floor function
inequalities
Irrational numbers

Prove that if $ \alpha$ and $ \beta$ are positive irrational numbers satisfying $ \frac{1}{\alpha}\plus{}\frac{1}{\beta}\equal{} 1$, then the sequences
\[ \lfloor\alpha\rfloor,\lfloor 2\alpha\rfloor,\lfloor 3\alpha\rfloor,\cdots\]
and
\[ \lfloor\beta\rfloor,\lfloor 2\beta\rfloor,\lfloor 3\beta\rfloor,\cdots\]
together include every positive integer exactly once.

[quote="Peter"]As far as I know, this is not an IMO question... certainly not from 1987.  :maybe:

But a proof of Beatty's theorem can be found [url=http://www.animath.fr/cours/arithm.pdf]here[/url] on page 6-7.[/quote]

Yeah, you're right. It is not from IMO problems.

Proof of Beaty's theorem from Cours d'Arithmetique I Part,par
P.Bornsztein, X. Caruso, P. Nolin, M. Tibouchi, 2004.
                          My translation.
Suppose $ \alpha$ and $ \beta$  be two irrational numbers such that
$ \frac{1}{ \alpha} \plus{} \frac {1}{\beta} \equal{} 1$.
Let $ k$ be a strictly positive  integer .
 $ k$ will  be an element of  $ S_\alpha$ if and only if  there exists an integer $ n$ such that
$ n \alpha \minus{} 1 < k <  n  \alpha$.

The RHS of inequality is strict one,  since $ \alpha$ is supposed to be  irrational.
The equation can be transformed into the following one:

$ \frac{k}{\alpha} < n < \frac{k}{\alpha} \plus{} \frac{1}{\alpha}$.
That means , the following. Namely $ k$ would be  element of $ S_ {\alpha}$
 if and only if the interval $ ] \frac{k}{\alpha}, \frac{k}{\alpha} \plus{} \frac{1}{\alpha} [$ contains an integer.
The same:   $ k$ would be element of $ S_ {\beta }$
if and only if  the interval $ ]\frac{k}{\beta}, \frac{k}{\beta} \plus{} \frac{1}{\beta}[$ contains an integer .
The lenght of the  interval $ ]\frac {k}{\alpha}, \frac{k}{\alpha} \plus{} 1[$  is equal to $ 1$ and  both of its end-points are irrational , so that is why it   consists one and only one integer $ n$.
If $ n < \frac{k}{\alpha} \plus{} \frac{1}{\alpha}$ then $ k \in S_{\alpha}$. But, if not we will have another inequality, namely:

$ \frac{k}{\alpha}\plus{}\frac{1}{\alpha} < n < \frac{k}{\alpha} \plus{}1$.
The LHS  of this inequality  is strict one since $ \frac {k\plus{}1}{\alpha}$ is irrational. Since $ \frac{k}{\alpha} \equal{} k \minus{} \frac{k}{\beta}$ we  have
$ \frac{k}{\beta}  < k \plus{} 1 \minus{} n  < \frac {k}{\beta} \plus{}  \frac {1}{\beta}$

so $ k \in S_{\beta}$. If $ k$ would be simultaneously element, both of
 $ S_{\alpha}$ and $ S_{\beta}$, then it would be an integer in the interval
$ ]\frac{k}{\alpha},\frac{k}{\alpha} \plus{} \frac{1}{\alpha}[$  and in the interval $ ]\frac{k}{\beta},\frac{k}{\beta} \plus{} \frac{1}{\beta}[$. So, in the same way as before it would be two elements
in the interval $ ]\frac{k}{\alpha}, \frac {k}{\alpha} \plus{}1[$, which is impossible.

Conversely, suppose that $ S_{\alpha}$, $ S_{\beta}$ partition the set  $ N$.
Let us consider a strictly positive  integer $ k$. So, there exist $ [\frac {k}{\alpha}]$ integers in the set $ {1,...,k}$ being  elements of $ S_{\alpha}$.
Similarly , there exist  $ [\frac {k}{\beta}]$ integers in $ {1,...,k}$ being  elements of $ S_{\beta}$.
Because of  the partition it appears that,
$ [\frac{k}{\alpha}] \plus{} [\frac{k}{\beta}] \equal{} k$ for all $ k$.
Let  $ k  \rightarrow  \infty$, so we have
$ \frac{1}{\alpha} \plus{} \frac{1}{\beta}\equal{} 1$
and that  proves the second condition.
Suppose now,   that $ \alpha$ be rational. So,  the same for $ \beta$ after the previous relations. Let us write $ \alpha \equal{} \frac{a}{b}$
and $ \beta \equal{} \frac{c}{d}$.
The integer $ a\cdot c$ is  element of $ S_{\alpha}$  (taking $ n \equal{} b\cdot c$)
and equally element of $ S_{\beta}$ (taking $ n \equal{} a \cdot d$ ) which is the contradiction.

let set $A=\[ \lfloor\alpha\rfloor,\lfloor 2\alpha\rfloor,\lfloor 3\alpha\rfloor,\cdots\]$

and set $B=\[ \lfloor\beta\rfloor,\lfloor 2\beta\rfloor,\lfloor 3\beta\rfloor,\cdots\]$

note that no two member are equal in $A$ and no two pair of members are equal in $B$ because $\alpha,\beta>1$

then we prove that there are exactly $n-1$ members of $A$ and $B$ that are less that $n$.

let $x$ be the greatest natural number for which $\lfloor x\alpha\rfloor <= n-1 \implies x=\lfloor n\alpha^{-1}\rfloor $ and we define $y$ the same way: $\lfloor y\beta\rfloor <= n-1 \implies y=\lfloor n\beta^{-1}\rfloor $

thereupon easily we get that $x+y=\lfloor n\alpha^{-1}\rfloor+\lfloor n\beta^{-1}\rfloor=n-1$ owing to the fact that $\beta^{-1}+\alpha^{-1}=1$ .

so we know there are exactly $n-1$ members of $A$ and $B$ that less that $n$ and there are exactly $n$ members of $A$ and $B$ that less that $n+1$ for any $n>1$ which mean there's exactly one member in $A$ and $B$  equal to $n$ we any natural number greater than $2$ and it's totally obvious that $1$ is in $A$ or $B$ and proving that happens only for one member of $A$ and $B$ is straightforward


modular arithmetic
number theory
Diophantine equation
number theory proposed

It is equavalent to $(4x+3)^{2}-48y^{2}=1$.
Therefore solution is $4x+3+y\sqrt{48}=(7+\sqrt{48})^{k}$.

these problems you post here should belong to the Number theory forum i think. [color=red][Done. -mathmanman][/color] anyways here's the solution:
the equation is equivalent to $(x+1)(2x+1)=6y^{2}$
so $x+1\equiv 0\pmod{2}$ Let $x=2k-1$ our equation becomes $k(4k-1)=3y^{2}$
since $\gcd(k,4k-1)=1$, by Euclid's theorem, either $3|k$ or $3|(4k-1)$

[b]case 1:[/b]$3|k$
Let $k=3m$. we have $m(12m-1)=y^{2}$
so $m=c^{2},12m-1=d^{2}$
hence $12c^{2}-d^{2}=1\Longrightarrow d^{2}\equiv-1\pmod{4}$, contradiction.

[b]case 2:[/b]$3|4k-1$
we have $k\equiv 1\pmod{3}\Longrightarrow k=3q+1$
so $(3q+1)(4q+1)=y^{2}\Longrightarrow 3q+1=s^{2}, 4q+1=t^{2}$
thus $4s^{2}-3t^{2}=1$ this is a generalised Pell equation with minimal soluiton $(1,1)$
hence it has infinitely many solutions $(s_{i},t_{i})$ where $2s_{i}+\sqrt{3}t_{i}=(2+\sqrt{3})^{2i+1}$

EDIT: well, thanks, i've corrected it. its just a typo, nothing serious. :D

check again case 1 :wink: just a little mistake.


I am going to start working backwards through past Balkan MOs soon. Anyone feel like joining in? I'll aim to sit 2006 and 2005 this week at some point. Also, does anyone (Tim?) know how long we get to address them? Is it 4 hours or 4 1/2 hours?

I'll probably just take up to 4 1/2 hours as required unless it's confirmed that it's 4.

It's 4 1/2 (or at least it was last year... I can't see that it would change). I might take some as well, though I have done some of the more recent ones before. Balkan MO papers tend to be slightly nicer than IMO ones in my opinion, even if some of the problems are rather second hand- compare 2000q1 with 1997q4  :lol:

Also, could you e-mail some of the old NSTs you mentioned a few months ago?

Yay, it appears that the Balkan questions are more open to exploitation of areal co-ordinates than IMO. 2006/2 dies fast to a bit of areal bashing, if anybody feels like trying it (I judged it suitable to mention given I doubt anyone would actually have otherwise tried this method).

How do you work out the co-ordinates of $A_{1}$ etc.? I presume you set D = (d,1-d,0), F = (f,1-f,0) or similar...

I did it by intersecting the parallels to m through A,B,C with the circumcircle, which gives a quadratic with one root zero.

Start by defining the line $m: qy+rz=px$ and this is your only degree of freedom, so everything else comes in terms of p,q,r and the triangle side lengths. Parallels are quite nice to figure out yourself.

Omg, or you can just do it geometrically and it's quite nice. Doh!  :mad:


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[hide]We can form a sequence.

1, 2, 3
2, 3, 4
3, 4, 5

4, 5, 6
5, 6, 7
6, 7, 8

Because each set of consecutive integers has 3 elements, clearly one of the have to be a multiple of 3.[/hide]

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Home » Mathematics » Middle School Classroom Math (Ages 9-13)

Divisibility
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small_unicorn
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Divisibility
Prove
Prove:
one of three consecutive odd numbers is divisible by 3 .
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AdithyaGanesh
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We can form a sequence.

1, 2, 3
2, 3, 4
3, 4, 5

4, 5, 6
5, 6, 7
6, 7, 8

Because each set of consecutive integers has 3 elements, clearly one of the have to be a multiple of 3.

Lets say that the smallest of the 3 is x, the others are x+2 and x+4. If x is a multiple of three then this is true. If x is one more than a multiple of 3 then x+2 is a multiple of 3. If x is 2 more than a multiple of 3 then x+4 is divisible by 3. If it is 3 more than a multiple of 3 then it is a multiple of three and so on.

Simply because consecutive multiples of three occur every 3 numbers.  Therefore, exactly 1 of any consecutive 3 numbers must be a multiple of three.

Three consecutive odd numbers may be written as $ 2n\minus{}1, 2n\plus{}1, 2n\plus{}3$.

$ n$ must be a multiple of $ 3$ or one more than a multiple of $ 3$ or two more than a moltiple of $ 3$, that is $ n$ must be $ 0, 1$ or $ 2 (mod 3)$.

If $ n \equiv 0 (mod 3)$ then $ 2n\plus{}3 \equiv 0 (mod 3)$

If $ n \equiv 1 (mod 3)$ then $ 2n\plus{}1 \equiv 0 (mod 3)$

If $ n \equiv 2 (mod 3)$ then $ 2n\minus{}1 \equiv 0 (mod 3)$.

all integers can be expressed as 3k, 3k+1, or 3k+2.
three consecutive numbers are one of these

3k+1, 3k+2, 3k+3 (which is divisible by three)
3k+2, 3k+3 (which is divisible by three), 3k+4
or 3k (which is divisible by three), 3k+1, 3k+2
therefore in three consecutive numbers one and only one is divisible by three.

this also holds true for in x consecutive numbers one and only one is divisible by x. proof isnt too complicated, and its easy to understand why.

its true because every 3 odd numbers contains a multiple of 3

For example,  3,5,7         5,7,9          115,117,119

if you dont believe me try to find a triplet that doesnt have a multiple of 3

After every 3 integers, of course there will be one being a multiple of 3. Example: 15, 16, 17.   17, 18, 19.

Look people, READ. smallunicorn said consecutive [u]odd[/u] integers.

Epicfails, that is not a proof.

FlyAgaric and AndrewTom have pretty nice proofs, though I think modular arithmetic might be beyond this forum.

[hide]3 times an odd is an odd.
E.G. 3*3=9, 3*5=15

three conscecutive odd numbers:
1,3,5
3*1=3
3,5,7
3*1=3
5,7,9
3*3=9
and etc.
Another way:
x= an odd number divisible by 3
One possibility
x+(x+1)+(x+2）
definitely x is the odd number divisible by 3
Another possibility
(x+1)+(x+2)+(x+3)
it is difinite that x+3 is the odd number divisible by 3
The last possibily
(x+2)+(x+3)+(x+4)
it is also definite that x+3 is still the odd number divisible by 3
and so on.....[/hide]
[/hide]


[hide]WLOG assume that $x\leq y\leq z$.  Then working backwards from chebyshev we only need to prove
\[\displaystyle xy^2+yz^2+zx^2 \geq x^2y +yz^2+xyz\]
This reduces to
\[\displaystyle y^2+z^2 \geq xy+yz\]
Since $y^2 \geq xy$ and $z^2 \geq yz$ this is obviously true and we are done.[/hide]

Well we can see that you can change[code]a[/code] for [code]a/d[/code] and so b and c so that [code]a+b+c=1[/code]  then multiplying by 1

[code]
\begin{eqnarray*}
3(x+y+z)(xy^(2)+yz^(2)+zx^(2))&\geq&\sqrt{3(x^(2)y^(2)+y^(2)z^(2)+z^(2)x^(2))}\\
\sqrt{\frac{(x^(2)y^(2)+y^(2)z^(2)+z^(2)x^(2))}{3}}&\geq&\frac{(xy+yz+zx)}{3}
\end{eqnarray*}
[/code]

and that is the cuadratic and aritmetic[/code]

sorry about that but i dont know yet to post the LaTex stuff but well here it is attached by the way if someone could explain me how to post right i would be really pleased

just put your code in between two dollar signs.
Btw, amd 's solution is wrong because the inequality is not symmetric.

[quote="shobber"]Can we assume that $x+y+z=1$ in this problem?[/quote]
yes, because it's homogeneous.

Homogenity -> assuming symmetric conditions
Symmetric -> assuming any order of variables
Cyclic -> assuming the largest/smallest variable

[quote="blahblahblah"]Say $x$ is largest.  If $x\geq y\geq z$, then we are done.  Otherwise, we have $x\geq z\geq y$.  Put $x=y+a+b$, $z=y+a$.  Then $3(xy^2-yz^2+zx^2)-(xy+yz+zx)(x+y+z)$ expands into something containing only positive terms, so we're done.[/quote]

You may assume without a loss of generality that x is the largest, y is second largest, and z is third largest.  Suppose, for example, that you had $x\geq{z\geq{y}}$.  Then, if you just switched y and z, the values on the left and right sides wouldn't change, and you'd then have $x\geq{y}\geq{z}$.  You can do this in any symmetric inequality (that is, one where f(a,b,c)=f(a,c,b)=f(b,c,a)=... and so on).

Edit: Ignore this.  I should have looked at the problem a bit harder first.

Magnara, your comment is off-base, because the inequality was cyclic.

Second, the inequality is false, take $x=1011,y=11,z=1$.  Cezar, check your inequalities a bit further before you post.  My original 'solution' is wrong, because I meant to work on the case $x\geq y\geq z$ instead of the case $x\geq z\geq y$.

[quote="blahblahblah"]Magnara, your comment is off-base, because the inequality was cyclic.[/quote]

That's a good point.... heh

[quote="shobber"]How do we kown if an inequality is homogenity or symmetric or cyclic?

This problem can be transformed into proving:
\[3(\frac{y}{z}+\frac{z}{x}+\frac{x}{y}) \geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\][/quote]

homogeneous means that the degree of each term is the same.
e.g. degree of x^2 y is 3, x^2/y is 1, and so on I think you know what i mean.

Symmetric means f(a,b,c)=f(p(a,b,c)) where p(a,b,c) is any permutation of a,b,c

Cyclic means f(a,b,c)=f(b,c,a)=f(c,a,b)

Same for more than three variables.

By using Weighted RMS-GM we have
$\sqrt{\frac{xy^2+yz^2+zx^2}{x+y+z}} \geq \frac{xy+yz+xz}{x+y+z}$
then by cauchy,AM-GM or whatever , one can prove the following easy & well-known inequality \[ 3(xy+yz+xz) \geq (x+y+z)^2 \]
so we are done
indeed nice problem cezar  :)

the "easy and well-known inequality" you have posted is of reversed sign.

How to prove the easy and well-known inequality?
For me, it is neither easy nor well-known.

as I check it, to prove the original ineuality we don't need this "easy.." just straightforward The W  RMS-AM.But if you want a prove of that "easy.." here
[hide] \[ (x+y+z)^2=x^2+y^2+z^ +2(xy+yz+xz) \] by cauchy we have \[ xy+yz+xz  \leq x^2+y^2+z^2 \] so (x+y+z)^2=x^2+y^2+z^ +2(xy+yz+xz) \geq 3(xy+yz+xz)  \]
[/hide]

OH my God , what happen to me!! :|  :huh:  :huh: I must go to bed right now :o .
That inequality is wrong. And also my computation :mad:  .
Silly me. I hate doing stupid mistakes :mad:
got to go to bed now...........

the inequality cezar posted is wrong (even with the sign reversed)

i was trying to solve it using the trick $y=x+a$, $z=x+b$, where $x=\min \{ x,y,z \}$, but got a expression which sometimes is true, sometimes is not

Do you guys not read the middle of threads?  I concluded that (with a counterexample) about 12 posts previously.

sorry... the posts were long, so i skipped your post (and the other 2 or 3 which were in the middle)  :blush:


group theory
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Galois Theory
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superior algebra unsolved

Prove that every procyclic group is isomorphic to $ \prod_p A_p$ where $ A_p \equal{} \mathbb{Z}/p^n$ or $ \mathbb{Z}_p$.

Is every proabelian group isomorphic to a product of procyclic groups?

the procyclic group $ \widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$ can be interpreted as the absolute galois group of some finite field $ F$. the closed subgroups then correspond to galois extensions of $ F$. these are composita of galois extensions of degree $ p^k$ for $ k \in \mathbb{N} \cup \{\infty\}$. so these subgroups look like $ \prod_p B_p$ where $ B_p \equal{} 0$ or $ B_p \equal{} p^k \mathbb{Z}_p$, and their quotient is $ \prod_p A_p$ with $ A_p \equal{} \mathbb{Z}_p$ or  $ A_p \equal{} \mathbb{Z}/p^k$. it remains to remark that every procyclic group $ G$ is a quotient of $ \widehat{\mathbb{Z}}$. in fact, choose a topological generator $ \sigma$, then $ G/G^n$ is finite (because $ \langle [\sigma] \rangle$ is finite and dense). conversely, if $ H$ is an open subgroup of $ G$ of finite index $ n$, we have $ n \equal{} |G : H| \leq |G : G^n| \leq n$, so $ H \equal{} G^n$. this shows $ G \equal{} \lim_{\leftarrow} G/G^n$ and the $ \sigma : \mathbb{Z}/n \to G/G^n$ induce $ \widehat{\mathbb{Z}} \to G$.

I'm curious about the second question, which probably has to be answered with "no", but I've got no counterexample :-).

By Pontrjagin duality, the question reduces to: Is every discrete abelian $ p$-torsion group isomorphic to a product of groups of the form $ \mathbb{Z}/p^n$ or $ \mathbb{Q}_p/\mathbb{Z_p}$.

how does the reduction exactly work? I think $ hom(\minus{},S^1)$ does not commute with products   :maybe:

by the way, $ \mathbb{Q}_p / \mathbb{Z}_p \cong \mathbb{Z}/p^\infty$, so your question is: is every $ p$-torsion reduced abelian group isomorphic to a product of cyclic groups? do you know the answer? :-)

Pontrjagin duality turns products in the category of compact abelian groups into [i]sums[/i] in the category of (discrete) abelian groups, not into products. The Pontrjagin dual reads:

Is every torsion abelian group a direct sum of ind-cyclic groups? By ind-cyclic I mean inductive limit of finite cyclic groups. As berti said, if we reduce the problem to the $ p$-primary case, the only summands we're looking for are finite cyclic and $ \mathbb Z_{p^\infty}$. An equivalent formulation would be: Is every reduced $ p$-primary abelian group a direct sum of finite cyclic groups?

The answer is 'no'. Try showing, for example, that the torsion of the group $ \prod_{n\ge 1}(\mathbb Z/p^n\mathbb Z)$ is not a direct sum of cyclic groups.


If we multiply across by $(x-5)^2$ we dont need to worry about having to change the inequality sign, because its positive. Simplifing it and factoring it leaves us with

$(x-7)(x-5) < 0$

$\rightarrow 5<x<7$

(PS sure this isnt your homework?)

[quote="TESoccer66"]Solve for x:
$\frac{x+3}{x-5}>5$.[/quote]

$\frac{x+3}{x-5}-5>0$
$\frac{x+3-5x+25}{x-5}>0$
$\frac{4(7-x)}{x-5}>0$
$(7-x)$ and $\frac{1}{(x-5)}$ have to either be both positive, or both negative,
and we find that the only area is when both are positive, is when $5<x<7$

An easier way is...

$\frac{x+3}{x-5}>5$
$x+3>5x-25$
$-4x>-28$
$x<7$

Since we have $\frac{x+3}{x-5}>5$, $x$ must be greater than 5 in order to satisfy a positive result and for the dinominator to be a number other than $0$.

Hence,

$5<x<7$

Masoud Zargar

Why can't you just clear the fractions and get x<7 and x is not equal to 5??

Because we have to consider the sign of $x-5$... If $x-5$ is negative, then we have to change $>$ to $<$ when multiplying both sides by $x-5$.

ok suppose x-5 is negative then... After you multiply you get $x+3 < 5(x-5)$ and $x<7$ you don't get the solution with the 5 in it

Well, from $x+3<5(x-5)$, you should get $x>7$ instead of $x<7$. But that doesn't satisfy $x-5<0$, so that's not the case.

When $x-5>0$ (or $x>5$), however, we have $x+3>5(x-5)\Longleftrightarrow x<7$. That gives us the solution: $5<x<7$.


5-Two circles $ S_1$ and $ S_2$ with equal radius and intersecting at two points are given in the plane.A line $ l$ intersects $ S_1$ at $ B,D$ and $ S_2$ at $ A,C$(the order of the points on the line are as follows:$ A,B,C,D$).Two circles $ W_1$ and $ W_2$ are drawn such that both of them are tangent externally at $ S_1$ and internally at $ S_2$ and also tangent to $ l$ at both sides.Suppose $ W_1$ and $ W_2$ are tangent.Then PROVE $ AB \equal{} CD$.

$ P \equiv \omega_1 \cap \omega_2$ and $ M, N$ are the tangency points of $ \omega_1$ and $ \omega_2$ with a common external tangent. Inversion with center $ P$ and power $ PB \cdot PD$ takes $ (S_1)$ and the line $ l$ into themselves, the circles $ \omega_1$ and $ \omega_2$ go to two parallel lines $ k_1$ and $ k_2$ tangent to $ (S_1)$ and the circle $ (S_2)$ goes to a circle $ (S_2')$  tangent to $ k_1,k_2.$ Hence, $ (S_2)$ is congruent to its inverse $ (S_2').$ Further, $(S_2),(S_2')$ are symmetrical about $ P$ $\Longrightarrow$ $ PC \cdot PA \equal{} PB \cdot PD.$

By Casey's theorem for $ (\omega_1,\omega_2,D,B,S_1)$ and $ (\omega_1,\omega_2,A,C,S_2),$ we get

$ DB \equal{} \frac {2PB \cdot PD}{MN} \ , \ AC \equal{} \frac {2PA \cdot PC}{MN}$

Since $ PC \cdot PA \equal{} PB \cdot PD$ $ \Longrightarrow$ $ AC \equal{} BD$ $ \Longrightarrow AB \equal{} CD.$


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Use search function! http://www.mathlinks.ro/Forum/viewtopic.php?t=45707


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mathfanatic wrote:Source: AHSME 1996 #25

Given that x:^2: + y:^2: = 14x + 6y + 6, what is the largest value that 3x + 4y can have?

[hide]i'll do this geometrically although there is probably a nice algebraic way too. We have the equation of a circle, and we 3x+4y=c is the equation of a line. To maximize c, we find the "upper" tangent to the circle with slope -3/4. Anyways, our circle is (x-7):^2:+(y-3):^2:=8:^2:. So the center is (7,3) and the radius is 8. Now, imagine the line with slope -3/4 going through the center. The y-intercept will be 33/4. Now, when we raise this line 8 units perpendicular to the line, we will raise it by the vertical distance 8(4/5)=32/5. So 33/4+32/5=293/20.[/hide]

so
anyone get it algebraiclly ??
it smells like a Cauchy to me, and maybe an AM-GM, but how do u go about it?

the answers are here if you want.

http://www.math.uncc.edu/~hbreiter/AHSME/1996so.pdf

I know that you're right, Mystic, but how do you "know" that the Maximum will occur at the point of tangency?


Prove that for each positive integer n, the number of  integer solutions (x,y) of the equation

x^2 -x.y+y^2= n   is finite and divisible by 6

$ N(x \plus{} y \omega) \equal{} x^2 \minus{} xy \plus{} y^2 \equal{} n$.  Finiteness is guaranteed by the fact that the LHS is positive-definite.  As for divisibility by six, the associates of $ x \plus{} y \omega$ are given by $ (x \plus{} y \omega) \omega^k, (x \plus{} y \omega^2) \omega^k, k \equal{} 0, 1, 2$.


In Canadia, people are taxed a certain amount of money per day. Jim makes 23 dollars after taxes as a lumberjack. If he would make four times as much after taxes if he made twice as much money before taxes, how much does he make before taxes?

[hide="a huge hint"]the answer is treething's favorite number.[/hide]

[hide="researched answer :rotfl: "]69[/hide]

[quote="jclarkemathL314159"]SPAM

And nutz, your answer is incorrect, and not funny at all.[/quote]
I'm pretty sure it is right. See the hint.
[hide="my solution"]
x=tax
$2(23+x)=92+x$
$x=46$
$46+23=69$
[/hide]
So did I misinterpret it or make a mistake?

[quote="jclarkemathL314159"]SPAM

And nutz, your answer is incorrect, and not funny at all.[/quote]

Actually it is correct.
And given the subject of the question, I'd say it's borderline funny.

[hide]$2(23+x)=92+x$.

So $x=46$

But it really is 23+46=69.

Sweet 69, sorry I had to say that.

$69$[/hide]

[hide]i know i sound dull so...isn't the answer 46?

 4 times 23 is 92 times 2 is 184 divided by 4 is 46? where did i go wrong?[/hide]

try expressing the thingy algebraically, i don't think you set it up right.

[hide]call $x$ jim's salary and call $y$ taxes.  $x-y=23$, so $y=x-23$.  We also have $4(x-y)=2x-y$ and $y=\frac{2}{3}x$.  Putting these together we have $x-23=\frac{2}{3}x$ and $x$ is Treething's favorite number :rotfl: $\boxed{69}$[/hide]


geometry
rectangle
combinatorics unsolved
combinatorics

A man wants to run from the point start to the point end inside all of concentric rectangular .

He want to count how many ways to do that??

He put some conditions on that :

1) He must pass all the ways in each time.

2) He can't pass the way more than one time.

3) He can pass the intersection points between the ways more than one time.

Could you tell him how many ways he could do it???

 :D  :D  :D

to be more easy to figure out ...

look at this way ( it's one of the ways)

Now, I do not fully understand this problem, but it seems to me that in your second post, the guy doesn't run through neither one of the halves of the inner-most rectangular.

You also seem to be slightly overusing the word "way", which makes the problem a little difficult to comprehend.

If I got it right, which I doubt, the answer would be $ 10! \equal{} 3628800$. If [i]"3) He can pass the intersection points between the ways more than one time."[/i] means that he can use the white line however he wants and not worry about intersecting his previous steps while on the blue line, than he can clearly go through these halves in what ever order he wants. Since there are $ 10$ halves ($ 5\cdot 2$), he can choose among $ 10$ halves to be the first one he runs through, then among $ 9$ halves to be the second one... That gives us the answer $ 10\cdot 9\cdot 8\cdot ...\cdot 1 \equal{} 10! \equal{} 3628800$.

However, I feel I misunderstood the problem and that there's more to the problem than this. Some further explanation of the problem may be necessary.

[quote="mornik"]
However, I feel I misunderstood the problem and that there's more to the problem than this. Some further explanation of the problem may be necessary.[/quote]

1st of all , let's change the word "way" to the word "path" , ...more suitable .

I put number on each path , he must pass all the paths in every time he wants to go to end point , and he couldn't pass any of them for more one time.Also, I put points on intersection points (probably ).Of Course  :wink:  he can use the white line . look at my previous post he passed the intersection point on path 6 more than one time as shown the path 6 in the figure below.

Is it more clear ???Any question ??? :huh:

Thank you for your concern dude.

[quote="mornik"]OK, wouldn't the solution then be 8! (since the innermost rectangle doesn't count)?[/quote]

I answered it as you did , but I don't know if our answer is true or not ?

So , I post it here to see your opinion guys  :roll:

Anyway , thank you dude  :wink:


I believe its the photons associated with microwaves, but I'm not 100% possitive.

Do you know any equations involving the energy of photons?  Might wavelength be a factor?

i would say microwaves because they can heat up your food faster than a flashlight can

Do you perhaps know a relationship between frequency and energy?

SirErnest -- which uses more energy: the microwave or the flashlight?

Exactly.  So you don't know how much the light would heat up food if you used the same amount of energy.  Also, there might be something special about light that made it convert into heat less easily than microwaves do.  It was a good thought, though.

quick question: why is white light comprised of the colors that its comprised of? why isnt it comprised of brown instead of blue light or something like that?

White light is a spectrum of different frequencies, and so includes specific colors as parts of the spectrum. And as for the microwave, microwave ovens can cook food well because their frequency enables them to vibrate water molecules which are almost always present in food and thereby heat them. Determining how much energy a type of light has depends more on frequency- towards this idea, gamma rays have more energy than radio waves because they have a much higher frequency.


Would appreciate any help on this problem.

Let p be prime. p does not divide a and p does not divide b.
Prove if a2   :equiv:  b2 mod p, then a :equiv:  :pm: b mod p.

[quote="amfulger"]What is this doing here?

$p|uv \Rightarrow p|u\ or\ p|v$.
$(a-b)(a+b)\equiv 0(\mod p)$. Apply the line above.[/quote]

Sorry, I thought I was in one of the other forums. I'll repost in the appropriate forum. What you have above is what I thought at first, but it seems too easy. Where do you use p doesn't divide a and p doesn't divide b? Do those 2 facts matter?

No, these facts don't matter, but in this case you have $a \equiv b \equiv -b \equiv 0 \mod p$, so only one 'solution' (the same happens for $p=2$ for all numbers).


I ask the moderators to please close discussion of the AIME from sometime late on Monday evening, March 17, 2008, until Saturday morning, March 23, 2008.  Students outside the USA are taking the AIME later  this week, and have asked that answers not be posted until they are finished.  Please cooperate and keep the contest interesting for everyone.

To those of you anxious to discuss the problems I apologize, but note that the rules allow you to discuss the problems within your school, after the AIME is finished.  Just don't communicate with others outside your school.  Instead, work with your teachers, coaches and fellow students on the problems.

Also, note that, as it is printed on the back page of the Problems Pamphlet:
"Your Exam Manager will receive a copy of the 2008 AIME Solution pamphlet with the scores."
This means that we will be sending the Solutions Pamphlets when we send the scores, about 10 days after the AIME.  The fact that the solutions are not included with the AIMEs is NOT an oversight on our part.

As for scoring and reporting scores, the most critical factor is to get the answer forms to the AMC office quickly.  We can score quickly once the packages are here, but if teachers and coaches delay sending, or send the packages through slow mail, we can't score them.  Receiving the packages in a timely fashion is the most critical factor affecting our scoring.

And of course, we won't even begin to think about USAMO cutoff levels until after the AIME II has been scored and we can compare statistics.  See the AIME Teachers' Manual, Section V, page 5.

Steve Dunbar
MAA Director, American Mathematics Competitions

As always, we will be honoring the AMC's request.  The AMC forum will be locked later today and will be reopened on the morning of Sunday, March 23.  This also means no discussion whatsoever of the AIME on any other part of the forum, on the wiki, or in blogs.

Also, this means that we have rescheduled the AIME Math Jam for [b]Monday, March 24[/b] at [b]7 PM Eastern / 4 PM Pacific[/b].  During the Math Jam we will discuss the solutions to all 15 problems on the AIME.


Note from one of the moderators of this forum:

[i]PLEASE never, ever ask for help here on a problem that is part of the qualifying process for a summer program, or that is part of an online math contest. I can't judge someone's motives, but I have to be no-tolerance about protecting the reputation of the AoPS site so that it doesn't become known in the mathematical community as a cheat site. You are on your own when applying to summer programs, so don't post the problems anywhere here on AoPS.[/i]

(The level placement of the problem I have edited out is also doubtful for this Getting Started forum.)

This problem is the exact problem from the mathcamp qualifying quiz. I don't think it's a good idea to post solution here because people will get free solutions.


Welcome to the Wisconsin forum. Here we can discuss some WI math competitions, maybe some Wisconsin news, and anything else Wisconsinish. If you want to be a member, please subscribe!

We are far too cool to have mods. I'm the senior member though. There doesn't really seem to be a need for a mod yet.

[quote="kstan013"]Or is there? Take a look at your last message mustafa. :rotfl: MWAHAHAHAHA[/quote]

OH wow.

Anyways, hear how the Bucks are getting owned this year?

[quote="kstan013"]Or is there? Take a look at your last message mustafa. :rotfl: MWAHAHAHAHA[/quote]

What was wrong with my last message? Are you making fun of me because I don't know what the date is? I know what the day is now because Christmas is on Monday, and so it's the 22nd today.

[quote="mathgeniuse^ln(x)"]WEll, I just look at the counter on the screen to figure it out.[/quote]

I feel stupid now :stink: . I'll never forget the date again!

That's not what I meant. I guess you couldn't take a hint that I'm the moderator. (I deleted your rotfl) :wink:

:rotfl: I joined before kstan013 somehow.   :rotfl:  :rotfl:

...and I have 653 more posts than footballrocks41237 somehow :huuh:


A vertical line divides the triangle with vertices $(0,0)$, $(1,1)$, and $(9,1)$ in the $xy\text{-plane}$ into two regions of equal area. The equation of the line is $x=$

$\textbf {(A) } 2.5 \qquad \textbf {(B) } 3.0 \qquad \textbf {(C) } 3.5 \qquad \textbf {(D) } 4.0\qquad \textbf {(E) } 4.5$

[hide]The area of triangle is $9 - \frac{9}{2} - \frac{1}{2} = 4$

Now consider the right angled triangle formed by the line with the given triangle. Its area must be $2$.

Hence $b \cdot h = 4$. It's similar to the triangle with side lenghts $4.5$ and $0.5$.

So $\frac{4.5}{0.5} = \frac{b}{h}$ and $b^2\frac{0.5}{4.5} = 4$ $\Longrightarrow$ $b = 6$.

Hence the line is $x = 3$.[/hide]


Prove that, for any positive integer [i]n[/i], $ \tau (n)\leq 2\sqrt{n}$.

$ \tau(n)$ is defined as the number of divisors $ n$ has.

[hide="Hint"] If $ d$ is a divisor of $ n$ such that $ d\le\sqrt{n}$, then $ \frac{n}{d}$ is a divisor of $ n$ such that $ \frac{n}{d}\ge\sqrt{n}$. [/hide]


Does anyone know how good (or bad?) did Belgium?

What were the scores?  How many medals?  Any honorable mention?  What's Belgium's ranking? ...

indead!!

(my first post here)

how did they do

my guess is quite optimistic : one silver medal, one wallonian bronze medal (at least) , one flemish bronze medal at least and perhaps an HM too!!

i was asked :D to see some of their training this year and it to be workin' out just fine :)

also, flemish and wallonians contestants have been training together too this year, which is quite new but certainly not a bad idea

so plz bring in the belgian results!!

I am really curious.  Why do you expect a silver?  I hope Cedric got one (wallonian), but I'm not sure, and I certainly don't expect it.  All I know is that the wallonian part is quite good this year.  What about the flemish?

And, I didn't hear anything about the training together.  Is that true?  How come?  In may, they didn't tell me anything about that.  I guess they spent one weekend together...

it's true

i heard about it after the proclamation on may 23rd

some time after that, they would all come together in a place called wpion for training :D

well , there's no harm in being optimistic right?

but what's the deal? last year the scores were out on 29 july, and we only came back the day after?

come on, how difficult is it to just post the results  :?

Here is the scoop :

BEL1 = Bram Gaasbeek
BEL2 = Dries Geebelen
BEL3 = Timothee Marquis
BEL4 = Arne Smeets
BEL5 = Antony Trinh
BEL6 = Cedric Troessaert

      P1 P2 P3 P4 P5 P6 total result
BEL1   2  3  0  0  0  0     5       -
BEL2   0  1  0  7  0  0     8      HM
BEL3   0  0  0  7  1  0     8      HM
BEL4   0  3  0  7  0  0    10      HM
BEL5   7  3  0  7  0  0    17      BM
BEL6   7  7  0  7  1  0    22      SM

Thresholds were 13 for BM, 19 for SM and 29 for GM.
Top three countries are Bulgaria (6 GM), China (5 GM + SM) and
USA. One German guy got his fourth GM (in addition to a BM he got earlier), which makes him the record medal owner.

By the way, who are you fredbel6 and gery ?

Franois G.

Wow, Cedric got a silver.  Not bad at all.  And a bronze for Anthony.
Looks like this team's year was not bad.  Belgium got 70 points.

I guess you are Franois Glineur.  In that case, you gave me an arithmetic class at Wpion, while I was in the group A, in the beginning of 2000.  I was selected to go to the 41st IMO, along with Franois Foucart and Pierre Gramme.  You came at Troessaert's home for a weekend to give us a problem drill, with Sbastien Leroy if I'm right.

Right now, I'm a student at the UCL.  I just finished my "candidatures" in mathematics and in engineering.  I may have you as professor in one or two or three years., who knows?

there is some stuff about me in the IMO participants--> Freddy topic as well

briefly : I participated last year in the IMO in Glasgow, together with Koen Struyve en William Vandenberghe (Flemish contestants), and the three others were : Vincelt Malmedy, Antony Trinh and Cedric Troessaert

I knew the boost in the training program would pay off!

Guess i wasn't exaggerating about one bronze and one silver after all!!! :D

We're even in the best half this year! :)


I'm trying to help my sister understand the order of operations, because for several years now, problems involving factoring, fractions, and exponents have been tripping her up. I don't want to just teach her "PEMDAS" because she'll probably forget again, but I don't know how to explain things like why $ \cfrac{a^x}{b^x}\equal{}\left(\cfrac{a}{b}\right)^x$ or $ (a^x)^y\equal{}a^{xy}$.

I realize that this may be "just the way it is," and there's no such intuitive approach to it. Then what's a good way to teach someone the order of operations? How did you learn it?

Yes, but that's exactly what I want to avoid. It's how school math is taught, and apparently it's not working for her. :( If this is indeed merely a set of rules, then I guess there'd be no other way. Waiting for confirmation. :|

PEMDAS is just a human convention.  It just has to be memorized with the understanding that it is just a convention humans made up.

But the facts you mentioned are different than PEMDAS.  These are not facts that you have to just accept because "that's the way it is".  It is possible (and not that hard) to have a good intuitive understanding of these facts.

For example, why is it true that $ (a^x)^y \equal{} a^{xy}$?  This doesn't have to be memorized.  If you forget it, you can figure it out again.  Consider the case where $ x$ and $ y$ are positive integers.  The explanation will be more clear if you just look at a specific case, like $ x \equal{} 2$, $ y \equal{} 3$.

$ (a^2)^3 \equal{} a^2 \cdot a^2 \cdot a^2 \equal{} a^{2 \plus{} 2 \plus{} 2} \equal{} a^{2 \cdot 3}$.

Of course, this argument assumes you already know that $ a^m a^n \equal{} a^{m \plus{} n}$, if $ m$ and $ n$ are positive integers.  This is a very intuitive fact, perhaps best explained by looking at a special case, like $ m \equal{} 3$, $ n \equal{} 4$.

$ a^3 a^4 \equal{} (a \cdot a \cdot a) \cdot (a \cdot a \cdot a \cdot a) \equal{} a^{3 \plus{} 4}$.

You can give a similar intuitive argument for the other rule you mentioned.  If your sister ever forgets these rules, she can just derive them again.

Another very important thing to teach math students is that, if they aren't sure if they remember a rule correctly, they can find out by doing some experiments.  A secret for being good at math, that for some reason most normal students don't do, is doing little experiments.

For example, let's say a student thinks the rule is: $ a^x a^y \equal{} a^{xy}$.  She can test this with an experiment, just picking random values of the variables.  Try $ a \equal{} 2$, $ x \equal{} 1$, $ y \equal{} 1$.  Then the "rule" in question would say $ 2^1 2^1 \equal{} 2^{1 \cdot 1} \equal{} 2^1 \equal{} 2$, or in other words $ 4 \equal{} 2$.  Which is wrong.  So the experiment showed that she remembered the rule wrong, and so she knows the rule is something else.

You probably knew most of that already, but perhaps something there will be helpful.

Ah, yes I remember doing the experiments you mentioned when I was in middle school. :) Still do.

Thanks for that idea to expand the exponentiated term - I'd forgotten about that (isn't it how it's taught at school?). I guess it's pretty intuitive.

ehh I realize that I may have used the wrong term. I was wondering if there was a more visual way to show this (like the proofs without words on the AoPS pages), but that may not be intuition for everyone; maybe intuition comes in different forms to different people?

Thanks for your suggestions anyway. :)

I guess...
You could have a square and then cut off a smaller square "portion" to show $ \frac {a^2}{b^2}$.

For example, have a square with side length 5, and and then have a square of side length 3 within it. Look at one of the sides, where there are 3 out of 5 (of the 3x3), and the other side with the same thing, and then count the blocks or shaded units or something to show that $ \left( \frac {3}{5} \right) ^2 \equal{} \frac {3^2}{5^2}$.

Alternately, to show $ \left( \frac {a}{b} \right) ^n$, write out the fractions. $ \left( \frac {a}{b} \right)^n \equal{} \frac {a}{b}\cdot \frac {a}{b}\cdot \frac {a}{b}\cdot \frac {a}{b}\cdot \frac {a}{b}\cdot \frac {a}{b}\cdots$ n times. Thus, multiply all the numerator and denominators together to get $ \frac {a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a\cdots }{b\cdot b\cdot b\cdot b\cdot b\cdot b\cdot b\cdots }$ n times.

Which is the same as $ \frac {a^n}{b^n}$.

Textangle, when you write $ \frac{a}{b}^n$, it's a better idea to write $ \left(\frac{a}{b}\right)^n$ to avoid confusion.


Sa se arate ca pentru orice n natural n nenul exista un singur nr nat divizibil
cu 5ⁿ care in baza 10 se scrie cu n cifre din multimea {1,2,3,4,5}.

Daca gaseste cineva o rezolvare mai scurta va rog frumos sa ma ajutati!!![/code]

sunt exact $5^n$ numere de $n$ cifre din multimea $\{1, 2, 3, 4, 5\}$. deci este suficient sa aratam ca oricare doua din aceste numere difera $\pmod {5^n}$, pt ca atunci rezulta ca unul (si numai unul) dintre ele e divizibil cu $5^n$. (toate resturile posibile apar, inclusiv 0.) deci...

presupunem ca $A = a_{n-1}...a_{0}$ si $B = b_{n-1}...b_{0}$ (unde $a_i, b_i \in \{1, 2, 3, 4, 5\}$) si $5^n |A - B$. fie $c_i = a_i - b_i$. observa ca $c_i \in \{0, 1, 2, 3, 4\}$. atunci

$5^n|10^{n-1} c_{n-1} + 10^{n-2} c_{n-2} + ... + 10^0 c_0$

de unde rezulta

$5^{n+1}| 10^n c_{n-1} + 10^{n-1} c_{n-2} + ... + 10 c_0$,

din care

$5^n | 10^{n-1}c_{n-2} + ... + 10 c_0$.

si la fel,

$5^{n+1} | 10^n c_{n-2} + ... + 10^2 c_0$, adica $5^n | 10^{n-1} c_{n-3} + ... + 10^2 c_0$, etc.

in general,

$5^n | 10^{n-1} c_{i} + ... + 10^{n-i-1} c_0$, pt $i = 0, 1, ..., n-1$  (**)

deci obtinem pana la urma $5^n | 10^{n-1} c_0$, adica $5 | c_0$, adica $c_0 = 0$. apoi (**) cu $i = 1$ ne da $c_1 = 0$, etc. deci $c_i = 0$ pt fiecare $i$, adica $a_i = b_i$, si deci $A = B$.


How many positive integer values of $ x$ are solutions to the inequality $ 10 < \minus{}x \plus{} 13$?

Subtract 13 from both sides.
-3<-x
Divide by -1 and REVERSE THE SYMBOL
to get 3>x.
x=1 or 2, thus you get 2.


If you're at either the Penn State or Iowa ARML sites, please come to the Friday Night Talk!  They happen during the coaches meeting and are always a good time.  The details:

[b]Penn State:[/b] Friday, 8PM, Thomas Building Room 100
[b]Surprising Binary Sequences and Mysterious Expanding Tilings[/b]
[i]given by dr. sarah-marie belcastro[/i]
Binary sequences use only 0s and 1s but within this seeming simplicity some stunning surprises skulk.  From elementary definitions, we will confront complexity.  The same goes for tilings---from basic tiles, complicated patterns can be made.  Using mysterious powers of expansion, we will prove that some tilings---and some binary sequences---are aperiodic.

[b]Iowa: [/b] Friday, 8PM, Illinois Room of the Iowa Memorial Union
[b]The Euler Spoilers[/b]
[i]given by Professor David C. Kelly[/i]
Hear the stories of the courageous few who dared to challenge the man who was called “analysis incarnate.”  You’ll witness counterexamples annihilating conjectures, marvel at wildly divergent series, see typos before typewriters were invented, puzzle over things that make no sense, and laugh at some major booboos.

Hmmm..... I seem to sense a strong HCSSiM connection...

Yes, I'm definitely going to the Iowa one.  I'd go to the Penn State talk too, if it weren't so far away.

In my schedule (from Mr. Sullivan), the talk begins at 8:17 and ends at 9:17 for some reason.

But still, wouldn't these times be more appropriate?

*sigh* I seem to remember this schedule issue happening a couple of years ago, as well.  I was told by the site coordinator and the tech person that the Penn State talk is scheduled for 8pm (and that's what the posters say).  Mr. Sullivan, on the other hand, knows that the talk [i]should[/i] start at 8:17 in an ideal world, which is probably why he's written it that way on a schedule.

cosinator, is the schedule which Mr. Sullivan handed out just going to the WMass team, or to all of the Penn State folks?  If it's just to the WMass team, then it would be great if you could tell the team that the talk is at 8 rather than 8:17 (despite the greater appropriateness of the latter time).  If it's everyone, then this means that I should start talking late *grin*.

[quote="Ciaccona"]I'll see if I can get out of the end of our team practice that night.

And, yes, 8:17-9:17 makes [i]much[/i] more sense, if they can't be at prime time.[/quote]

Kinda off topic, but at MIT's Campus Preview Weekend, all the events hosted by Random Hall began at 17 or 47(?) minutes after the hour.  Just saying :wink: .


Let $ ABCD$ be a convex quadrilateral and let $ O \in AC \cap BD$, $ P \in AB \cap CD$, $ Q \in BC \cap DA$. If $ R$ is the orthogonal projection of $ O$ on the line $ PQ$ prove that the orthogonal projections of $ R$ on the sidelines of $ ABCD$ are concyclic.

Denote E,F,G,H be the orthogonal projs of R on AB,BC,CD,DA respectively.
With mod $ \pi$:
(EH,ER)=(PH,PR), (ER,EF)=(BR,BF), so (EH,EF)=(PH,PR)+(BR,BF)
Similarly, (GH,GF)=(DH,DR)+(QR,QF)
We knew that \angle PRD=\angle QRB
Hence (EH,EF)=(GH,GF), then E,F,G,H are concyclic.
[u]Note:[/u] If quadrilateral ABCD is cyclic, E,F,G,H are collinear.

[quote="mr.danh"][u]Note:[/u] If quadrilateral ABCD is cyclic, E,F,G,H are collinear.[/quote]
More exactly, when $ ABCD$ is cyclic, $ R$ coincides with the Miquel point of the complete quadrilateral $ ABCDEF$ and therefore $ E$, $ F$, $ G$, $ H$ all lie on the so called Simson line of $ ABCD$, which is parallel to the Aubert line and perpendicular to the Gauss line of the given quadrilateral.

[quote="plane geometry"]With mod \pi:
(EH,ER)=(PH,PR), (ER,EF)=(BR,BF), so (EH,EF)=(PH,PR)+(BR,BF)

what does it mean？sorry for my unwisdom[/quote]
Oriented angles,I guess...

any solutions without Oriented angles please

Dear Mathlinkers,
a similar problem intitled "The Neuberg-Mineur circle" can be found on my website
http://perso.orange.fr/jl.ayme
Sincerely
Jean-Louis

Sorry to revive, but the above post contains an error; the second $AB$ in the last line should actually be a $BC$.

Also, the $ARC$ and $BRD$ should actually be $BRC$ and $ARD$.

[quote="plane geometry"]With mod \pi:
(EH,ER)=(PH,PR), (ER,EF)=(BR,BF), so (EH,EF)=(PH,PR)+(BR,BF)

what does it mean？sorry for my unwisdom[/quote]

Basically, they're angles. The common point at the left is the vertex. For example, $(EH,ER)=(\angle HER)$.

I think that there is a typo here: $H$ and $G$ are switched.

@above, so does order matter? (i.e. are $(EH, ER)$ and $(ER, EH)$ the same thing?)

I noticed that $GE$ and $HF$ intersect on $RO$. Did anyone manage to prove something like this?

[quote=Heebeen, Yang]Lemma China TST 2002 Day 1 problem 1,
$ ABCD$ is quadrilateral, define $ AB \cap CD \equal{} P$, $ AC \cap BD \equal{} O$, $ AD \cap BC \equal{} Q$, orthogonal projection of $ O$ on the line is $ R$.
Then, $ \angle ARC \equal{} \angle BRD$

If we define $ X, Y, Z, W$ be orthogonal projections of $ R$ on the line $ AB, DC, DA, AB$ respectively, then it is easy to show $ \angle XYW \equal{} \angle XZW$ with lemma:)[/quote]

How is it easy?I am not getting it

[quote=Pluto1708]

How is it easy?I am not getting it[/quote]

while it's not obvious, it is easy in the sense that only some simple angle chasing will be needed to complete the proof

Let $W,X,Y,$ and $Z$ be the orthogonal projections of $R$ onto $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively.

[b][color=#EA9999]Lemma:[/color][/b] [i][color=#EA9999](China TST 2002 D1 P1)[/color][/i] Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD.$ The two diagonals meet at $P.$ Let $O$ be the foot of the perpendicular from $P$ to $EF.$ Show that $\angle BOC=\angle AOD.$

[i]Proof.[/i] Since $$-1=(C,D;\overline{FP}\cap\overline{BC},E)\stackrel{F}{=}(C,A;P,\overline{CP}\cap\overline{FE})$$and $\angle POE=90,$ we know that $\overline{OP}$ bisects $\angle COA.$ Similarly, $\overline{OP}$ bisects $\angle BOD$ and we are done. $\blacksquare$

Hence, \begin{align*}\measuredangle YZX&=\measuredangle WZA+\measuredangle QZX\\&=\measuredangle WRA+\measuredangle QRX\\&=\measuredangle CRX+\measuredangle WRP\\&=\measuredangle CYX+\measuredangle WYC\\&=\measuredangle WYX.\end{align*} $\square$


Someone told me a geometry problem:
      Consider a trapezoid ABCD with bases AB and CD inscribed in a circle. Take E the intersection between AD and BC  and take O the intersection between AC and BD. Trough O we take a line paralel to AB, which cuts the circle in M. Then EM is tangent to the circle.
    Is it well known? I don't know.

Neither do I, Harazi :( . But if ABCD inscribed(not neccesary a trapezoid) then we have a special case of Brianchon's theorem.

yes it's well known in every geometry book which contains pole geometry(excuse me I don't know what do say)
u can eaisly see that the pole of E pass through O ( assume that the line pass through O and paralel to AB is MM') so MM' is pole of E so we are done.

Basicly, we only need to show that if OM cuts ED in T, then (E,T;A,D)=-1 (or, in other words, E, T are harmonicly comjugated with A, D). This is easy to show with Thales and stuff. We have thus shown that the pole of E with respect to the circle is OM, so EM is tangent to the circle.

This is what sam-n said, but I wanted to make it clear :).

I think I misunderstood the problem. What do you mean the bases AB,CD inscribed in a circle.? Does that mean ABCD is cyclic ?

Yes treegoner, it means that ABCD is a cyclic trapezoid.

So it is very familiar problem.
It can be generalized as follow
          Let ABCD be cyclic. O,M is the intersection of AC and BD;AB and CD. Then the pole of M with respect to the circle passes O.  :D

[quote="treegoner"]Let ABCD be cyclic. O,M is the intersection of AC and BD;AB and CD. Then the pole of M with respect to the circle passes O.[/quote]

This indeed a well-known fact, and there was a discussion about it at [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=390&forum=DCForumID6]CTK Exchange[/url] (see Theorem 1 of "Quadrilateral Inversions").

  Darij


modular arithmetic
number theory proposed
number theory
polynomial congruence

Find all pairs $\left(m,n\right)$ of positive integers, with $m,n\geq2$, such that $a^n-1$ is divisible by $m$ for each $a\in \left\{1,2,3,\ldots,n\right\}$.

Let $p$ be a prime dividing $m$. We then have $x^n-1=(x-1)(x-2)\ldots(x-n)$ in $\mathbb Z_p[x]$. We then have $n<p$ and $1+2+\ldots+n=0\pmod p\iff p|\frac{n(n+1)}2\iff n=p-1$. This means that $m$ is a prime power, $p^t$, and $n$ is $p-1$. However, it's well-known that there is a primitive root $\pmod{p^2}$ in the set $\{1,2,\ldots,p-1\}$ (it's even been discussed on the forum), so unles $t=1$, this can't be possible, meaning that the pairs are $(m,n)=(p,p-1)$ with prime $p$ (it's easy to check that these pairs work).

Thaks grobber, I did the same thing you did. Another way to prove the last part is to notice that:
$(p-1)^2-1=p^2-2p=0(p)$, and different from $0$ $mod p^2$ (p=2 is dealed manually). so,
$(p-1)^{2\left(\frac{p-1}{2}\right)}=\left((p-1)^2-1\right)\times A$, where $(A,p)=1$.

Let $p$ be a prime dividing $m$. We then have $x^n-1=(x-1)(x-2)\ldots(x-n)$ in $\mathbb Z_p[x]$ so $1+2+\ldots+n=0\pmod p\iff p|\frac{n(n+1)}2\iff n=p-1$.($n=p+1$ is impossible)
A similar argument shows that $m=p^k$ so to show this we will show that $(p-1)^{p-1}+1 \equiv (-1)^{p-1}+(-1)^{p-2}(p-1)^2 \equiv -(p-1)^2+1 \neq 0 \pmod {p^2}$ so the only solutions are $(m,n)=(p,p-1)$
@below,Reduce all the coefficients of $F(x):=\prod_{d=1}^n (x-d) \pmod p$.If $F \not\equiv X^n-1$ in $[\mathbb{F}_p]$ then subtracting them gives a polynomial of degree $n-1$ having $n$ zeroes modulo $p$ contradiction by Lagrange's Theorem.So all the coefficients of $F$ except the coefficient of $X^n$ and the constant term is divisible by $p$.Vieta's theorem finishes.


ratio
limit
real analysis
real analysis unsolved

Since mathwizarddude asked for it:

Let $ a_n$ be a sequence of positive numbers.

Suppose $ \lim_{n\to\infty}\frac{a_{n\plus{}1}}{a_n}\equal{}L$ and $ \lim_{n\to\infty}\sqrt[n]{a_n}\equal{}M$ both exist.

Prove that $ L\equal{}M.$

the equality readily follows once for a sequence of positive terms, it is shown that

$ {\lim\; \text{inf} }_{n\to\infty} \; \frac{a_{n\plus{}1}}{a_n} \le  {\lim\; \text{inf}} _{n\to\infty} \; \sqrt[n]{a_n} \le$

[color=white].................................................[/color]$ {\lim\; \text{sup} }_{n\to\infty} \; \sqrt[n]{a_n} \le {\lim\; \text{sup} }_{n\to\infty} \; \frac{a_{n\plus{}1}}{a_n}$


sorry i could not understand the question beofre..

Hi salahcool.

I don't know if there is more than one solution. Luckily starting with $ 22^{3} \equal{} 10648$, we get to $ 27^{3} \equal{} 19683$ and $ 27 \equal{} 1 \plus{} 9 \plus{} 6 \plus{} 8 \plus{} 3$ so $ (1 \plus{} 9 \plus{} 6 \plus{} 8 \plus{} 3)^{3} \equal{} 19683$.

Nice puzzle, salahcool.

thank you very much AndrewTom

You are an intelligent man.

salah :!:  :!:  :!:

Really do you have to say IN THE NAME OF GOD everytime? It's annoying and now it has no effect on anyone, so just stop.

[quote="salahcool"]IN THE NAME OF GOD

Replace the same characters by the same numerals so that the mathematical operations are correct.

(L+O+G+I+C)^3=LOGIC

(L+O+G+I+C)*(L+O+G+I+C)*(L+O+G+I+C)=LOGIC

THANKS
SALAH[/quote]
Does that mean L times O times G times I times C or are those variables just in the digits?

I'm pretty sure the RHS means that they are digits, otherwise the answer wouldn't be what it is.


limit
integration
real analysis
real analysis unsolved

Let $ x\in\mathbb{R}^{n}$, and define $ \parallel x\parallel _{\infty}\equal{}sup\{|x_{i}|;i\in\{1,\dots,n\}\}$, and $ \parallel x\parallel _{q}\equal{}\sqrt[q]{\sum_{k\equal{}1}^{n}|x_{k}|^{q}}$. Prove that $ \lim_{q\rightarrow\infty}\parallel x\parallel _{q}\equal{}\parallel x\parallel _{\infty}$.

[hide="My solution"]
Lets assume $ |x_{0}|: \equal{}|x_{1}|\equal{}\dots\equal{}|x_{p}|> |x_{p\plus{}1}|\geq\dots\geq |x_{n}|,1\leq p\leq n$.Therefore  $ \lim_{q\rightarrow\infty}|x_{o}|\sqrt[q]{p\plus{}\sum_{k\equal{}p\plus{}1}^{n}(\frac{|x_{k}|}{|x_{0}|})^{q}}$. Since $ \lim_{q\rightarrow\infty}(\frac{|x_{k}|}{|x_{0}|})^{q}\equal{}0$ , we have $ \lim_{q\rightarrow\infty}\sqrt[q]{p\plus{}\sum_{k\equal{}p\plus{}1}^{n}(\frac{|x_{k}|}{|x_{0}|})^{q}}\equal{}1$, and $ |x_{0}|\equal{}sup\{|x_{i}|;i\in\{1,\dots,n\}$.
[/hide]

Any ideas how to prove it when in $ \mathcal{L}_{q}$, i.e. $ \parallel f\parallel _{q}\equal{}(\int_{X}|f|^{q}d\mu)^{\frac{1}{q}}$.

On your second question:

Make use of the fact that if $ a<\|f\|_{\infty}$ then there is some set $ E$ of positive measure such that $ |f|>a\chi_{E}.$


algebra polynomial calculus integration function

prove that No polynomial f(n) with integral coefficients, not a constant,
can be prime for all n.

so we can take any other cofficient, and if they are all 1 then we can put x=2.
ok thats wrong

Any other co-efficient? Sure that'll work?

this theorem might help:

let $a$, and $b$ be integers such that $a\ne b$, $b\ne 0$ and $P(x)$ be a polynomial with integer coefficients then:

$\frac{P(a)-P(b)}{a-b}$ is an integer

[hide]if $P(0)\ne 0$, then

$\frac{P(P(0))-P(0)}{P(0)-0}$ so $P(0)|P(P(0))$

otherwise, we factor out the x, and the rest of the polynomial cannot always be 1 unless the function is $P(x)=x$, a simple counter example, P(6) is composite, finishes the proof[/hide]


Diophantine equation
number theory unsolved
number theory

Show that there exist infinitely many pairs of positive integers $ (m, n)$ such that $ \binom{m}{n\minus{}1}\equal{}\binom{m\minus{}1}{n}$

[quote="outback"]Show that there exist infinitely many pairs of positive integers $ (m, n)$ such that $ \binom{m}{n \minus{} 1} \equal{} \binom{m \minus{} 1}{n}$[/quote]
[hide="Solution"]
This is $ \frac{m!}{(m\minus{}n\plus{}1)!(n\minus{}1)!}\equal{}\frac{(m\minus{}1)!}{n!(m\minus{}n\minus{}1)!}$, so $ mn\equal{}(m\minus{}n)(m\minus{}n\plus{}1)$, which means that $ mn\equal{}m^2\minus{}2mn\plus{}n^2\plus{}m\minus{}n$, which implies that $ m^2\minus{}m(3n\minus{}1)\plus{}(n^2\minus{}n)\equal{}0$.  The discriminant of this must be a perfect square, so $ 9n^2\minus{}6n\plus{}1\minus{}4n^2\plus{}4n\equal{}x^2$ for some $ x$.  Then, $ 5n^2\minus{}2n\plus{}1\equal{}x^2$, so $ 25n^2\minus{}10n\plus{}5\equal{}5x^2$, so $ (5n\minus{}1)^2\plus{}4\equal{}5x^2$.  We will let $ x$ and $ 5n\minus{}1$ be even, so $ n$ is odd, say $ 2k\plus{}1$ and $ x$ is even, say $ 2y$.  This then becomes $ 5y^2\minus{}(5k\plus{}2)^2\equal{}1$.  Let $ 5k\plus{}2\equal{}z$.  Then, we get that $ 5y^2\minus{}z^2\equal{}1$.  Let $ z\equal{}2$ and $ y\equal{}1$.  This is one solution.  It is well-known that the other solutions to the Pell Equation then must be in the form of $ (y, z)\equal{}(a,b)$ where $ a\sqrt{5}\plus{}b\equal{}(\sqrt{5}\plus{}2)^n$.  We must, however, ensure that there are infinitely many $ (y,z)$ in the form of $ z\equiv 2\bmod 5$ as that is the only condition on $ z$.  Notice that $ z\equal{}\frac{(2\plus{}\sqrt{5})^n\plus{}(2\minus{}\sqrt{5})^n}{2}$, which is the recursion $ z_{n\plus{}1}\equal{}4z_n\plus{}z_{n\minus{}1}$ wtih $ z_0\equal{}1$ and $ z_1\equal{}2$.  The next number will be (in modulo $ 5$), $ 4$, then $ 3$, then $ 1$, then $ 2$, then $ 4$, etcetera.  The period $ 1, 2, 4, 3\cdots$ contains a $ 2$, so there are infinitely many $ z\equiv 2\bmod 5$, so there are infinitely many solutions. [/hide]


Could anyone post the derivation of
$ V_d \equal{} \minus{}\frac{eE\tau}{m}$??
$ \tau$ is relaxation time.
Or some links...

In the derivation,do we assume that after a collision, the electron loses its drift velocity, but retains it's random velocity??

[url=http://en.wikipedia.org/wiki/Drift_velocity]Drift velocity of an electron.[/url]

Thanks.
But why is $ v_{\it avg} \equal{} \mu \cdot E$ ??

I saw this proof in a book:

Let $ \tau$ be the average time between successive collisions. Then distance $ l$ drifted in time $ \tau$ is given by
         $ l \equal{} a\tau^2$

            $ \equal{}(\frac{eE}{m})\tau^2$

where $ e$ is charge of the electron with mass $ m$.

Therefore,
        $ v_d \equal{} \frac{l}{\tau}$

              $ \equal{}(\frac{eE}{m})\tau$

What kind of collision happens?
Here,do we assume that after a collission,the drift velocity becomes 0,but the random velocity is unaffected.(ie),after the collission,the velocity of the electron is purely random (similar to that when no electric field was present).Why doesn't the electron come to rest after collision??


I think yours is wrong, hoang. The right side should be $ \min(1,\frac{n}{4})$ and $ \min(1,\frac{n}{5})$ only.

yes, you are right  :blush: Let solve the following inequailty with 4 variables:
$ \bigstar$ Problem 3:
Let $ a,b,c,d,e$ be positive real numbers such that $ abcde\equal{}1$. Prove that:
\[ N\ge 1\]
with
\[ N\equal{}\frac{1}{1\plus{}a\plus{}a^2\plus{}a^3\plus{}a^4}\plus{}\frac{1}{1\plus{}b\plus{}b^2\plus{}b^3\plus{}b^4}\plus{}\frac{1}{1\plus{}c\plus{}c^2\plus{}c^3\plus{}c^4}\plus{}\frac{1}{1\plus{}d\plus{}d^2\plus{}d^3\plus{}d^4}\plus{}\frac{1}{1\plus{}e\plus{}e^2\plus{}e^3\plus{}e^4}\]


Find all $F \in C(X)$ rational fraction $F=\frac{P}{Q}$
$P,Q$ polynomial in $C[X]$
such that
$F(X^2)-F(X)=\frac{X}{X^2 -1}$

cresus nice answer give your solution, thx.

:cool: The problem is rather trivial ! Why is it in the  Advanced Section  ??! :D


vector
analytic geometry
geometry
parallelogram
combinatorics proposed
combinatorics

Here's a cute one :).

Given a finite set of vectors and a set of marked lattice points, we know that for any marked point $P$, if we put all the vectors with their starting end in $P$, then more than half of all the vectors will end in a marked point. Show that we have infinitely many marked points.

suppose more than half of the vectors go from left to right, if the set of points is finite, pick the rightmost point $P$, then, more than half of the vectors are to the right of $P$, but then, those vectors cant coincide with other members of our set of points since they are to the left of $P$, contradiction! :P

Cool! :) There are some details to work on (taking care of vectors which go directly up, and so on; I think the shortest way to overcome these is to consider a sistem of coordinates $xOy$ s.t. none of the vectors has direction $Ox$ or $Oy$), but I think it's Ok.

I did something completely different: consider two marked points $A,B$ which are farthest apart (i.e. the distance between them is the diameter of the set). With the given condition, there must be a vector $v$ in the set and marked points $C,D$ s.t. $\overrightarrow{AC}=\overrightarrow{BD}=v$. However, one of the diagonals of the parallelogram $ABDC$ will have to be strictly larger than $AB$, so we have a contradiction.


ratio
geometry
geometric transformation
homothety

[Si dimostri che l’insieme delle omotetie con centro O fissato è un gruppo.]

I have to demonstrate that ,the whole of omotetias, their centre is O already fixed, is a group.
How can I demonstrate it? I found it in my final math exam at high school!!!

Could you explain what "omotetia" means?

'omotetia' is an italian word but I don't know how to translate it, so I will try to explain the concept
When I say omotetia I mean a trasformation that uses simple equations, and using an omotetia I can find a point that can be transformed in another one with which it had a connection:
 general equation of it are for example:
 (consider these equation put in a system)
    X=k(x-a)+a
    Y=K(y-b)+b
  O(a;b) is the centre of the omotetia
  and K is the omotetia ratio that is K=OA'/OA
  For example we have a point A(2;3) and we know k=2 and O(1;1)

  our omotetia is in this case:
  X=2(x-1)+1
  Y=2(y-1)+1             so we can find the point A' (3;5)

Ah.  According to Wikipedia, it means a homothetic transformation.  I'm not too confident about my knowledge of homothety, but a homothety of a given center can be uniquely represented by a scale factor $k$, and a composition of homotheties is just a multiplication of scale factors, and that's clearly a group.

Shouldn't this be moved to Intermediate Topics?


geometry
circumcircle
LaTeX
national olympiad

let ABC be an acute angled triangle; let D,F be the midpoints of BC ,AB respectively.Let the perpendicular from F to AC and the pependicular from B to BC meet in N.Prove that ND is equal to the circumradius of ABC.

as per the ques.draw a line perpendicular to BC passing from B.this line is going to meet a line which is perpendicular to AC passing through F.
we know circum-radius of triangle ABC=a/2sin A
now come to triangle BDFN
its a cyclic quad(see the sum of opp. sides)
so angle BND=angle A
Thus in triangle BND
we have
ND=a/2sinA
if u don't like the ans kindly rate it poor.I will delete it.

Area of triangle ABC is abc/4R where R is the radius of the circumcircle.
F and D are the mid points of AB and BC respectively. Therefore FD is parallel to AC and half of it i.e FD= b/2.
Area (BDF) = 1/4 area (ABC)  = 1/4 (abc / 4R) --------    (1) and BC = a/2  and AB = c/2.
Now BDFN are concyclic with ND being diameter of the circle of radius r circumscribing triangle BDF.
Area (BDF) = (a/2 .  b/2  . c/2 ) / 4 r -------    (2)

Therefore from (1) and (2)
ND= 2r = R    Hence Proved

thanks anurag for the help.

i think that using trigo you can make the answer much more precise and short and i have one simple and short answer here
i ask someone to convert it to latex so that it could be more easily read

the answer goes something like this
now using midpoint theorem on points F and D we have FD ll AC.
which makes angle A=angle BFD

Applying R(Circumradius)=a/2sin A we have
R=2.BD/2sin BFD
=BD/sin BFD
=BD/sin BND (since BNFD is cyclic because NBC and NFD are both right angles)
=ND/sin NBD (applying sin rule in triangle BND)
=ND/sin 90
=ND
''''''

See if you know the answer why do u post it i unsolved problems section and why do u plead others to give their solns and that too with a thread already existing?

First of all i didnt have the idea that someone else had already posted the question as generally the RMO are asked on this forum and not on the other as u can c on the other RMO qs

next u shud hve noticed dat ive posted this question very close to d other 1


search
calculus
integration
logarithms
real analysis
real analysis unsolved

prove the equality
\[ \sum_{n\equal{}1}^{\plus{}\infty}\frac{(n!)^2}{(2n)!} \equal{} \frac{9\plus{}2\pi\sqrt{3}}{27}\]
I have a hint that some arc sin power extentions should be used  but I don't have idea...

[quote="Leonard"]prove the equality
\[ \sum_{n \equal{} 1}^{ \plus{} \infty}\frac {(n!)^2}{(2n)!} \equal{} \frac {9 \plus{} 2\pi\sqrt {3}}{27}
\]
I have a hint that some arc sin power extentions should be used  but I don't have idea...[/quote]

pls search this forum, i have already posed this question before (and someone has provided a solution)... actually, there are several ways of getting the sum, the solution provided involved equating the sum to an integral and then solving that integral. the question i asked was to show

$ \sum_{n\equal{}1}^\infty\;\dfrac{1}{\dbinom{2n}{n}}\;\equal{}\;\boxed{\frac{1}{3}\plus{}\frac{2}{9}\cdot\frac{\pi}{\sqrt 3}}$

[quote="Leonard"]prove the equality
\[ \sum_{n \equal{} 1}^{ \plus{} \infty}\frac {(n!)^2}{(2n)!} \equal{} \frac {9 \plus{} 2\pi\sqrt {3}}{27}
\]
I have a hint that some arc sin power extentions should be used  but I don't have idea...[/quote]
yes you are right. misans proof is  interesting he use Gamma finction  and the relation that $ \beta \equal{} \frac {\Gamma (n \plus{} 1) \Gamma (n \plus{} 1)}{\Gamma (2n \plus{} 2)}$  , but bythis way he  consider only particular case of the expression when $ x \equal{} 1$:
$ \sum _{n \equal{} 1}^{ \plus{} oo}\frac {(n!)^{2}}{(2n)!}x^{n} \equal{} \frac {4}{4 \minus{} x} \plus{} \frac {4 \sqrt {x}}{(4 \minus{} x)^{\frac {3}{2}}}arcsin \frac{\sqrt {x}}{2}$ if $ x\geq 0$ and is $ \frac {4}{4 \minus{} x} \minus{} \frac {4\sqrt {|x|}}{(4 \minus{} x)^{\frac {3}{2}}}\ln \frac {\sqrt {|x|} \plus{} \sqrt {4 \minus{} x}}{2}$ if $ x\leq 0$
last expression is no hard it's standart problem which ealsy solved if you consider this sum as $ f(x)$ and consider $ f'(x)$ and $ f''(x)$ and after some calculation you will be have differential equation , try do it yourself, if you can't then later I will post full solution. :wink:


AMC AIME probability articles blogs USA(J)MO USAMO \/closed

Both of these classes seem like they would be useful to me and wouldn't be too hard for me, although due to time/money, I can probably only take one of them.

Background info:
AMC 12 scores: 120-135
AIME scores: 7-9
Also taking AIME Problem Series A and WOOT
I plan to read AoPS 2.

For each of the courses, I basically know enough to do them, but I'm unfamiliar with most of the problems I see in the post-tests.
In NT I basically know all the basics about modular arithmetic, and I do know about FLT. In CP, I know the basic facts about probability and combinatorics.

So which one should I take? Which one would be of greater value to me? Which one might I learn by other means? Also, do you highly recommend that I stretch time and money and do both of them? Or does anyone even think that I don't really need to do either of them?

I think Intermediate Number Theory might be more useful, since it shows up more on the AIME.

Personally, I think my number theory skills really went up after doing the course. It helped me get a hard number theory problem on the AIME. Unfortunately, this class does not teach you to add and multiply properly (The two reasons I missed two problems on the AIME). I think AoPS should add a lesson to the course plan titled "Addition, Subtraction, Multiplication, and Division: Easier Said Than Done", and have homework problems be like miltiply $678$ and $3197$ by hand.  :D

Seriously, either class will be good for you, but Intermediate Number Theory's content will probably show up more during the AIME. Just remember to do the problems. That's the most important thing. I slacked off quite a bit when I took this course, and now I'm trying to do the problems (long after the class has ended) to catch up.

I believe there actually is an article on AoPS on how not to make stupid mistakes. It's probably in the resource section.

[quote="1234567890"]I think AoPS should add a lesson to the course plan titled "Addition, Subtraction, Multiplication, and Division: Easier Said Than Done", and have homework problems be like miltiply $678$ and $3197$ by hand.  :D
[/quote]

I included a chapter on Number Sense in the Intro NT book.  I've written a couple of helpful articles about how to take an algebraic approach to arithmetic in my blog and hope to one day compile a much larger article about the topic.  For now, I'm spreading the basic ideas around in pieces, partially because it's hard to convince some people that it's a mathematically useful topic beyond just achieving scores on tests (I view most of number sense as extensions of elementary discrete math).

[quote="PenguinIntegral"]How hard and advanced is the NT Seminar? Would it be more practical to just go through the NT chapters in AoPS 2 and the Sato packet, or is there a bunch of unique stuff in there?

On another topic, is it the Intermediate Algebra or CP that deals with constructing sets and recursive sequences? I see those a bunch on the USAMO.[/quote]

While AoPS 2 covers a lot of the material, it's expanded to greater detail in the class.  Also, there is a great deal of time spent in the class developing an eye for ways in which algebra is helpful in solving number theory problems.

intermediate counting and probability deals with recursion.

[quote="calc rulz"]Meh, guess I'll do the NT one. Also that one costs less :)

Though does anyone have anymore comments on usefulness of the two in competitions/other math and on how much I'd learn of them from other courses (AIME Problem Series, WOOT, AoPS 2)?[/quote]

There isn't much overlap between either of the courses and the books/courses you mention.  Some of the topics in each course are in AoPS 2, but we go into more detail (and into more advanced problems) in the course than in AoPS Vol 2.

As for which to take, it depends on your weaknesses, and where you are in contest math already.  Seeing as you're doing pretty well on the AIME already, and that so many problems on the last two USAMOs were number theory, I'd guess number theory is probably a good choice for you.  If you want to brush up on advanced counting techniques, we should have the Intermediate Counting book finished early next year.


Let $ a,b,c$ be positive real numbers. Prove that:

$ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \ge \frac{{a \plus{} b \plus{} c}}{{\sqrt[3]{{abc}}}}$

$ 3\sum\frac {a}{b} \equal{} \sum(\frac {a}{b} \plus{} \frac {a}{b} \plus{} \frac {b}{c})\geq 3\sum\sqrt [3]{\frac {a^2}{bc}} \equal{} 3\frac {\sum a}{\sqrt [3]{abc}}$


What is the triangle that has Morley and inner Napoleon  triangles coincide ?

  M.T.

[quote="armpist"]What is the triangle that has Morley and inner Napoleon  triangles coincide ?[/quote]

Nice question, but I believe that there is no such triangle. Below is my proof; please check it carefully since I am not that reliable when solving this kind of problems.

Well, let ABC be our triangle. Assume that its Morley triangle coincides with its inner Napoleon triangle.

Let A'B'C' be the Morley triangle of triangle ABC, labeled in the most natural way, i. e. such that we have $\measuredangle BAC^{\prime}=\measuredangle CAB^{\prime}=\frac{A}{3}$, $\measuredangle CBA^{\prime}=\measuredangle ABC^{\prime}=\frac{B}{3}$ and $\measuredangle ACB^{\prime}=\measuredangle BCA^{\prime}=\frac{C}{3}$. Now, by the assumption of the problem, this Morley triangle A'B'C' coincides with the inner Napoleon triangle of the triangle ABC; of course, this doesn't mean that each vertex of triangle A'B'C' coincides with the [i]corresponding[/i] vertex of the inner Napoleon triangle of the triangle ABC, but at least every vertex of triangle A'B'C' must coincide with [i]some[/i] vertex of the inner Napoleon triangle.

So, for instance, the vertex A' of triangle A'B'C' coincides with some vertex of the inner Napoleon triangle of the triangle ABC.

- If A' coincides with the vertex corresponding to A, then we must have < CBA' = 30° and < BCA' = 30° (just remember how the inner Napoleon triangle is constructed). But $\measuredangle CBA^{\prime}=\frac{B}{3}$ and $\measuredangle BCA^{\prime}=\frac{C}{3}$; thus $\frac{B}{3}=30^{\circ}$ and $\frac{C}{3}=30^{\circ}$, yielding B = 90° and C = 90°. But this is a contradiction, since a triangle cannot have two right angles.

- If A' coincides with the vertex corresponding to B, then we must have < A'CA = 30°. But $\measuredangle A^{\prime}CA=\frac{2C}{3}$. Thus, $\frac{2C}{3}=30^{\circ}$, so that C = 45°.

- If A' coincides with the vertex corresponding to C, then we similarly get B = 45°.

Hence, we have B = 45° or C = 45°. Similarly, considering the vertex B' of triangle A'B'C', we obtain that C = 45° or A = 45°, and considering the vertex C', we get A = 45° or B = 45°.

Hence, at least two of the angles A, B, C of triangle ABC must equal 45°. Let's WLOG assume that B = 45° and C = 45°. Then, by the sum of angles in triangle ABC, we have A = 180° - B - C = 180° - 45° - 45° = 90°. Thus, the triangle ABC is an isosceles right triangle, and hence, by symmetry, the angles < A'AB and < A'AC must be equal. Since these two angles add up to < A'AB + < A'AC = < CAB = A = 90°, they are therefore both equal to 45°.

Now, we know that the vertex A' of the triangle A'B'C' coincides with one vertex of the inner Napoleon triangle of triangle ABC. Moreover, it must either be the vertex corresponding to B, or the vertex corresponding to C (since in case it is the vertex corresponding to A, we obtained a contradiction). Assume WLOG that A' is the vertex of the inner Napoleon triangle corresponding to B. Then, by the construction of the inner Napoleon triangle, we must have < A'AC = 30°. But this contradicts < A'AC = 45°.

Hence, we obtained a contradiction, and the required triangle ABC cannot exist.

  Darij

Thank you Darij Grinberg for finding time in your busy schedule and responding with

 a nice solution. Unfortunately these triangles do not  coincide.

 Is there a remote chance that one of Napoleon's will match one of the 18(?)  Morley's?

 Is it possible  for one of Brocard's (1st, 2nd, 3rd, ...) to be equilateral  and match

 of Morley's or inward Napoleon's?

  Thank you.

   M.T.

[quote="armpist"] Is there a remote chance that one of Napoleon's will match one of the 18(?)  Morley's?[/quote]

Well, actually I don't feel like checking the (at least, I think there are far more Morleys) 36 possible configurations, but as a matter of principle, one should be able to handle each of these cases in a similar way.

[quote="armpist"] Is it possible  for one of Brocard's (1st, 2nd, 3rd, ...) to be equilateral  and match

 of Morley's or inward Napoleon's?[/quote]

Well,

- the first Brocard triangle is similar to the triangle ABC, and thus it is equilateral only if triangle ABC is equilateral; but if triangle ABC is equilateral, the first Brocard triangle actually degenerates to a point.
- the second Brocard triangle is similar to the pedal triangle of the centroid of the triangle ABC; it is, too, equilateral only in the case when the triangle ABC is equilateral, and in this case it collapses to a point;
- the fourth Brocard triangle is similar to the pedal triangle of the symmedian point of the triangle ABC or, equivalently, to the triangle whose sidelengths are the medians of triangle ABC; it also is equilateral only when the triangle ABC is equilateral, and degenerates to a point in this case.

As for the third Brocard triangle, I don't really know any properties of it remarkable enough to draw a similar conclusion (maybe I knew some two years ago, when I was more into triangle geometry).

Note that in the above, I assumed that your definition of the third and fourth Brocard triangles is the same as on [url=http://perso.wanadoo.fr/bernard.gibert/gloss/brocardtriangles.html]Bernard Gibert's site[/url].

  Darij


[quote="zaya_yc"]$ a_{1},a_{2},...,a_{n} > 0$ and $ a_{1}^{2} \plus{} ... \plus{} a_{n}^{2} \equal{} 1$
find min : $ \frac {a_{1}}{1 \minus{} a_{1}^{4}} \plus{} ... \plus{} \frac {a_{n}}{1 \minus{} a_{n}^{4}}$[/quote]
by Cauchy-Schwarz
\[ \sum \frac {a_i}{1 \minus{} a_i^4}\geq \frac {1}{\sum a_i \minus{} a_i^5}
\]
and if i using EV-method
\[ a_1 \plus{} \cdots \plus{} a_n \equal{}constant
\]

\[ a_1^2 \plus{} \cdots \plus{} a_n^2 \equal{} 1
\]
then the sum $ a_1^5 \plus{} \cdots \plus{} a_n^5$ is minimal when $ 0 < a_1\leq a_2 \equal{} a_3 \equal{} \cdots \equal{} a_n$
I dont know what i must do next. :blush:
maybe Vasile Cirtoaje helped me...

[quote="Sunjee"]by Cauchy-Schwarz

$ \sum \frac {a_i}{1 \minus{} a_i^4}\geq \frac {1}{\sum a_i \minus{} a_i^5}$

[/quote]

Wrong  :(


Find the smallest number which is the sum of 2002 numbers (not necessarily different) with the same sum of digits and also the sum of 2003 numbers with the same sum of digits.

easy.
let x be the desired number:
we claim that x=20007 is the smallest possible such number.
first note that 20007=110(27) + 1(18)+ 1891(9)
and that 20007= 109(27)+2(18)+1892(9)
now let:
x=x1+...+x2002
s(x)=s(yi)=S(xj)
x=y1+...+y2003
clearly the sum of the digits of x is divisible by 9. hence S(x)>=9
and therefore x>=2003*S(x) >=18027
the first possible number x>=18027 which has sum of the digits being 9
is 20007 and we know that it works. It remains to see if there are some x between 18027 and 20007 with S(x)>=18.
but if S(x)>=18 then x>=2003*s(x) =36054
hence x=20007 is the smallest.

I have one question harazi:
is the question saying that all the 2002 numbers are having the same sum as x.
because if not then the above solution is completely useless.

"is the question saying that all the 2002 numbers are having the same sum as x."

I meant sum of digits

Actually, I think the problem is clear: the 2002 numbers need not have the same sum of digits as x.


geometry
parallelogram
geometry unsolved

If X and Y are variable points on sides CA , AB of triangle ABC such that CX/XA+   AB/AY =1 prove that XY passes through a fixed point.

[quote="sophie germain"]If X and Y are variable points on sides CA , AB of triangle ABC such that CX/XA+AB/AY =1. Prove that XY passes through a fixed point.[/quote]
If Y=B, then X=C. If X is symmetric to A wrt C, then Y is symmetric to A wrt B, so YZ and BC don't have common point.
I think you wanted to say that $ \frac{CX}{XA}\plus{}\frac{BY}{YA}\equal{}1$, then XY always passes through the fourth vertex D of the parallelogram ABDC.


I need a proof for a basic property, the fact that you can add a multiple of any scalar of another row to a row in a square matrix, such as in row reduction and the determinant will not change.

To put it formally:
Let A be a square nxn matrix.
Prove that $ det(A)\equal{}det(A_r)$ where $ A_r$ is the matrix A with $ row_k(A)$ replaced by $ row_k(A)\plus{}c*row_j(A)$ with $ c \in \mathbb{R}$.

What kind of proof do you want?  Explicitly computing the determinant of the corresponding elementary matrix works.  You can also expand along row $ k$ and use the fact that replacing row $ k$ with row $ j$ gives a determinant of zero (since the corresponding matrix has linearly dependent rows).

I was looking for the proof that explicitly computes the determinant of the corresponding elementary matrix. Thanks for your help.


Prove for $x,y,z\geqslant 0$,

\[\left(\frac{a}{a+2b}\right)^{2}+\left(\frac{b}{b+2c}\right)^{2}+\left(\frac{c}{c+2a}\right)^{2}\geqslant\frac{1}{3}.\]

[quote="amcavoy"]Prove $\left(\frac{x}{x+2y}\right)^{2}+\left(\frac{y}{y+2z}\right)^{2}+\left(\frac{z}{z+2x}\right)^{2}\geqslant\frac{1}{3}$ for $x,y,z\geqslant 0$.[/quote]
Using the Holder inequality, we have

$\left(\sum_{cyc}x\right)^{3}\sum_{cyc}\left(\frac{x}{x+2y}\right)^{2}=\sum_{cyc}{\frac{x+2y}{3}}\sum_{cyc}{x(x+2y)}\sum_{cyc}\left(\frac{x}{x+2y}\right)^{2}\geq\frac{1}{3}\left(\sum_{cyc}x\right)^{3}$.


limit
real analysis
real analysis unsolved

Let $X$ be a Banach space. By the Cantor theorem we know that if $(A_{n})_{n \in \mathbb{N}}$ is decreasing sequence of nonempty sets such that $\mathrm{diam}\, A_{n}\to 0$ then $\bigcap A_{n}\neq \emptyset$.

Can we omit assumtion $\mathrm{diam}\, A_{n}\to 0$ if we consider decreasing sequence of closed balls?

Yes. Let $B(a_{n},r_{n})$ be a decreasing sequence of closed balls, and let $r=\lim_{n}r_{n}>0$. Since $\parallel a_{n}-a_{m}\parallel \le r_{n}-r_{m}$ for $n\le m$, the sequence $(x_{n})$ is Cauchy, hence has a limit $a$. We have $B(a,r/2)\subset B(a_{n},r_{n})$ for sufficiently large (and therefore for all) $n$.


solve this equation in IR:
$ x\equal{}12\minus{}\sqrt{12\minus{}\sqrt{x}}$

Setting$ t \equal{} \sqrt {x}$
equation$ \Leftrightarrow t^2 \equal{} \minus{} \sqrt { \minus{} t \plus{} 12} \plus{} 12$
Setting$ y \equal{} \sqrt { \minus{} t \plus{} 12}$
$ \Rightarrow$ y^2 = - t + 12
                                   t^2 = - y + 12,OK :)

$ x \equal{} 12 \minus{} \sqrt {12 \minus{} \sqrt {x}}$ is equivalent to
$ x \equal{} 12 \minus{} \sqrt{12\minus{}\sqrt{12\minus{}...}}$.  So we should have
$ x \equal{} 12 \minus{} \sqrt{x}$ or $ \sqrt{x} \equal{} 12\minus{}x$
or $ x \equal{} x^2 \minus{} 24x \plus{} 144$.  The resulting quadratic is easy to solve.

I'd like to see a formal treatment of the fact that $ x \equal{} 12 \minus{} \sqrt {12 \minus{} \sqrt {x}} \iff x \equal{} 12 \minus{} \sqrt{12\minus{}\sqrt{12\minus{}...}}$

hello, obviously is $ x\equal{}9$ the solution.
Sonnhard.


calculus
derivative
function
calculus computations

[hide="My sorution"]

Well if I remember correctly, If $f^{-1}(15)=y$, where $y$ is the number we are looking for, then $f(y)=15$. So this gives us
$y^{2}+2\text{ln}(\frac{4}{y})-e^{y-4}=15$
So you just have to solve for y. Am I supposed to use differentiatio somewhere since this topic is on differentiation or is that not a differentiation function, not an inverse?
[/hide]


geometry
geometric transformation
rotation

A configuration is made up of congruent regular hexagons, where each hexagon shares a side with another hexagon. What is the largest integer k such that the figure cannot have k vertices?

Hah! Geometry problem.

[hide]
There are two seperate cases in which we should consider, both springing from the 3rd hexagon connection. No matter how you connect 2 hexes, there will always be 10 vertices, because any connection is just a rotation of another. Now, on the third connection you can add either 3 or 4 vertices to the figure depending on how you connect it (I still haven't figured out how to post pics). From the figure with 13 vertices, there is no way to make 14 or 15 vertices, but there is a way to make 16 or 17 vertices. From the figure with 14 vertices, there is no way to make 15 vertices, but there is a way to make 16, 17, and 18 vertices. Just from adding 4 vertices to each of the possible vertices after the 3rd connection, all integers can be created. 15. [/hide]


calculus
integration
logarithms
calculus computations

can you show  :|

$ \int_{0}^{1}\int_{0}^{1}\;\;\frac{x-1}{(1\;-\;xy)\cdot(\ln\;x\;+\;\ln\;y)}\;\;\textbf dx\;\;\textbf dy\;\;=\;\;\boxed{\gamma}$

( where $ \gamma$ is euler's constant )

For the inner integral substitute $ x\equal{}\frac{u}{y}$ then $ I\equal{}\int_{0}^{1}\int_{0}^{y}\frac{\frac{u}{y}\minus{}1}{(1\minus{}u)\ln u}\frac{du}{y}dy$

After that reverse the integration order $ I\equal{}\int_{0}^{1}\int_{u}^{1}\frac{\frac{u}{y}\minus{}1}{(1\minus{}u)\ln u}\frac{dy}{y}du$ and calculate the inner integral.

$ I\equal{}\int_{0}^{1}\frac{\ln u\plus{}1\minus{}u}{(1\minus{}u)\ln u}\, du\equal{}\int_{0}^{1}\left(\frac{1}{1\minus{}u}\plus{}\frac{1}{\ln u}\right) du$. The last integral should represent $ \gamma$ .


What are the last two digits of $3^{{3}^{3}...}$ where it is a tower of 2007 threes.  Note $3^{{3}^{3}}=3^{27}$ not $27^{3}$

Yea it does..but I dont think it would matter for me because the number seems a [i]little big[/i] big so if it was 2006 or 2007 3's i still coundlt solve it.

I tried using that 3^20 thing that you said and that 3^20 mod 100 is 01 but the number is too big for me to mod 20..

[hide]
$2007 = 3 mod 4 \Rightarrow$ the units digit is $7$; I do not know if that is how modulus works  :blush: .
From $3^{1}=3$,$3^{2}=9$,$3^{3}=27$,$3^{4}=81$, the unit digits being $3,9,7,1$ that keeps iterating.
Every power is a multiple of three so it is implied that for powers of multiples of $3$ has a units digit of $7$.
For the tens digit, $2007 = 4 mod 10$, making it eight from the sum of the ten digits from the first 3 powers of $3$ times $4$, $2 \cdot 4 = 8$.
Making it $87$.
This solution maybe have just seemed to work without being truely valid, though I get the same answer as above.
[/hide]

[quote="indianamath"]I got 87.  All you need to do is find the exponent mod 20, because the last 2 digits of powers of 3 repeat every 20 powers. becuase 3^3=7(mod 20), the last 2 digits is equal to the last 2 digits of 3^7=[b]87[/b].

I admit, This solution does seem a bit awkward.[/quote]

I think i should've added taht 3^7=7 (mod 20), so it's an ongoing cycle until the very last step.

[quote="AstroPhys"]
$2007 = 3 mod 4 \Rightarrow$ the units digit is $7$; I do not know if that is how modulus works  :blush: .
From $3^{1}=3$,$3^{2}=9$,$3^{3}=27$,$3^{4}=81$, the unit digits being $3,9,7,1$ that keeps iterating.
[b]Every power is a multiple of three so it is implied that for powers of multiples of $3$ has a units digit of $7$.[/b]For the tens digit, $2007 = 4 mod 10$, making it eight from the sum of the ten digits from the first 3 powers of $3$ times $4$, $2 \cdot 4 = 8$.
Making it $87$.
This solution maybe have just seemed to work without being truely valid, though I get the same answer as above.
[/quote]
Sorry, I dont really get what you mean by every power is a multiple of three so it is implied that for powers of $3$ has a units digit of $7$.  For example, 6 is a multiple of 3, but $3^{6}$ does not have a unites digit of 7, neither does $3^{9}$..

And although $3^{27}$ does end in a 7 because 27 is 3 mod 4, but how do you know that for bigger numbers?

Oh yea just as a note, the answer key says it was E) none of the above, and one of the other choices was D..but they never gave a solution so I dont really know..

[quote="AstroPhys"]I meant for every $3^{3}$, since it is a tower of $3^{3^{3^{3^{3}}}}$, that the units digit would be $7$ as is from $3^{3}$.[/quote]
Oh I kind of get what you are saying, but I still dont get why the answer key says 87 was not the answer...
EDIT: whoa Im sorry, the answer key does say D was the answer..I was looking at the answer key for another test..sorry about that..

\phi\phi..(n) goes to one eventually...so the end ceases to matter after a while

like you have

mod

100,
40
16
8
4
2
1

so like [this is a power tower]

p.s. is this like Illinois ictm or like indiana ictm?

3 100
3 40
3 16
3 8
3 4 -> 3
3 2 -> 1
3 1 -> 0

then
3 100
3 40
3 16
27 8 -> 3

then
3 100
3 40
27 16 -> 11

then
3 100
177147 40 -> 27

3^27=7625597484987

87 is our answer

a lot of writing...but this is just routine computation...not so difficult on paper

also we can mod it 4,25 and crt

3 mod 4 [obviously]

then 25
20
8
4
2
1

[also that other theorem gives a faster decrease [i forget how to spell it but it is like charmicael function]]

[quote="1=2"][hide="solution"]We just need to find $3^{3^{3^{\cdots^{3}}}}\pmod 20$

Where there are 2006 3's.

Since 3^3=7mod20,  we can just use the descent relationship to get 3^7. 3^7=2187, so the last two digits are indeed 87.[/hide][/quote]

huh isn't that what i said?

I just added that we could use descent. I don't like reading a lot of solutions before I figure it out myself.

Also, I think that if there are n threes, n>2, the last two digits are always 87...

[quote="Altheman"]\phi\phi..(n) goes to one eventually...so the end ceases to matter after a while

like you have

mod

100,
40
16
8
4
2
1

so like [this is a power tower]

[b]p.s. is this like Illinois ictm or like indiana ictm?[/b]

3 100
3 40
3 16
3 8
3 4 -> 3
3 2 -> 1
3 1 -> 0

then
3 100
3 40
3 16
27 8 -> 3

then
3 100
3 40
27 16 -> 11

then
3 100
177147 40 -> 27

3^27=7625597484987

87 is our answer

a lot of writing...but this is just routine computation...not so difficult on paper

also we can mod it 4,25 and crt

3 mod 4 [obviously]

then 25
20
8
4
2
1

[also that other theorem gives a faster decrease [i forget how to spell it but it is like charmicael function]][/quote]

indiana, this years comprehensive tests was way harder than any of the previous years.


analytic geometry
graphing lines
slope
calculus
derivative
calculus computations

Find the slope of the line that is tangent to the line $ y\equal{}x^2$ at the point $ (5,25)$
Anyways, how would you do this?

[quote="ZhangPeijin"]Find the slope of the line that is tangent to the line $ y \equal{} x^2$ at the point $ (5,25)$
Anyways, how would you do this?[/quote]

don't... post... basic calculus questions... in HSB (well, ok, it's one of the easier ones, and could be solved non-rigorously without calc, but you get the idea)

and make a serious attempt at a question before asking

I did make a serious attempt.
It's just that graphing is one of my weaker points.

[quote="ZhangPeijin"]Find the slope of the line that is tangent to the line $ y \equal{} x^2$ at the point $ (5,25)$
Anyways, how would you do this?[/quote]

[hide="Solution"]
Apply the derivative  :wink: .

$ \frac {d}{dx} x^2 \equal{} 2x$.

Substituting $ 5$ into this, we have $ y \equal{} 10$.

Using the point-slope formula, we can say

$ y \minus{} y_1 \equal{} m (x \minus{} x_1)$

$ y \minus{} 25 \equal{} 10(x \minus{} 5)$

$ y \equal{} 10x \minus{} 25$

Therefore the slope of the line is $ \boxed{10}$.
[/hide]

This really should be in the Calculus forums. You should know better ZhangPeijin  :wink:

or, if you're extremely bored, you can go ahead and show that $ \frac{d}{dx}x^2\equal{}2x$ before actually doing the problem

You could try using the two nice methods shown [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194721]here[/url]


[hide][quote="vanityLAWL"]A large solid cube is assembled by gluing together identical unpainted small cubic blocks. All six faces of the large cube are then painted red. If exactly 27 of the small cubic blocks that make up the large cube have no red paint on them, how many small cubic blocks make up the large cube?[/quote]

We that 3^3 cubic are under  the large cube .
ANd We want know how many cubic are in the face of the large cube .
so we have: 5² * 2=50 (2faces) and the rest is:3*5*2=30 +3²*2=48and

SO THe NUmber Of small cubic blocks we want for make up the large cube its:
3²+50+48=125 :) [/hide]
simple :(

Or, another way to think about it is:

Take the 3x3x3 and add a layer to every side...

$(1+3+1)^{3}= 125$


geometry
incenter
analytic geometry
power of a point
radical axis
geometry proposed

Let $ ABC$ be a triangle and $ (O)$ an arbitrary circle. Let $ AA_1$, $ AA_2$, $ BB_1$, $ BB_2$, $ CC_1$, $ CC_2$ tangent to $ (O)$ in $ A_1$, $ A_2$, $ B_1$, $ B_2$, $ C_1$, $ C_2$, respectively. And denote as $ A_1'$, $ A_2'$, $ B_1'$, $ B_2'$, $ C_1'$, $ C_2'$ their intersections with $ BC$, $ CA$, $ AB$, respectively. Prove that points $ A_1'$, $ B_1'$, $ C_1'$ are collinear if and only if points $ A_2'$, $ B_2'$, $ C_2'$ are collinear.

[b]EDIT.[/b] Thank you [b]Moonmathpi496[/b], and [b]yetti[/b] for pointed out my mistake  :lol: Thank you dear [b]yetti[/b] for your solution.

Hi! Welcome back to Mathlinks :lol:

But, are you sure that your problem statement is correct? :maybe:

Because, if I didn't misunderstand $ A_2,B_2,C_2$ are on $ (O)$. So they can't really be collinear.(if you don't say $ A_2\equiv B_2 \equiv C_2$ collinear)
And order of the points $ A_i,B_i$ should be cleared. And it will be best, if you can attach a diagram :)

[b]April[/b] obviously means that the points $ A_2', B_2', C_2'$ are collinear.

Assume $ A_1', B_1', C_1'$ are collinear. Label the tangents $ t_A \equiv AA_1A_1',$ $ t_B \equiv BB_1B_1', t_C\equiv CC_1C_1'.$ Project the circle $ (O)$ to a circle $ (O^*)$ and the line $ A_1' B_1' C_1'$ to infinity, The projected tangents $ t_A^*, t_B^*, t_C^*$ become parallel to the projected lines $ B^*C^*, C^*A^*, A^*B^*,$ forming a $ \triangle XYZ$ with incircle $ (O^*),$ centrally similar to the the $ \triangle A^*B^*C^*$ with $ A^* \in t_A^* \equiv YZ,$ $ B^* \in t_B^* \equiv ZX,$ $ C^* \in t_C^* \equiv XY.$ This is possible only if the $ \triangle A^*B^*C^*$ is the medial triangle of the $ \triangle XYZ.$ Thus the proposition is reduced to the following theorem:

Let $ (O)$ be the incircle or any excircle of a $ \triangle XYZ$ and $ \triangle ABC$ its medial triangle. Tangents to the incircle $ (O)$ from $ A, B, C$ other than $ YZ, ZX, XY$ cut the midlines $ BC, CA, AB$ at points $ P, Q, R.$ Show that these intersections are collinear.

Assume $ (O)$ is the incircle of the $ \triangle XYZ$ touching $ YZ, ZX, XY$ at $ U, V, W$ and let $ AP, BQ, CR$ touch $ (O)$ at $ U', V', W'.$ Let $ AO, BO, BO$ cut $ BC, CA, AB$ at $ D', E', F'.$ $ \angle PD'A = \angle D'AZ = \angle D'AP,$ the $ \triangle PAD'$ is isosceles with $ PA = PD'.$ Similarly, $ QB = QE', RC = RF'.$ The incenter $ O$ of the $ \triangle XYZ$ is the Nagel point of its medial $ \triangle ABC$ and $ D', E', F'$ are tangency points of its excircles with $ BC, CA, AB.$ Take a circle $ (P)$ with radius $ PA = PD'$ and shrink its radius by $ AU' = AU,$ so that the new radius is $ PU'.$ The shrinked circle $ (P)$ then cuts $ BC$ at a point $ D,$ such that $ D'D = AU = AU',$ therefore, $ D$ is the tangency point of the incircle $ (I)$ of the medial $ \triangle ABC$ with $ BC.$ Similarly, (shinked) circles $ (Q), (R)$ with radii $ QV', RW'$ cut $ CA, AB$ at the tangency points of the incircle $ (I)$ with $ CA, AB.$ Power of $ O$ to the circles $ (P), (Q), (R)$ are equal, $ OU'^2 = OV'^2 = OW'^2,$ and powers of $ I$ to the circles $ (P), (Q), (R)$ are also equal, $ ID^2 = IE^2 = IF^2.$ Therefore, $ OI$ is the common radical axis of the circles $ (P), (Q), (R),$ they are are coaxal, and their centers $ P, Q, R$ collinear.

[hide="Proof for an excircle"] Or else, assume $ (O)$ is the X-excircle of the $ \triangle XYZ$ touching $ YZ, ZX, XY$ at $ U, V, W$ and let $ AP, BQ, CR$ touch $ (O)$ at $ U', V', W'.$ Let $ AO, BO, BO$ cut $ BC, CA, AB$ at $ D', E', F'.$ $ \angle PD'A = \angle D'AZ = \angle D'AP,$ the $ \triangle PAD'$ is isosceles with $ PA = PD'.$ Points $ Q, R$ are not obtained by cyclic exchange from $ P.$ $ \angle QE'B = \angle E'BZ = \angle E'BQ,$ the $ \triangle QBE'$ is isosceles with $ QB = QE'.$ Similarly $ RC = RF'.$ The problem is that now $ O$ is not the Nagel point of the $ \triangle ABC.$

The X-Gergonne line $ XK$ of the $ \triangle XYZ$ cuts the X-excircle $ (O)$ at a point $ K'$ diametrically opposite to the tangency point $ U$ with $ YZ.$ $ OAD' \parallel XKK'$ is a midline of the $ \triangle UKK'$ and the A-Gergonne line of its medial $ \triangle ABC.$ Take a circle $ (P)$ with radius $ PA = PD'$ and expand its radius by $ AU' = AU,$ so that the new radius is $ PU'.$ The expanded circle $ (P)$ then cuts $ BC$ at a point $ D,$ such that $ D'D = AU = AU',$ therefore, $ D$ is the tangency point of the A-excircle $ (I_a)$ of the medial $ \triangle ABC$ with $ BC.$

According to the lemma proved below, $ E', F'$ are tangency points of the C-, B-excircles $ (I_c), (I_b)$ of the $ \triangle ABC$ with $ CA, AB.$  Let $ E, F$ be the tangency point of the A-excircle $ (I_a)$ with $ CA, AB.$ Take a circle $ (Q)$ with radius $ QB = QE',$ intersecting $ CA$ again at $ L.$ Expand its radius by $ LE,$ so that the new radius is $ QE,$ and let the expanded circle cut $ CA$ again at $ L'.$ Then $ BV = E'E = L'L = BV',$ therefore, the expanded circle $ (Q)$ passes through $ V', E.$ Similarly, take a circle $ (R)$ with radius $ RC = RF',$ intersecting $ AB$ again at $ M.$ Expand its radius by $ MF,$ so that the new radius is $ RF,$ and let the expanded circle cut $ AB$ again at $ M'.$ Then $ CW = F'F = M'M = CW',$ therefore, the expanded circle $ (R)$ passes through $ W', F.$

Power of $ O$ to the circles $ (P), (Q), (R)$ are equal, $ OU'^2 = OV'^2 = OW'^2,$ and powers of $ I_a$ to the circles $ (P), (Q), (R)$ are also equal, $ I_aD^2 = I_aE^2 = I_aF^2.$ Therefore, $ OI_a$ is the common radical axis of the circles $ (P), (Q), (R),$ they are are coaxal, and their centers $ P, Q, R$ collinear.
______________________________________

Lemma: Let $ D$ be the tangency point of the incircle $ (I)$ of a $ \triangle ABC$ with $ BC.$ Let $ E, F$ be tangency points of the C-, B-excircles $ (I_c), (I_b)$ with $ CA, AB.$ Let $ \triangle XYZ$ be the anticomplementary triangle of the $ \triangle ABC$ with $ A \in YZ, B\in ZX, \C \in XY$ and $ YZ \parallel BC, ZX \parallel CA, XY \parallel AB$  The lines $ AD, BE, CF$ concur at the X-excenter of the $ \triangle XYX.$

Let $ a = BC, b = CA, c = AB$ be sides lengths and $ s$ semiperimeter of the $ \triangle ABC.$

$ \frac {\overline{DB}}{\overline{DC}} \cdot \frac {\overline{EC}}{\overline{EA}} \cdot \frac {\overline{FA}}{\overline{FB}} = - \frac {s - b}{s - c} \cdot \frac {s}{s - b} \cdot \frac {s - c}{s} = - 1$

By Ceva's theorem, $ AD, BE, CF$ are concurrent at a point with barycentric coordinates $ [ - s, s - c, s - b] \equiv [a + b + c, - a - b + c, - a + b - c],$ which is clearly anticomplement of the A-excenter $ I_a = [ - a, b, c]$ of the $ \triangle ABC,$ therefore the concurrency point is the X-excenter of the anticomplementary $ \triangle XYZ.$
______________________________________[/hide]

The solution works only if the line $ A_1'B_1'C_1'$ does not intersect the circle $ (O).$ I do not know yet how to do it when it does.