j yields k+1 . so we have this recursion $a_{n,k}=(k+1)a_{n-1,k}+(n-k)a_{n-1,k-1}$
sorry for delay (because i don't view Mathlinks every day)


inequalities
induction
inequalities proposed

thas right, i've posted this inequality that you said before:
http://www.mathlinks.ro/viewtopic.php?t=157913
, the only difference here was that we had the sum of $ x_{i}$'s :)


this is the exercise 19-9 on AOPS volume 2.

soppose only one percent of the population has the disease mathphoberia. we design a test which diagnoses the disease successfully in afflicted patients 90% of the time. unfortunately, 20% of the time it reports that the person has mathphoberia, when in fact they don't. the test reports that i have mathphoberia. what is the probability that the test is correct? how can we improve the accuracy of our testing procedure?

use complementary counting

1-(.01*.1+.99*.2)=81.2%  I'm pretty sure it's right

81.2% seems intuitively far too large. Almost everyone does not have mathphoberia but 20% of them test positive, whereas less than 1% of the population will legitimately test positive. Thus there is approximately a $ 20: 1$ ratio so we expect the probability to be about $ \frac{1}{21}$. I get

$ \frac{(0.01)(0.9)}{(0.01)(0.9)\plus{}(0.99)(0.2)}\equal{}\frac{1}{23}$.

[quote="unimpossible"]but it is given that you are tested postive. It asks for the probability that the test is correct.[/quote]

Yes . . . and?


Find the sequence:
$a_1=1$
$a_{n+1}=5a_n+1$
What is the nice way of doing such things, expect notice that: :)

Use the reccurence. Well, the equation of this sequence is $r^{2}-5r-1=0.$ After that you will find its solutions. It is clear how to proceed. ;)

No Cezar, that is not the correct equation.

yes, you are right. excuse me for beeing so dumb. Actually this is an equation of order I. :oops:

This is a geometry series
\begin{eqnarray*} a_{n+1} &=& 5a_n +1 \\ &=& 5(5a_{n-1} +1 ) +1 \\ &=& 5^2 a_{n-1}+5 +1 \\ &=& 5^{k+1} a_{n-k}+5^k +5^{k-1}+....+1 \\ &=& 5^n a_1 + 5^{n-1}+ .... +1 \\ &=& \frac{5^{n+1} -1}{4} \\ \end{eqnarray*}
so we need to prove (formally) by induction $a_n = \frac{5^{n} -1}{4}$
FIRST STEP: if $n=1$ then we have $a_1 =1$ which is true
INDUCTIVE STEP : Suppose that it is true for all positive integers less than $k$
then
\begin{eqnarray*} a_{k+1} &=& 5a_{k} + 1 \\ &=& 5 \left ( \frac{5^{k} -1}{4}\right ) +1 \\ &=& \frac{5^{k+1} -1}{4} \\ \end{eqnarray*}

or maybe renaming sequence in such a way that we have $a_{n+1} + \frac14 = 5\left(a_n + \frac14\right)$, so $a_n + \frac14$ turns out to be geometric, and $a_n + \frac14 = 5^{n-1}\left(a_1 + \frac14\right)$...


[quote="math92"]Use the quadratic formula to solve:
$d^2-4d-6=0$[/quote]

Do you want us to factor it out?

[quote="kyndig"]d= 7.16 or -0.84.

i never leave it in form.[/quote]
well what if you don't have a calculator


For clarification, 66236 is the only 5-digit number and must start at E.
The rest of the square is easily built off of this realization.

Seriously... how do you solve this in 45 seconds? I try Ebob's method but I get like 2 letters and times up. Is there no faster way?

[quote="AceOfDiamonds"]How to do this in 45 sec other than mem?[/quote]

I agree completely.  If the problem was presented on paper, perhaps it could be done that quickly.

Perhaps a few have AMAZING working memories and don't have to copy the grid or the numbers to paper before they even start to solve the problem?  Otherwise, in the context of FTW, it seems to do little more than reward memorizing answers.

Speaking of FTW devolving into memorizing answers, do you think it would be possible to have many questions that are very similar but with little twists, to make memorizing without understanding less rewarding in the game?

It reminds me of a math teacher here who traveled to India and taught there right after college.  One week before the end of the school year, we was drafted into replacing a Geometry teacher who left without warning and without writing the final exam.  In India, the final exam isn't 20% of the year's grade, it's 100%.   The students came to the American substitute and admitted that they had expected the final exam to be given to them a day or two before they sat for it and for the exam problems to all be straight out of the textbook.  After consulting with the headmaster (principal) the substitute teacher decided to write an original exam, but one third of the questions would be from the textbook except the vertices would be relabeled.

The substitute teacher was AMAZED when he graded the exams.  Many of the students wrote solutions (geometric proofs) that would have been perfect, if the vertices were not relabeled!  Not only had those students successfully memorized the proofs in a 900 page textbook full of geometry proofs.  They had done all of this memorization while hardly understanding any of it!!