I just wanted to know, What extracurricular activities are guys or girls going to pursue in college next year?

If you're already in college, what else do you do other than attend class and do your homework?

I play intramural sports. It's a way to play organized sports games at a level that's not way too competitive for me.

I also participate in a group that has dinner discussions with professors once a week. (But I am not sure if that's the sort of thing that you'd find everywhere.)

I am active in the Math Club too. We do social stuff and math stuff.

Generally, in college, being active in an extracurricular organization or a performance group requires much more time and effort than the equivalent in high school. For that reason, most people only have one or two extracurricular activities that they're serious about.

[quote="mdk"]I assume that you play basketball then (aren't you like 6'5''). Can you dunk? Ever dunk in a game?

Also, how is your school's varsity basketball team ?[/quote]
My favorite sports are soccer and ultimate frisbee. I also play volleyball and basketball, but I only play basketball because my roommate is a fanatic and he always asks me to go play with him. I can't dunk but I'm really close. I think if I did plyometric exercises I'd be able to.

Princeton's varsity basketball team has been really dismal the past few years. We set an NCAA record for lowest-scoring game since the 3-point line was instated vs Wyoming State or something my freshman year. (We also tied the lowest score by a single team with 22 or something).

Personally, I can't wait to join an a capella group.

I am going to Worcester Polytechnic Institute, and I plan on doing something that will get me exercise -- probably some intramural sport like soccer or baseball.  I REALLY like academics and games, so I'll probably join a math club and join/create a chess and/or games club.  I'm really looking forward to doing some of the extracurricular activities available at WPI.  I'm also going to take a Physical Education class on swimming.  I'm hoping to get exercise, social interaction, and academic stimulation out of the activities I participate in at WPI.  Maybe I'll get into some sort of science/engineering society?

I personally can't wait to try out for the basketball team, be in math club, and write for the school's newspaper.

Well, it looks like my extracurriculars will be similar to what they were last year.  I'll do cross country in the fall, which is a huge commitment.  I also plan on going to some of the UMA events (undergraduate math association).  I'll be sure to play in many intramural sports games for my fraternity such as basketball, football, softball, volleyball, and frisbee.    I also intend on doing some work on the side (either AoPS or a continuation of my current summer job, or both).


Let $a,b,c,d$ be nonegative real numbers, find the greatest positve number $k$ such that \[\frac a{\sqrt{a+b+c}}+\frac b{\sqrt{b+c+d}}+\frac c{\sqrt{c+d+a}}+\frac d{\sqrt{d+a+b}}\leq k\sqrt{a+b+c+d}.\]

Another one that was posted in MnF forum a long time ago: Find the greatest $ k$ such that the following is true for all non-negative real numbers $ a,b,c,d$
\[ (a\minus{}b)(b\minus{}c)(c\minus{}d)(d\minus{}a)(a\minus{}c)(b\minus{}d)\leq k(a\plus{}b\plus{}c\plus{}d)^6.\]


[i] You will enjoy this nice and simple problem ![/i]

Let $ ABCD$ a rhombus and a point $ E$ on side $ AB$ such that $ CE \equal{} AD$.If line $ AC$ meets $ DE$ at $ F$ ,

prove that         $ FB \equal{} FC$

[b][i]See the figure below ![/i][/b]
 Babis

[quote="stergiu"][i] You will enjoy this nice and simple problem ![/i]

Let $ ABCD$ a rhombus and a point $ E$ on side $ AB$ such that $ CE \equal{} AD$.If line $ AC$ meets $ DE$ at $ F$ ,

prove that         $ FB \equal{} FC$

 Babis[/quote]

I don't believe that is correct.. if E was on side AB such that CE = AD, E would be point B, implying that F is the intersection of the diagonals of the rhombus (which means that FB = FC isn't necessarily true)

To moutain.dew: There are two points on $ AB$ that are at a distance of $ AD$ from $ C$.  One is $ B$ and we can find the other.  If you don't want to take this for granted (which you definitely shouldn't), then construct a rhombus (or even a parallelogram).  Then use a compass with center $ C$ and radius $ CB$ and see where it hits $ AB$.  I hope this helps.

To MellowMelon: Whoops, sorry... my mistake... I will complete my solution.  Yes, $ E\ne B$ is needed for the problem.  I don't think an angle restriction is necessary for $ \angle C$, though.

Well, if it is a square, then the circle with center $ C$ is tangent to $ AB$... so, $ B \equal{} E$ in a square.  In a square, though, the diagonals are equal in a square and they bisect each other, so the solution follows...

Here is the solution to this problem.
[hide="Solution"]
Notice that $ CE \equal{} AD$ and $ AE\parallel CD$ and thus $ AECD$ is an isoceles trapezoid.  Thus, $ CF \equal{} DF$.  Also, since $ F$ is on $ AC$, which is the perpindicular bisector of $ BD$, we have that $ FB \equal{} FD\implies CF \equal{} FB$.  Sorry for the incomplete solution earlier.  [/hide]

Maybe the problem needs to stipulate that $ F \neq B$, though for this to be possible we need $ 90 < \angle C \leq 120$.

And uh, I'm totally confused about the solution above. I thought it was $ FB \equal{} FC$. Also even if you extend $ AB$ in both directions (which I don't think you're supposed to), there is no second intersection if $ ABCD$ is a square.

[quote="mountain.dew"]well, the problem was a little ambiguous about needing to extend side AB.. I took it to mean [b]on[/b] the segment AB[/quote]

 [i]Sorry ,
 I wanted to send the figure last night , but it was late for me and I was tired. I put  now the figure on my first message and I hope to help you.[/i]

Babis

Use SAS similarity with $ \triangle ADC$ and $ \triangle BFC$ with angle $ C$.

$ \angle FCB\equal{}\angle ACB\equal{}\angle DAC\equal{}\angle DCA$...


when i did this problem

$ \text{How many ways can 14 people be split into 7 pairs of two?}$

i thought that the answer was

$ \dbinom{14}{2}\dbinom{12}{2}\dbinom{10}{2}\cdots\dbinom{4}{2}\dbinom{2}{2}\equal{}681080400$

but the solutions said that every pair is scounted $ 7!\equal{}5040$ ways so the answer is

$ \dfrac{681080400}{5040}\equal{}135135$

can someone tell me which answer is right and why?

thanks so much

The second one, because the ordering of the pairs is irrelevant.

Another method: Designate one person in each pair as the A person and the other as the B person.  There are

$ \binom{14}{7}\equal{}3432$

ways to choose the A people, and then $ 7!\equal{}5040$ ways to assign the B people to them.  There are then 2 ways to permute each pair.  The answer is

$ \frac{3432*5040}{2^7}\equal{}135135$.

So if I want to choose 2 groups of 2 each from 10 people is the answer

a) (10C2) * (8C2) = 45 *28 = 1120 or
b) (10C2) * (8C2) /2! (allowing for overcount)

b.  The groups aren't ordered, so you need to divide out by the permutations.


Seven dwarfs $ D_1, D_2, \ldots, D_7$ have to choose from the set of seven works numbered $ W_1,W_2,\ldots W_7$. Dwarf $ D_1$ can't pick up $ W_2, W_3$, $ D_2$ can't take $ W_1, W_2$ and $ D_4$ can't take $ W_3$ and $ W_6$. How many ways can they arrange work for themselve?


geometry
trigonometry
AMC
AIME
angle bisector

In the diagram below, angle $ABC$ is a right angle.  Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$.  Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10.$  Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCFG.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=406[/img]

[hide]Let $BD = x$ so $BC = 35 - x$.

For angle bisector theorem we have $\frac{12}{x} = \frac{37}{35 - x}$ $\Longrightarrow$ $x = \frac{60}{7}$.

$\left[ABD\right] = 12 \cdot \frac{60}{14} = \frac{360}{7}$.

$\left[AEF\right] = \left[AEG\right] + \left[AGF\right]$

Let $AG = k$.

$\frac{1}{2} \cdot 30 \cdot \frac{35}{37} = \frac{1}{2}k \cdot \sin \frac{a}{2} \cdot 13$.

So $\left[AGF\right] = \frac{1}{2} \cdot 10k \cdot \sin \frac{a}{2} = \frac{5250}{481}$.

Hence $\left[DCFG\right] = 210 - \frac{360}{7} - \frac{5250}{481}$.

So the integer closest is $148$. [/hide]

I hope in Aime is allowed the calculator  :)

there is no calculators on the aime...

this problem is pretty straight forward if you bash it with trig, but no calculator makes that more difficult

It's a #10, and looking at how easy #11 and #12 are, people should have had no problem working that all out on paper. :P

the fastest solution that i see is find the sine and cosine of $\angle CAB$, then use the half angle formulas, then find EF, and use the length of an angle biscector theorem that was in aops

then use 1/2absinC to find the areas of ABC, ABD, AEF, and AEG, and add/subtract then solve

To joml88:

Just for sidenote, I posted this problem before (as a question). I don't think it matters much but this was just a link.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62899

 :)

Can someone post a solution using mass points? It doesn't  seem to work  :(


number theory
greatest common divisor
number theory proposed

It is easy to chek, that $ 2*3*7|a(n) \ \forall n\in N$. Consider $ gcd(a(1),a(2))$.

But we must cacurlas.
If we don't need some value?


geometry
trigonometry
function
inequalities
invariant
area of a triangle
algebra

The altitudes  AD, BE and CF of an acute-angled triangle  ABC cut at H.

IF [AFH]  = [BDH] = [CEH] ,   where [AFH] denotes area of triangle AFH,

Prove that triangle ABC is equilateral.

For the areas to be equal the altitudes have to be medians as well.

   lt. Pestich

Both solutions posted are wrong:

[quote="mecrazywong"]$[AFH]=2R^2\cos A\cos B\cos C\sin B$ :D[/quote]

No, we actually have $\left[AFH\right]=2R^2\cos^2 A\cos B\sin B$, and this is a term quite difficult to handle (so there is no direct solution using trigonometry).

[quote="pestich"]For the areas to be equal the altitudes have to be medians as well.[/quote]

No, these are not those areas whose equality immediately implies that the altitudes are medians.

___________________________________________________________________

Anybody interested in a correct solution of the problem? Well, here is one which I hope is okay:

The configuration of the problem is not symmetric, but at least it's cyclic. So we can WLOG assume that A is the smallest among the angles A, B, C of the triangle ABC. Thus, $B\geq A$ and $C\geq A$.

In the right-angled triangle ADB, we have < DAB = 90° - < ABD; in other words, < HAF = 90° - B. Hence, in the right-angled triangle AFH, we have $AF=HF\cdot\cot\measuredangle HAF=HF\cdot\cot\left(90^{\circ}-B\right)=HF\cdot\tan B$. Similarly, $BF=HF\cdot\tan A$. But since $B\geq A$, since both angles B and A lie in the interval ]0°; 90°[ (this is because the triangle ABC is acute-angled), and since the tangens function is monotonically increasing on the interval ]0°; 90°[, we have $\tan B\geq\tan A$, so that $AF=HF\cdot\tan B$ and $BF=HF\cdot\tan A$ yield $AF\geq BF$.

[[b]EDIT:[/b] There is a simpler way to see that $AF\geq BF$: In fact, since in a triangle, the greater angle lies opposite to the longer side, the inequality $B\geq A$ yields $CA\geq BC$, and thus $CA^2\geq BC^2$. But in the right-angled triangles AFC and BFC, the Pythagoras theorem yields $AF^2=CA^2-CF^2$ and $BF^2=BC^2-CF^2$, so that $CA^2\geq BC^2$ yields $AF^2=CA^2-CF^2\geq BC^2-CF^2=BF^2$. Thus, $AF\geq BF$.]

The triangle AFH is right-angled, and hence its area [AFH] equals $\frac12$ times the product of the catets:

$\left[AFH\right]=\frac12\cdot AF\cdot FH$.

Similarly, $\left[BFH\right]=\frac12\cdot BF\cdot FH$. Since $AF\geq BF$, we thus have $\left[AFH\right]\geq\left[BFH\right]$.

Now, consider the triangle DEF; this triangle is the orthic triangle of triangle ABC. Since the triangle ABC is acute, the angles of its orthic triangle DEF are given by < FDE = 180° - 2A, < DEF = 180° - 2B and < EFD = 180° - 2C (see Lemma 1 in http://www.mathlinks.ro/Forum/viewtopic.php?t=23696 post #2). Since $C\geq A$, we have $A\leq C$, so that $180^{\circ}-2A\geq 180^{\circ}-2C$, and thus $\measuredangle FDE\geq\measuredangle EFD$. Since, in a triangle, the greater angle lies opposite to the longer side, this implies $EF\geq DE$.

Let V and W be the orthogonal projections of the points F and D on the line BH.

It is well-known that the altitudes of a triangle bisect the angles of its orthic triangle. Thus, the line BE, being an altitude of triangle ABC, bisects the angle DEF of the orthic triangle DEF. Therefore, < FEV = < DEW. Also, < FVE = < DWE (since < FVE = 90° and < DWE = 90°). Thus, the triangles FEV and DEW are similar, and we conclude that $\frac{FV}{DW}=\frac{EF}{DE}$.

Now, the segment FV is the altitude of triangle BFH dropped onto the side BH; thus, since the area of a triangle equals $\frac12\cdot\text{ side }\cdot\text{ corresponding altitude}$, the area [BFH] of triangle BFH is $\left[BFH\right]=\frac12\cdot BH\cdot FV$. Similarly, $\left[BDH\right]=\frac12\cdot BH\cdot DW$. Thus,

$\frac{\left[BFH\right]}{\left[BDH\right]}=\frac{FV}{DW}=\frac{EF}{DE}$.

Since $EF\geq DE$, we thus have $\left[BFH\right]\geq\left[BDH\right]$. Together with $\left[AFH\right]\geq\left[BFH\right]$, this implies $\left[AFH\right]\geq\left[BDH\right]$. But by the condition of the problem, we must have [AFH] = [BDH]. Thus, both of the inequalities $\left[BFH\right]\geq\left[BDH\right]$ and $\left[AFH\right]\geq\left[BFH\right]$ must become equalities. But, studying again the proofs of these inequalities, we can easily see that the first of them becomes an equality if and only if A = C, and the second one becomes an equality if and only if A = B. Hence, since both of the inequalities $\left[BFH\right]\geq\left[BDH\right]$ and $\left[AFH\right]\geq\left[BFH\right]$ must be equalities, we must have A = C and A = B. Thus, A = B = C, and the triangle ABC is equilateral. $\blacksquare$

  darij

[quote="mecrazywong"]How careless I am :([/quote]

No problem, such kind of errors often happens at such problems...

[quote="pestich"]For each of the 3 given triangles consider their neighbours accross
   the altitude. The pair has the same height,[/quote]

Why that?

[quote="pestich"]the ratio of areas equals to the ratio of
   sides.  If 3 given triangles are equal the neighbours are equal as well by none
   other than Ceva.[/quote]

???

[b]PS.[/b] You won't succeed to solve this problem without using inequalities (or something comparable with inequalities, like arrangement considerations etc.) for the simple reason that it is not valid on $\mathbb{C}^2$.  :P

  darij

Hello Darij,

   I wonder what happens in the math camps the ones you visit so often?
   You return from there quite belligerent and confused, I have noticed.
   Is it still a policy of Bundesrepublic to keep its Youth confused via summer
   camp programms?

   You see, you mixed-up the altitudes of triangle ABC. It is obvious from the
   fact that triangles have the same height I was talking about the triangles
   that have the bottom part of altitude in common. Triangles accross the
   part of altitude from vertex to orthocenter have common side that is not
   their altitude any way you twist it, because it is their hypothenuse.

   And  why do I need 'neravenstva'?  As you can see there are no
   'nerovnye linii', all of them are pretty straight, not even a circle in sight.
   By the way,  what is big C squared?   I know  CO2,  carbon 12 and 14,
   in combinatorics C with 2 little numbers.

    Maj.  Pestich

[quote="pestich"]   You see, you mixed-up the altitudes of triangle ABC. It is obvious from the
   fact that triangles have the same height I was talking about the triangles
   that have the bottom part of altitude in common.[/quote]

Hmm, you have correctly observed that I misunderstood your post, but it was you who didn't define the points properly. For any of the three triangles, there are two "neighbours accross the altitude", and it is not directly clear which ones you meant. Anyway, your observation about the areas doesn't really yield anything, since your next argument ("If 3 given triangles are equal the neighbours are equal as well by none other than Ceva.") is plain nonsense.

[quote="pestich"]   And  why do I need 'neravenstva'?  As you can see there are no
   'nerovnye linii', all of them are pretty straight, not even a circle in sight.[/quote]

Must the use of inequalities always be indicated by circles? For instance, the well-known fact (Steiner-Lehmus theorem) that if, in a triangle, two interior angle bisectors are equal in length, then the triangle is isosceles, can only be proved with the help of inequalities. And still there are no circles directly present on the picture.

[quote="pestich"]   By the way,  what is big C squared?[/quote]

The complex plane. I just wanted to say that, if you reduce the condition [AFH] = [BDH] = [CEH] to a system of equations for the sidelengths a, b, c of the triangle ABC, then this system has not only the solution a = b = c corresponding to the equilateral triangle ABC, but also some solutions which yield non-acute triangles or triangles with complex sidelengths (that's why I went to the complex plane). This shows that the solution of the problem cannot be obtained by equalities only, but one must also use inequalities to honor the fact that the triangle ABC is real and acute-angled.

  Darij

Hello Darij,

   It was jolly nice of Mike to identify the point as orthocenter, so you were able to
   use easy formula for the area.  But it looks like even triangle centers not yet in the
   Book have this property - they all have to be located in the  center of a correct triangle.
   If Ceva Theorem  is 'plain nonsense' there is not much
   left to employ for a proof, is there? Do you want to mess around with Heron?
   Maybe there is something behind that big C squared?

   Sgt.  Pepper

[quote="pestich"]   It was jolly nice of Mike to identify the point as orthocenter, so you were able to
   use easy formula for the area.  But it looks like even triangle centers not yet in the
   Book have this property - they all have to be located in the  center of a correct triangle.
   If Ceva Theorem  is 'plain nonsense' there is not much
   left to employ for a proof, is there? Do you want to mess around with Heron?
   Maybe there is something behind that big C squared?[/quote]

Very typical for you, Pestich: Lots of nonsense, and some interesting observations inbetween - and, as usual, you leave me the job of filtering the latter out of the former.

To be brief, what is indeed a good observation is that one can replace the orthocenter by an arbitrary point inside the triangle, and, if the areas are equal, this point must coincide with the triangle's centroid.

I formulate this as a theorem:

[color=blue][b]Theorem 1.[/b] Let P be a point inside a triangle ABC, and let the lines AP, BP, CP meet the sides BC, CA, AB of triangle ABC at the points D, E, F, respectively. Assume that [AFP] = [BDP] = [CEP]. Then, the point P must be the centroid of triangle ABC.[/color]

Hereby, the notation $\left[P_1P_2P_3\right]$ means the [i]non-directed[/i] area of a triangle $P_1P_2P_3$.

[i]First proof of Theorem 1.[/i] By Lemma 3 from http://www.mathlinks.ro/Forum/viewtopic.php?t=6134 post #2, there exists an affine transformation mapping the triangle ABC to a triangle A'B'C' and the point P to a point P' such that the point P' is the orthocenter of triangle A'B'C'. Since equality of areas and centroids of triangles are affinely invariant notions, it is therefore enough to prove Theorem 1 for the case when the point P is the orthocenter of triangle ABC. In this case, the triangle ABC must be acute-angled (since its orthocenter P lies inside the triangle ABC), and the orthocenter of a triangle coincides with its centroid if and only if the triangle is equilateral; thus, Theorem 1 is just an easy consequence of the initial problem. Hence, we have proven Theorem 1.

[i]Second proof of Theorem 1.[/i] There is also a direct proof of Theorem 1, without any reference to the initial problem.

In the following, I will repeatedly use a well-known fact:

[color=blue][b]Lemma 2.[/b] If X, Y, Z are three collinear points and R is another point in the plane, then $\frac{\left[RXY\right]}{\left[RYZ\right]}=\frac{XY}{YZ}$.[/color]

Now, let x = [BPC], y = [CPA], z = [APB]. Since the configuration of Theorem 1 is cyclically invariant, we can WLOG assume that x is the smallest among the numbers x, y, z. In other words, $x\leq y$ and $x\leq z$.

By Lemma 2, applied to the three collinear points A, F, B and the point P, we have $\frac{\left[AFP\right]}{\left[BFP\right]}=\frac{AF}{FB}$, so that $FB\cdot\left[AFP\right]=AF\cdot\left[BFP\right]$. By Lemma 2, applied to the three collinear points A, F, B and the point C, we have $\frac{\left[CFA\right]}{\left[BFC\right]}=\frac{AF}{FB}$, so that $FB\cdot\left[CFA\right]=AF\cdot\left[BFC\right]$. Thus,

$FB\cdot y=FB\cdot\left[CPA\right]=FB\cdot\left(\left[CFA\right]-\left[AFP\right]\right)=FB\cdot\left[CFA\right]-FB\cdot\left[AFP\right]$
$=AF\cdot\left[BFC\right]-AF\cdot\left[BFP\right]=AF\cdot\left(\left[BFC\right]-\left[BFP\right]\right)=AF\cdot\left[BPC\right]=AF\cdot x$,

so that $\frac{AF}{FB}=\frac{y}{x}$. Combining this with $\frac{\left[AFP\right]}{\left[BFP\right]}=\frac{AF}{FB}$, we get $\frac{\left[AFP\right]}{\left[BFP\right]}=\frac{y}{x}$. But we know that $x\leq y$; in other words, $y\geq x$. Thus, we have $\left[AFP\right]\geq\left[BFP\right]$ with equality only for x = y.

Since $\frac{\left[AFP\right]}{\left[BFP\right]}=\frac{y}{x}$, we have

$\frac{\left[APB\right]}{\left[BFP\right]}=\frac{\left[AFP\right]+\left[BFP\right]}{\left[BFP\right]}=\frac{\left[AFP\right]}{\left[BFP\right]}+1=\frac{y}{x}+1=\frac{x+y}{x}$,

so that $\frac{\left[BFP\right]}{\left[APB\right]}=\frac{x}{x+y}$. Since [APB] = z, this becomes $\frac{\left[BFP\right]}{z}=\frac{x}{x+y}$, and thus $\left[BFP\right]=\frac{zx}{x+y}$. Similarly, $\left[BDP\right]=\frac{zx}{y+z}$. Since $x\leq z$, we have $x+y\leq y+z$, so that $\frac{zx}{x+y}\geq\frac{zx}{y+z}$. In other words, we have $\left[BFP\right]\geq\left[BDP\right]$, with equality only for x = z. Combining this with the above obtained inequality $\left[AFP\right]\geq\left[BFP\right]$, with equality only for x = y, we see that

$\left[AFP\right]\geq\left[BFP\right]\geq\left[BDP\right]$.

On the other hand, we are given that [AFP] = [BDP]. Thus, we must have equality in both of the inequalities $\left[BFP\right]\geq\left[BDP\right]$ and $\left[AFP\right]\geq\left[BFP\right]$. But the first of these inequalities becomes an equality only for x = z, and the second one becomes an equality only for x = y. Thus, we must have x = z and x = y, so that x = y = z.

Since $\frac{AF}{FB}=\frac{y}{x}$ and x = y, we have $\frac{AF}{FB}=1$. Thus, the point F is the midpoint of the segment AB, and the line CF is the C-median of triangle ABC. Therefore, the point P, lying on the line CF, must lie on the C-median of triangle ABC. Similarly, the point P also lies on the A-median and on the B-median of triangle ABC. Thus, the point P is the intersection of the three medians, i. e. the centroid of triangle ABC. And Theorem 1 is proven.

Of course, since Theorem 1 is a generalization of the initial problem, and since the second proof of Theorem 1 didn't make use of the initial problem, we thus have obtained a new solution for the initial problem.

[[b]EDIT:[/b] A simpler proof of Theorem 1 was given in http://www.mathlinks.ro/Forum/viewtopic.php?t=35200 .]

  Darij

Hello Darij,

  Using Ceva proved to be not a complete nonsense  idea  here:

          http://www.mathlinks.ro/Forum/topic-3865.html

  Maj.  Pestich


For the nonzero numbers $ a$, $ b$, and $ c$, define \[(a,b,c)\equal{}\frac{abc}{a\plus{}b\plus{}c}.\] Find $(2,4,6)$.

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 4 \qquad
\textbf{(D)}\ 6 \qquad
\textbf{(E)}\ 24$

[hide]48/12=4, C.[/hide]

I don't suppose you read the flooding announcement?


Let $(X,\Gamma,\mu)$ be a measurable space and $f_{i},g_{i},f,g\in L^{p}(\mu),i=1,2,\cdots$ and
\[h_{i}(x)=\max\{f_{i}(x),g_{i}(x)\},h(x)=\max\{f(x),g(x)\},\]
as $x\in X$. Prove that if $f_{i}\to f$ and $g_{i}\to g$ in $L^{p}(\mu)$ then $h_{i}\to h$ in $L^{p}(\mu)$.

[quote="MeKnowsNothing"]Let $(X,\Gamma,\mu)$ be a measurable space and $f_{i},g_{i},f,g\in L^{p}(\mu),i=1,2,\cdots$ and
\[h_{i}(x)=\max\{f_{i}(x),g_{i}(x)\},h(x)=\max\{f(x),g(x)\}, \]
as $x\in X$. Prove that if $f_{i}\to f$ and $g_{i}\to g$ in $L^{p}(\mu)$ then $h_{i}\to h$ in $L^{p}(\mu)$.[/quote]Use the fact that $max(a,b) = \max(a-b,0)+b = |a-b|/2+(a+b)/2$.


calculus integration function derivative logarithms parameterization trigonometry

Can anyone solve this problem integrate (x = 0 to x =  :pi:) (1 - 2u cos x + u 2 )

What is u? A constant, a function in x or something else?

Misha

[quote]no one can solve this problem.?[/quote]

No, the reason no one has answered you is that no one can figure out why it's a problem at all. Either it's an extremely elementary integral - a routine Calculus I homework exercise - or you haven't explained what you meant.

:lol: What a misnomer!!! I fell for it thinking this would be a hard problem according to the subject title, but then... it's all laughs.

Seriously, what is the variable $u$ doing in your problem? Obviously you are integrating the function with respect to $x$. Like [u]Misha[/u] said, is $u$ a constant, a function of $x$ like $u(x)$, or something else?

This is either a joke, a mistakenly written simple calculus homework problem, or some integration formula using $u(x)$. In any case, I believe the former two choices.

3X.lich

Oh. very sorry. I miss ln.

integrate (x = 0 to x =  :pi: ) ln (1 - 2u cos x + u 2 )dx

Ok. Can anyone solve this Problem. !!!

It's not an elementary integral. It would need to have the derivative of cos(x) outside the natural logarithm to be so, in the simplest case. You can check that it's not elementary with [url=http://integrals.wolfram.com/]The Integrator[/url]. The answer involves the [url=http://mathworld.wolfram.com/Polylogarithm.html]polylogarithm function[/url].

But it's a definite integral; we don't need the antiderivative. That, and the parameter $u$, give us some extra handles to work with. Here's an approach worth trying. Start by differentiating with respect to $u$:

$\frac{d}{du}\int_0^{\pi} \ln(1-2u\cos x+u^2)dx=\int_0^{\pi}\frac{-2\cos x+2u}{1-2u\cos x+u^2}dx$.

I'll be back tomorrow to see where that leads.

I think the answer is $2\pi\ log| x|$.

I will post  later.

kunny

[quote="Kent Merryfield"]But it's a definite integral; we don't need the antiderivative. That, and the parameter $u$, give us some extra handles to work with. Here's an approach worth trying. Start by differentiating with respect to $u$:
[/quote]

Yes,  this is quite a common exercise of integrals depending on a parameter.

The answer kunny gave is partly right; the full answer is this:

If $I(u)=\int_0^{\pi}\ln(1-2u\cos x+u^2)dx$, then

$I(u)=0$ for $|u|\le 1$ and $I(u)=2\pi\ln|u|$ for $|u|>1$.

From higher in this thread, $I'(u)=\int_0^{\pi}\frac{2u-2\cos x}{1-2u\cos x+u^2}dx$

(by symmetry) $= \frac12\int_0^{2\pi}\frac{2u-2\cos x}{1-2u\cos x+u^2}dx$

$=\frac12\int_0^{2\pi}\frac{2u-e^{ix}-e^{-ix}}{(1-ue^{ix})(1-ue^{-ix})}dx$

Now consider $z=e^{ix}$ to be a parameterization of the unit circle $C$. We can write $dz=ie^{ix}dx$ or $dx=\frac1i\frac{dz}z.$ Note also that $e^{-ix}=\frac1z.$

$I'(u)=\frac12\int_C\frac{2u-z-\frac1z}{(1-uz)(1-\frac uz)}\cdot\frac{dz}z= \frac12\int_C\frac{2uz-z^2-1}{z(1-uz)(z-u)}dz$

Now calculate this by the residue theorem. One pole is always at $z=0$, and it has residue $\frac1u$. If $|u|<1$, there is one other pole inside the circle at $z=u$ and the residue there is $-\frac1u$. Thus for $|u|<1$, the sum of the residues is zero and thus $I'(u)=0.$

If $|u|>1,$ the other pole is at $x=\frac1u$. The residue there turns out to be $\frac1u$. Thus, for $|u|>1, I'(u)=\frac{2\pi i}{2i}\left(\frac1u+\frac1u\right)=\frac{2\pi}u.$

The derivative does not exist when $|u|=1$, but we can show that $I(u)$ is continuous at those points. Also, we can see by inspection that $I(0)=0$. Integrating this, we get the result at the top.

Two checks to see that the solution looks right for large $u$ and for small $u$.

For large $u$: $\ln(u^2-2u\cos x+1)=\ln(u^2)+\ln(1-\frac2u\cos x+\frac1{u^2})$
$=2\ln|u|-\frac2u\cos x+O\left(\frac1{u^2}\right)$. Integrate that from $0$ to $\pi$, noting that the cosine integrates to zero, and we get $I(u)=2\pi\ln|u|+O\left(\frac1{u^2}\right)$ for large $u$.

By a very similar argument, we can show that for small $u, I(u)=O(u^2)$. So the result above matches what it should asymptotically.

Oh. Many thanks kent. !

I will try to understand it.

I'm very sorry to have kept you,everyone for waiting,especially for Kent Merryfield

I will show you my solution as follows.

Let $I(u)=\int^{\pi}_0 \ln(1-2u\cosh+u^2) dx$,we have $I(-u)=\int^{\pi}_0 \ln(1+2u\cosh+u^2)dx$.

Set $\pi-x=v,x=\pi-v$.as $\frac{dx}{dv}=-1,x:0\rightarrow\pi,v:\pi\rightarrow0$,

we have $I(-u)=-\int^{0}_{\pi}\ln\left[1+2u\cos(\pi-v)+v^2\right]dv=\int^{\pi}_0\ln(1-2u\cosh+v^2)dv=I(u)$  so,in using half-angle formula

$2I(u)=I(u)+I(-u)=\int^{\pi}_0\ln(1-2u\cosh+u^2)(1+2u\cosh+u^2)dx$

$=\int^{\pi}_0\ln(1-2u^2\cos2x+u^4)dx$,set $2x=\alpha$ ,$2I(u)=\frac{1}{2}\int^{2\pi}_0\ln(1-2u^2\cos\alpha+u^4)d\alpha$

$=\frac{1}{2}\int^{\pi}_0\ln(1-2u^2\cos\alpha+u^4)d\alpha+\frac{1}{2}\int^{2\pi}_{\pi} \ln(1-2u^2\cos\alpha+u^4)d\alpha$

Similarly in the second definite integral,set $\alpha=2\pi-\beta$,the integral is

$\frac{1}{2}\int^{0}_{\pi}\ln[1-2u^2\cos(2\pi-\beta)+u^4](-d\beta)$

$=\frac{1}{2}\int^{\pi}_{0}\ln(1-2u^2\cos\beta+u^4)d\beta$.

Therefore we obtain $2I(u)=I(u^2)$.In using this,$2I(1)=I(1),2I(-1)=I(1)$i.e. $I(-1)=I(1)=0$

Then,suppose that n is positive number,$I(u)=\frac{1}{2}I(u^2)=\frac{1}{2^2}I(u{}^{2{}^2})=\frac{1}{2^3}I(u{}^{2{}^3})=\cdot\cdot\cdot=\frac{1}{2^n}I(u{}^{2{}^n})$

If $|u|<1$,then$\lim_{n\to\infty}I(u{}^{2}{}^n)=0$,yielding

$I(u)=0$,and if $|u|>1$,then set $u=\frac{1}{w}$,as $|w|<1$,therefore

$I(u)=\int^{\pi}_0 \ln(1-2u\cos x+u^2)dx=\int^{\pi}_0\ln\left(1-\frac{2\cos x}{w}+\frac{1}{w^2}\right)=\int^{\pi}_0\ln\left(\frac{w^2-2w\cos x+1}{w^2}\right)dx$

$=\int^{\pi}_0\ ln(w^2-2w\cos x+1) dx-\int^{\pi}_0\ln{w^2}dx=I(w)-\pi\ln{w^2}=0-\pi\ln\left(\frac{1}{u}\right)^2 $

$=\pi\ln{u^2}=2\pi\ln|u|$

Thank you!

kunny

While we're waiting for kunny to finish, let me suggest that his work can be combined with the asymptotic results of my most recent post.

kunny has shown that $I(u)$ is an even function, so we need only consider $u>0.$

We have that $I(u)=\frac1{2^n}I\left(u^{2^n}\right)$. Suppose $u>1$. Then by one of my asymptotic statements, $I(u)=\frac1{2^n}\left(2\pi\ln\left(u^{2^n}\right)+ O\left(\frac1{u^{2^{n+1}}}\right)\right)$

$=2\pi\ln u + O\left(\frac1{2^nu^{2^{n+1}}}\right)$.

Take the limit as $n\to\infty$ and we see that for $u>1, I(u)=2\pi\ln u.$

Repeat this for $0<u<1$. By the other asymptotic statement, $I(u)=O\left(\frac{u^{2^{n+1}}}{2^n}\right).$ Again, taking the limit as $n\to\infty$, we see that $I(u)=0.$

The case of $I(1)$ can be addressed by arguing that $I(u)$ must be continuous; the case for negative values of $u$ rests on kunny's proof that the function is even.

[quote]The case of $ I(1) $ can be addressed by arguing that $ I(u) $ must be continuous[/quote]

That's unnecessary. kunny used his functional equation for $I$ to show directly that $I(1)=0.$

I like kunny's solution here; quite elegant. I needed the residue theorem from complex variables, while his solution was much more elementary. Both of our solutions give good reasons why the solutions are different for $|u|<1$ than for $|u|>1.$

Thank you very much kunny for your solution. I get much knowledge from this problem.

I also have another problem about simultaneous equation. If anyone want to try to solve. please go to this link. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=124915#124915


I got some ink on my 100% cotton T-shirt. How can I get rid of it??

[quote="amirhtlusa"]$NaClO$ is what is used in commercial bleach.
i think that would help, if the cotton is white..
it is  VERY STRONG[/quote]
Can you explain why??

Why can I use NaClO to get rid of the ink on my T-shirt??

DONT :wallbash_red:
that was just a guess,
see, i KNOW THAT $NaClO$ is a strong bleach, but you might wanna talk to some else like your chem teacher, before, doing it

[quote="amirhtlusa"]why, what? :?:
if you mean why $NaClO$ is strong,
the reason is that $ClO^-$ is a conjucate base of a weak acid.[/quote]

It's also a strong oxidizing agent, with the Cl at an inconvenient oxidation state.

I think.

So $Cl_2$ and $H_2O_2$ will also do??

What are in the ink?? $C$??

Ordinarily, black ink is made from nutgalls and a solution of some salt of iron, and consists essentially of a tannate or gallate of iron; sometimes indigo sulphate, or other coloring matter,is added. Other black inks contain potassium chromate, and extract of logwood, salts of vanadium...


Find all integers $(x,y)$ such that $y/x^2+1$ and $x/y^3+1$

See
http://www.mathlinks.ro/Forum/viewtopic.php?t=1586

@pleurestique: ¿

I have seen many links about this problem but there are no complete solutions :(

ah, well now i see. thing is, "/" is a *fraction sign*, "|" is what you meant, so you probably want to fix it.

Possibly, but surely not here ; post solutions and everything else in the older ones please. (well, know there's only one older to post in ;) )
Having three threads where the same is discussed is pure chaos!

Look exactly: the linked problem contains one more square.

:oops:, you're right, sorry..  :blush:
Then, I'll try to solve this problem...


modular arithmetic
quadratics
number theory unsolved
number theory

Prove or disprove this statement:
There exist integer solution (x,y) except (4,6) such that
$ x^2 \plus{} x^3 \equal{} 2^y \plus{} 16$

Since the RHS is divisible by $ 4$ the LHS must also be divisible by $ 4$. Therefore $ x$ is even(if $ x$ were odd, the RHS would be divisble with $ 2$ but not with $ 4$). So we can write:
$ x \equal{} 2t$
$ 4t^2 \plus{} 8t^3 \equal{} 2^y \plus{} 16$
$ t^2 \plus{} 2t^3 \equal{} 2^{y \minus{} 2} \plus{} 4$
$ t^2 \equal{} 2^{y \minus{} 2} \plus{} 4 \minus{} 2t^3$
Therefore $ t^2$ is even $ \equal{} >$ $ t$ is even.
We write
$ t \equal{} 2m$
$ 4m^2 \plus{} 16m^3 \equal{} 2^{y \minus{} 2} \plus{} 4$
$ m^2 \plus{} 4m^2 \equal{} 2^{y \minus{} 4} \plus{} 1$
$ m^2 \minus{} 1 \equal{} 2^{y \minus{} 4} \minus{} 4m^3$
$ (m \minus{} 1)(m \plus{} 1) \equal{} 4\cdot (2^{y \minus{} 6} \minus{} m^3)$
Since the RHS is divisible with $ 4$ the LHS is also divisible with $ 4$ and therefore m must be odd. But if m is odd $ (m \minus{} 1)(m \plus{} 1)$ will be divisible with $ 8$ which is not the case for the RHS because $ 2^{y \minus{} 6}$ is divisible with $ 2$ but $ m^3$ is not. Therefore there are no other solutions other than $ (x,y) \equal{} (4,6)$

[quote="panag"]if $ x$ were odd, the RHS would be divisble with 2 but not with 4[/quote]
let's take x=3
$ 3^2\plus{}3^3\equal{}36$ which is divisible by 4

Ooops my bad... :) So I have proved that $ x$ is odd. Also, $ x\equiv 15mod16$ because:
$ x \equal{} 16m \plus{} a$ with $ 1 \leq a \leq 16$
$ (16m \plus{} a)^3 \plus{} (16m \plus{} a)^2 \equal{} (16m \plus{} a)^2(16m \plus{} a \plus{} 1)$ must be divisible with $ 16$ so $ a$ is either $ 16$ or $ 15$. It can't be $ 16$ because then $ x$ would be even.
Don't know how to proceed from here though... :(

Hint:Try to prove that y is a multiple of 3.

if $ y\leq 0$ the rest is trivial and checking through  $ 6 \geq y\geq 1$ we got that only $ y \equal{} 6$ works
for $ y\geq 7$ Let $ y \equal{} z \plus{} 7$ for $ z\geq 0$
case x=even from
$ x^2(x \plus{} 1) \equal{} 16(2^{z \plus{} 3} \plus{} 1)$ we got that $ 4|x$ Let $ x \equal{} 4t$ so
$ 4t^3 \plus{} t^2 \equal{} 2^{z \plus{} 3} \plus{} 1$ but $ 4t^3 \plus{} t^2 \equal{} 0,4,5 (mod 8)$ contradiction
case x=odd from
$ x^2(x \plus{} 1) \equal{} 16(2^{z \plus{} 3} \plus{} 1)$ we got that $ 16|x \plus{} 1$ so $ x \equal{} 16m \plus{} 15$ so
$ (16m \plus{} 15)^2(m \plus{} 1) \equal{} 2^{z \plus{} 3} \plus{} 1$ since $ 16m \plus{} 15 \equal{} 3 (mod 4)$ so there must exist $ p \equal{} 3 (mod 4)$ such that $ p|2^{z \plus{} 3} \plus{} 1$
so $ z$ must be even then checking through $ (mod 5)$
from $ (16m \plus{} 15)^2(m \plus{} 1) \equal{} 0,1,2 (mod 5)$ we have a contradiction since $ 2^{z \plus{} 3} \plus{} 1 \equal{} 3,4 (mod 5)$ for $ z \equal{}$ even number
so only $ (x,y) \equal{} (4,6)$ satisfies the condition

[quote="Rose_joker"] ... so there must exist $ p \equal{} 3 (mod 4)$ such that $ p|2^{z \plus{} 3} \plus{} 1$
so $ z$ must be even ... [/quote]

Quite nice :)
Congrats.

[quote="turbo taskytojas"]Another way would be to check the equation mod 7. Then y=3k and you can check that x^3<x^3+x^2-16<(x+1)^3 for x>4.  :)[/quote]

Nicer :)

[quote="Rose_joker"]$ p \equal{} 3 (mod 4)$ such that $ p|2^{z \plus{} 3} \plus{} 1$
so $ z$ must be even[/quote]

Could you prove this part, please?  :maybe:

That solution with divisibility of 7 is really nice  :10:

[quote="Bachukas"][quote="Rose_joker"]$ p \equal{} 3 (mod 4)$ such that $ p|2^{z \plus{} 3} \plus{} 1$
so $ z$ must be even[/quote]

Could you prove this part, please?  [/quote]

$ 16m\plus{}15|2^{z\plus{}3}\plus{}1$
All prime factors of $ 16m\plus{}5$ are odd, so are equal to $ 1$ or $ 3$ mod 4.
If all were equal to $ 1$ mod 4, then their product would be equal to $ 1$ mod 4, which is wrong since $ 16m\plus{}15\equal{}3\pmod 4$.

So at least one prime divisor $ p$ of $ 16m\plus{}15$ is such that $ p\equal{}3\pmod 4$

But then Legendre symbol $ \left(\frac{\minus{}1}{p}\right)\equal{}(\minus{}1)^{\frac{p\minus{}1}{2}}\equal{}\minus{}1$ and so $ \minus{}1$ is not a quadratic residue mod p.

But, if $ z$ is odd, we have $ 2^{z\plus{}3}\plus{}1\equal{}0\pmod p$ $ \implies$ $ (2^{\frac{z\plus{}3}{2}})^2\equal{}\minus{}1\pmod p$ and $ \minus{}1$ would be quadratic residue mod p.

So $ z$ is even.


-1 divided by -(-2)^-2  I think its positive 4 but the book says -4 who's right

hello, we have $ \minus{}(\minus{}2)^{\minus{}2}\equal{}\minus{}\frac{1}{(\minus{}2)^2}\equal{}\minus{}\frac{1}{4}$, so we get
$ \frac{\minus{}1}{\frac{\minus{}1}{4}}\equal{}4$.
Sonnhard.

i'm pretty sure you're right

the book might've multiplied the negative with the negative 2 first, which is incorrect due to PEMDAS
the exponents are first

:winner_first: CONGRATULATIONS :winner_first:
your smarter than a book =)

Does your book write it $ \frac{\minus{}1}{\minus{}(\minus{}2)^{\minus{}2}}$

i would guess it does, but does it make a difference?

Well no, but he could have misposted the problem (since he didnt do latex)

Yeah most likely the equation was what mewto posted. Solving would give -4.


this is a problem i found to do with vectors/linear algebra. help is greatly appreciated, i get nowhere with this.

A median of a triangle is a line that connects a vertex tot he midpoint of its opposite side. Show that the medians of any triangle, intersect at a point. Also show that this point divides each median in the ratio 2:1 (measured from the vertex to the midpoint of the opposite side)

solutions are appreciated

hello, here you can find the proof
http://in.answers.yahoo.com/question/index?qid=20090205033128AACRsmT
Sonnhard.

thank you, greatly appreciated.

alternative proofs using vector algebra are welcome aswell :)

thank you dr. sonnhard :)

hello, here you can find a proof using vector algebra
http://www.textbooksonline.tn.nic.in/Books/11/Maths-EM/Vol1/2%20Vector%20Algebra.pdf
Sonnhard.

hmm, but that assumes that the ratio is 2:1 to prove they are concurrent? i am interested in a vector proof of the medians intersecting at a point, and showing the point divides the medians in the ratio of 2:1. maybe im misunderstanding the proof supplied by that site, but i need some help understanding :P

thank you :)

Let the three vectors be $ \vec A,\vec B,\vec C$. The midpoints of each side opposite $ A,B,C$, respectively, are $ \frac {\vec B \plus{} \vec C}2,\frac {\vec C \plus{} \vec A}2,\frac {\vec A \plus{} \vec B}2$. If some $ \vec G$ exists such that all three medians intersect it, then it is necessary and sufficient for there to exist some scalar constants such that $ c_1 \equal{} \frac {\left(\frac {\vec B \plus{} \vec C}2\right) \minus{} \vec G}{\vec G \minus{} \vec A},c_2 \equal{} \frac {\left(\frac {\vec C \plus{} \vec A}2\right) \minus{} \vec G}{\vec G \minus{} \vec B},c_3 \equal{} \frac {\left(\frac {\vec A \plus{} \vec B}2\right) \minus{} \vec G}{\vec G \minus{} \vec C}$ (which represents collinearity of a vertex, the centroid, and the median), which is equivalent to $ \vec G \equal{} \frac {2c_1\vec A \plus{} \vec B \plus{} \vec C}{2(c_1 \plus{} 1)} \equal{} \frac {\vec A \plus{} 2c_2\vec B \plus{} \vec C}{2(c_2 \plus{} 1)} \equal{} \frac {\vec A \plus{} \vec B \plus{} 2c_3\vec C}{2(c_3 \plus{} 1)}$. Matching $ \vec A$ coefficients (since it must hold for any value of $ \vec A$) gives $ \frac {c_1}{c_1 \plus{} 1} \equal{} \frac1{2(c_2 \plus{} 1)}\implies c_1 \plus{} 1 \equal{} 2c_1c_2 \plus{} 2c_1\implies1 \equal{} 2c_1c_2 \plus{} c_1$, and similarly, matching $ \vec B$ coefficients gives $ \frac1{2(c_1 \plus{} 1)} \equal{} \frac {c_2}{c_2 \plus{} 1}\implies1 \equal{} 2c_1c_2 \plus{} c_2$, and subtraction yields $ c_1 \minus{} c_2 \equal{} 0\implies c_1 \equal{} c_2$. By symmetry, $ c \equal{} c_1 \equal{} c_2 \equal{} c_3$. Plugging back in results in $ c \plus{} 1 \equal{} 2c^2 \plus{} 2c\implies2c^2 \plus{} c \minus{} 1 \equal{} (2c \minus{} 1)(c \plus{} 1) \equal{} 0$. Since $ c$ clearly can't be $ \minus{} 1$ which would result in a zero denominator, $ c \equal{} \frac12$, and the rest follows (in particular, $ \vec G \equal{} \frac {\vec A \plus{} \vec B \plus{} \vec C}3$).


hi
Mango(my real hero) was caught by some of the Aliens.they took him to a rectangular place which is on a cliff.
the place was full of vegetation.The cruel alien lit fire from a northern corner and wind is flowing in the direction from North to south.Our hero was standing on the middle .He couldn't jump from cliff as the whole place is covered with lava.He was no connection with our main office so I couldn't send anyone to save him.It will take about 1 hour to search him in this planet by the help of satellites.But in 50 min the whole vegetation will be burnt.
what would our genius Mango have done to save his life?

big hint :
just see that wind is flowing in a single direction .
going to post the answer tomorrow

[hide]He lights his own fire and follows behind it as the wind blows it to the cliff's edge.  The other fire will run out of fuel... thus destroying all other inhabitants to save himself.[/hide]


search
Let s keep it this way.
PROBABLY OLDEST UNLOCKED THREA
\/closed
super old unlocked thread
NOT THE OLDEST STOP SPAMMING

If the question is about forum posting rank (which Millennium Prize problem appears by your screen name), that's easy: it's just number of posts. I wish this site had quality ratings by selected users, a la slashdot.org, but that is hard to implement.

[quote="Yasha"]Just curious, what is the exact formula for calculating rank? Or at least what are all the factors that go into the calculation?[/quote]

I'm not sure what you mean by "rank". Moderators are appointed. But if u mean post rank, it's just based on the number of MEANINGFUL posts you make (ie ones that arent deleted due to spam, etc). You can check "Top Ten Posters" under "Memberlist" for listings.

I mean the green question mark rank. It seems 20 is the cutoff for 1 question mark, 50 is the cutoff for 2, 100 is the cutoff for 3, 250 is the cutoff for 4, 500 is the cutoff for 5, 1000 for 6, and 2500 is the cutoff for 7. Thanks for the tips.

By the way, there's a math thing above the number of ?'s. What are the Navier-Stokes equations? What is the Poincare conjecture? etc.

Those are 7 problems known as the "Millenium Problems" (or something like that, dubbed by the Clay Institute I think).  They are 7 unsolved math problems that would be directly applicable to real-world situations with powerful implications if solved.  If you want to get into more details, visit mathworld.wolfram.com and search for the various problems.

Or go to claymath.org and read their descriptions and watch the videos.

Now that AoPS and Mathlink have merged, the ranks seem to be of the Mathlink origin. Is it possible that we restore the AoPS ranks? They're more whimsical. ;)

yeah, i think question marks are cooler than stars

I was talking more about "Hodge Conjecture" v.s. "Hyper Active".

Yes, I would rather have the AoPS rank names (Millennium Prize problems). I am neutral about stars versus question marks.

[quote="JSRosen3"]I'll "fourth" this motion. I think the millenium problems as rank indicators are a lot better than "Maniaq", "PsychoMath", etc.[/quote]

Same here. Restore the millenium problems as rank names, but keep the stars rather than the question marks.

I liked the old names better to.

The ones we have now are really cheesy.

I'm seeing this "should the ranks be changed back" in many forums, so I will just post a poll about it in Fun and Games.

not only are the ranks not as good names, but the cutoffs are also different! is that also the mathlinks system?

[quote="takuanitromars36"]What were the old AoPS rankings like?[/quote]instead of the stars there were question marks. The stars are taken from MathLinks, and the names of the ranks from the old AoPS forum.


PLEASE HELP ME WITH THIS ONE

Solve   dx/dy = x^2 +y^2.

hello,i think your equation hase no solution in elementary functions. Maple 12 gives
$ y \left( x \right) \equal{}\minus{}{\frac {x \left( {\it \_C1}\,
{{\rm J}\left(\minus{}3/4,\,1/2\,{x}^{2}\right)}\plus{}
{{\rm Y}\left(\minus{}3/4,\,1/2\,{x}^{2}\right)} \right) }{{\it \_C1}\,
{{\rm J}\left(1/4,\,1/2\,{x}^{2}\right)}\plus{}
{{\rm Y}\left(1/4,\,1/2\,{x}^{2}\right)}}}$
Sonnhard.


Solve the equation $(x^2-5x+5)$^ (x^2-5x)=1

Only true if
$x^2-5x+5=1 \iff x=1 \vee x=4$
or if
$x^2-5x=0 \iff x=0 \vee x=5$
or if
$x^2-5x+5=-1$ and $x^2-5x$ is even.
$x^2-5x+5=-1 \iff x=2 \vee x=3$.  $4-10$ is even and so is $9-15$.  Both are solutions.


Thanks for the help in getting geometry_dev running smoothly; I'm really enjoying the package so far  :)

I've gotten hung up with bisectorpoint. With a triangle t, I want to find the foot of the angle bisector from A to side BC. From the documentation, that should be bisectorpoint(t.BC), since BC is the side opposite angle A. However, after a lot of incorrect-looking pictures, it seemed that in order to generate said point I needed to call bisectorpoint(t.AB). This... doesn't seem right. Here is my full code (in which I am trying to draw bisectors to sides BC and AC, but wind up using t.AB and t.BC (in that order):

import geometry_dev;

unitsize(1pt);
size(200,200);
point X, Y;
triangle t=triangleabc(60, 80, 100);
draw(t);
X=bisectorpoint(t.AB);
Y=bisectorpoint(t.BC);
segment AA=segment(t.A, X);
segment BB=segment(t.B, Y);
draw(AA); draw(BB);
label("$ A$", t.A, W);
label("$ B$", t.B, E);
label("$ C$", t.C, E);
label("$ X$", X, E);
label("$ Y$", Y, W);

I couldn't get it to compile on AoPS, so I am not enclosing it in brackets. Does AoPS not recognize geometry_dev, or am I making an additional error that prevents this from compiling on the forum?  :huh:

The problem is now fixed in [url=http://svnweb.tuxfamily.org/listing.php?repname=piprim%2Fasymptote&path=%2Fbranches%2F#path_branches_]the latest svn version[/url].
(click on the corresponding tarball to download the repository of "branches")

Thanks.


King MAth, the brackets mean rounded down.  [96/7=13+5/7]=13


Let $ x, y, z$ be real numbers greater than $ 1$ such that $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \leq 2.$ Prove that

$ \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1}\geq \frac {9}{4}\sqrt {\frac {3}{xyz}},$

with equality if and only if $ x\equal{}y\equal{}z\equal{}\frac{3}{2}.$

[quote="Ji Chen"]Let $ x, y, z$ be real numbers greater than $ 1$ such that $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \leq 2.$ Prove that

$ \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1}\geq \frac {9}{4}\sqrt {\frac {3}{xyz}},$

with equality if and only if $ x \equal{} y \equal{} z \equal{} \frac {3}{2}.$[/quote]
$ \sum_{cyc}\sqrt {x \minus{} 1}\geq\sum_{cyc}\sqrt {x \minus{} \frac {2}{\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}}} \equal{} \sum_{cyc}\sqrt {\frac {x(xy \plus{} xz \minus{} yz)}{xy \plus{} xz \plus{} yz}}.$
$ \frac {9}{4}\sqrt {\frac {3}{xyz}}\leq\frac {9}{4}\sqrt {\frac {3}{xyz}}\cdot\left(\frac {2xyz}{xy \plus{} xz \plus{} yz}\right)^2 \equal{} \frac {9\sqrt {3x^3y^3z^3}}{(xy \plus{} xz \plus{} yz)^2}.$
Hence, it remains to prove that $ \sum_{cyc}\sqrt {x(xy \plus{} xz \minus{} yz)}\geq9\sqrt {3\left(\frac {xyz}{xy \plus{} xz \plus{} yz}\right)^3}.$
Let $ xy \plus{} xz \minus{} yz \equal{} a,$ $ xy \plus{} yz \minus{} xz \equal{} b$ and $ xz \plus{} yz \minus{} xy \equal{} c,$ where $ a,$ $ b$ and $ c$ are positives.
It gives $ x \equal{} \sqrt {\frac {(a \plus{} b)(a \plus{} c)}{2(b \plus{} c)}},$ $ y \equal{} \sqrt {\frac {(a \plus{} b)(b \plus{} c)}{2(a \plus{} c)}}$ and $ z \equal{} \sqrt {\frac {(a \plus{} c)(b \plus{} c)}{2(a \plus{} b)}}.$
Hence, $ \sum_{cyc}\sqrt {x(xy \plus{} xz \minus{} yz)}\geq9\sqrt {3\left(\frac {xyz}{xy \plus{} xz \plus{} yz}\right)^3}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt {a\sqrt {\frac {(a \plus{} b)(a \plus{} c)}{2(b \plus{} c)}}}\geq9\sqrt {3\left(\frac {\sqrt {(a \plus{} b)(a \plus{} c)(b \plus{} c)}}{\sqrt8(a \plus{} b \plus{} c)}\right)^3}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt {\frac {a}{b \plus{} c}}\geq\frac {9\sqrt3}{4}\cdot\sqrt {\frac {(a \plus{} b)(a \plus{} c)(b \plus{} c)}{(a \plus{} b \plus{} c)^3}},$ which  follows from
$ \sum_{cyc}\sqrt {\frac {a}{b \plus{} c}}\geq2\sqrt {\frac {(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)}}.$
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221359
Id est, we need to prove that $ 64(a \plus{} b \plus{} c)^4(ab \plus{} ac \plus{} bc)\geq243(a \plus{} b)^2(a \plus{} c)^2(b \plus{} c)^2.$
Let $ a,$ $ b$ and $ c$ are non-negative numbers, $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ where $ v\geq0$ and $ abc \equal{} w^3.$
Hence, $ 64(a \plus{} b \plus{} c)^4(ab \plus{} ac \plus{} bc)\geq243(a \plus{} b)^2(a \plus{} c)^2(b \plus{} c)^2\Leftrightarrow w^3\geq9uv^2 \minus{} 8u^2v.$
But $ w^3$ gets a minimal value when two numbers from $ \{a,b,c\}$ are equal or when maybe $ w^3 \equal{} 0.$
Thus, for the proof it remains to check two cases:
Case 1: $ b \equal{} c \equal{} 1.$
In this case we need to prove that $ 16(a \plus{} 2)^4(2a \plus{} 1)\geq243(a \plus{} 1)^4,$ which is equivalent to
$ f(a)\geq0,$ where $ f(a) \equal{} 4\ln(a \plus{} 2) \plus{} \ln(2a \plus{} 1) \minus{} 4\ln(a \plus{} 1) \plus{} 4\ln2 \minus{} 5\ln3.$
But $ f'(a) \equal{} \frac {2a(a \minus{} 1)}{(a \plus{} 2)(2a \plus{} 1)(a \plus{} 1)},$ which gives $ a_{min} \equal{} 1$ and $ f(a)\geq f(1) \equal{} 0.$
Case 2: $ c \equal{} 0,$ $ b \equal{} 1.$
Here we need to prove that $ 64(a \plus{} 1)^4a\geq243(a \plus{} 1)^2a^2,$ which is obvious.


A sequence $ (a_n)$ is given by $ a_0\equal{}2$,$ a_1\equal{}1$ and $ a_{n\plus{}1}\equal{}a_n\plus{}a_{n\minus{}1}$, $ n\geq1$.
Prove that $ a_{59}$ divides $ (a_{30})^{59}\minus{}1$.

The first easy to check that :
$ a_n \equal{} (\frac {1 \plus{} \sqrt {5}}{2})^n \plus{} (\frac {1 \minus{} \sqrt {5}}{2})^n$
[u]Lemma[/u] If p is a prim such that  $ (\frac {p}{5}) \equal{} 1$ ,exist $ x\in N$ such that
$ x^2\equiv 5 (\mod p)$  and x is odd then
$ a_n \equiv (\frac {x \plus{} 1}{2})^n \plus{} (\frac {1 \minus{} x}{2})^n (\mod p )$
Prove this lemma is quite easy . You only use the expansion of $ a_n$
Now consider your problem :
Easy to check that $ 51^2\equiv 5 (\mod 59)$
So $ a_{30}\equiv {25^{30}} \plus{} 26^30 (\mod 59 )$
Because $ (\frac {25}{59}) \equal{} 1,(\frac {26}{59}) \equal{} \minus{} 26$ so
$ 25^29\equiv 1 (\mod 59), 26^{30}\equiv \minus{} 1 (\mod 59)$
So $ a_{30} \equiv 25 \minus{} 26\equiv \minus{} 1 (\mod 59 )$
So $ 59|a_{30} \plus{} 1$
But from Fermat's theory
$ a_{30}^{59}\equiv a_{30 } (\mod 59 )$ so $ 59|a_{30}^{59} \plus{} 1$

[quote="Inequalities Master"]A sequence $ (a_n)$ is given by $ a_0 \equal{} 2$,$ a_1 \equal{} 1$ and $ a_{n \plus{} 1} \equal{} a_n \plus{} a_{n \minus{} 1}$, $ n\geq1$.
Prove that $ a_{59}$ divides $ (a_{30})^{59} \minus{} 1$.[/quote]
The general problem : Komal A. 330.

[quote="N.T.TUAN"]Use Google , we have http://www.komal.hu/verseny/2003-11/A.e.shtml.  :P[/quote]
I solve other problem  :blush:
Sorry I don't read it carefully.


Joseph and Cody race for 100 meters, and Cody wins by 10 meters. When they race again, Cody begins 10 meters behind the original starting line. If they both run the race at the same constant speed as the first time, by how many meters will Cody win?

Source: Go Figure! Math Challenge

[hide]It is obvious he goes 10/9 the speed of the other guy, so make his speed 10 and the others 9.  Now he has to run 110 meters.
110=10t
t=11
d=9*11
d=99

He will win by 1 meter.[/hide]

i agree with nat mc[hide]He will win by 1 meter[/hide]

Yeah, I recognize this exact problem (maybe not the names) being a Chapter problem from 1995 I think.

[quote="4everwise"][hide]The answer is 99[/hide][/quote]
how did you get 99

uhh....i agree w/ my twin nat mc

99?
no


hello, did you meant
${ \frac{\left(A+2B\right)-\left(A-2B\right)}{\frac{8A+4B}{2\left(2A+B\right)}-\left(2-B\right)}}-3$?
Sonnhard.

[quote="mathfunguyman"]Yes, sorry I don't know how to format like that  :blush:[/quote]
you should use $ \text{\LaTeX}$ to use latex you should put a dollar sign at the begining and at the end of your code...
for furthermore information visit [url]http://www.mathlinks.ro/Forum/index.php?f=123[/url]

[hide]

$ \frac{\left(A\plus{}2B\right)\minus{}\left(A\minus{}2B\right)}{\frac{8A\plus{}4B}{2\left(2A\plus{}B\right)}\minus{}\left(2\minus{}B\right)}\minus{}3$
$ \equal{}\frac{A\plus{}2B\minus{}A\plus{}2B}{\frac{8A\plus{}4B}{4A\plus{}2B}\minus{}2\plus{}B}\minus{}3$
$ \equal{}\frac{4B}{B}\minus{}3$
$ \equal{}4\minus{}3$
$ \equal{}\boxed{1}$

[/hide]

[quote="Nexmus"][hide]

$ \frac{\left(A\plus{}2B\right)\minus{}\left(A\minus{}2B\right)}{\frac{8A\plus{}4B}{2\left(2A\plus{}B\right)}\minus{}\left(2\minus{}B\right)}\minus{}3$
$ \equal{}\frac{A\plus{}2B\minus{}A\plus{}2B}{\frac{8A\plus{}4B}{4A\plus{}2B}\minus{}2\plus{}B}\minus{}3$
$ \equal{}\frac{4B}{B}\minus{}3$
$ \equal{} 4\minus{}3$
$ \equal{}\boxed{1}$

[/hide][/quote]

So 2A + 2B - 2 +B = 4B?

and then 4B / B = 4?

That seems odd to me.

I think you are talking about the numerator right?

So, you have $ (A\plus{}2B)\minus{}(A\minus{}2B)$ Multiply by negative 1 to get $ A\plus{}2B\minus{}A\plus{}2B \equal{} A\minus{}A\plus{}2B\plus{}2B\equal{}4B$

Does that make sense?

yes that does.  it's the final steps that are confusing me.  I always thought that with:

$ \equal{}\frac{4B}{B}\minus{}3$

It would be simplified to $ 3B\minus{}3$.  But it looks as if the B's just canceled out and then the final step is $ 4\minus{}3 \equal{} 1$.  Just want to make sure that that is in fact correct.


How I can make a hyperbond in Latex? , I mean, if I want to press a link, it directs to me immediately to any part of the document.

For example, in this document

http://theoval.cmp.uea.ac.uk/~nlct/latex/novices/novices_scr.pdf

In contents, I click on 'Overview' and it directs to me directly to the page in where is the content.

Use the hyperref package (included in MiKTeX). In the [url=http://www.tug.org/applications/hyperref/manual.html]manual[/url] look for [i]hyperlink[/i] and [i]hypertarget[/i].

Thanks, but, can you help me a little bit more?

For example, I have

\documentclass[11pt]{article}
\usepackage[latin1]{inputenc}
\usepackage{polynom}
\usepackage{principal}
\usepackage{multicol}
\usepackage{color}
\usepackage{cancel}
\usepackage[dvips]{graphicx}
\usepackage{hyperref}
\textheight=23cm
\definecolor{rojo}{rgb}{1,0,0}
\definecolor{verde}{rgb}{0,0.7,0}
\definecolor{azul}{rgb}{0,0,1}
\definecolor{naranjo}{rgb}{1,0.5,0}
\definecolor{cyan}{rgb}{0,0.5,1}
\definecolor{fucsia}{rgb}{1,0,0.5}
\definecolor{negro}{rgb}{0,0,0}
\definecolor{marron}{rgb}{0.8,0,0}
\definecolor{morao}{rgb}{0.5,0,0.7}
\begin{document}
[color=darkblue]History[/color]

I live in Italy (I am sorry for my horrible english. Correct me if you want please.) and unforunately I do not like how things are going in my country. Obviously we have got beautyful places to visit and other touristic things of this kind. But I do not care of them. I would like going studying in a foreign country because I feel that here I have no oportunities. I would like to become a physiscian but in Italy it is really hard to find a work in this field.
I think that the problem comes from Italian politicians that I think are seen, in the rest of the world, but also here in Italy, like a funny joke and probably they are.[color=red]\newpage[/color] If you have any question about me, just [color=indigo]see[/color] the front part of the content
\end{document}

This is an example, but, if I want to click on [color=indigo]see[/color] and I want to direct me to [color=darkblue]History[/color], what should I do?

All you need to know is that the hyperlink is 'see' and the hypertarget is 'history'
So replace
'History' by [code]\hypertarget{history}{History}[/code] (the first is the name of the target and the second is the actual text)
'see' by [code]\hyperlink{history}{see}[/code]so it links to the target called history and shows the text 'see'

There are lots of formatting options for the colour of the link etc etc so ask if you want to change them.

Yes, of course, I wanna change them, tell me how (I prefer a blue link).

Another question... is there any way to remove

[img]http://img258.imageshack.us/img258/959/asdasdcc6.png[/img]

I just want only appears the document without him…

Instead of \usepackage{hyperref} use [code]\usepackage[bookmarks=false,colorlinks=true,linkcolor={blue}]{hyperref}[/code] [i]bookmarks=false[/i] gets rid of the bookmark section on the left and [i]colorlinks=true,linkcolor={blue}[/i] gets rid of the box round the link and colours the link in blue.

Thanks man, but I have another problem...

Once that I've compiled the PDF, appears in 53%

Is there a way to make it appears in 118%?

Add as an option pdfstartview={XYZ null null 1.18} so you get [code]\usepackage[bookmarks=false,colorlinks=true,linkcolor={blue},pdfstartview={XYZ null null 1.18}]{hyperref}[/code]

[quote="K®izalid"]http://theoval.cmp.uea.ac.uk/~nlct/latex/novices/novices_scr.pdf[/quote]
In this file

[img]http://img516.imageshack.us/img516/6203/inza9.png[/img]

Those hyperlinks always are set in all pages, how is made that?

It's all in the [url=http://www.tug.org/applications/hyperref/manual.html]hyperref manual[/url] under 5 Acrobat-speciﬁc behavior. Put in the footer ([url=http://www.ctan.org/tex-archive/macros/latex/contrib/fancyhdr/fancyhdr.pdf]fancyhdr package[/url] is best for that) something like:
[code]\Acrobatmenu{PrevPage}{Previous}\ \Acrobatmenu{NextPage}{Next}\ \Acrobatmenu{FirstPage}{First}\ \Acrobatmenu{LastPage}{Last}\ \Acrobatmenu{GoBack}{Back}\ \Acrobatmenu{GoForward}{Forward}\ \hyperlink{index}{Index}}[/code]
You will also need to add [code]menucolor={blue},pdfpagelayout={SinglePage}[/code] to the \usepackage code or hyperref.

The author of the pdf, Nicola Talbot, may also be able to help with the fine detail as the document properties of the pdf say she used [i]LaTeX with hyperref package[/i].


Hi
I wonder if it's possible to download a TeXer or any similar to the one you've got available here in this Forum. I use TeXniCenter distribution

LatexRender is a publicly available php package.  You can download it here:
[url]http://www.mayer.dial.pipex.com/tex.htm[/url]


factorial
floor function
superior algebra
superior algebra unsolved

Show that $ \left(n!\right)^{\frac{1}{n}} \leq \prod_{p|n!} p^{\frac{1}{p\minus{}1}}$.

A simple corollary of Legendre's formula for the exponent $ \alpha$ of a prime $ p$ that divides $ n!$
$ \alpha \equal{} \sum_{k\equal{}1}^{\infty} \left \lfloor \frac {n} {p^k} \right \rfloor < \sum_{k\equal{}1}^{\infty} \frac {n} {p^k} \equal{} \frac {n} {p\minus{}1}$. Therefore $ n! < \prod_{p|n!} p^{\frac{n}{p\minus{}1}}$.


calculus
derivative
inequalities
inequalities proposed

Find the maximum value of this expression:
$ A=$ $ \sqrt{\frac{x-3}{x^2}}$ $ +\sqrt{\frac{y-4}{y^2}}$ $ +\sqrt{\frac{z-5}{z^2}}$

[quote="Unidranix"]Find the maximum value of this expression:
$ A =$ $ \sqrt {\frac {x - 3}{x^2}}$ $ + \sqrt {\frac {y - 4}{y^2}}$ $ + \sqrt {\frac {z - 5}{z^2}}$[/quote]

perhaps there is some geometric significance to the 3,4,5 Pythagorean triple?

If im right, then the maximum would be the maximum value of each because they are all positive. So, using calculus,

$ \frac{d}{dx}\left(\sqrt{\frac{x-3}{x^2}}\right) = \frac{1}{2} \cdot \left(-1 \cdot x^{-2} + 6x^{-3}\right) \cdot \left(\frac{x-3}{x^2}\right)^{-\frac{1}{2}}$

For max/min values $ \frac{dy}{dx} = 0$

$ \frac{1}{2} \cdot \left(-1 \cdot x^{-2} + 6x^{-3}\right) \cdot \left(\frac{x-3}{x^2}\right)^{-\frac{1}{2}} = 0$

$ \left(-1 \cdot x^{-2} + 6x^{-3}\right) \cdot \left(\frac{x-3}{x^2}\right)^{-\frac{1}{2}} = 0$

Now, either,

$ -x^{-2} + 6x^{-3} = 0$

ie

$ \frac{6}{x^3} - \frac{1}{x^2} = 0$

$ 6 - x = 0$

$ x = 6$

or

$ \frac{x-3}{x^2} = 0$

$ x= 3$

Noting that

$ \sqrt {\frac {x - 3}{x^2}} \geq 0$

and

$ f(3) = 0$,

therefore x = 3 is a minimum,

it can be shown that x = 6 is a maximum

It can be shown that the derivative of

$ \sqrt {\frac {y - 4}{y^2}}$

is

$ \frac{1}{2} \cdot \left(-y^{-2} + 8y^{-3}\right) \cdot \left(\frac{y-4}{y^2}\right)^{-\frac{1}{2}}$

and

$ \sqrt {\frac {z - 5}{z^2}}$

is

$ \frac{1}{2} \cdot \left(-z^{-2} + 10z^{-3}\right) \cdot \left(\frac{z-5}{z^2}\right)^{-\frac{1}{2}}$

Furthermore, the minimum is given by $ \left(\frac{z-5}{z^2}\right)^{-\frac{1}{2}}$

hence to find the maximums we must only find when

$ -y^{-2} + 8y^{-3}= 0$

and

$ -z^{-2} + 10z^{-3} = 0$

thus

$ y = 8$

and

$ z = 10$

Thus the maximum of A is

$ \sqrt {\frac {6 - 3}{6^2}}$ $ + \sqrt {\frac {8 - 4}{8^2}}$ $ + \sqrt {\frac {10 - 5}{10^2}}$

$ = \boxed{\frac{\sqrt{3}}{6} + \frac{\sqrt{5}}{10} + \frac{1}{4}}$

[quote="kurt.math"][quote="Mathias_DK"][quote="Unidranix"]Find the maximum value of this expression:
$ A =$ $ \sqrt {\frac {x - 3}{x^2}}$ $ + \sqrt {\frac {y - 4}{y^2}}$ $ + \sqrt {\frac {z - 5}{z^2}}$[/quote]
$ \frac {t - a}{t^2} \le \frac {1}{4a} \iff 4at - 4a^2 \le t^2 \iff (t - 2a)^2 \ge 0$. Hence:
$ \sqrt {\frac {x - 3}{x^2}}$ $ + \sqrt {\frac {y - 4}{y^2}}$ $ + \sqrt {\frac {z - 5}{z^2}} \le \sqrt {\frac {1}{12}} + \sqrt {\frac {1}{16}} + \sqrt {\frac {1}{20}}$[/quote]

nice solution compaired to my calculus bashing.  :lol:

How did you get the first inequality ie $ \frac {t - a}{t^2} \le \frac {1}{4a}$   ?[/quote]
The derative of $ \frac{t-a}{t^2}$ is $ \frac{2a-t}{t^3}$. So we can guess that the max is $ \frac{2a-a}{(2a)^2} = \frac{1}{4a}$. So not really a different solution, just shorter :P

Thank you kurt.math and Mathias_DK.
@kurt.math: you noticed that the maximum would be the maximum value of each, that's the key
@Mathias_DK: very beautiful solutions :D


conics
ellipse
symmetry
geometry
3D geometry
sphere
parabola

Are ellipses truly round? For example, if you draw an ellipse on a computer with pixels, in some places if you zoom in the pixels will appear more like a straight line (especially if the ellipse is really flat). But if you look at it when drawn on a piece of paper, it looks round.

This also depends on how you define 'round'. What do you guys think, are ellipses actually round or not?

Is a circle round? Everything you just said applies equally to a circle, as a special case of an ellipse.

They have continuous strictly positive curvature, which is more or less what our visual apparatus perceives as "roundedness". If you scale the ellipse, the curvature is inversely proportional to the scaling factor, so when the latter tends to $ \plus{}\infty$, the curvature tends to $ 0$ and the ellipse (or even a circle) is perceived as "locally almost flat". You do not need a computer to see this flatness effect: just look at the surface of a small pond in calm weather. Does it look flat or spherical for you?

As an aside, I have another query: What's the primary difference between an Ellipse and an Oval? Does there exist any precise distinction?  (I have been told something like: All ellipses are ovals, but all ovals need not be ellipses, but I'm seeking a more precise articulation)

"Ellipse" is a technical term with several (equivalent) formal definitions.  "Oval" is not a technical term -- a close formal approximation of its (natural language) meaning would be any curve that can be approximated closely by a simple closed curve in the plane of continuous, positive curvature and the symmetry of a rectangle.

(Minor edit for using wrong word.  Also, I don't think I agree with yenlee about the egg -- my girlfriend and I both believe that you need rectangular symmetry, i.e., the egg (having only one mirror symmetry) is out.)

"Ellipse" is a standard technical term.  "Round" is not.  The OP seems to understand this point, and the question seems to be what meaning (not necessarily technical) the word "round" *ought* to have.

My contribution is that I've seen the word "round" used in a quasi-technical way in only one context:  Since most mathematicians use the word "sphere" to mean a "topological sphere," differential geometers often use the phrase "round sphere" to mean an actual geometric sphere, i.e. a standard sphere sitting in R^n.  In this context, an ellipse is not "round."

Restricting to the case of curves in the plane, this definition of "round" means *constant* positive curvature.  Fedja proposed a definition that uses only continuous positive curvature.  This doesn't seem right to me, since this would say that parabolas are round, and that even the profile of a needle is round (after smoothing out the pointy end).  I'd personally go with a *quantitative* measure of "roundness" such that a circle has infinite "roundness" (i.e. a circle is perfectly round).  If you want a purely local definition, one could define "roundness" at a point of the curve to be the square of the curvature divided by the derivative of the curvature (so that large "roundness" means that the curvature is not changing too quickly).  The "roundness" of the entire curve could then be the minimum roundness of points along the curve.  But I actually think that a global definition (involving the diameter of the curve) might be better.

Fun but pointless exercise: Formulate desired properties for "roundness" and come up with a definition that satisfies those properties.  For example, the circle should maximize roundness, and the definition should be scale-invariant.  An ellipse whose radii have a ratio of 2 should be more round than one with a ratio of 3.  An ellipse should be more round than a square with rounded edges, etc, etc.  Of course, the properties you choose would be a matter of taste.

Regarding ovals:  Yes, ovals also do not have a precise defintion, but I disagree with JBL's take on ovals.  I think that most people would regard the profile of an egg to be an oval.

Try using vector graphics software. An ellipse (or is it an oval?) stays curved even when zoomed in.


geometry
circumcircle
Euler
parallelogram
geometry solved

Let the altitude $AD$ of a triangle $ABC$ which intersects the circumcircle at the point $A_1$. prove that the distance from the center of circle of Euler (of the triangle ABC) to the $BC$ is $\frac{AA_1}{4}$

The distance between N and BC equals half of the sum of HD and the distance between O and BC.
The result follows since $HD=DA_1$ and the distance beteween O and BC equals half of AH.

OK... This solution is certainly wrong... But I post it anyway ^^

Let $A_2, B_2, C_2$ be the midpoints of the Sides $BC, AC$ and $AB$. Then the circle of Euler is the Circumcircle of $A_2B_2C_2$. Obviously, $ABC$~$A_2B_2C_2$ and $|A_2B_2|=2|AB|$, $|A_2C_2|=2|AC$, $|B_2C_2|=2|BC|$. Furthermore, $B_2C_2$ is parallel to $BC$.
Now, from the Center of the Euler-Circle (name it $N$) you have to look at the perpendicular (its length is the distance, name the foot $O$). Now draw a parallel to it through $A_2$ and a parallel to $BC$ through $N$, name the intersection-point of these parallels $X$.

Now, $A_2XNO$ is a parallelogram, and hence $A_2X$ is a perpendicular to $BC$, it is also Perpendicular for $B_2C_2$ and so it is the Altitude for $A_2B_2C_2$ through $A_2$. Let $Y$ be the point where the circle of Euler intersects $A_2X$. Since $A_2B_2C_2$~$ABC$ with factor 2, and since $A_2X$ is now similar to $AA_1$, $2|A_2X|=|AA_1|$.

It is obvious that $|YX|=|XA_2|$ and $|NO|=|XA_2|$, and so
\[
|AA_1| = 4|ON|
\]

I note the midpoints $S,M$ of the segments $[AH],\ [BC]$ respectively and the distance $x$ of the Euler's center $E$
 to the line $BC$. We know that the center $E$ is midpoint of the segment $[SM]$ (a diameter of the Euler's circle)
and $DH=DA_1$. Thus, $2\cdot SD=AD+HD=AD+DA_1=AA_1$ and $x=\frac 12 \cdot SD=\frac 14\cdot AA_1,\ q.e.d.$

Silouan ,
 I've just made a solution but it is the same as Post 2 , so I do not write it again.
 [b]Please , you must [/b]note that these  relations between(  AH  and OM , that is  AH = 2OM   )and  (  HD  =  $DA_1$ ) are very basic. The center E of Euler circle  is the midpoint of OH(or SM) .But there you have a trapez  OMDH ect( AD is the altitude , M is the midpoint od BC , H is the orthocenter , S the midpoint of AH).

 Babis


Here, I'll introduce you to a "magic trick".
Ask a friend of yours to choose a 3 digits number (at random; the hundreds digit has to be different from the units digit).
Then, tell him to write this number backwards and to substract the lowest from the greates.
Now, ask him to write this last number backwards and to add it to the previous one.

You're now able to tell him the final result.

What is this result, and why?

;)

Assume the chosen number is $\overline{abc}$. Then the backwards number is $\overline{cba}$. We can assume $a>c$ (for $a<c$ the solution is similar). Then $\overline{abc}-\overline{cba}$ has units digit $10+c-a$ (because $c<a$), tenths digit $9$ (because of the carryover) and hundredths digits $a-c-1$ (because of the carryover). Reversing the order, we get $\overline{(10+c-a)9(a-c-1)}$. Adding the units digits of the two numbers, we get $10+c-a+a-c-1=9$. Adding the tenths digits, we get $9+9=18$, with the carryover $1$, and adding the hundredths digits with the carryover we get $a-c-1+10+c-a+1=10.$ Therefore, the final result is $1089$ no matter what the initial number is.

Yes, that's it! ;)

I thought it was funny :)

There was an old classic trick based on this  which found "the name of your wife"  (if unmarrid then name of the future wife) ...
Wonder if anyone remembers that here.. it was really  neat.

[quote="lingomaniac88"]I've heard of that.  It takes your last name and puts it after "MRS".[/quote]

Or "Lady" (If you are a British Lord) or "Smt" (Short for "Shrimati") if you are are an Indian Shriman  :)
(Funny thing  was  that I played that trick on some body in a party - The culprit was sure that I did not even know if he was married , let alone his wife's name  ( someone has passed that info to me, but he had no clue  as he never met me before). I had plenty of time  to  calculate but supply what looked like  a random "magic number"  with his own   calculation on seemingly his choice of numbers.  (after selecting numbers depending on his favorite actress's name etc)  and lo and behold  his  actual wife's name turned up :) ... I don't think he ever knew how I pulled  that one .:)

(Actually I was very proud of my  (plagiarised/modified) version of the trick which I made it really funny.)


In this figure, each pair of adjacent sides is perpendicular. The figure is to be divided into 12 equal pieces having the same size and shape. The area of one of these pieces can be expressed in the form axⁿ. What is the product of a and n, expressed as a common fraction?

problem no 42  please see the figure in  the following pdf link

http://www.mnspe.org/mathcounts/2007tests/510-07_C_Countdown.pdf

answer is 1/2

Hi.

The $ L\minus{}$shaped figure is made up of a recagtangle of area $ 2x \cdot x \equal{} 2x^{2}$ on top of a square of area $ x^{2}$, giving a total area of $ 3x^{2}$.

If this is divided into $ 12$ identical shapes, say little squares, each of area $ y$, then $ 12y \equal{} 3x^{2}$ and so  $ y\equal{}\frac{1}{4}x^{2}$.

Finishing off the solution, $ \frac{1}{4}\times{2}\equal{}\boxed{\frac{1}{2}}$.  :D

Thanks. I didn't read the question fully.


function
modular arithmetic
algebra
domain
combinatorics proposed
combinatorics

Let $ n\geq 3 $ be an integer and let $ p\geq 2n-3 $ be a prime number. For a set $ M $ of $ n $ points in the plane, no 3 collinear, let $ f: M\to \{0,1,\ldots, p-1\} $ be a function such that

(i) exactly one point of $ M $ maps to 0,
(ii) if a circle $ \mathcal{C} $ passes through 3 distinct points of $ A,B,C\in M $ then $ \sum_{P\in M\cap \mathcal{C}} f(P) \equiv 0 \pmod p $.

Prove that all the points in $ M $ lie on a circle.

Well, let us suppose for the sake of contradiction that there exist points $B,C$ such that no other point $\in S$ lies on the circle $C(A,B,C)$.
This implies that $f(B) + f(C) \equiv 0\pmod p $.
Therefore, we can say that if we have $f(B) = i$, then we must have $f(C) = p-i$.
Now, let us denote $L= \sum_{X \in S} f(X)$.

Consider $b$ circles passing through $A, B$ and other points  of $S$ and $c$ circles passing through $A,C$ and other points of $S$.  Thus applying the given condition for all such circles,  we have : $L + (b-1) i \equiv 0\pmod p $ and $L + (c-1)(p-i) \equiv 0\pmod p $. This implies, $a+b - 2 \equiv 0\pmod p $.
Now, remember that $b,c \ge 1$ and $b,c \le n-2$.  This implies $2\le b+c \le 2n-4 <2n-3 \le p$       (Given that $p \ge 2n-3)$
This would imply that $a=b=1$, which is clearly a contradiction.

Therefore, we get that for any two points $B,C \in S$, there will exist atleast one more point being to $C(A,B,C)$  lying in $S$.
Therefore, all points of $S$ lie on a circle.

We can indeed  consider an inversion $I$ of center $A$. The set of points $S- {A}$ is transformed to the set of points $(S- {A} )'$ under the inversion. Let us call it as $M$.  Also, circles $C(A, P, Q)$ with $P,Q \in S$  get transformed to  lines $I_P I_Q$ using points of $M$.  Thus , the condition is equivalent to :
For two points of $I_P, I_Q$ in $M$, there should exist another point of $M$ lying on $I_PI_Q$. But this is true and hence we prove the inversive images are collinear thereby making the original domain concyclic.

Dear Agr_94_Math,
What are S and A?  Please define your variables.

I am very sorry Batominovski, I have used $S$ for $M$ and $A$ is the point of $ M$ for which $f(A)= 0 $.


geometry
3D geometry
probability
parameterization
conics
ellipse
rotation

Consider a unit cube $ ABCDEFGH$, where $ ABCD$ and $ EFGH$ are squares and vertices $ E$, $ F$, $ G$, $ H$ are respectively adjacent to vertices $ A$, $ B$, $ C$ and $ D$ in the cube. On face diagonals $ AF$ and $ BG$, a point is randomly and independently chosen on each of those diagonals. A line segment is then formed by connecting the two points. What is the probability that this line segment is longer than $ \frac {\sqrt{3}}{2}$?

[hide="Remark to problem writer"]
I strongly suggest you change $ d \geq \frac {\sqrt {3}}{2}$ to $ d \geq j$ for some $ \frac {1}{\sqrt {2}} \geq j > \frac {1}{\sqrt {3}}$, so that it's computationally reasonable without calculus. Personally, I suggest $ \frac {1}{\sqrt {2}}$.

This is because parametrization of the points as (0,t,t) and (s,1-s,0) results in an ellipse rotated 45 degrees in the s-t plane, and we want to find the area contained within this ellipse and the unit square. It would be computationally favorable if the ellipse were entirely contained within the unit square =)
[/hide]

[hide="Solution to suggested variant"]
[b]Setup and Parametrization:[/b] We have by the distance formula that $ s^2 \plus{} t^2 \plus{} (s \plus{} t \minus{} 1)^2 \geq \frac {1}{2}$ using the parametrization in the remark above, in which FG is the x direction, FE is the y direction, and FB is the z direction, with point F the origin.

[b]Preparation:[/b] Note that this equation is that of an ellipse, symmetric in s and t. So one axis will be along the line s=t, and the tangents and the endpoints of the other axis will be of the form s=t+k for some k.

[b]Minor Axis Endpoints:[/b] If s=t, we can solve $ 2s^2 \plus{} (2s \minus{} 1)^2 \equal{} \frac {1}{2}$ to get $ s \equal{} \frac {1}{2} , \; \frac {1}{6}$, so the length of the minor axis is $ \sqrt {2} \cdot \left( \frac {1}{2} \minus{} \frac {1}{6} \right) \equal{} \frac {\sqrt {2}}{3}$

[b]Major Axis Endpoints:[/b] Now, if we let $ s \equal{} t \plus{} k$, we have $ t^2 \plus{} (t \plus{} k)^2 \plus{} (2t \plus{} k \minus{} 1)^2 \equal{} \frac {1}{2} \implies t^2 \plus{} \left( k \minus{} \frac {2}{3} \right) t \plus{} \left( \frac {k^2 \minus{} k}{3} \plus{} \frac {1}{12} \right) \equal{} 0$.

We want $ k$ such that there is only one solution in $ t$. Clearly, then, we must have $ \left( \frac {k}{2} \minus{} \frac {1}{3} \right)^2 \equal{} \frac {k^2 \minus{} k}{3} \plus{} \frac {1}{12}$, to create a perfect square situation. Solving, $ k \equal{} \pm \frac {\sqrt {3}}{3}$, so the length of the major axis is $ \frac {1}{\sqrt {2}} \cdot \left( \frac {\sqrt {3}}{3} \minus{} \frac { \minus{} \sqrt {3}}{3} \right) \equal{} \frac {\sqrt {6}}{3}$

[b]Area:[/b] Thus the area of the ellipse is $ \frac {1}{4} \cdot \pi \cdot \frac {\sqrt {6}}{3} \cdot \frac {\sqrt {2}}{3} \equal{} \frac {\sqrt {3}}{18} \pi$

[b]Probability:[/b] Clearly, we want the area outside of this ellipse, since (s,t) = (1,1) results in a distance of $ \sqrt {3}$, and the denominator is the area of the unit square, or 1.

Hence, $ \boxed{P \equal{} 1 \minus{} \frac {\pi \sqrt {3}}{18}}$

By similar means, if we have $ d\geq j$ instead of $ d \geq \frac{1}{\sqrt{2}}$, with $ j$ in the bounds discussed in the remark, then $ P \equal{} 1 \minus{} \frac{\sqrt{3}}{9} (3d^2\minus{}1) \pi$
[/hide]

[quote="The Zuton Force"][hide="Remark to problem writer"]
I strongly suggest you change $ d \geq \frac {\sqrt {3}}{2}$ to $ d \geq j$ for some $ \frac {1}{\sqrt {2}} \geq j > \frac {1}{\sqrt {3}}$, so that it's computationally reasonable without calculus. Personally, I suggest $ \frac {1}{\sqrt {2}}$.

This is because parametrization of the points as (0,t,t) and (s,1-s,0) results in an ellipse rotated 45 degrees in the s-t plane, and we want to find the area contained within this ellipse and the unit square. It would be computationally favorable if the ellipse were entirely contained within the unit square =)
[/hide][/quote]

Thanks for the tip  :D Neat solution, it's going to take me a while to do the calculations.


calculus
integration
function
limit
trigonometry
derivative
real analysis

well here is the question:

evaluate: $\displaystyle\int\left(\sqrt{x^3+1}\right)dx$

this was not part of the test but was a sorta extra credit.

Put $\displaystyle \left(\sqrt{x^3+1}\right)=\sqrt{x^3}+t$

Then solve for $\sqrt{x^3}$.

Hope have helped

i am not quite sure that i follow what you are saying. :blush:

[quote="jmerry"]Don't worry, it's useless.

This is an elliptic integral, and the answer is not an elementary function. See [url=http://mathworld.wolfram.com/EllipticIntegral.html]here[/url] for more information. These functions are of course well-studied; they just aren't something you could write down in closed form with functions you already know.[/quote]
Following up with what jmerry said, there's no way this would be on a Calculus AB test.  It wouldn't make sense for the instructor to give calc. 1 students an integral they couldn't evaluate or expect them to write it in the form of an elliptic integral.  Are you sure you copied it correctly?

in the instructions it said
"dont bother to find the original fucntion...its really a waste of time and you wont be able to any way"
but she said that she would give extra credit to anyone who was able to do it.
by the way i am in Calc BC, thats prob. y my friend asked me.  :P

I can see asking a test question like this:

Suppose $f(x)$ is a function such that $f'(x)=\sqrt{x^3+1}$ and $f(2)=-7.$ Compute

$\lim_{x\to 2}\frac{f(x)+7}{x-2}$

Or maybe the question is something like this:

Find $\frac{d}{dx}\int_{-x}^{\sin x}\sqrt{t^3+1}\,dt$

My point in asking the question that way is that I really want you to demonstrate that you know the definition of the derivative, or the fundamental theorem of calculus, or whatever it is that I'm asking about, and I want to make sure that you cannot do the problem by some sort of automatic calculation without thought.

Rule of thumb: if you can't see a way to find an antiderivative (primitive) of some function, the odds of it even having an elementary antiderivative are very, very poor - and probably not worth your time to chase.

yeah u could do by parts, but that really doesnt get you anywhere.


Alcumus videos algorithm probability Support \/closed

[b]Alcumus[/b] is a new adaptive learning program from Art of Problem Solving.  It consists of a database of problems with and video lessons backed by an algorithm that chooses problems for students based on their past performance.  Alcumus currently complements our [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202]Introduction to Counting & Probability[/url] textbook.  We plan to add material in support of our other texts in coming years.

Early reviews for Alcumus include:

[quote="[url=http://albanyareamathcircle.blogspot.com/]Mary O'Keeffe[/url], educator, homeschool parent, state champion MATHCOUNTS coach"][i][b]All I can say is wow! I found myself thinking a lot - as I worked through the problems and then watched the videos - this is so cool! I wish I'd had access to something like when I was in the middle school grades. I think access to something like Alcumus could literally be a life-transforming experience for students![/b][/i] [/quote]

[quote="[url=http://www.AGMath.com]Jason Batterson[/url], middle school teacher, state champion MATHCOUNTS coach"][i][b]This is most incredible teaching resource I have ever found online. I can not think of a better way to learn how to solve really difficult problems than what has been created in Alcumus. Everything has been considered. Extraordinary! [/b][/i][/quote]

Please give Alcumus a try, and share it with your friends.  You can find a link to Alcumus on the sidebar.

You can also [url=http://www.artofproblemsolving.com/Alcumus/Introduction.php]click here to get started with Alcumus[/url]!

[quote="mathswimmer"]For how long do you plan to let Alcumus be free?[/quote]

We haven't decided.  We anticipate offering the Introduction to Counting & Probability portion for free indefinitely.  Whether or not future subjects will also be free depends a lot on how successful the program is and what its impact on our other offerings is.  If it brings in a lot of people who enroll in classes and buy books, then that would be a strong argument to keep it free.  So, tell your friends and teachers about it ;)

Will there be another program for a different subject?

when will the algebra part be integrated?

How come the alcumus introductory page says like 60 video lessons, when there's only about 7 or 8?

[quote="came"]How come the alcumus introductory page says like 60 video lessons, when there's only about 7 or 8 [u][i][b]that you are currently allowed to watch[/b][/i][/u]?[/quote]

You have to pass subjects before you can watch related videos.

no, you don't. i can see about 31 videos and it says the ones i have passed and the ones i haven't.

I believe math154 means the subjects you are working on, plus the ones you have passed, plus the ones you have mastered.

what different subjects are going to be introduced with algebra? Like square roots and other subjects like that.

[quote="Dojo"]Algebra will probably be the material covered in the [url=http://www.artofproblemsolving.com/Books/IntroAlgebra/toc.pdf]Introduction to Algebra[/url] book.[/quote]

That's correct.  We'll also cover radicals and fractional exponents, which are not thoroughly covered in the first edition of the Intro Algebra book, but will be added to the second edition.

As for timing, that's not as clear -- we have a lot on our plate, and the timing of producing the next portion of algebra will depend on a lot of factors, including the success of this first release of Alcumus.

Keep up the amazing work. This website, most definitely including alcumus, is awesome.

How soon is expected for Alcumus to have other subjects besides C&P? Will there be an option to let you choose 1 subject to work on at a time?

[quote="rrusczyk"]We expect to expand Alcumus over the next few years, first covering the rest of our Intro series, and then covering both more challenging material, and material for younger students.[/quote]

Intro to Algebra will be next.


modular arithmetic
algebra
binomial theorem

I read this somewhere:
If $ p$ does not divide $ a,b$ and $ a \equiv b \pmod p$
then $ a^{p} \equiv b^{p} \pmod {p^{2}}$

Does $ p$ have to be a prime number in this case ? Does this hold if $ p|a$ or $ p|b$ ?

If the result is true (I haven't tried to prove it), it won't work if $ p$ isn't prime: let $ p\equal{}6$, $ a\equal{}13$, and $ b\equal{}7$.

My proof:

Since $ a \equiv b \pmod{p}$, we have $ a \equal{} b \plus{} kp$ for some $ k$.

We want to show that $ a^p \equiv b^p$, so we take the $ p$ power of the first equation:

$ a^p \equal{} (b \plus{} kp)^p$.

From the binomial theorem, every term except for the first two is equivalent to 0 mod $ p^2$.

$ a^p\equiv b^p \plus{} p\cdot b^{p \minus{} 1}p\equiv b^p$, so it holds.

_______________________________________________________________________
in this, $ p$ must either divide neither $ a$ nor $ b$ or both $ a$ and $ b$.

[quote]It seems it has to be a prime number[/quote]

I dont get it. Doesnt your proof imply that there are no restrictions on $ p$?
 Can you please tell me where you have used the fact that $ p$ is prime ?

JRav, i think $ 36|13^{6} \minus{} 7^{6}$

yeah it does divide it...also, i agree; i don't see any reason for it just to be for primes (my bad on that)

[quote="JRav"]If the result is true (I haven't tried to prove it), it won't work if $ p$ isn't prime: let $ p \equal{} 6$, $ a \equal{} 13$, and $ b \equal{} 7$.[/quote]

$ 13\equal{}7 mod 6$
$ 13^6\equal{}1 mod 36$
$ 7^6\equal{}1 mod 36$

Yes, he's right.  I made an arithmetic mistake before.  Ignore what I wrote before.

[quote="lindlar"]I read this somewhere:
If $ p$ does not divide $ a,b$ and $ a \equiv b \pmod p$
then $ a^{p} \equiv b^{p} \pmod {p^{2}}$
[/quote]

This is a specific instance of a more general principle. You can read more about it here: reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf


Prove that for any distinct positive integers $ x,y$,the number $ x^4\minus{}x^2y^2\plus{}y^4$ is not a perfect square.

not sure if this is a correct solution

[hide]
$ x^4 \minus{} x^2y^2 \plus{} y^4 \equal{} (x^2 \minus{} y^2)^2 \plus{} x^2y^2$ which is a sum of two squares supposed to be equal to a square. That means that for some m,n of different parity either $ x^2 \minus{} y^2 \equal{} m^2 \minus{} n^2$ and $ xy \equal{} 2mn$ which of course leads to a contradiction or $ x^2 \minus{} y^2 \equal{} 2mn$ and $ xy \equal{} m^2 \minus{} n^2$ which implicates $ (x^2 \plus{} y^2) \plus{} 2n^2 \equal{} (m^2 \plus{} n^2) \plus{} 2y^2$. As m, n are of different parity we can claim that also x and y are of different parity, but that would lead us to a contradiction with $ x^2 \minus{} y^2 \equal{} 2mn$.

That means there are no such numbers.
[/hide]

Could anyone good check that?


Does anyone know what the usual cutoff score is on the USAMO in order to qualify for MOP?

u can check out the percentage and rank blocks on the AMC website...it's usually not too high...

I think it's not cutoff but rather AMC/AIME index + 9*USAMO.  I'd estimate the cutoff (for MOP index) to be around the 400s and possibly more, but it is doable if you prepare.

7*USAMO, actually.  Anyhow, it varies greatly year-to-year depending on the difficulty of the USAMO (and probably on the relative quality of students taking it).

AMC/AIME index determines MOP qualification? really? i thought it was just USAMO score...

Yeah, according to the AMC site, anyway.  By looking at the scores and AIME stats, you can have a rough estimate.

well for seniors, it's just top 12...which is like impossible...for me at least

ack....my amc/aime scores are coming back to haunt me...


Αν Δ,Ε είναι οι προβολές του ορθοκέντρου Η ενός τριγώνου ΑΒΓ στις διχοτόμους της γωνίας  Α (εσωτερικής και εξωτερικής ) και Μ είναι το μέσο της ΒΓ , να αποδειχθεί ότι τα σημεία Δ,Ε , Μ είναι συνευθειακά.

[quote="b555"]which language is this? :?:[/quote]

You're in the Greek part of AoPS :D

If you were looking for a translation, the problem reads:

If $ D,E$ are the projections of the orthocenter $ H$ of a triangle $ ABC$ on the bisectors (interior and exterior) of the angle $ A$ and if $ M$ is the midpoint of $ BC$, show that $ D,E,M$ lie on the same line.

Cheerio,

Durandal 1707

Εν συντομια αρκει να δουμε οτι  το ΔΗΕΑ ειναι ορθογωνιο.Αρα οι ΔΕ,ΑΗ εχουν κοινο μεσο ,εστω το Κ.Αν θεωρησουμε το περικεντρο Ο του

  τριγωνου ΑΒΓ τοτε απο την πολυχρησιμοποιημενη προταση το ΑΚΜΟ ειναι παραλληλογραμμο.Μετα ειναι ευκολο.


linear algebra
matrix
induction
inequalities proposed
inequalities

Let $ \left(a_{n}\right)^{n}_{i=1}$ and $ \left(b_{n}\right)^{n}_{i=1}$ be $ 2$ sequences of positive integers and for $ 1\leq i\leq n$ we have $ b_{i}\leq a_{i}$. Prove that:

$ \prod^{n}_{i=1}\begin{pmatrix}a_{i}\\ b_{i}\end{pmatrix}\leq\begin{pmatrix}a_{1}+a_{2}+\cdots+a_{n}\\ b_{1}+b_{2}+\cdots+b_{n}\end{pmatrix}$

We need only prove $ \binom{a_{1}}{b_{1}}\cdot\binom{a_{2}}{b_{2}}\leq \binom{a_{1}+a_{2}}{b_{1}+b_{2}}$ because induction on $ n$.


1)How do you get from $ xyz \equal{} x \plus{} y \plus{} z \plus{} 2 \rightarrow \dfrac{1}{1 \plus{} x} \plus{} \dfrac{1}{1 \plus{} y} \plus{} \dfrac{1}{1 \plus{} z} \equal{} 1$.
2)If $ abc \equal{} 1$
   Then let $ x \equal{} a \plus{} \dfrac{1}{a}, y \equal{} b \plus{} \dfrac{1}{b}, z \equal{} c \plus{} \dfrac{1}{c}$

     $ xyz \equal{} \displaystyle \left(a \plus{} \dfrac{1}{a} \right)\displaystyle \left(b \plus{} \dfrac{1}{b} \right)\displaystyle \left(c \plus{} \dfrac{1}{c} \right) \equal{} \displaystyle \left(a^2 \plus{} \dfrac{1}{a^2} \right)\plus{}\displaystyle \left(b^2 \plus{} \dfrac{1}{b^2} \right)\plus{}\displaystyle \left(c^2 \plus{} \dfrac{1}{c^2} \right) \plus{} 2 \equal{} \displaystyle \left(x^{2} \minus{} 2 \right) \plus{} \displaystyle \left(y^{2} \minus{} 2 \right) \plus{} \displaystyle \left(z^{2} \minus{} 2 \right) \plus{} 2$

I get lost after the $ \displaystyle \left(a \plus{} \dfrac{1}{a} \right)\displaystyle \left(b \plus{} \dfrac{1}{b} \right)\displaystyle \left(c \plus{} \dfrac{1}{c} \right)$ part.
Could someone please explain how it was reached?

[quote="rootspie"]1)How do you get from $ xyz \equal{} x \plus{} y \plus{} z \plus{} 2 \rightarrow \dfrac{1}{1 \plus{} x} \plus{} \dfrac{1}{1 \plus{} y} \plus{} \dfrac{1}{1 \plus{} z} \equal{} 1$.
2)If $ abc \equal{} 1$
   Then let $ x \equal{} a \plus{} \dfrac{1}{a}, y \equal{} b \plus{} \dfrac{1}{b}, z \equal{} c \plus{} \dfrac{1}{c}$

     $ xyz \equal{} \displaystyle \left(a \plus{} \dfrac{1}{a} \right)\displaystyle \left(b \plus{} \dfrac{1}{b} \right)\displaystyle \left(c \plus{} \dfrac{1}{c} \right) \equal{} \displaystyle \left(a^2 \plus{} \dfrac{1}{a^2} \right)\displaystyle \left(b^2 \plus{} \dfrac{1}{b^2} \right)\displaystyle \left(c^2 \plus{} \dfrac{1}{c^2} \right) \plus{} 2 \equal{} \displaystyle \left(x^{2} \minus{} 2 \right) \plus{} \displaystyle \left(y^{2} \minus{} 2 \right) \plus{} \displaystyle \left(z^{2} \minus{} 2 \right) \plus{} 2$

I get lost after the $ \displaystyle \left(a \plus{} \dfrac{1}{a} \right)\displaystyle \left(b \plus{} \dfrac{1}{b} \right)\displaystyle \left(c \plus{} \dfrac{1}{c} \right)$ part.
Could someone please explain how it was reached?[/quote]
For the first one, I can show you how to reach the first statement from the second...but I don't know how you would ever get from the first to the second...seeing as you basically multipled everything out from the second equation to get to the first one.
For the second problem, I am also a bit lost from the step $ (a \plus{} \frac {1}{a})(b \plus{} \frac {1}{b})(c \plus{} \frac {1}{c})$ to the next one, but I do see the 2nd step to the last step. Notice that because $ x \equal{} a \plus{} \frac {1}{a}$, if we square both sides we have that $ x^{2} \equal{} a^{2} \plus{} \frac {1}{a^{2}} \plus{} 2$. So $ a^{2} \plus{} \frac {1}{a^{2}} \equal{} x^{2} \minus{} 2$, and do the same for $ y$ and $ z$.  Hope this helps in any way possible.

Also are you sure for #2 that it isnt $ (x^{2}\minus{}2)(y^{2}\minus{}2)(z^{2}\minus{}2)\plus{}2$?

Sorry about that. It was a typo. I left out the plus signs in between
$ \left(a^{2}\plus{}\frac{1}{a^{2}}\right) and \left(b^{2}\plus{}\frac{1}{b^{2}}\right) and \left(c^{2}\plus{}\frac{1}{c^{2}}\right)$

where'd you get these problems?

for the 1st one:

Add $ xy \plus{} yz \plus{} zx \plus{} x \plus{} y \plus{} z \plus{} 1$ to both sides.  We can then factor it as

$ (x \plus{} 1)(y \plus{} 1) \plus{} (y \plus{} 1)(z \plus{} 1) \plus{} (z \plus{} 1)(x \plus{} 1) \equal{} (x \plus{} 1)(y \plus{} 1)(z \plus{} 1)$

Dividing both sides by (x+1)(y+1)(z+1),

$ \dfrac{1}{z \plus{} 1} \plus{} \dfrac{1}{x \plus{} 1} \plus{} \dfrac{1}{y \plus{} 1} \equal{} 1$, which is the desired equation.

@ckck: (#1) all you have to do is work backwards, no matter how improbable or weird the steps are :)

@rootspie: (#2) then i guess use algebraic manipulation to arrive to that step


function
calculus
derivative
advanced fields
advanced fields unsolved

Could someone help me solve these phase portraits problems:

1. Consider the motion of a pendulum subject to a constant torque, described by the equation: $ x'' \equal{} 1 \plus{} 2sin(x)$
Sketch the phase portrait. Which motions of the pendulum correspond to the different kinds of curves?

What I've done so far:
[hide]My problem: I know how to sketch phase portraits for a system with x and y, but this one doesn't have a y. So are the curves in the phase portrait all going to just flow in an x direction? If this is the case, then do I just integrate this and solve for $ x'$ and then sketch the phase portrait?[/hide]

2. Show that the system $ x'' \equal{} a \minus{} e^x$, where $ a$ is a constant, is conservative. Find the Hamiltonian and sketch the phase portrait in the three cases $ a < 0$, $ a \equal{} 0$, and $ a > 0$

What I've done so far: [hide]$ H(x,x) \equal{} ax \minus{} e^x \plus{} U(x) \equal{} constant$ Is this right so far? If so, what am I supposed to graph for the phase portrait? ($ U$ vs. $ x$? or $ x'$ vs. $ x$? or something else?)[/hide]

Thanks in advanced :)

Every system in the form $ x''\equal{}f(x)$ is conservative. In fact, that system is equivalent to the following:
$ x'\equal{}v$
$ v'\equal{}f(x)$

and the function $ E(x,v)\equal{}\frac{1}{2}v^2\plus{}V(x)$, where $ V'(x)\equal{}\minus{}f(x)$, is a costant for the motion of the system, that is, the Lie derivative of that function is 0. So when you have to sketch the phase portrait of a system, it is meant to be done in the plane $ x\minus{}v$, cause $ v$ represent the speed of the object you are considerating. So your phase portrait is just the set of curves in the $ x\minus{}v$ plane of the kind $ E(x,v)\equal{}k$ with $ k \in \mathbb{R}$.

Thanks so much! After reading what you said, I think I've managed to work out #1; however, I was still unsure of how to solve #2.

This is what I did for #1, most if not all should be correct:

[hide]Using the system
$ x'\equal{}v$
$ v'\equal{}1\plus{}2sin(x)$

We determine the equilibrium points to be $ (\frac{7\pi}{6}, 0)$ and $ (\frac{11\pi}{6}, 0)$ for $ (x, v)$

We then draw in the x-nullcline $ v\equal{}0$ and y-nullclines $ x\equal{}\frac{7\pi}{6}$ and $ x\equal{}\frac{11\pi}{6}$ and end up with something like this, keeping in mind that at $ (\frac{7\pi}{6}, 0)$, we get a center, and at $ (\frac{11\pi}{6}, 0)$ we get a saddle:
[img]http://img138.imageshack.us/img138/434/abcdefzy3.png[/img][/img][/hide]

For #2, I had a question:

[hide]Is it sufficient enough to show the system is conservative just by saying it is equivalent to the system:

$ x'\equal{}v$
$ v'\equal{}a\minus{}e^x$

and that the Hamiltonian is $ H(x,v) \equal{} \frac{1}{2}v^2 \minus{} ax \plus{} e^x \equal{} k$ in which $ k$ is constant, and thus the system is conservative? Or would I have to say more in order to prove this is conservative?

For the other part of the question, I'm pretty sure I know how to graph the phase portraits for $ a < 0$, $ a \equal{} 0$, and $ a>0$. I just wanted to make sure I proved this is conservative correctly.[/hide]

Do you agree with my answers?

For question #2, I'm sorry but I don't know what Hamiltonian is, but I'm sure that the function $ H$ you wrote is enough to say that your system is conservative, cause it is in the form $ x''\equal{}f(x)$. For the first part, you would have to say that that systyem has an infinite number of equilibrium points, cause all points in the form $ \left( \frac{7}{6}\pi \plus{} 2k\pi,0 \right)$ and $ \left( \frac{11}{6}\pi \plus{} 2k\pi,0 \right)$, with $ k \in \mathbb{Z}$ are equilibrium points.
The other fact is that when you have a non linear system, as in this case, you have to be careful in saying that a point is a center. Here you're right, but why? It is not enough just watching at the jacobian of your system, cause a point that is a center for the jacobian might not be that in the non linear case.
When you have a system of the form $ x''\equal{}f(x)$, the best thing to do to sketch the phase portrait is to skecth the function $ V(x)$ that I told you in my other post, and then skecth orizontal lines that represent the "levels of energy of the function $ E$, and sketching then the orbits of your system in the $ x\minus{}v$ plane remembering that for any fixed level $ E_0$ the quantity $ \frac{1}{2}v^2\plus{}V(x)$ os constant, equal to $ E_0$