:D A sequence of numbers 49, 4489, 444889, ………. In which every number is made by inserting 48 in the middle of the previous number.  Prove that each number is the square of an odd integer.

 ;)

Every time you insert 48 into the middle of the number, you add a six to the beginning of the square root(ex. 7 to 67, 67 to 667). Therefore each time the square root always has a seven at the end, so it is always odd.

You didn't really prove it... how do we know 44444444444444444444444888888888888888888888889 is a perfect square? I think this should be moved to number theory...

We have that $7^2=49$, $67^2=4489$, $667^2=444889$,.. So we have to proove that
$\underbrace{66..6}_{n-1}7^2=\underbrace{44..4}_{n}\underbrace{88..8}_{n-1}9$.

We have $\underbrace{66..6}_{n-1}7^2=\left(\frac{6(10^n-1)}{9} +1\right)^2=\left(\frac{2\cdot 10^n+1}{3}\right)^2=\frac{4\cdot 10^{2n}+4\cdot 10^n+1}{9}=\frac{4\cdot (10^{2n}-1)}{9}+\frac{4\cdot (10^n+1)}{9}+1=\underbrace{44..4}_{n}\underbrace{88..8}_{n-1}9$.
Thus, the problem is solved :icecream:
If you like this kind of problems see also problem one - Bulgaria
http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=17&year=1967


Let$ A\in\mathcal{M}_{4}(\mathbb Z)$.
Prove that if $ a_{11},a_{22},a_{33},a_{44}$ have the same $ parity$ which is different from the parity of $ a_{ij}$, $ i\neq {j}$ than $ detA\neq0$


How many integers with four different digits are there between 1,000 and 9,999 such that the absolute value of the diference between the first and the last digit is 2?

[hide]
If the first digit is an integer from 2 to 7, there are 2 possibilities for the last digit, since adding or subtracting 2 from any integer in this range yields a single-digit positive integer. In this range, there are 6 possibilities for the first integer (there are 6 integers in the range from 2 to 7), 8 for the second, 7 for the third, and 2 for the last (for every unique first integer). Therefore, this yields 6*8*7*2 = 672 possibilities. If the first integer is 1, 8, or 9, there is one possibility for the last integer, since subtracting 2 from 1 yields a negative number, and adding 2 to 8 or 9 yields an integer with more than 1 digit. Therefore, in this range, there are 3 possibilities for the first integer, 8 for the second, 7 for the third, and 1 for the last (for each unique first integer). This gives us 3*8*7*1 = 168 possibilities.
Taking the sum of the number of possibilities, we get
672 + 168 = 840 possibilities[/hide]

i left out that possible solutions can be

A, 672
B, 784
C, 840
D, 896
E, 1,008

[hide]There are 15 first-last digit pairs: (1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9) and their reverses, and (2,0).

Each of these pairs can have 8*7=56 numbers, considering the digits in between them (remember that the digits are distinct), for a total of 840 possibilities. [/hide]


Hi, if someone could help me with this, it would be greatly appreciated.

I need to find L for

L = 3 [ ( 4+8+12+16+L) / 5]

I know to add them so that L = 3 [ (40 + L) / 5 ]

but after that, I'm lost... thanks in advance :)

I think [x] means the greatest integer not exceeding x.
From L = 3 [ (40 + L) / 5 ]
 we obviously get  L is an integer.
L=3[40/5+L/5]=3[8+L/5]=24+3*[L/5] now  noting that 3|L and by trial we get  27<L<30 hence there's no integer L divisible by 3 and satisfying this.
I think that's correct. :)

Actually I made a big mistake in  "3|L and by trial we get 27<L<30 "
 :blush:
In  L=24+3*[L/5]  put L=3M that gives after dividing by 3, M=8+[3M/5]
Put M=5x+y,y=0,1,2,3 or 4.
5x+y=8+[15x/5+3y/5]=8+3x+[3y/5],
2x+y=8+[3y/5],from here we get cases:(1)if y=0,then x=4 hence M=20 and L=60.(2)if y=3,then x=3 hence L=54.(3)if y=4 then x=3 hence L=57,the other cases don't give solutions.


Let $n\geq 4$ be an integer, and let $a_1,a_2,\ldots,a_n$ be positive real numbers such that \[ a_1^2+a_2^2+\cdots +a_n^2=1 . \] Prove that the following inequality takes place \[ \frac{a_1}{a_2^2+1}+\cdots +\frac{a_n}{a_1^2+1} \geq \frac{4}{5}\left( a_1 \sqrt{a_1}+\cdots +a_n \sqrt{a_n} \right)^2 . \] [i]Bogdan Enescu, Mircea Becheanu[/i]

instinct tells me cauchy will work :D  :D
(i'm trying the prb neway)

apply cauchy schwarz to
\sum [(a_i)^2*{(a_(i+1)^2+1] and
 \sum [ (a_i)/{(a_(i+1)^2 + 1] and multiple  these  two to get their product greater than the rhs of the inequality.so it suffices to prove
\sum [(a_i)^2*{(a_(i+1)^2+1] <= 5/4 which is equivalent to
 \sum [ (a_i)^2*{a_(i+1)}^2] <= 1/4
denote by x_i = (a_i)^2
and one has to prove
x_1*x_2 +x_2*x_3 + ..... + x_n*x_1 <= 1/4. now i havent figured this out clearly but wot i did was to fix n-2 of the x_i's and vary the other 2 and get some conditions for maxima.check that method and give me a sign if u need more help or if something is wrong with my method

I know two methods of solving this inequality (one of them I found during the actual exam) and another was told to me by someone else.

When you want a solution, let me know.

I guess that the more solutions we have the better it is, because perhaps one of them will work on another problem.. so Valentin Vornicu.. I hope you won't mind posting your solutions! :D

thank you!

Here's my solution(although I participated in that exam and I got 0 for this question :D ):
 \sum (ai/(a(i+1)^2+1))=  \sum (ai^3/(a(i+1)^2*ai^2+ai^2). By using the inequality (easy to prove, just prove it for 2 numbers and then extend)   \sum (ai^2/xi)>=   (\sum ai)^2/ \sum xi where ai and xi are positive numbers, we can prove that the left part is>=  ( \sum ai^3/2)^2/ \sum (a(i+1)^2*ai^2+ai^2). To prove that this one, in turn, is >= 4/5 etc, etc, we get to the same thing as you found, that is: if   \sum xi=1 then   \sum xi*x(i+1)<= 1/4. Here's how we can prove this:

For n even we have  \sum xi*x(i+1)<= (  \sum x(2i))*  (\sum x(2i-1))<= 1/4 from AM-GM. Just do something similar for n odd.

Hope it's correct!

I think my sol for the inequality is wrong. Sorry!

The $LHS$ can be rewritten as
$\frac{a_1^3}{a_1^2a_2^2+1} + \cdots + \frac{a_n^3}{a_n^2a_1^2+1}$
By Titus Lemma it is
$\geqslant \frac{\left( a_1 \sqrt{a_1}+\cdots +a_n \sqrt{a_n} \right)^2}{1+a_1^2a_2^2+\cdots +a_n^2a_1^2} $
Simplifying it remains to prove that
$\frac{1}{4} \geqslant a_1^2a_2^2+ \cdots +a_n^2a_1^2$
which is true by chebyshev as $n\geqslant4$

Let $n=2,3$ , and let $a_1,a_2,\ldots,a_n$ be positive real numbers such that \[ a_1^2+a_2^2+\cdots +a_n^2=1 . \] Prove that the following inequality takes place \[ \frac{a_1}{a_2^2+1}+\cdots +\frac{a_n}{a_1^2+1} \geq \frac{n}{n+1}\left( a_1 \sqrt{a_1}+\cdots +a_n \sqrt{a_n} \right)^2 . \]


[quote="xxxyyyy"]Find the largest number less than $10^{9}$ satisfying the following conditions:
$a.$ has all different single digits ( no repeat digit )
$b.$ is divisible by $11$
$c$. is the smallest number using all the digits of the above largest number only once which is also divisible by $11$[/quote]

Here's a hint.
[hide]
The rule for divisible by 11 is that when you sum the alternate digits of a number (|(a1+a3+a5+a7...)-(a2+a4+a6+a8...)| and find the absolute value of the difference, if it is divisible by 11, then the whole number is divisible by 11.[/hide]

[quote="lingomaniac88"]I do not understand what you mean in part C.[/quote]

First he said find the largest number such that the conditions are ture, I think he means to find the smallest number in part C.

[quote="lotrgreengrapes7926"][quote="lingomaniac88"]I do not understand what you mean in part C.[/quote] The most number of digits in the number is 9, but there are 10 digits. And the number is the least number satisfying $a$ and $b$ with those 9 digits.

EDIT: I have a question. $110111111$ is divisible by $11$, but is it considered the smallest number with those digits, or is the smallest number $011111111$?[/quote]

$110111111$, not $011111111$.

[hide] I have no idea if this is right.
OK, for a number to be divisible by 11, we alternate adding and subtracting digits. So we can split the digits into 2 sets, one adding and one subtracting, whose sum should be equal. Since 0 cannot be the first digit, 1 must be the next digit to satisfy condition $c$. But we're trying to make it as large as possible, so we remove 1, making the first digit 2. (Since we can only use 9 digits). From the digits $0,2,3,4,5,6,7,8,9$, we need to construct the least number divisible by $11$. We can start with $203\_\_\_\_\_\_$ The alternating sum and subtract gives $5$, and the remaining digits sum to $39$, so the subtracting set must have sum $22$ and the adding set must have sum $17$. The next digit is from the subtracting set, so it should be 4, but $4+8+9=21$ and it is too small. So the next digit is $5$. That means the subtracting set is $\{5, 8, 9\}$ and the adding set is $\{4, 6, 7\}$. So the smallest number we can construct is $\boxed{203548697}$. [/hide]

To make the proof rigorous, we must add the lemma:
Given that the first (left-most) digit of a nine-digit number is $1$, it is possible to create a number divisible by 11, using [b]any[/b] combination of 8 of the remaining 9 digits.


calculus integration trigonometry geometry

I am stuck... How to integrate this integral.

[img]http://cube.ema.lv/i.png[/img]

Thank you very much.[/img]

The intermediate board doesn't allow calculus problems, so this one obviously won't, either.

However, if you know the definition of an integral graphically and understand what the notation means, you can do the rest without calculus and notice that...

[hide]$ y\equal{}\sqrt{25\minus{}x^2}$ is simply the top half of a circle of radius 5 centered at the origin.[/hide]

Do a trig substitution.
Set $ x\equal{}5\sin(t)$
      $ dx\equal{}5\cos(t)dt$
thus $ \sqrt{25\minus{}25\sin^2{t}}\equal{}5\sqrt{\cos^2{t}}\equal{}5 \cos{t}$
equals the integral: $ \int_0^\frac{\pi}{2}{25cos^2{t} .dt}$

[quote="Fraenkel"]Do a trig substitution.
Set $ x = 5\sin(t)$
      $ dx = 5\cos(t)dt$
thus $ \sqrt {25 - 25\sin^2{t}} = 5\sqrt {\cos^2{t}} = 5 \cos{t}$
equals the integral: $ \int_0^\frac {\pi}{2}{25cos^2{t} .dt}$[/quote]

[hide]Made a little mistake, but you have the right idea.

Let $ x = 5\sin \theta$, then $ dx = 5\cos \theta d\theta$

We have $ \sqrt {25 - (5\sin \theta)^2}dx = \sqrt {25(1 - \sin^2 \theta)}dx = 5\cos \theta dx = 5\cos \theta (5\cos \theta)$
$ = 25\cos^2 \theta$

The new limits of integration will be found with the formula $ \theta = \sin^{ - 1} (\frac {x}{5})$ which follows from $ x = 5\sin \theta$

for $ x = 0$ we have $ \theta = \sin 0 = 0$
for $ x = 5$ we have $ \theta = \sin 1 = \frac {\pi} 2$

The new integral is $ \int_0^\frac {\pi}{2} 25\cos^2 \theta d\theta$

This is useful to get the improper integral of anything in the form $ \sqrt {a^2 - x^2}$. In this case the best aproach is to notice that $ \sqrt {25 - x^2} = y$ implies
$ x^2 + y^2 = 25$ which is the equation of a circle centered at the origin with radius 5.

So, notice $ \int_0^5 \sqrt {25 - x^2}$ is simply the area of a quarter of the area of the whole circle so the answer is $ \frac 1 4 (5^2\pi) = \frac {25\pi} 4$[/hide]

This should have been under a calculus forum, such as Pre-Olympiad or International Contests and Programs.

[quote="gauss1181"]This should have been under a calculus forum, such as Pre-Olympiad[/quote]

Pre-Olympiad doesn't allow calculus, either.  This should have been in [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]Calculus Computations and Tutorials[/url].


geometry
circumcircle
Euler
trapezoid
incenter
rectangle
geometric transformation

If you che k out the topic "2 circles" posted by pascual, you'll find a very interesting property which helps a lot here: given 2 circles intersecting at $C$ and $P$ (I used this notation to make the connection with this problem), the locus of the midpoint of $ST$ where $S$ and $T$ are on the two circles and $ST$ passes through $C$ is a circle passing through $C$ and $P$ and centered at the midpoint of the segment having the centers of the two circles as extremities.

Apply this to the circumcircles of $I_aCP$ and $I_bCP$. The circumcircle of $ABC$ is that locus because it's the Euler circle of the triangle $I_aI_bI_c$ and it passes through the midpoint of $I_aI_b$.

I think your problem is very interested .We see that  the center of the segment which connects the circumcenters of the triangles  IaCP and IbPC coicides with
  the center of the circle . I can try solve it  :)

if angleA>angle B
let H is a center IaIb bisector of angle ACB cut (O) at M let KH//CM (K on (O))
   CH=KM=2Rsin[(A-B)/2]
     we have IaC=r(a)/cos(C/2)=ptg(A/2)/cos(C/2) IbC=ptg(B/2)/cos(C/2)
   then [Ia-Ib]/2= psin[(A-B)/2]/2 cos(C/2)cos(A/2)cos(B/2)= R *2sin[(A-B)/2] then 2CM =CIa-CIb
   let P1,P2 is a center circumcircle PIbC,PCIa let P1Q1//KH,P2Q2//OO1 then O1Q1=Q2Q1 then O is center P1P2 :?

Ok, I'll prove that property I was talking about. Assume two circles intersect at $C$ and $P$ and a variable line through $C$ cuts the circles in $S$ and $T$. We're looking for the locus of $M$, the midpoint of $ST$.

Consider the line through $P$, parallel to the line connecting the centers of the circles. Let this line cut the circles in $S'$ and $T'$ ($S,S'$ are on the same circle). Take $N$ to be the midpoint of $S'T'$. It's obvious that $SS'||TT'$, so $S'T'TS$ is a trapezoid so $MN||SS'||TT'$, so by a quick angle chase we find that $MNPC$ is cyclic, so $M$ is on the circumcircle of $CPN$.

Let $O_1, O_2$ be the centers of the two circles and let $O$ be the circumcenter of $CPN$. $O_1$ is the midpoint of $CS'$, $O_2$ is the midpoint of $CT'$ and $O$ is the midpoint of $CN$, so $O$ must be the midpoint of $O_1O_2$, which is exactly what we wanted: the locus of $M$ is the circle through $C,P$ whose center is the midpoint of $O_1O_2$.

I found this solution, quite different (I think) from those posted here.

Let $I_a$ $I_b$ be the excentres opposite to the vertices $A$ $B$. If $P$ is on $\omega$, the perpendicular bisector of $CP$ will pass through $O$ (the circumcentre of $ABC$). It's known that $\omega$ is the Feuerbach circle of $I_aI_bI_c$, therefore it meets $I_aI_b$ in $C$ and in $M$ (midpoint of $I_aI_b$). The circumcentres of $I_aCP$ and $I_bCP$ are the intersection of the perpendicular bisector of CP with the perpendicular bisectors of $I_aC$ and $I_bC$ respectively.
Let $H,K$ be the midpoints of $I_aC$ and $I_bC$ and $N$ the midpoint of $MC$.
It's easy to proof that $N$ is the midpoint also of $HK$. But $MC$ is a chord of $\omega$, therefore its midpoint is the projection of $O$ on $MC$ and $H$ and $K$ are the projection of the circumcenters of $I_aCP$ and $I_bCP$. Hence, for Talete's theorem, O is the midpoint of the two circumcenters.

[quote="grobber"]$ O_1$ is the midpoint of $ CS'$, $ O_2$ is the midpoint of $ CT'$ and $ O$ is the midpoint of $ CN$, so $ O$ must be the midpoint of $ O_1O_2$, which is exactly what we wanted: the locus of $ M$ is the circle through $ C,P$ whose center is the midpoint of $ O_1O_2$.[/quote]

 :huh:  Is it true?

Here's a new approach :

1) $O,O_1,O_2$ are collinear because they share a common radical axes $CP$.
2) let the foot of perpendiculars from $O,O_1,O_2$ on $I_AI_B$ be $S,M,N$ respectively($O_1$ is the circumcenter of $I_ACP$),all we need to prove is that $S$ is the center of $MN$, let $I$ be the incenter of $ABC$ and $II_A$ and $II_B$ intersect $(O)$ at $U,V$ respectively, so we get $OU=OV$ and $UV$ is the midline of $I_AII_B$. so $MNVU$ is a rectangle and $O$ lies on perpendicular bisector of UV=MN.
:)

second solution ([b]metric[/b]) :it's enough to prove that the power of points $M,N$ with respect to $(O)$ is equal,so that we'll get $MO^{2} -R^{2}=NO^{2} -R^{2}$ which $R$ here is the radius of $(O)$, by the fact that $M,N$ are the midpoints of $CI_B,CI_A$, and the fact that $(O)$ is the nine-points circle of $I_AI_BI_C$, it follows that the second intersection point of $(O)$ with $I_AI_B$ is the midpoint of $I_AI_B$,so all we left to prove is that $MS.MC=NS.NC$ where $S$ here is the midpoint of $I_AI_B$ equal to the second intersection of $(O)$ and $I_AI_B.$ we have $CI_A=b.sin(A/2)sin(B/2)^(-1)$,another equality holds for $CI_B$,so we can get the length of $I_AI_B$ by adding them up,and get $MS,NS$. all else is just some computations( It works :D).

third solution ([b]metric[/b]): let $S$ be the foot of perpendicular from $O$ to $I_AI_B.$ known result that $S$ is the midpoint of $MC$ where $M$ is the 2nd intersection of $(O)$ and $I_AI_B. $
(*) : easy to compute $OS^{2}$ by Casey's Theorem(the difference of power of a point wrt 2 crcles) for circles $(I_A) and (I_B)$.
Now let $S_0$ be the midpoint of UV where U,V are midpoints of $CI_A and CI_B.$
(**) : easy to compute $S_{0}I_A ^{2}$ by these lots of midpoints(it's not a so long equality).

applying Euler Equality for $OI_A ^{2},$ you can use Phythagorean Theorem and get that $OS-0=OS$ ,and as we know OS is perpendicular to $I_AI_B,$ we'll get $S=S_0.$

plus at the end of 3rd solution one may get this equality to prove :

$\frac{rp -2Rcos^{2}\frac{C}{2}sin^{2}\frac{B}{2}}{ (  p-c)}=\frac{rp -2Rcos^{2}\frac{B}{2}sin^{2}\frac{C}{2}}{ (  p-b)}, $

and its proof is in the link below(it will come soon :) ) :
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=519336&p=2924184#p2924184

[quote="orl"]Let $ I(A)$ and $ I(B)$ be the centers of the excircles of a triangle $ ABC,$ which touches the sides $ BC$ and $ CA$ in its interior. Furthermore let $ P$ a point on the circumcircle $ \omega$ of the triangle $ ABC.$ Show that the center of the segment which connects the circumcenters of the triangles $ I(A)CP$ and $ I(B)CP$ coincides with the center of the circle $ \omega.$[/quote]

Note that the circle $w$ is the $9$ point circle of $I_aI_bI_c$ because $ABC$ is the orthic triangle of $I_aI_bI_c$ (the internal and external bisectors are obviously perpendicular.  Hence, the midpoint $M$ of $I_aI_b$ is on $W$.

Note both $I_aCP$ and $I_bCP$ have their cirumcenters on the bisector of $CP$, so the midpoint of the segment connecting them is also on $CP$.  Now, the circumcenter of $I_aCP$ is on the perpendicular bisector of $I_aC$ and that of $I_bCP$ is on the perpendicular bisector of the parallel $I_bC$.  It follows that the midpoint of their circumcenters is on the perpendicular bisector of $CM$, since the midpoint of $CM$ is the midpoint of the segment $RS$ where $R$ is the midpoint of $CI_a$ and $S$ is the midpoint of $CI_b$.  This can be seen because the image of $RS$ under a homothety of scale $2$ centered at $C$ is $I_aI_b$, which has midpoint $M$.  This means that the midpoint of the segment connecting the cirumcenters of $I_aCP$ and $I_bCP$ is on the perpendicular bisectors of both $CM$ and $CP$, so it must be $O$.

Note that the centre one circle lies on the perpendicular bisector of $I_BC$. If $H$ is the othocentre of $\triangle I_AI_BI_C$ and $I_BH \cap \odot ABC = X$ then $XM$ is that perpendicular bisector where $M$ is midpoint of $I_CI_B$. Apply similar thing for the other. Note that since $\odot ABC$ and the circles are co-axial, their centres are collinear. Hence, through a homothety with centre $O$, the circumcentre of $\triangle ABC$, with ratio $-1$ we take the centres of the circles to eachother. So, done.

Dynamic geometry! :)

Let us consider a point $X$ on the perpendicular bisector $\ell_1$ of $CP$ and define point $f(X)$ as the second intersection point of the circle with centre $X$ and radius $r=XC=XP$ with the external bisector $\ell_2$ of angle $C$. Then, clearly as $X$ moves with a uniform velocity along the line $\ell$ point $f(X)$ moves with a uniform velocity along the line $\ell_2$. Since when $X$ is the centre of $I_ACP$, $f(X)=I_A$ and since when $X$ is the centre of $I_BCP$, $f(X)=I_B$. Thus, when $X$ is the midpoint of these two centres $f(X)$ is the midpoint $M$ of $I_AI_B$ and so, the midpoint of the lines joining these two centres passes through the centre of $CPM$ which is actually the centre of $\omega$.

Notice that we used uniformity of motion of $f(X)$. This is indeed, pretty obvious either by geometric continuity or since this intersection is the reflection of $C$ in the projection of $X$ on $\ell_2$ and this is obviously continuous and uniform in motion.

[b]Lemma:[/b]
In the triangle $\triangle ABC$  let $F$ be the feet of the perpendicular from $A$,let $B_{1}$,$C_{1}$,$S$ be the midpoints of $BF$,$CF$,projection of nine-point center on $BC$ $\implies$ $SB_{1}=SC_{1}$
Proof:
Let the nine-point center be $E_{9}$ i will prove $E_{9}B_{1}=E_{9}C_{1}$
$P_{\omega}(B_{1})=B_{1}F\cdot MB_{1}=\frac{BF\cdot CF}{4}=P_{\omega}(C_{1})$ which implies the result
here $\omega$ is the nine-point circle ,$M$ is the midpoint of $BC$

The circumcenters of $\odot I_{A}CP$,$\odot ABC$,$\odot I_{B}CP$ are colinear  on the perpendicular bisector of $CP$.Let the projections of circumcenters of $\odot I_{A}CP$,$\odot ABC$,$\odot I_{B}CP$ on $I_{A}I_{B}$ be $A_{1},O_{1},B_{1}$ respectively. It suffices to prove $A_{1}O_{1}=O_{1}B_{1}$ whish just [b]Lemma[/b].$\square$

I like this problem!  Here's a solution I don't think has appeared in this thread before.  I'm going to use $A$-indexing instead of $C$-indexing, and let $I_b$ and $I_c$ denote the $B$ and $C$ excenters, respectively.

Let $M$ and $N$ be the midpoints of minor and major arc $\widehat{BC}$, respectively.  Furthermore, let $O_b$ and $O_c$ denote the circumcenters of $\triangle I_cAP$ and $I_bAP$, respectively.  By a well-known spiral similarity lemma, $P$ is the center of spiral similarity $\mathcal S$ sending $\overline{I_bI_c}$ to $\overline{O_bO_c}$.
[asy]
size(400);
defaultpen(linewidth(0.7)+fontsize(10));
pair A = dir(110), B = dir(210), C = dir(-30), M = dir(270), P = dir(285), O = origin;
pair X = rotate(90,B)*bisectorpoint(A,B,C), Y = rotate(90,C)*bisectorpoint(A,C,B), Z = rotate(90,A)*M;
pair Ic = extension(B,X,A,Z), Ib = extension(C,Y,A,Z), Oc = circumcenter(Ic,A,P), Ob = circumcenter(Ib,A,P), Q = intersectionpoint(A--M,Ob--Oc);
draw(Ob--Oc,gray(0.7));
draw(circumcircle(A,B,C),rgb(0.1,0.5,0.1));
draw(A--B--C--A,rgb(0.1,0.2,0.8));
draw(B--Ic--Ib--C,rgb(0.8,0.1,0.8));
draw(P--Q--A--P--N--O--P,rgb(0.9,0.1,0.1));
draw(Q--M--O,orange);
dot("$A$",A,dir(O--A));
dot("$B$",B,dir(O--B));
dot("$C$",C,dir(O--C));
dot("$I_c$",Ic,NW);
dot("$I_b$",Ib,NE);
dot("$N$",((Ib+Ic)/2),dir(90));
dot("$O_c$",Oc,dir(Ob--Oc));
dot("$O_b$",Ob,dir(Oc--Ob));
dot("$O$",O,dir(330));
dot("$P$",P,S);
dot("$Q$",Q,dir(235));
dot("$M$",M,S);
[/asy]
Now let $Q$ be the intersection point of the $A$-angle bisector with $O_bO_c$; observe that $O_bO_c$ is the perpendicular bisector of $\overline{AP}$, and hence $AQ = QP$.  Now, since $A,Q,M$ and $N,O,M$ are collinear triplets of points,
\[
\angle QAP \equiv\angle MAP = \angle MNP \equiv\angle ONP,
\]
so triangles $QAP$ and $ONP$ are similar as well.  Thus $\mathcal S$ also sends $N$ to $O$.

But $N$ is the midpoint of $\overline{I_bI_c}$, so $O$ is the midpoint of $\overline{O_bO_c}$.  Done.

I'll use the diagram and $A$-indexing from above. Let $N$ be the midpoint of arc $BAC$.

Since $(ABC)$, $(API_b)$, $(API_c)$ are coaxial, points $O$, $O_b$, $O_c$ are collinear. Now let $X$, $X_b$, $X_c$ be the projections of $O$, $O_b$, $O_c$ onto $\overline{I_bI_c}$, and $Y$, $Y_b$, $Y_c$ the images of $X$, $X_b$, $X_c$ in homothety $(A,2)$.

Then $Y_b=I_b$, $Y_c=I_c$ since $O_bA=O_bI_b$, $O_cA=O_cI_c$. Similarly $Y=N$ since $OA=ON$. Then $Y$ is the midpoint of $\overline{Y_bY_c}$, so $X$ is the midpoint of $\overline{X_bX_c}$, and $O$ is the midpoint of $\overline{O_bO_c}$.

[hide = solution]
Obviously, we animate $P$ projectively along $\odot(ABC)$. Let $O_1, O_2$ be the respective circumcentres in question. It's easily computed that both are tethered to the perpendicular bisectors of $I_AC$ and $I_BC$ respectively.  Further, the line through $P$ perpendicular to $CP$ has degree $2$ since $P$ has degree $2$. A homothety centred at $C$ with factor $1/2$ sends this line to $CO_1$, therefore \[\deg O_1 = \deg CO_1 = \deg CP = 2\]
We need to show that $O$ is the midpoint of $O_1O_2$. Reflecting $O_1$ about $O$ preserves it's degree therefore the previous condition is a degree $2$ conditions. It therefore suffices to verify for $3$ choices of $P$. These are as follows:
[list]
[*] $P = B$. In this case the circumcentre of $I_ACB$ is the midpoint of arc $BC$ not containing $A$. On the other hand, the circumcentre of $I_BCB$ is the same as that of $I_BCBI_C$ where $I_C$ the $C$-excentre. But the circumcentre of this triangle is clearly the midpoint of arc $BC$ containing $A$ and so this case is proven.
[*] $P = A$. Follows from the above analogously.
[*] $P$ tends to $C$. In this case the two circles limit to the ones passing through $I_B, I_C$ and tangent to $\odot(ABC)$ at $A$ respectively. Let $N$ denote the midpoint of $I_AI_B$. From homothety we have the length conditions: \[\frac{CI_A}{CN} = \frac{CO_1}{CO} \text{ and } \frac{CI_B}{CN} = \frac{CO_2}{CO} \]
Adding and remembering that $N$ is the midpoint of $I_AI_B$ we obtain that $CO_1+CO_2= 2CO$. Since these four points are collinear along the common normal it follows that $O$ is the midpoint of $O_1$ and $O_2$, as desired.
[/list]
All in all, since we've checked $3$ cases it follows that the problem is true for all $P$ as well, as required.
[/hide]


Prove that for any odd integer number n, we have:
$ 128 | (n^6\minus{}n^4\minus{}n^2\plus{}1)$

$ n^6\minus{}n^4\minus{}n^2\plus{}1$
$ n^6\minus{}n^4\minus{}(n^2\minus{}1)$
$ n^4(n^2\minus{}1)\minus{}(n^2\minus{}1)$
$ (n^2\minus{}1) (n^4\minus{}1)$
$ (n\plus{}1) (n\minus{}1) (n^2\plus{}1)(n^2\minus{}1)$
$ (n\plus{}1) (n\minus{}1) (n^2\plus{}1)(n\plus{}1)(n\minus{}1)$
$ (n\plus{}1)^2 (n\minus{}1)^2 (n^2\plus{}1)$

we know that for whatever value of n, n^2+1 will always be divisible by 2. We also know that for any odd integer the number either above it or below it must be divisible by 4.

we can know see that we have

(2*2)^2  * 2^2 * 2 this is the same as

$ 2^4*2^2 *2$ or $ \boxed{2^7}$


I have absolutely no idea if someone else brought this issue up, but here goes:
I recently saw an advertisement (on the top of the window, it is [by default] in a light blue color and advertises AoPS products) about Alcumus, but it is still under construction! I would say that it is not necessary to keep this ad because those who have not known about alcumus cannot now benefit from it and those who have known about it probably cannot benefit too much except for entertainment from the videos.

Apparently Alcumus will be up soon (possibly tommorow at least that is what I heard )and disabling the Alcumus add would delay Alcumus release by the admins editing the add and stuff which is time they would not spend working on Alcumus so getting rid of that add wont be worth it.

Alcumus is up now, so this thread should be locked.


I play puzzle frenzy on my calculator, and I'm stuck on level 55. Here is the [hide="password"]b0N[/hide]

Please help me!

It's a calculator game. It's pre-installed on a TI-84+ Silver Edition calculator. Other games pre-installed are Dino Puzzle, Block Dude, and Pegs. I have the password for Level 7 on Block Dude, if you want it.

[quote="perfect628"]I'm bad at all of those. I do have level 9 block dude, though.[/quote]

Oh, yeah, I'm stuck on that too.

I once beat Pegs, but then I accidentally cleared my RAM and the calculator forgot that.

For the level 55 pf
[hide="solution"]you just (in this order) move the 2nd row (from bottom), 4th one on the right down, 3rd row 4th down, and finally, 2nd row 4th down[/hide]
And on blockdude, you can get to all the levels by going into block dude then pressing: "alpha", right, "alpha", right, "alpha", right, "alpha".  :wink:

I just got to level 57. Here's the [hide="password"]kPZ[/hide]


search
function
inequalities proposed
inequalities

Show that for any real numbers $a,b,c>0$ and any integer n > 1,
$(\frac{a}{a+b})^n+(\frac{b}{b+c})^n+(\frac{c}{c+a})^n\geq \frac{3}{2^n}$.
Determine the condition for the equality sign to hold.

Sorry for typing error! The expression has been edited.

As you state it, it's false. ($n=2, b\to0, a=1,c=2$) It might be true though for $n\ge 3$ but in that case not very interesting, as it's completely asymmetric.
Anyway I guess you mean the first term to be $\frac a{a+b}$. If so, I think it has been in the forum not a long time ago, but I can't find it.  :huh: Searching for the expression yields  16351 results. :mad: (How does the search function for expressions work?)


Well they are celebrating 10 years in the browser bussiness and they decided to do some party favors:

[quote="Opera People"]Get a free Opera registration code by clicking the "go free" button below. We're giving away registration codes for as long as the party lasts![/quote]The link is http://my.opera.com/community/party/

Cool :D

I got one. :D

How'd you find out about this, Valentin?

[quote="hwenterprise"]How'd you find out about this, Valentin?[/quote]A friend of mine told me ...  :)

OK, no more free licenses, so topic is closed.


inequalities
induction
inequalities unsolved

Here are some more Inequalities I want help for :

A]  $ 2 < \left(1\plus{}\frac{1}{n}\right)^{n} < 3$

B]  $ \sqrt{n^{n}}<\ n!$  for $ n>2.$

C]  $ (2n)!\ <\ \left[n(n\plus{}1)\right]^{n}$

D]  $ n\displaystyle \left(\sqrt[n]{n\plus{}1}\minus{}1\right)\ <\ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}....\plus{}\frac{1}{n}\ <\ n\left(1\minus{}\frac{n\minus{}1}{\sqrt[n\minus{}1]{n^{n}}}\right).$

E]  $ n^{n}\ >\ 1.3.5....(2n\minus{}1).$

F]  $ \displaystyle \prod_{k\equal{}0}^{n}\left(\begin{array}{l}
n\\
k
\end{array}\right)\ <\ \left(\frac{2^{n}\minus{}2}{n\minus{}1}\right)^{n\minus{}1}.$

G]  $ \displaystyle \sum_{k\equal{}1}^{n}\sqrt{\left(\begin{array}{l}
n\\
k
\end{array}\right)}\ \leq\ \sqrt{n(2^{n}\minus{}1)}.$

[quote="Shanku"]Here are some more Inequalities I want help for :

A]  $ 2 < \left(1 \plus{} \frac {1}{n}\right)^{n} < 3$

B]  $ \sqrt {n^{n}} < \ n!$  for $ n > 2.$

C]  $ (2n)!\ < \ \left[n(n \plus{} 1)\right]^{n}$

D]  $ n\displaystyle \left(\sqrt [n]{n \plus{} 1} \minus{} 1\right)\ < \ 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} .... \plus{} \frac {1}{n}\ < \ n\left(1 \minus{} \frac {n \minus{} 1}{\sqrt [n \minus{} 1]{n^{n}}}\right).$

E]  $ n^{n}\ > \ 1.3.5....(2n \minus{} 1).$

F]  $ \displaystyle \prod_{k \equal{} 0}^{n}\left(\begin{array}{l} n \\
k \end{array}\right)\ < \ \left(\frac {2^{n} \minus{} 2}{n \minus{} 1}\right)^{n \minus{} 1}.$

G]  $ \displaystyle \sum_{k \equal{} 1}^{n}\sqrt {\left(\begin{array}{l} n \\
k \end{array}\right)}\ \leq\ \sqrt {n(2^{n} \minus{} 1)}.$[/quote]
set $ a_n\equal{}{\left(1\plus{}\frac{1}{n} \right)}^{n}$ $ (n\epsilon N)$
we have:
$ a_n\equal{}{\left(1\plus{}\frac{1}{n} \right)}^{n}\equal{}\sum_{k\equal{}0}^{n}C_{k}^{n}\frac{1}{{n}^{k}}\equal{}C_{0}^{n}\plus{}\frac{1}{n}C_{1}^{n}\plus{}\frac{1}{{n}^{2}}C_{2}^{n}\plus{}...\plus{}\frac{1}{{n}^{n}}C_{n}^{n}$
$ \equal{}2\plus{}\frac{1}{2!}\left(1\minus{}\frac{1}{n} \right)\plus{}...\plus{}\frac{1}{k!}\left(1\minus{}\frac{1}{n} \right)\left(1\minus{}\frac{2}{n} \right)...\left(1\minus{}\frac{k\minus{}1}{n} \right)$
(because $ \frac{1}{{n}^{k}}C_{k}^{n}\equal{}\frac{1}{k}.\frac{n\left(n\minus{}1 \right)...\left(n\minus{}k\plus{}1 \right)}{k!}\equal{}\frac{1}{k!}\left(1\minus{}\frac{1}{n} \right)\left(1\minus{}\frac{2}{n} \right)...\left(1\minus{}\frac{k\minus{}1}{n} \right)$ )
so we have:
$ a_n\leq 2\plus{}\frac{1}{2!}\plus{}...\plus{}\frac{1}{n!}\leq 2\plus{}\frac{1}{2}\plus{}\frac{1}{{2}^{2}}\plus{}...\plus{}\frac{1}{{2}^{n\minus{}1}}$
$ \leq 1\plus{}\left(1\plus{}\frac{1}{2}\plus{}\frac{1}{{2}^{2}}\plus{}...\plus{}\frac{1}{{2}^{n\minus{}1}}\right)\equal{}1\plus{}\frac{1\minus{}\frac{1}{{2}^{n}}}{1\minus{}\frac{1}{2}}\leq 3\minus{}\frac{1}{{2}^{n\minus{}1}}<3$

[b]Random Question:[/b] Can you solve those problems by induction??? If yes can we gave a try

We know $ \sum_{k \equal{} 0}^{n}\ C(n,k) \minus{} C(n,0)\equal{}\sum_{k \equal{}1}^{n}\ C(n,k) \equal{} 2^n \minus{}1$

so from CS we have :

$ \sum_{k \equal{} 1}^{n}\sqrt{\left(\begin{array}{l}n\\ k\end{array}\right)}\ \leq\ \sqrt{(1\plus{}1\plus{}...\plus{}1)(\sum_{k \equal{} 1}^{n}\ C(n,i)))}\equal{}\sqrt{n(2^{n}\minus{}1)}$

Q.E.D
hasan4444 I don`t think Induction be a good way for these problems .
For part A you can use Bernoulli Inequality and this fact : [img]http://upload.wikimedia.org/math/6/3/9/639ff4900f0b8fac424e099445c50a38.png[/img] $ \equal{}e\equal{}2.71...$

Sorry MeKnowsNothing, for [E] it will be $ \geq$, equality for $ n\equal{}1$
Inductive proofs are absolutely Ok, if suitable.

[quote="Shanku"]B]  $ \sqrt {n^{n}} < \ n!$  for $ n > 2.$[/quote]
we have:
$ \left(n\minus{}k \right)\left(k\minus{}1 \right)\geq 0\Rightarrow k\left(n\minus{}k\plus{}1 \right)\geq n$
So:
$ 1.n\geq n$
$ 2.\left(n\minus{}1 \right)> n$
...
$ k\left(n\minus{}k\plus{}1 \right)> n$
...
$ n.1\geq n$
$ \Rightarrow {\left( n! \right)}^{2}>{n}^{n}\Leftrightarrow n!> \sqrt{{n}^{n}}$
Q.E.D


vector
abstract algebra
topology
function
real analysis
real analysis unsolved

Previously I have posted a message about disconnecting a n-dimensional space.

Now my question is that how (if it is possible) to disconnect an infinite-dimensional vector space.

I am eagerly waiting for your replies.[/b]

[quote="jmerry"]The kernel of any continuous linear functional will do.[/quote]
Sorry I didnt understand what you meant :blush:
Can you please elaborate.

Here's what jmerry meant.

Let $V$ be an infinite dimensional topological vector space. We need a topology in order for "connected" to make any sense.

Now IF (and this is a very big if) there are any nontrivial continuous linear functionals on $V$ we proceed as follows.

Let $g : V\mapsto\mathbb{R}$ be a nonzero linear functional. Since it is nonzero, there is a point $x\in V$ such that $g(x)\ne0.$ Then by considering both $g(x)$ and $g(-x)$, we have at least one point at which $g(x)>0$ and at least one point at which $g(x)<0.$

The kernel of $g$ is the set $K=\{x\in V : g(x)=0\}.$ Then $g$ is a continuous function that maps $V\setminus K$ into $(-\infty,0)\cup (0,\infty).$ By the comments above, the range of this function must contain elements of both halves of the codomain, and therefore the range is a disconnected set.

The continuous image of a connected set must be connected, so $V\setminus K$ must fail to be connected.

However, this proof does contain one flaw: it presumes the existence of a nontrivial continuous linear functional. There exist infinite dimensional topological vector spaces with no nontrivial continuous linear functionals. You may very well be able to disconnect them as well, but the proof will need some modification.

Hmmm. I understand.Thanks a lot for explaining.


Triangle $ ABC$ is inscribed in circle $ (O)$ having bisectors $ AD,BE,CF$ intersect $ (O)$ at $ A',B',C'$.The lines passing through $ O$ and paralleling $ AD,BE,CF$ intersect $ BC,CA,AB$ at $ A",B",C"$ ,respectively.Prove that $ A'A",B'B",C'C"$ are concurrent.  :roll:

$ \triangle A_0B_0C_0$ is the tangential triangle of $ \triangle ABC$ and $N$ is  the midpoint of $ BC.$ Tangent line of $ (O)$ at $ A'$ cut $ A_0C_0$ and $ A_0B_0$ at $ X,Y,$ respectively. By Newton's theorem for the tangential quadrilateral $ XYB_0C_0,$ we have $ D \equiv XB_0 \cap YC_0 \cap BC.$ Let  $ R$ be the midpoint of $ DA_0.$ Since $ A'$ is the midpoint of $ XY,$ due to obvious symmetry about $ OA_0,$ it follows that $ R,A'$ and the midpoint $ M_1$ of $ B_0C_0$ are collinear on the Newton's line of the complete quadrangle $ XDYA_0.$ On the other hand, let the parallel to $ RA''$ from $ A_0$ cut $ BC,$ $ XY,$ $ DA'$ at $ M,P,Q.$  $ A'A''$ and $ PA''$ become midlines of $ \triangle MQD,$  thus $ \frac {_{PA_0}}{^{PQ}} \equal{} \frac {_{PA_0}}{^{PM}} \equal{} \frac {_{A'A_0}}{^{A'N}}.$  But since $ BC$ is the polar of $ A_0$ WRT $(O),$ we have $ \frac {_{A'A_0}}{^{A'N}} \equal{} \frac {_{OA_0}}{^{OA'}}$ $ \Longrightarrow$ $ \frac {_{OA_0}}{^{OA'}} \equal{} \frac {_{PA_0}}{^{PQ}}$ $ \Longrightarrow O$ lies on the line $ PA''.$ Thus, $ A'A''$ passes through $ M_1$ and similarly $ B'B''$ and $ C'C''$ pass through the midpoints $ M_2,M_3$ of $ C_0A_0$ and $ A_0B_0$ $\Longrightarrow$ $ A'A'',$ $B'B'',$ $C'C''$ concur at the [url=http://xtec.net/~qcastell/ttw/ttweng/definicions/d_mig_arc2_p.html]Second mid-arc point[/url] of $ \triangle A_0B_0C_0.$


Factorise :
$ (x\minus{}6)(x\plus{}1)(2x\plus{}3)(x\plus{}6)\plus{}8x^4$


inequalities
geometry
area of a triangle
inequalities solved

Prove that

(a+b)(a+c) \geq 2 \sqrt (abc(a+b+c)) for all positive reals.

First of all, we can consider a to be 1: we divide the whole thing by a^2 and then we make the substitutions b/a->b and c/a->c. We are left with (1+b)(1+c)>=2(bc(1+b+c))^1/2. We square this and we get (1+b+c+bc)^2 >= 4bc(b+c+1). Now denote x=bc and y=b+c+1 and we have to prove (x+y)^2 >= 4xy iff (x-y)^2 >= 0, which is true.

to prove that
(a+b)(a+c)  >= 2  sqrt(abc(a+b+c))
it suffices to prove
(a+b)^2(a+c)^2 - 2*abc(a+b+c) >=0

After calculation it is:
a^4 +2*a^3*c+ a^2*c^2+ 2*a^3*b+ 2*a^2*b*c+ b^2*a^2+ b^2*c^2>=0

which is obviously true.
(i know it's ugly but it works ;-))

It's really simple. $(a+b)(a+c)  = a(a+b+c) + bc \geq 2 \sqrt {abc(a+b+c)}$ by AM-GM.

This is a neat solution a Brazilian student gave at the time: let $x = a + b$, $y = a + c$ and $z = b + c$. Since $a,b,c$ are positive, $x,y,z$ are side of a triangle. The inequality becomes
\[{xy\over 2} \geq \sqrt{p(p-x)(p-y)(p-z)},\]
where $p = {x+y+z\over 2}$.

This inequality is true because the area of a triangle with sides $x,y,z$ is $\sqrt{p(p-x)(p-y)(p-z)}$ and does not exceed ${xy\over 2}$.


inequalities
geometry
trigonometry
triangle inequality
area of a triangle
Heron\u0027s formula

In a triangle ABC, prove that \[4(ab+bc+ca) > (a+b+c)^{2}\]

[hide]Make a substitution $a=x+y$, $b=y+z$, $c=z+x$[/hide]

EDIT

Ok, here is much better proof

[hide]By triangle inequality \[a<b+c \implies a^{2}<a(b+c)\]  \[b<a+c \implies b^{2}<b(a+c)\]  \[c<b+a \implies c^{2}<c(b+a)\] By summing these up and adding $2(ab+bc+ca)$ to both sides of equation \[a^{2}+b^{2}+c^{2}<2ab+2bc+2ca\]  \[a^{2}+b^{2}+c^{2}+2ab+2bc+2ca<4(ab+bc+ca)\]  \[(a+b+c)^{2}<4(ab+bc+ca)\] [/hide]

Or rewrite it as \[2ab+2bc+2ca-a^{2}-b^{2}-c^{2}> 0,\] or \[\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}-\sqrt{a}\right)\left(\sqrt{c}+\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right) > 0.\] Since $a+b > c$, we have that $\sqrt{a}+\sqrt{b}> \sqrt{c}$, etcetera, so we're done.

nice proof ArcticMonkey!!  :lol: ...i had a think about $a+b > c$ but never went as far as to construct the question through it...wow!! that's nice stuff. Anyway, how would you approach this one -

Given Heron's Formula: the triangle area $A = \sqrt{p(p-a)(p-b)(p-c)}$ , where $p = \frac{a+b+c}{2}$. Prove that \[a^{2}+b^{2}+c^{2}\geq 4\sqrt{3}A\]

$a^{2}+b^{2}+c^{2}\geq ab+bc+ac=2A\left(\frac{1}{\sin\alpha}+\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\geq\frac{6A}{\sin\frac{\pi}{3}}=4A\sqrt{3}$.
Jensen and AM-GM.

Masoud Zargar


geometry
geometric transformation
rotation

If 100 seconds were a minute, 100 minutes were an hour, and 10 hours were in a day,
a.) What will the difference between our "year" and this "year"?
b.) Would base 10 be actually useful in real life? Why, or why not.
c.) If you can change the clock (or how time is measured), what will it be like and why?
For b and c, please state [b]your[/b] opinion, not why the things are the way they are.

I know this is an opinion question, but I think it's important to wonder why things are the way they are, and what we can to improve upon it.

a.  There days would be 13600 seconds longer, so their new year would be shorter than ours.

b.  No.  People are used to our current clock, and it fits the way our lives are done very well.  If you changed the length of the day, you would also have to change the length of a second aswell, so that a day still takes one rotation on our axis.  Also, look at our current keyboard.  A better one has been invented, but no one uses it because we are too accustomed to the old keyboards.

c.  I would not, because what we have now works fairly well, even though it is not the best possible.

it would be almost impossible to change the system of time. The scientific community uses time as a factor in almost all inquiries - if we changed the lenght of a second it would destroy their work. If we change the lenght of the hour/minute but leave the second as it is, we would no longer have our days being equivalent to a rotation of the earth, years as a revolution around the sun,. etc.. It would be difficult to change the system and rather unecessary. I guess you are thinking about a "metric" time system. The metric time system exists, for seconds and part of seconds (centi,milli, micro, nano, picoseconds...). As for the larger unit times, only the hour and the minute do not directly correspond to astronomical means, so they are the only onew worth changing. The day would still have to have 86400 seconds.

I agree. Changing the system of time does not seem to confer a great advantage - compared to the disadvantages, anyway. No one wants to go back and redo all the work that's been done just for this new system of measuring time.


Let $p_{n}$ be the sequence such that $p_{1}=3,\ p_{n}=1+\alpha^{n-1}+\beta^{n-1}\ (n=1,\ 2,\ 3,\ \cdots)$ for complex numbers $\alpha ,\ \beta \ (\alpha ,\ \beta \neq 0).$ If $p_{n}p_{n+1}=0,$ then find the values of $\alpha ,\ \beta ,\ p_{n}.$

i guess an answer is the 3rd roots of unity....and p_n=0,3

[quote="JBL"]How about an actual [hide="solution"]We know $p_{2}= 1+\alpha+\beta = 0$.
Divide into two cases: if $p_{3}= 1+\alpha^{2}+\beta^{2}= 0$, we can substitute in from the equation for $p_{2}$ to get $2\alpha^{2}+2\alpha+2 = 0$ so $\alpha$ is one of the complex 3rd roots of unity, and $\beta =-1-\alpha = \alpha^{2}$ is the other, giving us the sequence $3, 0, 0, 3, 0, 0, \ldots$.
If $p_{3}\neq 0$ then $p_{4}= 1+\alpha^{3}+\beta^{3}= 0$, and we can do the same thing, and the quadratic we get is $-3\alpha^{2}-3\alpha = 0$.  Then $\{\alpha, \beta\}= \{0,-1\}$ and the sequence is $3, 0, 2, 0, 2, 0, \ldots$.[/hide][/quote]

$\alpha\ne 0$

so just the first solution...and yes...sorry i was lazy


In AMM este data problema 3225, care contine o relatie, care are drept consecinte
teorelele I si II ale lui PTOLEMEU (din patrulaterul inscriptibil).
Eu am gasit problema fara solutie, undeva pe internet(nu mai stiu unde).
Nu stiu daca autorul problemei (C. N. Schmall) si-a dat sau nu seama ca relatia sa
este mai generala decat cele 2 teoreme ale lui Ptolemeu.

In attachment gasiti problema mentionata si o solutie(a mea) a problemei lui Schmll.

[color=darkblue][b]Foarte interesant ![/b]

[b]Remarca.[/b] Notam $ I\in AC\cap BD$ si $ AB \equal{} a$, $ BC \equal{} b$, $ CD \equal{} c$, $ DA \equal{} d$, $ AC \equal{} e$, $ BD \equal{} f$.

Se arata usor ca $ \frac{IA}{ad}\equal{}\frac{IB}{ab}\equal{}\frac{IC}{bc}\equal{}\frac{ID}{cd}\equal{}\frac{AC}{ad\plus{}bc}\equal{}\frac{BD}{ab\plus{}cd}\equal{}\sqrt{\frac{\minus{}p_{w}(I)}{abcd}}\equal{}\lambda$,

unde $ p_{w}(I)$ este puterea punctului $ I$ fata de cercul circumscris $ w$ al patrulaterului $ ABCD$. Se observa ca

([b]relatia lui C.N. Schmall[/b]) $ PA^{2}\cdot [BCD]\plus{}PC^{2}\cdot [ABD] \equal{} PB^{2}\cdot [ACD]\plus{}PD^{2}\cdot [ABC]$

este echivalenta cu relatia $ \boxed{\ \frac{bc\cdot PA^{2}\plus{}ad\cdot PC^{2}}{bc\plus{}ad}\equal{}\frac{cd\cdot PB^{2}\plus{}ab\cdot PD^{2}}{cd\plus{}ab}\ }\ \ (*)$.

[b][u]Demonstratia sintetica a relatiei[/u][/b] $ (*)$. Notam $ IA \equal{} x$, $ IB \equal{} y$, $ IC \equal{} z$, $ ID \equal{} t$. Aplicam [b]relatia lui Stewart[/b] cevienei $ PI$ in :

$ \left\{\begin{array}{cc}\triangle PAC\ : & zPA^{2}\plus{}xPC^{2}\equal{} (x\plus{}z)PI^{2}\plus{}xz(x\plus{}z)\\ \\ \triangle PBD\ : & tPB^{2}\plus{}yPD^{2}\equal{} (t\plus{}y)PI^{2}\plus{}yt(y\plus{}t)\end{array}\right\|$ $ \implies$ $ \left\{\begin{array}{c}bc\cdot PA^{2}\plus{}ad\cdot PC^{2}\equal{} (bc\plus{}ad)\cdot PI^{2}\plus{}abcd(bc\plus{}ad)\cdot\lambda^{2}\\ \\ cd\cdot PB^{2}\plus{}ab\cdot PD^{2}\equal{} (ab\plus{}cd)\cdot PI^{2}\plus{}abcd(ab\plus{}cd)\cdot\lambda^{2}\end{array}\right\|$ $ \implies$

$ \frac{bc\cdot PA^{2}\plus{}ad\cdot PC^{2}}{cd\cdot PB^{2}\plus{}ab\cdot PD^{2}}\equal{}\frac{bc\plus{}ad}{ab\plus{}cd}\equal{}\frac{e}{f}$. Vezi problema propusa $ 8.60$ de la pag. $ 111$ din cartea[/color]

[color=darkred][b]Geometrie plana (sintetica, vectoriala, analitica). Culegere de probleme., Editura GIL, 2002[/b][/color]


Given that two triangles $ ABC,DEF$ are circumscribed in the same circle with $ O$ as its center, and the lines connecting the corresponding vertices of the triangles $ ABC,DEF$ concur at $ P$. Let $ Q_{1},Q_{2}$ be the isogonal conjugates of $ P$ with respect to triangles $ ABC,DEF,$ respectively. Show that $ \angle OQ_{1}P \equal{} \angle OQ_{2}P.$

I am sorry dear [b][size=100]Fang-jh[/size][/b] if I am mistaken but, I think that the result of $ \angle OQ_{1}P \equal{} \angle OQ_{2}P$ as the problem states, is not correct.

I see by the drawing that the circumcircles of the triangles $ \bigtriangleup OQ_{1}P,\ \bigtriangleup OQ_{2}P,$ are not congruent.

I can only see by the drawing, another interesting result :

[b][size=100][color=DarkBlue]'' The line segment $ OP,$ bisects the segment $ Q_{1}Q_{2}$ ''[/color][/size][/b]

Best regards, Kostas Vittas.

It's right！  give some Conclusion，  next time I wil give a proof .

（1）：If：$ OQ2\parallel{}PQ1$ or $ OQ1\parallel{}PQ2$，Then $ P,Q_1$ must be Circumcenter and orthocenter of triangles $ ABC$， and $ PQ1,OQ2$ are coincide .

（2）：Let：$ Q3$ be the isogonal conjugates of $ Q1$ with respect to triangles $ LMN$，then $ OQ2\equal{}OQ3$，$ Q2Q3\parallel{}PQ_1$，

$ angel OQ2Q3\equal{}|90\minus{}angel ABP\minus{}angel BCP\minus{}angel CAP|\equal{}m$，and also can prove $ angel OTP\equal{}m$，So $ ahgel ORP\equal{}angel OTP$，

So $ angel OQ2P\equal{}angel TPR\plus{}m\equal{}angel Q1OR\plus{}m\equal{}angel OQ1P$，$ angel OPQ1\equal{}angel OQ3Q1$，$ angel OPQ2\equal{}angel OQ1Q3$

通过作垂足三角形或用位似可以证明上面的结论。


Can anybody do a 4 by 4 rubik's cube? How long does it take you compared to a regular 3 by 3? Do the same general techniques work?

Er...Rubi[b]k's[/b] Cube...

Yeah.  The way I do it is just by first forming 2x2 centers and 2-edge edges, so I have something analogous to a 3x3x3.  Then I just solve that.

i can do 1 face of the 4x4, because it's like the first layer of hte rubiks cube. you make the middle edges first and then do the same ting as the 3x3 (make the edges and then the corners). I nver really tried the second layer, but maybe it's like the 2nd layer of th3 3x3 cube...

I need about five till seven minutes for a $5 \times 5 \times 5$ (don't have the smaller one), but that just guessed, I never stopped.

And the techniques for a normal one are not enough, but the techniques from a $4 \times 4 \times 4$ suffice to solve all bigger ones normally.

But you should try the $3 \times 3 \times 3 \times 3$ one in four dimensions :D

[quote="ZetaX"]But you should try the $3 \times 3 \times 3 \times 3$ one in four dimensions :D[/quote]

:rotfl: I heard that there was actually an applet that lets you do that... :?

In fact there's a beautiful 5-D one :P
http://www.gravitation3d.com/magiccube5d/

[quote="moogra"][quote="funcia"]wow i can't even do a 3*3*3 without directions(really a bit off a push but that means i can't do it by myself)[/quote]

well just try the algos a bit and you'll get them. IF you do it the beginner's way (I usually still use this) you can still get pretty fast.

Btw I tried the 4x4 cube, I got 3.5 layers :) for 5x5, I got 4 layers[/quote]

Doing layers on larger cubes will be extremely tough, trust me...

[quote="Ciaccona"]I expect that trying to solve by layers would be difficult. I've only had a 4x4 for 48 hours, but it seems easier to me to do the centers, pair the egdes, and then solve as for a 3x3.

Of course there's always parity errors...I just finished my first full solve, but I did two almost-solves this morning. Both times I had a double-edge flipped, and no algorithm for fixing it.[/quote]

yeah that's the suggested way. There are parity solutions though. I saw them

[quote="fireforce"]Hehe 4x4? I can do them in my sleep...

I almost brook a minute but it turned out that I had not oiled my Rubik's cube, all the torques built up pressure while I was turning so fast that it exploded...now I have no Rubik  :([/quote]

nice, I can only do the centers of the 4x4, pairing up edges takes too long and I didn't learn the algos.

I can do the 4x4x4, however, I’m quite slow at it; I solve the top layer, then the bottom, and then do the remaining edges and then the centers on the remaining four faces. Probably not the best way to go; my best time is something along the lines of 10 minutes. Well, I don’t really go as fast as I do on the 3x3x3, because my first 4x4x4 exploded a few turns from finishing my first solve because I was turning it way too fast.

I did it in 4 min. 38 seconds today! :D  :D  :D

wow.. :P

What was ur fastest time on the 4x4, moogra?

Oh, and now it's down to 4 min. 34 seconds!

A little under 2 but that was lucky (but with a bad cube). I'm getting a better one though.

is it eastsheen? my cube is an official rubik's.

was ur bad cube a rubik's cube?

[quote="moogra"]Yeah it's eastsheen, but it's the cheap one. I've never seen a really good rubik's. Well at least not as good as my 3x3. My bad cube was rubik's brand and it could turn ok but the centers kept breaking. I think 1 center died after only 3 months.[/quote]

I'm now down to 4 minutes! With a rubik's brand. :D

4 minutes is unranked. I'm averaging 2 minutes now. I could have gotten 1:20 something but I had 2 parities so it was lame. My final time was 1:37. 4x4 is hard

omg you're getting close to international level :P

I CAN DO IT IN 1 MINUTES AND 34 SECONDS!!!

you're better than moogra and you're international too :D

Hi Guy's ... Did anybody have a 4x4x4 Rubik's Revenge Cube solution ? ( In PDF file or u can email me at kofxi123@gmail.com ) Thank's a million.... :blush:  :P

...[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=295366[/url]
Please refrain from posting the same thing twice.


What is the sum of the prime factors of 85,085?

85085/85=1001. If you wish to do well at math competitions, you should know this factors to 7x11x13. 85 factors to 5x17. So the prime factors are 5+7+11+13+17=53.


Prove that for any x in R, we have:

sin (sinx+cosx)<cos(pi/24).

[b] Solution [/b]: because cos(pi/24)=sin(pi/2-pi/24)=sin (11pi/24) the ineq becomes like so:  sin (sinx+cosx) < sin (11pi/24) [b] (1) [/b].

because |sinx+cosx|<=sqrt(2) (Cauchy) and sqrt(2)<pi/2 we get that:
-pi/2 <sinx+cosx<pi/2. we notice that sinx+cosx<11pi/24 (sqrt(2)<11pi/24) and the function sinus is striclty increasing on [-pi/2,pi/2] we get that the ineq (1) is true.

So problem solved!! :D

[b]cheers! [/b] :D


Given sequence $ \{ c_n \}$ satisfying the conditions that $ c_0\equal{}1$, $ c_1\equal{}0$, $ c_2\equal{}2005$, and  $ c_{n\plus{}2}\equal{}\minus{}3c_n \minus{} 4c_{n\minus{}1} \plus{}2008$, ($ n\equal{}1,2,3, \cdots$). Let $ \{ a_n \}$ be another sequence such that $ a_n\equal{}5(c_{n\plus{}1} \minus{} c_n) \cdot (502 \minus{} c_{n\minus{}1} \minus{} c_{n\minus{}2}) \plus{} 4^n \times 2004 \times 501$, ($ n\equal{}2,3, \cdots$).

Is $ a_n$ a perfect square for every $ n > 2$?

hello, i think no, i have calculated $ a_3\equal{}5(c_4\minus{}c_3)(502\minus{}c_3\minus{}c_2)\plus{}4^3\cdot 2004\cdot501$ where
$ c_3\equal{}2004$
$ c_4\equal{}\minus{}4007$
and so is $ a_3\equal{}109428921\equal{}3^3\cdot7\cdot167\cdot3467$
and this is not a perfect square.
Sonnhard.


algebra
polynomial
coefficients
number theory
Divisibility
IMO Shortlist

Let $ f(x) \equal{} x^8 \plus{} 4x^6 \plus{} 2x^4 \plus{} 28x^2 \plus{} 1.$ Let $ p > 3$ be a prime and suppose there exists an integer $ z$ such that $ p$ divides $ f(z).$ Prove that there exist integers $ z_1, z_2, \ldots, z_8$ such that if \[ g(x) \equal{} (x \minus{} z_1)(x \minus{} z_2) \cdot \ldots \cdot (x \minus{} z_8),\] then all coefficients of $ f(x) \minus{} g(x)$ are divisible by $ p.$

Anyone has a good idea? I'm very confused about this problem.
I think the answer given in the Compendium is not right.

Please forgive me to replace some $\equiv$ with $=$.
  Notice that $f(z)=(z^4+2z^2-1)^2+32z^2$,and $p|f(z)$.We can conclude $p$ has quadratic residue $-1$,let $m^2=-1(modp),n^4=2(modp),q^4=-2(modp)$,then $((x-m)^4+q^4)((x+m)^4+q^4)=(x^2+1)^4+q^4((x-m)^4+(x+m)^4)+q^8=f(x)(modp)$.
  Take $z_1=m+n,z_2=m-n,z_3=m+mn,z_4=m-mn,z_5=-m+n,z_6=-m-n,z_7=-m+mn,z_8=-m-mn$,which could be easily checked satisfy the condition.Done


So I've never received a particularly satisfying answer to this question:

[i]Why are solubility products constant (at constant temperature, pressure, etc.)?[/i]

A solubility product is an equilibrium constant, and equilibrium constants are thermodynamic functions of the system's variables of state: temperature, pressure, volume, and so on (which, of course, are not independent). So, if all these are constant, so are equilibrium constants. Notice that equilibrium constants are defined more rigorously in thermodynamics as quotients of activities (raised to some exponent), not concentrations.


calculus integration ratio quadratics function

Let {x} = x - [x] denote the fractional part of x.
For example, {5.61} = 0.61, {2} = 0 and {-1.34} = 0.66.
Determine the number of real numbers x which satisfy :
(1) -100 :le: x :le: 100 ;
(2) {x} = {x}.

[hide]I believe that saying {x:^2:}={x} is equivalent to saying x:^2:-x is an integer. So we factor x:^2:-x as x(x-1). Okay, we know this works for all integral x, and we know it cannot work for irrational x. Let's see if it works for any rational x. Let x=a/b, gcd(a,b)=1.  Then we want b:^2:|a(a-b). Since (a,b:^2:)=1, this is equivalent to b:^2:|a-b, which cannot be true since (a,b)=1. So therefore it only works for integral x. (so the answer would be 201). Correct? Or did I mess up in my reasoning when it said it couldn't work for rational x?[/hide]

x can be irrational.  For example, x = golden ratio works.

Yep ... so think of the question again MT :)

[hide]I only counted x >= 0, since, probably, it is similar for x < 0 (and I'm too lazy   )

Let 0 <= a <= 100  IR, a = n + x with n  IN, 0 <= x < 1. a = n + 2nx + x

{a} = x, {a} = {2nx + x}

{x+y} = {x} + {y} or {x+y} = {x} + {y} - 1, for x,y >= 0

If k/2n <= x < (k+1)/2n, with 0 <= k < 2n, then {a} = 2nx - k + x or {a} = 2nx - k - 1 + x.

The solutions are then N + (-(2*n-1) + sqrt ((2*n-1)^2 + 4*k))/2 for 0 <= N <= 99, 0 <= k < 2*N

It makes 9900 of them.

Arn, is-it correct for >= 0 x  :?: [/hide]

Yep ... but it can be done in an easier way.

[hide]

If x and x have the same fractional part then x - x is in N. Consider the quadratic function f(x) = x - x. Since it is monotonous and strictly increasing on [0, 100], all the integer values between f(0) and f(100) will be assumed by f exactly once. For each positive integer m, the equation x - x = m has precisely one positive solution. Now f(0) = 0 and f(100) = 9900. So the 9901 possible values of m are 0, 1, 2, ..., 9900. The answer for 0:le:x:le:100 is 9901. [/hide]

[quote="Arne"]Yep ... but it can be done in an easier way.
[/quote]

Dude, that's really nice  :oops:  :!:

Arne wrote:Yep ... but it can be done in an easier way.

[hide]
If x and x have the same fractional part then x - x is in N. Consider the quadratic function f(x) = x - x. Since it is monotonous and strictly increasing on [0, 100], all the integer values between f(0) and f(100) will be assumed by f exactly once. Now f(0) = 0 and f(100) = 9900, so there are 9901 such values : 0, 1, 2, ..., 9900.  Hence the answer for 0 :le: x :le: 100 is 9901. [/hide]

agreed belenos. Wow I got a whole fifth of the way to the solution! sigh...


$2.3^{x}+x=1$
2.3^x+x is increse. Then this equation has only one root. How do u find it :(

Rewrite the equality as $2\cdot 3^{x}=-x+1$and sketch the graphs of $y=2\cdot 3^{x}$ and $y=-x+1.$

[quote="The Zuton Force"]Use the Lambert function $W(x) \; : \; W(x e^{x})=x$ and go through some plowing.

[hide="You should end up with"]
$x=1-\frac{W(6 \ln 3)}{\ln 3}$
[/hide][/quote]
Hm, any solution ??? :(. I don't unterstand W :(.


geometry
MATHCOUNTS
calculus
integration
number theory
relatively prime

which is the biggest number that can't be written like so 5x+8y? where x, y are natural!

59
23
27
717
1023?

This problem is an example of the "greatest sum not possible" problem.

A similar rewording of the problem would be something like this:

You throw darts on a dartboard.  The dartboard has two areas, one which is worth 5 points and one which is worth 8 points.  What is the largest score you CANNOT get?

One way to solve this is to make a list of all possible scores:
5, 8, 10, 13, 15, 16, 18, 20, 21, 23, 24, 25, 26, 28, 29, 30, 31, 32, 33, 34,...

If you were to extend this list a lot, you'd see that every positive integer after this can be made.  But for our purposes, this should do.  The biggest number not on this list is 27. <-- answer

I believe there is a formula for this, which goes:
The greatest number that can't be written in the form of ax + by for natural x, y, is ab - a - b.

In this case, we have 5*8 - 5-8 = 27

I think you are reffering to Sylvesters theorem, It says:
If a and b are positive integers (a,b) = 1, consider the set S of integers {ax+by} where x , y range over all positive integers. Then the set S contains all integers greater than c = ab - a - b but not c itself. I believe that it's hard to prove this.

I don't remember it being toooo hard, but I'm not sure. First you'd have to prove that ab-a-b cannot be expressed in such a way, that equals a(b-1) + b(-1) and since b and a are coprime, you have to keep adding multiples of b to the a one and the opposite to the b bit. Anyway, to make the b number positive you'd have to make the a number negative. Very formal speech that is. But then you have to prove (ab-a-b)+1, all the way up to (ab-a-b)+a are expressible, thats the hard bit..

[quote="MysticTerminator"]It's not that hard. This sort of problem was on Mathcounts a lot, so in my 8th grade year, I got tired of having to solve it individually each time and actually sat down, and dervived that formula. So trust me, it's not that hard.

BTW, just my luck, that sort of problem did not show up on nats that year, which is what I had been preparing for.[/quote]

I was referring to proving sylvesters teorem. If you have a proof of sylvesters theorem please post it here, I would like to see it.

I'm sorry; I wasn't replying to you, but I think the my replies equally well to your post, so it's my fault, shoulda been more clear, sorry.

I'm afraid I didn't actually prove the formula, but if you want it, here is my thinking on the matter (as best as I can recall):

Let's start with a concrete example, i.e. 5 & 8. Well, then, for every number that can be expressed as 5x+8y, I will express it as the ordered pair (x,y) (there may be more than one ordered pair for each number).

4*8=32 is obviously in S. So is 4*8+1=33, because there are 5 nonnegative multiples of 8 less than 33, and they all have different residuals mod 5, so at least one of those residuals will be 0, therefore it is in S. You can use the same argument for any number>32.

Now, any number that can be expressed as an ordered pair (x, y) can also be expressed as (x-8k,y+5k), where k is any integer, as long as this does not make those values negative.

Ok, now obviously, the ordered pair of 32 will have x=0. The ordered pairs of the 4 numbers (namely 33-36) greater than 32 will not have x=0 (since none of them are multiples of 8). Therefore, if we can subtract 5 from each of them, and the resulting number will still be in S (since we can just subtract one from the x-value of their ordered pair). Therefore, we know that 28-31 are all in S.

So far, we have proved that every number greater than 5*8-5-8 is is S. We just need to show that the number itself is [b]not[/b] in S, and we will be done.

Ok, now if a number can be expressed as the order pair (x, y), it can also be expressed as (x+8k, y-5k) for integral k. Now, the number 5*8-5-8 can be expressed as (7+8k, -1-5k) for integral k. However, it will be in S iff there exists a k such that both those values are positive. There obviously is not, therefore 5*8-5-8 is not in S, and we are done.

This can be generalized to any relatively prime a and b.

I'm sorry, I've never been good at explaining, but this should give you my general idea. Of course, there may be some flaw, it's been a while since I thought about this; if there is, please point it out.

[quote="MysticTerminator"]This sort of problem was on Mathcounts a lot, so in my 8th grade year, I got tired of having to solve it individually each time and actually sat down, and dervived that formula. So trust me, it's not that hard.

BTW, just my luck, that sort of problem did not show up on nats that year, which is what I had been preparing for.[/quote]

*groans*  As I was prepping for mathcounts nationals, my dad and I saw this too.  I admit my thinking wasn't half as deep, more like "let's find a pattern here - oh, this might work".  But I memorized that formula too.  And sadly, it hasn't been on national targets since 1998.  :(


[quote="rajudragon"]This is my first time doing USAMTS and I was wondering whether the problems were arranged in progressive difficulty. As I was doing them it seems that they are except for parts of some.[/quote]

I believe they are for the most part problem 5 on a round will generally be harder than problem 1 and round 4 would be harder than round 1

I will never think of tetrahedrons the same way again  :wallbash:

EDIT: I have a question: if you have figured out the steps to figure out the solution but you don't know how to do a key step, can you say what you would have to do from there and provide plenty of reasoning and still get a few points?  Would it be good to include a guess of some sort?

the tetrahedron problem has boggle with my mind for days

please do not talk about ANYTHING related to the round before permission is granted.


Find all continuous functions, such that: $ g: [0,1]\rightarrow R,$ ${ g(x)=\frac{1}{3}g(x^2)+\frac{1}{4}g(1-x^2})$.

For x=0, nad x = 1 we taking a system that give us, g(0)=g(1)=0
Since g is continuous at [0,1] then has a maximum M and a minimum value m
So
$ m \le g\left( x \right) \le M,\forall x \in \left[ {0,1} \right]$

If we put x=0 we take $ m \le 0 \wedge M \ge 0$
If we put $ x \equal{} {x^2}$, then we will take $ m \le g\left( {{x^2}} \right) \le M \Leftrightarrow \frac{m}{3} \le \frac{1}{3}g\left( {{x^2}} \right) \le \frac{M}{3}$
Similarly we take $ m \le g\left( {1 \minus{} {x^2}} \right) \le M \Leftrightarrow \frac{m}{4} \le \frac{1}{4}g\left( {1 \minus{} {x^2}} \right) \le \frac{M}{4}$
Adding
$ \frac{{7m}}{{12}} \le g\left( x \right) \le \frac{{7M}}{{12}}$
Since
m is minimum value, then $ m \ge \frac{{7m}}{{12}} \Leftrightarrow m \ge 0$
Similarly $ M \le 0$
That means m=M=0 and function g(x)=0


[hide="i"]
Let's manipulate $x^{y}= y^{x}$ a bit:
$x^{y}=y^{x}\iff y \log x = x \log y \iff \frac{ \log x}{x}= \frac{ \log y}{y}$ for non-negatives $x$ and $y$, which I assume we're dealing with.

Plugging this in: $\frac{ \log \left( \frac{n+1}{n}\right) \cdot n}{ \left( \frac{n+1}{n}\right)^{n}}= \frac{ \log \left( \frac{n+1}{n}\right) \cdot (n+1) }{ \left( \frac{n+1}{n}\right)^{n+1}}\iff \frac{n}{1}= \frac{ n+1}{ \left( \frac{n+1}{n}\right)}\iff n=n$, which obviously is true. [/hide]

[hide="ii"] Straightforward by an induction argument and $(n+1)^{2}= 2(1+2+\cdots+n)+(n+1)$.[/hide]


topology
real analysis
real analysis unsolved

Let
\[{\bar D}\]
be the closed unit disc in the complex plane. Let g be a continuous mapping of
\[{\bar D}\]
into the unit circle T.

Prove that g(z) = -z for at least one z in T.

If not, then $g_{t}(z)=tg(z)+(1-t)z$ is a homotopy between $g_{|\mathbb T}$ and $\mathrm{id}_{|\mathbb T}$ in $\mathbb C\setminus \{0\}$. But $g_{|\mathbb T}$ is nullhomotopic in $\mathbb C\setminus \{0\}$.

I am not familiar with the term homotopy or nullhomotopic, can you expand a little on your reasoning for why that works with this problem. Thanks.


linear algebra
matrix
quadratics
algebra
system of equations

For each $a \in \mathbb{R}$ determine all pairs $(x;y) \in \mathbb{R}^2$ for which the system of equations:

$
\begin{gather*}
\begin{matrix} x^2 + y^2 = 25 \\ x + y = a \end{matrix}
\end{gather*}
$

is satisfied.

Quote:For each  determine all pairs

 for which the system of equations:  is satisfied.[hide]

Substitute y = a - x  into x2 + y2 = 25 . . . x2 + (a - x)2 = 25

which simplifies to the quadratic: 2x2 - 2ax + (a2 - 25) = 0

which has roots: x = [a  sqrt(50 - a2}]/2

and hence: y = [a  sqrt{50 - a2}]/2

These are the real roots when |a| < 5sqrt{2}

[Complex roots otherwise.][/hide]


trigonometry
inequalities
inequalities unsolved

$ f(x) \equal{} \frac{7\minus{}3 \cos x\minus{}6 \sin x}{9\minus{}3 \cos x\minus{}8 \sin x}$

Prove that $ \frac{1}{2} \leq f(x) \leq 1.$

[quote="jimi"]$ f(x) \equal{} \frac {7 \minus{} 3 \cos x \minus{} 6 \sin x}{9 \minus{} 3 \cos x \minus{} 8 \sin x}$

Prove that $ \frac {1}{2} \leq f(x) \leq 1.$[/quote]
$ f(x) \le 1 \iff \sin x \le 1$ which is true.
$ f(x) \ge \frac {1}{2} \iff 5 \ge 3 \cos x \plus{} 4 \sin x$ Which is true by cauchy schwartz: $ 5 \equal{} \sqrt {\cos^2 x \plus{} \sin^2 x}\sqrt {3^2 \plus{} 4^2} \ge 3 \cos x \plus{} 4 \sin x$

Is it a basic inequality problem in trigonometry? I suppose to.
By the way, we can solve this problem algebraically.
We have
$ asin(x) \plus{} bcos(x) \equal{} c$ holds the important inequality
$ |c| \le \sqrt {a^2 \plus{} b^2}$

P/s: to Mathias DK, it seems your solution has some critical mistakes! Can you guess what..? Be careful with  the signs of the inequality (+) or (-)
.Anyway, it is good when you gave it a try

If $ f(x)$ takes some real value $ k$, the point $ (\cos x,\ \sin x)$ on the unit circle is satisfies the equation $ \frac {7 \minus{} 3X \minus{} 6Y}{9 \minus{} 3X \minus{} 8Y} \equal{} k$ and $ X^2 \plus{} Y^ 2 \equal{} 1$ on the $ XY$ plane, that is to say, the line $ l: 3(k \minus{} 1)X \plus{} 2(4k \minus{} 3)Y \minus{} (9k \minus{} 7) \equal{} 0$ intersects the circle. Thus the condition is given that the distance between the center $ O(0,\ 0)$ and the line $ l$ is less than or equal to the radius $ 1\Longleftrightarrow \frac{|\minus{}(9k\minus{}7)|}{\sqrt{\{(3(k\minus{}1)\}^2\plus{}\{2(4k\minus{}3)\}^2}}\leq 1$ arranging this, we have $ 2k^2 \minus{} 3k \plus{} 1\leq 0$, yielding $ \frac {1}{2}\leq k\leq 1$. Q.E.D.

[quote="great math"]P/s: to Mathias DK, it seems your solution has some critical mistakes! Can you guess what..? Be careful with  the signs of the inequality (+) or (-)
.Anyway, it is good when you gave it a try[/quote]
Don't be patronizing!
Instead check that $ 9 > \sqrt {73} \equal{} \sqrt {\cos^2 x \plus{} \sin^2 x} \sqrt {3^2 \plus{} 8^2} \ge 3 \cos x \plus{} 8 \sin x$
And $ 7 > \sqrt {45} \equal{} \sqrt {\cos^2 x \plus{} \sin^2 x} \sqrt {3^2 \plus{} 6^2} \ge 3 \cos x \plus{} 6 \sin x$

So please tell me if there really is a mistake  :!:


conics
parabola
inequalities unsolved
inequalities

Let a,b be positive real numbers satisfy [tex](a+b)^2= \frac{a}{b}[/tex].
Show that:
[tex](a-2)^2 \ge 2b-1[/tex]

hmm...this is somewhat weird:

rewrite the given condition as a 2 b + a(2b 2 - 1) + b 3 = 0. so we must have (2b 2 - 1) 2 - 4b 4 >=0 (otherwise equation doesn't hold since the upward-opening parabola doesn't intersect with x-axis.) This reduces to 2b - 1 <= 0 since 2b + 1 > 0. Hence result.  :D


What does 1/n(A1+..+An) mean? Is it (A1+..+An)/n or 1/(n*(A1+..+An)) ? I mean: is the sum  \sum Ai at the numerator or denominator?

My belief is that we're in the first situation, because otherwise it would be wrong: we take A1=1 and all the other A's really large and we obtain a contradiction. However, I thought I should ask.

It means (A1+....+An)/n. Sorry for the lack of precision in the test.

The given inequality is equivalent with (1+(a1+a2+...+an)/n)^n-(1-(a1+a2+...+an)/n)^n \geq (1+a1)(1+a2)...(1+an)-(1-a1)(1-a2)...(1-an), or after computations, 2[ (n,1)(S1/n)+(n,3)(S1/n) 3+(n,5)(S1/n) 5+...]\geq 2(S1+S3+S5+...), where Si denotes the sum of all products of ak, k=1,...,n taken i at a time, and (n,k) denotes the binomial coefficient.  We only have to prove now that (n,2k+1)(S1/n) 2k+1\geq S(2k+1), or (S1/(n,1)) 2k+1\geq  S(2k+1)/(n,2k+1), which is true by the MacLaurin inequalities.


function
trigonometry
calculus
integration
calculus computations

This was confusing me so any help or explanation of how this type of problem is solved would be appreciated.

Find the function y(t), given that:

y' = e^-y * cos(t)

and y(0) = 0.

Also, verify that the solution satisfies both the differential equation and the initial condition.

You can use the separation of variables.

$y'=e^{-y}\cos t\Longleftrightarrow e^{y}y'=\cos t.$

Integrate both sides, $e^{y}=\sin t+C.$

Well, you haven't used the initial condition yet, so it's rather unlikely that your solution will satisfy it.   :wink:   The initial condition allows you to solve for $C$; after you do that, you'll have a proper function (well, sort-of; at least, you'll get a proper function defined on $(-\pi, \pi)$), and you can then verify (i.e., plug in) to the differential equation to make sure it works.

(Edit: when I wrote this, the original post had been edited to include a partial solution on which I was commenting; that has since disappeared.)

Well, this is what I did to find what y is equal to:

dy/dt = e^-y cos(t)
dy/dt = 1/e^y * cos(t)
dy/dt = cos(t)/e^y
integral of e^y dy = integral of cos(t) dt
e^y = sin(t) + C
taking the natural log of both sides, we get ln(e^y) = ln(sin(t)+C)

So, if you plug in 0 for t and 0 for y, you get C = 1.

Using this, how would one find verify the solution?

Okay, so the thing I was responding to now appears after my post ;-)

So, you have $C = 1$.  Then you have the function $y$.  Now you must verify that your function satisfies the original conditions, i.e. calculate $y'$, etc., etc.

Incidentally, the question would be much nicer if you had $y(0) = c$ for any $c > 0$.


[quote="leftblame"]I have a problem, I am trying to find the equation that produces this output.

x= 0     y = 100;
x=0.5    y = 66 & 1/3;
x=1      y = 50;
x= 1.5   y = 40;
x=2      y=33 & 1/3;
x=2.5    y= (approx.)28.58
x=3       y= 25;
x=4       y=20;
x=5    y=16 & 2/3;
x=6    y= (approx.) 14.29;
x=7    y=12.5;
x=8    y = (approx.) 11.12;
x=9    y = 10;

Can someone find the equation that produces these values and also solve x=10 y= ? ?

Thanks and goodluck![/quote]

$f(x) = \frac{100}{x+1}$
$f(10) = \frac{100}{11}\approx 9.09$

Excellent! That was quick. I appreciate the help Treething.. you're awesome!

[quote="myspacesucks"]easier, use the vandettian formula.[/quote]

what is that? a google search yields nothing...


I am currently an undergraduate student. I am going to attempt the GRE in almost 3 weeks time; I need some suggestions/ advice regarding preparation, I am pretty confident of securing a good score, but I would to ask people in this community to help me streamline my preparation by providing valuable advice/suggestions. ( with respect to only the Verbal ability section, AWA section).

I managed to learn almost all the 'famous' word lists of Barron's.

Can anyone who secured a very high score on the GRE advise me how he/she went about in the last 3 weeks of his/her preparation? What are the steps one should follow to secure a  1500+ GRE score ( in the last 3 weeks time)?

I ask all the members who are willing to help me to post their suggestions/advices over here; The advices should be only with respect to the verbal ability section and if possible, the AWA section.

[b] TO THE MODERATOR/MODERATORS: I ask them not to "shift" this post somewhere else because I ain't getting any constructive/productive replies, so please don't do it!

[/b]

I thank everyone in advance.

I would not spend too much time preparing for the verbal section. I took the test over the summer and I thought the verbal section was downright senseless. I hate how they find the most ridiculous, useless words and try to make you differentiate fine shades of meaning between them or test you on them completely out of context like in analogies. I admit I spent some time memorizing definitions from a list of "GRE words" and I wish I had done less of that. There is such a vast collection of obscure words that it is almost futile to try learn any significant portion of them in a crash course before you take the GRE.

I thought the reading comprehension questions were better than the vocabulary questions, but still they were really annoying. On some of them I felt like I could easily make an argument for why each of the choices is the "best".

So, basically, I think it is a bad idea to try to "master" the verbal section. I think if you have in general been doing your work in school and have a healthy interest in knowing how to read and understand text, then you should do fine on the test. But you should do some practice exams to get an idea of what kind of questions are asked.

The Analytical Writing Assessment (AWA) section was actually a good assessment, I thought. Learning how to write critically or analyze an argument is an important skill if you are going to graduate school. Also, it involves a lot of the problem-solving skills that are stressed here on AOPS such as organizing the given data in your mind and building an argument based on that. Unlike the verbal section, you are not bound to one of five choices all the time, and can really let your creativity flow.

The only thing I did to prepare for the Writing section was to read the sample essays in the thing you download from the GRE website. Again, if you have you've written essays in high school English or in liberal arts classes in college, the Writing section should not require much preparation. If you still want to practice, I think a good way to do that would be to just write lots of responses to threads here at "The Round Table". That will give you some motivation to write well and perhaps get some responses to your writing. I thought about writing out responses to some of the prompts for practice, but I realized that since no one would read them, I wouldn't have much idea how good my responses were and I would probably get lazy as a I wrote them.

As long as you do reasonable, I think your GRE scores, and especially your verbal ones, are a pretty small factor in graduate school admissions. The admissions agents are hopefully aware of how silly some of the verbal questions can be. So, don't let studying for the GRE take over your life.

And don't waste too much money on test-prep books published by private companies. I think its really abusive how much those companies take advantage of the poor high school and college kids trying to make sure they get into a good college or graduate school.

I had more to stay about this than I thought. :lol: Anyways, good luck.

Okay, I THANK you for your suggestions, besides, I have not purchased any study-material, etc  :lol: ; I just borrowed some from my sister, but since the verbal score plays an important role in getting admissions to a putative school, I was more focused on it, and yes, some of the questions that crop up on the GRE can be termed as 'risible' as well as ridiculous.

Lol, One of the questions in an analogy I had to answer in some practice test was distinguishing the finer shade of meaning of 'Speed' which also means --  'an amphetamine' drug!.

I am asking this out of curiosity, what was your GRE score?

For what its worth, I got 800 on math, 670 on verbal, and I don't remember my writing score.

I only got a 510 on the verbal, and basically every school I had a shot at getting into accepted me. The two that rejected me probably had much better reasons to reject me than a silly standardized test on random words. I did do the memorizing thing, but it didn't help. You could spend months memorizing all the "top" or "most frequent" words, but in the end you could end up seeing none of them. I think the best thing is to learn techniques about how to answer the questions in the way they want, and not to do much more than that.

Yeah Hilbert, your suggestion is correct, between is it true that the threshold of GRE score at MIT is 1500?

I would appreciate if I receive more response, advice/suggestions to my query, anyone there to provide their suggestions?  :lol:

First of all, you gain nothing by worrying about your chances for admission to a certain school. At most, you would work yourself into a panic and perform less than your best in the GRE. Your future is not doomed if you do not get into your choice school, and the GRE score may not even be the deciding factor for your admission.
I would say just get familiar with the current GRE format by looking at sample tests. That would eliminate the surprise factor. On a short advance notice, there is not much more that you could do.
If you participate in and enjoy the math portions of this forum, you will get about 800 in the math part.
The verbal part will measure how you have enjoyed reading over the years and how wide your vocabulary is. If you know a foreign language (Spanish, Latin, Italian), you will benefit from cognates. If you "savor" the words while you are reading, and read a large quantity and variety of materials, you probably will get a high score. If you don't like reading, or if you mostly read for meaning, figuring that you "caught the drift" without paying much attention to the words and nuances in the shades of meaning, you are at a disadvantage. The way you read has been part of you for years. You cannot improve too much on what you've learned from a lifetime of reading in just a few weeks.
One strategy you can use is to start paying attention to the words, and getting into the habit of analyzing unknown words for clues to their meaning. See if you can identify a root of an unknown word from the similarities with other words you know in any language. Figure out if it could be a noun from the ending (e.g. if it ends in "age" or "dom" or "ation" it may be a noun). You can practice "word meaning guessing" on a sample test.
Also, get used to the stupid way the test makers think when asking questions. In my time, for "analogies" they thought in terms of filling the blanks. If [b]house:wood[/b] was the "model" given, they were thinking of "A [b]house[/b] can be made of [b]wood[/b]." Then, any pair of words that could be substituted in the same sentence and made sense was a reasonable answer. It may not seem logical to you, but you get points by thinking the way the test makers do.

Can anyone provide some suggestions regarding Analogies, and Reading comprehension in the GRE? I feel these two sections of Verbal Ability are hard.

What kind of grad program do you want to attend?  Different programs place different levels of weight on the GRE.  For example, math PhD programs don't care a lick about GRE verbal scores.  I guess they don't want to admit illiterates, but beyond that, they really don't care.  It might matter more for international students, but that's what TOEFL is for.


function
integration
calculus
real analysis
real analysis unsolved

prove: any monotonic function is the sum of a singular fuction and an absolutely continous function

If $ f(x)$ is the monotonic function, then the absolutely continuous function is  $ \int_a^x f(t) \, dt$


quadratics
absolute value
number theory unsolved
number theory

for $ a_{1}\equal{}a_{2}\equal{}0,a_{3}\equal{}9$ and $ a_{n\plus{}3}\equal{}\frac{a_{n\plus{}2}\plus{}a_{n\plus{}1}\plus{}a_{n}}{3}$
prove that exist $ lim$ {$ a_{n}$}

The characteristic equation is $ 3x^3 \equal{} x^2 \plus{} x \plus{} 1$ which has a root $ x \equal{} 1$.  We can factor it as $ (x \minus{} 1)(3x^2 \plus{} 2x \plus{} 1)$.  Clearly the roots of the quadratic have absolute value less than $ 1$.  The conclusion follows.

Alternately:  we can show that $ d_n \equal{} \text{max}(a_{n \plus{} 2}, a_{n \plus{} 1}, a_n) \minus{} \text{min}(a_{n \plus{} 2}, a_{n \plus{} 1}, a_n)$ is strictly decreasing.


Prove that $ \left (\sum_{i\equal{}0}^m a_ix^i\right) \left(\sum_{j\equal{}0}^nb_jx^j\right)\equal{}\sum_{r\equal{}0}^{m\plus{}n}\left(\sum_{k\equal{}0}^ra_kb_{r\minus{}k}\right)x^r$

[quote="mathwizarddude"]Prove that $ \left (\sum_{i \equal{} 0}^m a_ix^i\right) \left(\sum_{j \equal{} 0}^nb_jx^j\right) \equal{} \sum_{r \equal{} 0}^{m \plus{} n}\left(\sum_{k \equal{} 0}^ra_kb_{r \minus{} k}\right)x^r$[/quote]

$ \left (\sum_{i\equal{}0}^{m}a_ix^i\right )\cdot\left (\sum_{j\equal{}0}^{n}a_jx^j\right )\equal{}\sum_{i\equal{}0}^{m}\left (a_ix^i\sum_{j\equal{}0}^{n}b_jx^j\right )\equal{}\sum_{i\equal{}0}^{m}\sum_{j\equal{}0}^{n}a_ib_jx^{i\plus{}j}\equal{}\sum_{r\equal{}0}^{m\plus{}n}\sum_{i\plus{}j\equal{}r}a_ib_jx^{i\plus{}j}\equal{}\sum_{r\equal{}0}^{m\plus{}n}\sum_{k\equal{}0}^{r}a_kb_{r\minus{}k}x^r.$

It's basically saying that $ (i,j)$ is uniquely determined by $ (i,i\plus{}j)$


a wiec zadanie brzmi nastepujaco:
zaba skacze po osi liczbowej, zaczynajac od punktu n, gdzie n jest dodatnia liczba calkowita. i skacze do n+1 lub do n-1. natomiast przy kazdym pozniejszym ruchu, pewien potwor, obraca zabe w taki sposob, ze nie wie ona w ktora strone skacze ( ale zawsze albo o jeden w ta albo we wte). jesli zaba doskoczy do 0, to wraca do punktu n, i znow rozpoczyna sie ten proces. jesli k jest liczba calkowita wieksza od n, jaka jest srednia liczba ruchow potrzebna do osiagniecia punktu k?

Poszukaj coś może pod hasłem ,,błądzenie losowe na prostej" :)
A tak ogólnie to zadanie jest niedoprecyzowane, trzeba by jeszcze podać to, jak ten potwór żabę obraca, tj. odpowiedzieć na pytanie, jakie jest prawdopodobieństwo, ze obróci ją w lewo, oraz jakie, ze w prawo.


modular arithmetic
LaTeX
number theory
least common multiple

I was wondering how do you do some problems that go like:
If a number is divided by
2, remainder is 1
3, remainder is 2
4, remainder is 3
5, remainder is 4.
What is the original number?
I am wondering about particularly this question.

If possible, could you please explain how it works for other similar questions such as:
If divided by
4, remainder is 1
5, remainder is 3

I heard that there was an equation for these sorts of problems.  Could you explain the equation and how to use it?
Thanks.

Have you heard of modular arithmetic?

May want to get familiar with the notation first, http://en.wikipedia.org/wiki/Modular_arithmetic

I'll show you how to solve your second problem.

Set up the equations, calling $x$ the number(s) you're looking for:
$x \equiv 1 \pmod 4$ ... (1)
$x \equiv 3 \pmod 5$ ... (2)

From the first equation,

$x = 4k+1 \text{\,\,\,for\,integer\,k}$

Subbing into equation (2).

$4k+1 \equiv 3 \pmod 5$
$4k \equiv 2 \pmod 5$
Multiplying by 4, the inverse of 4 in mod 5:
$4\cdot 4 k \equiv k \equiv 2\cdot 4 \equiv 3 \pmod 5$
Therefore, $k = 5a+3 \text{\,\,\,for\,integer\,a}$

So, $x = 4k+1 = 4(5a+3)+1 = 20a+13$

So one solution is 13, another is 33, another is -7 and so on..

How do you leave large gaps in latex, so i can write $a = b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (2)$?

edit: damn, im no teacher..  :oops:

so would you use the same method for:
Prove that $2^{55}+1$is divisible by 3?
So how would you make the equation to start off with?
Maybe first find $2^{55}mod 3$?

$2^{1}\equiv 2 \pmod 3$
$2^{2}\equiv 1 \pmod 3$
$2^{3}\equiv 2 \pmod 3$
$2^{4}\equiv 1 \pmod 3$

Can you see the pattern?
$2^{2k+1}\equiv 2 \pmod 3$

So,
$2^{55}\equiv 2 \pmod 3$
$2^{55}+1 \equiv 3 \equiv 0 \pmod 3$

So $2^{55}= 3k+0$ meaning it's divisible by 3.

[quote="max1546"]
If a number is divided by
2, remainder is 1
3, remainder is 2
4, remainder is 3
5, remainder is 4.
What is the original number?
I am wondering about particularly this question.

[/quote]
[hide]
To solve one like this, you can instead note that it is [b]1 less[/b] than a multiple of 2, 3, 4, and 5. So find the lcm of 2, 3, 4, 5, and you can subtract 1 to get the answer (59)[/hide]


Given that $ x^3\equal{}y^3\plus{}z^3\plus{}3xyz$, show that $ x\equal{}y\plus{}z$

[quote="Sashsiam_2"]Given that $ x^3 \equal{} y^3 \plus{} z^3 \plus{} 3xyz$, show that $ x \equal{} y \plus{} z$[/quote]

Consider $ (\minus{}1,1,1)$.  :wink:

[hide]
This uses the pretty well-known factorization $ 0 \equal{} \minus{}x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz \equal{} \frac 12(\minus{}x \plus{} y \plus{} z)((x\plus{}y)^2 \plus{} (x\plus{}z)^2 \plus{} (y\minus{}z)^2)$, so either $ x \equal{} y\plus{}z$, or $ \minus{}x \equal{} y \equal{} z$.
[/hide]

hello, note that
$ x^3\minus{}y^3\minus{}z^3\minus{}3xyz\equal{}(x\minus{}y\minus{}z)(x^2\plus{}y^2\plus{}z^2\plus{}xy\minus{}yz\plus{}xz)$.
Sonnhard.

[quote="Sashsiam_2"]Given that $ x^3 \equal{} y^3 \plus{} z^3 \plus{} 3xyz$, show that $ x \equal{} y \plus{} z$[/quote]

Well, if we cube both sides we get x^3 = y^3 + z^3 + (3yz^2 + 3zy^2)

So apparently the contents inside the parenthesis has to equal 3xyz which we can not guarantee, so it is not true. Isn't it?

[quote="t0rajir0u"]Try again.  $ 3yz^2 \plus{} 3zy^2 \equal{} 3xyz$ is equivalent to $ 3yz(x \minus{} y \minus{} z) \equal{} 0$.  Of course, the problem with that approach is that you're starting from the conclusion you want.[/quote]

If 3yz(x - y - z) = 0, then that means that either y, or z is equal to zero, OR x = y + z. So in order to prove that x = y + z, we need to prove that y or z are not necessarily equal to one. I think that is right.