geometry
inradius
conics
hyperbola
geometric transformation
reflection
circumcircle

Find D on the side AC of the triangle ABC: If incircle of triangle ABD radius has $ r_1$ and incircle of triangle BCD has radius $ r_2$ then $ r_1 \plus{} r_2$ has maximum value.

Let $ r$ be inradius and $ h_b$ the B-altitude of $ \triangle ABC.$ When $ D \longrightarrow A$ or $ D \longrightarrow C,$ then $ r_1 \plus{} r_2 \longrightarrow r.$ Inradii $ r_1, r_2$ are constrained by $ 2r_1r_2 \equal{} (r_1 \plus{} r_2 \minus{} r)h_b > 0,$ therefore $ r_1 \plus{} r_2 > r.$ Since both the inradii sum and the constraint are symmetrical WRT $ r_1, r_2,$ the sum is maximum for $ r_1 \equal{} r_2.$

[color=red]1st solution:[/color] Let $ (I), (I_1), (I_2)$ be incircles of $ \triangle ABC, \triangle ABD, \triangle CBD.$ Let $ (I)$ touch $ BC, CA, AB$ at $ U, V, W.$ Let $ I_1I_2$ cut $ BD$ at $ S.$ The common internal tangent of $ (I_1), (I_2)$ other than $ BD$ goes throught $ V.$ Quadrilaterals $ ABSV, CBSV$ are tangential $ \Longrightarrow$ $ AB \minus{} AV \equal{} SB \minus{} SV \equal{} CB \minus{} CV$ $ \Longrightarrow$ $ S$ is on hyperbola branch with foci $ B, V$ passing through $ A, C.$ Since the center line $ I_1SI_2$ bisects the angle $ \angle BSV,$ it is tangent to the hyperbola at $ S.$ When $ r_1 \equal{} r_2,$ this hyperbola tangent is parallel to $ AC.$

Let $ a \equal{} BC, b \equal{} CA, c \equal{} AB$ be sides and $ s$ semiperimeter of $ \triangle ABC.$ Midpoint $ M$ of $ BV$ is the hyperbola center and $ AB \minus{} AV \equal{} c \minus{} (s \minus{} a) \equal{} s \minus{} b$ is length of its major axis. Let $ K, L$ be midpoints of $ UV, VW.$ $ \triangle MKL \sim \triangle BUW$ are centrally similar with center $ V$ and coefficient 1/2 $ \Longrightarrow$ $ MK \equal{} ML \equal{} \frac {_1}{^2} BU \equal{} \frac {_1}{^2} BW \equal{} \frac {_1}{^2}(s \minus{} b).$ Circle $ (M)$ with center $ M$ and radius $ MK \equal{} ML$ is the hyperbola pedal circle. Perpendicular $ VI \perp AC$ at $ V$ cuts $ (M)$ at $ N$ closer to $ AC.$ Parallel $ n \parallel AC$ through $ N$ is tangent of the hyperbola branch through $ A, C.$ Let $ B'$ be reflection of the hyperbola focus $ B$ in the hyperbola tangent $ n.$ $ B'V$ cuts the tangent $ n$ at the tangency point $ S.$ $ BS$ cuts $ AC$ at $ D.$

[color=red]2nd solution:[/color] $ BD$ cuts the circumcircle $ (O)$ of $ \triangle ABC$ again at $ Y.$ Let $ s_1, s_2$ be semiperimeters of $ \triangle ABD, \triangle CBD.$ Let $ \varrho_1, \varrho_2$ be inradii and $ \sigma_1, \sigma_2$ semiperimeters of $ \triangle YAD, \triangle YCD.$ From similarity $ \triangle ABD \sim \triangle YCD,$ $ \triangle CBD \sim \triangle YAD,$

$ \frac {r_1}{s_1} \equal{} \frac {\varrho_2}{\sigma_2},\ \frac {r_2}{s_2} \equal{} \frac {\varrho_1}{\sigma_1}\ \Longrightarrow \frac {r_1}{r_2} \cdot \frac {s_2}{s_1} \equal{} \frac {\varrho_2}{\varrho_1} \cdot \frac {\sigma_1}{\sigma_2}.$

On the other hand,

$ \frac {r_1}{r_2} \cdot \frac {s_1}{s_2} \equal{} \frac {|\triangle ABD|}{|\triangle CBD|} \equal{} \frac {AD}{CD} \equal{} \frac {|\triangle YAD|}{|\triangle YCD|} \equal{} \frac {\varrho_1}{\varrho_2} \cdot \frac {\sigma_1}{\sigma_2}$

Multiplying the 2 equations,

$ \left(\frac {\sigma_1}{\sigma_2}\right)^2 \equal{} \left(\frac {r_1}{r_2}\right)^2 \equal{} 1\ \Longrightarrow \sigma_1 \equal{} \sigma_2.$

This means that $ YD$ splits in half perimeter of $ \triangle AYC,$ it is its Y-Nagel line. Parallel to $ AC$ through $ B$ cuts the circumcircle $ (O)$ again at $ Q.$ $ YB, YQ$ are isogonals WRT the angle $ \angle AYC$ $ \Longrightarrow$ $ Q$ is tangency point with $ (O)$ of the Y-mixtilinear incircle of $ \triangle AYC.$ Let $ X$ be midpoint of the circumcircle arc $ AC$ opposite to $ Y,$ $ (X)$ a circle with center $ X$ and radius $ XA \equal{} XC.$ Perpendicular to $ XQ$ at $ Q$ cuts the arc $ AC$ of $ (X)$ inside $ (O)$ at the incenter $ J$ of $ \triangle AYC.$ $ JX$ is internal bisector of the angle $ \angle AYC,$ cutting $ (O)$ again at $ Y.$ $ BY$ cuts $ AC$ at $ D.$

[quote="yetti"]$ 2r_1r_2 \equal{} (r_1 \plus{} r_2 \minus{} r)h_b$[/quote]
Could you explain how you got this?

[quote="math_explorer"][quote="yetti"]$ 2r_1r_2 \equal{} (r_1 \plus{} r_2 \minus{} r)h_b$[/quote]Could you explain how you got this?[/quote] See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=123243]An "Easy" Sangaku[/url], [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=270926]The Nicest Sangaku[/url], etc.

[quote="boroo12"]Find $D$ on the side $AC$ of the triangle $ABC$: If incircle of triangle $ABD$ radius has $r_1$ and incircle of triangle $BCD$ has radius $r_2$ then $r_1^2 + r_2^2$ has maximum value.[/quote]
$a=BC,b=CA,c=AB.$ $\Sigma, p, r$ are area, semiperimeter and inradius of $\triangle ABC$ and $h_b$ its B-altitude. $F(r_1, r_2)=r_1^2+r_2^2$ is subject to constraint $G(r_1,r_2) = r_1+r_2-\frac{2r_1r_2}{h_b} = r = \text{const}.$ $H(r_1,r_2) = F(r_1,r_2)-\lambda G(r_1,r_2).$ $F$ has local extrem for
$\frac{\partial H}{\partial r_1} = 2r_1 - \lambda\left(1 - \frac{2r_2}{h_b}\right) = 0$
$\frac{\partial H}{\partial r_2} = 2r_2 - \lambda\left(1 - \frac{2r_1}{h_b}\right) = 0$
$\Longleftrightarrow$ $\frac{r_1}{r_2}= \frac{h_b-2r_2}{h_b-2r_1}$ $\Longleftrightarrow$ $h_b(r_1-r_2)=2(r_1^2-r_2^2)$ $\Longleftrightarrow$ $r_1=r_2$ or $2(r_1+r_2)=h_b.$ The 2nd local extrem does not exist for very acute $\angle B.$
If $r_1=r, r_2=0,$ then $F_0=F(r,0) =r^2$.
If $2(r_1+r_2)=h_b,$ then from the constraint $G(r_1,r_2)=r$ $\Longrightarrow$ $2r_1r_2 = \frac{h_b^2}{2}-rh_b$ and $F_2 = F(r_1,r_2)=r_1^2+r_2^2=(r_1+r_2)^2-2r_1r_2= rh_b- \frac{h_b^2}{4}.$
$F_0 > F_2$ $\Longleftrightarrow$ $r^2 >  rh_b- \frac{h_b^2}{4}$  $\Longleftrightarrow$ $\left(r-\frac{h_b}{2}\right)^2>0,$ which is always true. Therefore, 2nd extrem is always a mninimum (as long as it exists) and abs. maximum can be only at $r_1=r_2$ or at the boundary $r_1=r, r_2=0$ or $r_1=0, r_2=r$.
If $r_1=r_2$, then from the constraint  $G(r_1,r_1)=r$ $\Longrightarrow$ $2r_1^2 - 2r_1h_b - rh_b = 0$, $r_1= \frac{h_b-\sqrt{h_b^2 -2rh_b}}{2}$ and $F_1=F(r_1, r_1)=2r_1^2= 2r_1h_b-rh_b=h_b^2-h_b\sqrt{h_b^2-2rh_b}-rh_b$.
$F_0 <F_1$ $\Longleftrightarrow$ $r^2 < h_b^2-h_b\sqrt{h_b^2-2rh_b}-rh_b$  $\Longleftrightarrow$ $h_b\sqrt{h_b^2-2rh_b} < h_b^2 -rh_b - r^2$.
LHS of the last inequality is always positive and so is RHS = $h_b^2 -rh_b - r^2 = \frac{h_b^2}{4} + \left(\frac{h_b}{2}- r\right)\left(\frac{3h_b}{2} + r\right) > 0$, the inequality can be squared.
Therefore, $F_0 <F_1$ $\Longleftrightarrow$ $0<r^2(r^2+2rh_b-h_b^2)$   $\Longleftrightarrow$ $0<\frac{\Sigma^2}{p^2}+ \frac{4\Sigma^2}{pb}-\frac{4\Sigma^2}{b^2}$  $\Longleftrightarrow$
$0<b^2+2b(a+b+c)-(a+b+c)^2$  $\Longleftrightarrow$ $(c+a)^2 < 2b^2.$

As a result, if $c+a<b \sqrt 2$, $F(r_1,r_2)=r_1^2+r_2^2$ is max. for $r_1=r_2$, otherwise it is max. for $r_1=r, r_2=0$ or $r_1=0,r_2=r$.


calculus
integration
linear algebra
matrix
group theory
abstract algebra
Ring Theory

Prove that $ \begin{pmatrix}\minus{}8&3\\5&8\end{pmatrix}$ is not the product of $ 2$ integral symmetric matrices.

Sorry, but isn't this straightforward in the $ 2 \times 2$ case?  The symmetric matrices form a subgroup (in fact a subring) here.

[quote="t0rajir0u"]The symmetric matrices form a subgroup (in fact a subring) here.[/quote]
Do they really?    :o

Why is the order of the prime ideal $ P: \equal{} (3,1 \plus{} 2\sqrt {79})$ in the ideal class group of $ \mathcal{O}: \equal{} \mathbb{Z}[\sqrt {79}]$ greater than $ 2$?
I don't know how to show that.  :huh:

EDIT
I found $ P^3 \equal{} (17 \minus{} 2\sqrt {79})$ principal.
So we only need $ P$ not principal.
But how to show that?

Whoops.  I was thinking of the hyperbolic rotations.

If $ C: \equal{}\begin{pmatrix}\minus{}8&3\\5&8\end{pmatrix}\equal{}AB$ with integral symmetric $ A,B$,
then $ \minus{}79\equal{}\det C\equal{}\det A\cdot\det B$ gives $ \det A\equal{}\pm 1$ or $ \det B\equal{}\pm 1$.
But if $ \det B\equal{}\pm 1$, then by Cayley-Hamilton we get also $ C\equal{}79C^{\minus{}1}\equal{}B^{\minus{}1}\cdot 79A^{\minus{}1}$
with $ B^{\minus{}1}, 79A^{\minus{}1}$ also integral symmetric and $ \det B^{\minus{}1}\equal{}\pm 1$.

So we may assume $ \pm 1\equal{}\det A\equal{}ad\minus{}b^2$ for $ A\equal{}\begin{pmatrix}a&b\\b&d\end{pmatrix}$.
Then $ CA\equal{}ABA\equal{}A(AB)^T\equal{}AC^T$ gives $ d\equal{}{5a\plus{}16b\over 3}$
and together this gives $ c^2\minus{}79a^2\equal{}\pm 9$ for $ c: \equal{}3b\minus{}8a$.
But $ c^2\minus{}79a^2\equal{}\minus{} 9$ would give $ c^2\plus{}a^2\equiv 3\pmod{4}$, which is impossible.

So $ 9\equal{}c^2\minus{}79a^2\equal{}N_{\,\mathbb{Q}}^{\mathbb{Q}(\sqrt{79})}(c\plus{}a\sqrt{79})$.

Also $ 1\equal{}80^2\minus{}9^2\cdot 79\equal{}(80\plus{}9\sqrt{79})(80\minus{}9\sqrt{79})$ gives the unit
$ 1<\zeta: \equal{}80\plus{}9\sqrt{79}\in \mathbb{Z}[\sqrt{79}]^\times$ with $ N_{\,\mathbb{Q}}^{\mathbb{Q}(\sqrt{79})}(\zeta)\equal{}1$.

Then clearly for every $ r\in\mathbb{R}_\plus{}$ we find some $ n\in\mathbb{Z}$ with $ \alpha: \equal{}\zeta^n\cdot |c\plus{}a\sqrt{79}|\in [r,\zeta r)$.
Then also $ \alpha\equal{}\pm\zeta^n(c\plus{}a\sqrt{79})\in\mathbb{Z}[\sqrt{79}]$, so $ \alpha\equal{}x\plus{}y\sqrt{79}$ with some $ x,y\in\mathbb{Z}$.
And then also $ x^2\minus{}79y^2\equal{}N_{\,\mathbb{Q}}^{\mathbb{Q}(\sqrt{79})}(\alpha)\equal{}N_{\,\mathbb{Q}}^{\mathbb{Q}(\sqrt{79})}(c\plus{}a\sqrt{79})\equal{}9$
and $ y\equal{}{(x\plus{}y\sqrt{79})\minus{}(x\minus{}y\sqrt{79})\over 2\sqrt{79}}\equal{}{\alpha\minus{}9/ \alpha\over 2\sqrt{79}}\in \left[{r\minus{}9/ r\over 2\sqrt{79}} ,{\zeta r\minus{}9/ \zeta r\over 2\sqrt{79}}\right)$.
And then for $ r\equal{}{3\over \sqrt{\zeta}}$ we get $ y\in \left[\minus{}{3\sqrt{\zeta}\minus{}3/\sqrt{\zeta}\over 2\sqrt{79}} ,{3\sqrt{\zeta}\minus{}3/\sqrt{\zeta}\over 2\sqrt{79}}\right)\subseteq (\minus{}3,3)$.
So $ y\equal{}0,\pm 1,\pm 2$ and $ x\equal{}\pm\sqrt{79y^2\plus{}9}\in\mathbb{Z}$ forces $ y\equal{}0$ and $ \alpha\equal{}3$.

But this gives $ c\plus{}a\sqrt{79}\equal{}\pm 3\zeta^{\minus{}n}\equal{}\pm 3(80\pm 9\sqrt{79})^{|n|}$ and $ 9\nmid c,9|a$.
This contradicts $ 9|9d\equal{}3(5a\plus{}16b)\equal{}143a\plus{}16c$.

Welldone Olorin.
It's a result of Olga Taussky, the "torchbearer for matrix theory". This proves that $ A$ is decomposable in a field and not necessarily in an euclidean ring.
His idea is as follows (a sketch):
$ det(A\minus{}xI)\equal{}x^2\minus{}79$, an irreducible polynomial (over $ \mathbb{Q}$). $ 79$ is squarefree and $ 79\equal{}3mod(4)$.
An eigenvector with coordinates in $ \mathbb{Z}(\sqrt{79})$ is $ [3,\sqrt{79}\plus{}8]^T$. Let $ \mathcal{I}$ be the ideal in $ \mathbb{Z}(\sqrt{79})$ the basis of which is: $ \{3,\sqrt{79}\plus{}8\}$.
FACT: If $ A$ is decomposable then the order of this ideal (in the class-group of the ring of integers) must be $ 1,2$ or $ 4$.
Here (with MAGMA) the number class of $ \mathbb{Q}(\sqrt{79})$ is $ 3$ (or, like Olorin did, show that $ I^3$ is principal). and it remains to show that $ \mathcal{I}$ is not principal (then its order divides $ 3$ and is not $ 1$). With MAGMA (I am lasy) we obtain that its order is $ 3$.      QED

Hi Olorin, you consider a slightly different idal; can you post the details of your first proof ?

Actually you see my first proof, just basic calculations.
Maybe you could explain that theory, why the ideal coming
from an eigenvector must have order $ 1,2,4$.

We must have $ (3)\equal{}P\bar{P}$ with some conjugate prime ideals $ P,\bar{P}$ of the [url=http://en.wikipedia.org/wiki/Dedekind_ring]dedekind ring[/url] $ \mathbb{Z}[\sqrt{79}]$. (if $ 3$ is not prime)
Then $ P,\bar{P}$ are the only proper ideals of $ \mathbb{Z}[\sqrt{79}]$, which contain $ (3)$ properly. (see [url=http://en.wikipedia.org/wiki/Real_quadratic_field]quadratic field[/url])
So for any $ \alpha\in\mathbb{Z}[\sqrt{79}]$ with norm divisible by $ 3$ and not by $ 9$
we have $ (3)\subsetneq (3,\alpha)\subsetneq\mathbb{Z}[\sqrt{79}]$, and therefore $ (3,\alpha)\equal{}P$ or $ \bar{P}$.

In this case we have $ P\equal{}(3,1\plus{}2\sqrt{79})\equal{}(3,\sqrt{79}\plus{}8)$. So it's actually the same ideal.

$ c^2\minus{}79a^2\equal{}\pm 9$ gives $ (c\plus{}a\sqrt{79})(c\minus{}a\sqrt{79})\equal{}(9)\equal{}P^2\bar{P}^2$,
and then $ (c\plus{}a\sqrt{79})\equal{}P^2$ or $ \bar{P}^2$ or $ P\bar{P}\equal{}(3)$.
In the first two cases we get $ P^2$ and $ P$ principal, say $ P\equal{}(\alpha)$ with $ N_{\,\mathbb{Q}}^{\mathbb{Q}(\sqrt{79})}(\alpha)\equal{}\pm 3$,
and in the last case we get $ c\plus{}a\sqrt{79}\equal{}3\alpha$ with some unit $ \alpha\in\mathbb{Z}[\sqrt{79}]^\times$ with $ N_{\,\mathbb{Q}}^{\mathbb{Q}(\sqrt{79})}(\alpha)\equal{}\pm 1$.

So I had to find all $ \alpha$ with $ N_{\,\mathbb{Q}}^{\mathbb{Q}(\sqrt{79})}(\alpha)\equal{}\pm 1,\pm 3$.
Instead I saved time by finding all $ \alpha$ with $ N_{\,\mathbb{Q}}^{\mathbb{Q}(\sqrt{79})}(\alpha)\equal{}9$.
But the method is the same and standard.

Hi Olorin,
"Maybe you could explain that theory, why the ideal coming
from an eigenvector must have order . "
The answer is in
Taussky, Acta arithmetica 24, 151-156, 1973
This paper is in free access here:
http://matwbn.icm.edu.pl/tresc.php?wyd=6&tom=24


algebra
polynomial
number theory proposed
number theory

Consider $ n>1$
$ P(x)\equal{}a_nx^n\plus{}....\plus{}a_1x\plus{}a_0$
$ a_{k}\equal{}a_{n\minus{}k},\forall k\equal{}0,..,n$
Prove that exist infinite $ (x,y)\in Z$ such that
$ y|P(x),y|P(x)$
It is follow while has infinite $ (x,y)$ such that
$ xy|x^p\plus{}y^p\plus{}1$ where $ p>1$

It is an nice problem but no one reply.
Consider :$ y,\frac{P(x)}{y}$


Hi everyone,

Does anyone know where can I get some pointers to coaching my kid for mad hatter type problems, that are quite fast calculations. Is it similar to mathcounts or harder than mathcounts?
I have tried to look on the web for mad hatter and leap frog type of events but there is hardly any info out there.

Any help is much appreciated.

Thanks
Krishnalekha

Could you please explain what you mean by "mad hatter" problem? Do you have any examples?
Thanks


This is a target, so you could just program it, but...

Our cases are:

$ \overline a \plus{} \overline b \plus{} \overline c \plus{} \overline d \equal{} a \plus{} b \plus{} c \plus{} d$: the only possible total is $ 10$. ($ 1$)
$ \overline{ab} \plus{} \overline c \plus{} \overline d \equal{} 10a \plus{} b \plus{} c \plus{} d$: we can see that all that matters is the choice of $ a$, which there are $ 4$ of. ($ 4$)
$ \overline{ab} \plus{} \overline{cd} \equal{} 10(a \plus{} b) \plus{} c \plus{} d$: we are only concerned with the number of non-ordered pairs $ \{a,b\}$, which there are $ \dbinom42 \equal{} 6$ of. However, there are $ 3$ overlaps with the case $ \overline{ab} \plus{} \overline c \plus{} \overline d$. ($ 3$)
$ \overline{abc} \plus{} \overline d \equal{} 100a \plus{} 10b \plus{} c \plus{} d$: all that matters is the choice of $ a$ and $ b$, which there are $ 4\cdot3$ of. ($ 12$)
$ \overline{abcd} \equal{} 1000a \plus{} 100b \plus{} 10c \plus{} d$: all digits matter, so there are $ 4!$ sums. ($ 24$)
Finally, our answer is $ 1 \plus{} 4 \plus{} 3\plus{}12 \plus{} 24  \equal{} \boxed{44}$.

Out of curiosity, how would you program it?

On a TI-84, a really inelegant program might be (that is pretty easy to come up with during a competition)

[code]
:ClrHome
:{0->L_1
:For(A,1,4
:For(B,1,4
:For(C,1,4
:For(D,1,4
:{A,B,C,D->L_2
:SortA(L_2
:If prod(L_2={1,2,3,4
:augment(L_1,{A+B+C+D,10A+B+C+D,100A+10B+C+D,10A+B+10C+D,1000A+100B+10C+D->L_1
:End
:End
:End
:End
:DelVar XSortA(L_1
:For(A,2,dim(L_1
:If L_1(A)≠L_1(A-1
:X+1->X
:End
:Output(1,1,X
[/code]

Can you explain me the three overlaps with earlier part for the 3rd case below?

[quote="math154"]This is a target, so you could just program it, but...

Our cases are:

$ \overline a \plus{} \overline b \plus{} \overline c \plus{} \overline d \equal{} a \plus{} b \plus{} c \plus{} d$: the only possible total is $ 10$. ($ 1$)
$ \overline{ab} \plus{} \overline c \plus{} \overline d \equal{} 10a \plus{} b \plus{} c \plus{} d$: we can see that all that matters is the choice of $ a$, which there are $ 4$ of. ($ 4$)
$ \overline{ab} \plus{} \overline{cd} \equal{} 10(a \plus{} b) \plus{} c \plus{} d$: we are only concerned with the number of non-ordered pairs $ \{a,b\}$, which there are $ \dbinom42 \equal{} 6$ of. However, there are $ 3$ overlaps with the case $ \overline{ab} \plus{} \overline c \plus{} \overline d$. ($ 3$)
$ \overline{abc} \plus{} \overline d \equal{} 100a \plus{} 10b \plus{} c \plus{} d$: all that matters is the choice of $ a$ and $ b$, which there are $ 4\cdot3$ of. ($ 12$)
$ \overline{abcd} \equal{} 1000a \plus{} 100b \plus{} 10c \plus{} d$: all digits matter, so there are $ 4!$ sums. ($ 24$)
Finally, our answer is $ 1 \plus{} 4 \plus{} 3 \plus{} 12 \plus{} 24 \equal{} \boxed{44}$.[/quote]

Sorry, I actually made a small mistake (but the answer came out the same because of another mistake :P ).

In case $ 2$, we get $ 19,28,37,46$.

In case $ 3$, we get $ 37,46,55,55,64,73$, so we see that two repeat with case $ 2$ and that there are two duplicates in case $ 3$. Thus, there are only $ 6\minus{}3\equal{}3$ new ones for case $ 3$.

Basically the motivation to check for duplicates in cases $ 2$ and $ 3$ is that they are both within $ 10\minus{}99$, which makes us suspect that the duplicates exist. The other cases are sufficiently large/small, so we don't need to check them.

[quote=GameBot]Using each of the digits $ \{1, 2, 3, 4\}$ exactly once and using
zero or more $ \plus{}$ signs, how many different totals can be
obtained? No digit can be used as a power, root, etc. Only
addition can be done. When $ n$ digits are written next to each
other, they represent an $ n$-digit number and not the product of
$ n$ factors. Four of the totals to be included are $ \underline{1234}$,
$ \underline{55} \equal{} 12 \plus{} 43$, $ \underline{127} \equal{} 123 \plus{} 4$ and $ \underline{37} \equal{} 31 \plus{} 2 \plus{} 4$.[/quote]

[hide]We will have $5$ cases to solve the problem. Our cases and ways for each cases are as follows:

[b]Case $1$:[/b] Each of the numbers in the sum is one digit. There is just one such way, namely, $1+2+3+4=10$. Note that any orientation of the numbers of the numbers has the same total; for example, $1+2+4+3=10$. Hence, there is [color=#f00]1[/color] totals for this case.

[b]Case $2$:[/b]There is one two digit number and two one digit numbers. For this case, just the tens digits of the two digit number determines the total. After we have chosen the tens digit, the sum of the $3$ digits that remain in the set ${1,2,3,4}$ make the units digit of the total. Hence, there are $4$ possibilities, namely, $19,28,37,$ and $46$. Therefore, there are [color=#f00]4[/color] totals for this case.

[b]Case $3$:[/b]There are two two-digit numbers. The total is determined by the tens digits of the two numbers because when we have the tens digits, the units digit of the total is determined. Hence, for any of the $\binom 4 2=6$ ways to choose an unordered pair of $2$ numbers from the given set, there is one possibility. We notice that we may be overcounting because the numbers in case $2$ are in the set $S\in [10,99]$ and the numbers in case $3$ are in $S$, too. Too try to not overcount, we can actually write the numbers since there is a maxima of $6$ totals that satisfy case $3$. The sets of tens digits that we must consider are ${1,2},{1,3},{1,4},{2,3},{2,4},$ and ${3,4}$. First we notice that $2+3=1+4=5$ in other words, the two totals are equal. Therefore, now we have a count of $5$ totals. The sets of ten digits yield the totals $37,46,55,$ and $64$. We have two overcounting totals, namely, $37$ and $46$. It follows that there are [color=#f00]3[/color] totals that satisfy case $3$ exclusively.

[b]Case $4$:[/b]There is one $3$ digit number and one $1$ digit number. The factors that determine the total are the hundreds place of the $3$-digit number and the tens place of the $3$-digit number; after this, the ones digit of the total is determined. Hence, there are $4\times 3=\color{red}12.$ totals.

[b]Case $5$:[/b]There is one four-digit number. The total is 'uniquely determined' by the order of the numbers in the four-digit number. It follows that there are $4!=\color{red}24$ totals that satisfy case $5$.

Note that we have counted the totals that satisfy the condition once and only once because we did cases by the number of digits in each number. In other words, we started from each number having one digit, and then 'merged' digits together until we had a four digit number. Therefore, the sol is $1+4+3+12+24=\boxed{44}$ ways.[/hide]


Let $ a,b >0$
$ a^3 \plus{}b ^6 \equal{}a^4 \plus{}b^9 \equal{}a^5 \plus{} b^{12}$  Prove that $ a\equal{}b$

__________________________
Azerbaijan Land of Fire :ninja:

we have $ a^3(1 \minus{} a) \equal{} b^6(b^3 \minus{} 1)$ and $ a^4(1 \minus{} a) \equal{} b^9(b^3 \minus{} 1)$. note that $ a \equal{} 1\Leftrightarrow b \equal{} 1$.  so we may assume neither $ a$ nor $ b$ is 1. dividing the two equations, we have $ a \equal{} b^3$. substitution gives $ b^9 \plus{} b^6 \equal{} b^{12} \plus{} b^9 \equal{} b^{15} \plus{} b^{12}$, so $ b^6 \equal{} b^{12}$ and $ b^9 \equal{} b^{15}$, i.e. $ b^6 \equal{} 1$, which means $ b \equal{} 1$. but then $ a \equal{} 1$.


modular arithmetic
induction
number theory
number theory proposed

1) Let $ n \ge 2$ and $ k$ be any positive integers. Prove that $ (n\minus{}1)^ {2} divides (n^{k} \minus{} 1)$ if and only if  $ \frac {k}{(n \minus{} 1)}$

2) Prove that $ (a,a\plus{}2)$ = $ 1$ $ or$ $ 2$ for every integer $ a$

[quote="anurags92"]if and only if  $ \frac {k}{(n \minus{} 1)}$[/quote]
What?

[quote="anurags92"]2) Prove that $ (a,a \plus{} 2)$ = $ 1$ $ or$ $ 2$ for every integer $ a$[/quote]
$ (a,a\plus{}2)\equal{}(2,a)\equal{}1 or 2$ :lol:

1) Note that $ (n \minus{} 1)^2\mid n^k \minus{} 1\Leftrightarrow n \minus{} 1\mid 1 \plus{} n \plus{} n^2 \plus{} \dots \plus{} n^{k \minus{} 1}$. But $ 1 \plus{} n \plus{} \dots \plus{} n^{k \minus{} 1}\equiv 1 \plus{} 1 \plus{} \dots \plus{} 1\pmod {n \minus{} 1}\equiv k\pmod{n \minus{} 1}$. Therefore $ (n \minus{} 1)^2\mid n^k \minus{} 1$ iff $ n \minus{} 1\mid k$.

Alternate solution for 2): Let $ g \equal{} (k,k \plus{} 2)$. Then $ g\mid k, g\mid k \plus{} 2$ hence $ g\mid k \plus{} 2 \minus{} k \equal{} 2$, which implies $ g \equal{} 1$ or $ g \equal{} 2$.

Prove that any positive integer $ a$ can be uniquely expressed in form
$ a \plus{} 3^{m} \plus{} b_{m\minus{}1}3^{m\minus{}1} \plus{} b_{m\minus{}2}3^{m\minus{}2} \plus{}......................\plus{} b_{0}$

where each $ b_{i} \equal{} 0, 1 , or \minus{}1$

this is a form of a number in binary system.
so, u want that it be proved that every number can be expressed in binary system?

well i didnt know about this... u are talkin about the second question haina?
then leave that one it sounds to be useless

[quote="anurags92"]Prove that every positive integer is uniquely expressible in the form
$ 2^{j_{0}} \plus{} 2^{j_{1}} \plus{} 2^{j_{2}} \plus{} 2^{j_{3}} \plus{} ............... \plus{} 2^{j_{m}}$
where $ m\ge0$ and $ 0 \le j_{0} < j_{1} < j_{2} < j_{3} ............. < j_{m}$[/quote]

i am talking of this one.

:oops:
oks anyways i find both useless..... try the problems i postedunder the topis some tough ones..... they are good

[quote="anurags92"]Prove that every positive integer is uniquely expressible in the form
$ 2^{j_{0}} \plus{} 2^{j_{1}} \plus{} 2^{j_{2}} \plus{} 2^{j_{3}} \plus{} ............... \plus{} 2^{j_{m}}$
where $ m\ge0$ and $ 0 \le j_{0} < j_{1} < j_{2} < j_{3} ............. < j_{m}$[/quote]
Induction (if it is true),anurags92
and the second one,it is a basic knowledge about base $ 3$.I think it is popular,you can find in every number theory book

oks thanks allnames....
i found this question in niven - an intro to number theory...
but there are no solutions in that book  :(  :(


Could anybody explain the trigonometric solution more clearly? I didn't understand the Math Jam solution very well, especially the part where an explicit formula was found.

I was not at Math Jam, but the double angle identity for cosine is
$ \cos 2 \theta \equal{} 2 \cos^2 \theta \minus{}1$, which looks a bit like $ a_n \equal{} 2a^2_n \minus{}1$, does it not?

So, if you set $ c \equal{} a_1 \equal{} \cos \theta$, $ a_2 \equal{} \cos 2 \theta$, $ a_3 \equal{} \cos 4 \theta$, $ a_4 \equal{} \cos 8 \theta$, $ \dots$.  The pattern is $ a_n \equal{} \cos (2^{n\minus{}1} \theta)$.

Does that help?

Thanks. Yes, I understand it now.

I never memorized the double-angle formula. If I did, I think I would have thought of this solution. Darn it.
Maybe next time.


I'm curious about what scores people in Ohio got on the AMC. Post  your scores if you want. By the way, I got a 138 on the 10A.

I got a 112.5 on the AMC 12A.  I'm a little miffed about it, because we took it on a snow day, and I think some things like that made me not do as well as I could have, but I'm not gonna rant about that here.  I made AIME at least...

[quote="mwpl11"][quote="mathgeniuse^ln(x)"]144 on both 12A/10B[/quote]

ppl in [i]ohio[/i]. aren't u from michigan?[/quote]

Sorry!!!

I didn't read the ohio part, just post your scores.

Dang, these scores are good compared to those in Michigan.

[quote="ehehheehee"]150 10A
121.5 12B[/quote]

Same with me.  Except, I didn't take the 10A because we had a snow day.

[quote="mathgeniuse^ln(x)"]

Sorry!!!

I didn't read the ohio part, just post your scores.
[/quote]

i didn't mean it that way! it's just that there's already a thread for [i]everyone[/i] to post their scores in the AMC forum. this thread is just meant to give ppl in ohio a sense of where they stand, so ya. oops :blush:

[quote="roadnottaken"]I'm pretty sure I got a 123 on the 12B, although my teacher seems to think it's a 118.5.  Maybe she counted one as blank that was filled in, or maybe I thought I filled one in that I actually left blank.  We had school off due to weather on the day of the AMC A[/quote]

 :oops: Apparently I really did leave it blank, because AMC has my score as a 118.5.

Also, seeing what other Ohio AoPSers got, is not a particularly accurate way to estimate your standings.  You are looking only at the upper crust.

im not from indiana but

Amc 10A 117

AIME: [b]please do not discuss anything about the AIME -- (edited by Administration)[/b]

[quote="cognos599"]im not from indiana but...[/quote]
It's good that you're not from Indiana, because this isn't the Indiana forum :roll: .

I don't think we're supposed to discuss the AIME yet :maybe: , but we can all talk about it tomorrow :jump: !

Yeah.  I'm really curious to see what other people thought of it...

I answered nine of the problems, but I'm almost positive I missed a few :( . The AIME is a difficult test, but I almost certainly did better on it this year than last year (my score last year was a 2 :huh: ).

I just looked at the answers for the AIME, and I got a 2 again! :wallbash_red:  :fool:  :surrender:

Does anyone have any tips on how to not make a fool of myself again next year? (I already have AoPS volume 2, have gone through it, and am working my way back through it while doing the harder problems--also, next year is my last chance to make it to USAMO [unless this year's cut-off is below 140 :noo: ])

Grr I got a 126 on the amc 10a

AIME: I got a miserable 8

AIME

Answered 9
Correct: 8
This is assuming I did not make a bubbling mistake.

I got a 25 on AMC 8.

Never did anything else.

Well, since I moved here a month or so ago, might as well...

I got 109.5 on AMC 10A; I made a lot of silly errors... didn't even make AIME :(
Oh, well, maybe this year...

Edit: No offense (since most of you aren't eligible for AMC 8) but Ohio's AMC 8 scores aren't *that* great. Only 2 perfects, and the highest 6th grade score a 20?

Ooooh. That 6th grader was from our school I think.

I got a 24. I think I missed some easy problem

I got a 23 :(
Missed #s 7 and 13, ju7st 'cause I rushed.


function
real analysis
real analysis unsolved

Having this relation x+y=z+w
I want to prove that x^3 +y^3-z^3-w^3 >0.

Try $ x\equal{}\minus{}1, y\equal{}\minus{}4, z\equal{}\minus{}3, y\equal{}\minus{}2$.

I forgot to say that x. y, z, w are positive:((

Sorry!
So it is working? I tried with Lagrange multipliers but no luck....
Thanks for trying!
If you have any idea please let me know!

mikiduta

Ok I figure out it is not true...
Sorry for the spam:D

Thanks!


My answers to this problem differ from what my teacher gave me...so here it is:

Given: P(1,2,3), Q(-4,1,5) R(2,-2, 1)

write parametric equations for the line through R perpendicular to line PQ

For any given line, there are many different ways to parametrize it.  Are you sure that your parametrization and your teacher's aren't equivalent?  The most likely possibility is that if your parameter is $t$ and your teacher's is $s$, then your two parametrizations are equivalent under a substitution of the form $s = at+b$ for some constants $a$ and $b$.


Twenty-six people, numbered consecutively 1 through 26, are seated equally space around a circular table. Which numbered person is seated directly across from person number 9?

1 sits directly across from 14. Thus, 9 sits directly across from $ (14\plus{}8)\equal{}\boxed{22}$

I don't get it...

I answered 22 because they're equally spaced, so we should add 26/2=13 to any person to find the person across from him...so I got 9+13=22

However, I reviewed it and the answer said 14...

yeah, i got 14 as the correct answer under review too. if you just try to imagine the circle, 9 and 14 aren't even close to being directly across from each other. 22 would make more sense....

I guessed AOPS changed it, now the answer is 22.


geometry
trapezoid
rectangle
MATHCOUNTS
trigonometry
Pythagorean Theorem

next weekend we are going to a wedding shower and my mom was told to design the sticker that they are putting on the hand lotion they will be giving out. she asks her daughter, (moi) to tell her what the angles should be at the corners of the isosceles trapezoid-shaped sticker. The two parallel sides have lengths 1.5 and 1.25 inches and the other 2 sides both are 2.125 inches. What are the angles?

If we drop perpendiculars from the shorter side to longer side, we get a rectangle and two right triangles.

So we want to figure out the angles of the triangles, because from there we can get the angles of the trapezoid. The most practical way to do that would probably be to get a calculator and use trigometry to get the angles. However, that's not Mathcounts level, although basic trig is taught in middle schools.

[hide="If we were to use that solution..."]The base of each of the right triangles is (1.5-1.25)/2=.125, or 1/8. The hypotenuse of the right triangle is 2.125=17/8. By the pythagorean theorem, the height of the triangle, which is the height of the trapezoid, is
$\sqrt{\left(\frac{17}{8}\right)^2-\left(\frac 18\right)^2}=$
$\sqrt{\frac{289}{64}-\frac{1}{64}=}$
$\sqrt{\frac{288}{64}}=$
$\frac{12\sqrt{2}}{64}=$
$\frac{3\sqrt{2}}{16}=$

And what you would actually type into the calculator:

$\text{sin}^{-1}\left(\frac{3\sqrt{2}}{34}\right)$
and
$\text{sin}^{-1}\left(\frac{1}{17}\right)$

The second expression gives the triangle's smallest angle, which we have to add 90 degrees to in order to get the angle in the trapezoid.[/hide]


induction
conics
ellipse
analytic geometry
probability
USAMTS
AMC

I liked this round, for the most part. 1-3 were sort of ugly, but 4 and 5 were really cool.

My solution to number 4 was just epic. And I wrote a program for #3 lol.

Ha, I didn't even finish number 5, yet I still turned it in.  I only got numbers 1-3 fully.

Hmm... For #2 my solution was:

2.

Define a positive integer $ n$ to be “winning” if, with $ n$ stones in the middle, the 1st player (A) has a winning strategy.  Define it as “losing” if the second player (B) has a winning strategy.

Lemma 1:

If $ n$ is a “losing” number, $ n \minus{} 7$, $ n \minus{} 4$, and $ n \minus{} 3$ are all “winning” numbers.

Proof:

Assume that one of these is a “losing” number. Thus, whoever is left with this number will lose. From our number $ n$, we can take $ 3,4,7$, giving the second player a “losing” number, meaning the first person will win. Thus, by definition, $ n$ is a “winning” number.

Lemma 2:

If $ n$ is a “winning” number, at least one of $ n \minus{} 3,n \minus{} 4,n \minus{} 7$ must be a “losing” number.

Proof:

Assume that $ n \minus{} 3$,$ n \minus{} 4$,$ n \minus{} 7$ are all winners. Thus, however many A takes,  B will have a winning number, making $ n$ “losing” by definition.

Lemma 3: 1 and 7 are both winning.

Proof: In both cases, just take them all!

*End Lemmas*

Solution to problem:

Assume that $ n$ is a "`losing"' number. By Lemma 1, this means $ n \minus{} 4$ is a "`winning"' number.

By Lemma 2, one of $ n \minus{} 7,n \minus{} 8,n \minus{} 11$ must be "`losing"'.

By Lemma 1, $ n \minus{} 7$ is winning, so at least one of $ n \minus{} 8,n \minus{} 11$ must be "`losing"'. Assume the

first case. $ n$ is "`losing"', and $ n \minus{} 8$ is "`losing"' for any $ n$, so $ n \minus{} 8,n \minus{} 16,n \minus{} 24,\hdots$

are all "`losing"' numbers. Thus, letting $ n \equal{} 2009$, $ 1$ is a "`losing"' number. But, by Lemma

3, we know that $ 1$ is a "`winning"' number! Thus, one of our assumptions must be false, and

$ n$ is "`winning"' number, or $ n \minus{} 8$ is "`winning"', making...

$ n \minus{} 11$ is a "`losing"' number.

Thus, $ n \minus{} 11k$ for any $ k$ is a "`losing"' number.

Letting $ n \equal{} 2009$ and $ k \equal{} 182$ gives $ 7$ is a "`losing"' number. However, by Lemma 3, this is false.

If $ 2009$ is "`losing"', then our second assumption in both cases must be false, making

$ n \minus{} 8,n \minus{} 11$ "`winning"' However, this is impossible by Lemma 2, as there would be no "`losers"',

and $ n$ would be "`winning"'. Thus, $ 2009$ is "`winning"', and Player A will win.

This is pretty much what they said, is this a 5?

#3 was perfect, #1 I failed, I sorta had the right solution but ended up with $ \frac{11}{567}$.

I didn't do #4,5.

1. Wrong answer.
3. Wrong answer, didn't read problem correctly...

Very quick solution to 5: Consider 1/a in base b. This solution is very quick...

4 was right, 2 was right.

In problem 2, is it ok that I didn't write what the first player should do in the first turn.(Because the problem doesn't ask for it) but I did prove that what's the strategy and who will win.

I'm feeling good about 2-5. Not sure about 1. I though 4 and 5 were pretty easy.

[quote="pythag011"]
Very quick solution to 5: Consider 1/a in base b. This solution is very quick...[/quote]

Thanks, Engel.  :P

[quote="BOGTRO"]#1 I failed, I sorta had the right solution but ended up with $ \frac {11}{567}$.[/quote]A good check was to see if your number was relatively close to $ \frac{1}{11}$.

[quote="236factorial"]My expectations:

1: 5 - I'm pretty optimistic
2: 5 - Nothing much could have gone wrong here, maybe a flaw in my induction
3: 3 - Right answer but convoluted logic, so not sure how much mercy they'll show
4: 2 - Good ideas but not even close
5: 3 - You have to give me credit for my effort...

Expected total - 18[/quote]

You used induction on #2?

*Gulp* I failed this round.

1: Perfect, 5.
2: I'm pretty sure this is perfect, but I overkilled it :P. 5.
3: Argh, I forgot that I could use 1 as a factor (random part of my solution), so I got the wrong answer. Right idea though? 2 or 3.
4: Ha, I forgot that every point inside gives two points. Also messed a bit of other stuff up. Hopefully 2.
5: I randomly used the hermite identity. Hopefully 1 :P
So 5+5+2/3+1/2+0/1, so 13 at worst, 16 at best. Actually, I might get a bit more for #5. Dunno.

Can someone look at my #1 and tell me what I did wrong?  :(

[quote="pythag011"]Induction is pretty much the only way...

btw, bogtro, i believe your solution to 2 is incorrect. You assume that if there exists a winning number n such that n - 8 is losing, then this is true for all n...[/quote]

Sorry?

1. got 11 sets of numbers that work and concluded that if we orient the set with 0 on the evens we have $ 4*4!$ ways and if on the odds then $ 5!*5!$, There are 11 sets so $ 11(4*4! \plus{} 5!5!)/9*9! \equal{} 11/126.$ What did I do wrong?
2. Used induction to show moving at stones congruent to 0,5,6 mod 11 is losing. Thus the first player wins by taking 7.
3. Perfect; maybe a little convoluted logic, used $ gcd(a,b)[a,b] \equal{} ab$ and $ (a,b) \geq (a,b,c)$
4. Perfect solution until the last line where I divided wrong and forgot to square the $ x_2$ in the ellipse
$ \frac {x^2}{2x_2} \plus{} \frac {y^2}{x_2} \equal{} 1.$ So 4 or 5, we will see.
5. Meh some BS, maybe 2.
so $ 3/4 \plus{} 5 \plus{} 5 \plus{} 4/5 \plus{} 1/2 \equal{} 17 \minus{} 21.$
I think my total score is about:
$ 19 \plus{} 20 \plus{} 18 \equal{} 57,$ so 18 for a silver medal. That would be a nice accomplishment for the old trophy case.

wtf, i got #1 even after i ran a program to check it...(dang, i must be both a bad mathematician and a programer...epic liaf)

Perfect (hopefully) on all others.

Masterstroke of fail on #4, but other than that, I'd be willing to bet on perfect.

is this a valid proof for #3? :
$ (m,n)[m,n] \equal{} mn \implies abc \equal{} (a,b)(a,c)(b,c)(a,b,c).$ Let $ (a,b,c) \equal{} p$ then because $ (a,b) \geq (a,b,c)$ we have that the minimum value of $ abc \equal{} p^4$. Also because $ (a,b),(a,c),(b,c) \equal{} p$ the minimum values of a,b,c are $ 2p,3p,5p$ yielding $ a \plus{} b \plus{} c \equal{} 300$. We are assured this is the minimum because if a,b,c are greater than $ 2p,3p,5p$ respectively then$ p > 30$ and $ a \plus{} b \plus{} c > 300.$

[quote="dnkywin"]Prediction: 5 5 4/5 5 5, so 24-25.
My proof to #3 was awkward...

[quote="pythag011"]
Very quick solution to 5: Consider 1/a in base b. This solution is very quick...[/quote]
Doesn't work if 1/a is a terminating fraction in base b  :wink: .[/quote]

Then $ \frac {b^k}{a}$ is an integer for infinitely many $ k$ and $ \left \lfloor \frac {b^k}{a}\right \rfloor a$ wins instantly.

See my solution. I think it's pretty good.

[quote="Duelist"]Another mistake on my part...I thought I got a perfect score until I saw the official solution to #4, and then I when I went back to my paper I realized that I magically changed a $ 2a^2$ to $ a^2$. So a coefficient of the ellipse should be 2 instead of 1, and I really hope this is only a one point deduction.

I also got that the endpoints did not work, which although I admit it seemed a little tricky, if I caught it, it probably wasn't. But maybe I was compensating for my stupid mistake earlier, so...you never know.

Also, I love my solution for #5 (I think everyone likes theirs; it can be solved in many creative ways). Very hard problem, though. Got it after I let $ b \equal{} 2$. My solution merely generalizes the proof for the $ b \equal{} 2$ case.

[hide="Solution to 5"]
We will prove the desired result as follows. Let

$ A \equal{}$ ``There is an infinite amount of powers of $ b$ of the form $ \lfloor ka\rfloor$, where $ k\in Z$.''

$ \neg A \equal{}$ ``There is a finite amount of powers of $ b$ of the form $ \lfloor ka \rfloor$, where $ k \in Z$.''

$ B \equal{}$ ``$ \exists$ an $ m\in Z$ such that $ \forall \beta \geq m, \, 2^\beta, 2^{\beta \plus{} 1}, ...$ are not of the form $ \lfloor ka \rfloor$.''

Clearly, $ \neg A \implies B$. We will prove $ \neg B$, proving by modus tolens $ \neg (\neg A)$, or $ A$, completing the proof.

Suppose statement $ B$ is true. We can immediately say $ 1 < a < \frac {b}{b \minus{} 1}$, since $ \frac1a \plus{} \frac1b > 1 \implies a < \frac {b}{b \minus{} 1}$, and $ a\leq1 \implies \{\lfloor a \rfloor, \lfloor 2a \rfloor, ...\}$ contains the set of natural numbers, so we would be done.

Now we take a $ \beta \geq m$. Since $ b^\beta$ is not of the form $ \lfloor ka\rfloor$ for some $ k$, we can say $ b^\beta \minus{} 1$ is, since $ 1 < a < \frac {b}{b \minus{} 1} \leq 2$. So let

$ b^\beta \equal{} \lfloor ja\rfloor \plus{} 1$
$ b^{\beta \plus{} 1} \equal{} \lfloor ka\rfloor \plus{} 1$

Now we write the following identity, substitute, and play with the equation a bit. I ask the reader to remember that for $ x\in \Re$, $ \lfloor x \rfloor \equal{} x \minus{} \{x\}$, where $ \{x\}$ will denote the fractional part of $ x$.
\[ b^{\beta \plus{} 1} \equal{} b(b^\beta)
\]

\[ \lfloor ia \rfloor \plus{} 1 \equal{} b(\lfloor ja \rfloor \plus{} 1)
\]

\[ ia \minus{} \{ia\} \plus{} 1 \equal{} b(ja \minus{} \{ja\} \plus{} 1)
\]

\[ ia \minus{} bja \equal{} \{ia\} \minus{} b\{ja\} \plus{} b \minus{} 1
\]

\[ (i \minus{} bj)a \equal{} \{ia\} \plus{} (b \minus{} 1) \minus{} b\{ja\}\quad\quad\quad\quad\quad (1)
\]
Now we will analyze $ \{ja\}$ and $ \{ia\}$. Suppose these were less than or equal to $ \frac {b \minus{} 2}{b \minus{} 1}$. Then $ (j \plus{} 1)a$ and $ (i \plus{} 1)a$ would be greater than $ 2^\beta$ and $ 2^{\beta \plus{} 1}$, respectively, since $ a > 1$, but less than $ 2^\beta \plus{} 1$ and $ 2^{\beta \plus{} 1} \plus{} 1$, since $ a < \frac {b}{b \minus{} 1}$, so $ \frac {b \minus{} 2}{b \minus{} 1} \plus{} a < \frac {b \minus{} 2}{b \minus{} 1} \plus{} \frac {b}{b \minus{} 1} \equal{} 2$. This reaches contradiction, since $ \lfloor (j \plus{} 1)a \rfloor$ and $ \lfloor (i \plus{} 1)a \rfloor$ are $ 2^\beta$ and $ 2^{\beta \plus{} 1}$. So $ \{ja\} > \frac {b \minus{} 2}{b \minus{} 1}$ and $ \{ia\} > \frac {b \minus{} 2}{b \minus{} 1}$. It follows that either:

$ \frac {b \minus{} 2}{b \minus{} 1} < \{ja\} \leq \frac {b \minus{} 1}b \text{ or } \frac {b \minus{} 1}b < \{ja\} < 1$

So

$ \{ia\} \plus{} b \minus{} 1 \minus{} b\{ja\} < \{ia\} \text{ for } \frac {b \minus{} 1}b < \{ja\} < 1$ and
$ \{ia\} \plus{} b \minus{} 1 \minus{} b\{ja\} < \{ia\} \plus{} \frac {(b \minus{} 1)^2 \minus{} b(b \minus{} 2)}{b \minus{} 1} \equal{} \{ia\} \plus{} \frac1{b \minus{} 1} \text{ for } \frac {b \minus{} 2}{b \minus{} 1} < \{ja\} \leq \frac {b \minus{} 1}b$

$ (i \minus{} bj)a < \{ia\} \implies i \minus{} bj \equal{} 0$
$ (i \minus{} bj)a < \{ia\} \plus{} \frac1{b \minus{} 1} \implies i \minus{} bj \equal{} 1 \text{ or } 0$,

since $ a > 1$. So we now have only two cases to consider:

Case I: $ i \equal{} 2j \plus{} 1$

Using (1), we get

$ a \plus{} b\{ja\} \equal{} \{ia\} \plus{} b \minus{} 1$
$ 1 \plus{} \{a\} \plus{} b\{ja\} \equal{} \{ia\} \plus{} b \minus{} 1$
$ \{a\} \plus{} \{ja\} \plus{} (b \minus{} 1)\{ja\} \equal{} \{ia\} \plus{} b \minus{} 2$

Now we take a moment to prove $ \{a\} \plus{} \{ja\} \geq 1$:

$ 2^\beta \minus{} 1 \equal{} \lfloor ja \rfloor$
$ 2^\beta \plus{} 1 \equal{} \lfloor (j \plus{} 1)a \rfloor$
$ (2^\beta \plus{} 1) \minus{} (2^\beta \minus{} 1) \equal{} 2 \equal{} (j \plus{} 1)a \minus{} ja \minus{} \{(j \plus{} 1)a\} \plus{} \{ja\}$
$ 2 \equal{} a \plus{} \{ja\} \minus{} \{(j \plus{} 1)a\}$
$ 1 \plus{} \{(j \plus{} 1)a\} \equal{} \{a\} \plus{} \{ja\}$
$ 1 \leq \{a\} \plus{} \{ja\}$

Now we continue where we left off, appplying this inequality.

$ \{a\} \plus{} \{ja\} \plus{} (b \minus{} 1) \{ja\} \geq 1 \plus{} (b \minus{} 1) \{ja\}$
$ 1 \plus{} (b \minus{} 1)\{ja\} \leq \{ia\} \plus{} b \minus{} 2$
$ (b \minus{} 1) \{ja\} \leq \{ia\} \plus{} b \minus{} 3 < 1 \plus{} b \minus{} 3 \equal{} b \minus{} 2$
$ \{ja\} < \frac {b \minus{} 2}{b \minus{} 1}$,

which is a contradiction to $ \{ja\} > \frac {b \minus{} 2}{b \minus{} 1}$. So $ i \equal{} 2j \plus{} 1$ is impossible.

Case II: $ i \equal{} bj$

We have

$ b^\beta \equal{} \lfloor ja\rfloor \plus{} 1$
$ b^{\beta \plus{} 1} \equal{} \lfloor ia\rfloor \plus{} 1$

Plugging $ i \equal{} bj$ into (1) gives:

$ 0 \equal{} \{ia\} \plus{} b \minus{} 1 \minus{} b\{ja\}$
$ b\{ja\} \equal{} \{ia\} \plus{} b \minus{} 1$
$ \{ia\} \equal{} b\{ja\} \minus{} (b \minus{} 1) \quad\quad\quad (2)$

So, since Case II is the only possiblity,

$ b^\beta \equal{} \lfloor ja\rfloor \plus{} 1$
$ b^{\beta \plus{} 1} \equal{} \lfloor 2ja\rfloor \plus{} 1$
$ b^{\beta \plus{} 2} \equal{} \lfloor 4ja\rfloor \plus{} 1$
.
.
.

It follows that $ \forall \alpha$ in the nonnegative integers,

$ b^{\beta \plus{} \alpha} \equal{} \lfloor 2^\alpha ja \rfloor \plus{} 1$

Now we create a sequence $ u_0, u_1, ...,u_\alpha,...$ of the fractional parts of $ 2^\alpha ja$. Since they must satisfy (2), we have:

$ u_0 \equal{} \{ja\}$
$ u_1 \equal{} b\{ja\} \minus{} b \plus{} 1$
$ u_2 \equal{} b^2\{ja\} \minus{} b^2 \plus{} 1$
.
.
.
$ u_\alpha \equal{} b^\alpha\{ja\} \minus{} b^\alpha \plus{} 1 \equal{} b^\alpha(\{ja\} \minus{} 1) \plus{} 1$

Since $ \{ja\} \minus{} 1$ is a fixed negative number, and $ b^\alpha$ can become arbitrarily large, we can certainly find an $ \alpha$ such that $ b^\alpha(\{ja\} \minus{} 1)$ is as small as we desire, let's say less than -1. But that means $ u_\alpha < 0$, but fractional parts of real numbers must be in $ [0,1)$. We have reached a contradiction, so all cases have been ruled out, and $ \neg B$ is proven.

By the argument at the beginning of the proof, we have shown $ A$ is true, as desired.
[/hide]

Edit: Just saw your question (and solution) mathemonster. Okay, I just about to say I didn't get it/didn't think it was right, but now I see what you mean and agree. I think it works, and it's a nice, short solution, unlike mine and the official one which are frightingly identical. Of course, I think you have to give a specific values of $ a,b,c$ (or $ p$) to show that equality is valid, but I'm assuming you put that in there and it's not that big of a deal.[/quote]
yes I did  :P , I also stated that if $ (a,b) > p \equal{} kp$ we have a higher value of p because 2 factors of $ kp$ go onto the left side, while 1 goes to the right side. Hopefully, this succint proof will get me a 5.

I think i got:
#1-perfect:5
#2-no time
#3-almost perfect. right answer, half-rigorous proof. 3-5?
#4-almost perfect (though i can't see anywhere where the graders could take off points, i'm sure they can find a way to give me a 4) 4-5
#5-no time

So total=12-15. Epic fail. BIGTIME. (just like round 2, which i epic failed BIGTIMER.)

On number 4, I was not sure if a "point at infinity" would be a valid point in a plane (since then the circle's center could be a point at infinity and the endpoints of the ellipse would be included - i.e. the center of a circumcircle of a degenerate triangle which is actually a line is a point at infinity...maybe...I was not sure). Luckily, I decided to guess that the point at infinity would not be a possible point in the plane.

On number 2, I forgot to check to see what the question asked for, and did not say which person won; all I did was explain how to win with any given number of stones (I also forgot to do any work related to the number 2009). How much do you think this would count off in USAMO? (USAMTS grading may still give me a 5 hopefully, but most likely I get a 4).

In USAMO I would say you'd get a 4, since you didn't exactly answer the question.

1. Same idea as official, only apparently 45-24=11, so hoping for a 4.

2. Same as official.

3. Probably on the long side, but seems to be correct.

4. Really short solution. Used only areas and law of cosines.

(Originally found a dissatisfying solution using coordinates.)

5. Quite similar to official solution, but used an interval argument.

Hopefully a 24, which with a hopeful 22 on Round 2 still could give me a chance for 96.

[quote="Yongyi781"][quote="hurdler"]On number 4, I was not sure if a "point at infinity" would be a valid point in a plane (since then the circle's center could be a point at infinity and the endpoints of the ellipse would be included - i.e. the center of a circumcircle of a degenerate triangle which is actually a line is a point at infinity...maybe...I was not sure). Luckily, I decided to guess that the point at infinity would not be a possible point in the plane.
[/quote]

If you did the problem correctly, there shouldn't be any "points at infinity" in your answer... only the points at the ends of the major axis of the ellipse, which would cause the circle to be degenerate, and thus those points are not included.

Is that what you meant?[/quote]

what I meant was that as we consider the points A  in the locus of the ellipse, as the points approach the ends of the major axis, the center of the circumcircle of the cyclic quad moves further and further away from the midpoint of BC (on the perpendicular bisector of BC), so that I thought when we finally consider the 2 points A at the endpoints of the major axis, the center should be a point at infinity on the perpendicular bisector of BC, or maybe I don't know a point at infinity really is.....which is probably why I got confused.

Also, on this problem, I noticed that the official solution did not show that all points in the locus worked (I thought locus problems were two part problems - show that the only points possible are the ones you claim them to be, and show that all points that meet the criteria you claim work,...or at least you should mention something about the second part of the proof in your solution if it clearly easily follows some fact you proved earlier). I spend about a page and a half each on both parts, so I was wondering if analytic geometry covers both parts of the proof somehow. For instance, I guess it sort of makes sense intuitively that using coordinate geo prevents the possibility that only part of the ellipse is in the locus

1~4. Very similar to official solutions.
5. I will probably get a 1...


limit
geometry
3D geometry
logarithms
real analysis
real analysis unsolved

Solve differential equation $ y'\equal{}x\plus{}\frac{x^3}{y}.$

Well, of course the solutions $ y \equal{} x^2$ and $ y \equal{} \minus{} \frac {x^2}{2}$ jump out at you, but there are others.  In fact, even if you require the initial condition $ \lim_{x\to 0}y(x) \equal{} 0$, there are still infinitely many solutions!
[hide="More generally"]
\[ y \equal{} \frac {1}{2}\left ( x^2 \plus{} \frac {x^4}{u(x)} \plus{} u(x)\right )
\]
where $ u(x) \equal{} \left ( c \minus{} x^6 \pm \sqrt {c^2 \minus{} 2cx^6}\right) ^{\frac {1}{3}}$, and $ c$ is an arbitrary constant (note the cube root is also branched, so this represents potentially up to six distinct families of solutions).

$ c \equal{} 0$ with the real cube root corresponds to $ y \equal{} \minus{} \frac {x^2}{2}$, while $ c \equal{} 0$ with either of the complex cube roots gives the solution $ y \equal{} x^2$. Note also that if we take the negative square root and any $ c>0$, it's easy to verify that $ u(x) \equal{} \minus{}2x^2\plus{}\mathcal{O}\left( x^4 \right)$ in a neighborhood of $ 0$ and therefore $ \lim_{x\to 0}y(x) \equal{} 0$, so there are infinitely many solutions to that initial value problem.[/hide]

Let $ y\equal{}x^2 u$, where $ u\equal{}u(x).$ Now
$ 2xu\plus{}x^2 u'\equal{}x\plus{}\frac{x^3}{x^2 u}$ or $ 2u\plus{}x\frac{du}{dx}\equal{}1\plus{}\frac{1}{u}.$
$ \frac{udu}{u\plus{}1\minus{}2u^2}\equal{}\frac{dx}{x}$
$ \minus{}\frac{1}{3}\left(\ln |u\minus{}1|\plus{}\frac{1}{2}\ln |2u\plus{}1|\right)\equal{}\ln\frac{|x|}{C}$
Final answer:
$ (y\minus{}x^2)^2 (2y\plus{}x^2)\equal{}C.$


Consider oriented,strongly connected clique $ G$,i.e any two vertexes $ A,B$ are connected with oriented edge and if $ B\rightarrow A$,then there exist $ C_1,C_2,\dots,C_k\in G$,such that $ A\rightarrow C_1\rightarrow C_2\dots\rightarrow C_k\rightarrow B$.
Prove that $ G$ contains cycles of length $ 3,4,\dots,n$,where $ n$ is the number of vertexes in $ G$.

[hide="hint:"]
Prove,at first, that there exist a cycle of a length $ 3$ and after that, use induction.
[/hide]

I think that either nobody interested or just everyone who solved it was just too lazy to post a solution,as the problem is simple, even without hint... :maybe: anyway it is unsolved here for a month,so I think it is time to post a solution,also I think I won't be here until IMO.
[b]Solution:[/b]
As I said,at first we will prove that there is a cycle of a length $ 3$.
Suppose to the contrary that there is no cycle of a length $ 3$.
Let $ A,B$ be two arbitrary vertices of $ G$ and assume that $ B\rightarrow A$,therefore,there exist $ C_1,C_2,\dots,C_k$,such that
$ A\rightarrow C_1\rightarrow C_2\dots\rightarrow C_k\rightarrow B$.
It is easy to understand that $ B\rightarrow C_1$,because otherwise there is a cycle $ A\rightarrow C_1\rightarrow B\rightarrow A$.
The same reason gives us that $ B\rightarrow C_i$,for all $ 1\leq i\leq k \minus{} 1$.
But then $ C_{k \minus{} 1}\rightarrow C_k\rightarrow B\rightarrow C_{k \minus{} 1}$,contradiction.
Now we will prove that if there exist a cycle of a length $ k < n$,then there is a cycle of a length $ k \plus{} 1$,too.
Let $ A_1\rightarrow A_2\rightarrow \dots\rightarrow A_k\rightarrow A_1$ be our cycle and suppose to the contrary that there is no cycles of a length $ k \plus{} 1$.
Since $ k < n$,it follows that there is at least one vertex not in the cycle,denote this vertex as $ B$.
Let $ B\rightarrow A_k$.Then $ B\rightarrow A_{k \minus{} 1}$,because otherwise we would obtain a new cycle of a length $ k \plus{} 1$:
$ B\rightarrow A_k\rightarrow A_1\dots\rightarrow A_{k \minus{} 1}\rightarrow B$.
Similarly we conclude that $ B\rightarrow A_i$ for all $ 1\leq i\leq k$.
We will call such vertices [i]"good"[/i]
There must exist a path between any vertex of our cycle to $ B$,therefore,there must be at least one vertex $ C$,such that $ A_i\rightarrow C$.As we did it before we conclude that $ A_i\rightarrow C$ for all $ 1\leq i\leq k$, and we will call such vertixes [i]"bad"[/i]
If there exist at least one[i] "bad"[/i] vertex then there exist at least one[i] "good" [/i]vertex.
Therefore,we can divide all vertices ,except those that in cycle, into two groups $ B \equal{} B_1,B_2,\dots,B_m$ and $ C \equal{} C_1,C_2,\dots,C_l$,of [i]"bad"[/i] and [i]"good"[/i] vertices.
If there exist at least one vertex $ C_i\in C$,such that $ C_i\rightarrow B_j$ for some $ 1\leq j\leq m$,then we can find a cycle of a length $ k \plus{} 1$,contradiction.
Therefore,for all $ 1\leq i\leq m$ and $ 1\leq j\leq l$,all edges directed as $ B_i\rightarrow C_j$,but then there is no path between any [i]"good" [/i]and [i]"bad" [/i]vertices.Contradiction.
$ \blacksquare$.


limit
function
calculus
calculus computations

Obviously does not exist. The function behaves completely differently on e.g. $t \rightarrow (t,t^{3})$ and $t \rightarrow (t,t)$


geometry
3D geometry
calculus
calculus computations

I get the shape of cone on this problem but I'm not so confident. Can anyone help me?

A lighthouse is on a small island 3 km away from the nearest point P on a straight shoreline and its light makes four revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from P?

[quote="quantum leap"][hide=" hint:"]
$\frac{d\theta}{dt}=4$rev/min$\implies(4\cdot2\pi)rad/60sec=\frac{2\pi}{15}$rad/sec
$tan\theta=\frac{1}{3}$[/hide][/quote]

I don't really understand what you're trying to say.  :(


What is the 2007th digit after the decimal point of 1/2007?

[quote="13375P34K43V312"]The following will be helpful: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145946[/url][/quote]

I dont get how that helps?

Once you know the digits repeat with period $ 222$ (which is the hardest part of the problem), then the $ 2007$th digit must be the same as the $ 9$th digit ($ 2007=9+9\times 222$), which I bet you can compute with your long division skills.


2 circles intersect at 2 points A & B .ST is a line which passes through B.PQ is a line that passes through B .P & S are located in the circumference of circle 2nd.Prove that SP=BQ.

Dear Mathematicswizard,
can you precise the position of your points on the circles for a more comprehension of your figure. Thanks in advance.
Sincerely
Jean-Louis


geometry
circumcircle
geometric transformation
reflection
trigonometry
geometry proposed

Let $ \Delta ABC$ be a triangle with its Fermat point $ F$ and the circumcenter $ O$.
Let $ A'$ be the intersection of $ AF,BC$.define similarly $ B',C'$.
Let $ A"$ be a point which forms an equilateral triangle with the side $ BC$ and $ A,F,A"$ are collinear.
define similarly $ B",C"$.
Let $ O_1$ be the cicumcenter of $ \Delta BA"C$.Define similarly $ O_2,O_3$ for triangles $ AB"C,AC"B$,respectively.
Define $ F_1$ such that $ F_1B\equal{}FB,F_1C\equal{}FC$ and $ F_1 \neq F$.define similarly $ F_2,F_3$ for triangles $ AB"C,AC"B$,respectively.
Prove:
the concurrency of $ F_1A',F_2B',F_3C'$.(tot,spring 2007,A-Level,senior,prob.8)

their intersection will be called $ F'$.
i have some other questions.
do we have the concurrency of $ AF_1,BF_2,CF_3$?
Do we have the concurrency of $ O_1A',O_2B',O_3C'$?if yes , let's call their intersection $ S$.
finally, do we have the collinearity of $ F',S,O$?
thx for your help.

Let $ R$ be the intersection of $ R$ and $ FF', A'',B'',C''$ be the reflections of $ F'$ across $ BC, CA, AB$
From http://www.artofproblemsolving.com/viewtopic.php?t=304719 , you can see the nice proof of [b]Luis[/b], it told that $ F$ is the circumcenter of equilateral triangle $ A''B''C''$. Denote $ R$ the radius of $ (A''B''C'')$
Applying Menelaus's theorem for triangle $ FF'A'$ we have:
$ \frac {RF}{RF'} \equal{} \frac {F_1A'}{F_1F'}.\frac {AF}{AA'} \equal{} \frac {FA'}{FA''}.\frac {AF}{AA'} \equal{} \frac {FA'}{AA'}.\frac {AF}{R}$
$ \equal{} \frac {AF.S_{BFC}}{S_{ABC.}R} \equal{} \frac {AF.BF.CF.\sin 120^o}{S_{ABC}.2R}$
Similarly we get $ AF_1, BF_2, CF_3$ concur at $ R$ which lies on $ FF'$.

+The concurrency of $ O_1A', O_1B', O_1C'$:
Let $ S$ be the intersection of $ O_aA'$ and $ OF'$.
We have $ \frac {SF'}{SO} \equal{} \frac {F'J}{OO_a}$
See the names of points on the figure. We see that $ F'Q \equal{} A_2Q$
$ \frac {QA_2}{QJ} \equal{} \frac {RA_1}{O_aR} \equal{} 3$
Therefore $ F'J \equal{} \frac {2}{3}F'A_2$.
Denote $ A_2B_2 \equal{} B_2C_2 \equal{} C_2A_2 \equal{} x,O_aO_b \equal{} O_bO_c \equal{} O_cO_a \equal{} y$.
Since $ O_aO_b//A_2B_2, O_bO_c//B_2C_2, O_cO_a//A_2C_2$ we get $ \frac {F'A_2}{OO_a} \equal{} \frac {A_2B_2}{O_aO_b} \equal{} \frac {x}{y}$
So $ \frac {F'J}{OO_a} \equal{} \frac {2x}{3y}$. Similarly we are done.

[b]Remark:[/b] Denote $ K$ the intersection of the lines through $ O_a, O_b, O_c$ and perpendicular to $ B'C', C'A', A'B'$, respectively. Prove that $ S,F,K$ are collinear and $ FF'//OK$

[quote="mathson"]
which software do you use to make these diagrams?[/quote]
It's [b]Geometer's Sketchpad[/b]


calculus
integration
inequalities
real analysis
real analysis theorems

can anyone tell me where can i find integral inequalities on the net, or recommend me some books (romanian)?
4 ex. holder. but others, too not only those "famous"
10x a lot.

well Cauchy is one of the most used integral inequalities. It is proved right here on the forum.

About other inequalities I will post them when I have time, or maybe someone else you care to do it :).

well here you have a beautiful link:
http://mathworld.wolfram.com/topics/Inequalities.html


algebra
polynomial
calculus
integration
functional equation
algebra proposed

Find all polynomials $P(x)$ with integral coefficients such that $P(P'(x))=P'(P(x))$ for all real numbers x.

Clearly $P(x) = 0$ is a solution and no other constant polynomial is a solution.
Now assume that $P$ is a solution and that $P$ has degree $n \geq 1$ and leading coefficient $a_n$. Thus both $P(P'(x))$ and $P'(P(x))$ have degree $n(n-1)$. Identifying the leading coefficient, we obtain $a_n \cdot (na_n)^n = na_n \cdot (a_n)^{n-1}$ so that $n^{n-1}a_n = 1$.
Since $a_n$ is an integer, it follows that $n=a_n=1$ and $P(x)=x+b$.
Back to the functional equation, we deduce that $b=0$.

Conversely, it is easy to verify that $P(x)=x$ is indeed a solution.

Pierre.

These arguments don't work in case $n=1$, but work for all $n>1$.
I would like to see a solution for non-integral coefficients.

[quote="peregrinefalcon88"]why does P(x) = -x-2 work? I cant find a flaw in your argument but it clearly works.[/quote]
The flaw is quite subtle, indeed: [quote="pbornsztein"]$a_n \cdot (na_n)^n = na_n \cdot (a_n)^{n-1}$[/quote]
Consider a polynomial $Q(x)=q_nx^n + q_{n-1}x^{n-1} + \ldots + x_0x^0$. If we compose it with a polynomial $R(x)$, we get $Q(x)=q_nR^n(x) + q_{n-1}R^{n-1}(x) + \ldots + x_0R^0(x)$. If the degree of $R(x)$ is equal to $m$, the degree of the first term in the sum is $nm$, the second one is $(n-1)m$ and so on, all the way down to $0m$. If $m>0$, the degree of the first term is greater than all the others, so it's the only one which affects the leading term (and leading coefficient) and it's thus just $q_n$ which plays any role. This is not the case if $m=0$: all of the terms will be of degree zero too, so they all will contribute to the "leading coefficient" (which happens to be the only coefficient in this case).
Thus, the quantity on the right-hand side was correct (since we assumed that $\deg P(x)\geq 1$ but the one on the left-hand side is correct only when $\deg P'(x)\geq 1$ or, equivalently, $\deg P(x)\geq 2$.

The easiest approach to correct the flaw is to consider polynomials of degree 1 and zero at the same time: Plugging $P(x)=ax+b$ into the original equation gives us $b = a(1-a)$. In other words, all polynomials of the form $P(x)=ax + a(1-a)$ are solutions (this covers both the solutions found by [b]pbornsztein[/b] and the one you provided.


Let a,b,c be positive real numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$. prove that inequality
$ \sum_{cyc}\frac{a}{a^3\plus{}2bc}\geq \frac{ab\plus{}bc\plus{}ca}{3}$

[quote="lian109"]Let a,b,c be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. prove that inequality
$ \sum_{cyc}\frac {a}{a^3 \plus{} 2bc}\geq \frac {ab \plus{} bc \plus{} ca}{3}$[/quote]
We have:
$ \sum_{cyc}\frac {a}{a^3 \plus{} 2bc} \equal{}\sum_{cyc} \frac{a^2}{a^4\plus{}2abc} \geq \frac{(a\plus{}b\plus{}c)^2}{a^4\plus{}b^4\plus{}c^4\plus{}6abc} \geq \frac{3(ab\plus{}bc\plus{}ca)}{a^4\plus{}b^4\plus{}c^4\plus{}6abc}$
We need prove that $ \frac{3(ab\plus{}bc\plus{}ca)}{a^4\plus{}b^4\plus{}c^4\plus{}6abc}\geq \frac {ab \plus{} bc \plus{} ca}{3}$
or $ 9 \geq a^4\plus{}b^4\plus{}c^4\plus{}6abc$
or $ (a^2\plus{}b^2\plus{}c^2)^2 \geq a^4\plus{}b^4\plus{}c^4\plus{}6abc$
<=>$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2 \geq 3abc$
<=>$ \frac{ab}{c}\plus{}\frac{ac}{b}\plus{}\frac{cb}{a} \geq 3$
This is a inequality in France Pre-MO 2005


sin 25 :^o: *cos 25 :^o:

How is this done with no calculator (and in less than 6 minutes)?

I ran into this on a math contest today and can't figure it out. Help!

Well, it's (sin 50o)/2, but that doesn't have a really nice value.

What were you supposed to do with this value?

The problem was simply: sin25cos25

It was a "target" round question, that I really didn't know what to do with it.  There were no words or questions along with it.  I think I'll eventually get the test back.

Well, the answer is that it was a poorly written question.  My guess is that they wanted you to write down sin50 / 2.  You could perhaps appeal it as horribly unclear, although it's a little late for that.

sin25cos25 (exact value) is sin25cos25.  You cant get more exact than that & they probably wanted it to the nearest 1000th or sumthing.  Don't worry, itll never be in MC.[/i]

Yeah will someone please post what will and will not be on the MATHCOUNTS? I think that may be helpful...(along with the formula sticky in MATHCOUNTS Problem Series if you're in it, I recommend copy&pasting that to some file on your HD before the board is deleted.)


I'd like to give you another "game". We also got it at the IMO training.

But I think nobody solved it back then ... I'm not sure though.

Two guys are playing a game. They are given 4 boxes, containing candies.
Each player is allowed to take - on his turn - as much candies he wants, from precisely one box. (He can't take candies from the other three boxes.) The one who has to take the last remaining candy out of one of the four boxes loses.
A and B agree on the following : A will start the game (he can move first) but B can choose the number of candies in the fourth box, given that the first box contains 9 candies, the second one contains 18 candies and the third box contains 34 candies.
What's the number of candies B has to choose such that he can guarantee his victory ?

Do you by any chance have the answer (not necessarily a proof). I'm asking because an idea just stroke me: what if the conditions for A and B to win or lose are something like in Nim (a well known game, look it up on the Net). The main idea in that game is this: we have some groups of let's say candies  :D The players start taking candies. On his turn a player can only take candy from one group. The player to take the last candy wins (so it's a bit different).

It can be shown (although I don't know how) that if we write the numbers of candies in each group in base 2 and add the numbers without passing 1 from right to left (I mean like this 110+010=100, for example, so it's a different kind of addition), player A wins iff there is at least a 1 in the final result-this is for Nim. I've checked a few small cases for our game and it seems to work the same way here: for 1 1 1 1 B wins (1+1+1+1=0), for 2 1 1 1 A wins (2=10, so 10+1+1+1=11, so there is a 1 in the final result)
and I've checked a lot more cases (with boxes having 3, 4, candies) and it seems to work here too!

If this is the case, the number we're looking for is that number which, when added in that funny way to the other 3 in base 2 will give only 0's. 34=100010, 18=10010, 9=1001. By "adding" them (in the way shown above) we get 111001, so the number B should choose is 111001 in base 2, so it's 1+2^3+2^4+2^5=57. That's why I asked you if you knew the result. If it was 57 it would be a strong suggestion I was right  :D

Sorry abt this LONG post. I might be completely wrong..

No, I don't have the answer. But we got a hint at the training : look at the situation "modulo 2".


Let  $ (x_n) , n\equal{}1,2..$  be a sequence given by the following recursive formula:

$ u_1\equal{}12 ,u_2\equal{}29 ,u_{n\plus{}1}.u_n\equal{}u^2_{n\plus{}1} \plus{}(\minus{}1)^{n\minus{}1}$ .Prove that exist infinite $ n$

such that $ u_{n\plus{}2}\minus{}2$ and $ u_{n\minus{}1}$ divisible by 2008

[quote="Flakky"][quote="vuthanhtu_hd"]Let  $ (x_n) , n \equal{} 1,2..$  be a sequence given by the following recursive formula:

$ u_1 \equal{} 12 ,u_2 \equal{} 29 ,u_{n \plus{} 1}.u_n \equal{} u^2_{n \plus{} 1} \plus{} ( \minus{} 1)^{n \minus{} 1}$ .Prove that exist infinite $ n$

such that $ u_{n \plus{} 2} \minus{} 2$ and $ u_{n \minus{} 1}$ divisible by 2008[/quote]
There is something wrong with the indexes in the recursive formula because for $ n \equal{} 1$ we get that $ u_{1} \equal{} u_2 \plus{} \frac {1}{u_2}$ which is not thue![/quote]

Sorry ,$ u_{n \plus{} 1}.u_n \equal{} u^2_{n \plus{} 1} \plus{} ( \minus{} 1)^{n \minus{} 1}$ with $ n\geq 2$  :D


This inequality was proposed by Orl in "Other Problem" section.

I think it is better to have it also in this section.

[quote="orl"]Prove for $n > 2$ that $(n!)! > n[(n-1)!]^{n!}.$[/quote]

This is way too weak (unless I'm making a mistake somewhere, which is highly probable :)).

Anyway, it's equivalent to $(n!)!>n^{1-n!}(n!)^{n!}\iff n^{n!-1}>\frac{(n!)^{n!}}{(n!)!}\iff n^{\frac{n!-1}{n!}}>\frac{n!}{\sqrt[n!]{(n!)!}}=f(n!)$, where $f(n)=\frac n{\sqrt[n]{n!}}$.

$f(n)=\sqrt[n]{\frac {g(2)}{g(1)}\ldots \frac{g(n)}{g(n-1)}}$, where $g(n)=\frac{n^n}{n!}$. However, $\frac {g(n+1)}{g(n)}=(1+\frac 1n)^n<e$, so $f(n)<e,\ \forall n\in\mathbb N^*$, and this means that we only have to check $2$ or $3$ cases by hand, because $n^{\frac{n!-1}{n!}}$ increases like $n$.

I think it works. :lol:

Congratulations, Grobber.

I would like to know if this is the official solution.


[quote="anurags92"]In $ \bigtriangleup ABC$  bisectors $ AA_{1}$  and $ BB_{1}$ are drawn. Prove that the distance from any point $ M$ of  $ A_{1}B_{1}$ to line  $ AB$ is equal to the sum of distances from $ M$  to $ AC$ and $ BC$ .[/quote]
[quote="chien than"][quote="anurags92"]In $ \bigtriangleup$ $ ABC$ bisectors $ AA_{1}$ and $ BB_{1}$ are drawn. Prove that the distance from any point $ M$ of $ A_{1}B_{1}$ to line $ AB$ is equal to the sum of distances from $ M$ to $ AC$ and $ BC$.[/quote]
$ MN \perp AB;MV \perp BC;MU \perp AC;MN \equal{} z;MV \equal{} x;MU \equal{} y$
We need prove that $ z \equal{} x \plus{} y$
Suppose that $ MB_1 \equal{} p;MA_1 \equal{} q$;$ B_1S;A_1K$ are perpendicular to $ AB$;$ A_1L;B_1T$ are perpendiculars to $ AC$ and $ BC$
Suppose that $ SB_1 \equal{} B_1T \equal{} b;A_1K \equal{} A_1L \equal{} a$,we have
From $ B_1S//MN//A_1K$ we have $ z \equal{} \frac {ap \plus{} bq}{p \plus{} q}$
In triangle $ B_1A_1T$ we have $ MV //B_1T$ so
$ \frac {x}{b} \equal{} \frac {q}{p \plus{} q}$=>$ x \equal{} \frac {bp}{p \plus{} q}$
Similar we have $ y \equal{} \frac {ap}{p \plus{} q}$
So $ z \equal{} x \plus{} y$
Done![/quote]

Let us to rewrite the notations of the problem, considering as $ P$ an arbitrary point on $ B'C',$ where $ BB',\ CC'$ are the angle bisectors of $ \angle B,\ \angle C$ respectively, of the given triangle $ \bigtriangleup ABC$ and let $ D,\ E,\ F$ be the orthogonal projections of $ P,$ on $ BC,\ AC,\ AB,$ respectively.

We will prove that $ PD \equal{} PE \plus{} PF$ $ ,(1)$

$ \bullet$ Let be the point $ M\equiv CC'\cap PD$ and we denote the point $ K\equiv PE\cap DN,$ where $ N$ is the orthogonal projection of $ M,$ on $ AC.$

It is easy to show that $ PK \equal{} PD$ $ ,(2)$ from $ \angle PKD \equal{} \angle MND \equal{} \angle PDK.$

So, it is enough to prove that $ EK \equal{} PF$ $ ,(3)$

Let $ X,\ X'$ be, the orthogonal projections of $ B',$ on $ BC,\ AB,$ respectively and also let $ Y,\ Y'$ be, the ones of $ C',$ on $ BC,\ AC,$ respectively.

Through the point $ B',$ we draw the line parallel to $ PK\parallel C'Y'$ and we define the point $ L$ on this line, such that $ B'L \equal{} B'X \equal{} B'X'$ $ ,(4)$

From $ C'Y \equal{} C'Y',\ PD \equal{} PK,\ B'L \equal{} B'X$ and $ \angle YC'Y' \equal{} \angle DPK \equal{} \angle XB'L$ and because of the collinearities of $ Y,\ D,\ X$ and $ C',\ P,\ B',$ we conclude that the points $ Y',\ K,\ L,$ are also collinear.

So, from $ EK\parallel B'L$ $ \Longrightarrow$ $ \frac {EK}{B'L} \equal{} \frac {Y'K}{Y'L}$ $ ,(5)$

From $ PF\parallel B'X'$ $ \Longrightarrow$ $ \frac {PF}{B'X'} \equal{} \frac {C'P}{C'B'}$ $ ,(6)$

From $ C'Y'\parallel PK\parallel B'L$ $ \Longrightarrow$ $ \frac {Y'K}{Y'L} \equal{} \frac {C'P}{C'B'}$ $ ,(7)$

From $ (5),$ $ (6),$ $ (7)$ $ \Longrightarrow$ $ \frac {EK}{B'L} \equal{} \frac {PF}{B'X'}$ $ \Longrightarrow$ $ EK \equal{} PF$ $ ,(8)$ and the proof is completed.

Kostas Vittas.

[quote="anurags92"] [color=darkred]In $ \triangle ABC$ the bisectors $ [BE$  and $ [CF$ are drawn, where $ E\in (AC)$ and $ F\in (AB)$ . For a point $ M\in [EF]$

denote the distancies $ x$ , $ y$ , $ z$ from $ M$ to the lines $ BC$ , $ CA$ , $ AB$ respectively. Prove that $ x \equal{} y \plus{} z$ .[/color]  [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Observe that $ ax \plus{} by \plus{} cz \equal{} 2S\ (*)$ , where $ S$ is the area $ [ABC]$ of $ \triangle\ ABC$ .

Since $ \frac {[AEF]}{[ABC]} \equal{} \frac {AF}{AB}\cdot\frac {AE}{AC} \equal{} \frac {b}{a \plus{} b}\cdot\frac {c}{a \plus{} c}$ obtain $ [AEF] \equal{} \frac {bcS}{(a \plus{} b)(a \plus{} c)}$ .

But $ 2\cdot [AEF] \equal{} y\cdot AE \plus{} z\cdot AF$ $ \implies$ $ \frac {2bcS}{(a \plus{} b)(a \plus{} c)} \equal{} \frac {bcy}{a \plus{} c} \plus{} \frac {bcz}{a \plus{} b}$ $ \implies$

$ y(a \plus{} b) \plus{} z(a \plus{} c) \equal{} 2S$ $ \implies$ $ a(y \plus{} z) \equal{} 2S \minus{} (by \plus{} cz)$ $ \stackrel {(*)}{\implies}$ $ a(y \plus{} z) \equal{} ax$ $ \implies$ $ y \plus{} z \equal{} x$ .

[b]Remark.[/b] Prove [u]reciprocally[/u], i.e. the geometrical locus of the point  $ M$ which belongs to the interior of a given $ \triangle ABC$

and for which its distance to the line $ BC$ is equally to the sum of its distancies to the lines $ AB$ , $ AC$ is $ (EF)$ , where

 $ E\in (AC)$ , $ F\in (AB)$ and the rays $ [BE$ , $ [CF$ are the bisectors of the angles $ \widehat {ABC}$ , $ \widehat {ACB}$ respectively.[/color]

[quote="Virgil Nicula"][color=darkred][b][u]Generalization.[/u][/b] Given are the triangle  $ ABC$ and two positive numbers $ \beta$ , $ \gamma$ .  Consider

the points $ \left\|\ \begin{array}{c}
F\in (AB)\ ,\ \frac {FA}{FB} \equal{} \beta\\\\
 E\in (AC)\ ,\ \frac {EA}{EC} \equal{} \gamma\end{array}\ \right\|$ . For a point $ M\in [EF]$ denote the distancies

$ x$ , $ y$ , $ z$ from $ M$ to the lines $ BC$ , $ CA$ , $ AB$ respectively. Prove that $ ax \equal{}\frac {by}{\beta} \plus{} \frac {cz}{\gamma}$ .  [/color][/quote]


linear algebra
matrix
search
linear algebra solved

Prove that real $n \times n$ matrix $A$ is positive semi-definite if and only if $trAB\geq0$ for every real symmetric positive semi-definite $n \times n$ matrix $B$.

THis statement is so strange. It this really what we should prove?

[quote="alekk"]what is semi-definite ?[/quote]
$A$ is called semi-positive definite if $x^TAx\geq 0$ for all $x$.

Suppose that $ tr(AB) \geqslant 0 $ for every $ B $
positive semi-definite. Denote $ C:= A + A^{t} $. Then also
for every such $ B $, $ tr(CB) = 2tr(AB) \geqslant 0 $
Then $ C $ is symmetric, and $ C \geqslant 0 $. Therefore there exists
an orthogonal matrix $ Q $, such that $ D = Q^{-1}CQ $ is diagonal
with entries $ \lambda_{1}, \lambda_{2}, \ldots , \lambda_{n} $.
If we take arbitrary $ x_{1}, x_{2}, \ldots, x_{n} \geqslant 0 $, then
$ B:= Q diag(x_{1},x_{2}, \ldots , x_{n} ) Q^{-1} $ is symmetric positive
semi-definite and $ tr(CB) = \sum_{i=1}^{n} x_{i}\lambda_{i} $.
So we get that for every $ x_{1}, x_{2}, \ldots, x_{n} \geqslant 0 $ that
$ \sum_{i=1}^{n} x_{i}\lambda_{i} \geqslant 0$, therefore $ C \geqslant 0 $,
so $ A \geqslant 0 $.

Now if $ A \geqslant 0 $, then denote as before $ C:= A + A^{t} $, so also
$ C \geqslant 0 $. As before, there exists an orthogonal matrix
$ Q $, such that $ D = Q^{-1}CQ $ is diagonal
with entries $ \lambda_{1}, \lambda_{2}, \ldots , \lambda_{n} \geqslant 0 $.
Now if we take a positive semi-definite symmetric $ B \geqslant 0 $, then
denote $ E :=  Q^{-1} B Q $, so E is also positive semi-definite symmetric, and
$ tr(CB) = tr(DE) = \sum_{i=1}^{n} \lambda_{i} (Ee_{i},e_{i}) \geqslant 0 $,
where $ e_{1},e_{2}, \ldots , e_{n} $ is a standard basis.
So $ tr(AB) = \frac{1}{2}tr(CB) \geqslant 0 $.

[quote="harazi"]In fact, the hard implication follows from one of the problem posted by alekk: the Hadamard product of two semipositive matrices also has this property.[/quote]
harazi,could you please give the link with this topic, because i can't find any matches neither for Hadamard product nor something like
with search option


Hi, I don't really have many songs for my new Ipod Video (like 200), so, I'm counting on you guys to tell me some good songs. Please post all good songs you know.

Thanks! :lol:

[b]Eric Clapton[/b]:
You Look Wonderful Tonight
[b]Bob Dylan[/b]:
I Want You
Knockin' On Heavens Door
If Not For You
Maggie's Farm
[b]Sting & The Police[/b]:
englishman in new york
every breath you take
fragile
[b]Michael Jackson[/b]
Another Day Is Gone
and many many other sBeautiful Songs that i'll post them later for you cuz i dont have time now... :lol: Good Luck

Anigma!!!!!!!

srry don't know who its by

i think it is a cd, but o well.

Depends what genre you like.  I like Progressive Rock and the best band for that is RUSH.
Top Songs:
YYZ
Tom Sawyer
Limelight
Far Cry

For any rock these songs are good:
Trippin' on a hole in a paper heart,  Stone Temple Pilots
Sweet Child O' Mine, Guns N' Roses

I also like Jump from Van Halen

Michael Jackson: Thriller, Billie Jean, You are not alone, and more

Guns and Roses
Sweet Child of Mine
November Rain
Patience

All American Rejects
Move Along
It Ends Tonight

[u]Red Jumpsuit Apparatus[/u]
i like the album
[i]Dont you Fake it[/i]

[u]Boys Like Girls[/u]
[i]The Great Escape[/i]

[u]Avril Lavigne[/u]
[i]When Youre Gone[/i]

those are my faves, but depends on yr taste

The Great Escape (great)

Umbrella (okay...)

Because of you

Thanks for the memories

It ends tonight

Hey there delilah.. ha

Party like a rockstar

Summer love

Beautiful girls

Wait for you (Awesome!!!)

Buy you a drank

Girlfriend

Unfaithful

Numb

In the End

Grillz (not that new...)

You're beautiful (ha)

Over my head

How to save a life

Ridin'

Me and U

I write sins not tragedies

Tryin ta win

Too little too late

Lips of an angel

This aint a scene, its an arm race

Cupid's chokehold

Face down (<3)

Icebox

Better than me (good)

Last night

Like a boy

Hotline.... or was it On the hotline (pretty good)

I have a lot of songs I like, so I'll post bands. I hope you like rock/alternative/punk.

AC/DC
Aerosmith
Augustana
Boston
Depswa
Green Day
Seether
Sum 41
Snow Patrol (but not Chasing Cars, it's really overrated; get some stuff from the album "Final Straw".)
Nickelback
Guns 'n' Roses
Ozzy Osbourne

John Lennon:
Imagine

Bob Dylan:
Blowin' in the Wind
Like a Rolling Stone
The Times They are a Changin'
Just Like a Woman
Mr. Tambourine Man

Pete Seeger:
We shall overcome
Little Boxes
Where Have all the Flowers Gone

Beatles:
Hey Jude
Yesterday
Let it Be

Pink Floyd:
Hey You
(and many others)

Led Zeppelin:
Stairway to Heaven

Michael Jackson:
Heal the World

Boyzone:
No Matter What
Words
Everyday I Love You
You Needed Me
I Never Wanna Say Goodbye

Westlife:
I wanna grow old with you
Flying Without Wings
Bap Bap Baby

INXS:
Afterglow

Akon:
Lonely

here r some good bands:

scorpions
all american rejects
ac/dc
guns and roses(awesome)
Pink floyd
Foreigner
green day
simple plan

[quote="SoxFan34"]Nickelback[/quote]

why post a band in this case? just post their one song

[quote="mz94"]bands:

all american rejects
red jumpsuit apparatus
ac/dc
boys like girls(still wondering what is meaning of their name)
the fray
fall out boy
my chemical romance
AVRIL LAVIGNE(rules)

dont make fun of me b/c i like avril!

whos guns&roses?[/quote]

everything on that list sucks except for like 1 all american reject song and some ac/dc

ANYWAY

gwen stefani
The Sweet Escape
Hollaback Girl
Rich Girl
Wind It Up
What You Waiting For?
Cool
Luxurious
Crash
4 in the Morning

no doubt
Hey Baby
Don't Speak
Bathwater
Just a Girl
Spiderwebs

Uhh.... let me filter out......

An Garhar Ban - Clannad
Northern Lighhts - Stella Maris
Tarnima - Nasser Kilada
Carol of the Bells     Christmas music
Offering - Ravi Shankar & Philip Glass

I really need to update my i-Pod.

also I have to say that every song of [i][u]Scorpions[/u][/i] from the album [u]Moment of glory [/u]  is worthy

ok, updated.

Top rated:

An Gabhar Ban
Northern Lights
Tarnima
Carol of the Bells
Offering
Dona Crei Tun
Glass from Jasper
Akwaba
Earth Meets Sky(both parts)
Dancing for a Vision
Bajya Sahita
Descent to the Lowerworld
Hamana Nale
Emerging Power
Atma Victu
Africa in Your Head
Bhupali Tabla Duet
Flowering Tree
Deliverance
Chinese Dance

top 20.

PINK FLOYD!!!!!!!!!!!!!!!!!!!!

Pigs on the Wing (1 and 2)
Dogs
Pigs (Three Different Ones)
Sheep

Wish You Were Here
Have a Cigar
Welcome to the Machine
Shine on You Crazy Diamond

Breathe
Time
Money
Brain Damage
Eclipse

Echoes
Pillow of Winds
Fearless
San Tropez

Wots...uh the Deal?
The GOld it's in the...
Burning Bridges
Stay

If
Fat Old SUn
Summer '68

Set the COntrols for the Heart of the SUn
Corporal Clegg
A Saucerful of Secrets

Careful With That Axe Eugene
Arnold Layne
See Emily Play

Cymbaline
Green is the Colour
Nile Song
Ibiza Bar

The POst War Dream
Not Now John
When the Tigers Broke Free
Two Suns in the Sunset

Learning to Fly
On the TUrning Away
Dogs of War

Marooned
High Hopes
Cluster One

Bike
Astronomy DOmine
Interstellar Overdrive
The Gnome

Another Brick in the Wall (1,2,3)
Comfortably Numb
Hey You
Mother
The Trial
In the FLesh

etc, etc, etc.

Oh, and don't forget, RIDERS ON THE STORM, BY THE DOORS

Sympathy for the Devil
One (Metallica and U2)
Sweet Child O'Mine
Sweet Home Alabama
Smells Like Teen Spirit
Where the Streets Have No Name
Paradise City
Smoke on the Water
Kashmir
My Generation
Born to Be Wild
Born to Run
Fourtunate Son
Lamb of God
We Are the Champions

I like everything except for country..(no offense..)

The Bohemian Rhapsody
Anything by the Beatles or the Who (btw I am only 13, but I am a HUGE Beatles fan!)

My favorite band is Westlife!!! :D :D

[url=http://www13.nhac.vui.vn/Music/#Play,32834][b]You must have had a broken heart[/b][/url]

YOU MUST HAVE HAD A B.HEART

The way you say the things you do
The softness of the words you choose
The times that you can read my mind
And take my worries out of sight

Your fingers touching on my lips
And say a kiss is still a kiss
And when you look at me I see
I see the pain that you had to feel

You must've had a broken heart, to love me the way do
Must've been so torn apart, I can see it when I look at you
All the meaning that is in your eyes, the love you give will never die
And I knew right from the start, you must've had a broken heart

You seem to have that certain smile
I can't forget after a while
The day you walked into the room
At once I knew the hurt that you'd been through

Sometimes when its late at night and I see your face in the fire light
Showing all the love you have for me, well I love you as much

Sometimes, sometimes
Must've had a broken heart
Yeah
All the meaning that is in your eyes, the love u give will never die
And I knew right from the start, you must've had a broken heart
Sometimes.....

Favorite, in order:

One World One Dream (Wang LeeHom)
Stand Up (Jackie Chan)
Beijing Welcomes You
As A Deer
Not Be Shaken
(A LOT of classical songs)
(A bunch of other christian songs (Note: I'm not christian, but I still like the songs :)))

No good taste of music =/

And this is actually serious.

[quote="7h3.D3m0n.117"]Numb by Linkin Park
In the End by Linkin Park
It's Not My Time by Three Doors Down
Blow Me Away by Breaking Benjamin
Mr. Brightside by The Killers[/quote]

In the End and Numb are both really not good. My preference is Blvd of Broken Dreams, Holiday, Somewhere I Belong, Weasel Stomping Day (Because it still makes me laugh like a maniac) and Surround.

[quote="crazychinesemaniac"]My preference is Blvd of Broken Dreams, Holiday, Somewhere I Belong, Weasel Stomping Day (Because it still makes me laugh like a maniac)[/quote]

Green Day is okay. And Weird Al is ridiculous.


quadratics
inequalities
function
algebra unsolved
algebra

For any positive integer n > 3, let Qn be the set of quadratic functions f : R to R, f(x) =
ax2 + bx + c where a; b; c are elements of set (1,2,3,...,n). Let P(n) be the number of members, f(x), of Qn
such that f(x) = 0 has integer roots.
Prove that

                                     P(n) <n^2/4.

Have a look here :

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=integer+roots&t=29126

Pierre.


Choose any three-digit number; write it twice in a row (e.g. 123123). Add $105$; then
divide by $7$; then subtract $4$; then divide by $11$; then subtract $1$; then divide by $13$. What
is the result? Why?

[hide]Taking a 3 digit number and repeating it multiplies it by 1001. So we have (((1001x+105)/7-4)/11-1)/13). Then do the algebra.[/hide][/hide]

[hide]Let the three-digit number be $abc$. Then $abcabc = 1001(abc)$.
Adding $15$ we get $1001(abc)+105$.
Dividing by $7$ we get $143(abc)+15$.
Subtracting $4$ we get $143(abc)+11$.
Dividing by $11$ we get $13(abc)+1$.
Hence the result is $abc$.[/hide]


You can use Cauchy-Schwarz and yielding the answer is $ 15\sqrt {5}$. The equality is left for you.

[quote="Kono"]11.) Find the maximum value of $ a(x) \equal{} 3f(x) \plus{} 4g(x) \plus{} 10h(x)$ given that $ f(x)^2 \plus{} g(x)^2 \plus{} h(x)^2 \leq 9.$[/quote]
Note that $ a(x) \equal{} (3,4,10)\cdot(f(x),g(x),h(x))$.
Using $ \parallel{}\vec{a}\parallel{}\cdot\parallel{}\vec{b}\parallel{}\ge \vec{a}\cdot \vec{b}$ we get $ \sqrt {9 \plus{} 16 \plus{} 100}*\sqrt {9}\ge a(x)$.

[quote="nsato"]Note that you can't find the maximum value of $ a(x)$, you can only find an upper bound on $ a(x)$.[/quote]
If you mean to say that the inequality is strict, please explain why. (as I could not figure it out). :blush:
[quote="peine"]be Cauchy Schwarz,
$ a(x) \equal{} 3f(x) \plus{} 4g(x) \plus{} 10h(x)\leq \sqrt {(9 \plus{} 16 \plus{} 100)((f(x))^2 \plus{} (g(x))^2 \plus{} (h(x))^2)}\leq 3\sqrt {125} \equal{} 5\sqrt {15}$[/quote]
There is a small typo in the last, but, anyways, it is a very nice solution!

Well, the functions are arbitrary. So Cauchy-Schwarz can give only give us an upper bound, which is optimal, because we can have equality e.g.:

$ (f(x),g(x),h(x)) \equal{} \sqrt {\frac {9}{3^2 \plus{} 4^2 \plus{} 10^2}}\cdot (3, 4, 10) \equal{} \left(\frac {9\sqrt {5}}{25},\frac {12\sqrt {5}}{25},\frac {6\sqrt {5}}{5}\right)$

In other words, the inequality is sharp and non-strict in the extremal (equality) and therefore general case. For cases such as $ (f(x),g(x),h(x))\equal{}(0,0,0), \dots$, the inequality is not so sharp. :P


X^3+aX +b = 0 has real roots h,k and L
with h<=k<=L

X^3 +cX+d =0 has real roots p,q and r
with p<=q<=r .d is non-Zero

If h/p<=k/q<=L/r Prove that (a/c)^3 =(b/d)^2

if h,k,l are roots of f(x) = x:^3:  + ax + b
then f(x) = (x - k)(x - l)(x - h)

expanding this product we find that
           - hkl = b
           hk + kl + hl = a
and     - (h + k + l) = 0

similarly
           - pqr = d
           pq + qr + rp = c
           - (p + q + r) = 0

applying the AM >= GM inequality to (hk + kl + hl = a), we find that

(hk + kl + hl)/3  >=  (hkl)^(2/3)

hence, applying the AM >= GM to (pq + qr + rp) and dividing,

(hk + kl + hl) / (pq + qr + rp) >= (hkl)^(2/3) / (pqr)^(2/3)

i.e. (a/c):^3:  >=  (b/d):^2: ,  d  <>  0.

to finish off the question, determine when equality occurs and use the given inequality.

I hope this was helpful.

This "solution" is completly wrong!!
1) You cannot apply AM-GM!!!!!!!!!!!
2) You cannot divide inequalities!!!!!!!!

The solution of this is trivial since one get h+k+l=0,hkl=-b,hk+hl+kl=a, e.t.c


Use sin(a+b) = sinacosb + sinbcosa

So,

say x = pi since i don't know how to use LAtex :P

sin(2x/7) = 2sin(x/7)cos(x/7)
You can use this all the way down, but sin (7x/7) is a problem.

isn't sin(pi) = 0???

[quote="silouan"]Prove that
$sin\frac{\pi}{7}sin\frac{2\pi}{7}sin\frac{3\pi}{7}sin\frac{4\pi}{7}sin\frac{5\pi}{7}sin\frac{6\pi}{7}sin\frac{7\pi}{7}=\frac{7}{64}$[/quote]
That is incorrect. $\sin{\pi}=0$

Sorry it was a typo .It is corrected now

Similar to this (and has been asked before:  find the value of:

$\sin 1^{o}\sin 2^{o}\sin 3^{o}\dots \sin 179^{o}$

(A little easier is $\sin 1^{o}\sin 3^{o}\sin 5^{o}\dots \sin 179^{o}$)

Enjoy!

(Or in general: if $a= (\pi /n)$  find the value of:
$\sin a \sin (2a) \sin (3a) \dots \sin ((n-1)a)$

[hide](As I said, the question has been asked a few times here)  [/hide]

A nice and beautiful  formula to know:
$\sin (nx) = 2^{n-1}\sin (x) \sin (x+\frac{\pi}{n}) \sin (x+\frac{2\pi}{n}) \sin (x+\frac{3\pi}{n}) \dots \sin (x+\frac{(n-1)\pi}{n})$

[hide]If  you put n=7, and  take the limit when x tends to zero you get what you want. [/hide]

[quote="Altheman"]let $z=e^{\frac{\pi\cdot i}{7}}$

$\frac{1}{(2i)^{6}}\prod_{k=1}^{6}(z^{k})(1-z^{2k})$

$-\frac{1}{64}\cdot (z^{21})*\prod_{k=1}^{6}(1-z^{2k})$

the second part can be evaluated by plugging in $x=1$ into $f(x)=x^{6}+x^{5}+...1$ and the factorization of $f(x)$ giving $\frac{7}{64}$ as desired[/quote]

Could you write the solution in all the details ??Thank you

Let $w = e^{ix}$. Then we have that $\sin x = \frac{e^{ix}-e^{-ix}}{2i}= \frac{w-\frac1{w}}{2i}= \frac1{2i}\cdot \frac{w^{2}-1}{w}$ (this is sometimes taken as the definition of $\sin$).

Now let $z = \exp \left( \frac{\pi\cdot i}7 \right)$. Using the above formula we get that your product is equal to $\frac1{(2i)^{6}}\prod_{k=1}^{6}z^{k}\left( 1-z^{2k}\right)$.

The evaluation of $\prod_{k=1}^{6}\left( 1-z^{2k}\right)$ can be done by noticing that $z^{2}, z^{4}, \ldots$ are the seventh roots of unity.

Thank you very very much. You are so helpful .

oh, dear It's nearly also has been disscus completely at:
http://www.mathlinks.ro/Forum/viewtopic.php?t=142639
 http://www.mathlinks.ro/Forum/viewtopic.php?t=137400

These trig formulas are very useful. Most can be derived using the $e^{i \theta}$ notation.


trigonometry
integration
calculus
calculus computations

Solve

$ y" \plus{} \sin x \ y \equal{} 0$

I think it has no solutions in elementar functions. You can find it this way: $ y(x) \equal{} \sum_{k \equal{} 0}^{\infty}{a_k x^k}$
Then $ y''(x) \equal{} \sum_{k \equal{} 2}^{\infty}{k(k \minus{} 1)a_k x^{k \minus{} 2}} \equal{} \sum_{k \equal{} 0}^{\infty}{(k \plus{} 1)(k \plus{} 2)a_{k \plus{} 2} x^{k}}$.
Alsow $ \sin x \equal{} \sum_{m \equal{} 0}^{\infty}{\frac {( \minus{} 1)^k x^{2m \minus{} 1}}{(2m \minus{} 1)!}}$
Now get: $ \sum_{k \equal{} 0}^{\infty}{(k \plus{} 1)(k \plus{} 2)a_{k\plus{}2} x^{k}} \equal{} \left( \sum_{m \equal{} 0}^{\infty}{\frac {( \minus{} 1)^k x^{2m \minus{} 1}}{(2m \minus{} 1)!}}\right)\cdot \left(\sum_{k \equal{} 0}^{\infty}{a_k x^k}\right)$
Ans continiously find $ a_k$.

hello, yes, that's right. The solution contains the MathieuC and the MathieuS function.
Sonnhard.

Another way:

If for example $ y(0) \equal{} 1$ and $ y'(0) \equal{} 0$. Then equation will be:
$ y \equal{} 1 \minus{} \int_{0}^{x}dx\int_{0}^{x}y\sin x\;dx$
Put here $ y \equal{} 1$, we get second approximation
$ y_{1}(x) \equal{} 1 \minus{} \int_{0}^{x}dx\int_{0}^{x}\sin x\;dx \equal{} 1 \minus{} x \plus{} \sin x$
Third approximation will be
$ y_{2}(x) \equal{} 1 \minus{} \int_{0}^{x}dx\int_{0}^{x}\left(1 \minus{} x \plus{} \sin x\right)\sin x\;dx \equal{} 3 \minus{} x \plus{} \sin x \minus{} 2\cos x \minus{} x\sin x \plus{} \frac {1}{4}\sin^2 x \minus{} \frac {1}{4}x^2$
etc.

Anyway, for small $ x$ we can write $ \sin x\approx x$ and get equation
$ y \equal{} 1 \minus{} \int_{0}^{x}dx\int_{0}^{x}xy\;dx$
Solution will be
$ y\approx 1\minus{}\frac{1}{3!}x^3\plus{}\frac{1\cdot 4}{6!}x^6\minus{}\frac{1\cdot 4\cdot 7}{9!}x^9\plus{}\ldots\equal{}1\plus{}\sum_{k\equal{}1}^{\infty}(\minus{}1)^{k}\frac{1\cdot 4\cdot \ldots\cdot (3k\minus{}2)}{(3k)!}x^{3k}$


Is there a winner for each general test part, or the combined score of the first and second general tests?


vector
parameterization
parametric equation
linear algebra

11. For what values of [i]k [/i]will the line $ \frac{x - k}{3} = \frac{y + 4}{2} = \frac{z + 6}$ intersect the plane $ x - 4y + 5z + 5 = 0$?

a) in a single point?

*b) in an infinite number of points? [color=blue](I was able to do this part)[/color]

c) in no points?

----

For b) to be true, the line must lie on the plane. Therefore, I substituted the parametric equation of the line into the scalar equation of the plane.

$ \frac{x - k}{3} = \frac{y + 4}{2} = {z + 6}
\Rightarrow \{\begin{array}{cc}x = 3t + k \\y = 2t - 4\\ z = t - 6\end{array}$

Substitution:

$ (3t + k) - 4(2t - 4) + 5(t - 6) + 5 = 0
\Rightarrow k = 9$.

However, I am not sure how to complete parts a) and c). Could someone please offer starting advice?

I think you had the right Idea...

Try writing the line as a point and a vector, how does the line change as $ k$ varies
[hide="answer"]
If $ l \equal{} (k, \minus{} 4, \minus{} 6) \plus{} t(3,2,1)$ then by varying $ k$ we only vary the initial point of the ray, but the direction of the vector wont change. Hence, this line will always be parallel to the plane and should never intersect it at exactly one point. Furthermore it will have no solutions for $ k\not \equal{} 9$

This can also be done algebricly the way you did it

$ (3t \plus{} k) \minus{} 4(2t \minus{} 4) \plus{} 5(t \minus{} 6) \plus{} 5 \equal{} 0 \Leftrightarrow k \minus{} 9 \equal{} 0\qquad(1)$

If the line intersects the plane at one point, then there should be one value $ t$ such that (1) is true, but since all the $ t$'s cancel this is impossible.
We will have no solutions if $ k\not \equal{} 9$[/hide]


algebra polynomial AMC AIME USA(J)MO USAMO factorial

Same rules as other marathons. AIME to USAMO problems.

Problem 1:
Find 1/(2! 17!) + 1/(3! 16!) + ... + 1/(9! 10!).

[hide]The factorials in the denominator, which each sum to 19, remind us of $\binom{19}{k}$. Unfortunately, there is no 19! in the numerator. Not a problem...multiply the whole expession, let's call it $x$, by 19!, to get \[ 19!x = \binom{19}{2} + \binom{19}{3} + \cdots + \binom{19}{9}.  \] Add $\binom{19}{0}+\binom{19}{1}=20$ to both sides to get \[ 19!x + 20 = \binom{19}{0} + \binom{19}{1} + \binom{19}{2} + \binom{19}{3} + \cdots + \binom{19}{9}.  \] Remember that \[ \binom{19}{0} + \binom{19}{1} + \cdots + \binom{19}{19} = 2^{19}.  \] The terms that are in $19!x+20$ but not this are \[ \binom{19}{10}+\binom{19}{11}+\cdots+\binom{19}{19}, \quad \text {or} \quad \binom{19}{9}+\binom{19}{8}+\cdots+\binom{19}{0},  \] which is the same as $19!x+20$. Hence, $2(19!x+20)=2^{19}$, and $x=\frac{2^{18}-20}{19!}$.[/hide]

[size=150]Problem 2[/size]

Let $x$, $y$, and $z$ be distinct real numbers that sum to 0. Find the maximum possible value of \[ \frac{xy+xz+yz}{x^2+y^2+z^2}. \] (HMMT 2005)

Am I right?
[hide]11!-1[/hide]

Not sure how to explain it, though. :(

Yes, that is correct. You can prove this using induction:

Base Case: $1(1!)=2!-1$

Assume $1(1!)+\ldots+n(n!)=(n+1)!-1$. Then: \begin{eqnarray*}1(1!)+\ldots+(n+1)(n+1)!&=&(n+1)!-1+(n+1)(n+1)!\\ &=&(n+2)!-1\end{eqnarray*}

$\sum k(k)!=\sum ((k+1)-1)k!=\sum (k+1)!-k!$.

So the sum telescopes into $11!-1!$, which is $39916799$.

Although during the actual tournament, unless you immediately saw that algebraic manipulation it would be quicker to just multiply it out.

[quote="seamusoboyle"][quote="PenguinIntegral"]Find all positive real solutions to $x^3 + y^3 + z^3 = x + y + z$and$x^2 + y^2 + z^2 = xyz.$[/quote]

Are these similtanious equations?[/quote]
Yes. But post a new problem. That is really difficult.

[quote="PenguinIntegral"]Find all positive real solutions to $x^3 + y^3 + z^3 = x + y + z$ and $x^2 + y^2 + z^2 = xyz.$[/quote]

[hide]
$x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2-xy-xz-yz)+3xyz \forall x,y,z \in \mathbb{R}$
$x+y+z = (x+y+z)(xyz-xy-xz-yz)+3xyz$
I already found earlier that $xy+xz+yz=0$ (at least I think I did...), so
$x+y+z = (xyz)(x+y+z) + 3xyz$
$0=(xyz-1)(x+y+z)+3xyz$
I also found earlier that $(x+y+z)^2=xyz$, so if we let $x+y+z=a$ and $xyz=b$, then we get
$0=a(b-1)+3b$
$0=a(a^2-1)+3a^2$
$a(a^2+3a-1)=0$
$a=0,\frac{-3\pm \sqrt{9+4}}{2}=0,\frac{-3}{2} \pm \frac{\sqrt{13}}{2}$

Are you sure this problem is right? Or did I make a mistake somewhere?

What's the source of this problem, anyway? :?:
[/hide]

OK then, someone just post a different problem. :roll:

[hide]\[ \frac{P(2x)}{P(x+1)}=\frac{8x}{x+7},  \] \[ P(2x)(x+7)=(8x)P(x+1).  \] $x=0$ is a root of the RHS, so it must be a root of the LHS. That is, $x|P(x)$. We cannot consider $x=-7$ for the LHS, since that would make the original equation undefined on one side. Hence, say $P(x)=xQ(x)$ and continue. \[ (2x)Q(2x)(x+7)=(8x)(x+1)Q(x+1),  \] and $x=-1$ is a root of the RHS, so $Q(-2)$ must be a root of the LHS. Say $Q(x)=(x+2)R(x)$ and continue. \[ (2x)(2x+2)R(2x)(x+7)=(8x)(x+1)(x+3)R(x+1).  \]$x=-3$ is a root of the RHS, so $R(-6)=0$. Let $S(x)=(x+6)R(x)$. \[ (2x)(2x+2)(2x+6)S(2x)(x+7)=(8x)(x+1)(x+3)(x+7)S(x).  \] All roots of RHS are roots of LHS already. Thus, we try to find the degree of $S(x)$. Let $\deg(P(x))=d$. Then in the original expression, we see that $\frac{2^d}{1^d}=8$, so $d=3$. Since $\deg(R(x))=\deg(P(x))-3$, $R(x)$ must just be a constant. Plug in $x=1$ for \[ P(x)=x(x+2)(x+6)S(x),  \] to find that $S(1)=S(x)=\frac{1}{21}$. Finally, \[ P(x)=\frac{1}{21}x(x+2)(x+6),  \] and $P(-1)=-\frac{5}{21}$.[/hide]Next problem.
Source: Richardson 1977

If $x,y,z$ represent real numbers and $x^2+y^2+z^2=8x+4y+6z-29$, find the numerical value of $x+y^2+z^3$.

[hide]The equation rearranges to $(x-4)^2+(y-2)^2+(z-3)^2=0$ after completing the square.

Using trivial inequality, the only solution is $(4,2,3)$ making $x+y^2+z^3=35$.[/hide]

Find all real $x$ that solve the equation $\sqrt{5+\sqrt{5-x}}=x$.

[hide]After squaring and re-arranging we get:

$x^4-10x^2+x+20 = 0, \Rightarrow  (x^2-x-4)(x^2+x-5)=0$

From which we'll get 4 values, but only $\frac{1}{2}+\frac{1}{2}\sqrt{17}$ works[/hide]

Are people trying to solve the question ?  :huh:

[quote="Dannyboy990"]Prove that no positive $x,y$ can be found for the following equations:

$x^{19}y^{92}=2^{1992}$

$19x+92y=5752$

Source : Gillis Competition 1992, first question.[/quote]

Nice problem. Solution.
[hide]
Let $x^{19}y^{92}=2^{1992}=A$. Let $19x+92y=5752=B$.
Assume that $A$ and $B$ have a set of positive solution(s).
So $B=(x+x+x+x...x)$(Repeated 19 times)$+(y+y+.......y)$(repeated 92 times.
By AM-GM, we have:
$x^{19/111}y^{92/111}\le 5752/111\Rightarrow x^{19}y^{92} \le (5752/111)^{111}$.
Notice that $(5752/111)^{111}<2^{1992}$ because $5752/111<64=2^6$.
Thats makes the above inequality true because $64^{111}=2^{666}<2^{1992}$.
Having established that $A \le (5752/111)^{111}$, there is no way for $A=2^{1992}$.
We have reached the contradiction.

QED

[/hide]

JimHu.. congrats.. yeah .. i also like that problem..though you can get stuck on it for some time if you don't go in the right path :)

Hmmm, i will start it again.

[hide="Problem 9"]Let $n$ be a positive integer greater than two. Prove that the Fermat Number $f_n$ has a prime divisor greater than $2^{n+2}(n+1)$[/hide]

For problems posted here:

Do they have to be from one of the listed competitions, or do they just have to be of that level???

The problem must Algebra, nt, and polynomials and the problem must on the HS Intermediate Level.


Yesterday evening, something changed for me, making the forum very hard to use. Most text on the forums is tiny, especially the box I'm entering this in. There are also a lot of places with dark blue bars that completely hide any text on them unless it's selected; I can't tell who last posted in a forum or in what thread unless I select or highlight the link.

This is all in Firefox 3 on Mac OS X. Either switching to Safari or to Mathlinks style eliminates the problem.

Weird; have you tried deleting your cache/cookies?

All right, I just cleared the cache and that did it. I must have had a bad version of something stored.

At one point, I was getting both good and bad pages seemingly at random; I must have had both a good version and a bad version in the cache then.


Find all solutions in integers of the equation:
$\large x^3+(x+1)^3+\ldots+(x+7)^3=y^3$.

As a first step,[hide]we expand out the left to get $y^3 = 8x^3 + 84x^2 + 420x + 784$ so that $(2x+10)^3 - y^3 = 36x^2 + 180x + 216 = 36(x + 2)(x + 3)$ which is non-negative for all integers $x$.  This gives us two solutions $(x, y) = (-2, 6), (-3, 4)$ and gives us a nice upper bound on $y$.[/hide]

As a second step,

[hide]Show $y \geq 2x + 4$ for all $x$.[/hide]

perhaps the closed form for $\sum n^3$ would help some?

[quote="Altheman"]perhaps the closed form for $ \sum n^3$ would help some?[/quote]
More we can find the general of
$ \sum_{k \equal{} 1}^nk^p$
Where $ p\in N$

That question has been addressed repeatedly on the forums.  Go do a search.