Many of you have been reporting some issues with percentages declining and some odd behavior. We hope we've fixed these issues. If you run into problems, please post them here and be very specific! Tell us exactly how you ran into that issue.

We want to make the Alcumus experience as smooth and painless as possible, so any input is greatly appreciated :)

When I first came on, I was at 98.08 rating.  I got a problem correct on my first try, then I checked the statistics and my rating was 89 or somewhere around there.  Is this a glitch?

Ratings have decreased dramatically without any activity. I know this has occurred for multiple people. For example, my rating is 98.21, but my report says 90.09. I have not played for quite a while now, and my rating just dipped. Fortunately, for me, I did no more problems, and thus my 98.21 is on the Statistics page. However, for others, notably CountdownKing and prezcoin (in the top 10), their ratings dropped sharply and this is reflected on the Statistics page (because they did a problem recently, after the glitch).

It appears that the leaderboard page probably has the correct Alcumus score. I did a few problems, and it shot up, but when I returned, the old Alcumus score was on my report, while the new one was on the leaderboard. Glitch?

The Alcumus rating is not intended to be stable when you are not playing -- the Alcumus overall rating is a measure of each person compared to everyone else using the system.  So, if you achieve a high rating, and then don't use Alcumus for a long time, others will "catch up" to you, and thereby bring your comparative rating down.  (Conversely, if a lot of people join Alcumus and do poorly, your rating will drift up.)

I think Paul was working on the percentages in each subject that are used for passing/mastery.  These, I think, should be stable (at least, passing and mastery should be).

I understand that, but whenever I do a problem and raise my score, whenever I refresh, my score before I do a problem correctly appears.

I don't understand what you're saying.  Do you mean you have score X on one page and score Y on a different page?  And what score are you referring to?

I'm referring to Alcumus Score.
First, lets say my score was 74.9. I did some Alcumus in the subject of Pascal's Triangle, trying to master it. I answered all questions correctly, and may have tried two times on a few problems. When I started, it showed that I had answered 9 Pascal's Questions. When I ended, it says I have done 20 questions. I stopped doing Alcumus and took a break. My score was 75.8.
When I came back, not only had my Alcumus score dipped back to 74.9, the questions it said that I had answered was 9 again. Everything was reset.
The leaderboard, or stat page, displays my Alcumus score 75.8. My report says 74.9. I start another problem, and the stat page goes back to 74.9. Glitch?

On the stats page for me, it says 98.21, my actual rating. Then, on the report section, not only does it show a much lower 90.04, but it says I've done very few problems. For example, in one topic, it says I have gotten 2/2 problems correct, whereas before the glitch, i was around 98/100.

Ratings don't really matter, but I'm fairly certain there is a glitch. For some people, they did one problem and dropped, for example from 97 to 89.

Thank you for your detailed replies. This is not a problem we had looked into, but it seems that this should be a straightforward fix. We'll let you guys know when this is fixed.


function
Pythagorean Theorem
geometry
algebra proposed
algebra

Find all function $f: \mathbb{R}\rightarrow [0,+\infty)$ s.t. $\forall x,y\in\mathbb{R}$
$f(x^2+y^2)=f(x^2-y^2)+f(2xy)$

Well, because of the Pythagorean theorem, $F(x)=x^2$ is right and also $F(x)=0$ (constant).I suppose that's all, but I'll see that later, now I'm busy  :lol:  :lol:  :roll:

Defining $f(x) = g(x^2)$ helps a lot, since then $g$ is an additive function (satisfying the Cauchy equation, but with some restrictions) and hence, under the conditions of the problem, $g$ must be linear (it's not immediate though).

So, I think all solutions are given by $f(x) = ax^2$ where $a \in \mathbb{R}$, and $a$ can be 0.

yeah, i forgot the coefficent $a$ :blush:  :blush:


number theory
prime numbers
number theory unsolved

[quote="iura"]Given are prime numbers $p_1,\cdots,p_r$ and exponents $a_1,\cdots,a_r$.
Prove that there are at most $2^{r-1}$ pairs $(x,y)$ for which
$x^3+y^3=\prod_{i=1}^r p_i^{a_i}$[/quote]
mmmm, not sure, if you consider $r=1$ and number $2^3$ then we have the solutions $(2,0)$ and $(0,2)$, so more that $2^0$ solutions. Please, specify HOW two choose those pairs $(x,y)$.

RX

I think $(x,y)=1$ right Iura? anyway here is what i found:

we work with numbers $Z+Zw$ where $w^3=1$ and $w$ is not $1$.

We have that every prime of the form $3k+2$ is irreducible in this field.
We have that every prime of the form $3k+1$ is of the form $a.a'$ where $a$ and $a'$ are conjugates, and irreducibles.

Now $x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)(x+yw)(x+yw^2)$ So we have, iff $p_i=3k+2$, then $p_i^{a_i}$ divides $x+y$. Now iff $p_i=3k+1$ then $p_1^{a_i}$ divides $x+y$  or $x^2-xy+y^2$. If it divides $x+y$ there is no mistery, If it doesnt, write $p_i=a.a'$ and we are left with 2 options: $a^{a_i}$ divides $x+yw$ and $a'^{a_i}$ divides $x+yw^2$ or backwards (2 options)because their common factor can only possibly be $1-w$ and then $p_i=3$.

now if $p_i=3$ then we have that $p_i=(1-w)(1-w^2)$, if 9 divides $x^2-xy+y^2$ then $3$ divides $x$ and $y$, contradiction, so we must have 2 options, $1-w$ divides $x+yw$ and $1-w^2$ divides $x+yw^2$ or backwards.

Anyway, our choice of $x+yw$ and $x+yw^2$ determiens $x+y$. there are at most $2^r$ ways of building this pair by choosing wich irreducible divisor of every prime goes where, but obviously, there are $2^{r-1}$ different ways, becuase for every way, another one (the one constructed changing every irreducible by its conjugate) give the same paor $x,y$, so result follows.

There are many things i didnt explain formaly, but they are known results.

I don't know Pascual.. I have some doubts.
Firstly in the condition is not mentioned $gcd(x,y)=1$.
Secondly, I think you produce $3^r$ not $2^r$ cases because you investigate if $p$ goes to $x+y$-first case;
$a$ goes to $x+yw$ and $a'$ goes to $x+yw^2$ -second case and viceversa - third case

damn, anyway, thats the best i cuold give you

I still think $(x,y)=1$, but i dont know why i did all that staff when there were easier things to do.

Put $x+y=a$ $xy=b$ and $x^2-xy+y^2=c$. We have that $ac$ is the product of some primes, first lets do an easy case $(a,c)=1$. then there are at most $2^r$ ways to distribute the primes in $a$ or $c$. since $a^2-c=xy$ we have that $4c-3a^2>0$ is a square, but then $4a-3c^2<0$, So for every asignation, the oposite one, dont gives good couples $(a,c)$ (look that iff $4c-3a^2>0$ and $4a-3c^2>0$, We would need $a=c=1$ contradiction). So there are at most $2^{r-1}$ good asignations. Now the hard part is when 3 is between those primes, because we may have $(a,c)=3$ so in this case there are at most $2^{r-1}3$ initial asignations, becuase $9$ cant divide $c$. iff 3 divides $c$ and $a$ we get $4c-3a^2=9t^2$ so 9 divides $c$, contradiction, so this cannot happen, and we have the same analizis from the easy case, I hope this is ok.

[quote="Pascual2005"]...but i dont know why i did all that staff when there were easier things to do.[/quote]  :P  :P  :P Unfortunately, I didn't see it, so I invented exactly the same solution.

Anyway, who knows nontrivial examples of numbers, which can be represented as sum of two cubes in many ways?

jaja, my proof is full of mistakes! jaja but thew concept is allright...

here i write the correct proof:

let $a=x+y$, $b=x^2-xy+y^2$ then $ab$ is a product of a lot of primes, since $a^2-b=3xy$ we have that if 3 divides 1 of them then it divides both and moreover 9 doesnt divide $b$. So considere 2 cases:

case 1: $p_i=3$. then for the other $r-1$ primes there are $2^{r-1}$ of putting them either on $a$ or in $b$.

case 2: $p_i$ is not $3$, then we have $2^r$ assignments, and "oposite" or "complementary" asignments cant both give good values for $x$ and $y$, using some inequalities. (maybe some trivial cases fail).

For the thing we dont now about if $(x,y)=1$ i presume it must be like that, otherwise, non of waht we have done works.

[quote="kueh"]The smallest one as famously exclaimed by Ramanujan (did I misspell?): $1729 = 10^3 + 9^3 = 12^3 + 1^3$[/quote]
His name is spelled Ramanujuan, Srinivasa Ramanujuan in full.

[quote="Myth"]Anyway, who knows nontrivial examples of numbers, which can be represented as sum of two cubes in many ways?[/quote]
this is one curious entry in mathworld: http://mathworld.wolfram.com/TaxicabNumber.html

Is there anyone who has a correct solution orthe official one??

The problem statement is so wrong in so many ways.
First, $gcd(x,y) = 1$, and the bound is $2^{r+1}$

Case 1. $x+y \neq 0 \pmod {3}$
$(x+y,x^2-xy+y^2) = 1 \implies$ we can partition $p_i$ and solve for $x,y$
We can partition the primes in $2^r$ ways, and since if $(x,y)$ is a solution, $(y,x)$ is a solution, we have at most $2^{r+1}$ solutions.

Case 2. $x+y \equiv 0 \pmod {3}$
$x+y = 3^sk$ and $x^2-xy+y^2=3^{2s}k^2-3xy$
$x^3+y^3 = 3^{s+1}k(3^{2s-1}k^2-xy)$
Since $k(3^{2s-1}k^2-xy) \neq 0 \pmod {3}$, we can let $p^{r_1}_1 = 3^{s+1}$
Partition the other primes in $2^{r-1}$ ways, and since if $(x,y)$ is a solution, $(y,x)$ is a solution, we have at most $2^r$ solutions.

Note that Case 1 and Case 2 cannot happen at the same time.
By Case 1 and Case 2, we are done $\blacksquare$


Ilian will randomly choose a value for p from the integers 0 through 13, inclusive, and will make $\frac{p}{13}$ the coordinate of point A. The coordinate of point B is $\frac{2}{3}$. Points A and B will be on the same number line. What is teh probability that point A and point B will be less than $\frac{1}{5}$ of a unit from each other? Express your answer as a common fraction.

[hide]Only 7, 8, 9, 10, and 11 satisfy the conditions, so the probability is $\frac5{14}$.[/hide]

[quote="superwizard188"][hide]We have: 0<= p =>13 Case 1. 2/3 > p/13

(26-3p)/39 < 1/5
Simplify:

130-15p < 39
15p > 91
p > 6 and 1/15

p = 7, 8 ....13
Case 2. p/13 > 2/3.

(3p-26)/39 < 1/5
Simplify:

15p - 130 < 39
15p < 169
p < 11 and 4/15
p = 11, 10 .... 0

The intersection of p = 7, 8, 9, 10, 11. (actually, we could have set more restrictions earlier I think, but this works too)

so 5/14 (since it's inclusive) ;)[/hide][/quote]
Why would you subtract the two fractions $\frac{2}{3}$ and $\frac{p}{13}$ in the first case?


I've learned new technique, but I'm first interested in your solutions.

For x,y,z reals prove that:

$(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})+(x^{2}+y^{2}+z^{2})+(xy+yz+xz) \geq 2(x^{2}y+y^{2}z+z^{2}x)+(xy^{2}+yz^{2}+zx^{2})$

[b]EDIT[/b]: Sorry, it is not valid, I was wrong.

[quote="Vihang"]Is there some typo

You said for a,b,c reals
There are no a,b,c

Also both sides can be divided by 2[/quote]

Thanks Vihang, I've edited.

[quote="Davron"]OK it is time look for your solution ...  :)

Davron[/quote]

Actually not. I'm waiting for some other solution. And I don't want to influence you by writing my solution.

Noone interested in this problem? I have two solutions yet. Do you want a hint?


In the given diagram, we need to get from the top-left corner to the bottom-right corner. Each of the planks is $5 \times 1m$ and you can not travel on the same plank twice. You are allowed to travel left and right on horizontal planks and up and down on vertical planks. Also, You don't have to travel the whole length of a plank as you will find out once you start experimenting.
a) How many ways are there for going from the top-left corner to the bottom-right corner.
b) How ways are there if we are allowed to travel on the same plank twice.

[quote="Altheman"]i don't understand the rules for moving...[/quote]

Me neither. If you travel say left on one plank, can't you go back right and say its travelling on the plank next to the previous one?

[quote="white_horse_king88"]Are we traveling along the plank edges or the whole plank? Also if we could travel on a plank more than once, could we travel on it, say, back and forth forever?[/quote]
[quote="seamusoboyle"]Its seems you walk along the middle of the plack. But for example if you walk along the lowest horizontal plank in a set of 5 you can get off halfway and go down.[/quote]
As seamusoboyle said, we are travelling in the middle. And no, you can't go back and forth forever, you just go in a direction. Here is an example.

[quote="Inspired By Nature"]Any guesses as to how to approach it?[/quote]

What I did was I took the four top left corner boxes to simplify the case. I think a way to do this is to number as you go, how many ways to get to this plank, and then multiply my the number of ways to get to the next plank, etc, but this is sort of tedious, I'll try to get an answer.

[quote="Inspired By Nature"]That certainly is a way to go but it may take a while. Is there a more ingenious approach, i.e., comparing it to say squares (in place of the rectangular planks)? I'm just brainstorming, so don't take it too seriously. :lol:[/quote]

Hmmm, Interesting. Maybe you can take all the planks off and put suqares and have just square walking, except you can go left, right, and up, down, but you can only go on one edge one time (just like you can only walk on one plank one time). But I haven't figured out an eligant solution for this yet...

Won't that take a while? Could you please explain more, just in case I'm not following properly.

[quote="seamusoboyle"]Or you can draw in all the places you can actually walk, transforming it into a graph where you trying to count the routes from one end to the other[/quote]

This route may actually work, but it could be confusing if you use the same color to draw it.

So did anyone get a solution (partial,etc)? I'm having some trouble with another question right now, I'll try this one again later.


analytic geometry
geometry
incenter
symmetry
circumcircle
angle bisector
AMC

#18

A point (x,y) is to be chosen in the coordinate plane so that it is equally distant from the x-axis, y-axis, and the line x+y=2. Then x is:

A) $\sqrt {2} - 1$
B) $\frac {1}{2}$
C) $2 - \sqrt {2}$
D) 1
E) Not uniquely determined

[hide]
This is how i would do it
the resulting triangle has legs along x and y axis with side lengths 2
the hypotenuse is 2sqrt2.
the circumscribed circle has radius r :
r = (a+b-c)/2 by the greek geometer.
r = 2 - sqrt2.

you'll notice one radius is perpendicular to the y axis, therefore x is
2 - sqrt2
[/hide]

Hint:

[hide]The three lines form a triangle.  What special point(s) in a triangle is equidistant from the three sides?[/hide]

The answer key said they had four coordinates. But I have no idea how to find them, and determine their coordinate.

The four answers $(x,y)$ are the centers of the incircle and three excircles. (The three excircles of a triangle are the circles outside of the triangle that are tangent to the extensions of the sides of the given triangle.) The incenter $I$ is easy to find, the excenters can be found via some triangle-geometry calculations. Call the three points of intersection between the lines $A$, $B$, and $C$. The center $O_A$ of the excircle opposite $A$ has the property that $A$, $I$, and $O_A$ are collinear (along the angle bisector of $\angle{BAC}$) and $\frac{IA}{O_AA} = \frac{s-a}{s}$, where $s$, the semiperimeter of $ABC$, is given by $\frac{a+b+c}{2}$, and $a, b, c$ are the sides of $ABC$. We'll compute the solution in the fourth quadrant.

$A = (0,2)$
$I = (2 - \sqrt{2}, 2 - \sqrt{2})$
$\vec{AI} = <2 - \sqrt{2}, -\sqrt{2}>$
$a = 2$
$s = 2 + \sqrt{2}$.

Hence, $\vec{AO_A} = \frac{2 + \sqrt{2}}{\sqrt{2}}\vec{AI} = <\sqrt{2}, - 2 - \sqrt{2}>$, giving the solution $O_A = (\sqrt{2}, -\sqrt{2})$. By symmetry, the final solution is $(-\sqrt{2},\sqrt{2})$.


trigonometry
calculus
calculus computations

Need help on my homework :blush:

Replace the Polar equation by an equivalent Cartesian equation

$ r\equal{}8sinx$

rrr..... x is actually theta, but i don't know how to write that...sorry

Given $ r\equal{}8\sin\theta$
multiply both sides by $ r$:
$ r^2\equal{}8r\sin \theta$
which means $ x^2\plus{}y^2\equal{}8y$.

[quote="charlestreykang"]x is actually theta, but i don't know how to write that...sorry[/quote]
For $ \theta$, use \theta. Also, $ \sin$ (\sin) is nicer; there are similar commands for most standard functions.


geometry
3D geometry
probability
number theory
prime numbers

[quote="ckck"][quote="Quevvy"][quote="ckck"]Two cubes, each with faces numbered 1,2,3,4,5 and 6 are tossed. Each cube lands wiith a prime number on top. What is the probability that the sum of these prime numbers is less than 9? Express your answer as a common fraction.[/quote]
[hide]The prime numbers are 2, 3, and 5 so the sums are 5, 7 and 8 so 1/1 is the probability[/hide][/quote]
But what happens when the first cube is a 5 and the second cube is a 5? The sum is 10. That changes the probability.[/quote]
Oh yeah, I forgot they could be different.....
[hide="so"] The sums are 4, 5, 6, 7, 8 and 10 and 5 of those  work. so then the probability is $\frac{5}{6}$[/hide]

[quote="Quevvy"][/quote][quote="ckck"][quote="Quevvy"][quote="ckck"]Two cubes, each with faces numbered 1,2,3,4,5 and 6 are tossed. Each cube lands wiith a prime number on top. What is the probability that the sum of these prime numbers is less than 9? Express your answer as a common fraction.[/quote]
[hide]The prime numbers are 2, 3, and 5 so the sums are 5, 7 and 8 so 1/1 is the probability[/hide][/quote]
But what happens when the first cube is a 5 and the second cube is a 5? The sum is 10. That changes the probability.[/quote][quote="Quevvy"]
Oh yeah, I forgot they could be different.....
[hide="so"] The sums are 4, 5, 7, 8, 9 and 10 and 4 of those  work. so then the probability is $\frac{4}{6}=\frac{2}{3}$[/hide][/quote]

where did 9 come from??

I get 8/9, only 5 5 is greater than 9 and there are 3*3 possibilities

[quote="slamchess"][/quote][quote="Quevvy"][/quote][quote="ckck"][quote="Quevvy"][quote="ckck"]Two cubes, each with faces numbered 1,2,3,4,5 and 6 are tossed. Each cube lands wiith a prime number on top. What is the probability that the sum of these prime numbers is less than 9? Express your answer as a common fraction.[/quote]
[hide]The prime numbers are 2, 3, and 5 so the sums are 5, 7 and 8 so 1/1 is the probability[/hide][/quote]
But what happens when the first cube is a 5 and the second cube is a 5? The sum is 10. That changes the probability.[/quote][quote="Quevvy"]
Oh yeah, I forgot they could be different.....
[hide="so"] The sums are 4, 5, 7, 8, 9 and 10 and 4 of those  work. so then the probability is $\frac{4}{6}=\frac{2}{3}$[/hide][/quote][quote="slamchess"]

where did 9 come from??[/quote]
I multiplied 3*3. sorry, I'll edit.

Art of Owna: How are there 3*3 possibilities?

ok three primes so 3 choices for first die, 3 for second, 3*3=9
only one pair that is great than 9, 5+5
so 1-1/9=8/9

[quote="Quevvy"][/quote][quote="slamchess"][/quote][quote="Quevvy"][/quote][quote="ckck"][quote="Quevvy"][quote="ckck"]Two cubes, each with faces numbered 1,2,3,4,5 and 6 are tossed. Each cube lands wiith a prime number on top. What is the probability that the sum of these prime numbers is less than 9? Express your answer as a common fraction.[/quote]
[hide]The prime numbers are 2, 3, and 5 so the sums are 5, 7 and 8 so 1/1 is the probability[/hide][/quote]
But what happens when the first cube is a 5 and the second cube is a 5? The sum is 10. That changes the probability.[/quote][quote="Quevvy"]
Oh yeah, I forgot they could be different.....
[hide="so"] The sums are 4, 5, 7, 8, 9 and 10 and 4 of those  work. so then the probability is $\frac{4}{6}=\frac{2}{3}$[/hide][/quote][quote="slamchess"]

where did 9 come from??[/quote][quote="Quevvy"]
I multiplied 3*3. sorry, I'll edit.

Art of Owna: How are there 3*3 possibilities?[/quote]
When doing probability you need to count all the combinations, including order.  so, 3 choices for the first die, and 3 for the 2nd.  Multiply to get 9.  Now, only one of those combos doesn't work (5,5) so it's 8/9.

The key thing to remember is that (x.y) and (y,x) anre different combinations.


I figured this out couple of days ago..

You can use asy(program) to draw pictures!

Some examples:
[asy]size(100); draw(unitcircle,dotted+blue+linewidth(1));[/asy]

[asy]size(100); draw(arc((0,0),2,0,180)); draw((-2,0)--(2,0));[/asy]

[asy]size(100); pair A,B,C, A=(0,0); B=(10,0); C=(0,10); draw(A--B); draw(B--C); draw(C--A);[/asy]

Isn't it awesome?
Learn more at http://www.artofproblemsolving.com/Forum/index.php?f=123

I would recommend [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Basics]this[/url] instead.

[asy]draw((0,0)--(1,0)); draw((1,0)--(0,1)); draw((0,1)--(0,0));[/asy]


On a coordinate graph, the points $ (0,6)$ and $ (\minus{}3,2)$ are joined by a line segment. By extending this line segment, find the point where it intersects the $ x$-axis. Express your answer as an ordered pair in decimal form.

[quote="GameBot"]On a coordinate graph, the points $ (0,6)$ and $ ( \minus{} 3,2)$ are joined by a line segment. By extending this line segment, find the point where it intersects the $ x$-axis. Express your answer as an ordered pair in decimal form.[/quote]

Equation to the line joining the points is $ \ 3y \equal{} 4x \plus{} 18\implies\left( \minus{} \frac {9}{2},0\right)\equal{}(\minus{}4.5,0)$


I thought it was an interesting problem from mathcounts.

A circle of radius 9 cm is inscribed in an equilateral triangle and is tangent at three points. Three smaller circles are inscribed so that they are each tangent to two sides of the triangle and to the larger circle. The process is continued with circles of ever decreasing diameters. What is the number of centimeters in the sum of the circumferences of all such circles? The answer is x Pi . Find x .
a.  50
b.  55
c.  30
d.  45
e.  60

[hide] Let R=9 cm be the radius of the big circle. The next step will give three smaller circles having each R/3 as a radius. The next step after that will give three smaller circles having each R/32 as a radius and so on. The answer is the sum of an infinite geometric series as follows.
2 Pi R + 3* ( 2 Pi R) ( 1/3 + 1/32 + 1/33 +.... )=
2 Pi R + 3* ( 2 Pi R) *1/3 *( 1 + 1/3 + 1/32 + 1/33+...)=
2 Pi R + 3* ( 2 Pi R) * 1/3 *(1/(1-1/3)=2 Pi R + ( 2 Pi R) *3/2= 5 Pi R= 45 Pi
The answer is 45.
[/hide]

Hello,

Refer to Round 4 Problems and Round 4 Solutions [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=246134&start=120]here[/url]; the problem is essentially the same, just with a radius of $ 10$.  It is problem number $ 6$.

--AIME15--

You actually don't even need a geometric series. Note that all the diameters of the small circles make up the segment from the center of the triangle to vertices. Excluding the big circle, this is a distance of 9 in the three directions. So it's 27 pi + the circumference of the big circle, totaling 45 pi.

EDIT:Yeah, I meant circumference.

You meant the circumference of the big circle, right?  :roll:


What is the largest 4-digit palindrome that is the sum of 2 different 3-digit palindromes?

Thanks

[hide] The greatest sum which two three-digit integers can have is $ 999\plus{}999\equal{}1998$. So we know our palindrome will be $ 1xx1$, where $ x$ is a digit. Our three-digit palindromes will either be of the form $ zyz$ or $ zzz$ where $ z$ and $ y$ are digits. Note also that since $ 1$ is the units digit of our $ 4$-digit palindrome, the units digit of our three-digit palindromes will be one of of $ (9,2)$, $ (8,3)$, $ (7,4)$, $ (6,5)$. After some trial and eror, we see that $ \boxed{1221}$ is our four-digit palindrome ($ 999\plus{}222$, $ 888\plus{}333$, $ 777\plus{}444$, $ 666\plus{}555$).[/hide]

Trial and error is not necessary.

[hide="Solution"]
$ \overline{aba}\plus{}\overline{cdc}\equal{}\overline{effe}$. If $ a,b,c,d$ are maximized, we'd have $ 999\plus{}999\equal{}1998$, so clearly $ e\equal{}1$. Then:

$ 101(a\plus{}c)\plus{}10(b\plus{}d)\equal{}1001\plus{}110f$

From this we have that $ a\plus{}c\equiv 1\bmod 10$. We can't have one of $ a$ or $ c$ be $ 0$, and $ a\plus{}c\le 18$, so it's clear that $ a\plus{}c\equal{}11$.

Then $ 101(a\plus{}c)\equal{}101\cdot 11\equal{}1111$, so we manipulate our previous equation to get $ 110(f\minus{}1)\equal{}10(b\plus{}d)\implies b\plus{}d\equal{}11(f\minus{}1)$. Again, $ b\plus{}d\le 18$, so either $ 11(f\minus{}1)\equal{}0$ or $ 11(f\minus{}1)\equal{}11$ (can't be $ 22$). Then $ f$ must be either $ 1$ or $ 2$, and choosing $ f\equal{}2$ gives the larger palindrome.

Answer: $ \boxed{1221}$.
[/hide]


geometry
circumcircle
projective geometry
geometry proposed

[color=darkred]The tangents from a point $ A$ to the circle $ w(O)$ touch it in the points $ B$, $ C$. Denote : the point $ D\in w$ for which $ O\in (AD)$ ; The projection $ X$ of the point $ B$ to the line $ CD$ ; the middlepoint $ Y$ of the segment $ [BX]$ ; the second intersection $ Z$ of the line $ DY$ with the circle $ w$. Prove that $ ZA\perp ZC$.[/color]

Trivial problem:

Let $CH$ be diameter, then $HD\parallel XY$. We have to show: $X,Z, A$ are collinear.
Let $A' = AC\cap HZ$ and $B' = BC\cap HZ$. Now we have:
\begin{eqnarray*}(A',B;'Z,H) &=& (CA',CB'; CZ,CH)\\ &=& (CC,CB; CZ,CH)\\ &=& (C,B; Z,H)\\ &=& (DC,DB; DZ,DH)\\ &=& (DX,DB; DY,DH)\\ &=& (X,B; Y,\infty)\\ &=&-1 \end{eqnarray*}
So, $A'$ and $B'$ are harmonical conjugate. Since $B'\in BC$= polar line for $A$ so are $A$ and $B'$. This means $AA'$ is polar for $B'$ if $A'\ne A$. But $AA'\bot OC$ and $AA'\bot OB'$. So $O,B'C$ are collinear.  Contradiction! Thus $A'=A$ wich proves your statement.

Let $E$ be the midpoint of $BC$. So $EY // CD$ and $EY \bot BX$
$\angle YEB=\angle DCB=\angle DZB$ so $Y,E,Z,B$ are cyclic
Thus $\angle ZBC=\angle ZEA$ (because $OA$ is the tangent of circle $(YEZB)$)
We also have $\angle ZBC=\angle ZCA$ so $\angle ZCA= \angle ZEA$
This means $C,A,Z,E$ are cyclic. So $\angle CZA=\angle CEA=90$ $q.e.d$

[color=darkblue]Thank you very much for the your nice problem, [b]Cosmin Pohoatza[/b] !

Thank you very much for the your nice and cool proof, [b]Huyen Vu[/b] !

Thus, using the [b]Huyen's idea[/b] of proof, I generalized the [b]Pohoatza's problem.[/b][/color]

[quote="Virgil Nicula"][color=darkred][b][u]Generalization.[/b][/u] The tangents from a point $A$ to the circle $w(O)$ touch it in the points $B$, $C$. For a point $E\in (BC)$ denote : the point $D\in w$ for which $E\in (AD)$ ; the second intersection $X$ of the line $CD$ with the circumcircle of the triangle $BED$; the point $Y\in BX$ for which $EY\parallel CD$; the second intersection $Z$ of the line $DY$ with the circle $w$; the intersection $T\in AC\cap BX$; the second intersection $H$ of the line $AZ$ with the circle $w$. Prove that the quadrilaterals $AZEC$, $AZYT$ are cyclically and $BH\parallel \overline{AED}$.[/color][/quote]
[color=darkblue][b]Proof.[/b]

$1.\blacktriangleright\widehat{ZBE}\equiv$ $\widehat{ZBC}\equiv$ $\widehat{ZDC}\equiv$ $\widehat{ZYE}$ $\Longrightarrow$ $\widehat{ZBE}\equiv\widehat{ZYE}$ $\Longrightarrow$

the quadrilateral $ZBYE$ is cyclically.

$2.\blacktriangleright\widehat{EBY}\equiv$ $\widehat{EBX}\equiv$ $\widehat{EDX}\equiv$ $\widehat{EDC}\equiv$ $\widehat{YED}$ $\Longrightarrow$ $\widehat{EBY}\equiv \widehat{YED}$ $\Longrightarrow$

the line $\overline{AED}$ is tangent to the circumcircle of the quadrilateral $ZBYE$.

$3.\blacktriangleright\widehat{ZEA}\equiv$ $\widehat{ZBE}\equiv$ $\widehat{ZBC}\equiv$ $\widehat{ZCA}$ $\Longrightarrow$ $\widehat{ZEA}\equiv$ $\widehat{ZCA}$ $\Longrightarrow$

the quadrilateral $AZEC$ is cyclically.

$4.\blacktriangleright\widehat{ZYB}\equiv\widehat{ZEB}\equiv\widehat{ZAC}\equiv\widehat{ZAT}$ $\Longrightarrow$ $\widehat{ZYB}\equiv\widehat{ZAT}$ $\Longrightarrow$

the quadrilateral $AZYT$ is cyclically.

$5.\blacktriangleright$ $\widehat{AEC}\equiv\widehat{AZC}\equiv\widehat{CDH}\Longrightarrow\widehat{AEC}\equiv\widehat{CDH}$ $\Longrightarrow$ $BH\parallel \overline{AED}$.

[b][u]Remark.[/u][/b] In the particular case $AE\perp BC$ obtain $AZ\perp ZC$ si $O\in CH$.[/color]

[quote="Number1"]Trivial problem:

Let $CH$ be diameter, then $HD\parallel XY$. We have to show: $X,Z, A$ are collinear.
Let $A' = AC\cap HZ$ and $B' = BC\cap HZ$. Now we have:
\begin{eqnarray*}(A',B;'Z,H) &=& (CA',CB'; CZ,CH)\\ &=& (CC,CB; CZ,CH)\\ &=& (C,B; Z,H)\\ &=& (DC,DB; DZ,DH)\\ &=& (DX,DB; DY,DH)\\ &=& (X,B; Y,\infty)\\ &=&-1 \end{eqnarray*}
[/quote]

Very nice proof Number1 ! from here it is really obvious, because $A'C$ is tangent to $w$, and $(A'ZB'H)=-1$, so $BC$ is the polar of $A'$ w.r.t to $w$, therefore $A=A'$, very nice!


Hello :)

Recently, I saw this problem that I have not been able to solve. That is why I seek some help from you guys. The problem can be found [url=http://undernetmath.wordpress.com/potd/]here[/url] . It is the one from 2008-2-1 by kmh, where one should prove that the green area of the circle is as big as the white area of the circle.

I assume that the problem statement is correct but I do not know if solving problem requires substantial mathematical knowledge or if it is a problem requiring only basic math yet ingenuity beyond average.

I look forward to hearing if you have a solution and would like to post it :)

--
MarF

http://www.maths.unsw.edu.au/~mikeh/webpapers/paper57.pdf
The problem is also given at Russian competition - The tournment of towns.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="Mathstunned"]Show that if H and N are subgroups of a group G, and N is normal in G, then H intersect N is nromal in H. Show by an example that H intersect N need not be normal in G.
-----------------

The answer is.. Let G=N=S3 and H={Ro, mew1}. Then N is normal in G, but H intersect N=H is not normal in G.

Please explain these results!!!!
 :?:[/quote]

What are Ro and mew1? I've never seen this notation...

I consider $ H \equal{} \{e, (1, 2)\}$. $ H \cap N \equal{} H$ that is not normal in $ G$.

Let $ h \in H \cap N$. $ \forall a \in G, aha^{ \minus{} 1} \in N$ but it isn't necessarily in $ H$.

$ \rho\equal{}(123)$ and $ \mu_1\equal{}(12)$ if I remember this notation correctly.

[quote="HilbertThm90"]$ \rho \equal{} (123)$ and $ \mu_1 \equal{} (12)$ if I remember this notation correctly.[/quote]

$ H \equal{} \{\rho, \mu_1\}$ is not a subgroup. It’s not closed, it have not the neutral elements and $ \rho^{ \minus{} 1}$ and $ \mu_1^{ \minus{} 1}$ aren't in $ H$...

Maybe you mean $ H \equal{} \left < \rho, \mu_1 \right >$ but in this case $ H \equal{} D_3 \equal{} S_3$

because

$ H \equal{} \left < \rho, \mu_1 \right > \equal{} \{e, \rho, \rho^2, \mu_1, \rho \mu_1, \rho^2 \mu_1\} \equal{} \{id, (123), (132), (12), (13), (23)\} \equal{} S_3$


I just noticed that the before 1986, the scoring formula is 30+4R-W, where R=# of right answers, W=#of wrong answers, which is basically 5R+B, where B=# of blank answers.  But after 1986, the scoring formula 5R+2B...

So do they make the tests harder after 1986 because basically u get B extra points after 1986 on AHSME....

also what was the cutoff score for AIME before 1986???

According to [i]The Contest Problem Book V[/i] by Berzsenyi and Maurer, the purpose of the change was to discourage random guessing.  Before 1986, the scoring system was guessing-neutral.  From 1986 on, the system was guessing-negative.

From 1983 to 1985, the AIME cutoff was 95.  In 1986, the AIME cutoff changed to 100.

oh, I got 121 according to the new formula on AHSME 1984, which is, unfortunately, only 113 according to the old formula... :( , so trick is you could guess and don't need to spend the time to check the problems you did....

I'd also note that some of the AHSMEs just before the introduction of the AIME were notably harder than the ones just after they introduced the AIME, I think.

Back when I took the AHSME (1969-70-71), there was no AIME and the guessing penalty was set so that a blank answer was zero points and a wrong answer was some negative number. The highest possible score was 150, but negative scores were possible, and I heard of negative scores in my school. I think my own scores at the time were in the 70's and 80's. I have no idea how to translate that into a more recent score.


i know there have already been posts on this topic but i couldn't access the video.. :(
can someone tell me where to get a video that was premiered on tv on some show?
i would also like the one of kevin chen on the regis show thingy.

thanks!

No part of the competition was aired on television. Every year, Mathcounts releases a ~3 minute video to the public, containing clips of interviews and countdown. It has not been released for this year yet.

I don't think the video's out yet, like 13375P34K43V312 said...
But, the pictures are out. Just click [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=1023&z=108]here[/url].


trigonometry
function
limit
logarithms
complex analysis
complex analysis unsolved

Prove that $\sin{\pi(z+\alpha)}= e^{\pi z \cot{\pi\alpha}}\prod_{n \in \mathbb{Z}}(1+\frac{z}{n+\alpha})e^{-\frac{z}{n+\alpha}}$, where $\alpha \in \mathbb{C}-\mathbb{Z}$.

Got part of it:

If f(z) is entire with m the order of the zero of f(0) and $\{z_{k}\}$ the list of zeros, then for some set $\{p_{k}\}$ and another entire function h(z), we can use the Weierstrass Factorization Theorem to write f(z) as:

$f(z)=e^{h(z)}z^{m}\prod_{k=1}^{\infty}E_{p_{k}}\left[\frac{z}{z_{k}}\right]$

where:

$E_{p_{k}}[z]=(1-z)Exp[z+z^{2}/2+...+z^{p_{k}}/p_{k}]$

for any sequence $\{p_{k}\}$ such that:

$\sum_{k=1}^{\infty}\left|\frac{r}{z_{k}}\right|^{p_{k}+1}<\infty,r>0$

Since $Sin[\pi(z-\alpha)]$ has zeros of order 1 at $-(n+\alpha)$ for each $n\in\mathbb{Z}$ and:

$\sum_{n\in\mathbb{Z}}\frac{1}{|n+\alpha|^{2}}$ converges then let $\{p_{k}\}=\{p_{k}=1\}$

Therefore:

\[\begin{eqnarray*}Sin[\pi(z+\alpha)] &=&e^{h(z)}\prod_{n\in\mathbb{Z}}E_{1}\left[-\frac{z}{n+\alpha}\right] \\ &=&e^{h(z)}\prod_{n\in\mathbb{Z}}\left(1+\frac{z}{n+\alpha}\right)e^{-z/(n+\alpha)}\end{eqnarray}\]

can't figure out how to compute h(z) though.

I found by accident the following theorem in Schaum's:

If $f(z)$ be entire with simple zeros $\{a_{n}\}$ and $\lim_{n\to\infty}|a_{n}|=\infty$ then $f(z)$ can be expressed as:

$f(z)=f(0)e^{zf'(0)/f(0)}\prod_{n=1}^{\infty}\left\{(1-z/a_{k})e^{\frac{z}{a_{k}}})\right\}$

so that I could write:

$Sin[\pi(z+\alpha)]=Sin[\pi\alpha] e^{\pi zCot[\pi \alpha]}\prod_{n\in\mathbb{Z}}\left\{(1+\frac{z}{n+\alpha}) e^{-\frac{z}{n+\alpha}})\right\}$

Now, when I check this numerically, it agrees well with $Sin[\pi(z+\alpha)]$.  Also, when I check it exactly with Mathematica's "FullSimplify" the right side indeed reduces to $Sin[\pi(z+\alpha)]$.

Can anyone suggest where a proof of this theorem might be found?

Sorry, there was a mistake in the problem.It should be

Prove that $\sin{\pi(z+\alpha)}= e^{\pi z \cot{\pi\alpha}+\log{\sin{\pi \alpha}}}\prod_{n \in \mathbb{Z}}(1+\frac{z}{n+\alpha})e^{-\frac{z}{n+\alpha}}, where \alpha \in \mathbb{C}-\mathbb{Z}$.


integration
trigonometry
calculus
function
geometric series
real analysis
real analysis solved

[hide="Proof by Fourier Series"]
We wish to find the Fourier Series of $f(x) = x$.  We have $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos nx \, dx = 0$ for all $n$ and $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin nx \, dx = -\frac{2 \cos (n \pi)}{n}$, or in other words, $\displaystyle\frac{2}{n}$ for odd $n$ and $-\displaystyle\frac{2}{n}$ for even $n$.

$x = 2( \sin x - \frac{\sin 2x}{2} + \frac{ \sin 3x}{3} -+ ...)$

Substitute $x = \frac{\pi}{2}$.  We then have

$\frac{\pi}{2} = 2(1 - \frac{1}{3} + \frac{1}{5} -+...)$
$\pi = 4(1 - \frac{1}{3} + \frac{1}{5} -+...)$

Q.E.D.
[/hide]

I tought of placing this in teh calculus sections, so woudl someone please move this?  Thanks in advance. :)

Taylor series for $\arctan t$.  Wrong forum.

It´s just a expand of a Taylor Series of the function $f(x)=arctan(x)$. You have to calculate $f^k(0)$ for all k natural and you can do this by induction. You won´t calculate all the expression, but just the part that you are interest. Ok?

or use term-by-term integration (which I think you can do because it converges absolutely) on $\frac{1}{1+x^2} = \sum_{k = 0}^{\infty} (-1)^k x^{2k}$ by a geometric-series argument.

[quote="blahblahblah"][quote="t0rajir0u"][hide="Proof by Fourier Series"]
We wish to find the Fourier Series of $f(x) = x$.  We have $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos nx \, dx = 0$ for all $n$ and $b_n = \frac{1}{\pi} \int_{\pi}^{\pi} x \sin nx \, dx = -\frac{2 \cos (n \pi)}{n}$, or in other words, $\displaystyle\frac{2}{n}$ for odd $n$ and $-\displaystyle\frac{2}{n}$ for even $n$.

$x = 2( \sin x - \frac{\sin 2x}{2} + \frac{ \sin 3x}{3} -+ ...)$

Substitute $x = \frac{\pi}{2}$.  We then have

$\frac{\pi}{2} = 2(1 - \frac{1}{3} + \frac{1}{5} -+...)$
$\pi = 4(1 - \frac{1}{3} + \frac{1}{5} -+...)$

Q.E.D.
[/hide][/quote]

How do you know the Fourier series converges to the function?

[hide]$f(x)=x$ satisfies a Lipschitz condition [/hide][/quote]

I think that if a function is "locally Lipschitz", i.e. for every point $x$, there exists a $c$ (which may depend on $x$) such that $\forall\,y, |f(y) - f(x)| \le c|y-x|$, then the Fourier series converges; this is of course true for $f(x) = x$.


geometry
incenter
geometric transformation
homothety
power of a point
radical axis

We shall first prove that P belongs to the radical axis of the circles [code]w_1[/code] an d [code]w_2[/code], using inversion.
Let's call D and E the contact points of the circles [code]w_1[/code] and [code]w_2[/code] with the circle [code]w[/code], while F and G are the contact points of the same circles and BC.
It is well known that DF is the angle bisector of <BDC (due to homothety, the tangent to the circle [code]w[/code] at the point of contact of DF with circle [code]w[/code] is parallel to BC, therefore it touches the circle at the midpoint of the arc BC ( P ) ).
Similary, EG passes through P.
An inversion of pole P and power [code]BP.BP[/code] takes the circle [code]w[/code] to the line BC, therefore the images of D and E will be F and G, consequently the circles [code]w_1[/code] and [code]w_2[/code] remain unchanged through the said inversion.
Since [code]PF.PD = PB.PB = PG.PE[/code], we can conclude that [code]P [/code] belongs to the said radical axis, so PM is the common tangent of the 2 circles. As [code]PF.PE = PM.PM[/code], we conclude that [code]PB = PM = PC[/code], i.e. M is the incenter of triangle [code]ABC[/code]. (M is on the angle bisector of <BAC at the same distance as B and C of the middle of the arc BC).

Best regards,
sunken rock

[quote="jbmorgan"]Let $ w_1$ and $ w_2$ be two externally tangent circles that are tangent at $ M$ and both are internally tangent to a circle $ w$. Let a common tangent (not passing through $ M$) of $ w_1,w_2$ cut $ w$ at $ B,C$, and let $ P$ be the midpoint of arc $ BC$ opposite of $ M$. Let $ PM$ cut $ w$ again at $ A$. Prove that $ M$ is the incenter of triangle $ ABC$.[/quote]

[hide="Hint."]1. Prove that $ PM$ is a common tangent of $ \omega_1$ and $ \omega_2$.
2. Let $ E$, $ F$ the tangency points of $ \omega_1$ with $ BC$ and $ \omega$, respectively.
Prove that $ PB^2\equal{}PE\cdot PF\equal{}PM^2$, i.e., $ PB\equal{}PC\equal{}PM$.
Hence, we conclude that $ M$ is the incenter of triangle $ ABC$.

You can find some more problems of this type in the article (collection): Son Hong Ta, "A metric relations and its applications" available at http://reflections.awesomemath.org/2008_2.html
[/hide]


For given initial conditions $a,b,c\in \mathbb{R^{+}}$ and a suitable function $f: \mathbb{R}^{2}\to\mathbb{R}^{2}$ such that for $x>y$, $f(x,y)=(z,z)$ and $x<z<y$, we want to find the limiting effect of $f$ on $a,b,c$. Perform a sequence of 'corrections', in which two variables (perhaps $(a_{i},b_{j})$) are replaced with $f(a_{i},b_{j})=(z,z)=(a_{i+1},b_{j+1})$. Clearly, no matter which two variables are chosen at each step, $a,b,c$ will approach some limit $L$. In general, I am not even sure that this limit is well defined. However, the case of two of the three variables being equal is clearly well defined and seems like it should be simple.

For instance, when $f: =(\frac{x+y}{2},\frac{x+y}{2})$ we know that for $n$ initial numbers $a_{i}$ the sum of all the $n$ elements is an invariant and it follows that the common limit is the arithmetic mean of all the numbers.

However, I am considering the function $f: =(\frac{2x+y}{3},\frac{2x+y}{3})$ and can't make any progress on the limit, even in the case of initial conditions $(a,a,b)$. Any ideas?

For instance, consider the triplet $(1,1,0)$. Prove that the limit $L=\frac{6}{7}$

Interesting... well yeah as you mentioned for one thing I'm not sure that the limit is always the same. Also, there need not be a limit (it could go on forever without approaching anything). However, we can make progress on your problem.

We basically have with a > b, (a, a, b) --> ((2a+b)/3, (2a+b)/3, a) --> ((4a+5b)/9, (4a+5b)/9, (2a+b)/3). Now it should be obvious, as

(2a+b)/3 - b = 2/3(a-b)
a - (4a+5b)/9 = 5/9(a-b)

so actually I'm getting the result 6/11. Maybe I miscalculated, but anyways you get the idea.

Sorry could you outline your steps? Are you just getting an infinite sequence with terms of powers of three, it seems like that's what's happening...

I don't quite understand how you're applying a function from $\mathbb{R}^{2}\to \mathbb{R}^{2}$ onto three initial inputs...

Anyway, if you can show the limit exists, isn't the limit just whatever triplet is invariant under the transformation?

You apply it to two variables at a time :wink:
Okay so for each of the invariant triplets (which are all of the form $a,a,a$) we know the answer - but about the non-degenerate case of different variables?

The question can be rephrased with somewhat less technology.  Note that the case where $a \neq b \neq c \neq a$ reduces to the case where two terms are equal after one application of $f$, so we might as well just solve the problem where we start with two equal numbers.  In that case, we really only need to act as if we have two numbers, not three.  Furthermore, there is no reason to have the function $f$ going into ${\bf R}^{2}$; $\bf R$ works fine.  With these changes we get the following:

Choose a function $f$ with the property that $x \leq y$ implies $x \leq f(x, y) \leq y$.  Begin with two real numbers $a, b$.  At each step, replace one of the two current numbers with one of the numbers $f(a, b)$ or $f(b, a)$.  Continue indefinitely.  (This actually allow more freedom than the original setup -- can you see why?)  Here are some questions you can ask about this:
1) Suppose that I choose a particular, systematic method for deciding which of my two numbers to toss and which order to place them as the arguments of $f$.  (Or, even nicer would be if I just choose an $f$ symmetric in its arguments like a power mean, but it doesn't really matter.)  Then, what will the limiting value of this process be?
2) For a fixed $f$ and fixed initial values $a, b$, given that I give you free choice about either the order you take $a, b$ as arguments of $f$ or which value you replace or both, what are all possible limiting values?

The most obvious "fixed" strategies in part 1 are to say $a = a_{0}, b = a_{1}$ and $a_{n+1}= f(a_{n}, a_{n-1})$ for $n \geq 2$.  Other possibilities are two have $f$ act on the $a_{i}$ in the opposite order (which is equivalent just to taking the function $g(x, y) = f(y, x)$ instead), or to have the order alternate back and forth, or whatever else you like.  (The most general approach here would be to fix an infinite sequence $\alpha$ of bits and let $a_{n+1}= f(a_{n}, a_{n-1})$ if the $n$th bit of $\alpha$ is a 0, and $a_{n+1}= f(a_{n-1}, a_{n})$ if the $n$th bit of $\alpha$ is a 1.  Then you could ask for the answer in terms of $f, \alpha$.)

Also note that since applying the function to two equal values gives you the same value, every end position (with $a = b$) is a stable state.  T0r's suggested analysis doesn't give us anything here.


OMG!!!!!
I haven't been here in a long while.
Listen guys.
AFter AMC,
anyone want to discuss the problems here?

[quote="Syntax Error"]OMG!!! sure[/quote]

ROFL.

Yeah, we'll discuss them here, but not until after a certain time, discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10087]here[/url].

I thought about 12 questions from the amc 10 and 12 overlap.  Does those questions only overlap for each testdate?

yea
obviously
questions on the amc 10 A won't appear on the amc 12 B
 :wink:

wait, does this also apply for AMC 12? so im taking AMC 12 A, so i just memorize the questions for last year's AMC 12 A??

Seriously??? No kidding? That's [b]whacked [/b]man! :lol:

oy.  there is often an overlap between questions on the AMC 10 and AMC 12 that are given at the same time.  That overlap is roughly 10-13 questions.  Questions from last years test will not be on this years test; questions from this year's A exam will not be on this year's B exam.

Oy indeed.  Each question is only used on one test date.

OK guys.
AMC-A is tomorrow.

Good Luck to all. GOod luck to myself, hahah. Hope careless mistakes won't kill any of us...(like they did to me last year)

lol i'm taking none of the amc's at churchilljrhigh (the school i go to), amc12 at skylinesrhigh and amc10 at wasatchjrhigh.


What value of $ n$ makes the following sentence true?
\newline
$ \frac{n}{n} \minus{} \frac{n}{12} \equal{} \frac{n}{24}$

$ \frac{n}{n}=1$, so we have
\begin{align*}
1-\frac{n}{12}=\frac{n}{24} &\Rightarrow 1=\frac{n}{12}+\frac{n}{24} \\
&\Rightarrow 1=\frac{2n+n}{24} \\
&\Rightarrow 1=\frac{n}{8} \\
&\Rightarrow \boxed{8}=n \\
\end{align*}


geometry
circumcircle
symmetry
geometry proposed

The vertices of a right triangle $ABC$ inscribed in a circle divide the circumference into three arcs.  The right angle is at $A$, so that the opposite arc $BC$ is a semicircle while arc $AB$ and arc $AC$ are supplementary.  To each of the three arcs, we draw a tangent such that its point of tangency is the midpoint of that portion of the tangent intercepted by the extended lines of $AB$ and $AC$.  More precisely, the point $D$ on arc $BC$ is the midpoint of the segment joining the points $D'$ and $D''$ where the tangent at $D$ intersects the extended lines $AB$ and $AC$.  Similarily for $E$ on arc $AC$ and $F$ on arc $AB$.  Prove that triangle $DEF$ is equilateral.

[quote="silouan"]Let a right angled triangle $ABC$ (right at A) and its circumcircle. Let a tangent at the <<small>> arc $AB$ which intersect AC at M and BC at N and let K the point of tangency such that K is the midpoint of MN. Simiarly we define L, S at the arcs $BC$ and $CA$ . Prove that the triangle $KLS$ is equilateral .[/quote]

This cannot possibly be true. For one thing, there is no smaller arc BC, so that the point L cannot be defined similarly as the point K. Even supposing that L is on the arc BC opposite to the vertex A (otherwise $\triangle KLS$ could not be equilateral), the problem is incorrect.

Consider an isosceles right triangle $\triangle ABC$ with $\angle A = 90^\circ$ and circumcenter $O$, the midpoint of the hypotenuse $BC.$ By symmetry, $OL \perp BC.$ Assuming $\triangle KLS$ is equilateral, $\angle LOK = 120^\circ,\ \angle NOK = 30^\circ.$ Since $OK \perp MN,$ $\angle ONK = 60^\circ.$ Let the tangent to the circumcircle $(O)$ at $S$ meet $AB, BC$ at $M', N',$ respectively and let the tangents $MM, M'N'$ meet at $X.$ Similarly as above, $\angle ON'S = 60^\circ,$ so that $\triangle NN'X$ is equilateral. Hence $NK = N'S = \frac{NN'}{4}= \frac{NX}{4}= \frac{N'X}{4}.$ Let $D, D'$ be the midpoints of $NX, N'X.$ Then $K, S$ are the midpoints of $ND, N'D'$ and $DD' \parallel BC$ is a tangent to $(O)$ at $A.$ But $AB, AC$ clearly intersect the tangents $NX, N'X$ at $M, M'$ ouside the segments $ND, N'D',$ so that $K, S$ are not midpoints of $MN, M'N'$ for the equilateral $\triangle KLS.$

Clearly, $L$ must be on the arc $BC$ opposite to $A$ and perhaps the hypotenuse line $BC$ is to be ignored in defining the tangent segment $MN$ with the tangency midpoint $K,$ i.e., $M \equiv t_{K}\cap AC, N \equiv t_{K}\cap AB,\ MK = NK.$ Similarly, the endpoints of the tangent segments $t_{L}, t_{S}$ with the tangency midpoints $L, S$ all lie on $AC, AB,$ never on $BC.$ Then the problem would be correct. Such problem has been posted and solved previously in the unsolved geometry problems, which is a contradiction.

Yetti

Could you send the figure you worked on ?? :|
I think I write something non-understandable at the problem in the first post since my English are not so good..... :blush:
Please send the figure you worked on

The first figure is the counter-example that I worked on. The second figure is the suggested correct problem.

Indeed, the old geometry unsolved topic is http://www.mathlinks.ro/Forum/viewtopic.php?t=25900 . See also http://www.mathlinks.ro/Forum/viewtopic.php?t=81435 (this was problem 5 of the Canada MO 2006).

  darij

My apologies for the typos...... :blush:  .
The original problem is just the one on Canada MO
Thanks for the correction.......


Να βρεθουν οι πραγματικοι αριθμοι $ x, y, z$ που ειναι λυσεις του συστηματος:

$ \begin{cases} 12x^2 \minus{} 18y^2 \plus{} 2z \equal{} 16x \minus{} 48 \\
54y^2 \minus{} x^2 \minus{} 24y \equal{} 6xy \plus{} 16 \plus{} z \end{cases}$

Προσπαθησα να το κανω δυσκολο, ελπιζω να τα καταφερα και να μη βγαινει με κανεναν απλο τροπο :P

edit: το αλλαξα λιγο, νομιζω ειναι καλυτερα τωρα

Η λυση σε hide: [hide]$ (x, y, z) \equal{} (1, \frac {1}{3}, \minus{} 21)$[/hide]

[quote="Protonios"]Βγαίνει σε 1,5 λεπτό [/quote]

M areseis...

[quote="Protonios"]Βγαίνει σε 1,5 λεπτό (ίσως πρέπει να το βελτιώσεις λίγο γιατί όντως βγαίνει με εύκολο τρόπο). :)

Απλά απλοποιείς την πρώτη σχέση, προσθέτεις κατά μέλη και SOS. :)[/quote]

θα το καθομουν λιγο να το κανω λιγο δυσκολοτερο αλλα τωρα εχω πολυ διαβασμα, απο Τριτη παννεληνιες και αν και ειμε 100% ετοιμος να δωσω μαθηματικα φυσικη κλπ, δεν εχω τελειοσει τις Αρχες Οργανωσης και εχω μεχρι την αλλη τριτη να μαθω αλλες 10 σελιδες και να κανω και μια γερη επαναληψη σε ολα... πραγματικα μου εχει βγαλει τη πιστη αυτο το μαθημα

SOS=sum of squares...είναι μέθοδος που χρησιμοποιείται κυρίως στις ανισότητες αλλά και σε ισότητες όταν δείχνεις ότι κάποιο άθροισμα τετραγώνων ισούται με 0 κι άρα τα τετράγωνα ισούνται με 0 :wink:

Παντως εδω δεν βγαινει με SOS...
Protonie εισαι σιγουρος για την λυση σου?

Σωστος ειναι ο προτονιος αλλα αυτο που κανεις στο τελος για να φτασεις στη λυση δεν ειναι ακριβος SOS (δεν το εχω κοιταξει κι ολας, ισος να βγαινει και με SOS), πιο συγκεκριμενα αυτο που κανεις στο τελος για να βγαλεις τη λυση και με βαση το οποιο εφτιαξα την ασκηση αυτη ειναι:

[hide]Σχηματιζεις ενα τελειο τετραγωνο 3 προσθετεων και μεταφερεις οτι απομενει στο αλλο μελος, μετα παιρνεις B.C.S και περιπτωση ισοτητας και ετσι βρησκεις το x και το y[/hide]

Με αυτό που λέω η τελική σχέση είναι ισοδύναμη με αυτό:

$ (x \minus{} 3y)^2 \plus{} 4[(x \minus{} 1)^2 \plus{} (3y \minus{} 1)^2] \equal{} 0$

οπότε βγαίνει μόνο με SOS και δε χρειάζεται Cauchy Scwarz. :wink:

Α! και λέγομαι Κώστας. :lol:


Define the sequence by $ a_1 \equal{} 1, a_2 \equal{} b, a_{n\plus{}2} \equal{}2a_{n\plus{}1} \minus{} a_n \plus{} 2$, where $ b$ is a positive integer. Show that $ a_na_{n\plus{}1} \equal{} a_m$ for some $ m$.

Find second differences.
See also [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=181105[/url]


So im reading engel and the vector section of geometry (Ive realized lately that im weakest in geometry...) and he presents this question of Napoleonic Triangles.
Aside 1: Napoleonic Triangles are formed with taking a triangle, then adding a regular triangle onto each side. Then take the centroid of each of these new regular triangles and connect them. Label the vertices of the original triangle a, b, c, the vertice of the reg triangle built onto a side opposite a = a_0 and so forth for the rest of the letters, and the centroid opposite a = a_1, and so forth again.
The goal was to prove that these new centroids are vertices of a reg triangle.
Aside 2: x = e^i(60  :^o: ).
First he says that b_0 = a + (c-a)x, c_0 = b + (a-b)x, a_0 = c + (b - c)x.
So far, so good.
But then he goes on to say that
3(a_1 - c_1) = c_0 - b_0 + c - a      HMM? How did this happen?
                     = 2c - a - b + (2b - a - c)x ... This expression here I thought would be equal to (c-a) + (c-b) + (b-c) + (b-a) = 2(c-b) + (c-a) + (b-a). Anyone see how he got this? Btw, he then continues listing equations of this form (but 3(b_1 - c_1) and then 3(a_1 - c_1)x, which he then shows = 3(b_1 - c_1).)
(10 points if you understand the topic name)

[quote="jelyman"]So im reading engel and the vector section of geometry (Ive realized lately that im weakest in geometry...) and he presents this question of Napoleonic Triangles.
Aside 1: Napoleonic Triangles are formed with taking a triangle, then adding a regular triangle onto each side. Then take the centroid of each of these new regular triangles and connect them. Label the vertices of the original triangle a, b, c, the vertice of the reg triangle built onto a side opposite a = a_0 and so forth for the rest of the letters, and the centroid opposite a = a_1, and so forth again.
The goal was to prove that these new centroids are vertices of a reg triangle.
Aside 2: x = e^i(60  :^o: ).
First he says that b_0 = a + (c-a)x, c_0 = b + (a-b)x, a_0 = c + (b - c)x.
So far, so good.
But then he goes on to say that
3(a_1 - c_1) = c_0 - b_0 + c - a      HMM? How did this happen?[/quote]

Don't wonder. Just one of the tons of typos in Engel's book. Read

  3(a_1 - c_1) = a_0 - c_0 + c - a = 2c - a - b + (2b - a - c)x;
  3(b_1 - c_1) = b_0 - c_0 + c - b = a + c - 2b + (b + c - 2a)x;
  ....

  Darij

as darij said, don't wonder, it's engel. he's a really good mathematician, but he's got even worse with typos...he's now about 75 and still does imo training. it is real fun with him because every second letter or number of his formula is a typo :D

Peter


function
Putnam
modular arithmetic
algebra
floor function
number theory unsolved
number theory

There was a problem posted in the Intermediate Topic.  Here is a more chalanging problem:
If $p$ is prime number,  and $k = \left[ \frac{2p}{3}\right]$  (where   [x]     is the greatest integer function)
Prove that $p^2$ divides:
$\binom p 1 + \binom p 2 + \binom p 3 + \dots + \binom p k$

It's problem A5 of the 1996 Putnam
[url=http://www.kalva.demon.co.uk/putnam/putn96.html]www.kalva.demon.co.uk/putnam/putn96.html[/url]

I'm moving this to the Olympiad forums...

Dividing by $p$, we want to show that the sum $\frac{(p-1)!}{x!(p-x)!}$, with $x$ running from $1$ to $k$, is $0 \pmod{p}$. From here on, we work $\pmod{p}$. Obviously we can throw the common factor $(p-1)!$ out.

Now note that $p-x=(-1)x, p-x-1=(-1)(x+1), \ldots, 1=(-1)(p-1)$. Then $x!(p-x)!=(-1)^{p-x}(p-1)!x=(-1)^{p-x+1}x$.

Let $i(a)$ be such that $i(a)a=1$. Then our original sum is $(-1)^x \times i(x)$. Now, the important observation is $i(2k)=i(2)i(k)$. Suppose that $p=6m+1$, so that $k=4m$. Then the sum runs from $x=1$ to $x=4m$. Consider the sum $-(i(1)+\ldots+i(4m))+2i(2)(i(1)+\ldots+i(2m))$. If we multiply the $i(2)$ with each term in the second bracket, we see that this is the original sum. But since $2i(2)=1$, it is also equal to $-i(2k+1)-\ldots-i(4m)$. Now we can look at this sum as a sum of pairs of the form $i(a)+i(b)$ with $a+b=p$, and each such pair adds up to $0$, since $i(a)+i(b)=bi(a)i(b)+ai(b)i(a)=(a+b)i(a)i(b)$. Similarly, if $p=6m+5$, then $k=4m+3$, and doing the same trick we get $-(i(1)+\ldots+i(4k+3)+2i(2)(i(1)+\ldots+(2k+1)$, which is $-i(2k+2)-\ldots-i(4k+3)$, where the same trick applies.


inequalities
function
inequalities proposed

Let $ n>2$. Find the largest constant k such that if $ x_1,x_2,...,x_n $ are positive real numbers whose sum of squares is 1, then $ (1-x_1)(1-x_2)...(1-x_n)\geq kx_1\cdot x_2\cdot...\cdot x_n $ . This problem is a generalization created by Vasc and harazi of a problem discussed on this forum: if $ a^2+b^2+c^2+d^2=1 $ then $ (1-a)(1-b)(1-c)(1-d)\geq abcd$.

using jensen for the function $ln(\frac{1}{\sqrt{x}}-1)$, we can easily force n-1 of the variables to be equal(the function is convex for $x\leq \frac{4}{9}$(i used mathematica to derive, so it is unlikely that there's a mistake) and if two of the variables are greater than 4/9, the others are all smaller than 1/9(i substituted the squares as my variables) which is easy to exclude. then we get an inequality in one variable which shouldn't be very hard to handle:
${(\frac{1}{\sqrt{x}}-1)(\frac{\sqrt{n-1}}{\sqrt{1-x}}-1})^{n-1}\geq (\sqrt{n}-1)^n=k$

Peter

What is matematica? I saw some references of people about it on ML but I don't know what is it.
where can I get it from?
What other programs do you use?

I never see Matematica, but I use MathCad (about 8 years). Another excellent program is Geometr SketchPad (unfortunatelly, I have old version :(  ).
All these programms are not freeware.

Dear Peter, I wonder if you could please post a complete solution, since your idea is completely different from the very hard solution I gave to the problem. I have some doubts that this idea can practically solve the problem and I would like to see a solution. Also, I really don't understand why you say the case when 2 of the numbers are greater than $\frac{4}{9} $. Maybe I'm too tired now and I don't see obvious things. Thanks

I still miss your idea. Please, explain it to me.

Any idea? Believe me, this is really fantastic.


What is the degree measure of the angle between the hour hand and the minute hand of a clock at exactly 2:30 p.m. on a 12-hour analog clock?

At $ 2$:$ 30$, the hour hand is half way between $ 2$ and $ 3$.  From $ 12$:$ 00$, this forms a $ 75^o$ angle.  The minute hand forms a $ 180^o$ angle with $ 12$:$ 00$.  Thus, the answer is $ 180 \minus{} 75$, or $ \boxed{105^\circ}$.

[color=blue]Fixed degrees symbol. :) -Izzy[/color]

Or you could use the clock formula.

$|5.5m-30h|$, where $m$ is minutes and $h$ is hours.

This works because the minute hand gains $5.5^\circ$ per minute, and every hour is $30^\circ$

$165^\circ-60^\circ = \boxed{105^\circ}$


stiti cumva de unde as putea lua baremele de la onm 2005-cls a 9-a?

Andyciup, acelea sunt baremele pentru anul 2006.Anca a cerut pentru O.N.M.2005.In acest caz, poti vizita http://www.olimpiade.ro, sectiunea 2005. :D


Find all polynomials $W$, $x,y\in R$, that:
$W(0)=3$
$W(x+y)=W(x)+W(y)+2y-3$

Are you sure that's the question? Because for all $x,y\in\mathbb{R}$, we would get that $2x = W(x+y)+3-W(x)-W(y) = 2y$, i.e. $x=y$...


Show $2^b-1|2^{ab}-1$ for positive integers a,b

huh? because 2 ab - 1=(2 b - 1) * (2 (a-1)b + 2 (a-2)b + ... + 2 b + 1), so it's obviously true? besides, generalizations could be easily made...sure this is what you mean?  :?


Group II, Problem 1.3 (adapted)

"A sample of hydrogen, $ H_{2}(g)$, contained in a closed vessel of variable capacity, at 10.0ºC and at a pressure of 1.0 atm, occupies a volume of 100 L. At a pressure of 1.0 atm but at a temperature of 25ºC the volume of the same sample must be 250 L, because volume and temperature are directly proportional and as the temperature has raised by a factor of 2.5, the same must happen with the volume."

Without doing any calculations, explain if this statement is true or false.

False, because the volume increases not proportionally to the Celsius temperature, but to the Kelvin temperature. It increases by $ \frac{298}{283}$, not 2.5.


On the sides AB and AC we take points M and N respectively such as: BM = MN = NC.

MX and NY are bisectors in triangle AMN. Prove that XY || BC

[b]First proof:[/b] (using Thales's theorem)
$ \frac{AX}{AM}\equal{}\frac{XN}{MN}\equal{}\frac{AX\plus{}XN}{AM\plus{}MN}\equal{}\frac{AN}{AB}\Rightarrow AX\cdot AB\equal{}AM\cdot AN$
Analogously, $ AY\cdot AC\equal{}AM\cdot AN$. Hence result.
[b]Second one:[/b]
Denote O,I be the circumcenter of triangle AMN.
---OI is perp to BC
---OI is perp to XY

[quote="dgreenb801"]I got the first proof, can you explain the second? O and I are two points, and you only listed one, the circumcenter of AMN.[/quote]
These in the second are two well know results (?) and it's a open proof (can help contact with other problems)


Let a,b,c be positive numbers such that:$ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$.find the minimum the value of the expression:
P=$ a \plus{} b \plus{} c \plus{} 12abc$

hello, i think we have $ a\plus{}b\plus{}c\plus{}12abc\geq4$ for all positive reals and $ ab\plus{}bc\plus{}ca\plus{}abc\equal{}4$.
Sonnhard.

[quote="nsato"]Given the condition $ ab \plus{} ac \plus{} bc \plus{} abc \equal{} 4$, try the substitution $ a \equal{} 2x/(y \plus{} z)$, $ b \equal{} 2y/(x \plus{} z)$, $ c \equal{} 2z/(x \plus{} y)$.[/quote]
yes,it is true :)


geometry
3D geometry
sphere
pyramid
geometric transformation
reflection
articles

[u]2003: Mathematicians Face Uncertainty[/u] from [i]Discover Magazine[/i], Jan. 04

This will surely be remembered as the year mathematicians finally had to agree that their prized notion of absolute proof is an unattainable ideal  an excellent goal to strive for, but achievable only in relatively simple cases. Moreover, they were forced to make this adjustment under the harsh glare of the media, following three major news stories about so-called mathematical proofs.

Early in the year, American mathematician Daniel Goldston and his Turkish colleague Cem Yildirim announced a proof of the twin prime conjecture, which says there are an infinite number of prime numbers differing by two, such as 3 and 5, or 11 and 13. Although experts around the world initially agreed that the new proof was correct, a few weeks later an insurmountable error was discovered.

In late 2002 the Russian mathematician Grigori Perelman posted on the Internet what he claimed was an outline for a proof of the Poincare conjecture, a famous century-old problem of the branch of mathematics known as topology. If Perelman is correct, he will pocket a $1 million prize offered for a solution by the Clay Mathematics Institute. But after months of examining the argument, mathematicians are still unsure whether it is right or not.

Never mind a delay of weeks or months  pity poor Thomas Hales, an American mathematician who has been waiting for [i]five years[/i] to hear whether the mathematical community has accepted his 1998 proof of the astronomer Johannes Keplers 390-year-old conjecture that the most efficient way to pack equal-size spheres (such as cannonballs on a ship, which is how the question arose) is to stack them in the familiar pyramid fashion that greengrocers use to stack oranges on a counter. After examining Haless argument for five years, in the spring of 2003 a review panel of world experts appointed by the prestigious journal [i]Annals of Mathematics[/i] finally declared that, whereas they had not found any irreparable error in the proof, they were still not sure that it was correct. The journal agreed to publish Haless proof, but only with a disclaimer saying they were not sure that it was right.

What all three episodes reflect is the complexity and abstraction of many modern proofs. Even the experts find it almost impossible to be sure if some arguments are correct.

So where does all this leave the field of mathematics? Ever since the idea of proof was introduced by the ancient Greeks around 600 B.C., it has played a major role in the subject. In his mammoth work [i]Elements[/i], written around 300 B.C., Euclid began by writing down axioms  basic assumptions whose truth was assumed to be self-evident  and using them, with logically sound arguments, to deduce the theorems of geometry. As mathematicians from the 19th century onward ventured into ever greater heights of abstraction, the axiom-proof approach became an indispensable tool for handling concepts that were frequently counterintuitive.

So what exactly is a proof? There are two very different answers. One answer, which I will call the right-wing (the right or wrong or rule of law) definition, is that a proof is a logically correct argument that establishes the truth of a given statement. The other answer, the left-wing definition (fuzzy, democratic, and human centered), is that a proof is an argument that convinces a typical mathematician.

The right-wing concept of a proof is valid in an idealistic sense. The problem is that, except for trivial cases, it is not clear that anyone has ever seen such a thing. The examples familiar to most people from their high school math classes are the geometric arguments Euclid presents in [i]Elements[/i]. But as the German mathematician David Hilbert pointed out in the late 19th century, many of those arguments are logically incorrect. Euclid made repeated use of axioms that he had not stated, without which his arguments are not logically valid.

It took some effort  following a delay of more than 2,000 years  for Hilbert to present correct proofs. Today most mathematicians regard his arguments as valid right-wing proofs, myself included. But if you push me to say how I know this, I will end up mumbling that his arguments convince me and have convinced all the other mathematicians I know. But thats the left-wing definition of proof, not the right-wing one. After all, like everyone else I thought Euclids proofs were correct until I learned otherwise.

And theres the problem. The right-wing definition of mathematical proof is an unrealistic ideal, unattainable in the real world. Actual mathematical proofs are all left wing. An argument becomes a proof when the mathematical community agrees it is such. But at what point does that happen? The three cases I started with highlight this problem. All three arguments are far too long and complicated for anyone to seriously believe these are anything more than left-wing proofs. But are they even that? How can we ever be sure?

Mathematicians have always said that proofs are to mathematics what experiments are to scientists: the way of telling right from wrong. After the events of 2003, it is clear that the comparison is closer than most mathematicians would care to admit. Just as all scientists must live with the ever-present possibility of experimental error, so too mathematicians, like our judicial system, may sometimes have to settle not for absolute proof but rather proof beyond a reasonable doubt.

							-- Keith Devlin

I wasn't really impressed with this article at all (read it in the magazine when it came out) -- first because I was annoyed that it was the only bit about mathematics out of the top 100, and secondly because I don't think it's really that important.  I mean, mathematics is complicated.  And I don't see why that makes so much difference.  I mean, in terms of our practical ability to prove things, maybe, but we already know that that is limited by undecidability.  So what's the big deal?

I thought it was an interesting take on the story. This is a fundamentally different issue than decidability. These questions are formally decidable in the logical sense. So this is not about the inherent limitations of mathematics itlself, it's about the limitations of translating math into human language and the inherent limitations of man to understand it.

I suppose the story of the discovery of the 40th Mersenne Prime could have been included in the list, but I can't think of many other newsworthy events in mathematics from the last year. Groundbreaking news in math tends to come slowly compared to the other sciences. This managed to synthesize sort of three stories into one though.

I felt like it was more a question of, I don't know, sociology or psychology than mathematics.

I thought the Mersenne prime deserved to be in, and I also thought they pushed Perelman's proof to the side and replaced it with this non-mathematical question of our ability to decide if proofs are good.  Maybe if they'd labelled it meta-mathematics, I would have been happy.

I think that was the whole point - that mathematics [i]is[/i] sociological. That mathematics isn't as universal as we might like to believe; that it is in fact an outgrowth of humanity. And that the process of deciding if proofs are good or not is becoming more and more relevent in mathematics.

Dr. Keith Devlin is Executive Director of CSLI and a Consulting Professor in the Stanford University Department of Mathematics.


function
logarithms
limit
algebra solved
algebra

Solve the equation:

3^x+4^(1/x)=7.

ciao!

This is nice!

Hint: if b>0 and a>1 study then the monotony of the function f: (0,+oo)-->R, if f(x)=a^x+a^(b/x).

cheers!! :D

And how about the other real solution, at around $1.26186$ ? It is not its value that's relevant, but the fact that no analysis has been done whatsoever on the equation, and just a wishful guess that the obvious solution is the only one ...

Did you find it by using logarithms (therefore cancelling out exponents)?

You cannot take logarithms, since it's a sum of powers ... Denote $f(x) = 3^x + 4^{1/x} - 7$, defined for ease of analysis on just $(0,\infty)$. Then $\lim_{x\to 0} f(x) = \lim_{x\to \infty} f(x) = \infty$, $f(1) = 0$, and $f'(x) = 3^x\ln 3 -\dfrac {1} {x^2} 4^{1/x} \ln 4$. Equating $f'(x) = 0$ yields $3^x \ln 3 = \dfrac {1} {x^2} 4^{1/x} \ln 4$, with LHS increasing and RHS decreasing, hence having exactly one positive real solution $\rho > 1$. Therefore $f$ is decreasing on $(0,\rho]$, with one zero at $1 < \rho$, and increasing on $[\rho,\infty)$, with one zero at some point $r > \rho$.


[quote="shfdfzhjj"]No one?
please help me![/quote]
Any idea? :maybe:

[quote="shfdfzhjj"]please help me![/quote]
Can we see a proof now? :maybe:


Compute $ \tan^2 1^\circ \plus{} \tan^2 3^\circ \plus{} \tan^2 5^\circ \plus{} ... \plus{} \tan^2 87^\circ \plus{} \tan^2 89^\circ$.

hello, after an ugly computation i have found that your sum $ \sum_{i\equal{}1^{\circ}}^{89^{\circ}}\tan(i)^2\equal{}\frac{15931}{3}$.
Sonnhard.


Prove that there are many infinitely positive integers $ n$ such that the sum of its digits divides $ n$.

Either I cant' translate on myself the problem, or there is a missed condition, because
$ n\equal{}100000000000000000000000000000...00$ has a sum of its digits 1.

Consider the elements of the set $ \{1,10, 10^{2}, \ldots\}$. Done. :D


[quote="Victory.US"]$ (1,1,1,1,2)$ is one of the roots[/quote]

Your solution shows too many variable positions.
You gave an answer *as if* the equation were
$ x^x \plus{} y^y \plus{} z^z \plus{} w^w \equal{} t^t$.

[hide]
There are no solutions. Since $ t$ is the greatest number, this can be shown by maximizing $ x^x,y^y,z^z$ and proving that it is less than $ t^t$, i.e. proving that $ (n \plus{} 1)^{n \plus{} 1} > 3n^n$.
$ n \equal{} 1: 2^2 \equal{} 4 > 3 \equal{} 3\cdot 1^1$.
For $ n > 1$,
$ (n \plus{} 1)^{n \plus{} 1} \equal{} (n \plus{} 1)\cdot (n \plus{} 1)^n > 3\cdot n^n$.
[/hide]


trigonometry
number theory proposed
number theory

Suppose that $a,b,c$ are real numbers in $(0,\frac{\pi}{2})$ such that $a+b+c=\frac{\pi}{4}$ and $\tan{a}=\frac{1}{x},\tan{b}=\frac{1}{y},\tan{c}=\frac{1}{z}$ , where $x,y,z$ are positive integer numbers. Find $x,y,z$.


Solve this!

Carlos is making 5-digit numbers with the digits 2, 3, 4, 5, and 7. Only the digit 2 can be used more than once. What's the probability that he'll make an even number?

[hide]The first four numbers don't matter, only the last one does. Out of the $ 5$ possible digits, $ 2$ are even. Therefore, the probability is $ \tfrac{2}{5}$.[/hide]

Not necessarily so.
[quote] The first four numbers don't matter, only the last one does. Out of the $ 5$ possible digits, $ 2$ are even. Therefore, the probability is $ \tfrac{2}{5}$.[/quote]

If there were 3 numbers, 1,2,3, (2 is allowed to be used more than once) then to calculate all the possibilities, assume 2 wasn't allowed to repeat. Then there would be $ 3\times2\times1$ possibilities. But we have to add:
221
222
223
212
232
122
322

There are 13 combinations. Only numbers with 2 at the end are even so the possibility is $ \frac{7}{13}$

If we used I_like_pie's method, there would be $ \frac{1}{3}$ not $ \frac{7}{13}$.

Although this is an example, it can help in finding the answer where the digits are 2,3,4,5,7.


geometry
geometric transformation
rotation
real analysis
real analysis unsolved

The question is

Given a curve C whose equation is give by y=(12/(x + 3))- 4,
find the region bounded by C, the line x=0, the line x=3 and the line y=8 is rotated through 360 degree about the y-axis. Calculate the exact value of the volume generated.

My solution is here
http://www.geocities.com/myjunkmail31/Volume.jpg

However, this solution is not the answer in the textbook. The answer in the textbook is
[b]36pi(1+2ln2).[/b]
Can someone show me where am I doing wrong?

Thanks, the error is

v1 should be v1=pi(3^2)(2) - pi(54 - 72ln2)

The volume I calculated is that of the solid generated by rotating the region between the given curve and the y-axis about the y-axis, but I wanted to rotate the region between the given curve and x=3 about the y-axis. So just subtract what I got for V1 from the volume of the cylinder of radius 3 with height 2.


combinatorics
Sperner
Partial Orders
IMO Shortlist

Forty-nine students solve a set of 3 problems. The score for each problem is a whole number of points from 0 to 7. Prove that there exist two students $ A$ and $ B$ such that, for each problem, $ A$ will score at least as many points as $ B.$

Look here for a very similar problem:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2439466&sid=cacae6af292a61637bc1f88b8bfe0bbe#p2439466[/url]

hmm,I cannot see any relevance between the link above and pro itself.
      By the way,solution:
      Represent everyone's points with littices such that $(x,y)$ represents one's points in the 1st and the 2nd problem.Partition the $7*7$ points set into several groups.
      $A_1=\{ (x,y)|0\le x\le 7,y=0,or,x=7,1\le y\le 7 \},A_2=\{ (x,y)|0\le x\le 6,y=2,or,x=6,2\le y\le 7 \},A_3=\{ (x,y)|0\le x\le 5,y=2,or,x=5,3\le y\le 7 \},A_4=\{ (x,y)|0\le x\le 4,y=3,or,x=4,4\le y\le 7 \},A_5=\{ (x,y)|x=2,3,4\le y\le 7 \},A_6=\{ (x,y)|x=0,1,4\le y\le 7 \}$.
      By pigeonhole,there must be $\lfloor \frac {49-1}{5} \rfloor+1=9$ points in some group.If $9$ points in $A_5$ or $A_6$,there are $2$ points coincide,take the higher points on 3rd pro be $A$ is ok.If $A_i,i=1,2,3,4$ has $9$ points,observe there is one students whose $1$ and $2$ pro is better than the other one,also the 3rd pro has only $8$ possible points,so by pigeonhole,there are $2$ has same points on 3rd pro.Done.

[quote=mathmaths]hmm,I cannot see any relevance between the link above and pro itself.[/quote]
Change the bounds to $0\leq a,b,c\leq 7$ and let $a$ be the score on the first problem, $b$ the score on the second, etc...

Phrasing mathematically, we have $49$ triples $(x, y, z)$ of integers from $0$ to $7$. Thus, it suffices to show that there exist two triples $(x, y, z)$ and $(x', y', z')$ such that $x \ge x'$, $y \ge y'$, and $z \ge z'$. To make this a divisors condition, we can reformulate the condition as \[2^{x'} \cdot 3^{y'} \cdot 5^{z'} | 2^x \cdot 3^y \cdot 5^z.\]

    Suppose for the sake of contradiction that among any subset of the positive divisors of $N = 2^7 \cdot 3^7 \cdot 5^7$ with $49$ elements, there exist no two divisors such that $d_1 | d_2$. This is equivalent to there existing an antichain of length $49$ of $N$. However, by the Maximal Antichain Lemma, the maximal size of an antichain is the number of solutions to the equation $x + y + z = \left\lfloor\frac{7 + 7 + 7}{2}\right\rfloor = 10$ with $0 \le x,y,z \le 7$. By Stars and Bars with PIE or pure bash, we get that the number of solutions is $\binom{12}{2} - 3\binom{4}{2} = 66 - 18 = 48$. This is a contradiction, so there must always exist two divisors such that $d_1 | d_2$.

[quote=aryabhata000]Would you be willing to give a citation for this lemma?[/quote]

It’s essentially a generalization of Sperner’s Theorem, where we consider antichains of positive divisors instead of sets.

Rough sketch: First, prove the BTK Theorem, which says that the set of positive divisors of $n$ can be partitioned into symmetric chains. A simple induction on the number of distinct prime factors of $n$ will suffice. Then, construct a bijection between the number of symmetric chains and the number of solutions to the particular equation, by considering the middle term in every symmetric chain. Then, note that any antichain can pick only one element from each symmetric chain by definition, so we get the conclusion.


Find either a closed form or a recursive formula (in terms of fibonacci numbers) for $S_{n}$ if

$S_{n}=F_{1}+2F_{2}+3F_{3}+\dots+nF_{n}$

By Binet's formular we only need computation $x+2x^{2}+3x^{3}+...+nx^{n}$. :)

Here's a nice "simple" recurrence relation for it.

$S_{n}=S_{n-1}+nF_{n}$

(hey, its what you asked for)


What are the accepted rules for rounding numbers such as $ 1.35, \minus{}2.05$ etc?

Sorry for my poor English.

For $ 1.35$, you will get 1.0, and for $ 1.5$ you will get 2.0.

How do I should say this situation in English?

Oh, that's not "to the first decimal place", but rather "to integer" :)

If you do $ 1.43\rightarrow 1.4$, that would be to the first decimal place.

So, fractional part of $ 0.5$ and higher is rounded up, and everything strictly less than $ 0.5$ is rounded down?

[quote="Farenhajt"]
So, fractional part of $ 0.5$ and higher is rounded up, and everything strictly less than $ 0.5$ is rounded down?[/quote]

Yes, that's right.

[quote="i_like_pie"]I think the problem should say:

Find a real number $ x$ for which $ \left[(x \plus{} 1)(x \minus{} 2)\right] \equal{} 5x \plus{} 1$, where $ [n]$ denotes rounding $ n$ to the nearest integer.

If this is the case, then shouldn't $ x \equal{} 1.35$ be $ \left[(2.35)( \minus{} 0.65)\right] \equal{} \left[ \minus{} 1.5275\right] \equal{} \minus{} 2$?[/quote]

Thanks alot.

[quote="Farenhajt"][quote="i_like_pie"]I think the problem should say:

Find a real number $ x$ for which $ \left[(x \plus{} 1)(x \minus{} 2)\right] \equal{} 5x \plus{} 1$, where $ [n]$ denotes rounding $ n$ to the nearest integer.

If this is the case, then shouldn't $ x \equal{} 1.35$ be $ \left[(2.35)( \minus{} 0.65)\right] \equal{} \left[ \minus{} 1.5275\right] \equal{} \minus{} 2$?[/quote]

I was NOT proposing a solution to the equation - I was merely asking for the rules of rounding, in order to compose correct auxilliary equations :)[/quote]
I didn't say anything about you. :ninja:

I was inquiring whether $ f(1.35)\equal{}\minus{}2$ was true to make sure I understood the problem.

[hide="Solution"]Since $ 1\plus{}5x$ must be integer, we can put $ x\equal{}{n\over 5}$, where $ n$ is integer.

By the rules of rounding, we have

$ 1\plus{}n\minus{}{1\over 2}\leqslant{n^2\over 25}\minus{}{n\over 5}\minus{}2<1\plus{}n\plus{}{1\over 2}$

Solving the LH part (in integers) gives $ n\leqslant \minus{}2\lor n\geqslant 32$

Solving the RH part (in integers) gives $ \minus{}2\leqslant n\leqslant 32$

Hence the only possible values for $ n$ are $ \minus{}2$ and $ 32$, and that gives $ x\in\left\{\minus{}{2\over 5},{32\over 5}\right\}$[/hide]


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

Let p be a prime and G be a group of order p^e, where e > 0. prove that G has a subgroup of order p^f
for every 0<= f <= e. Hint use Induction.

Let $ \cal{C}$ be a set of representatives of conjugation classes of elements in $ G$. Then a counting argument gives

$ |G|\equal{}|Z(G)|\plus{}\sum_{c\in\cal{C}\setminus Z(G)}|G: C_{G}(c)|\Longrightarrow 0\equiv|Z(G)|\bmod{p}\Longrightarrow Z(G)>1$

Choose an arbitrary element $ 1\not\equal{} x\in Z(G)$ and by potentially switching to some appropriate power of $ x$ we can already assume that $ x$ has order $ p$. Then $ N\equal{}<x>$ is a normal subgroup of order $ p$ in $ G$ and now apply induction in $ G/N$.


trigonometry
ARML
LaTeX
sine double angle

Here is the question: (Chicago ARML 2004)

Evaluate
(sin50°)(sin170°)(sin250°)

Answer
[hide]They give the answer: -1/8
I don't know how they got there.[/hide]

I have been trying to use product to sum.
(Sorry I don't know LaTeX)

sinA sinB = 1/2 [cos (A-B) - cos (A+B)]

sin170 sin50 = 1/2 [cos120 - cos220]

sin 170 sin 50 = 1/2 [-1/2 - cos 220]

sin 170 sin 50 = -1/4 - (cos220) /2

cos 220 :huh:

this product is equivalent to
-(sin 10)(cos 20)(cos 40)
now we recognize this classic product: sine double angle formula
-2*2*2(sin 10)(cos 10)(cos 20)(cos 40)/8cos 10
like dominoes this cancels
-2*2(sin 20)(cos 20)(cos 40)/8cos 10
-2(sin 40)(cos 40)/8cos 10
-sin 80/8cos 10
-cos 10/8 cos 10
=-1/8, as desired. yay!

A little differently with product-to-sum.
First simplify to
\[ \sin 50^\circ \cdot \sin 170^\circ \equal{} \sin 50^\circ \cdot \sin 10^\circ \equal{} \frac12 (\cos 40^\circ \minus{} \cos 60^\circ) \equal{} \frac14 ( \minus{} 1 \plus{} 2 \cos 40^\circ),
\]
multiply it by $ \sin 250^\circ \equal{} \minus{} \cos 20^\circ$,
\[ \frac14 ( \minus{} 1 \plus{} 2 \cos 40^\circ) ( \minus{} \cos 20^\circ) \equal{} \frac14 (\cos 20^\circ \minus{} 2 \cos 40^\circ \cos 20^\circ )
\]
and use double angle formula,
\[ \equal{} \frac14 \left(\cos 20^\circ \minus{} 2 (2 \cos^2 20^\circ \minus{} 1) \cos 20^\circ \right) \equal{} \frac14 \left(3 \cos 20^\circ \minus{} 4 \cos^3 20^\circ \right).
\]
Here recognize the triple angle formula to simplify to
\[ \minus{} \frac14 \cos 60^\circ \equal{} \minus{} \frac18.
\]


Dear friends,
Two years ago, I met a combinatorical problem in geometry:
[i]In an unit square (whose sides equal to 1), chosen 101 arbitrary points. Prove that there always exists a triple of those points form a triangle whose area is not greater than 1/100.[/i]
It's an interesting problem and can be easily solved with Dirichlet lawer principle. However, the extensive problem is not easy:
[i]In an unit square (whose sides equal to 1), chosen [b]n[/b] arbitrary points. Find the minimum value of [b]n[/b] such that there always exists a triple of those points form a triangle whose area is not greater than 1/100.[/i]
Now I am only successful in showing that $ n\le 59$. Do you have a strong result? Can you introduce your ideas to me and other Mathlinkers? If this problem was discussed in the past, please post links.
I hope to receive the final value (answer) of n. Thank you very much.
Sincerely, mr.danh

I expect a solution. Any ideas? (you can post your comment)


Point $ S_0 $ starts at $ (0,0) $. A sequence of line segments is drawn on the plane. Each segment has length 1, and for 1 <= n <= 179,  the measure of angle $ S_nS_{n+1}S_{n+2} $ is 180-n-1. What is the distance between S_1 and S_{181} when:

1) It is in the shape of a spiral, the angles always turned in the same direction

2) It is a jagged line, the angles always alternating direction.

Sorry if it's a little vague, it's hard to find the correct math terms for what I'm trying to say.  :blush:

Don't worry. You have written it very clear.


Let $ X$ be a set of $ k$ vertexes on a plane such that no three of them are collinear. Let $ P$ be the family of all $ {k \choose 2}$ segments that connect each pair of points. Determine $ \tau(P)$.


Let $X$ be any point on the altitude $CP$ of $\triangle ABC$. $AX,BX$ meet $BC,AC$ at $M,N$. $CP$ meets $MN$ at $Y$. An arbitrary line through $Y$ meets the sides of quadrilateral $CNXM$ at $D,E$. Show that $\angle EPC=\angle DPC$.

Part [b]b)[/b] of http://www.mathlinks.ro/Forum/viewtopic.php?t=20809 . Can you be more precise about your source - which year did this problem appear in?

  darij

It appeared in 2003, and yeah, it has part a),b) like in that link.

Denote ${Q}$∈$DY$∩$AB$

Fistly,we have seen that the points $C,Y,X,P$ are a harmonic set easily,so if we regard $A$ as a perspective center and $DQ$ a transversal ,we can know that $D,Y,E,Q$ are also a harmonic set!
Hence,we have $YD/YE=QD/QE$.

Secondly,we think of the circle of Apollonius.If we draw a circle with diameter $YQ$,we can obtain that any point (for instance,$W$)on the circle must satisfy $WD/WE=YD/YE=QD/QE$.Because of ∠$YPQ=90$°,P is on the circle,and then $PD/PE=YD/YE$,so  ∠$EPC$=∠$DPC$.

[color=darkblue]Nice-looking, that's all ![/color]

[color=darkblue][b][u]Proof.[/u][/b] Define the intersection points $R\in MN\cap AB$ and $S\in DE\cap AB$.
Denote $\mathrm{h.d.}$- a [b]harmonical division[/b] and $\mathrm{h.f.}$- [b]a harmonical fascicle (pencil)[/b].
Is well-known that the division $(A,B,P,R)$ on the line $AB$ is harmonically. Therefore,
$C(A,B,P,R)\--\ \mathrm{h.f.}$ and the line $MN$ $\implies$ $(N,M,Y,R)\--\ \mathrm{h.f.}$ $\implies$
$A(N,M,Y,R)\--\ \mathrm{h.d.}$ and the line $DE$ $\implies$ $(D,E,Y,S)\--\ \mathrm{h.d.}$
In conclusion, $\left\{\begin{array}{c}(D,E,Y,S)\--\ \mathrm{h.d.}\\\ PY\perp PS\end{array}\right\|$ $\implies$ $\widehat{YPD}\equiv\widehat{YPE}$ $\Longleftrightarrow$ $\widehat{CPD}\equiv\widehat{CPE}$.[/color]


logarithms
real analysis
real analysis solved

Let:
\[x(x+y)dy -y^2dx=0\]
Find $y=f(x)$

This is a "homogeneous" (not to be confused with linear homogeneous) first order differential equation. The key is to make the substitution $y=xu$. Then $dy=x\,du+u\,dx$ The equation becomes:

$x^2(1+u)(x\,du+u\,dx)-x^2u^2dx=0$

$(1+u)x\,du+(1+u)u\,dx-u^2dx=0$

$(1+u)x\,du+u\,dx=0$

This is now a separable first order equation:

$\frac{1+u}u\,du+\frac{dx}x=0$

$u+\ln u+\ln x = C$

$xy+\ln(xy)+\ln x = C.$

This implicitly defines $y$ in terms of $x$. You can't get that as a closed form. I've also been implicitly assuming $x$ and $y$ are both positive.


We throw $ N$ dices where $ N$ is a natural number, with $ P(N\equal{}i)\equal{}\frac{1}{2^i}, i \in \mathbb{N}$. Denote by $ S$ the sum of faces of those $ N$ dices. Find the probability for the following events:

1.) $ N\equal{}2$, if $ S\equal{}3$;

2.) $ S\equal{}4$ knowing that $ N$ is even;

3.) $ N\equal{}2$ knowing that $ S\equal{}4$ and the first die shows the  number $ 1$;

4.)   the greatest number that appears is $ 4$, knowing that $ N\equal{}3$.

For 1), one can can write $ P\left(N \equal{} 2|S \equal{} 3\right) \equal{} \frac {P\left(S \equal{} 3|N \equal{} 2\right). P(S \equal{} 3)}{P(N \equal{} 2)}$
and $ P\left(S \equal{} 3|N \equal{} 2\right) \equal{} 2.\frac {1}{6}.\frac {1}{6} \equal{} \frac {1}{18}$, and  $ P(N \equal{} 2)$ is given.
 $ P(S \equal{} 3) \equal{}P(S \equal{} 3, N \equal{} 1)\plus{}P(S \equal{} 3,N \equal{} 2)\plus{}P(S \equal{} 3,N \equal{} 3)\equal{}\frac {1}{6}\plus{}\frac {1}{18}\plus{}\frac {1}{216}$.

4) In general, given $ n$ dice, the probability that the maximum number shown is $ m$ is given by
\[ P \equal{} \left(\frac {m}{6}\right)^n \minus{} \left(\frac {m \minus{} 1}{6}\right)^n.\]
So, in your case we have $ m \equal{} 4, n \equal{} 3$ and hence $ P \equal{} \frac {37}{216}$.


Find the solutions of the eqation :
x^n=2^n+y^2,for x not odd.
In the book where i found this the author only says that this pb is very
difficult and he is giving the solution x=6,y=88,n=5.
It's pretty easy to solve for x odd,but for x not odd how did the author
came up with this solutions. :?  :?  :?

I just had enough time to prove that n should be odd.
I will think on it later this afternoon.

 Case 1. n=even=2m
hence x^m -y =2^s
and x^m+y= 2^(n-s)
hence x^m=2^(a) +2^(b)  , m>=1, a+b=n-2 and a,b>=1 and a=b (mod 2)
let a>=b
hence x^m=2^b( 2^(a-b) +1)
hence m=b.
(x/2)^b = 2^(a-b) +`
t^b = 2^(a-b) +1 hence b is even and a is even.
b=2e
(t^e-1)(t^e+1) =2^(a-b)
t^e-1 =2 and hence t^e=3 which implies e=1 and t=3.
hence b=2 , x=6. a=5. absurd since a=b mod 2.
so n is odd.
i


Find the number of units in the sum of the lengths of the edges of a $ 2\times3\times4$ unit rectangular block.

The block is made up of two $ 2\times4$ rectangles, two $ 3\times4$ rectangles and two $ 2\times3$ rectangles. However, if we calculate the perimeter of each rectangle and add them, we would count the $ 2\times4$ rectangles twice. Thus, lengths of the edges of the rectangular block is $ 2(2\times3)\plus{}2(3\times4)\plus{}2(2\times4)\minus{}2(2\times4)\equal{}36$


limit
induction
logarithms
real analysis
real analysis unsolved

Let $x_{1}\in (0,\infty)$ and $x_{n+1}=x_{n}+(\frac{x_{n}}{n})^{2}$.
a)Prove that $\{x_{n}\}$ has finite limit.
b)Prove that $\lim_{n\to\infty}n(L-x_{n})=L^{2}$, where $L=\lim_{n\to\infty}x_{n}$.

May be $x_{1}\in (0,1)$? Otherwise, if $x_{1}>1$ we can prove by induction $x_{n}>n$
1. $x_{1}>1$
2. $x_{n}>n \ \rightarrow \ x_{n+1}=x_{n}+\left(\frac{x_{n}}{n}\right)^{2}>n+1$.
If $x_{1}\in (0,1)$ we can choose $\varepsilon>0$ such that $x_{n}<n^{1-\varepsilon}$. Let's prove it by induction:
1. $x_{1}<1$, $x_{2}<2$
2. $x_{n}<n^{1-\varepsilon}\ \rightarrow \ x_{n+1}<n^{1-\varepsilon}+\frac{1}{n^{2\varepsilon}}<(1+n)^{1-\varepsilon}$ for small ${\varepsilon}$ and for $n\geq 2$.
Now write
\[x_{n+1}=x_{n}(1+\frac{x_{n}}{n^{2}})=x_{n-1}(1+\frac{x_{n-1}}{(n-1)^{2}})(1+\frac{x_{n}}{n^{2}})=\ldots=x_{1}(1+x_{1})\ldots(1+\frac{x_{n}}{n^{2}})\]
And, as far as $\sum\frac{x_{n}}{n^{2}}<\infty$, $\lim x_{n}<\infty$.

Sorry, I edited it

[quote="N.T.TUAN"]Let $x_{1}\in (0,1)$ and $x_{n+1}=x_{n}+(\frac{x_{n}}{n})^{2}$.
a)Prove that $\{x_{n}\}$ has finite limit.
b)Prove that $\lim_{n\to\infty}n(L-x_{n})=L^{2}$, where $L=\lim_{n\to\infty}x_{n}$.[/quote]

I forgot i didnt post the second part solution :) Here is the sketch of the proof:
$\lim_{n\to\infty}n(L-x_{n})=\lim_{n\to\infty}n(\lim_{i\to\infty}x_{i}-x_{n})$
Note that $x_{n+1}-x_{n}=\frac{x_{n}^{2}}{n^{2}}$, so
\[x_{m}-x_{n}=\sum_{i=n}^{m-1}\frac{x_{i}^{2}}{i^{2}}\]
Now we will write lower and upper bounds for the sum above:
1) $n(x_{m}-x_{n})\leq nx_{m}^{2}\sum_{i=n}^{m-1}\frac{1}{i(i-1)}$
2) $n(x_{m}-x_{n})\geq nx_{n}^{2}\sum_{i=n}^{m-1}\frac{1}{i(i+1)}$
Both of the limits above are equal to $L^{2}$  :roll:

[quote="Yustas"]

And, as far as $\sum\frac{x_{n}}{n^{2}}<\infty$, $\lim x_{n}<\infty$.[/quote]
Can you concrete?  :maybe:

It is true because \[x_{n+1}=x_{1}\prod_{i=1}^{n}(1+\frac{x_{i}}{i^{2}})\]
The sufficient condition for convergence of $\prod (1+a_{k})$ is absolute convergence of series $\sum a_{k}$, it is a well-known fact.

I misunderstood you. Sorry. I didn't read everything  :blush:

[quote="N.T.TUAN"][quote="Yustas"]
The sufficient condition for convergence of $\prod (1+a_{k})$ is absolute convergence of series $\sum a_{k}$, it is a well-known fact.[/quote]
Can you post prove of this statement?  :oops:[/quote]
Let $a_{k}>0$, $k_{0}: \forall \ k>k_{0}\ a_{k}<1$, then \[\ln \prod_{k=k_{0}}^{\infty}(1+a_{k})=\sum_{k=k_{0}}^{\infty}\ln (1+a_{k})=\sum_{k=k_{0}}^{\infty}(a_{k}+o(a_{k}))\]
 which converges if and only if $\sum a_{k}$ converges.

Ok, this is solution i know for part a).
By induction on $n$ we have $nx_{1}>x_{n}\forall n>1$, therefore for some $m$ : $m-1>x_{m}$. For all $n>m$ by $\frac{1}{x_{n}}-\frac{1}{x_{n+1}}<\frac{1}{n^{2}}<\frac{1}{n-1}-\frac{1}{n}$ and telescopic sum we have $\frac{1}{x_{m}}-\frac{1}{x_{n}}<\frac{1}{m-1}-\frac{1}{n-1}<\frac{1}{m-1}$, also $\{x_{n}\}$ is bounded, therfore it conveger( because it increasing).


function
number theory unsolved
number theory

Find all positive integer $n$

such that: $3^{n-1} + 5^{n-1}$  divides $ 3^n + 5^n.$

Since $3(3^{n-1}+5^{n-1})<3^n+5^n<5(3^{n-1}+5^{n-1})$, we must have $\frac{3^n+5^n}{3^{n-1}+5^{n-1}}=4$. This is true for $n=1$, and the function $f(n)=\frac{3^n+5^n}{3^{n-1}+5^{n-1}}$ is strictly increasing, so that's the only solution.

Another way, to prove that $n=1$ is the only possibility without using analysis, is to see that $3^n + 5^n = 4(3^{n-1} + 5^{n-1})$ is equivalent to $3^{n-1} = 5^{n-1}$ since $4=3+1 = 5-1$.

Pierre.

You should try this:
FInd all $n$ positive integers such that:
$(a-1)^{n-1}+(a+1)^{n-1}$ divides $(a-1)^n+(a+1)^n$.

That's pretty much the same, isn't it? Oh, and the monotonicity need not use analysis: we can simply apply AM-GM to show $\frac{(a-1)^{n+1}+(a+1)^{n+1}}{(a-1)^n+(a+1)^n}>\frac{(a-1)^n+(a+1)^n}{(a-1)^{n-1}+(a+1)^{n+1}}$.

...and what about:
$a^{n-1}+b^{n-1}$ divides $a^n+b^n$.

$a^{n-1}+b^{n-1}$ divides $a^n+b^n$, so:
$a^{n-1}+b^{n-1}$ divides $a^n+a b^{n-1}$ and
$a^{n-1}+b^{n-1}$ divides $a^{n-1} b+b^n$. Then:
$a^{n-1}+b^{n-1}$ divides $a b (a^{n-2}+b^{n-2})$.
In the case of $(a,b)=1$ and $n$ greater than $1$  follows that:
$a^{n-1}+b^{n-1}$ divides $a^{n-2}+b^{n-2}$, impossible.
If $n=1$ then $2$ divides $a+b$, so if $a,b$ have the same parity $n=1$ is always a solution, otherwise not.


calculus
derivative
vector
function
calculus computations

how does one prove this
if $ \vec r(q_{1},q_{2}.....,q_{n},t)$ be a the position vector of a point then
$ \frac{d \frac{\delta \vec r}{\delta q_{i}}}{dt}=\frac{\delta \frac{d \vec r}{dt}}{\delta q_{i}}$

[url=http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives]To prove the equality of mixed partials[/url], we can forget all but two variables and consider a scalar-valued function $ f(x,y)$. (It does not matter that you have a vector-valued function, because derivatives are taken component-wise.) Assuming that both $ f_{xy}$ and $ f_{yx}$ are continuous, you can deduce their equality by repeated application of the mean value theorem. The proof is not very long, but I don't feel like typing it, so I suggest that you consult a multivariable calculus textbook.

See blahblahblah's link from [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=mixed&t=72477]here[/url].


How many positive three-digit integers less than 500 have at least two digits that are the same?

Another solution suggestion:
There are four-hundred 3-digit numbers less than 500.
Lets find the numbers with three different digits.  The hundreds digit can be one of four numbers, leaving 9 digits to choose from for the tens position and 8 for the ones.

400 - (4x9x8) = 112

Yeah complementary counting kills this problem, casework is too slow

Isn't the topic "complementary counting"? I mean, if it was in the topic "casework counting," then it would make sense to use casework, but it should seem rather obvious to use complementary counting because of the topic this problem is in...

[hide]We use complementary counting. The number of three-digit integer less than $500$ that have $0$ digits that are the same is $4\cdot 9\cdot 8=288$. Therefore, the number of possibilities is $\boxed{400-4\cdot 9\cdot 8=112}$.[/hide]

I wanted to make my sol as I made it with casework (why is this in complementary counting?), but I saw that this was the alcumus sol 1.
EDIT: 222 posts!
EDIT March 2022: ([size=50]get it, 222nd post, 2022 yeah whatever[/size])Why would I write this? This post has no content whatsoever.

How are there 400 3 digit numbers less than 500? Or am I just being dumb. can someone explain

EDIT: NEVERMIND I WAS BEING DUMB WOOPS

[quote=PM.MATHLEAGUE.VA.CHAMP]How are there 400 3 digit numbers less than 500? Or am I just being dumb. can someone explain
[/quote]
$100, 101, 102, ..., 499$ = $499-100 + 1$ = $399+1$ = $400$ :P

yea I know lmao @above my brain registered single and double digit numbers as 3 digit numbers  :wallbash_red:

how is this level 24

we complementary count.

the amount of ways to have no digits the same is $4 \cdot 9 \cdot 8 = 288$.

the total amount of ways is $499-100+1=400$

the answer is $400-288=\boxed{112}$.

[hide=Casework Supremacy]
You have 4 cases:
SSD: $4\times1\times9=36$.
SDS: $4\times9\times1=36$.
DSS: $9\times4\times1=36$.
SSS: $4\times1\times1=4$.

(S is same, D is different)

Thus the answer is $36\times3+4=108+4=\boxed{112}$.
[/hide]


limit
advanced fields
advanced fields unsolved

In R prove the sequence is Cauchy, and
*therefore* it converges.

1, 1+1, (1+1/2)^2, (1+1/3)^3, (1+1/4)^4, ...

Hmm.. I wonder if the limit of this sequence is e (~2.71) :)

If he is talking about the sequence $e_{n}=(1+\frac{1}{n})^{n}$, then the following results are known:

 1) $2 <e_{n}<3$
 2) $e_{n}$ is strictly increasing
 3) $\lim_{n \to \infty} e_{n}=e$. ;)

About the first one: Let $\sum_{n\geq 1} x_{n}$ a serie of real numbers. Then $\sum_{n\geq 1} x_{n}$ is convergent if and only if for any $\epsilon >0$ there is $n_{\epsilon}\in\mathbb{N}$ such that $|x_{n}+x_{n+1}+...+x_{n+p}| < \epsilon$ for any $n\geq n_{\epsilon}$ and for any $p\in\mathbb{N}.$
is this what are you reffering to? :?


Let  a  be greatest root of  4x^3 - 2x^2 - 15x + 9, and  b be greatest root of 12x^3 + 6x^2 - 7x + 1. Show that  a^2 + 3b^2 = 4.

Namdung

i remember i posted this problem a long time ago (meaning before sept 11 :D ) in the proposed problems section.i solvd it in a week's time,but my solution wasnt natural,it seemed pickd out of thin air.anyways,lagrangia(fast as he is :D ) later posted the solution within 2 hours.i dont know whether the problem is in the solved problems archive