Sa se construiasca cu ajutorul unei rigle si al unui compas triunghiul [b]ABC[/b] daca se stiu pozitiile a trei puncte: [b]D[/b] , mijlocul inaltimii [b]AH[/b], [b]E[/b], mijlocul inaltimii [b]CP[/b] si [b]M[/b] mijlocul laturii [b]AC[/b]

@mihneaA: double post :wink: Ar trebui sa stergi unul din ele
Problema pare interesanta :)

[hide="cheia problemei"]diametralul opus lui $M$ in cercul $MDE$ este ortocentrul lui $\triangle{ABC}$ si unghiul $\angle{DME}=\angle{ABC}$, de aici constructia se usureaza considerabil.[/hide]

[hide="solutie completa"]acum deci se cunosc punctele $D$, $E$, $M$ si $X$, cu $X$ ortocentru, deci se cunosc dreptele suport ale inaltimilor din $A$ si $C$, adica $DX$, respectiv $EX$., dar $M$ mijloc lui $AC$, deci putem gasi $A$ si $C$: fie $Y$ simetricul lui $X$ fata de $M$, si astfel prin $Y$ ducem paralela la $DX$ care taie $EX$ in $C$, si paralela la $EX$ care taie $DX$ in $A$ (fapt dovedit de faptul ca $AXCY$ este paralelogram , prin constructie).

si s-a cam terminat acum, ducem paralele din $A$ respectiv $C$ la $EM$ , respectiv $DM$ , care se vor intersecta in $B$. astfel se respectva si conditia ca $D,E$ sunt mijloace, caci sunt acelea linii mijlocii, si ca $\angle{ABC}=\angle{DME}$.
PS: in rezolvare am folosit ca $DM \perp DX$, $EM\perp EX$ , rezultat evident.[/hide]


Find the sum of all 3-digit natural numbers which contain at least one odd digit and at least one even digit.

denote them by $ a_1a_2a_3$
now try to find how many times each integer appears as  $ a_1$ , $ a_2$ and $ a_3$ the rest is easily done since $ a_1a_2a_3\equal{}10^2a_1\plus{}10a_2\plus{}a_3$

Sum of all $ \ 3$ digit numbers $ \equal{}\frac {900}{2}(100\plus{}999)\equal{}494550$

Sum of all $ \ 3$ digit numbers in which all digits are even is

$ 10^2 \times 25\times (2\plus{}4\plus{}6\plus{}8)\plus{}10 \times 20\times (2\plus{}4\plus{}6\plus{}8)\plus{}20\times (2\plus{}4\plus{}6\plus{}8)\equal{}54400$

Sum of all $ \ 3$ digit numbers in which all digits are odd is

$ 10^2 \times 25\times (1\plus{}3\plus{}5\plus{}7\plus{}9)\plus{}10 \times 25\times (1\plus{}3\plus{}5\plus{}7\plus{}9)\plus{} 25\times (1\plus{}3\plus{}5\plus{}7\plus{}9)\equal{}69375$

Hence required sum is

$ \equal{}494550\minus{}54400\minus{}69375\equal{}\boxed{370775}$
 :)

I got 370775 as the answer too
I summed up all 3 digit numbers and subtracted all numbers with 3 even and 3 odd.
I did by listing down every number with 3 even/odd digits out of each 100 numbers i.e 100 - 199 n 200 - 299 n so on in the RMO anwer sheet. Then for summing up I made into an AP with additional numbers n summed it up. Then I subtracted from TOTAL sum.

The question is will I get Marks for this trial n error sort of method?

[quote="aadil"][quote]Hey makar i am new to all this combinatorics so could you help an ignorant guy like me understand how you summed up all those numbers with only odd digits. It would really help. thanks
I skipped that problem anyway on the exam[/quote]
required sum of numbers= sum of all 3 digit numbers-sum of 3 numbers with only odd digits-sum of 3 digit numbers with only even digits[/quote]

see i got that far i just didnt get how to calculate the sum of all the numbers with only odd digits and only even digits. :oops:

arrange the numbers with only even digits in a column.there are 4*5^2 such numbers.now if you consider the sum of the one's digits,it will contain the same no of 2s,4s,6s,8s,0s.therefore the sum is 4*5(2+4+6+8+0).similarly the sum of the ten's digits is 4*5(2+4+6+8+0) and sum of the hundred's digits is 5*5(2+4+6+8).
so the number is 54400

[quote="Manmather"]Hey makar i am new to all this combinatorics so could you help an ignorant guy like me understand how you summed up all those numbers with only odd digits. It would really help. thanks
I skipped that problem anyway on the exam[/quote]

To sum all three digit number in which all digits are odd:

Odd digits can be $ \  1,3,5,7,9$

Now calculate how many times 1 will occur at unit place  :idea:
As $ \ 100^{th}$ place can be filled in $ \ 5$ ways (by using 1 or 3 or 5 or 7 or 9) and $ 10^{th}$ place also can be filled in $ \ 5 $ ways $ \implies 1$ will occur at unit place $ 5\times 5\equal{}25$ times

Similarly $ \ 3,5,7,9$ each will occur $ \ 25$ times at unit place

Hence sum of digits at unit place in all such numbers will be $ 25\times (1\plus{}3\plus{}5\plus{}7\plus{}9)$ :idea:
Similarly we can find sum at $ \ 10^{th}$ place and $ \ 100^{th}$ . obviously it will be same as  $ 25\times (1\plus{}3\plus{}5\plus{}7\plus{}9)$

Finally sum of all such numbers in which all digits are odd is

$ 10^{2}\times 25\times (1\plus{}3\plus{}5\plus{}7\plus{}9)\plus{}10\times 25\times (1\plus{}3\plus{}5\plus{}7\plus{}9)\plus{}25\times (1\plus{}3\plus{}5\plus{}7\plus{}9)\equal{}69375$

Similarly you can find sum of numbers in which all digits are even
in this case because of digit $ \ 0$ you have to take some precaution basically $ \ 0$ can not occur at $ \ 100^{th}$ place

I think this will clear your doubt
 :wink:


Compute the area of the circle that passes through all the intersection poits of 4x 2 +11y 2 =29 and x 2 -6y 2 =6.

Thanks!

[hide]
Add the equations to get $x^2 + y^2 = 7$, so the area is $7\pi$
[/hide]


real analysis
advanced fields
advanced fields unsolved

Let $ m^*$ be the Lebesgue outer measure. Then one can show that it is "subadditive" or that $ m^*(\bigcup A_i)\leq \sum m^*(A_i)$. Is there an easy case where equality doesn't hold? My book does not have any example of this and intuitively it seems as if it equality [i]should[/i] hold...

For an "easy" case in which equality does not hold, take $ A_i$ that are not pairwise disjoint (and some of the overlaps have positive measure).

If you add "pairwise disjoint" as a hypothesis, then equality holds when the $ A_i$ are measurable. To get a case in which equality does not hold, you will need non-measureable sets.

Oh duh! I was thinking that in the hypothesis they were disjoint for some reason. Thanks.


function
topology
algebra unsolved
algebra

Let $f: R \to R$ be a continuous function such that : $\forall a<b \in R$, we have $c \leq d \in R$ such that $f((a,b))=[c,d]$. Prove that $f= const$ .

It is not true let $f(x)=sinx$, (a,b)=(0,1000), [c,d]=[-1,1].
Iam sorry I don't see $\forall a<b.$ For this case, I think give contradition function f(x): If $x=a_{1}a_{2}...a_{k},a_{k+1}a_{k+2}...$ were $a_{i}$ are digits x we define $f(x)=0,a_{k+1}a_{k+3}a_{k+5}....$. It give periodic (f(x+1)=f(x)) continiusly function, and f((a,b))=[c,d], were $a=A,a_{1}a_{2}..., b=B,b_{1}b_{2}...., c=0,a_{1}a_{3}....,d=0,b_{1}b_{3}....$

I can't understand? So you think it isn't true?

I am sorry, this function give not contradition when 0<a<b<1 and
$a=0,a_{1}9a_{3}9a_{5}9.... \ \ a_{2i}=9 \forall i$ or $b=0,b_{1}0b_{3}0b_{5}0... \ \ b_{2i}=0 \forall i$.
I think it is true.
Let $f((a,b))=[c,d]$ and $c\not =d$, then $f^{-1}(c,d)$ is open set in (a,b). It give $a\le a_{1}<b_{1}\le b$ with $f((a_{1},b_{1}))=(c,d)$.


Polynomial $P(x)$ has integer coefficients. Prove, that if polynomials $P(x)$ and $P(P(P(x)))$ have common real root, they also have a common integer root.

Let $W(x)=P(P(P(x)))$ we know that exists a $t\in\mathbb{R}$ such $P(t)=W(t)=0$. From this it's trivial to conclude, that $P(0)$ is a common integer root for polynomials $P$ and $W$.

If k is the common root of $P(x)$ and $P(P(P(x)))$
So $P(P(P(k)))=P(P(0))=0$ it meas $P(0)$ be root of $P(x).$
We know that $P(x)$ has integer coefficients.
So $P(0)$ be interger.

Let $x_0$ be a common real root. Denote $P(x)=\sum_{k=0}^{n}a_kx^k$.Then $P(x_0)=0\wedge 0= P^2(P(x_0))=P(P(0))=P(a_0)$. So $P(P(P(a_0)))=P(P(0))=P(a_0)=0$. Because $P(a_0)=0\wedge P(P(P(a_0)))=0\wedge a_0\in Z$ we have finished the proof.


geometry
circumcircle
3D geometry
tetrahedron
geometry unsolved

Let $ \ G$ be the centroid of any triangle $ \ ABC$ . The lines $ \ AG , BG$ and $ \ CG$ meets the circumcircle resp. at  $ \ A_1$  , $ \ B_1$  and $ \ C_1$  .
Show that  $ \frac{GA}{GA_1}\plus{}\frac{GB}{GB_1}\plus{}\frac{GC}{GC_1}\equal{}3$


Show that if  $2^n-1$ is a prime number than n is a prime number too. $n \in z$

Why does one of the factors need to be one?

 :huh:

because the whole thing needs to be a prime

[quote="Bruno Bonagura"][hide="My solution"]$2^n - 1 = (2^{\frac{n}{2}} - 1)(2^{\frac{n}{2}} + 1)$
One of the factors must be equal $1$.
$2^{\frac{n}{2}} + 1$ can't be $1$, then...

$2^{\frac{n}{2}} - 1 = 1$
$2^{\frac{n}{2}} = 2$
$n = 2$[/hide][/quote]
There are more than one value of $n$. Example: $n=5$

Proof below is one way of proving if $2^n-1$ is prime then $n$ is also prime.

Proof by contrapositive: if $n$ is composite, then $2^n-1$ is also composite.
Suppose that $n$ is composite. Then, let $n=ab$ for some positive integers $a,b>1$.

\[ \begin{aligned} 2^n-1=2^{ab}-1&=(2^a-1)(2^{a(b-1)}+2^{a(b-2)}+\cdots+2^{a(b-(b-1))}+1)\\ &=(2^b-1)(2^{b(a-1)}+2^{b(a-2)}+\cdots+2^{b(a-(a-1))}+1)\\ \end{aligned} \]

Hence, $2^n-1$ is composite.

In the problem, he states that if $n$ is composite, it can be represented by the product of two factors $a*b$.  He then writes $2^n - 1$ as $2^{ab} - 1$ which can be factored as he demonstrated, meaning that $2^n-1$ cannot be prime.  This means that for $2^n-1$ to be prime, $n$ must also be prime.

And the way he factored it, was basically using the fact that
$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \dots - b^{n-1})$

So you can write
$(2^{a})^b - 1^{b} = (2^{a}-1)((2^{a})^{b-1} - (2^{a})^{b-2}(1) + \dots - 1^{b-1})$


See http://www.mathlinks.ro/Forum/viewtopic.php?t=13943 .

  darij


LaTeX
inequalities unsolved
inequalities

for all $ a,b,c,d$ in $ (0,1/2]$
pf that
$ abcd$/$ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)$ $ < \equal{}$ $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4/(1 \minus{} a)^4 \plus{} (1 \minus{} b)^4 \plus{} (1 \minus{} c)^4 \plus{} (1 \minus{} d)^4$
srry for my bad latex skill


hmmmm. im soo bored that im now posting questions. what a sin. lool.

on to the question in the not so exact words

x+1/x=4

what does x:^3:+1/x:^3: equal

it caused quite a hustle bustle on the late mathcounts forum. i have no idea why, it wasnt hard. especially since i think its in the aops book 1. at least i think. im pretty sure. if you want to know the answer, highlight below

[hide](x+1/x):^3:=x:^3:+3x+3/x+1/x:^3:=64=x:^3:+3(x+1/x)+1/x:^3:=

x:^3:+3(4)+1/x:^3:=64

so...x:^3:+1/x:^3:=64-12=52. cool beans.[/hide]

There was a similar problem at NYSML once.

If cos(x)+sec(x)=3, find (sec^10(x)+1)/sec^5(x). It has a solution like Syntax Error's question, and it also has a very elegant solution (that I found because I didn't want to multiply that much).

Use your boredom as a time to learn mathematics. That's one of my favorite time-spending methods.

i am doing math. its not the math thats boring me. very interesting indeed. i just am a mix of tired hungry stressed angry, and i should do some other math problems. cause im doing recycled ones. hmmmmm. maths very fun, dont get me wrong. i love math

Let x be a real number such that x + 1/x is an integer. Prove that x^n + 1/x^n is an integer, for all positive integers n. (Germany, 1995). This shouldn't be tough for most of you to show.

How very interesting... yeah... I would use [hide]induction[/hide] <-- Highlight to find out... That should be all you need...

Peace.

IW.OAV

In mindspa's question, doesnt x have to be 1? If it does, then that is really easy and doesnt even need induction.

No, that is not necessary. Suppose that x+1/x=5, for example. Then (by solving a quadratic), we can have x=(5+sqrt(21))/2.

[quote="Syntax Error"]
x+1/x=4
what does x:^3:+1/x:^3: equal
[/quote]

[hide="solution"]
(x + 1/x)^3 = 4^3
x^3 + 3x + 3x^-1 + x^-3 = 64
x^3 + x^-3 + 3(x + 1/x) = 64
x^3 + x^-3 + 12 = 64
x^3 + x^-3 = 52
[/hide]

I love these kinds of questions in math competitions, as I know that some of my competitors will do it the long way xD .


trigonometry
Asymptote
function
MIT
college
inequalities
algebra proposed

Problem 1: Prove that the equation:
\[ 4^x(4x^2\plus{}1)\equal{}1\]
have only 3 real different roots  :)

Problem 2: Solve
\[ \frac{1}{1\minus{}x^2}\plus{}1>\frac{3x}{\sqrt{1\minus{}x^2}}\]

Obviously $ x\equal{}0$ is one root.

A quick sketch of $ y\equal{}4^{x}$ and $ y\equal{}\frac{1}{4x^{2}\plus{}1}$ shows that there seem to be 2 other roots and that both of them are negative (about $ \minus{}\frac{1}{2}$ and $ \minus{}2$, I think.)

Using Rolle's theorem. I don't think that -2 is root.

[quote="zaizai-hoang"]Problem 1: Prove that the equation:
\[ 4^x(4x^2 \plus{} 1) \equal{} 1
\]
have only 3 real different roots  :)
[/quote]
There exist $ t \in (\minus{}\frac{\pi}{2},\plus{}\frac{\pi}{2})$ s.t. $ x\equal{}\frac{\tan(x)}{2}$. It implies $ 2^{\tan(t)}\equal{}\cos^2(t)$.
May be helpful?

[quote="zaizai-hoang"]Problem 2: Solve
\[ \frac {1}{1 \minus{} x^2} \plus{} 1 > \frac {3x}{\sqrt {1 \minus{} x^2}}
\]
[/quote]

Since $ |x| < 1$ there exist $ t \in [ \minus{} \pi,\pi)$ s.t. $ x \equal{} \cos(t)$. Then $ \frac {1}{1 \minus{} x^2} \plus{} 1 > \frac {3x}{\sqrt {1 \minus{} x^2}} \implies \frac {1 \plus{} \sin^2(t)}{\sin^2(t)} > 3 cotan(t)$. So $ 1 \plus{} (\frac {1 \minus{} \cos(2t)}{2}) > \frac {3}{2}\sin(2t) \implies \cos(2t) \plus{} 3\sin(2t) \minus{} 3 < 0$ that can be rewritten as $ \sin(2t \plus{} arsin(\frac {1}{\sqrt {10}})) < \frac {3}{\sqrt {10}}$,from here it is straightforward.

Again I've done a rough sketch.

$ y\equal{}\frac{1}{1\minus{}x^{2}} \plus{} 1$ has asymptotes $ x\equal{}\pm1$ and $ y\equal{}1$ and has a local minimum at $ (0, 2)$.

$ y\equal{} \frac{3x}{\sqrt{1\minus{}x^{2}}}$ is only defined for $ \minus{}1<x<1$.

The curves meet when $ x\equal{}\frac{1}{\sqrt{2}}$ and $ x\equal{}\frac{2}{\sqrt{5}}$.

So I think the answer is

$ \minus{}1<x<\frac{1}{\sqrt{2}}$ and $ \frac{2}{\sqrt{5}} < x <\infty$.

Hi Christiano89.

Just spotted your post: the function $ y \equal{} (4^{x})(4x^{2} \plus{} 1) \minus{} 1$ is continuous. My calculator gives $ y( \minus{} 2) \equal{} \plus{} 0.0625$ and $ y( \minus{} 2.2) \equal{} \minus{} 0.035625348...$, so I guess there is a root which is a little gretaer than $ \minus{} 2.2$.

It is $ \minus{}2.1287$ to 4 decimal places.

hello, to problem 1, i have found $ x_1\approx -2.128730957$, $ x_2=-0.5$ and $ x_3=0$.
Sonnhard.

[quote="AndrewTom"]Hi Christiano89.

Just spotted your post: the function $ y \equal{} (4^{x})(4x^{2} \plus{} 1) \minus{} 1$ is continuous. My calculator gives $ y( \minus{} 2) \equal{} \plus{} 0.0625$ and $ y( \minus{} 2.2) \equal{} \minus{} 0.035625348...$, so I guess there is a root which is a little gretaer than $ \minus{} 2.2$.

It is $ \minus{} 2.1287$ to 4 decimal places.[/quote]

Hi AndrewTom,
you are right. But the aim of this problem is to ''prove'' the number of roots. Somebody, like Herr Sonnhard Graubner can find a better approximation. :) So try to prove it without calculator. :lol:

hello, you can consider the function $ f(x)\equal{}4x^2\plus{}1\minus{}4^{\minus{}x}$.
Sonnhard.

Hello, Sonnhard.

Habe gerade Ihre Antwort gesehen und gemerkt, dass in meine Loesung ein Widerspruch war. Statt $ \infty$ sollte, wie bei Ihnen, $ 1$ sein.

Mit Ihre Idee, habe ich eine Skizze von $ f(x) \equal{} 4x^{2} \plus{} 1$ und $ f(x) \equal{} 4^{\minus{}x}$ gemacht und wenn man die Steigungen (gradients?) vergleicht ,sieht man dass es nur drei reelle Loesungen gibt: $ 0$, $ \minus{}\frac{1}{2}$ und eine dritte die kleiner als $ \minus{}\frac{1}{2}$ ist.

Hello, darf ich Ihnen mal eine Frage stellen, wo haben Sie eigentlich so gut deutsch sprechen gelernt?
Viele Grüße,Sonnhard.

Hello Sonnhard,

Ich habe grosses Interesse an Sprachen und meine Freundin ist Deutsche und hilft mir.

(Uebrigens, wie schreiben Sie Umlaute?)

Freundliche Gruesse,

AndrewTom.

[hide="Problem#2"]
For both sides to be defined, $ x \in ( \minus{} 1,1)$
So, let $ x \equal{} \sin\theta$ for some $ \theta \in \left( \minus{}\dfrac{\pi}{2},\dfrac{\pi}{2} \right)$
Then, $ \sqrt {1 \minus{} x^2} \equal{} \cos\theta$

So, the equation becomes:

$ \dfrac{1}{\cos^2\theta} \plus{} 1 > \dfrac{3\sin\theta}{\cos\theta}$

$ \sec^2\theta \plus{} 1 > 3\tan\theta$

$ \tan^2\theta \minus{} 3\tan\theta \plus{} 2 > 0$

$ (\tan\theta \minus{} 1)(\tan\theta \minus{} 2) > 0$

$ \tan\theta < 1$ OR $ \tan\theta > 2$

$ x \equal{} \sin\theta < \dfrac{1}{\sqrt {2}}$ OR $ x \equal{} \sin\theta > \dfrac{2}{\sqrt {5}}$

So, the values of $ x$ that satisfy the inequality are $ \boxed{x \in \left( \minus{} 1,\dfrac{1}{\sqrt {2}}\right) \bigcup \left(\dfrac{2}{\sqrt {5}},1\right)}$
[/hide]


conics
ellipse
trigonometry
advanced fields
advanced fields unsolved

Points P,Q and R lie on the ellipsoid of equation$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$  such that segments OP, OQ, OR are pairwise perpendicular, where O denotes the centre of the ellipsoid. Prove that the distance of plane PQR from O is independent of the position of the points P, Q, R.

oh ! why I posted it in this topic ,moderator plz replace it :(  :blush:

Basicly, we do this (no details for now :)):

We show that a similar thing happens on the ellipse: if it's centered at $O$ and $M,N$ are points on the ellipse s.t. $OM\perp ON$, then the distance from $O$ to $MN$ is constant. Then we look at the intersection of a plane with the ellipsoid: this is an ellipse. Now take $Q,R$ on this ellipse s.t. $OQ\perp OR$. In this case $OP$ is fixed because it passes through $O$ and it's $\perp (OQR)$. The locus of $T$, the projection of $O$ on $QR$, is a circle (we use the property for the ellipse). $OP$ is perpendicular to the plane of this circle and passing through its center, so the locus of the projection of $O$ on $PT$ is a circle.

This means that each plane cuts the locus of the projection of $O$ on $(PQR)$ in a circle. It's well-known that this means that the locus is, in fact, a sphere.

In order to prove the property for an ellipse, we just use the polar equation of the ellipse: $r^2=\frac {a^2b^2}{b^2\cos \theta+a^2\sin \theta}$. It's easy to see that $\frac 1{OM^2}+\frac 1{ON^2}$ is constant, where $M,N$ are points on an ellipse centered at $O$ s.t. $OM\perp ON$, and this, in turn, is equivalent to the stateed property: the distance $d(O,MN)$ is constant.


quadratics
modular arithmetic
number theory unsolved
number theory

Let $p$ be a odd prime number and $m>1,h$  be natural numbers such that $h<2^m,n=2^mh+1$ if $n^{\frac{p-1}{2}}\equiv -1$ $(\mod p)$ .prove that if  $n$  wants to be a prime number It's nessecery and suffient that:

           $p^{\frac{n-1}{2}} \equiv -1$  $(\mod n)$

$n=4k+1$. If $n$ is a prime, then the quadratic reciprocity law says that $\left(\frac pn\right)\left(\frac np\right)=1$. Since $\left(\frac np\right)=-1$, we get $\left(\frac pn\right)=-1$, as desired.

Conversely, let $q|n$ be a prime. We have $p^{n-1}\equiv 1\pmod q$, and $p^{q-1}\equiv 1\pmod q$, so $p^{(n-1,q-1)}\equiv 1\pmod q$. If $2^m\not|\ q-1$, then $(n-1,q-1)=\left(\frac{n-1}2,q-1\right)$, so $p^{\frac {n-1}2}\equiv 1\pmod q$, contradiction. This means that $2^m|q-1$, so $q=2^mt+1$. Since this is true for all prime factors of $n$, if $q'|\frac nq$ is another prime factor, then $n\ge qq'>(2^mt+1)\cdot 2^m>(2^mt+1)\cdot h\ge 2^mh+1=n$, so $n=q$, and we're done.

What is the Pocklinton's theorem? Can you explain to me

Pocklinton's theorem as follow:For n be interger and s.t.: [tex]n-1=FR(R<F)[/tex] and [tex]F[/tex] has form:[tex]F=p_1^{e_1}p_2^{e_2}...p_t^{e_t}[/tex].If there exist [tex]x_i (i=1,2..t)[/tex] such that:[tex]1/x_i^{n-1}\equiv 1(mod n)[/tex]
[tex]2/gcd(x_i^{\frac{n-1}{p_i}}-1,n) =1[/tex]
then n is a prime number.
And Proth's theorem is a its particular case:If[tex] n=2^m.h+1[/tex] and[tex] h<2^m[/tex].If there exist a integer such that:[tex]a^{\frac{n-1}{2}}\equiv -1 (mod n)[/tex] then n is prime number

I can't find Poclinton's theorem in any documnet ! Can you tell me where to find it ?

[quote="al.M.V."]keira_khtn forgot a "g", it's here:
http://mathworld.wolfram.com/PocklingtonsTheorem.html :D[/quote]
What did i forgot,al.M.V??

[quote]What did i forgot,al.M.V??[/quote]

Well, it's nothing significant, just a letter:

[quote]Pocklinton's theorem as follow:For n be interger and s.t.:[/quote]

It should be Pocklin[b]g[/b]ton, that's why b2 couldn't find anything.  :)


AMC
AMC 10
AMC 10 B
AIME
USA(J)MO
USAMO
search

[quote="Trotwood"]I am a Chinese student who has passed the AMC10B and I have been invited to take part in AIME. But in fact, a lot of students in our school have been qualified.( There are 5 students who got perfect marks.)
I decided to take part in AMC10B because I want to have my Maths ability testified and get some benefit. But if there are a lot of people who passed it, the AMC10B prize won't be valuable.
I was astonished that the problems in AMC are so easy and I think the ease of its problems gets rid of the value of the contest itself.[/quote]

Keep in mind that the AMC 10 is only the first step. The AIME is much harder, and should you make USAMO (although unfortunately you will not be able to take it), this should pose quite a challenge to all but the best.

AMC 10's are not meant to be exceedingly difficult. If you were a student, took the AMC 10, and couldn't do any of the problems, would you be encouraged to go on with math?

Also, difficulty vaires by year. I thought that these tests were comparably easier, although I didn't do very well on them.

Once you take the AIME, if you find this very easy, try some USAMOs. These should pose a challenge. If they don't, then you've basically pwned most americans  :P .

But excellent job qualifying for AIME! many students do not make it to the AIME. Good luck!!

I got a perfect too (Same test, 09 AMC 10B), which was unexpected. The difficulty of the AMC's change every year, and this year was pretty easy (looking back and reviewing). However, at the actual competition I was so stressed I couldn't think straight.

My advice is... If you're dissatisfied with the level of the AMC10, do well on the AIME ;). The "benefit" in these math competitions in the end boils down to seeing where you place and being inspired to learn math. There's plenty of hard problems out there to try, and AIME has a good collection of feasible problems of all sorts of difficulties. Try check the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]AIME problems and solutions[/url] page on AoPS. Also, beyond that there's USAMO and all kinda of fun stuff (too advanced for me xD)

You might also want to take the AMC 12 next year instead.

Good luck on the AIME!  :)

(Edit: Btw... 5 perfect marks... Your school's good.)

The AMCs were really easy this year, but the AIME will be much harder, I promise. (probably the same difficulty as the Chinese "AMCs.")
Also, you might want to look into the International Mathematical Talent Search (IMTS).


calculus
integration
analytic geometry
calculus computations

Set up an integral to find the volume of the solid below the graph of f(x,y) = 64 - x2 - y2 and above the graph of z = 16.

I am having a hard time to visualize this, can someone help me out?

1) Find the intersection of both surfaces.

[hide]If $ f(x,y) \equal{} 64 \minus{} x^2 \minus{} y^2$ and $ g(x,y) \equal{} 16$ then $ f(x,y) \equal{} g(x,y) \equal{}> 16 \equal{} 64 \minus{} x^2 \minus{} y^2 <\equal{}> x^2 \plus{} y^2 \equal{} (4 \sqrt{3})^2$. The intersection is then a circunference with radius $ 4 \sqrt{3}$.[/hide]

2) The projection of the surface found in (a) on the xy plane is the region R in which you are to integrate.

3) With the geometric interpretation of a double integral as the volume of a solid in mind, set up the double integral, with the appropriate limits of integration, that will give you the volume of the solid. Use cilindrical or spherical coordinates if appropriate.

[hide]$ V \equal{} \int_{\minus{} \pi}^{\pi} \int_0^{4 \sqrt{3}} [(64 \minus{} \rho ^2) \minus{} 16] \rho \, d \rho \,d \theta$.[/hide]


Our students are learning to write geometry proofs. Some of them, however, find the statements that they're asked to prove to be so obvious that they don't appreciate what proofs are really for.

I'm intersted in finding some geometry statements that seem to be true, but that are actually false. I'm also looking to find some simple, yet beautiful, proofs of geometric facts that are surprising (or at least, not immediately obvious). The proofs and statements should be simple enough that they could be discovered and/or understood by an "average" student with no more than a few subtle hints, and presented by a teacher in say, under 10 minutes.

Don't know where to find counterexamples but this is good: http://www.amazon.co.uk/Crossing-Bridge-Gerry-Leversha/dp/1906001065

You probably know this (I haven't seen it but have come across some nice "Proof without words" on the internet: http://books.google.co.uk/books?id=Kx2cjyzTIYkC&dq=Proofs+without+words&printsec=frontcover&source=bn&hl=en&ei=ug_cStCeJZWF4QbkmI31Bg&sa=X&oi=book_result&ct=result&resnum=4&ved=0CBIQ6AEwAw#v=onepage&q=&f=false

Sorry I can't be more helpful, but I'll watch for other replies to your post as this seems quite interesting.

Around what level are they?

One example I can think of is the proof that Given a point P in an equilateral triangle, the sum of the perpendiculars from that point to the sides is equal to the altitude. If you want, I can write up a solution for it.


I found this test on http://www.geocities.com/grissom_math , GEOMETRY 2003

AT the end I lookd at the answer key, and I don't know where to find stuff about chords and inscribed things

I still have  problems with number   14,15,16,17,19,21,22, and 24. Can some people help me? I am terribly frustrated.

#10.

Make the center O. Draw a line from O to C. The arc formed by OC and OB is the same degree as that formed by AB and AC.

OB=OA=OC=radius=AC

Hence, triangle OAC is equilateral. Thus, <COB = 180-60 = 120.

Answer to #10 is 120.

Will post others later if you want.

#13:

Draw the set out.

We can divide the larger triangle into siz equal triangles using the radius of the circle as an altitude. The base equals            rcot(30) = r(3)^(1/2).

We can also divide the smaller triangle into 6 equal triangles with the radius beings from the center of the circle a vertex of the triangle. The altitude in this case is now r(sin30)=.5r. The base equals rcos30 = (3)^(1/2)r/2.

So, Area is proportional to base and height.

Abig/Asmall = (r(r(3)^(1/2)))/((.5r)((3)^(1/2)r/2))

= 4

Answer is A.

I hope you can post some later. I see how you got number 10 and 13 now. thanks

#25: Use Ceva: http://mathworld.wolfram.com/CevasTheorem.html
[hide="25"] If the answer is $\frac{1}{a}$, we can set $CD=z$ and $DB=(a-1)z$. We also have $AG=x$, $GB=2x$, $AE=3y$, $EC=y$.

By Ceva, we have $3y\cdot2x\cdot z=y\cdot x\cdot(a-1)z$
$6=a-1$
$a=7$
$\boxed{\frac{1}{7}}$ :D [/hide]
[hide="23"] Call the radius of circle $P$, $p$, and same for circle $O$. Draw a line perpendicular to $AO$ thru $P$. The length of that segment is equal to the length of $AB$. By the Pythagorean Theorem, We have $AB=6^{2}-(o-p)^{2}=-o^{2}-p^{2}+2op+36$. We know $o^{2}+p^{2}=28$ and $o+p=6$, which is enough information to find $AB$. :D [/hide]

[quote="now a ranger"]on number 23, how do you know o squared plus p sqaured equals 28? I missed out on that part[/quote]
$\text{Area of circle O+Area of circle P = 28}\pi$
$o^{2}\pi+p^{2}\pi=28\pi$
$o^{2}+p^{2}=28$

ohhhh snapp i didnt see that

CAn anyone help me with any others?

#14 is hard to explain without sitting right next to you since it requires a lot of drawing. A hint it to remember that tangents that come from the same point to the circle are the same length and that the bisector of the chord from the center is perpendicular to the chord.

how about the others?

Also, does anyone know where to find more practice tests on geometry? with solutions.

also, how do you put a picture underneath your posting name..

#15:

A^2= d^2 + s^2, where d is a diagnol of a square surface and s is the edge.

A^2 = (s^2 + s^2) + s^2 --> A = s(3)^(1/2)

(1/2B)^2 = (s/2)^2 + s^2 --> B= s(5)^(1/2)

Hence,

B-A = (5^(1/2)-3^(1/2))s

Answer is B.

I don't have much time. You'll have to wait. I'll post one or two a day.

okay thanks

I'll learn 1 or 2 a day, it will help a lot

started out with 10 i got wrong or something like that

 guessed on 2 of em and got right

Sorry, but the rest are kind of hard to explain here without drawing everything in front of you.

okay, im gonna make a new thread with lots of various problems from tests ive done that i dont get

Some advice to now a ranger, don't make consecutive posts, instead edit your previous posts.

practice=http://www.satmathpro.com/Practice.html
but only the free stuff  :wink:
other than that if you just want to do problems without much expl. try mathcounts.org's link saab!


I was wondering how u would count the number of paths from vertex A to vertex B on an a X b X c rectangular prism if points A and B are endpoints of one of the space diagonals (they are as far away as possible from each other). Is there a specific method?

Thanks

Simply extend the reasoning used to count paths on a 2-dimensional grid.  :)


USAMTS
calculus
Harvard
college
Princeton
AMC 10

Hi, I'm wondering what I should do to better my application for college and whether I will be admitted to any of the Ivies.

Credentials:

I haven't took the SAT yet but I will, (approx. scores are 470 for Critical Reading, 500 for Writing, and 770 for Math).  However, I did took three Subject Tests and is considering to taking a fourth one soon (770 for Mathematics I, 790 for Mathematics II, 700 for Chemistry, and considering Biology)

I have participated in sports like Indoor Track (with only three honors) and will participate in Tennis.

I play two instruments (violin, self taught piano), and I also sing in a selected choir.

My overall GPA so far is 4.0 (actual average is about 95.6) and I am ranked #1 in my graduating class.

I am involved in the musical (and will be involved in the play next year)

I am doing the USAMTS contest (hopefully I will achieve a bronze status) Current Score is 34/50 (not bad for a first timer)

I am currently taking AP Chemistry, and AP US History (and considering of taking BC Calculus, AP Biology, AP Microeconomics, AP Statistics, and maybe AP English or AP Psychology)

I have leadership (currently Treasurer in a selected orchestra and a Director in a voluntary activity.)  I am considering of being Treasurer for National Honor Society as well.

Also I have at least 50 hours of service from Freshman year to the middle of Junior year.

Lastly, I am involved in Spanish Club and will be taking the AMC 12 this year. (I have taken AMC 10 in the past two years and I'm unsure about my scores.)

Sorry about my lengthy achievements in high school... but I am really considering being admitted in the Ivies, especially Dartmouth and Cornell... please reply with your opinions :)

Sorry to tell you but I doubt it.  A 1740 is terrible for top 20 schools (<10 percentile I'm sure) and your EC's don't stand out.  Unless you're a recruited athlete or you have some amazing EC you didn't mention, I don't see much of a chance.

if you get better SAT scores (the above ones are really bad), then you have a decent chance of getting into one of the lower-ranked ivies.  you pretty much have no shot at harvard or princeton or any similarly competitive school though.

Are you a native speaker?  Those Critical Reading and Writing SATs really hurt.  As much as I don't like to emphasize test scores, it's very unlikely that you can get away with scores that low without something else spectacular, and nothing stands out.

well, I guess I should begin reading more books, and learn more SAT words before May... hopefully I will get something beneficial from it

Then you should take TOEFL.  I think TOEFL is more important and the SAT is considered only if it is above 680 or something.

You are required to take the TOEFL if English is not your native language, and to be on the safer side, it would be good if you can get your SAT's up by at least another 200-250 points.

Yeah, I took a practice test for the SATs and this is now what I got:

Critical Reading: 540 - 600
Math: 720 - 800
Writing: 580 - 710
Total: 1840 - 2110

With these scores, I hope I can get to one of the Ivies... any more tips on improving my credentials?

I guess a 2110 is decent if thats ure best. Take the ACT also. Plus, you should volunteer more and do more AP's. Im assuming ure a junior currently?

Have you done anything special with your EC's?  Otherwise, it just looks like you're a serial joiner and colleges hate that.

[quote="iironiic"]yeah, my first language is Chinese, and it was very hard for me to understand the English language =/[/quote]

Make sure that you spend time around people with good grammar. Getting the sound of correct language into your ears and grammatical intuition into your heart will help you a lot. Being with people who speak well also tends to lead to more useful social connections.


calculus
integration
logarithms
parameterization
real analysis
real analysis unsolved

Compute $ \int_0^\infty\frac{e^{\minus{}ax}\minus{}e^{\minus{}bx}}{\ln x}\,dx.$

Could it be computed applyin' double integration? I've tried to find a suitable parameter but I've couldn't.

I dunno the answer, that's why I haven't mentioned the constraint for $ a,b.$

This diverges unless $ a\equal{}b,$ and $ a\equal{}b$ isn't exactly an interesting case.

The problem arises not at zero or infinity but from the singularity at $ x\equal{}1.$


Does anyone know why they didn't send out an answer booklet for the teacher with the tests this year?

They will be mailed along with the AIME results for your school.

oh thanks.  I don't know why my teachers were expecting to have one now then.

For the AMC, a solutions booklet was actually mailed out with the test. So as soon as the test was over, students could basically check their answers against the official ones. Now, I'm not sure about the reasoning behind not attaching a solutions booklet to the AIME, but it does decrease the possibility of cheating. Qualifying for the USAMO is a highly respectable achievement, so I'm assuming there would be at least some temptation if answers were handed out.

[i]That's why [/i]people in other schools knew their scores the day after the test. Anyways i agree completely with Mits...

[quote="RC-7th"]Or you could just happen to go to a school with a good enough math team that they can figure out all the answers right after the test.[/quote]

Or you could come here =).

or you could be so smart that you already know all 15 answers..

[quote="RC-7th"]Or you could just happen to go to a school with a good enough math team that they can figure out all the answers right after the test.[/quote]

Yes that works. You could also befriend enuf USAMO HM, who are many times as good as the answer key is. :-) (Well, at least several of them put together are.)

THE committee :) (of the AIME) i believe


inequalities
trigonometry
geometry
incenter
geometric transformation
reflection
circumcircle

I have wondered this for a long time: what does you a.s.o stand for? Is it something along the lines of "and symmetric others"?

$\left\{\begin{array}{c}4\cdot IM^{2}=2\left(IB^{2}+IC^{2}\right)-a^{2}\\\\ IB^{2}=\frac{ac(p-b)}{p}\\\\ IC^{2}=\frac{ab(p-c)}{p}\end{array}\right\|$ $\Longrightarrow$

$4p\cdot IM^{2}=2ac(p-b)+2ab(p-c)-pa^{2}=a[c(a+c-b)+b(a+b-c)-ap]=$

$a\left[(b-c)^{2}+a(b+c)-ap\right]=a(b-c)^{2}+a^{2}(p-a)$ $\Longrightarrow$

$\left\{\begin{array}{c}4p\cdot\sum IM^{2}=\sum \left[a(b-c)^{2}\right]+\sum a^{2}(p-a)\\\\ \sum a^{2}(p-a)=4S(R+r)\end{array}\right\|$ $\Longrightarrow$

$4p\cdot \sum IM^{2}=\sum \left[a(b-c)^{2}\right]+4pr(R+r)$ $\Longrightarrow$ $\boxed{\ \sum IM^{2}\ge r(R+r)\ }\ .$

Forall $P$ on plane then $PA'^{2}+PB'^{2}+PC'^{2}\ge\frac{a^{2}+b^{2}+c^{2}}{12}\ge\frac{r(4R+r)}{3}\ge r(R+r)$ with $A',B',C'$ are midpoint of $BC,CA,AB$ resp
Thus the result of Virgil Nicula is true for all $P$ on plane! And more here we can show min of $\sum PA'^{2}$ with a fix triangle $ABC$ and really it is only simple result!

[quote="tsemng"][quote="Omid Hatami"]In triangle $ABC$, if $L,M,N$ are midpoints of $AB,AC,BC$. And $H$ is orthogonal center of triangle $ABC$, then prove that \[LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2})\][/quote]

i tried in complex co-ordinat  and the result became
$2abc+ \sum (a^2b+b^2a) \ge 0$ is this true?

$a,b,c$ are co-ordinat of $A,B,C$ and $h=a+b+c$ $l=(a+b)/2$ $m=(a+c)/2$ $n=(b+c)/2$[/quote]
Hmm, I am getting a different result.
Denote by $H$, the origin; and give $A,B,C$ the affixes $u,v,w.$ So we see that $l=\frac{v+w}{2}, m=\frac{w+u}{2},n=\frac{u+v}{2}.$
Therefore $AB^2+BC^2+CA^2=(u-v)^2+(v-w)^2+(w-u)^2$ and
$LP^2+MP^2+NP^2=\frac 14\left[(w+u)^2+(u+v)^2+(v+w)^2\right].$
Also, $u+v+w=0$; so that
\[\begin{aligned}\frac 14\left[AB^2+BC^2+CA^2\right]&=\frac 14\left[(u-v)^2+(v-w)^2+(w-u)^2\right]\\&=\frac 14\left[2(u+v+w)^2+(u-v)^2+(v-w)^2+(w-u)^2\right]\\&=\frac 14\left[2(u^2+v^2+w^2)+(u+v)^2+(v+w)^2+(w+u)^2\right]\\&\geq \frac 14\left[(u+v)^2+(v+w)^2+(w+u)^2\right];\end{aligned}\]
And we are done. $\Box$
:)

And so what is the equality case ? Is it when the triangle is reduced to a point ?
Thanks in advance.

Only if $ABC$ is acute or right. Equality holds if it is right.
We use the fact that reflection of the orthogonal center with the respect to the midpoints of the triangle lay on the circumcircle.
Denote $P,Q$ intersection of $HL$ with circumcircle. ($P$ is reflection of $H$ wrt $L$)
Looking at power of the point $L$ wrt to the circumcircle and noticing that $LP=HL \leq LQ$ we get
$\frac{AB^2}{4}=AL*BL=LQ*LP \geq HL^2$
Applying this to two other sides we get the desired result.

From Leybnits's identity it's equivalent to $ cosA \cdot cosB \cdot cosC \ge 0 $.It's true if and only if $ ABC $ is non-obtuse!

[quote=Virgil Nicula][size=117]Only if the triangle $ABC$ is acute or right ![/size][quote="Omid Hatami"][color=darkred]In [b][u]acute or right[/u][/b]  triangle $ABC$, if $L,M,N$ are midpoints of the its sides and $H$ is its orthocenter, then show that $\boxed{\ LH^{2}+MH^{2}+NH^{2}\leq\frac14\cdot \left(a^{2}+b^{2}+c^{2}\right)\ }$[/color][/quote][color=darkblue][b]Proof.[/b] It is well-known that $HA=2R|\cos A|$, $a=2R\sin A$ a.s.o. and $\Delta\equiv\sum \cos 2A=-1-4\prod\cos A\ .$ Therefore,
$4\sum HL^{2}-\sum a^{2}=$ $\sum\left(4\cdot HL^{2}-a^{2}\right)=$ $\sum\left[2\left(HB^{2}+HC^{2}\right)-2a^{2}\right]=$ $4\sum HA^{2}-2\sum a^{2}=$
$16R^{2}\sum \cos^{2}A-8R^{2}\sum \sin^{2}A=$ $8R^{2}\sum (1+\cos 2A)-4R^{2}\sum (1-\cos 2A)=$ $8R^{2}(3+\Delta )-4R^{2}(3-\Delta )=$
$4R^{2}(6+2\Delta-3+\Delta)=$ $12R^{2}(1+\Delta )=$ $-48R^{2}\cos A\cos B\cos C\le 0$ $\Longrightarrow$ $\sum HL^{2}\le\frac{1}{4}\cdot\sum a^{2}\ .$

[b]Remark.[/b] I posted long ago the following inequality : $\boxed{\ \sum IM^{2}\ge r(R+r)\ }\ ,$ where $I$ is the incenter of the triangle $ABC$. Try solve it ![/color][/quote]

Sorry,I don,t know why LH^2-a^2=2(BH^2+CH^2)-2a^2
Can you explain?

$ O $ the center of  $ \odot  (ABC) $ and WLOG let  $\odot (ABC)$ be unit circle vectors
By Euler line we get $ H=a+b+c $

$ LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2})  \Longleftrightarrow \sum_{cyc} (\frac{a+b+2c}{2})^2 \le \sum_{cyc} (\frac{a-b}{2})^2 $

We use that $ a^2=b^2=c^2=1 $

$ \Leftrightarrow -1 \ge ab+bc+ca $

 By scalar product ,$ ab= cos(2C) $,

$ \Leftrightarrow -1 \ge cos(2A)+cos(2B)+cos(2C) $

$ \Leftrightarrow -1 \ge 2cos(A+B)cos(A-B)+2cos^2(A+B)-1 $

$ \Leftrightarrow 0 \ge +4cos(A+B)cosAcosB $

$ \Leftrightarrow 0 \ge -4cos(C)cos(A)cos(B) $

This is true for acute or right triangle.


anyone already planning an april 2008 one?

Me; I am working on a website that will basically have some similar features as #H34N1's site did.
When the time comes, I will let everyone know how and where to signup.
Also, if anyone would like to run it this time, just send me a PM. :)
minicon

You may now register your team [url=http://aopsftw.freehostia.com/home/]here[/url].
But first, you must create an account there.  Mail server does not work, so if it is a long password, remember it or write it down. :D
This is an individual tournament.
Avatars must be private messaged to me.  Attachments currently do not work on that site.
Thank you for your cooperation,
minicon

So...
Mail server not working + required email validation = no account created :D.

just a curiousity, what's the point of the team name and avatar if it's individuals?

Incase you want to be called something different + user may not have avatar.  Its easier to see it with avatars.
[b]Everyone who has registered, please re-register.  I fixed the email settings[/b]

You may now register your team via a form located at
http://aopsftw.freehostia.com/home/?q=node/3
Thank you.
minicon
:)

For your information, [b]the site aopsftw.freehostia.com/home/ will be down tonight for maintenance[/b].  If you wish to register, please post here temporarily.  Hopefully the site will be back up and running tomorrow.  I will post when that happens.
Thank you for your cooperation.

Site is now running on 5.7 .
Everything should be same; no files were deleted.
:)

I accidentally signed up without for it without an avatar. :|
What should i do?

Just sign up again with all info you can fill out.
I won't add you twice.
:)

I think it will be max. 16 people, unless there is a giant rush before then, in which case it will be max. 32.

I am planning to have 16 people and then have a qualification round where 1/2 of the people will be eliminated.
:)

I will probably not make it to over half of the scheduled tournaments, so could I file as a substitute?

Yeah, if someone can't make it to the whole tournament (or part of it), you can fill in if you can make it;  if no-one can make it, the opponent gets an automatic bye.
 :P

The April Tournament is an individual tournament.  MustItAlwaysBeMe's tournament is a team tournament.  But all spots are taken for those.  If you want, try sending a private message to MustItAlwaysBeMe and he might try to add you in; all of his slots aren't full, it has just been past the deadline but no games have been played.
:)

Thanks.  :) I think I want to join the individual tournament, sounds more challenging. (As a substitute, of course) Where is the current player list?

Currently [url=http://aopsftw.freehostia.com/home/?q=node/10]here[/url].
It will probably move to another place as I am planning on changing the hosts tonight.  A link will be provided to the new site when up though. :lol:
Edit: Don't fill out any forms,  I can't access them as logging in doesn't work.   :(

Darn it, I keep on missing those new tournaments... Team April tourney, anybody?

For the tournament, if you don't play the three games who gets the bye? because I'm playing rd5493 and he's not responding to my private messages. :|

Yeah, AIMEman and rd5493 have not gotten my messages.  I'll take care of this, but first let's wait a couple of days.  If he doesn't respond by Wednesday night, he will be replaced.

When's the May tournament?  Because its May already.  And is it a team or individual tournament?

I am taking May off.  :(   But I will probably run another in June.  I don't exactly know what I'm doing yet though.

I've been busy, and there is going to be no May tournament.

Technically, I should be doing it and I should have wrapped up the April Tournament, so sorry guys.  As for the next one, I have no clue about it.

Wait, does that mean minicon's running two tournaments in a row? (April and June)

I don't know.  I might do the June tournament.  No matter what, the next one will definitely go smoother.  A lot smoother, in fact.

Um...It's June, and there's no tournament......Could one of you please run it?


The standard tip in a restaurant is $ 15 \%$ of the bill before tax. Many people in California find the tip by doubling the sales tax, which is $ 8.25 \%$. By how many cents does this exceed the standard tip on a bill of $ \$60$ before tax?

Tax is $ .0825\times 60\equal{}4.95$ so tip is $ 9.90$.  Alternatively, $ .15\times 60\equal{}9$, so the difference is 90 cents.

Faster way is:
$ .0825*2\equal{}.165\equal{}$ the actual % tipped.
$ .165\equal{}.15\equal{}.015\equal{}$ the extra % tipped
$ .015*\$60\equal{}\$.9\equal{}90$ cents = the extra amount tipped.


induction
strong induction
combinatorics proposed
combinatorics

Consider a convex polygon having $n$ vertices, $n\geq 4$. We arbitrarily decompose the polygon into triangles having all the vertices among the vertices of the polygon, such that no two of the triangles have interior points in common. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is also a side of the polygon and in white those triangles that have no sides that are sides of the polygon.

Prove that there are two more black triangles that white ones.

Here is my solution :

We prove by induction that there are at least two black triangles, and that $b = w+2$ where $b$ (resp. $w$ ) denote the number of black (resp. white) triangles.

First for $n=4$ there are two black triangles and no red and white. Thus the result holds.
Now, let $n > 4$ be given, and assume the result holds for any $k$ such that $4 \leq k < n$.

Let's consider a $n$-gon as colored as in the statement of the problem.
Let $d$ be any diagonal (not side) which has been drawn for the triangulation.
Then $d$ separates the $n$-gon into two disjoint polygons. If any of these polygons is a triangle then it is a black one in the original coloring. If none of these polygons is a triangle then use the induction hypothesis for each of the two subpolygons. They gives us at least 2 black triangles on each side of $d$. And, for each side at most one of these black triangles have $d$ as a side. Thus, in any case, there is at least one black triangle on each side of $d$, which ensures that there are a least two black triangles for the $n$-gon.

Now, let $T =ABC$ be a black triangle (which does exist from above), with vertices in that order on the $n$-gon. Consider the $n-1$-gon obtained by deleting $T$ and replacing it by the edge $AC$.
The coloring induced on this $(n-1)$-gon is the same as the initial one except for the triangle $T'$ having $AC$ as side. Note that since $AC$ is a diagonal of the $n$-gon, then $T'$ was red or white in the original coloring, and since it is a side of the $(n-1)$-gon it is now black or red, respectively.
From the induction hypothesis, we have $b'=w'+2$ for the $(n-1)$-gon.
- If $T'$ is black for the induced coloring then it was red for the initial one. Adding $T$, it leads to $b=b'-1+1=b'$ and $w=w'$.
- If $T'$ is red for the induced coloring then it is white for the initial coloring. Adding $T$ leads to $b = b'+1$ and $w=w'+1$.

Thus in any case, we have $b=w+2$ as desired, which ends the induction step and the proof.

Pierre.

Shortlist's solution of problem 4

Solution. Denote by b, r, w the number of black, red white triangles respectively. It is easy to prove that the polygon is divided into n-2 triangles, hence
b + r + w = n – 2.
each side of the polygon is a side of exactly one triangle of the decomposition, and thus
2b + r = n.
Subtracting the two relations yields w = b – 2, as needed.

This is a very interesting solution for pb 4, because it does not need induction to prove the result. :)

Pierre.

[quote="pbornsztein"]This is a very interesting solution for pb 4, because it does not need induction to prove the result.[/quote]

I don't think it is really induction-free. Actually, I don't know any proof of the "easy to prove" fact that the number of triangles used in the triangulation is n - 2 but the one using induction.

  Darij

Yet, this is easy to do :

The sum of the internal angles of the convex $n$-gon is $(n-2) \pi$, which is also the sum of the angles of the triangles of the subdivision according to the statement of the problem, hence there are $n-2$ triangles  :)

Moreover, the fact that the sum of the internal angles of the convex $n$-gon is $(n-2) \pi$ is easy to prove without induction again : Just divides the convex $n$-gon into $n-2$ triangles by all the diagonals from one arbitrary vertex, and use again that the sum of the internal angles is the sum of the angles of all these triangles.

It is nice that one particular triangulation (the last above) gives the key for any triangulation.

Pierre.

And here is my solution:

At first we will show a rather simple lemma:

[b]Lemma.[/b] Let $n\geq 4$ be an integer. Every triangulation of a convex $n$-gon contains at least $2$ black triangles.

[i]Proof.[/i] The proof will use induction after $n$:

[i]Induction base:[/i] The case $n = 4$ is not what most of you would call difficult...

[i]Induction step:[/i] Now assume that we have proved our lemma for all $n\leq k$, where $k$ is a certain number with $k\geq 4$, and now we are going to prove it for $n = k + 1$.

In fact, consider a triangulated convex $\left(k + 1\right)$-gon $P$. Assume that it doesn't contain $2$ black triangles. Every side of $P$ belongs either to a black triangle or to a red triangle. Since $k\geq 4$, we have $k+1\geq 5$, and our $\left(k + 1\right)$-gon $P$ has, at least, $5$ sides. Only $2$ of these sides can belong to black triangles (else, we would have at least $2$ black triangles, contradicting our assumption!). Hence, we will always find, at least, one side of $P$ which belongs to a red triangle $T$. If we remove this red triangle, we are left with two triangulated convex polygons $Q$ and $Q^{\prime}$ having one common vertex (namely the third vertex of the removed red triangle). Each of these two polygons $Q$ and $Q^{\prime}$ has $\leq k$ vertices. Now, by our induction assumption, each of these two polygons must have at least two black triangles. Hence we have four black triangles altogether. Now, if we put in the removed red triangle $T$ again, then at most two of these black triangles may possibly become red triangles (in fact, every of our two polygons $Q$ and $Q^{\prime}$ had a side bordering with the removed red triangle $T$; now, if such a side belongs to a black triangle in $Q$ or in $Q^{\prime}$, then, after putting in the red triangle $T$ again, this black triangle becomes a red triangle). But at least $4 - 2 = 2$ black triangles remain. Hence, we have found $2$ black triangles in our $\left(k + 1\right)$-gon $P$, contradicting our assumption. Hence, our $\left(k + 1\right)$-gon $P$ must have $2$ black triangles (at least). This completes the induction step, and the Lemma is proven.

Now to the solution of the problem:

By the [i]niveau[/i] of a triangle in our triangulation, we will mean the number defined according to the following rule: If the triangle is white, then its niveau is -1; if the triangle is red, then its niveau is 0; if the triangle is black, then its niveau is 1.

The [i]summary niveau[/i] of a triangulated polygon will denote the sum of the niveaus of all triangles involved in the triangulation. Equivalently, it is the number of the black triangles minus the number of the white triangles.

Now we will prove that the summary niveau is always equal to $2$ (when $n\geq 4$). This will be obviously enough to solve the problem.

I will prove my assertion by induction. For $n = 4$, it is more than trivial. Now we assume that it holds for a certain $n\geq 4$ (i. e. it holds for all $n$-gons), and we have to show that it holds for $n + 1$ (i. e. for all $\left(n + 1\right)$-gons).

Consider a triangulated $\left(n + 1\right)$-gon $P$. Clearly, $n\geq 4$ yields $n+1\geq 5$. On the other hand, after the Lemma, the triangulated $\left(n + 1\right)$-gon $P$ contains at least two black triangles. Let $D$ be one of them. Now remove this black triangle $D$. What you then get is a triangulated $n$-gon $P^{\prime}$. All black, white, red triangles in the triangulation of $P$ preserve their color in $P^{\prime}$, except for the initial black triangle $D$, which disappears, and the triangle $D^{\prime}$ directly bordering with $D$. In fact, if $D^{\prime}$ was a red triangle for $P$, then it becomes a black triangle for $P^{\prime}$, and if $D^{\prime}$ was a white triangle for $P$, then it becomes a red triangle for $P^{\prime}$. (You can easily figure out that $D^{\prime}$ could not have been a black triangle for $P$; since otherwise, all three sides of $D^{\prime}$ would be sides of $P^{\prime}$, and thus, the polygon $P^{\prime}$ would be a triangle, and $P$ would be a quadrilateral, contradicting $n+1\geq 5$.)

Now, we see that the triangulated polygons $P$ and $P^{\prime}$ have the same summary niveau. This is because when we remove the black triangle $D$, then on one hand, we lose a niveau of $1$ (since $D$ had a niveau of $1$), but on the other hand, we gain a niveau of $1$ again (in fact, if $D^{\prime}$ was a red triangle for $P$, then it becomes a black triangle for $P^{\prime}$, so that the niveau changes from $0$ to $1$, and if $D^{\prime}$ was a white triangle for $P$, then it becomes a red triangle for $P^{\prime}$, so that the niveau changes from $-1$ to $0$). Hence, the sum of the niveaus of all triangles remains constant, i. e., the summary niveaus of the triangulated polygons $P$ and $P^{\prime}$ are equal. But since $P^{\prime}$ has $n$ edges, then by our induction assumption, its summary niveau is $2$, and hence the summary niveau of $P$ is also $2$. Proof complete.

  Darij

We use strong induction on $n$.
Consider a white triangle in the middle of the polygon. It then divides the problem into three separate subpolygons.

Lemma: In each of the subpolygons, $B = W + 1$.
This is easy to see, because red triangles are counted as black and white polygons are counted as red. Then we apply the induction hypothesis.

Therefore, summing over the subpolygons, $B = W + 3$. But we must include the white triangle in the center, so $B = W +2$, as desired.

First we shall state a $lemma$:

$Lemma$: A convex polygon with $n$ sides can be arbitrarily decomposed into exactly $n-2$ triangles having all the vertices among the vertices of the polygon, such that no two of the triangles have interior points in common.

$Proof of Lemma$:
Sum of internal angles of a convex polygon with $n$ sides is $(n-2) \pi$.
Since triangles are using all of the vertices of the polygon, the aggregate sum of angles of all the
triangles must equal the sum of internal angles of a convex polygon = $(n-2) \pi$.
Thus, number of triangles = $(n-2) \pi / \pi = n-2$.

$Proof of problem$:
Let there be a total of $x Black$ triangles in $n$ sided convex polygon $P1$.

Now consider the polygon $P2$ formed after removing all the $x$ black triangles from polygon $P1$. It's not hard to see that $P2$ has exactly $(n - x)$ sides.

Now, in polygon $P2$, we are left with exactly $(n - x) - x = n - 2x$ sides which are still part of the original Polygon $P1$.

No two adjacent sides of these $(n - 2x)$ can be joined to form another $Black$ triangle (since we have assumed a total of $x Black$ triangles exist).

So that leaves us with $(n - 2x)$ $Red$ triangles in total.

So, now all the triangles that remain must be $White$ triangles.

From lemma, $P2$ has exactly $(n - x - 2)$ triangles satisfying the conditions of the problem, so the number of $White$ triangles = $(n - x - 2) - (n - 2x) = x - 2$

Thus total number of $White$ triangles = $x - 2$ which is $2$ less than total number of $Black$ triangles.

We induct on $n$, with trivial base cases $n=4$ and $n=5$. We also note that 0 white triangles imply 2 black triangles and the rest are red. Hence we assume the existence of at least 1 white triangle, which we draw in. The remainder of the polygon is split into 3 smaller polygons, from which we claim the following:

[color=#00f][b]Claim:[/b][/color] In each of these polygons, we have $B-W=1$.
[list]
[*] Obvious if the polygon is a triangle.
[*] Otherwise, we utilize our inductive hypothesis, which says $B-W=2$ with respect to the sides of our smaller polygon. However, we must consider the triangle with one side on our main white triangle.

If the triangle is black in our smaller polygon, it is red in our original polygon, and if it is red in our smaller polygon, it is white in our original polygon. Thus the difference $B-W$ decreases by 1. ${\color{blue} \Box}$
[/list]

Thus we conclude by summing over the 3 new polygons and adding back our white triangle. $\blacksquare$

[hide="Storage:"]

We'll prove by induction two statements at the same time : First, there exists a black triangle. Second, there are two more black triangles than white ones. For $n=4$ both are trivial, for we are forced to connect two diagonally opposite vertices of the quadrilateral.

Now take $n\ge5$ and consider a convex $n$-gon. Supposing that the first statements are true for a $k$-gon st $4 \leq k <
 n$, then we can take any arbitrary triangle in the triangulation of the $n$-gon. IF it's black, we are done. Otherwise, it divides the polygon in two. Take any half, use $1$st inductive statement on it. If the new black triangle is also black in the $n$-gon, we are done. Otherwise, it contains a side of the original triangle. Repeat this operation until we clearly eventually get to a black triangle. This proves the first statement.

For the second statement, take a black triangle in the $n$-gon.Consider the $(n-1)$-gon formed by removing the two sides of the original $n$-gon that were lying on said black triangle, with the vertex connecting them with it. By induction, there are $2$ more \emph{new} black triangle than \emph{new} white ones. Consider the triangle containing the third segment of the original black triangle. If this triangle is \emph{new} black, then its \emph{old} red. If it's \emph{new} red, then its \emph{old} white. In any case, taking in account the first black triangle, we notice that the difference stays constant.

So by induction, there are two more black triangles than white ones.

$$\mathbb{Q.E.D.}$$
[/hide]


[b]Pro1[/b] Let $ x,y,z\in N^*$. Solve equation
$ 12^x\plus{}4^y\equal{}2008^z$

[b]Pro2[/b] Find $ x,y\in N$ such that
$ (x^2\plus{}y)(y^2\plus{}x)\equal{}(x\minus{}y)^3$

[b]Pro3[/b] $ p$ is odd prime number. $ x,y\in N$
Solve equation
$ x^p\plus{}y^p\equal{}p.((p\minus{}1)!)^p$

It might be more convenient if these problems were posted as three separate topics.


AMC
USA(J)MO
USAMO
induction
number theory

Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Please use LaTeX for posting solutions. Thanks.

Don't post in solved problems, please. This (and all the others) should go in the proposed problems section.

Ooh and this was the very first problem I solved on an official olympiad ... yay :).

[hide]It follows by induction on $n$; clearly for $n = 1$, $5$ satisfies the given conditions. Suppose the proposition works for $n$, i.e. there exists some number $N$ satisfying the giving conditions. Consider the five numbers $i*10^n + N$, where $i = 1, 3, 5, 7, 9$. It is clear that these five numbers all have $n+1$ digits with all digits odd, from the inductive hypothesis. Also from the inductive hypothesis, they are all divisible by $5^n$. If one were to divide them by $5^n$ however, they would differ by $2^n$ between consecutive numbers, i.e. they would all have different residues mod $5$. Thus one exists whose residue mod $5$ is $0$, or in other words, the original number was divisible by $5^{n+1}$ and thus satisfies the conditions for $n+1$.[/hide]

It is easy to construct this number by induktion.
$k_0=1, N_0=k_0*5^n=a_{n-1,0}a_{n-2,0}\dots a_{0,0}, a_{0,0}=5 \ is \ odd.$
Let we have $k_l<2^{m_l+1},k_l*5^n=N_l=a_{n-1,l}a_{n-2,l}\dots a_{0,l}$ were m_l+1 digits $a_{m_l,l}\dots a_{0,l}$ are odd.
Let we have $m_{l+1}>m_l$, suth that $a_{m_{l+1},l}$ is even. We construct
$k_{l+1}=k_l+2^{m_{l+1},N_{l+1}=k_{l+1}*5^n}$ with property: digits $a_{i,l+1}$ are odd for $i\le m_{l+1}$.
This number is unique.

[quote="Rust"]It is easy to construct this number by induktion.
$k_0=1, N_0=k_0*5^n=a_{n-1,0}a{n-2,0}\dots a_{0,0}, a_{0,0}=5 \ is \ odd.$
Let we have $k_l<2^{m_l+1},k_l*5^n=N_l=a_{n-1,l}a_{n-2,l}\dots a_{0,l}$ were m_l+1 digits $a_{m_l,l}\dots a_{0,l}$ are odd.
Let we have $m_{l+1}>m_l$, suth that $a_{m_{l+1},l}$ is even. We construct
$k_{l+1}=k_l+2^{m_{l+1},N_{l+1}=k_{l+1}*5^n}$ with property: digits $a_{i,l+1}$ are odd for $i\le m_{l+1}$.
This number is unique.[/quote]

Thank you very much Rust please and is this the answer to my question ?

Davron

[quote="Rust"]I think you are wrong. Check it by programming. It is easy calculate for n<1000.[/quote]

nope but we have all the terms as odd numbers...?
 :)

the first number to satisfy ur condition is 375=3*5^3

We will proceed by induction to prove that there always exists a $n$ digit number with all odd digits $A_n$ such that $5^n|A_n$.

Base Case: $n=1$, let $A_1=5$ and $5^1|A_1=5$.

Inductive Hypothesis: There exists a $k$ digit number with all odd digits such that $5^k|A_k$.

Now let $a\in\{1, 3, 5, 7, 9\}$ and $A_k=b_k\cdot 5^k$, then let $A_{k+1}=a\times 10^{k}+bA_k$ and we shall show that you can always pick some $a$ and $b$ such that $5^{k+1}|A_{k+1}$.  Notice that $A_{k+1}$ is $A_k$ with an odd digit augmented to beginning of the number.  Thus, $A_{k+1}$ contains only odd digits.
\[A_{k+1}=5^{k}(a\cdot 2^k+b_k)\]
So if we can choose some $a$ such that $a\cdot 2^k+b_k$ is divisible by $5$, then $5^{k+1}|A_{k+1}$.  However, this is obviously true because $2^k a$ can take on any possible residue modulo $5$ since $a$ can take on any residue modulo $5$ and we are done. $\blacksquare$

Induction: Clearly true for $n=1$. Now assume true for $n=k$, where $d_i$ are the digits in the decimal representation, i.e. $5^k\mid{\overline{d_1d_2\cdots{d_k}}_{10}}$ with all $d_i$ odd. Then the next number in the list can potentially be any one of:
$\overline{(2a+1)d_1d_2\cdots{d_k}}_{10}$, where $0\leq{a}\leq{4}$; we want to show that one of these has to be divisible by $5^{k+1}$.
By the inductive hypothesis: $(2a+1)\cdot{10^k}+5^kt$; $t\in{\mathbb{Z^+}}$.
Factoring (I-V) we have: $5^k(2^k(2a+1)+t_1)=5^k(2^{k+1}a+2^k+t_1)$.
But wait! $(2^{k+1}a+2^k+t_1)$ forms a complete set of residue classes $\mod{5}$ as it runs across the $a$'s (this is obvious, see myticterminator's proof of it above), so we're done. $\blacksquare$

Quick
[hide=Solution]
We use induction on $n$. For $n=1$ pick $5$, completing the base case. Suppose the result is true for $n=k$, we will show it is for $k+1$.

Let $\overline{a_ka_{k-1}\cdots a_1}$ be an $k$ digit number such that all $a_i, 1 \le i \le k$ and $5^k|\overline{a_ka_{k-1}\cdots a_1}$. Suppose $\frac{\overline{a_ka_{k-1}\cdots a_1}}{5^k}=x$.
Consider the set of numbers $x_1=\overline{1a_ka_{k-1}\cdots a_1}, x_3=\overline{3a_ka_{k-1}\cdots a_1}, x_5=\overline{5a_ka_{k-1}\cdots a_1}, x_7=\overline{7a_ka_{k-1}\cdots a_1}, x_9=\overline{9a_ka_{k-1}\cdots a_1}$. Trivially, $5^n|10\cdots 0$ where there are $n$ zeroes.
Note that when $1 \le i \le 5, x_i=5^k(i+x)$. $x \pmod{5}$ can be $0, 1, 2, 3, 4$ but $i$ ranges through the whole set of residues$\pmod{5}$, so we can choose an appropriate $i$ such that $5^{k+1}|x_i$. This completes the inductive step so done.
[/hide]

We prove the set of $n$-digits integers whose digits are odd form a complete set of residue modulo $5^n$. There are $5^n$ such integers. It suffices to show they are distinct.

[hide=Proof]
Let $A,B$ be 2 distinct $n$-digits integers whose digits are odd. For sake of contradiction, suppose $A\equiv B\pmod{5^n}$. Let $0\le k\le n-1$ be the smallest integer s.t. $A$ and $B$ differs on the $k^{\text{th}}$ digit (from the right). I.e.
$$A=\overline{a_{n-1}\ldots a_ka_{k-1}\ldots a_0}\;\wedge\;B=\overline{b_{n-1}\ldots b_ka_{k-1}\ldots a_0}$$
Hence $5^n\mid A-B\iff 5^{n-k}\mid\overline{a_{n-1}\ldots a_k}-\overline{b_{n-1}\ldots b_k}=C\implies 5\mid C$.
So the last digit $c_0$ of $C$ is either $0$ or $5$. Note $c_0\equiv a_k-b_k\pmod{10}$ and both $a_k$ and $b_k$ are odd, so $c_0$ is even ie $c_0=0$. Finally, $a_k=b_k$. Contradiction.
[/hide]

PS : I suppose this problem inspired [url=https://www.artofproblemsolving.com/community/c6h148016p836869]Croatian National MO 2005[/url].

[hide=Solution]We shall induct on $n$.

Base Case: $n=1$.

Take the $1$-digit number $5$.

Inductive Step: Suppose the $k$ digit number $a_{k-1}a_{k-2}\cdots a_1a_0$ is divisible by $5^k$ with all odd digits.

Claim: One of $1a_{k-1}a_{k-2}\cdots a_1a_0$, $3a_{k-1}a_{k-2}\cdots a_1a_0$, $5a_{k-1}a_{k-2}\cdots a_1a_0$, $7a_{k-1}a_{k-2}\cdots a_1a_0$ or $9a_{k-1}a_{k-2}\cdots a_1a_0$ will be divisible by $5^{k+1}$. It is easy to see that all of these are still divisible by $5^k$ since $10^k\equiv 0\pmod{5^k}$. Say the odd number for the first digit we pick is $2m+1$, then let $a_{k-1}a_{k-2}\cdots a_1a_0=5^k\cdot n$. We have $(2m+1)\cdot 10^k+5^k\cdot n=5^k(2^{k+1}\cdot m+2^k+n)$. So it suffices to show that $2^{k+1}\cdot m+2^k+n$ is divisible by $5$ for one of $m=0,1,2,3,4$. However this is basically the same as showing that $2^{k+1}\cdot m$ each produce a different residue mod $5$ as $m$ varies which is obvious. Thus, at least one of these values for $m$ will give the expression to be divisible by $5$.

Thus, our induction is complete and our claim is true.[/hide]

We proceed by induction on $n$. When $n=1$, then number is $5$.
\\$~$\\
Suppose $\overline{c_{n-1}c_{n-2}\dots c_1c_{0}}=5^n\cdot k$ is the $n$-digit number described in the conditions, then we need \[5^{n+1}\mid c_n\cdot 10^n+5^n\cdot k=5^n(2c_n+k)\implies 5\mid (2c_n+k).\] Since $\gcd(2,5)=1$, there is a solution $c_n\equiv c\pmod 5$ where $0\le c\le 4$. Now, pick either $c$ or $c+4$ and one of them will be odd. Therefore, we are done.

We induct on $n$; the base case $n=1$ is obvious. Now, consider a valid number $\overline{a_1a_2\cdots a_n}$ with all odd digits divisible by $5^n$. Then, one of the five numbers given by $\overline{k a_1a_2\cdots a_n}$, with $k$ an odd integer between $1$ and $9$, must be divisible by $5^{n+1}$, as $5^{n+1} \nmid 2\overbrace{00\cdots 0}^{n \text{ zeroes}}$ by Pigeonhole, which completes the induction.

Use induction
when $n=1$, $5$ works
if $n$ works, then $a*10^n+\text{previous value that worked}$ is divisible by $5^n$. There are $5$ possible values and each of them has a different remainder when divided by $5^{n+1}$ because of pigeonhole, so one of the must divide $5^{n+1}$

USSR 1967 or All-Russian 1967
[url]https://artofproblemsolving.com/community/c893771h1859931p12577115[/url]

When $n = 1$, we see that $5$ works. We will now induct on $n$. Let $a_k$ be our desired positive integer with $k$ digits, where we write $a_k = 5^kr$ for some odd positive integer $r$. Then we claim that we can construct $a_{k + 1}$ by appending some odd digit $d$ to the start of $a_k$, so we want
$$a_{k + 1} = d \cdot 10^k + a_k = 5^k(d \cdot 2^k + r). $$
to be divisible by $5^{k + 1}$, or $d \cdot 2^k + r$ to be divisible by $5$. But since $2^k$ and $5$ are relatively prime for all positive integers $k$ and $1$, $3$, $5$, $7$, and $9$ all have distinct residues modulo $5$, we will always be able to find exactly one suitable $d$ such that $5 \mid (d \cdot 2^k + r)$. This completes the proof. \\

Failed on this some months ago. Tried again and did it!

[i]Solution:[/i] We will perform induction on $n$. Some base cases are
[list]
    [*] For $n = 1$, take 5,
    [*] For $n = 2$, take 25,
    [*] For $n = 3$, take 375,
    [*] For $n = 4$, take 9375.
[/list]
As an induction hypothesis, assume there exists a number (say $X$) with $k$ digits divisible by $5^k$ with all odd digits. Write $X$ as
    \[X = \sum_{n = 0}^{k}10^{n-1}\cdot a_n\]
where $a_i$ is a digit. For the induction step, consider number $X' = X + 10^kb$ where $b$ is a digit. We need to show that there exists an odd $b$ such that $X'$ is divisible by $5^{k+1}$. Reduce $X'$ modulo $5^{k+1}$ with the fact that $X = 5^k \cdot \alpha$ for positive integer $\alpha$.
    \[X' \equiv 0 \pmod{5^{k+1}} \iff b\cdot 2^k + \alpha \equiv 0 \pmod{5}\]
Since $5 \nmid 2^k$, there will always be a solution to the congruence for $b$. If the smallest positive residue of $b$ is odd, then we are done, otherwise add 5 to it to make it odd. Everything will obviously work by tracing the work backwards. $\blacksquare$

Sketch:

We can prove this inductively.

Base Case: $n = 5$ clearly abides.

Induction: Take the number $k$ digit number that is divisible by $5^k$. Then to construct a $k+1$ digit number that is divisible by $5^{k+1}$, we can either append $1, 3, 5, 7, 9$ to the left of the number. One of these selections is legal (we can see this through algebra and some basic principles of modular arithmetic).

proceed by induction. Base case n=1 is trivial. Now, we'll append an odd digit onto the left side of the old $n$ digit number $a_n$. $\frac{a_n}{5^n}=b_n\implies 5^n(2^nk+b_n)$. If $b_n$ is divisible by 5 we can just choose k=5; if not, noting that $$5\nmid 2^n,b_n\implies b_n=\{1,3,5,7,9\}\equiv 2^nk\pmod 5,$$ since 1,3,5,7,9 is a complete residue set and so is $2^nk$, hence we can always choose a $k\le 10$ such that this works.

We will show that if there exists a $k$ digit integer divisible by $5^{k}$ there exists a $k+1$ digit integer divisible by $5^{k+1}$.
If we add a digit $a$ to the far left of are $k$ digit integer we will show there exists an $a$ such that this is divisible by $5^{k+1}$
Let are $k$ digit integer be $c$ then we require
 $$c+a10^{k}\equiv  0 \pmod{5^{k+1}}$$
$$\frac{c}{5^{k}}+2^{k}a \equiv 0 \pmod{5}$$
Since we get a complete residue set mod $5$ as possibilities for $a$ the desired number exists.
Since $5$ has $1$ digit and is divisible by $5$ we have completed our induction.

We prove this statement by induction. The base case for $n = 1$ is trivial, as $5^1 \mid 5$.

Assume that for all $N \leq n$, we have $k \cdot 5^n$ has all of its $n$-digits as odd. (Observe that $k \in \{1, 3, 5, 7, 9\}$). Consider the $(n+1)$-digit integer $\ell \cdot 10^n + 5^n$. Factoring yields, \[5^n(\ell \cdot 2^n + k), \phantom{ccc}\ell \in \{1, 3, 5, 7, 9\}.\] Notice that the inside expression goes through all residues modulo $5$. Hence for every value of $\ell$, there is some $k$ for which  $(\ell \cdot 2^n + k)$ is divisible by $5$, so that the entire expression including the factor of $5^n$ is divisible by $5^{n+1}$, which finishes the induciton. $\blacksquare$

Consider induction on $n$. When $n=1$, the desired number is $5$, and when moving forward, consider a working $n$-digit number to be $N$. Then, simply notice that exactly one of the elements in

\[S = \{10^n+N, 3 \cdot 10^n+N, 5 \cdot 10^n+N, 7 \cdot 10^n+N, 9 \cdot 10^n+N\}\]

will satisfy the conditions for an $(n+1)$-digit number, so we are done. $\square$

We proceed with induction. The base case, $n=1$ is obvious by just taking $5$ as our number. For our inductive step, assume $k$ is an $n$-digit number divisible by $5^n$ with all odd digits. We know that $k\equiv 0 \pmod {5^n}$, so $k$ can be congruent to either $0, 5^n, 2(5^n), 3(5^n)$, or $4(5^n)$ modulo $5^{n+1}$. Notice that the $n+1$ digit number $x(10^n) = x\cdot 2^n (5^n)$ as $x$ ranges over $1, 3, 5, 7, 9$ can only be $0, 5^n, 2(5^n), 3(5^n)$, or $4(5^n)$ modulo $5^{n+1}$. In fact, each of these five residues appears exactly once due to $\gcd(5, 2^n) = 1$. Thus, given a $k$, we can always pick an $x(10^n)$ that makes $k+x(10^n)$ an $n+1$ digit number with all odd digits and $k+x(10^n)\equiv 0\pmod {5^{n+1}}$. This completes our induction, so we're done. $\blacksquare$

We will use induction.
$\newline$
Let our $n$-digit number be $k_n$.
Then our base case of $n = 1$ works for $k_1 = 5$.
$\newline$

Inductive Step:
$\newline$

Assume that $k_n$ works. Then we can prove that $k_{n+1}$ works.
Let $k_{n+1} = a \cdot 10^n + k_{n}$ for some $a \in \{1, 3, 5, 7, 9\}$.
And since all of $\{1, 3, 5, 7, 9\}$ are distinct modulo $5$, then there must be some $a$ that makes $k_{n+1} \equiv 0\pmod{5^{n+1}}$ since both $10^n$ and $k_n$ are divisible by $5^n$, so we are done.

[quote=Davron]Who can show that the first digit number of the k-digit number can be never equal to 1 ?

Davron[/quote]

Consider 193,359,375

we will use induction
we first have that there exists $(2x_0+1)*10^0$ that is divisible by $5^1$
next, assume there exists $(2x_0+1)*10^{n-1}+...+(2x_n+1)*10^0$ divisible by $5^n$
multiply by $10$ to get $2(x_1*10^n+...+x_n*10)+\frac{10^{n+1}-10}{9}$
this is divisible by $5^{n+1}$
add one to get $2(x_1*10^n+...+x_n*10)+\frac{10^{n+1}-1}{9} \equiv 1 (\text{mod }5^{n+1})$
we need $2(y_1*10^n+...+y_n) \equiv -1 (\text{mod }5^{n+1})$
the existence of this is trivial since $5^{n+1}-1$ is even
therefore, we are done


geometry
abstract algebra
complex numbers

If $a , b , c$ are dinstict complex numbers such that $a^2 + b^2 + c^2 = ab +bc +ca$ , prove that the images of $a ,b , c$ form a equilateral triangle.

  [b]NOTE[/b]

 Virgil has a nice solution there. Please , try to find - if possible - a solution using modules and simple algebra. Not roots of unit , or trigonometrical form. I repeat:If possible. Perhaps such a solution is impossible.I also try to find it.

  Thanks a lot !

 [u]Babis[/u]

Actually here is a nice solution to this one.

The relation that Stergiu wrote is equivalent with this one
 \[ (a-b)^{2}=(b-c)(c-a)\Leftrightarrow \]
 \[ |a-b|^{2}=|b-c||c-a|. \]
Writing the analogues, and by adding them up, we get \[ |a-b|^{2}+|b-c|^{2}+|c-a|^{2}=|a-b||b-c|+|b-c||c-a|+|c-a||a-b|. \]
 From here we easily get $|a-b|=|b-c|=|c-a|$. :)  ;)


Has the difficulty of the AMC 12 changed over the years or has it stayed relatively constant?  I don't know if it is just me but it seems as though the tests after 2007 are easier (because of no calculators allowed now?)...

This is to some extent, true(about the easiness thing). This works for all AMCs. For example, take the 2009 AMC10. You'll find it extremely easy compared to past years, mainly because they outlawed calculators.

[quote="MathWise"]I got a 126 on the 2006 A AMC 12, but I didn't use a calculator. Does anyone have an idea of to what score this would correspond on a non-calculator AMC 12?[/quote]There really is no set number, it depends on your mental math skills and how much hard mental math there was at that time.

[quote="MathWise"]I got a 126 on the 2006 A AMC 12, but I didn't use a calculator. Does anyone have an idea of to what score this would correspond on a non-calculator AMC 12?[/quote]

As Ihatepie said, this corresponds with your own mathematical ability. Basically, the advantage with calculators is that you save time, and considering that you got 126 without a calc, with a calc you could probably boost this score up by 6 points (maybe higher, again depending on your mathematical ability).

so would any of you agree that the AMC 12s after 2007, for the most part, are easier than the ones of 2007 and earlier?


Suppose we have $ 2n$ integers between $ \minus{}n$ and $ n$ whose sum equals one (Where $ n$ is a natural number bigger than one). Prove that the sum of a subset of these $ 2n$ integers equals zero.

[quote="theDeviation"]Suppose we have $ 2n$ integers between $ \minus{} n$ and $ n$ whose sum equals one (Where $ n$ is a natural number bigger than one). Prove that the sum of a subset of these $ 2n$ integers equals zero.[/quote]

First of all i assume you wanted to include -n and n while choosing 2n numbers, or else there are only 2n-1 numbers in between.
Next check that sum of all no. from -n to n = 0. also there are 2n+1 no. Hence we must choose all no. except -1. The subset {0} works. :wink:

I don't think he assumes the numbers to be distinct...

  darij

For the case $ n\equal{}1000$, this was the third problem of the Canadian Mathematical Olympiad in the year 2000.

Here's the solution: [url]http://www.math.ca/Competitions/CMO/solutions/sol_2000.pdf[/url]


Determine $ a,b\in \mathbb{Z}$ such that $ a^4\plus{}b^4$ is a prime number and $ 2a^4\plus{}2b^4\minus{}1$ is a perfect square .

There aren't any.

Suppose $ a^4\plus{}b^4 \equal{} p$, a prime number, and $ 2a^4\plus{}2b^4 \minus{} 1 \equal{} n^2$, a square. $ n$ is obviously odd so we may write $ n \equal{} 2m\plus{}1$ and then

$ 2a^4 \plus{} 2b^4 \minus{} 1 \equal{} (2m\plus{}1)^2 \equal{} 4m^2 \plus{} 4m \plus{} 1$

so

$ a^4 \plus{} b^4 \equal{} 2m^2 \plus{} 2m \plus{} 1$

but the LHS is just $ p$ so we have

$ p \equal{} m^2 \plus{} (m\plus{}1)^2$.

We now have $ p$ written as a sum of $ 2$ squares in two ways: $ p \equal{} a^4 \plus{} b^4$ and $ p \equal{} m^2 \plus{} (m\plus{}1)^2$. But it's well-known that if a prime number can be written as a sum of $ 2$ squares, then it can be so written in an essentially unique way. It follows that $ m\equal{}a^2$ and $ m\plus{}1\equal{}b^2$ or vice versa, which is absurd.

The most elegant and conceptual proof that $ p\equal{}a^2\plus{}b^2$ has a unique solution (if it has any solutions at all) relies on simple properties of the ring $ \mathbb{Z}[i]$. This must be written up here somewhere, but if you're not familiar with it I can provide a brief account.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $G$ be finite abelian group in which the number of solutions in $G$ of equation $x^n=e$ is at most $n$ for every positive integer $n$. Prove that $G$ must be cyclic group.

Just count how many of each order are there (for the abelian case it's simple to show that $x^d=e$ has at least $\gcd(d, |G|)$ solutions).
By http://www.mathlinks.ro/Forum/viewtopic.php?t=22490 the word abelian is not necessary.

I'm new about this, Can you write the proof ZetaX?

thanks

Claim: for $d$ dividing $|G|$ there are at least $d$ solutions to $x^d=e$.

The prove for all groups is in the link.

The prove for abelian groups only:

When $a,b$ are elements with coprime orders, then $ab$ has the order of their product (not hard to show).

Thus it suffices to show that for any prime power $p^n$ dividing $|G|$ there are at least $p^n$ solutions to ${x^{p^n}}=e$, since then for any $d=p_1^{v_1} \cdot p_2^{v_2} \cdot ... \cdot p_m^{v_m}$ dividing $|G|$ we can take products $x=x_1 x_2 ... x_m$ where $x_i^{p_i^{v_i}} = e$  and there are enough such products (they will all be different, since when we have two such produvts $x,y$, then by the order-of-product-theorem above $xy^{-1}$ has an order $>1$ when they are different).
So it is reduced to the prime powers, where it follows from the Sylow theorem (which can be shown a lot easier for abelian groups), because we then have subgroups of any prime power dividing $|G|$, and Lagranges theorem gives us that all these elements of that subgroup are what we want.

Another way to prove this would be to use that every finite abelian group is a product of cyclic groups.

So by the above and the statement we have that there are at most and at least $d$ solutions to $x^d=e$, thus exactly $d$, for any divisor $d$ of $|G|$.
Now start counting how many solutions are there for $x^{|G|}=e$ that are not solutions of 'smaller' equations.
Well, when we write $y(d)$ for the number of elements with order $d$, we have that $n= \sum_{d|n} y(d)$ for any $n$.
Thus the möbius inversion formula tells us that there are $y(d) = \sum_{n|d} \mu (d/n) n = \phi(d)$ elements with order $d$, and $y(|G|)$ is clearly $>0$, thus there is a generating element.

thanks for the proof ZetaX.

i don't remember that i've post this problem, until my prof gave this problem last week.
But i think there is another solution which is more elementary..
this problem appears in the early section, that is about order of group.

Yes you are right, it can be done more directly:

Let $ \psi(d)$ be the number of elements of order $ d$. Every single $ x$ with order $ d$ gives already $ d$ different $ y$ with $ y^{d}\equal{} e$, namely the powers of $ x$, and exactly $ \phi(d)$ of these powers have order $ d$ again. As a result, $ \psi(d)\leq\phi(d)$.
Now $ n \equal{}\sum_{d|n}\psi(d)\leq\sum_{d|n}\phi(d) \equal{} n$, thus we always have equality in  $ \psi(d)\leq\phi(d)$, especially $ \psi(|G|) \equal{}\phi(|G|) > 0$, which means that the group is cyclic.

Yes, that's what i mean, and that is also my solution after two day searching for  :blush:

thanks for the smart reply,


Howdy
I am both new to Latex and to Mac.
I have used Latex without too many problems but today i needed to use the combinations notation and discovered that I needed to install amsmath to do this.
I followed the instructions from the Ams link
ftp://ftp.ams.org/pub/tex/amslatex/install.txt
i managed to download the relevant files and stored them in my Applications folder. The compiler still doesn't recognise the \binom command however. I have a feeling i should be careful about what folders the downloaded files should be stored but haven't a clue as to what to do next.
Any help would be appreciated.
Mark

I'm not familiar with LaTeX on the Mac but I can't believe any LaTeX distribution comes without amsmath installed as it is such an important part of it.
When you put \usepackage{amsmath} in your document does the log say it can't find the package? Do any amsmath files (such as amsmath.sty) exist on your computer, apart from the ones you've downloaded?

Please post which LaTeX distribution you have installed and then maybe others with the same distro will be able to advise.

Problem solved...as i said i am new to Latex!...a friend spotted that i had put \usepackage{amsmath} after the begin{doc...} command instead of before!
Thanks anyway.
Mark


Find all functions $ f: \mathbb R\rightarrow\mathbb R$ satisfying $ f(x\plus{}f(xy)\equal{}x(1\plus{}f(y))$ for every $ x,y\in\mathbb R$

It is very easy.
Let $ P(x,y)$ be the mentioned property of f.
Then:
$ (P(0,y))\Rightarrow (f(f(0))\equal{}0)$ let us have $ a\equal{}f(0)$ then
$ f(a)\equal{}0$
$ (P(1,a))\Rightarrow (f(1)\equal{}1)$ and now
$ (P(a,1))\Rightarrow (0\equal{}f(a)\equal{}f(a\plus{}f(a))\equal{}2a)$ so $ f(0)\equal{}0$
Now from $ P(x,0)$ we conclude that $ f(x)\equal{}x$ :wink:


Let $ABCD$ be a square with diagonals $AC$ and $BD$, and $P$ a point in one of the sides of the square. Show that the sum of the distances from P to the diagonals is constant.

[hide="Fun solution"]
Let $AC$ be the positive $y$-axis and let $BD$ be the negative $x$-axis.  WLOG $P$ is on $AB$.  Then the line $AB$ is clearly the graph $y + x = \frac{AB}{ \sqrt{2} }$, and $x, y$ are the distances to the diagonals! [/hide]

Let the distance from $P$ to BD and AC be $x$ and $y$ respectively and $O$ be the centre of the square. Then:

$x=\frac{AO}{AB}\cdot PB$ and $y=\frac{BO}{AB} \cdot PA$. Sum up we get $x+y=BO=\frac{AB}{\sqrt{2}}$.


geometry
parallelogram
vector
invariant
geometry unsolved

How do you prove that the medians of a quadrilateral bisect each other? I think I have to prove that the quadrilateral created by connecting the midpoints of the original quad is a parallelogram (aka the Varignon quadrilateral). But I can't figure out how to prove this. Any suggestions?

Vectors!

Write each of the points of the original quadrilateral as vectors $\vec{a},\vec{b},\vec{c},$ and $\vec{d}.$

Then the midpoints of the sides are $\frac12(\vec{a}+\vec{b}), \frac12(\vec{b}+\vec{c}), \frac12(\vec{c}+\vec{d}),$ and $\frac12(\vec{d}+\vec{a}).$

It turns out that we're now setup to find the midpoints of the diagonals:

$\frac12\left[\frac12(\vec{a}+\vec{b})+\frac12(\vec{c}+\vec{d})\right]= \frac14(\vec{a}+\vec{b}+\vec{c}+\vec{d})$

$\frac12\left[\frac12(\vec{b}+\vec{c})+\frac12(\vec{d}+\vec{a})\right]= \frac14(\vec{a}+\vec{b}+\vec{c}+\vec{d})$

We didn't even go through the proof that the smaller quadrilateral is a parallelogram, but that is also straightforward in this setup.

Thank you Kent. I was anticipating a more "geometric" proof using Euclidean theorems, but now I see a different way to look at it.

I think that we don't need to solve the basic theorems by vectors, because we will lose, the meaning of Euclidean geometry.

$\bullet$ We denote as $K,$ $L,$ $M,$ $N,$ the midpoints of the sidesegments $AB,$ $BC,$ $CD,$ $DA$ respectively, of an arbitrary quadrilateral $ABCD.$

So, we have that $KL\parallel = \frac{AC}{2}\parallel = MN.$

Hence, the quqdrilateral $KLMN,$ is a parallelogram and the intersection point $Q,$ of it's diagonals, bisects them.

Similarly, if $P,$ $R,$ are the midpoints of $AC,$ $BD$ respectively, we have that $PL\parallel = \frac{AB}{2}\parallel = NR.$

That is the point $Q,$ bisects all the medians of $ABCD.$

Kostas Vittas.

One thing that the vector proof shows is that this is an affine-invariant theorem. That is, all of the points we're concerned with would retain their identities under any (invertible) affine transformation of the plane.


X,Y,Z are positive real numbers satisfying

$\frac{X(Y+Z-X)}_{logX}$ = $\frac{Y(Z+X-Y)}_{logY}$ = $\frac{Z(X+Y-Z)}_{logZ}$

Prove $Y^{Z}.Z^{Y}$ = $Z^{X}.X^{Z}$ =$X^{Y}.Y^{X}$

[hide]
Let us assume $ \frac{x(y+z-x)}{\log{x}}=\frac{y(x+z-y)}{\log{y}}=\frac{z(x+y-z)}{\log{z}}=k$
$\log{y}=\frac{y(x+z-y)}{k}=\frac{xy+yz-y^{2}}{k}$
$\log{x}=\frac{x(y+z-x)}{k}=\frac{xy+xz-x^{2}}{k}$
\begin{eqnarray*}\log{y}-\log{x}&=& \log{\frac{y}{x}}\\ &=& \frac{yz-xz-y^{2}+x^{2}}{k}\\ &=& \frac{(x-y)(x+y-z)}{k}\\ &=& \frac{(x-y)\log{z}}{z}\end{eqnarray*}
\begin{eqnarray*}\frac{(x-y)\log{z}}{z}&=&\log{\frac{y}{x}}\\ (x-y)\log{z}&=&z\log{\frac{y}{x}}\\ z^{x-y}&=&\left(\frac{y}{x}\right)^{z}\\ \frac{z^{x}}{z^{y}}&=&\frac{y^{z}}{x^{z}}\\ z^{x}x^{z}&=&y^{z}z^{y}\end{eqnarray*}
and similarly for the other cases
[/hide]


nice un  :rotfl:

btw im not able to post/send pms at all wen im on my pc. i have to go on my dads laptop every time, which is irritating coz i get that opportunity on once in a while :( even rohit had the same prob on his comp.
any ideas why, any1??

Is it because no one has replied or does it not let you send one alltogether? If it's the latter, you've probably been banned from pming.

Is there something like you have to send a min no of posts before you get PM'ing priveledges?

I do not believe there is no requirement for pming (there might be a wait period for new users).


grommit wrote:I wrote this problem while investigating interesting integer functions.  The "f(3n)=f(2n)" part below appears on the Mandelbrot Problem Corner.

f:{0,1,2....}->{0,1,2,...}
i) f(3n)=2f(n)
ii) f(3n+1)=f(3n)+1
iii) f(3n+2)=f(3n)+2, for all integers n.
Prove: f(3n):ge:f(2n)>f(n).
Also, when does f(3n)=f(2n), and when does f(2n) = f(n) + 1?

An "elegant" brute-forcing...actually its not elegant at all.

[hide]

This is a well-defined function. We know by i that f(0) = 0, so by the ii and iii, we get f(1) = 1 and f(2) = 2. We know f(3) = 2(1) = 2, and f(4) = 3, f(5) = 4.

Continuing, we see that f(6,7,8) = 4,5,6, f(9,10,11) = 4,5,6, f(12,13,14) = 6,7,8, f(15,16,17) = 8,9,10, f(18,19,20) = 8,9,10...

f(3n) thus retains a pattern of 0,2,4,4,6,8,8,10,12....

Namely, f(3n) = f(3(n+3))+4, about f(n) = 4/3*n.

f(2n) retains a pattern of 0,2,3,4,6,5,6,8,9,8,10,11,12,14...

f(2n) = f(2(n+10)) + 10, about f(n) = n.

f(n) gives a pattern of 0,1,2,2,3,4,4,5,6,4,5,6,6,7,8,8,9,10....

f(n) = f(n+9) + 4, about 4/9*n.

We can trivially prove the first few, and thus f(3n):ge:f(2n)>f(n)

f(3n) = f(2n) when n = 0,1,3,4, and nowhere else.

f(2n) = f(n) + 1 when n = 1,2,5 and nowhere else.

[/hide]

Ummm......masamune, your answers are false... for instance, there are infinitely many n such that f(3n) = f(2n)...  not sure what you did wrong, since there don't seem to be any proofs in you post... would you mind posting your reasoning so that we can work out what is wrong with it?  It is very tempting to assume that patterns here will continue in ways that they do not.

Having already seen this problem, I won't post any solutions unless someone wants me to.  A very nice problem, by the way.

Ah. The pattern doesn't continue...my bad. I'll come back later to fix this solution...

Hmm...can we generalize anything about f(p^k*3^n), where p is a prime?

[hide]

Lemma:  f(n) returns n, written in base 3, read in base 2 .

Proof is by induction. If n = 3k + c, with c < 3, we induct on k.

If k = 0:

f(0) = 2f(0) => f(0) = 0.

f(1) = f(0) + 1 = 1

f(2) = f(0) + 2 = 2, so the claim is satisfied for k=0.

In general,

f(3(k+1) + c) = f(3(k+1)) + c = 2f(3k) + c. Since f(3k) is the base 3 representation read in base 2, we see that 3(k+1) has a base 3 representation shifted one place to the left from 3k, and that f(3(k+1)) = 2f(3k), so f(3(k+1)) has its base 2 representation shifted one place to the left of f(3k). Therefore, the lemma is proven.

---

Let k = a_n a_(n-1) a_(n-2) ... a_0 in base 3. Then

f(k) =  :Sigma: 2i a_i .

Let 2k = b_(n+1) b_n ... b_0 in base 3. Then letting c_n represent potential carries in addition base 3...

b_0 = 2a_0 - 3c_0

b_1 = 2a_1 + c_0 - 3c_1

.

.

.

b_n = 2a_n + c_(n-1) - 3c_n

b_n+1 = c_n

f(2k) = :Sigma 2i b_i =  2:Sigma: 2ia_i  - :Sigma: 2i c_i = 2f(k) -  :Sigma: 2i c_i :le: 2f(k) = f(3k).

 Therefore, f(3k) :ge: f(2k), with equality iff  :Sigma: 2i c_i  = 0 => there are no carries in the multiplication of k by 2, in base 3 =>  there are only 1's and 0's in the base 3 representation of k.

Since we have f(2k) :le: f(3k), we now only need to show f(2k) > f(k).

f(2k) = :Sigma 2i b_i =  2 :Sigma: 2ia_i  - :Sigma: 2i c_i = :Sigma: 2ia_i + :Sigma: 2i(a_i - c_i) =  f(k) + :Sigma: 2i(a_i - c_i).

So, we want to show :Sigma: 2i(a_i - c_i) > 0.

Taking cases:

a_i = 0 => c_i = 0.

a_i = 1 => c_i = 0 , or c_i 1 if c_(i-1) = 1

a_i = 2 => c_i = 1.

So a_i :ge: c_i for all i, but we need strict inequality in at least one case.

If a_i = 2 for some i we are guaranteed strict inequality.

Suppose a_i = 0 or 1 for each i.

Then there are no carries => c_i never equals 1 => c_i = 0 when a_i = 1. This yields strict inequality.

Therefore,   f(2k) > f(k) , so we have our combined inequality:

 f(3k) :ge: f(2k) > f(k) .

Note also, that we can find when f(2k) = f(k) + 1 by this procedure.

Since f(2k) = f(k) + :Sigma: 2i(a_i - c_i), we want :Sigma: 2i(a_i - c_i) = 1, or:

a_0 = c_0 + 1 and a_j = c_j for all j > 0.

Let's take the two cases separately for a_0.

a_0 = 1 => c_0 = 0. Then a_j = c_j for larger j iff a_j = 0, so we get k = ...0013 = 1.

a_0 = 2 => c_0 = 1, so a_1 = 1 => c_1 = 1, and we can continue in a block of this fashion as far as we want. Once some a_j = 0, we can only have zeroes for any larger j.

 Therefore, f(2k) = f(k) + 1 when k = ...1123 (a block of 1's followed by a 2) or when k = 1.

[/hide]

Now that took a lot of time to write up!


induction
inequalities
inequalities unsolved

sorry for asking two problems concurrently.

Let $a_{1},a_{2},...,a_{n}$ be positive numbers, and let $A,G,H$ be an arithmetic, geometric, and harmonic mean of that numbers.
(a) If $n$ is even, show that
$\frac{A}{H}\leq-1+2 (\frac{A}{G})^{n}$

(b) If $n$ is odd, show that
$\frac{A}{H}\leq-\frac{n-2}{n}+\frac{2(n-1)}{n}(\frac{A}{G})^{n}$

On these types of problems, is the main way to go induction? The peculiar even/odd condition makes me think that...

The MacLaurin inequality says that:

$GM=\sqrt[n]{d_{n}}\leq \sqrt[n-1]{ d_{n-1}}\leq ... \leq \sqrt{d_{2}}\leq d_{1}= AM$

Where $d_{k}$ are the aritmetic mean of the products with $k$ elements. For example with 4 variabiles $d_{2}= \frac{1}{6}(ab+bc+cd+da+bd+ca)$.

Now we have that $HM=\frac{GM^{n}}{d_{n-1}}$ and so our inequality becomes:

${AM * d_{n-1}}{GM^{n}}\leq-1+2 \frac{ AM^{n}}{GM^{n}}$

$\frac{AM (AM^{n-1}+AM^{n-1}-d_{n-1})}{GM^{n}}\geq 1$

Now we know that $AM^{n-1}\geq d_{n-1}$ and so

$\frac{AM (AM^{n-1}+AM^{n-1}-d_{n-1}) }{GM^{n}}\geq \frac{AM *AM^{n-1}}{GM^{n}}= \left( \frac{AM}{GM}\right)^{n}\geq 1$

and so the case with $n$ even is done. The case with $n$ odd is the same calculations...

in fact, from MacLaurin it follows directly $\frac{A}{H}\leq (\frac{A}{G})^{n}$  :)


Note: There are six rows of buckets as shown on the picture.

I used recursion to get the answer (really long solution)... is there a faster way? Pascal's Triangle?

Answer: [hide][/hide]

Let $ x$ litres be poured into the top.

Now, consider the problem from a different viewpoint. The buckest will distribute the water the same way no matter how they receive it as long as there is the same amount of water given to it. So we start with the amount of water the top bucket will receive, $ x$. It takes 1, and distributes the remaining halves to the 2 buckets underneath. Continue in this fashion to get a lengthy equation for the bottom buckets. Then set them equal to 1, and pwn.

the problem is that some buckets will receive water at different rates. And at the same time another bucket will receive water at a different rate... setting up a lengthy equation would give me a headache keeping track of all the buckets

That's the beauty of it. It doesn't matter at what rate the buckets get the water, only the total amount of water received by each bucket.

i still don't quite understand your method.. are you assuming each bucket is 1L instead of 2 to simplify the problem?

starting with $ x$ litres from the top:
the two buckets underneath will get $ \frac {x \minus{} 2}{2}$ each..
bucket 1&3 of row 3 will get $ \frac {(\frac {x \minus{} 2}{2}) \minus{} 2}{2}$ and bucket 2 will get $ \frac {x \minus{} 2}{2}$...
bucket 1 of row 3 will get $ \frac {\frac {(\frac {x \minus{} 2}{2}) \minus{} 2}{2} \minus{} 2}{2}$

continuing this pattern will be a mess when it gets to the 4th row... (especially for the buckets in between)

is this what you are suggesting? :blush:

Looking at the arrangements of the buckets you can say for sure that bucket $ C$ and bucket $ D$ will both be the first to receive $ 2$ litres. (Pascal's method)
I'm not sure if that information would help  :|

Does sombody knew how we use the pascal triangle here? I really guess we have to use them to solve it. It's the secret of the bukets beetwen. If we concidere the do the thing manualy? I will se if I found an algoritem...


geometry
geometric transformation
rotation
trigonometry
integration

A non-conducting ring of mass m and radius r is lying at rest in the vertical XY plane on a smooth horizontal XZ plane. Charge +q and -q are uniformly distributed on the ring on the two sides of the vertical diameter of the ring. A constant and uniform electric field E is set up along the x direction. The ring is given a small rotation about the vertical diameter of the ring and released. Find the period of oscillation of the ring.

no diagram given  :noo:

[hide="Ans"] $ 2\pi \sqrt{\frac{mr}{4qE}}$ [/hide]

Let angle of rotation be $ \theta$.

Let $ dq$ be a small positive charge at a height $ h$ from the surface. The torque due to this charge is $ Edq\sqrt{h^2\plus{}2hr}\sin\theta\approx Edq\sqrt{h^2\plus{}2hr}\theta$

Now, $ dq\equal{}\frac{\lambda dh}{\sqrt{h^2\plus{}2hr}}$ ($ \lambda\equal{}\frac{Q}{2\pi}$) and hence, torque is $ E\lambda\theta dh$

Thus, total torque due to positive side is $ \tau_1\equal{}\int_0^{2r}E\lambda\theta dh\equal{}2E\lambda\theta r$ and total torque on the ring is $ \tau\equal{}4E\lambda\theta r\equal{}\minus{}\frac{2EQ\theta r}{\pi}$

Now, moment of inertia of the ring is $ \tfrac{1}{2}mr^2$ and hence period of oscillation is $ T\equal{}2\pi\sqrt{\frac{\pi mr}{4EQ}}$

If the origin is taken as the lowest point of the ring, we have the generic elementary torque as

 $ d\tau = 2\theta \sqrt {2yr-y^2} E {Q \over{\pi r}} {{r dy} \over {\sqrt{2yr - y^2}}}$

which, integrated from 0 to 2r leads to $ \tau = 4 \theta E {Q \over{\pi }} r$.

As the moment of inertia  with respect an diameter is  $ {{mr^2}\over 2} \rightarrow  \omega^2 = {{4 \pi^2} \over T^2} = {{8EQ} \over {\pi m r}}$,
and $ T = \pi \sqrt{{{\pi mr} \over {2EQ}}}$.


In the table below, you may switch the signs of all the numbers of a row, column, or parallel to one of the diagonals. In particular, you may switch the sign of each corner square. Prove that you cannot have all the entries as $ 1.$
\[ \begin{tabular}{|c|c|c|c|}\hline
1&1&1&1\\\hline
1&1&1&1\\\hline
1&1&1&1\\\hline
1&\minus{}1&1&1\\\hline\end{tabular}.\]

[hide="Solution"]Consider the eight "edge squares". These are the squares around the boundary which are not the corners. Each legal move switches an even number of these squares, so there remains an odd amount of $ \minus{} 1$'s in the edge squares. Therefore there can never be no $ \minus{} 1$'s which which means that not all the squares can be $ 1$.[/hide]

[quote="KrazyFK"][hide="Solution"]Consider the eight "edge squares". These are the squares around the boundary which are not the corners. Each legal move switches an even number of these squares, so there remains an odd amount of $ \minus{} 1$'s in the [b]edge squares[/b]. Therefore there can never be no $ \minus{} 1$'s which which means that not all the squares can be $ 1$.[/hide][/quote]

Corrected a slight error

Oh yes, thank you for spotting that.  :lol:


geometry
trigonometry
trig identities
Law of Cosines
Law of Sines

Here's a neat geometry problem to try.

An acute triangle ABC is inscribed in a circle. Two altitudes AM and CN are extended to meet the circle at P and Q respectively. Given that PQ:AC=3:2, find cos :ang: ABC.

[hide]Let BK be the other altitude. Let O be the orthocenter. Let QP=3, and AC=2. AngleABQ=angleQPA=angleQCA=angleABK.

THerefore QB=BO.

AnglePAC=anglePQC=anglePBC=angleCBK.

therefore BP=BO

and QB=BP. angleQBP=2 angleABC. Let cosABC be k, QB be x.

Law of cosine

x^2+x^2-2x^2(2k^2-1)=9

Also 2/x=NC/NB=tanABC= :sqrt: (1-k^2)/k

Solve this very nice system of equation, we have k=3/4 [/hide]

A different method:

[hide]Let a, b and c represent respectively the degree measures of angles CAB, ABC, and BCA. We know MAC = 90 - c, NCA = 90 - a, so in triangle HAC, angle AHC = 180 - (90 - a) - (90 - c) = 180 - b.  Thus, (mAC + mPQ)/2 = 180 - b, where mXY represents the measure of the arc XY.  Since mAC = 2b, mPQ = 360 - 4b.  Now construct the triangles OPQ and OAC, where O is the center of the circle.  The measure of the two central angles are (360 - 4b) and 2b.  Applying the law of cosines and dividing, we get [tex]\displaystyle(\frac{AC}{PQ})^2=\frac{1-\cos{2b}}{1-\cos{(360-4b)}}[/tex].  Substituting for the double-angle cosine formulae in terms of sin, we get [tex]\displaystyle{4/9 = \frac{\sin^2b}{\sin^2(2b)}=\frac{1}{4cos^2b}}[/tex], which, given the acuteness of B, gives us cosb = 3/4.[/hide]

Another different method:

Refering to the labels in JBL's picture.

[hide]

[tex]\angle{BCQ}=90^\circ-\angle{B}[/tex] and [tex]\angle{BCP}=\angle{BAP}=90^\circ-\angle{B}[/tex]. So [tex]\angle{PCQ}=180^\circ-2\angle{B}[/tex]

By the Law of Sines:

[tex]\displaystyle\frac{PQ}{AC}=\displaystyle{\frac{2R\sin(180^\circ-2{B})}{2R\sin B}= 2\cos{B}[/tex]

[/hide]