geometry
geometric transformation
reflection
circumcircle
homothety
symmetry
ratio

Consider an isosceles triangle $ABC$ with $AB=AC$, and a circle $\omega$ which is tangent to the sides $AB$ and $AC$ of this triangle and intersects the side $BC$ at the points $K$ and $L$. The segment $AK$ intersects the circle $\omega$ at a point $M$ (apart from $K$). Let $P$ and $Q$ be the reflections of the point $K$ in the points $B$ and $C$, respectively. Show that the circumcircle of triangle $PMQ$ is tangent to the circle $\omega$.

Cute little problem :).

Let $S,T$ be the points where $\omega$ touches $AB,AC$ respectively. The quadrilateral $SKTM$ is harmonic, so $SK,SM,ST,SA$ form a harmonic division. This means that the same holds for the intersection points between these four lines and $BC$, but since $ST\|BC$, we find that $MS\cap BC=P$. In the same way we get $MT\cap BC=Q$, and it follows that the triangles $MPQ,MST$ are homothetic with center $M$. The same must hold for their circumcircles, and we're done.

[hide="Here is a proof with a harmonical division (similarly with Grobber-s proof)."][color=darkblue]Denote the points $X\in AB\cap w$, $Y\in AC\cap w$, $P_1\in BC\cap MX$, $Q_1\in BC\cap MY$, $S\in AM\cap XY$.I will mean by : $\mathrm {h.p.}$ - harmonical pencil ; $\mathrm {h.d.}$ - harmonical division. From the well-known property we have $(A,M,S,K)\ -\ \mathrm {h.d.}$, i.e. $X(A,M,S,K)\ ,\ Y(A,M,S,K)\ -\ \mathrm {h.p.}\ (XY\parallel BC)$ $\Longleftrightarrow$
$(P_1,B,K,\infty )\ ,\ (Q_1,C,K,\infty )\ -\ \mathrm {h.d.}$ $\Longleftrightarrow$
$BK=BP_1\ ,\ CK=CQ_1$ $\Longleftrightarrow$ $P_1\equiv P\ ,\ Q_1\equiv Q$ a.s.o.[/color][/hide]

[quote="grobber"]so $SK,SM,ST,SA$ form a harmonic division. This means that the same holds for the intersection points between these four lines and $BC$, [/quote]

What does this mean?

This is my synthetic proof. However, it is essentially rooted in projective geometry.

[b]Solution[/b]

Let $\omega$ be tangent to sides $AB$ and $AC$ at $X$ and $Y$, respectively. Let $P'=MX \cap BC$ and $Q'=MY \cap BC$.

Since $XA$ and $YA$ are the tangents to $\omega$ at $X$ and $Y$, respectively, $KM$ is the symmedian of triangle $\triangle{KXY}$. This yields that
\[\frac{XK}{XM}=\frac{YK}{YM}\]

Since $\triangle{ABC}$ is isosceles, we have that $XY \| BC$. This yields that $\angle{YMK}=\angle{YXK}=\angle{XKP'}$. Further, we have that $\angle{P'XK}=180^{\circ} - \angle{KXM}=\angle{KYM}$. Therefore $\triangle{P'XK} \sim \triangle{KYM}$. Similarly, $\triangle{KXM} \sim \triangle{Q'YK}$. Therefore
\[\frac{XM}{XK}=\frac{YM}{YK}=\frac{XK}{P'X} \quad \Rightarrow \quad \frac{P'X}{XM}=\frac{XK^2}{XM^2}\]

Since $XA$ is tangent to $\omega$, we have that $\triangle{AXM} \sim \triangle{AKX}$. This yields that
\[\frac{AX}{AK}=\frac{AM}{AX}=\frac{XM}{XK} \quad \Rightarrow \quad \frac{AM}{AK}=\frac{XM^2}{XK^2}\]

Applying Menelaus Theorem to $\triangle{P'MK}$ combined with the collinearity of $A$, $M$ and $K$ yields that
\[\frac{P'B}{BK}=\frac{P'X}{XM} \cdot \frac{AM}{AK} = \frac{XK^2}{XM^2} \cdot \frac{XM^2}{XK^2}=1\]

Hence $P'B=BK$ and $P'=P$. By the same argument, $Q'=Q$. Therefore $P$, $X$ and $M$ are collinear and $Q$, $Y$ and $M$ are collinear. Since $XY \| BC$, there is a homothecy sending the circumcircle of $\triangle{PMQ}$ to the circumcircle of $\triangle{XMY}$ which is $\omega$. Therefore, the circumcircle of $\triangle{PMQ}$ is tangent to $\omega$.

[hide="A rather long solution with only rudimentary concepts, similar to SnowEverywhere's solution"]
 [geogebra]1942177509cd72a72fbb98f185ec82955de2a323[/geogebra]
Let the points where $\omega$ is tangent to $AB$ and $AC$ be $X$ and $Y$, respectively, and let $MX$ and $MY$ intersect $BC$ at $P'$ and $Q'$, respectively.
Now we find a bunch of similar triangles. Since clearly $XY\parallel BC=P'Q'$, we have $\bigtriangleup MXY\sim \bigtriangleup MP'Q'$. Also, $\angle MP'Q'=\angle MXY=\angle MYA$ (since AY is a tangent)$=Q'YC$, and $\angle YQ'C=\angle MYX=\angle MXA=\angle P'XB$, so $\bigtriangleup P'XB\sim\bigtriangleup P'Q'M\sim \bigtriangleup YQ'C$. Also, $\angle XMK=\angle XLK=\angle YKL=\angle YKQ'$ since $X$ and $Y$, $K$ and $L$, etc. are symmetric about the same line, and $\angle XKM=\angle XYM=\angle KQ'Y$. So, $\bigtriangleup XMK\sim\bigtriangleup YKQ'$, and by symmetry $\bigtriangleup YMK\sim \bigtriangleup XKP'$. Also, $\angle XYK=\angle XLK=\angle XLB$ and $\angle BXL=\angle XYL=\angle KXY$, so $\bigtriangleup BXL\sim \bigtriangleup KXY$.
It's logical that we have to somehow use the fact that $A,M,K$ are collinear. By power of a point, $AM\cdot AK=AX^2$, so $\bigtriangleup AMX\sim \bigtriangleup AXK$, and similarly $\bigtriangleup AMY\sim \bigtriangleup AYK$. So now, using length ratios from a bunch of our similar triangles together, $\frac{YK}{YQ'}=\frac{XM}{XK}=\frac{AX}{AK}=\frac{AY}{AK}=\frac{YM}{YK}=\frac{XK}{XP'}$, so $\frac{XP'}{YQ'}=\frac{KX}{KY}$. Now, $\frac{BX}{BL}=\frac{KX}{XY}=\frac{XP'}{YQ'}=\frac{MP'-MX}{MQ'-MY}=\frac{MP'}{MQ'}=\frac{BP'}{BX}$, so $BX^2=BL\cdot BP'$. Since by power of a point $BX^2=BK\cdot BL$, this means $BP'=BK$, and since clearly $P'$ is not $K$ since $M,X,K$ are concyclic and not collinear, We have $P'=P$. Then since $P'X\cdot Q'Y=BX\cdot CY=BX^2=BK\cdot BL$ by one of those similarities, $CQ'=BL=CK$, so $Q'=Q$. So $MP$ and $MQ$ pass through the tangency points $X$ and $Y$ of $\omega$ to $AB$ and $AC$, so since $\bigtriangleup MXY\sim \bigtriangleup MPQ$ and $XY\parallel PQ$, the two triangles are homothetic about $M$, and therefore so are their circumcircles. So the tangent line to $\omega$ at $M$, invariant under the homothety, is also the tangent line to the circumcircle of $MPQ$, so the two circles are tangent at that point, and we're done! $\blacksquare$
[/hide]

Let $\omega$ touch $AB, AC$ at $Y, X$. Obviously $YX || KL $ Let $N_P, N_Q$ be vampire points in $\omega$ other than $Y,X$ such that $Y, P, N_P$ and $Y, P, N_Q$ are colinear. Obviously the quadrilateral $MYKX$ is harmonic.

Now, $(M, K; Y, X) = -1 = (P, K; B, P_{\infty}^{BC} ) \overset{Y}{=} (N_P, K; Y, X) \Rightarrow N_P = M$
Also $(M, K; Y, X) = -1 = (Q, K; C, P_{\infty}^{BC} ) \overset{X}{=} (N_Q, K; X, Y) \Rightarrow N_Q = M$

Hence $\Delta MXY$ is homothetic to $\Delta MPQ$, so a homothety at $M$ maps $\omega_{MXY}$ to $\omega_{MPQ}$, hence they're tangent, as desired.

Let $X, Y$ be the touchpoints of $\omega$ with $AB, AC$ and $Z$ be the infinity point on $XY$. Let $N$ be a point on $BC$ with $NM$ tangent to $\omega$. Then $NM^2=NK \cdot NL$. Now $L$ is the midpoint of $PQ$ hence it suffices to show $(PQ, KN)=-1$. Let $F=MZ \cap AB, G=MZ \cap AC$ and $H=XY \cap KA$.

[b]Claim.[/b] $M=PX \cap QY$.

(Proof) Note $-1=(MK, HA) \overset{Z}{=} (FB, XA) \overset{M}{=} (ZB, P'K)$ hence $P'=P$. $\blacksquare$

Now project $(XY, KM) \overset{M}{=} (PQ, KN)$ and we're done. $\blacksquare$

Solution:
$R,S$ are the points where the circle touches $AB,AC$. Then $-1= (M,K;R,S) \stackrel{R}{=} (MR\cap BC,K;B, P_{\infty \parallel BC}) \Longrightarrow P \in MR$. So $MRS$ and $MPQ$ are homothetic and done.

Very cool problem
[b]S[/b]olved with [b]mxlcv[/b]
[hide = solution]
Let the circle touch $AB, AC$ at points $T_1, T_2$ respectively. We first show that $M, T_1, P$ are collinear. As $MK$ and the tangents at $T_1, T_2$ concur, we have $(MK;T_1T_2) = -1$. Projecting through $T_1$ from the circle to $BC$, we obtain: \[-1 = (MK;T_1T_2) \overset{T_1}{=} ((T_1M \cap BC)K;B\infty_{BC})\]
implying that $T_1M \cap BC = P$. Similarily, $M, T_2, Q$ are collinear. Let the tangent to $M$ at $\odot(MKL)$ meet $BC$ at $X$. This results in \[-1 = (MK;T_1T_2) \overset{M}{=} (XK;PQ)\]
As $L$ is the midpoint of $PQ$ (everything is symmetric wrt midpoint of $BC$), it follows that $XM^2 = XL \cdot XK = XP \cdot XQ$. Hence $XM$ is tangent to $\odot(MPQ)$ as well, finishing the problem as $\odot(MPQ)$ and $\odot(MKL)$ are now tangent to each other.
[/hide]


[quote="easternlatincup"]Can anyone post the Korean Math Olympiad for the past few years?[/quote]
I didn't find the previous years, but i did find these. Hope they help. :)

I can't download the files...  :(

Anyway, if you want KMO papers, you can look at

http://www.kms.or.kr/home/kmo/board/bulletin_view_subject.asp?bulletinid={2DDC4816-CD42-4D29-B151-FAD6574CB0FE}&boardname=%BD%C3%C7%E8%B9%AE%C1%A6&globalmenu=0&localmenu=4&CurrentPage=1&totalSubjectNum=30&id=3817

and maybe ask somebody to translate  :)

could anyone please translate the  problems.  especially the 2005/2006 korean  olympiads.  id really appreciate it. :)


function integration LaTeX algebra polynomial calculus derivative

Let $f:[0,1]\to R$ a function such that $f$ is $C^4[0,1]$ and $\int_0^1f(x)dx+3f(1/2)=8\int {1/4}^{3/4} f(x)dx$
 Prove that there is a point $c$ in $(a,b)$ such that $f^(4) (c)=0$

Sorry for my bad Latex.I hope you do understand what I've written.

Please,can someone edit my post?? I'm sure nobody understands what I wrote.

But then how could that someone edit if he doesn' understand? Please, try to edit yourself, there are only two minor problems in the LATEX formula I think.

cezar I edited is it the correct version of your problem ?

Let $f:[0,1]\to R$ a function such that $f$ is $C^4[0,1]$ and $\int_0^1f(x)dx+3f(1/2)=8\int_{1/4}^{3/4} f(x)dx$
 Prove that there is a point $c$ in $(a,b)$ such that $f^{(4)} (c)=0$

This problem is very nice (assuming $(a,b)=(0,1)$).

In order to pursue simmetry we replace $f(x)$ by $f(x-1/2)$ so that the given equality becomes
\[\int_{-1/2}^{1/2}f(x)dx+3f(0)=8\int_{-1/4}^{1/4}f(x)dx.\]

Every polynomial with degree $\leq 3$ satisfies this equality.  So we may assume, wlog, that $f$ and its first 3 derivatives are 0 in $x=0$. With this assumption and a variable change in the RHS, the equality finally reduces to
\[\int_{-1/2}^{1/2}f(x)dx=4\int_{-1/2}^{1/2}f(x/2)dx,\]
or equivalently
\[\int_{-1/2}^{1/2}(f(x)-4f(x/2))dx=0\]

Now we prove that if $f^{(4)}$ never vanishes in $(0,1)$, then $f(x)-4f(x/2)$  never vanishes in $(0,1)$, and therefore the equality cannot occur.

To this end, let us set $g(x)=f(x)-4f(x/2)$. Since $g(0)=g'(0)=0$, by Taylor's formula we have that
\[2g(x)=g''(\xi)x^{2}=(f''(\xi)-f''(\xi/2))x^{2},\]
for a suitable $\xi$ lying between 0 and $x$.

Assume now that $f^{(4)}$ is always positive.  Then $f'''$ is increasing, and so $f''(x)$ is increasing in $(0,1/2)$ and decreasing in $(-1/2,0)$ (remember we assumed that $f''(0)=f'''(0)=0$). This proves that $g(x)>0$ for $x\neq 0$.

If $f^{(4)}$ is always negative, the argument is the same.

I read your solution Xamog,and I think there's something wrong here.You assume that $f$ and its first 3 derivates
 are 0 in $x=0$.Why? Am I missing something here?

What he did is replace $f$ by the function obtained after subtracting off its third order Taylor polynomial. That is, write $f(x)=P(x)+g(x)$ where $P(x)=f(0)+f'(0)x+\frac{f''(0)}2x^2+ \frac{f'''(0)}6x^3$ and $g(0)=g'(0)=g''(0)=g'''(0)=0.$

The identity is linear, and it is true for $P(x)$ because it is true for all polynomials of degree no greater than 3. Hence the identity is true for $g(x).$ Since $P^{[4]}(x)=0$ for all $x$, the problem then comes down to finding an $x$ for which $g^{[4]}(x)=0.$

Thank you, Kant Merryfield: this was exactly what I did.


function
limit
real analysis
real analysis unsolved

Let $ I\equal{} [a,b]$ and let $ f : I \rightarrow R$ be a function with the property that for every $ x \in I$, the function $ f$ is bounded on a neighbourhood of $ x$. Prove that $ f$ is bounded on $ I$.

assume that $ f$ is not bounded on $ I$. assume that $ f$ attains arbitrarily large values. (the case where $ f$ attains arbitrarily small values is the same.) then for every natural $ k$, there exists $ x_k$ such that $ f(x_k) > k$. since $ I$ is compact, there exists a convergent subsequence $ x_{k_i}$ of $ x_k$, and its limit is in $ I$. let $ \lim_{i\to \infty}x_{k_i} \equal{} x\in I$. this means every neighborhood of $ x$ contains infinitely many $ x_k$'s. but since $ \lim_{k\to \infty}f(x_k) \equal{} \infty$, this means that $ f$ is unbounded on every neighborhood of $ x$, a contradiction.

Using the open cover version of compactness: Each point of $ I$ has a neighborhod on which $ f$ is bounded. Finitely many of these neighborhoods cover $ I$, and a function bounded on each of a finite collection of sets is bounded on the union of these sets.


$\left(\frac 14 \right)^{-\frac 14}=$

$\text{(A)}\ -16 \qquad \text{(B)}\ -\sqrt 2 \qquad \text{(C)}\ -\frac 1 {16} \qquad \text{(D)}\ \frac{1}{256} \qquad \text{(E)}\ \sqrt 2$

[hide]$\left(\frac14\right)^{-\frac14}=4^{\frac{1}{4}}=\sqrt{\sqrt{4}}=\sqrt2\implies\boxed E$[/hide]

4everwise says: Would you like to hide your post?  :P

Sorry, I forgot. :oops:

[hide]$(\frac14)^{-\frac14}$ is the same as $4^{\frac14}$ then $\frac{4}{4^{\frac34}}$ so finally $\frac{4}{2\sqrt 2}=\sqrt 2$[/hide]

everyone knows this... its E. :P

 [/hide]


i need help with molality

i know it is
mol of solute/Kg of solvent

but i have none of that info in this problem:

[size=150]Determine the molality of a 2.78 M aluminum sulfate solution that has a density of 1.47 g/mL.
[/size]

Okay. Here's how you do it.

Say we have 1 L of this solution. It has 2.78 mol of Aluminum Sulfate, and the entire mass of the 1 L solution is 1470 g

Now, what you can do is find the mass of aluminum Sulfate in this solution by multiplying 2.78 mol by the molar mass of Aluminum Sulfate (342.15 g/mol). This is 951 g.

The solution only contains water and aluminum sulfate. So we can find the mass of water (the solvent) by subtracting 951 from 1470. This equals 519 g.

Here's what we know now
Mass of water: .519 kg
Moles of Aluminum Sulfate: 2.78

Thus, the molality equals 2.78 / .519 = 5.36 m

Hope this helps, and correct me if I'm wrong.


Hello,

I know that as the temperature of a system increases, enzymes become more active because the substrates move faster and are more likely to attach to the binding site.  But is temperature directly proportional to enzymatic activity linearly, quadratically, exponentially, logarithmically, etc?  If possible, please cite a journal article.

Temperature does increase rate of reaction to some degree, but above the enzyme's optimal temperature the enzyme denatures and reaction rate drops off, making a mound-shaped curve. Several other factors affect the relationship between temperature and reaction rate, however, including the three-dimensional structure of the enzyme. I think, then, that a simple mathematical model like a Bell curve does not describe the association between temperature and reaction rate for most enzymes. I don't have a journal article to cite, but if you would like to know more look up enzyme kinetics.


[quote="ryan22"]$22.5 degress \tan a/2 = \sin a\1 + \cos A$[/quote]

Do you want to solve
\[ \frac{pi}{8}tan\frac{a}{2}=sina+cosa \] ? :ninja:


geometry
analytic geometry
geometry unsolved

Let $ P$ be a point in the plane of a triangle $ ABC$. Lines $ AP,BP,CP$ respectively meet lines $ BC,CA,AB$ at points $ A',B',C'$. Points $ A'',B'',C''$ are symmetric to $ A,B,C$ with respect to $ A',B',C',$ respectively. Show that: $ S_{A''B''C''}\equal{}3S_{ABC}\plus{}4S_{A'B'C'}$.

I am looking forward for the solution. Could anyone help me? :?: Thank you!

I wiil post the solution I have in mind later

Denote S(ABC) stands for the area of ⊿ABC
S(AB’’C’’)=Sa   S(A’’BC’’)=Sb  S(A’’B’’C)=Sc
S(ABC)=S1  S(A’B’C’)=S2   S(A’’B’’C’’)=S
C’,B’,A’ is the midpoint of CC’’,BB’’,AA’’  respectively  =>
S(A’B’C’’)+ S(A’B’’C’)+ S(A’B’C’’)=6S2-S(A’B’C)-S(A’C’B)-S(AB’C’)=5S2+S1  (1)
Also we have
S(A’B’C’’)+1/2S(S+Sb)+S(A’B’A’’)+S(A’A’’C’’)=S                           (2)
S(A’B’’C’)+1/2S(S+Sa)+S(A’C’C’’)+S(C’B’’C’’)=S                           (3)
S(A’B’C’’)+1/2S(S-Sc)+ S(B’B’’C’)+S(B’B’’A’’) =S                          (4)
Notice S(ABC’’)=S(ACB’’) =S(BCA’’)=S1
and 4S1+Sc-Sa-Sb=S                                                    (5)
(2)+(3)+(4)  and combine with (5)  =>
3S=S1+5S2+1/2(3S+4S1-S)+S(A’B’A’’)+S(A’A’’C’’)+S(A’C’C’’)+S(C’B’’C’’)+S(B’B’’C’)+S(B’B’’A’’)                                                                (6)
Yet, S(A’B’A’’)+S(A’A’’C’’)+S(A’C’C’’)+S(C’B’’C’’)+S(B’B’’C’)+S(B’B’’A’’)=S-S2  (7)
Consider (6)(7) we have
3S=S1+5S2+S+2S1+(S-S2)   =>  S=3S1+4S2
Thus we are done!

Suppose w.l.o.g. $ [ABC]\equal{}1$ . For $ P(x,y,z)$ with barycentrical coordinates $ (x,y,z)$ w.r.t. $ \triangle ABC\ ,\ x\plus{}y\plus{}z\equal{}1$

 we have $ A'\left(0,\frac {y}{y\plus{}z},\frac {z}{y\plus{}z}\right)$ a.s.o. and $ A''\left(\minus{}1,\frac {2y}{y\plus{}z},\frac {2z}{y\plus{}z}\right)$ a.s.o. Therefore,

$ [A'B'C']\equal{}\frac {1}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\cdot \left|\begin{array}{ccc}
0 & y & z\\\\
x & 0 & z\\\\
x & y & 0\end{array}\right|\equal{}\frac {2xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}$ .

$ [A''B''C'']\equal{}\frac {1}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\cdot \left|\begin{array}{ccc}
\minus{}(y\plus{}z)  & 2y & 2z\\\\
2x & \minus{}(z\plus{}x) & 2z\\\\
2x & 2y & \minus{}(x\plus{}y)\end{array}\right|\equal{}$ $ \frac {14xyz\plus{}3\cdot\sum xy(x\plus{}y)}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}$ .

Prove easily that $ [A''B''C'']\equal{}3\cdot [ABC]\plus{}4\cdot [A'B'C']$ .


a(1) =2

a(n+1) = a(n)^2 - 1

Can we find out general term of this sequence?

[quote="Larry Yoon"]If $ a(1) \equal{}2$ and $ a(n\plus{}1) \equal{} a(n)^2 \minus{} 1$, the find a formula for $ a(n)$.[/quote]

[hide="partial solution"][hide="Lemma"]Lemma: $ a(n)$ is not a polynomial.

Proof: Let $ a(n)$ be a have degree $ p$. Therefore, $ a(n)^2$ will have degree $ 2p$, and $ a(n\plus{}1)$ will have degree $ q\leq p$. We gather to one side to get a polynomial with degree $ 2p$ that is constant, but that is impossible unless $ p\equal{}0$, and $ a(n)$ is constant, which it isn't. Therefore, $ a(n)$ is not a polynomial.[/hide][/hide]

Can't get anywhere except there... sorry. :(


In any triangle ABC  show that
$  \frac{3}{8}<\frac{h_a+m_b+m_c}{a+b+c}<1  $

$ m_a+m_b+m_c < a+b+c $ because $ m_a < \frac{b+c}{2} $
That's all I can do now.

I think this source is not unknown.It's was created by Lo and Ting.It's is as similar as Garfunkel ineq.(was solved by Gardner)and it's is hard so I haven't solved it. :(


geometry
geometric transformation
rotation
function
integration
trigonometry
calculus

Find the volume of the body obtained by rotation over $Ox$ of the graph of the following function :
$f  : [1,3] \rightarrow R_{+}$ and $f(x)=\frac{\mid arcsin(x-2) \mid}{\sqrt{2^x+4}}$

As usual we are to consider $f^2(x)+f^2(4-x)=\frac{1}{4}\arcsin^2 (2-x)$. ;)

Let $V$ the desired volume .

$V=\int_1^3 \pi f(x)^2 dx$

$=\pi \int_1^3 \frac{(\sin ^{-1}(x-2))^2}{2^x+4}dx$

Finally we are to calculate $V=\frac{\pi}{4} \int_0^1 (\sin ^{-1}x)^2 dx=\frac{\pi ^3}{16}-\frac{\pi}{2}$

[quote="adidasty"]yeah seems to be a volumes of solids of revolutions problem, seems kunny's got it.[/quote]

Of course it is.  But you can't find an antiderivative of $f^2(x)$ in terms of elementary functions, and hence you have to note the symmetry.  This is no average Calculus II problem.


How do I delete a post? Cause occasionally, I make a post twice...

Oh, I see this edit thing with the little scissors on it, but I can't figure out how to delete it...

I have the feeling this was posted somewhere...

You can only delete your post if yours was the last one in the thread. If this condition is met, then there should be a small button with an x on it to the right of the edit button. This is the delete button (you should be able to figure it out from there). :P


MATHCOUNTS
geometry
probability
3D geometry
prism
search
number theory

Hi, this summer I will be training my little brother (will be a 6th grader next year) and a few other incoming 6th graders for competitions like Mathcounts and the AMC 8. In class they follow this learning path:

6th - Pre-Algebra
7th - Algebra I
8th - Geometry

What's the best way to prepare them if they only know basic equation solving and pre-algebra concepts; and would the AoPS Mathcounts class be above their level?

I took it and I think it could have been an idgensmidgen harder...it might be a bit hard for him as they go further in-depth

The MathCounts Problem Series, is predictably, for MathCounts. If that's what you guys are interested in preparing for specifically, I say go for it.

But as for books, I say go for Introduction to Algebra, then Counting/Probability, then Number Theory, reading Geometry concurrently with everything. If you don't want to buy all those books (they aren't exactly cheap, but it's worth it anyhow), AoPS 1 is a good alternative. Although for a sixth grader, the Introduction series is a little easier to read, and has a wider range of problems.

I say give them a bunch of old MathCounts competitions and find out what they dont know and what they do know and teach them the stuff they need to know.

That's all the advice I can give right now.

[quote="Aznwithabrain"]My Advice:

There are three goals for your brother:
Make Chapter
Make State
Make Nationals

If Chapter:
Get him to do all School Packets
Make sure if he does them again
He'll get them all right

If State:
Get him to do all School, Chapter, and State packets
Make sure if he does them again
He'll get only a few wrong

If Nationals:
Get him to do all School, Chapter, State, and National packets
Make sure if he does them again
He'll get them all right (except for Nationals. You can go easy on that but he should at least feel comfortable about doing a national packet)

AOPS Volume 1 is HIGHLY RECOMMENDED
Supplemental AOPS books should be bought if you have the money

If you can't get the money to buy packets:
I would recommend the "AllTime Greatest MATHCOUNTS Problems"[/quote]

I love how this kid always responds to things by making some really long list of EXACTLY how to do stuff, and says it in the way that it's the only way to do stuff.

I really don't see the point of doing "packets," (why not just call them tests?), twice, why not just re-do the problems you didn't know how to do...?

most importantly, learn the concepts of problems you miss in the tests ("packets" :P )

jorian

I'm a 6th grader and I took MathCounts, MOEMS, and the Amc 8 this year.

For MathCounts, our coach made us do a level highers, so for Chapter, we did State...

I would concentrate mostly on Geometry and Probability, a lot of thos questions come up, and our most people's weak points.

yeah, I guess I'm repeating what jhredsox said, but completely learn how to do EVERY problem. Even if you got it right. Iguess I'm quoting aznwithabrain also.

For a sixth grader? Umm. That is hard. Have them do the MC related things in Algebra I, then speed it up and read AoPS vol 2. Finally, a little Geometry should work well (angle formulas, etc.). Make sure that they learn all of this at least 1 month before the competition date. For that month, have them take previous chapter/state tests. Grade them and spot check any concepts before the competition.

[quote] I love how this kid always responds to things by making some really long list of EXACTLY how to do stuff, and says it in the way that it's the only way to do stuff.

I really don't see the point of doing "packets," (why not just call them tests?), twice, why not just re-do the problems you didn't know how to do...? [/quote]

I don't see how his post sounded like it was "the only way to do stuff" any more than yours did. You're just asking for a flame war, please use better judgement when criticizing people. Also, when you refer to aznwithabrain as "this kid" it comes off as EXTREMELY condescending.

It's obvious that you two have different styles. L33t's is book-based, azn's is practice-based. Is your 6th grade better at learning through books or by "hands on" practice tests?

[quote="Fanatic"]Is your 6th grade better at learning through books or by "hands on" practice tests?[/quote]

Well I think most sixth graders would benefit in MC from books. Hands on practice tests on stuff you haven't learned (factorizations, combinations, permutations) would be considered unbenefitial. Of course after a reading a collection of past problems related to the topic would help  :lol: .

Not necessarily.

Remember "way back when" when AoPS books and the AoPS community didn't exist?

people managed, quite well in fact.

Personal recommendation is the test taking approach. But obviously not everyone would agree

Warm ups, workouts, and the countdown round are your friends early on.

People also tend to forget that mathcounts is more of a speed competition than anything. That's why taking real tests (always timed reasonably, no matter what the quantity of problems is) is, to me, more beneficial.

Agreed
I think developing speed is important
Warm-ups and Workouts really help you alot

then again
I suck at mathcounts...

may i suggest, do what i did.

when you get a problem wrong or are stuck on it, keep it in your head! don't go to the answer sheet for immediate help!!! rather, keep on persuing it and think to yourself you can get 1 hint from someone else if you really need it, and so on. this way you learn the subject rather than memorizing it.

jorian

[quote="Fanatic"]Remember "way back when" when AoPS books and the AoPS community didn't exist?

people managed, quite well in fact.

[/quote]

Ah, but we should use any advantage we've got, as long as it's beneficial. You are right, but it would be silly to purposely avoid AoPS materials just because people didn't use them in the past.

But I see your pointk, yes.

Wow, thanks for all of the replies and suggestions. It's hard to say what they will learn better by as they really haven't done many math competitions. I would love to mix both approaches as I have AoPS Volumes I & II as well as access to many of the chapter, states, and nationals test through various website; however, I would like to use warmups & workouts more towards the beginning because giving them the chapter+ tests without them knowing too much might be intimidating. The problem is their middle school switched coaches and a various chain of events occured which basically rendered us with only the 2006-2007 Mathcounts Handbook; not to mention the coaches are relatively new and busy so I'm trying my best to help them out by giving them people to work with and helping them coach.
I'm certain my parents wouldn't mind buying AoPS Intro books (such as Algebra and Geo), but I'm just worried whether the kids (it actually will most likely turn out to be just my brother) will be able to handle the things and in the given time constraints; after all their elementary school math is rather weak (the Elementary school competition is watered down and cheesy) and they're 6th graders moving into pre-algebra. Maybe I'm underestimating them?

Bleh, I screwed up that part of my post. I'm trying to give something away without giving it away because there's still one month left and someone could still take my strategy [ok, my brother's strategy... and that has stock] and  :mad:

ahhhhhhhhhhhhhhh /dies.

[quote="7h3.D3m0n.117"][quote="Aznwithabrain"]Agreed
I think developing speed is important
Warm-ups and Workouts really help you alot[/quote]

They don't help you if you don't know the material  :wink: . To a sixth grader, finding the units digit of like $2007^{2007}$ could be annoying.

First, read up on MC related topics. Then practice.[/quote]

Yeah, once u get really good, u start to see patterns, and then youll find out that the answers

[hide]$3$[/hide]

[quote="kyyuanmathcount"]Welll, I doubt Volume II would help much if you haven't learned pre-algebra.[/quote]

Pre-algebra? All pre alg is is solving equations for x and other stuff. Learning pre alg in 4th or 5th grade would creatly speed things up. Then in 6th grade take Alg I while reading AoPS vol 1 and 2.

[quote="7h3.D3m0n.117"][quote="Aznwithabrain"]Agreed
I think developing speed is important
Warm-ups and Workouts really help you alot[/quote]

They don't help you if you don't know the material  :wink: . To a sixth grader, finding the units digit of like $2007^{2007}$ could be annoying.

First, read up on MC related topics. Then practice.[/quote]

Highly true, but there are enough problems that don't involve that kind of thing to do reasonably well in 6th grade.

I'd try to get some competition going between the 6th graders so they all want to do better than the others (i.e. motivation) or prizes or something like that.

I know when I was in 6th grade, I struggled to get 13 right on chapter sprints  :rotfl: but I managed to make nationals without doing any nationals practice.

Give him a round that should challenge him (preferably with a wide range of problems), then see which types of problems he misses & work from there.

[quote="kyyuanmathcount"]Welll, I doubt Volume II would help much if you haven't learned pre-algebra.[/quote]

Do you really need to learn pre alg before you analyze patterns in modular arithmetic? Sorry, but I find pre alg to be one of the most boring subjects of math ive ever been taught....it didn't evn help me understand the MC stuff in 6th grade  :maybe:

I agree with rcv. It seems pretty futile to give a 6th grader the difficult AOPS Volumes. They should be given at maximum Workouts and Warmups or maybe easy competitions.

it would be useless to give them hard AoPS books, without a tutor to help understanding, but i think they could be given more questions than Warmups and workouts, etc.   He/she should definitely be exposed to as much material as possible.

Sixth graders don't think like us. Ok well when I was in sixth grade I didn't think like I do now  :lol: . But I liked getting answers correct.

At this point, I think building interest & motivation is extremely key... make sure Mathcounts doesn't lose its FUN that early. The torture is in high school XD and competing against people around the nation.

Yes, to build motivation, they should be given easy, but various problems dealing with with different subjects.

If you're good at giving pep talks, they should be motivated. :roll:
but thats unrealistic

Make sure they DO NOT BURN OUT
If you have a buzzer system, let all the kids participate in some quick recall with countdown questions.
This is probably the only reason I elected to stay on math team in 6th grade (I was really dumb back then...even though I made the A Team)

Although all this is coming from somebody that has completely failed his goals, I'm trying to give advice based on what I would've done differently (I only found out about AOPS in 7th grade)

i would start with the intro to algebra, counting and prob, number theory, and the geometry books. that would really help them to understand the different concepts. (if necessary, for extra, go with the aops volume 1).

then, i would make them take every single mathcounts test (school, chapter, state, national) from like fifteen years ago. make sure they understand and know how to do each problem.

doing all of this could probably guarantee you a spot in nationals....cause i have a friend who only did the previous years mathcounts exams (he knew every problem though) and he made it to nationals

Definately thorrougly do Algebra I.

If you don't have enough money to buy the book off AoPS, then ask an 8th grade math teacher (or any other math teacher that teacher algebra) to lend you one for the year. Thats what I did, and my knowledge increased exponentially.

[quote="kyyuanmathcount"]Welll, I doubt Volume II would help much if you haven't learned pre-algebra.[/quote]

True...but Volume I pretty much prepares for Volume II, and it explains everything. No, i mean EVERYTHING!!!

For a incoming sixth grader, Volume I is best. It teaches a lot of the strategies, etc. and prepares for pretty much everything. After  reading AoPS, you will be a monster at math. :)

Ok maybe I'm just VERY. VERY. VERY. VERY. VERY. VERY dumb

But AoPS books KILLED me. I had taken ALGEBRA TWO (but not geo), it was the summer of 7th grade, I was 5th in Indiana but AoPS volume one DESTROYED me.

DO NOT GIVE BIG Math BOOKS TO A 6th GRADER GOING INTO PRE-ALGEBRA UNLESS he or she is VERY DEVOTED.

I think some of us are forgetting that the motivation isn't quite there yet when we're in 6th grade. Well maybe it was for some of you guys, but we're at a math forum... kinda attracts the elites.

when i was in 7th grade, i didn't understand any of the ways to get the geometry problems on sprint and target. it was very frustrating. so i may suggest to give your students some basic concepts, and work further into depth with geometry topics, as they will not learn *any* of that *in* school (which i had to learn by myself). make sure they are not expecting too much, and keep on telling them how good you have to be just to take a step in the mathcounts room (only us 8th graders have reasons to whine). just have fun! if you're doing it because you feel like you must come in 34th at states, then it will never work out for you. just try your best, and don't get brain-washed with this stuff and get it too deep in your head (what i did this year). make sure if you do want to win, etc., that you are commited. i would not recommend the aops books at first, or even until about 8th grade, unless they are way, way, past their grade level (and can obviously persue those problems). focus mainly on worksheets, and problems focusing in on what you know they need help on. challenge them with the problems, but do not go crazy. i say this, as you *must* explain *every* problem they did not get/did not understand (even if it's just re-inforcement and they just accidentally messed up). if you're the coach you must also be commited. make sure you help these kids and go *above* and *beyond*, espeacially if these kids aren't going to succeed (well, being in mathcounts is succeeding, rather, succeeding with in mathcounts. just wanted to get that off my toungue.) within mathcounts without any supervision and help. if you need more, i can help more.

jorian

I think http://www.yashar.biz/mathcounts has some workouts/warmups.

[quote="GrantCBHS"]Once again, thank you for all of your suggestions!

Addressing the issue of motivation and winning first, yes, that is a very important issue that I think I'll need to get across first. While winning is good, if you don't win, so what make sure you learned something and enjoy doing Mathcounts (not just winning it); and I intend to go a rather lax pace so they don't burn out. In terms of hard question, as long as I don't give them too many hard questions I believe they'll take the challenge fine.

About AoPS, in my opinion I'm more then qualified to teach them most of the stuff in Volume I and a lot in Volume II (I'd have to read through it before hand to make sure, but I'm competing in Geometry & Algebra II for MAO this year and have been doing very well). So if I am going to use AoPS I would definitely be their teaching the stuff to them and awaiting questions. Having gone through about 9 chapters of AoPS Volume I, it's evident that a lot of the stuff AoPS is hitting is actually Algebra II, so I might hold back on some of the material there and skip to the Geo. chapters.

With Warmups & Workouts, like I said I don't really have access to any but this year's ones. Is there a site that has previous ones? I want to use these first before going into the actual tests (like you've been suggesting).

I'm considering getting AoPS Intro to Algebra and Intro to Geo, can anyone help verify whether these books are slow and thorough enough for a 6th grader to understand on his own (I intend to coach them, but I have to prepare for my competitions and it'd be good if they did a little self-study).[/quote]

I think goodlymath has some warm-ups and workouts
though the site hasn't been working very well for me (due probably to my bad internet connection)

I have Intro to Geo., and I liked the book.
It really improved my geometry skills
You can skip the proof problems though (they're not needed for MATHCOUNTS)
Although reading over the solutions of the proofs wouldn't hurt

I also recommend Number Theory and Counting and Probability
Two key concepts that are in MATHCOUNTS but not in the normal school curriculum.

Is the actual url: http://www.goodlymath.com ? Or is Goodly Math, just the title.

AoPS book 1, the mathcounts problems books, and the class will all help.
Also, practice countdown with him to build up his speed.

i meant worksheets from anywhere...just search like probability worksheets

jorian

[quote="rcv"][quote="GrantCBHS"]...I'm just worried whether the kids (it actually will most likely turn out to be just my brother) will be able to handle the things and in the given time constraints; after all their elementary school math is rather weak (the Elementary school competition is watered down and cheesy) and they're 6th graders moving into pre-algebra. Maybe I'm underestimating them?[/quote]
I hope you realize the suggestions you are getting from MATHCOUNTS students who are still hanging around this Forum, in April, are not from your average Mathlete.  For the most part, these are the students headed for Nationals.  The top 228 from a starting pool of a hundred thousand.

If you are expecting three incoming sixth graders, you are going to need to guage their abilities.  Don't be surprised if they don't know basic formulas, such as the surface area of a rectangular prism or the area of a triangle.  Don't expect them to know *any* number theory, such as the divisibility rule for 9.  Don't even expect them to be able to read and interpret a graph.

I will strongly suggest you avoid AoPS Volume II.  It is likely to be so far beyond most incoming 6th graders, you are more likely to frustrate them than to help them.  I would also suggest you not start with tests.  Tests are a valuable resource, and there are too few of them to start with those.  They also test a wide variety of topics.  Most incoming 6th graders -- even pretty bright ones -- will struggle beyond the first page or so of a school or chapter competition.  You must give them some coaching.

I would suggest you start with the Handbook.  Review the problem-solving strategies with your students.  (Students of that age need to be reminded (frequently) of the problem solving strategies.)  After that, I would start with WarmUp 1.  You won't get three perfect papers.  Keep an eye on them as they do their worksheets -- you should get valuable information on their abilities.  If they are stuck, be prepared to help them.  You might ask them to work independently when they start, but allow them to work together after they've each attempted many of the problems.  Are you going to do the problems on the board?  Are you going to ask them to do the problems on the board?  Are you going to give lectures?  No two groups of students are the same.  Those who are self-motivated, and very, very smart, are the ones who will benefit from AoPS Volume I, although you might pick out specific problems and solution strategies to supplement the worksheets.

It's easier to ramp up the difficulty if you've aimed too low than it is to lower the difficulty if you've aimed too high.  If possible, ask an experienced middle school gifted math teacher to observe one of your first sessions.  They should be able to help you zero in on just where your students are, and what they need.  There really isn't a magic bullet to train a future Mathlete.  Remain flexible.[/quote]
I basically agree, but there might be a bit of exagerration on some part.  The general idea gets across though.

Ah yes; that would be me. =D
In chapter, I got each problem in less than 2 seconds. In chapter, I missed every problem because of careless errors. In chapter, I was screwed over. XD

But then at state, I tried going slow, and was eliminated in two questions.  :o

[quote="Karth"]One mistake I see people make in Countdown is they look at the first two numbers, words that stand out, and multiply/add/etc. this is a bad habit that may continue into high school tournaments. so, practice speed, but also practice accuracy a lot.[/quote]
My friend always skimming through lokking for the numbers, and he missed the number 4 spelled four.
So always read the problem :P


How do you show that $\mathbb{R}^{2}$ has the power of the continuum?

It suffices to find two injective maps: $\mathbb R\to\mathbb R^{2}$ and $\mathbb R^{2}\to\mathbb R$. The first one is easy. For the second, represent the numbers $x,y$ in $(x,y)\in\mathbb R^{2}$ by decimal expansions and interlace them to get a decimal expansion of some real number.


calculus
integration
trigonometry
calculus computations

You can use $\sin^{2}x=\frac{1-\cos 2x}{2}$.

$4\sin^{2}x=2\cdot 2\sin^{2}x=2(1-\cos 2x)$ and for $k\ne 0$, $\int\cos kx\ dx=\frac{1}{k}\sin kx+\mathcal C$ a.s.o.


Who have got Junior Balkan Math Olympiad shortlists I have a book but there are not all of them.
(Pls dont send 2002 and 2006)


Given that $a>b>c>0$, find the distance from the origin to the plane passing through $A(a,a^{2},a^{3}), B(b,b^{2},b^{3})$, and $C(c,c^{2},c^{3})$.

[quote="boxedexe"]Find the equation of the plane $Ax+By+Cz+D=0$, and use the formula $d=\frac{|D|}{\sqrt{A^{2}+B^{2}+C^{2}}}$, where $d$ is the required distance.[/quote]

Yeh...that can't be more obviously. But the algebra manipulation is a mess for me. :(

In order to find $A,B,C,D$, you can find $\vec{AB}\times \vec{AC}$.


conics
hyperbola
Asymptote
geometry
limit
symmetry
analytic geometry

For $ a>0$, the circle $ C$ centered the point $ A(0,\ a)$ has exactly two intersection points with the hyperbola $ H: x^2\minus{}y^2\equal{}1$. Let $ P$ be the intersection point which is in the first quadrant among the two curves.Let $ Q,\ R$ be the intersection points of $ C, H$ and the asymptote $ y\equal{}x$.Denote $ S(a)$ as area bounded by two segment $ AQ,\ AR$ and the arc $ QPR$ of $ C$, find the limit $ \lim_{a\rightarrow \plus{}\infty} \frac{S(a)}{a}$.

I'll start by changing the problem slightly; in the end, I'll come back to the original version. Define $ T(a)$ to be the area of $ \triangle AQR.$ I will show that $ \frac{T(a)}{a}\equal{}\frac{\sqrt{2}}2.$

By symmetry about the $ y$ axis, any point of intersection of $ C$ and $ H$ in the first quadrant must have a corresponding reflected point of intersection in the second quadrant. Hence the only way there can only two points of intersection is for $ C$ and $ H$ meet at a point of common tangency, which I'll call $ M.$

Let $ M$ be $ (b,c).$ The slope of $ H$ at $ M$ is given by implicit differentiation as $ \frac{dy}{dx}\equal{}\frac{x}{y}\equal{} \frac{b}{c}.$ The slope from $ A$ to $ M$ must be perpendicular to the tangent, since it is a radius of the circle. Hence $ \frac{c\minus{}a}{b}\equal{}\minus{}\frac{c}{b}$ from which $ a\equal{}2c.$ To get the radius of the circle, we compute the distance from $ A$ to $ M:$ $ r^2\equal{}b^2\plus{}(2c\minus{}c)^2\equal{}b^2\plus{}c^2\equal{}c^2\plus{}1\plus{}c^2\equal{}2c^2\plus{}1.$

That makes the equation of the circle $ C$

$ x^2\plus{}(y\minus{}2c)^2\equal{}2c^2\plus{}1.$
$ x^2\plus{}y^2\minus{}4cy\plus{}4c^2\equal{}2c^2\plus{}1$
$ x^2\plus{}y^2\minus{}4cy\plus{}2c^2\equal{}1.$

Now intersect that with $ x\equal{}y$ to find the points $ Q$ and $ R.$

$ 2y^2\minus{}4cy\plus{}2c^2\equal{}1$

$ (y\minus{}c)^2\equal{}\frac12,$ so $ y\equal{}c\pm\frac1{\sqrt{2}}.$

That gives the points $ Q$ and $ R$ as $ \left(c\minus{}\frac1{\sqrt{2}},c\minus{}\frac1{\sqrt{2}}\right)$ and $ \left(c\plus{}\frac1{\sqrt{2}},c\plus{}\frac1{\sqrt{2}}\right)$

Note that $ QR\equal{}2.$ Furthermore, the midpoint $ N$ of $ QR$ is $ (c,c).$ Note the the segment $ \overline{AN}$ from $ (0,2c)$ to $ (c,c)$ has slope $ \minus{}1$ and is perpendicular to $ x\equal{}y.$ That makes $ \overline{AN}$ an altitude of triangle $ AQR$ corresponding to base $ \overline{QR},$ and $ AN\equal{}\sqrt{2}c.$ Hence the area of $ \triangle AQR$ is $ T(a)\equal{}\frac12\cdot2\cdot\sqrt{2}c\equal{}\sqrt{2}c.$

Hence, $ \frac{T(a)}{a}\equal{}\frac{\sqrt{2}c}{2c}\equal{}\frac{\sqrt{2}}{2}.$

Now we return to the original problem, and consider the area $ S(a)$ of the circular sector. Let $ \theta\equal{}\angle QAR$ and $ r\equal{}\sqrt{2s^2\plus{}1}$ be the radius of the circle. We can write the triangle area as $ T(a)\equal{}r^2\sin\left(\frac{\theta}2\right)$ and the sector area as $ S(a)\equal{}\frac12r^2\theta.$ Then $ \frac{S(a)}{T(a)}\equal{}\frac{\theta/2}{\sin(\theta/2)}.$

We note that as $ a\to\infty,$ we have that $ r^2\equal{}2c^2\plus{}1,$ which tends to $ \infty.$ Since the area $ T(a)$ grows only as a linear multiple of $ c,$ then we must have that $ \sin\left(\frac{\theta}{2}\right)$ tends to zero and hence $ \theta\to0.$ That means that

$ \lim_{a\to\infty}\frac{S(a)}{T(a)}\equal{}\lim_{\theta\to0} \frac{\theta/2}{\sin(\theta/2)}\equal{}1.$

Since $ \frac{T(a)}{a}\equal{}\frac{\sqrt{2}}2$ is constant, we then have that $ \lim_{a\to\infty}\frac{S(a)}{a}\equal{}\frac{\sqrt{2}}{2}.$

That's correct answer.

How difficult do you think the problem is?

Difficult for whom?

On the one hand, I can claim that I used no ideas that lie beyond a typical American Calculus I class (or AP Calculus AB, which is more or less the same thing). On the other hand, if I were to actually assign this problem to a typical Calculus I class of 30 or 40 students, I would expect to see zero correct answers (and rather few serious attempts.)

Of course there are dozens of people who post here in the calculus section of AoPS/Mathlinks who genuinely understand calculus and aren't intimidated by a lengthy computation; all of most of them should have been able to do this.

I think I had two pieces of insight: the first was to work based on the the point of tangency $ M$ rather than directly on $ A$ - I think that reduced the algebraic complexity. The second insight was that the area of the triangle was easier to deal with than the area of the circular sector, but the triangle was nearly all of the sector. I was going to say something like "I ground through the algebra, doing what I had to do" but that wouldn't really apply to those two steps.

Thank you for your replies, Dr. Kent Merryfield.

An alternative way to start the problem: Fix $ a>0.$ Circle $ C$ has equation $ x^2\plus{}(y\minus{}a)^2\equal{}r^2$ for an $ r$ yet to be determined. Intersect this with $ x^2\equal{}y^2\plus{}1$ to get the equation $ 2y^2\minus{}2ay\plus{}a^2\minus{}r^2\plus{}1\equal{}0$ or $ y^2\minus{}ay\plus{}\frac{a^2\plus{}1\minus{}r^2}2\equal{}0.$

In order for this to have exactly one solution (for $ x>0$), this must be a perfect square. It must be the equation $ \left(y\minus{}\frac{a}2\right)^2\equal{}0.$ That in turn requires that $ \frac{a^2\plus{}1\minus{}r^2}2\equal{}\frac{a^2}4$ which simplifies to $ r^2\equal{}\frac{a^2}2\plus{}1.$ And we see that the point of tangency has $ y\equal{}\frac{a}2.$

Once we have that established, the rest of the problem will proceed more or less as in my previous post


inequalities geometry circumcircle trigonometry

If the triangle ABC is acute prove that CosA ,CosB ,SinC form the sides of another triangle

Well, the phrase another triangle is a little vague.

But, if you mean a triangle incongruent or one not similar to ABC, then by SSS and AA there doesn't exist one.

[quote="Magnara"]Well, the phrase another triangle is a little vague.

[/quote]

No it isn't.

If we can prove

cosa + cosb >= sinc

and the 2 similar triangle inequalities, then the side lengths cosa, cosb, sinc can form another triangle.

[hide]

sin(c) = sin(180-(a+b))=sin(a+b)

sin(c)=sin(a)cos(b)+cos(a)sin(b)<cos(a)+cos(b)

and

WLOG

for a,b<90

cos(b)>(1-sin(b))

(1+sin(a))>1>cos(a)

(1+sin(a))(cos(b))>(1-sin(b))(cos(a))

sin(a)cos(b)+sin(b)cos(a)+cos(b)>cos(a)

sin(a+b)+cos(b)=

sin(c)+cos(b)>cos(a)

[/hide]

[quote="Magnara"]Oh.  Woah.  I misunderstood that pretty badly.[/quote]

Don't worry, because I think there's a typo in the problem statement.

That is, it should be, in acute triangle ABC, prove that cosa, cosb, sinc can form a triangle.

Find the point D  such that  DA  is perpendicular to CA and DB is perpendicular to  CB. Now connsider the triangle  DAB . Now DAB is obviously a traningle (if A and B are both less than pi/2) with the  angles (pii/2-A),(pi/2-B),(pi-C)  and hence the sides  proportional to  (cos A, cos B, Sin C) obviously forms a traingle..
QED

In fact you can modify the problem: if neither A  nor B is pi/2  then for any such   triangle  (need not be acute)
|cos A|, |cos B|,Sin C will form a traingle ..

[quote="Gyan"]Find the point D  such that  DA  is perpendicular to CA and DB is perpendicular to  CB. Now connsider the triangle  DAB [...][/quote]

Nice observation! The point D is actually the other end of the diameter CD of the circumcircle of the triangle ABC.

Another way to look at this:

The triangle $\Delta{HBA}$ has angles $90^{\circ}-A$, $90^{\circ}-B$, $180^{\circ}-C$, where $H$ is the orthocenter of the triangle of $\Delta{ABC}$.

So the three sides are proportional to $\sin(90^{\circ}-A), \sin(90^{\circ}-B), \sin(180^{\circ}-C)$. That is, they are proportional to $\cos{A}, \cos{B}, \sin{C}$.

Therefore $\cos{A}, \cos{B}, \sin{C}$ form sides of a triangle similar to the triangle $\Delta{HBA}$.

to prove CosA +CosB>SinC we can use the identity

CosA)^2 + (CosB)^2 +(CosC)^2 +2CosACosBCosC=1 as follows:

Since all quantities are positive

CosA)^2 +(CosB)^2 +2CosACosB >1-(CosC)^2 which is

(CosA+CosB)^2 >(SinC)^2...............

It seems that the usefulness of the cosine  identity is not to be underestimated


[quote="123456789"]Mods, please remove this problem. It appeared on this month's MML, and some people have not taken the MML yet for this month.[/quote]

my bad


Hi everybody, need your help to prove something :

Let $ T\in\mathcal{L}(\textbf{V})$ be a normal operator, ie, $ TT^* \equal{} T^*T$.

Prove that $ \ker(T) \equal{} \ker(T^k), \forall k \in\mathbb{N}$.

Thanks

P.S Maybe one may need some of the following results: (some, only true if $ T$ is a normal operator)
1) $ \ker T^* \equal{} (im T)^{\perp}$
2) $ (T^*)^k \equal{} (T^k)^*T, \forall k \in\mathbb{N}$
3) $ T^kT^*\equal{} T^*T^k, \forall k \in\mathbb{N}$

Ooh I think I have it.

First $ \ker T \equal{} \ker T^*$ cause $ \ker (T \minus{} \lambda*Id_V) \equal{} \ker ( T^* \minus{} \bar{\lambda}*Id_V), \forall \lambda \in \mathbb{R}$ and just set $ \lambda \equal{} 0$.

Let $ \vec{v} \in \ker T^2 \Rightarrow T\vec{v} \in \ker T \equal{} \ker T^* \equal{} (Im(T)^{\perp})$.

So we have $ \forall \vec w \in \textbf{V} \Rightarrow \langle T\vec v, T \vec w \rangle \equal{} 0$.

Setting $ \vec w \equal{} \vec v$ it becomes $ \| T\vec v\| \equal{} 0 \Rightarrow T\vec v \equal{} 0 \Longrightarrow \vec v \in \ker T$.

So now $ \vec v \in \ker T^2 \Longrightarrow \vec v \in \ker T$, (of course $ \vec v \in \ker T \Rightarrow \vec v \in \ker T^2$ is easy).
And then by induction $ \forall k$...


$n^{7}-14n^{5}+49n^{3}-36n$
without the factor theorem


Hikaru79 why make it so complicated?

[hide]Write $11^{n}-4^{n}=(11-4)\left(\sum_{k=1}^{n}(11)^{n-k}(4)^{k-1}\right)$. Now proceed with induction.[/hide]

Ta da! ;)

[quote="Hikaru79"]I'm assuming you want this done by induction.

First, we prove for $n=1$
Then, $11-4=7=(7)(1)$.

Then, we assume true for $n=k$:
$11^{k}-4^{k}=7a$
This implies that $11^{k}=7a+4k$ and $4^{k}=11^{k}-7a$

Now we must prove true for $n=k+1$ \[11^{k+1}-4^{k+1}\]  \[=(11)(11^{k})-(4)(4^{k})\]  \[=(11)(7a+4^{k})-(4)(11^{k}-7a)\]  \[=77a+11(4^{k})-4(11^{k})+28a\]  \[=105a+44(4^{k-1}-11^{k-1})\]  \[=105a+44\left(\frac{7a}{k}\right)\]  \[=7\left(15a+\frac{44}{k}\right)\] So there is a factor of 7. QED.

I'm sure there are many more elegant ways to do this, but this is what I found. :)[/quote]

can you explain how  did $(4^{k-1}-11^{k-1})$ became $\frac{7a}{k}$

[quote="guile"]Prove that $\forall n \in N, 11^{n}-4^{n}$ is divisible by 7.[/quote]
can't you do it by binomial theorem?:
$11^{n}-4^{n}=(7+4)^{n}-4^{n}$
all terms of the expansion contain a power of $7$ except for the last one, which is canceled out by the $-4^{n}$
 :?:

[quote="minsoens"]can't you do it by binomial theorem?:
$11^{n}-4^{n}=(7+4)^{n}-4^{n}$
all terms of the expansion contain a power of $7$ except for the last one, which is canceled out by the $-4^{n}$
 :?:[/quote]

But this wouldn't be [b]induction[/b] :P

If you wanted a non-inductive proof, the only thing you would need is this [hide="very important lemma"] $a \equiv b \bmod m, c \equiv d \bmod m \Rightarrow ac \equiv bd \bmod m$ [/hide]

[quote="t0rajir0u"]If you wanted a non-inductive proof, the only thing you would need is this [hide="very important lemma"] $a \equiv b \bmod m, c \equiv d \bmod m \Rightarrow ac \equiv bd \bmod m$ [/hide][/quote]

Or perhaps
[hide="This not-so-important-anymore :blush: lemma"]
$a^{n}\equiv b^{n}\bmod m$ if and only if $a \equiv b \bmod m$
I don't know how to do the biconditional arrow thingy.
I hope this is true... :?
[/hide]

[quote="mathnerd314"]$a^{n}\equiv b^{n}\bmod m$ if and only if $a \equiv b \bmod m$
I don't know how to do the biconditional arrow thingy.
I hope this is true... :?[/quote]

\leftrightarrow $\leftrightarrow$
\Leftrightarrow $\Leftrightarrow$
\Longleftrightarrow $\Longleftrightarrow$
\longleftrightarrow $\longleftrightarrow$

It is most certainly not true.  Consider $a = 2, b = 4, n = 3, m = 8$.

Or $a = 1, b =-1, n = 2$ for any $m \geq 3$.

simple non-inductive:
[hide=" :D "]
$11^{n}-4^{n}\equiv 4^{n}-4^{n}\equiv 0 (\bmod\ 7)$
[/hide]

[quote="Hikaru79"]I'm assuming you want this done by induction.

First, we prove for $n=1$
Then, $11-4=7=(7)(1)$.

Then, we assume true for $n=k$:
$11^{k}-4^{k}=7a$
This implies that $11^{k}=7a+4k$ and $4^{k}=11^{k}-7a$

Now we must prove true for $n=k+1$ \[11^{k+1}-4^{k+1}\]  \[=(11)(11^{k})-(4)(4^{k})\]  \[=(11)(7a+4^{k})-(4)(11^{k}-7a)\]  \[=77a+11(4^{k})-4(11^{k})+28a\]  \[=105a+44(4^{k-1}-11^{k-1})\]  \[=105a+44\left(\frac{7a}{k}\right)\]  \[=7\left(15a+\frac{44}{k}\right)\] So there is a factor of 7. QED.

I'm sure there are many more elegant ways to do this, but this is what I found. :)[/quote]

This proof doesn't look right to me... Just to disprove it, try $k=1$. According to his proof, we get that $11^{1}-4^{1}= 7 = 7a$, so $a=1$. But at the end it states that $105 = 11^{2}-4^{2}= 7\left(15a+\frac{44}{k}\right) = 7(15+44) = 413$. That $\frac{44}{k}$ looks suspicious anyway because suppose $k\not| 44$, then $11^{k+1}-4^{k+1}$ would not be an integer.

Perhaps: \[\begin{aligned}4^{k-1}-11^{k-1}& = \frac{4^{k}}{4}-\frac{11^{k}}{11}= \frac{4^{k}}{4}-\frac{7a+4^{k}}{11}= 4^{k}\left(\frac{1}{4}-\frac{1}{11}\right)-\frac{7a}{11}\\ & = \frac{7\cdot 4^{k-1}}{11}-\frac{7a}{11}= 7\cdot\left(\frac{4^{k-1}-a}{11}\right)\end{aligned}\] Therefore, you would get that $11^{k+1}-4^{k+1}= 7\cdot\left(11a+4^{k}\right)$: Tadaaaaaaa!  :)

However, without induction $11^{n}-4^{n}\equiv 4^{n}-4^{n}\equiv 0\mod{7}$  :P


Solve $ \frac{1}{y} \frac{dy}{dx} \equal{} x^{n\minus{}1}y \plus{} \frac{x^{n\minus{}1}}{1\plus{}x^{n}}$ where $ n$ is an integer greater than $ 1$ and where $ y\equal{}1$ when $ x\equal{}0$.

hello, this is a Bernoulli equation with the solution with the solution
$ y(x)\equal{}\minus{}\frac{n\plus{}1}{1\plus{}x^n\minus{}n(1\plus{}x^n)^{\minus{}\frac{1}{n}}\minus{}2(1\plus{}x^n)^{\minus{}\frac{1}{n}}}$.
Sonnhard.

Thanks Sonnhard, but how do we solve it?

hello, see here
http://mathworld.wolfram.com/BernoulliDifferentialEquation.html
Sonnhard.


When you change the forum to another style, write a new message, then change it back, would the post that you wrote come up as a new post?
It happens to me a lot, and it's kind of confusing.

Oh, and I also noticed that the same thing happens when you move or split a thread into another forum right after you write a new message (which is written in the moved/split thread).


Find all functions $ f$ : (0,+oo) ->(0,+oo) such as $ f(f(x)) \plus{} f(x) \equal{} 2x$  ,$ x > 0$

[hide="Solution"]Suppose $ f(x) \equal{} x \minus{} k$ for some positive $ x,k$.  Then we have $ f(x \minus{} k) \plus{} x \minus{} k \equal{} 2x$, so $ f(x \minus{} k) \equal{} x \plus{} k$.  But then $ f(f(x \minus{} k)) \plus{} f(x \minus{} k) \equal{} 2(x \minus{} k)$, so $ f(x \plus{} k) \plus{} x \plus{} k \equal{} 2x \minus{} 2k$, and we get $ f(x \plus{} k) \equal{} x \minus{} 3k$.  Therefore, given $ f(x_{1}) \equal{} y_{1}$, where $ x_{1} > y_{1}$, we can find $ x_{2}$ such that $ f(x_{2}) \equal{} y_{2}$, and $ y_{2}$ is $ 2(x_{1} \minus{} y_{1})$ less than $ y_{1}$, and $ x_{2} \minus{} y_{2}$ is greater than $ x_{1} \minus{} y_{1}$.  It is clear that repeating this process a finite number of times will yield a nonpositive output, since $ \{y_{i}\}$ is decreasing and the positive difference between consecutive terms is increasing.  This gives us our contradiction.

If $ f(x) \equal{} x \plus{} k$ for some positive $ x,k$, then we have $ f(x \plus{} k) \plus{} x \plus{} k \equal{} 2x$, so $ f(x \plus{} k) \equal{} x \minus{} k$.  This gives us our $ f(x) \equal{} y$ with $ x > y$, so we can get the same contradiction.

Therefore $ f(x) \equal{} x$.[/hide]

[hide="A fairly standard approach to a problem like this"]Define $ a_0 \equal{} x, a_n \equal{} f(a_{n\minus{}1})$. The problem gives $ a_n \equal{} \minus{}a_{n\minus{}1} \plus{} 2a_{n\minus{}2}$. Such recurrences are easy to solve for explicitly using characteristic polynomials.[/hide]

[quote="MellowMelon"][hide="A fairly standard approach to a problem like this"]Define $ a_0 \equal{} x, a_n \equal{} f(a_{n \minus{} 1})$, for all $ n\in \mathbb{N}_0$. The problem gives $ a_n \equal{} \minus{} a_{n \minus{} 1} \plus{} 2a_{n \minus{} 2}$. Such recurrences are easy to solve for explicitly using characteristic polynomials.[/hide][/quote]

Is there a simpler way to finish what you started than this?

[hide]
=====
We find that for any given $ x\in \mathbb{R}_ \plus{}$, there exist $ c_1,c_2\in \mathbb{R}$ such that $ f^n(x) \equal{} c_1 \plus{} c_2( \minus{} 2)^n$. Using $ x \equal{} c_1 \plus{} c_2\implies c_1 \equal{} x \minus{} c_2$, we find $ f^n(x) \equal{} x \plus{} c_2(( \minus{} 2)^n \minus{} 1)$. But for any $ c_2\neq 0$, we can find some large $ n$ so that $ f^n(x) < 0$, contradiction. Thus $ c_2 \equal{} 0$, $ c_1 \equal{} x$, and $ f^n(x) \equal{} x\implies f(x) \equal{} x$ for all $ x\in \mathbb{R}_ \plus{}$.
=====
[/hide]

Not that I'm aware of; ironing things out afterward is often a little tedious. (Take e.g. the Putnam problem that was a generalized form of this; there are so many little steps and kinks that it's one of the ten Putnam problems with the lowest average score.) But it gets you on the right track, which is very good since one-variable functional equations are not easy to get started on.

EDIT: In response to below, I don't remember. I just recall seeing it on a "ten hardest problems" list given out at my school's Putnam seminar.


superior algebra
superior algebra unsolved

It is known that for elements $ u_1, u_2, v_1, v_2$ of group $ G$ holds following conditions:
\[ u_1v_1\equal{}v_1u_1\equal{}u_2v_2\equal{}v_2u_2, u_1^{p_1}\equal{}u_2^{p_1}\equal{}v_1^{p_2}\equal{}v_2^{p_2}\equal{}e,\]
where $ p_1, p_2\in N$ with $ GCD(p_1, p_2)\equal{}1$. Show that $ u_1\equal{}u_2$ and $ v_1\equal{}v_2$.


ratio
limit
geometric sequence
geometric series
absolute value

Hello i wondering if anyone could give me a hand solving this

Given
1. the sum of the geometric progression := S
2. the number of terms := n
3. the first term := t

Can you get and equation for the common ratio

finite G.P       S = t * ( 1 - r^n)/(1 - r)

the number i workin off are
S = 292.618
n = 15
t = 50

Anto!!!

Do you know the sum of geometric series formula?  If so, this should be fairly simple.  If not, you can derive it by multiplying by (1-r), where r is the common ratio, and dividing by that again later.

I think he even has the formula there already... so all you have to do is plug in the numbers and solve for $r$ I guess...

now that this is solved... i want to post a challenge:

prove that the sum of the infinite geometric sequence with the beginning term a, ratio r is
$\frac{a}{1-r}$.

of course, absolute value of r is less than 1.

it's not much of a challenge!!!
we know that $|r|<1$ for your formula to work.

if you let $r=\frac{p}{q}$ where $q>p$, then
$\lim_{n\rightarrow\infty}(\frac{p}{q})^n=0 $ so the original formula of
$\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}$

Another way... I guess it could be called more "conventional"

[hide]
$S = a+ar+ar^2+...$
$Sr = ar+ar^2+ar^3+...$

Subtracting the second from the first, we have

$S-Sr = a$
$S(1-r) = a$
$S = \frac{a}{1-r}$

Note that you can derive the formula for the sum of a finite geometric series in a similar fashion.[/hide]

i would use paladin8s method because like he said we can use the same convention to prove the formula for finite geometric sequence

In case none of you noticed, solving that equation requires you to solve something of the [b]50th degree[/b], so saying "just solve it" is not a very helpful thing to do at all.  (Actually, 49th degree, but whatever.)  In this particular case, the only real root is $r \approx 0.953472$, but I don't know a nice way to solve the equation in general.  Numerical approximation is probably the way to go for any n of reasonable size, say, 5 or larger.

Cheers JBL, :)
I used the numerical approximation method in the end and implemented it within a c++ for loop.

I dont know what paladin8 is on about with this " was under the impression" crap, cos i need the exactly float number to implement this for a GPS system.

[quote="anthonycronin"]I dont know what paladin8 is on about with this " was under the impression" crap, cos i need the exactly float number to implement this for a GPS system.[/quote]

Just FYI, most people don't appreciate condescending remarks around here, whether you mean it or not. I'll let it fly for now, but I think everyone would agree that a statement like that does not belong here. I will admit that maybe I did not go into detail about how to solve it like JBL said, but that does not warrant such a comment.

Sorry,

Indeed i didnt thank everyone else for their contribution. Thanks.

But your comment is still condescending

"I was under the impression that we have the benefits of a calculator... with a number like 292.618... I could be wrong."

thats why i posted that reply!

This argument needs to stop. Seriously. Moderator, please lock this, or this might get more serious. This is non-sense.

[quote="probability1.01"]As for the original topic, what was JBL referring to in his post?  I'm conjecturing that some posts were removed...[/quote]
He is refering to solving the equation  t * ( 1 - r^n)/(1 - r), where S = 292.618, n = 15, t = 50
and after you substituted the numbers, it turns out to be a 50 degree equation.


Prove that for every $ n>1$, ther exist $ n$ consecutive composite numbers.

The short answer is $ (n\plus{}1)! \plus{} 2, (n\plus{}1)! \plus{} 3, ... (n\plus{}1)! \plus{} (n\plus{}1)$.

The more interesting answer is to think about the Chinese Remainder Theorem.

How exactly would you ensure that those are $ n$ consecutive numbers?

In the long answer?  Find a large solution to the system of congruences

$ x \plus{} 1 \equiv 0 \bmod p_1$
$ x \plus{} 2 \equiv 0 \bmod p_2$
...
$ x \plus{} n \equiv 0 \bmod p_n$

for distinct primes $ p_i$.

that is using the chinese remainder theorem


Gigel is a big fan on the 70's music. He has an impressive colection of casettes from that time. Technology back then was rather rudimentary, but Gigel wants to bring back the soul of that times as much as he can.
Gigel has $n$ casettes $1,2,...,n$ an $n+1$ cassete winders $0,1,2,...,n$ of which one is empty. (Just imagine the cassetes strapped around a pencil each).
When Gigel listens to a cassete (i don't know how to say this, i hope that you will understand) the magnetic part of the tape passes from that winder to the empty one in the reverse order, thus the initial one becoming the empty one.
Gigel makes sure that after listening to each cassete he rewinds it an puts in back on its original winder.
His evil twin brother Gigle though has messed them up.

Find the minimum number of operations required to place every cassete on its omologue cassete winder.

In data
In the file role.in there will be the number $N$ , the number of cassetes and on each of the following $N$ lines there will be to numbers $W,I$ that the describe the position of the $k^{th}$ cassette . $W$ is the winder on which it is and $I$ is either $0$ or $1$ depending if the cassetes is in the correct order or backwards.

Out data
In the file role.out on each line there will be the number of the $i^{th}$ cassette that will be moved on the empty cassete winder during the $i^{th}$ operation.

Restrictions
$1\leq N\leq 10000$

edit:

This is what I understood and which I hope can make things clearer:
We have the number $1,2,3,...,n$ and the holes $0,1,2,...,n$ . Each number can be in only one hole either in it's normal form or reversed. A move consists of choosing a hole and replacing and putting the number in that hole in the empty hole AND multiping it by $-1$. Thus having an initial configuration we must place them in $(1,1),(2,2),...,(n,n)$ with the least number of moves.
THis is the example data which I forgot to post:
role.in
2
1 0
0 1

role.out
0

role.in
3
2 0
0 1
3 1

role.out
2
1
3
2
0

Time limit: 0,3 sec (16 mb of memory of which 1mb for the queve)

(Btw: theres a couple of typos in there: n should be up to 100000, and the second test case output has an extra 3 at the start).

[hide]Each cassette can be classified into one of four groups: a) Correct winder, correct type, b) Wrong winder, wrong type, c) Wrong winder, right type, d) Correct winder, wrong type. These four categories require at least 0,1,2, and 3 moves involving that cassette alone. Thus the minimum number of moves is at least the sum of that value for each cassette.

I haven't done a rigorous proof, but I'm 99% sure that a greedy algorithm achieves that minimum. At any stage, we are guaranteed that we can change a cassette from b) to a), or c) to b), or d) to c). If you assume that we can't, a parity argument shows you that in fact the situation is impossible to achieve.[/hide]


An $ a,b,c \in\mathbb{R} +$  ν.δ.ο
$ {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}}\geq{\dfrac{a + b + c}{ab + bc + ca}}$

Πολλαπλασιάζω με τους παρονομαστές και η ανισότητα γίνεται:

$ (ab\plus{}bc\plus{}ca)^2\geq abc(a\plus{}b\plus{}c)$ που ισχύει από την γνωστή $ (a\plus{}b\plus{}c)^2\geq 3(ab\plus{}bc\plus{}ca)\geq ab\plus{}bc\plus{}ca$

Μετά από πράξεις:

Έχουμε:

$ LHS \equal{} \frac {ab \plus{} bc \plus{} ca}{abc}$

Οπότε αρκεί ν.δ.ό.:

$ \frac {ab \plus{} bc \plus{} ca}{abc}\geq\frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}$

Ή

$ (ab \plus{} bc \plus{} ca)^2\geq abc(a \plus{} b \plus{} c)$

Ή

$ (ab \plus{} bc \plus{} ca)^2\geq (ab)\cdot(bc) \plus{} (bc)\cdot(ca) \plus{} (ca)\cdot(ab)$

$ (x \plus{} y \plus{} z)^2\geq 3(xy \plus{} yz \plus{} zx)\geq xy\plus{}yz\plus{}zx$

Χάνω κάπου ή είναι η ανισότητα τόσο χαλαρή; Αν δε χάνω, ένα 3 στον αριθμητή του RHS θα 'μαστε οκ.

Cheers,
Στέλιος

Νομίζω ότι η ανισότητα είναι πολύ χαλαρή:
πολλαπλασιάζοντας και τα δύο μέλη με $ ab\plus{}bc\plus{}ca$ έχουμε:

$ a\plus{}b\plus{}c\plus{}(something \ge 0)) \ge a\plus{}b\plus{}c \Leftrightarrow (something \ge 0)\ge0$ που ισχύει


calculus integration function geometry algebra polynomial Vieta

The functions $ f(x) ,\ g(x)$ satify that $ f(x) \equal{} \frac {x^3}{2} \plus{} 1 \minus{} x\int_0^x g(t)\ dt,\ g(x) \equal{} x \minus{} \int_0^1 f(t)\ dt$.
Let $ l_1,\ l_2$ be the tangent lines of the curve $ y \equal{} f(x)$, which pass through the point $ (a,\ g(a))$ on the curve $ y \equal{} g(x)$.
Find the minimum area of the figure bounded by the tangent tlines $ l_1,\ l_2$ and the curve $ y \equal{} f(x)$ .

First, we must observe that $ g(x)$ is linear, and of the form:

$ g(x) \equal{} x \plus{} \alpha$

Plugging this into the other equation:

$ f(x) \equal{} \frac {x^3}{2} \plus{} 1 \minus{} x\int^x_0(t \plus{} \alpha)\,dt \equal{} \frac {x^3}{2} \plus{} 1 \minus{} x\left(\frac {x^2}{2} \plus{} \alpha x\right) \equal{} 1 \minus{} \alpha\,x^2$

Now, from the beginning we have:

$ \alpha \equal{} \minus{} \int^1_0 f(t) \, dt \equal{} \minus{} \int^1_0\left(1 \minus{} \alpha\,t^2\right)\,dt \equal{} \frac {\alpha}{3} \minus{} 1$

Now, from this we have:

$ \alpha \equal{} \minus{} \frac {3}{2}$

$ f(x) \equal{} \frac {3}{2} x^2 \plus{} 1$

$ g(x) \equal{} x \minus{} \frac {3}{2}$

Now, as for the surface... Would it be right to say that the minimum area is the one that is gained by drawing tangents to $ f$ through the point on $ g$ that is the closest to $ f$?

for the bounded surface i found $ 2\sqrt 2$ but the solution is pretty big with vieta

\[ \text{Let }p(a)\text{ be the left-most contact point, and }q(a)\text{ be the right-most contact point then }\]

\begin{align*}
A &= \int_{p(a)}^{a}(f(x)-l_1(x))\,dx + \int_{a}^{q(a)}(f(x)-l_2(x))\,dx\\
But\,\; \frac{dA}{da} &= 0 \; \forall a, \text{ (by use of Leibniz Rule)}\\
\end{align*}

So any value of a should do, ie area is constant.

I get an area of 14/27 can someone please verify :?:


inequalities
function
calculus
inequalities proposed

[quote="ThAzN1"]Let $a,b,c$ be positive reals.  Prove that
\[\sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}} \geq \sqrt{4+\frac{3abc}{ab(a+b)+bc(b+c)+ca(c+a)}}.\]
Well, maybe it's not so well-known, but I've seen the RHS replaced with a $>2$.  Hmm...[/quote]

the problem with $>2$ is easy: use cauchy on the LHS, then weighted jensen on the new RHS (for $x \mapsto \sqrt{x}$) and get smth trivial

i'm not sure, but i think that the improvement  can be solved in the same way; but i can't prove that \[ \sum_{sym} x^7y^2 + \sum_{sym} x^5y^3z \geq \sum_{sym} x^6y^3 + \sum_{sym} x^6y^2z .\] (i'm not sure whether it's true :( )

any ideas?

[edit:] if i think about, it is just vornicu-schur applied twice in a convenient way;  :roll: ; tomorrow i'll post my complete soln if i don't find any mistake in it

I haven't given it much thought but
WLOG abc=1 and set u=a/(b+c),v=b/(c+a),w=c/(a+b).  Then

\[\displaystyle \sqrt{u}+\sqrt{v}+\sqrt{w} \geq \sqrt{4+\frac{3}{\frac{1}{u} +\frac{1}{v}+\frac{1}{w}}}\]
Using AM-HM we have to prove
\[\displaystyle \sqrt{u}+\sqrt{v}+\sqrt{w} \geq \sqrt{4+\frac{u+v+w}{3}}\]
After squaring simplifies to
\[\displaystyle \frac{1}{3}(u+v+w)+(\sqrt{uv}+\sqrt{vw}+\sqrt{wu}) \geq 2\]
I'm stuck for know but it's late.  I'll try more later.

[quote="silouan"]If we show that
$ \[\sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}} \geq3\frac{\sqrt2}{2}$ then we are done.
Any ideas in this?[/quote]
Obviously, it's wrong.

ok... here's my proof for:

[quote="ThAzN1"]Let $a,b,c$ be positive reals.  Prove that
\[\sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}} \geq \sqrt{4+\frac{3abc}{ab(a+b)+bc(b+c)+ca(c+a)}}.\]
Well, maybe it's not so well-known, but I've seen the RHS replaced with a $>2$.  Hmm...[/quote]

[b]Solution[/b]

By Cauchy we have \[ \sum \sqrt{\frac{a}{b+c}} \cdot \sum \sqrt{a(b+c)} \geq \left( \sqrt{a} + \sqrt{b} + \sqrt{c} \right)^2 ,\] so it suffices to prove that \[ \left( \sum \sqrt{a} \right)^2 \geq \sqrt{4+\frac{3abc}{\displaystyle \sum\limits_{\textrm{cyc}} ab(a+b)}} \cdot \sum \sqrt{a(b+c)} .\]

Let $x,y,z$ be positive reals s.t. $a=x^2$, $b=y^2$, $c=z^2$, so the inequality to prove becomes \[ \left( x+y+z \right)^2 \geq \sqrt{4+\frac{3x^2y^2z^2}{\displaystyle \sum\limits_{\textrm{cyc}} x^2y^2 \left( x^2+y^2 \right)}} \cdot \left( x \sqrt{y^2+z^2} + y \sqrt{z^2+x^2} + z \sqrt{x^2+y^2} \right) .\]

By weighted Jensen for the strictly concave function $x \mapsto \sqrt{x}$ we have \[ \sqrt{\frac{x \left( y^2+z^2 \right) + y \left( z^2+x^2 \right) + z \left( x^2+y^2 \right)}{x+y+z}} \geq \frac{x \sqrt{y^2+z^2} + y \sqrt{z^2+x^2} + z \sqrt{x^2+y^2}}{x+y+z} .\] Thus, it suffices to prove that \[ \left( x+y+z \right)^2 \geq \sqrt{4+\frac{3x^2y^2z^2}{\displaystyle \sum\limits_{\textrm{cyc}} x^2y^2 \left( x^2+y^2 \right)}} \cdot \sqrt{x+y+z} \cdot \sqrt{x \left( y^2+z^2 \right) + y \left( z^2+x^2 \right) + z \left( x^2+y^2 \right)} .\]

By squaring we obtain the equivalent inequality \[ \left( x+y+z \right)^4 \geq \frac{3x^2y^2z^2 + \displaystyle 4 \sum\limits_{\textrm{sym}} x^4y^2}{\displaystyle \sum\limits_{\textrm{sym}} x^4y^2} \cdot \left( x+y+z \right) \cdot \sum\limits_{\textrm{sym}} x^2y ,\] or, equivalently, \[ \left( x+y+z \right)^3 \cdot \sum\limits_{\textrm{sym}} x^4y^2 \geq \left( 3x^2y^2z^2 + \displaystyle 4 \sum\limits_{\textrm{sym}} x^4y^2 \right) \cdot \sum\limits_{\textrm{sym}} x^2y .\]
\[ \Leftrightarrow \left( \sum\limits_{\textrm{cyc}} x^3 + 3 \sum\limits_{\textrm{sym}} x^2y + 6xyz \right) \cdot \sum\limits_{\textrm{sym}} x^4y^2 \geq \left( 3x^2y^2z^2 + \displaystyle 4 \sum\limits_{\textrm{sym}} x^4y^2 \right) \cdot \sum\limits_{\textrm{sym}} x^2y \]
\[ \Leftrightarrow \sum\limits_{\textrm{cyc}} x^3 \cdot \sum\limits_{\textrm{sym}} x^4y^2 + 3xyz \cdot \sum\limits_{\textrm{sym}} x^4y^2 + 3xyz \cdot \sum\limits_{\textrm{sym}} x^4y^2 \geq 3xyz \cdot \sum\limits_{\textrm{sym}} x^3y^2z + \sum\limits_{\textrm{sym}} x^4y^2 \cdot \sum\limits_{\textrm{sym}} x^2y \]
\[ \Leftrightarrow \left( \sum\limits_{\textrm{sym}} x^4y^2 \right) \cdot \left( \sum\limits_{\textrm{cyc}} x^3 + 3xyz - \sum\limits_{\textrm{sym}} x^2y \right) + 3xyz \cdot \left( \sum\limits_{\textrm{sym}} x^4y^2 - \sum\limits_{\textrm{sym}} x^3y^2z \right) \geq 0 \]

Since $x^4y^2 + z^4x^2 \geq 2x^3yz^2$ and the kind hold by $AM-GM$, it follows that $\displaystyle \sum\limits_{\textrm{cyc}} x^4y^2 \geq \sum\limits_{\textrm{cyc}} x^3yz^2$. Similarly, $\displaystyle \sum\limits_{\textrm{cyc}} x^2y^4 \geq \sum\limits_{\textrm{cyc}} x^3y^2z$. Sum up and get $\displaystyle \sum\limits_{\textrm{sym}} x^4y^2 \geq \sum\limits_{\textrm{sym}} x^3y^2z$.

But $\displaystyle \sum\limits_{\textrm{sym}} x^4y^2$ is obviously positive. Furthermore, $\displaystyle \sum\limits_{\textrm{cyc}} x^3 + 3xyz - \sum\limits_{\textrm{sym}} x^2y$ is nonnegative by Schur, so the $LHS$ is nonnegative and the proof of the inequality is complete.

$\blacksquare$

[u]Equality.[/u] For equality to take place we must have $x=y=z$ (the Schur equality condition for positive reals). This implies $a=b=c$, in which case we see that we have equality.

Therefore equality holds iff $a=b=c$.

What do you think about this? Is it correct?

[quote="silouan"]$[ \sum_{sym} x^7y^2 + \sum_{sym} x^5y^3z \geq \sum_{sym} x^6y^3 + \sum_{sym} x^6y^2z ]$
Could we make something to prove this with Muihead?[/quote]

I don't think this is possible, because it's stronger than Muirhead. I arrived at that inequality by multiplying all out, but then I saw that a simpler argument was available.

[quote="TomciO"]I guess that this inequality (posted by silouan) can be proved by using calculus (without huge computations) but I haven't checked so I can be wrong.[/quote]

if you are reffering to the inequality with symmetric sums, just use vornicu-schur twice and it's done. i can't think of a shorter proof

Interesting ideas... my proof is as follows.  Holder's yields

$\left( \sum_{\text{cyc}} \sqrt{\frac{a}{b+c}} \right)^2 \left( \sum_{\text{cyc}} a^2(b+c) \right) \geq (a+b+c)^3$,

however, Schur's implies that

$\frac{(a+b+c)^3}{ \sum_{\text{cyc}} a^2(b+c) } = 3+\frac{a^3+b^3+c^3+6abc}{ \sum_{\text{cyc}} a^2(b+c)}$ $\geq 3+1+\frac{3abc}{ \sum_{\text{cyc}} a^2(b+c)}$,

so the result follows...

nice... your proof is shorter  :D

i'm not very used with holder and i never try this method when solving a problem (unless the ineq comes with a hint like : use holder  :P  ) .


Hey everyone,

I need an example of a second order ODE in the form

u''(x) + a^2 u(x) = f(x) - with f(x) being a nonlinear function...

that is exactly solvable...

So if anyone knows an example, please tell me the eqn and its solution..

Thanks alot

[url=http://en.wikibooks.org/wiki/Differential_Equations/Non_Homogenous_1]this[/url] page talks all about it, if you dont understand it I can explain

hehe... I am so sorry... I meant:

u'' + a^2 u = f(u) - [u][b]where f is NONLINEAR in u[/b][/u]

Sorry for that... That was lame...[/b]


[quote="Megus"]It is given a convex quadrilateral $ABCD$ in which $AC \perp BD$. Point $M,N,P,R$ are respectively midpoints of $AB,BC,CD,DA$. Points $W,X,Y,Z$ are orthogonal projections of $M,N,P,R$ onto respecpectively $CD,DA,AB,BC$. Prove that points $M,N,P,R,W,X,Y,Z$ lies on one circle.[/quote]

At first we have $MN\parallel AC \parallel RP, \; MR \parallel BD \parallel NP\Rightarrow$
$MNPR$ is a rectangle. Its circumcircle centre is on its diagonals. So the diagonals $MP$ and $NR$
are the diameters of this circle.
At second as $\angle NXR=90^\circ \Rightarrow X\in$ this circle.
Similarly $W,Y,Z \in$ this circle.
That is it.

V:)


Let p be a prime =1 mod 4. Prove that the equation x^2-py^2=-1 is necessarily soluble.

Bomb

the other discussion is here:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=x%5E2-py%5E2%3D-1&t=22977]link[/url]


If $a,b,c>0$  then prove that

$\frac{a^2-bc}{b+c+2a}+\frac{b^2-ca}{a+c+2b}+\frac{c^2-ab}{a+b+2c}\geq 0$

[hide]
$\sum_{cyclic} \frac{a^2 - bc}{b+c+2a}\ \geq \ \sum_{cyclic} \frac{a^2 - bc}{2b+2c+2a}$, so now we have to prove that $\sum_{cyclic} \frac{a^2 - bc}{2(a+b+c)} = \frac{a^2+b^2+c^2-bc-ac-ab}{2(a+b+c)} \geq 0$.
Because $2(a+b+c) > 0$, we can multiply both sides with it, so $a^2+b^2+c^2 \geq ab+bc+ac$, which is a well-known inequality which can be proven by AM-GM, rearrangement, trivial ineq. etc. $\square$
[/hide]

[quote="probability1.01"]Your first step is non-trivial.

Edit--I found a solution using [hide="wabbits"]the substitution $x = b+c, y = a+c, z = a+b$.[/hide][/quote]

Hmm, you're right. Thanks for noticing.

Jensen on $f(x)=\frac1x$ tells us that \[ LHS \ge (\sum_{cyc} (a^2-bc))^2\frac1{\sum_{cyc}((a^2-bc)(2a+b+c))}  \] the first part is a perfect square thus positive, the second part has denominator $2\sum_{cyc} (a^3 -abc)$ which is positive by AM-GM.

[quote="probability1.01"]But this is more fun!

[hide="nore"]
Use substitution I said.

$\frac{a^2-bc}{2a+b+c} = \frac{(y+z-x)^2 - (x+y-z)(x+z-y)}{4(y+z)}$
$=\frac{x^2 + (y+z)^2 - 2x(y+z) - x^2 + (y-z)^2}{4(y+z)}$
$=\frac{2y^2 + 2z^2 - 2x(y+z)}{4(y+z)}$

So we're wondering if

$\sum \frac{y^2 + z^2}{y+z} \ge x+y+z$

But we can just be all like whatev $y^2 + z^2 \ge \frac{1}{2} \cdot (y+z)^2$.
[/hide][/quote]
very fun. :)
and by the hint of your solution:
i'm sure,there exist a constant $m$ such that
$\frac{a^2-bc}{2a+b+c} \ge 2ma-mb-mc$ :)

[quote="zhaobin"]i'm sure,there exist a constant $m$ [/quote]

$m= \frac{1}{4}$ perhaps  ;)

Thank you very much for the interest. Nice solutions of all of you, but something smells wrong in the first solution .My solution is similar to zhaobin's one.

[quote="Silouan"]but something smells wrong in the first solution[/quote]

Well, I've read it three times :roll:  and finally figured out that we don't have $\frac{a^2 - bc}{b+c+2a}\ \geq \ \frac{a^2 - bc}{2b+2c+2a}$ unless $a^2-bc\geq 0$.

For instance, $\frac{-2}{3}<\frac{-2}{5}$, although $3<5$

[quote="Raluca"]Well, I've read it three times :roll:  and finally figured out that we don't have $\frac{a^2 - bc}{b+c+2a}\ \geq \ \frac{a^2 - bc}{2b+2c+2a}$ unless $a^2-bc\geq 0$.[/quote] He didn't claim that either. He claimed that there was inequality in the cyclic sum, not in every summand.

His claim is true, although not obvious as he initially thought. Equality iff a=b=c.

How about this?  Let's just WLOG $a+b+c = 1$ and $a \geq b \geq c$ for fun.

Note that (sums are cyclic)

$\sum \frac{a^2}{1+a} \geq \sum \frac{b^2}{1+a}$
$\sum \frac{a^2}{1+a} \geq \sum \frac{c^2}{1+a}$

because $a^2 \geq b^2 \geq c^2$ and $\frac{1}{1+a} \leq \frac{1}{1+b} \leq \frac{1}{1+c}$ so the above follows from rearrangement.  Now sum them up and use the fact that $x^2+y^2 \geq 2xy$.  :)


advanced fields
advanced fields unsolved

Hi,

I'm having a hard time with the 2 recurrence relations below, any help would be appreciated. Thanks!

1.) a_n = 3* a_n-1  + n^2  - 3 (n>= 1) , where a_0 = 1

2.) a_n = n* a_n-1  + n! (n>1) , where a_1 = 1

1. We have $\frac{a_{n}}{3^{n}}-\frac{a_{n-1}}{3^{n-1}}=\frac{n^{2}-3}{3^{n}}$, now use telescopic sum.
2. We have $\frac{a_{n}}{n!}-\frac{a_{n-1}}{(n-1)!}=1$, use telescopic sum, again.


Let's see how many psychics we have. Choose one of the above. Poll ends in 24 hours for obvious reasons...

[quote="cardwacko9999"]wat if theres a dispute (ex. Elections 2000, Florida)[/quote]

Doesn't change the fact that eventually one candidate will get 270 votes or it'll be the House that decides.

Aiighty. Poll closed.

Bush wins with 270 votes. - 11 votes
Kerry wins with 270 votes. - 7 votes
House decides President. - 3 votes.

Let's see how psychic you guys are. :-D

I'll un-stickify this later on tonight.

[quote="MithsApprentice"]Congratulations everyone who voted for Bush. And congrats to President Bush, 2004-2008.[/quote]

You can't be sure that Bush is going to win becuase the margin is less than the number of absentee ballots.  Kerry might win Ohio, making him president 2004-2008.

Kerry has already conceded, so Bush wins.

he won, so now the nation needs to come and together and support him
(that is going to be very hard!)

I don't understand why would people vote for Bush either.......

Just shows you the lack of understanding of things people have in America.

[quote="Athena"]BUSH WON! *sob* no thats not out of happiness.  why do ppl vote for him WHEN HE DECLARED WAR FOR SUM SO-CALLED WEAPONS OF MASS DESTRUCTIONS THAT DID NOT EXIST. now the world hates us. yay.[/quote]

Blahblah, hindsight.  Just a few years ago, the American people were clamoring for blood and retribution.

i gess we were wrong. the president is supposed to lead us away from these kind of things. al gore would have been president if it werent for the electoral college. whether he would have done the same thing is unknown, but if he didnt....wouldnt ppl like us more?

Might as well accept the fact the Bush is the president.

Well, my standard greeting for the next few weeks is gonna be, "Hey, how's it going?  The world's going to end, and it's all because of Bush's so-called lucky ears (and may I say, the people of Ohio and Florida)"

i'm so happy that Bush is going to try to make the world a better place. (hopefully...)

[quote="Jhyon8149"]Bush has screwed USA by going into Iraq. After 9/11, practically the whole world felt sorry but after Bush became president, practically the whole world hates us.....

But would Kerry have been a better president if he had won? We will never know because he was robbed of the chance just like Al Gore and Robert F. Kennedy and other people I have not named. But Kerry has no views whatsoever. Bush actually believes strongly about what he does, no matter how wrong it is. Kerry flip-flops on his views and that probably hurt him in the election.

Just think about it this way: it was a lose-lose situation. If Bush got elected, he would screw up the country more and cause the world to be angry and perhaps start a war  :( But if Kerry got elected, he would probably have let the country get screwed over because he has no views whatsoever.[/quote]

My only regret about this election is that the American people will not see for themselves what a screw-up Kerry was cruising to be.

hey! kerry never got a chance yet! u don't know if he'll do better

don't wory guys. the world will end in 200 more years. don't ask why. i asked a fortune teller.

[quote="Athena"]yea he cant do as bad as bush unless he makes us go into war with some other ocuntries but thats just STUDPID! so nothing can REALLY get worse. but the good thing is that hes still in the senate so its not ALL up to bush.[/quote]

War with some other countries is "just STUPID?"  One word: Holocaust.

another word: realy realy realy realy realy realy rela y realy realy realy relay relay realy relay relya ...dumb...u know, you guys don't have to worry.  if you live 200 more years, then you need to worry. the world shall end in 200 years...trust me...i'm physcic...FINE! DON'T BELIEVE ME! BE THAT WAY! I'M SERIOUS! I'M NOT JOKING AROUND HERE GUYS! THE ICE AGE IS COMING!!!!!!!!!!!

Thanks for spamming another topic with some potentioal gameworld. At least when other people debate politics they actually back up their claims with reasonable evidence instead of just saying something and leaving it at that.

[quote="RobinHood3000"][quote="Athena"]yea he cant do as bad as bush unless he makes us go into war with some other ocuntries but thats just STUPID! so nothing can REALLY get worse. but the good thing is that hes still in the senate so its not ALL up to bush.[/quote]

War with some other countries is "just STUPID?"  One word: Holocaust.[/quote]
huh? well yea were ALREADY at war. so hopefuly kerry or bush doesnt drag us into another one.  any ppl no about the thing called SELECTIVE SERVICE ASSIGNMENT??? its wat caused the 26th ammendment.  but yea about the psycic thing......the ice is actually melting because of global warming. are u just trying to get ur posts up or sumthin?

no, i'm trying to make a good conversation.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $p$ be a prime and $\left( G, \cdot \right)$ be a group with $2 p^{2}$ elements. Furthermore, let $H_{1}, H_{2}$ be subgroups of $G$ such that $H_{1}\cap H_{2}= \left\{ e \right\}$ and such that no matter how we choose $p$ elements from $G$, at least two of them are in $H_{1}\cup H_{2}$.
Prove that $H_{1}= G$ or $H_{2}= G$.

[i]Dana Heuberger[/i]


A rectangle has an area of 100 square units. Its dimensions are
whole units. What is the number of units in the greatest
possible perimeter?

The perimeter will be greatest when the numbers are as far apart as possible.

This will occur with a 100x1 rectangle, so the perimeter will be 2(100+1)=202.


I am not sure if the following exist

X = 2 * m + n
Y = 3 * m – 2 * n

Such that both X and Y are integers and at least one of m and n is a fraction.
Any help would be highly appreciated.

There are so many variables and only two equations, it might be imposible. With the conditions you gave below, it might be solvable, but I don't know.

After rewriting the equations, i don't find an answer.

An infinite number of solutions exist.
Getting two equations:
$2x+y=7m$
$3x-2y=7n$
Then you can start by picking any two integers x, y s.t. 2x+y/7 is fractional, so m is fractional.  Then n is still rational, so it fits the requirements.

For example, $x=1 y=1 m=\frac{3}{7} n={1}{7}$


linear algebra
matrix
induction
Gauss
linear algebra unsolved

[size=125]I have trouble understanding (the last part of) the proof of the theorem stating that every matrix has one and only one reduced row echelon form.

Proof:

Let A be m x n.  The proof uses induction on n.  For n = 1, the theorem is obvious.  Suppose, now, that the theorem has been established for m x (n-1) matrices.  Let A be an m x n matrix.  Denote by A_(n-1) the matrix obtained from A by removing its nth column.  Any sequence of row operations that brings A to reduced row echelon form also puts A_(n-1) in reduced row echelon form.  By the induction hypothesis, A_(n-1) has one and only one reduced row echelon form.  If B and C are reduced row echelon forms of A, they can differ only in the nth column.  Assume B =/= C.  Select i so that row i in B differs from row i in C.  Thus, we have b_in =/= c_in.  Because A ~ B ~ C, the homogeneous systems Ax = 0, Bx = 0, and Cx = 0 have the same solutions.  Let Bu = 0.  Then Cu = 0 and (B-C)u = 0.  The first n-1 columns of B-C are all zeros.  Hence, we have [(B-C)u]_i = (b_in - c_in)u_n = 0.  Since b_in =/= c_in, we must have u_n = 0.  Thus, any solutions of Bx = 0 or Cx = 0 must have x_n = 0.

It follows that the nth column of B and C must have pivots, for otherwise those columns would be associated with free variables, and we could choose x_n to be nonzero.  Because the first n-1 columns of B and C are identical, the row in which this pivotal 1 appears must be the same for B and C; namely, it is the row which is the first zero row of the reduced row echelon form of A_(n-1).  Because the remaining entries in the nth columns of B and C must be zero, we have B = C, which is a contradiction.
This proof follows the one given by Yuster [1984].

This proof appears in a single paragraph in my textbook, but I deliberately separate the proof into two when I type this.  This is because I don't understand the first sentence in the second paragraph.  I don't really see the reason given for having pivots, which states that otherwise those columns would be associated with free variables.  Why would those columns be associated with free variables?  this confuses me.

I understand the first paragraph, but it is the second paragraph that gets me... :D

Thank you for your help![/size]

If you are familiar with the Gauss-Jordan method, you should know that that method "partitions" the set of variables used in the system $ Ax \equal{} b$ into two disjoint sets, namely the set of free variables and the set of pivot variables, the latter of which is nonempty for all nonzero matrices $ A$. So, if a variable is not a free one, what should it be?


topology
function
advanced fields
advanced fields unsolved

Suppose that $X$ is a metric space with the metric $\rho$. Prove that $X$is compact iff $X$ is bounded with each metric on $X$ (e.g. $\rho'$) that induce the same topolgy on $X$ az $\rho$

For every $d$ equivalent to $\rho$ (from now on we'll write this as $d\sim\rho$), the identity map on $X$ is a homeomorphism from $(X,\rho)$ onto $(X,d)$. If $(X,\rho)$ is compact, this means that any $\forall d\sim\rho,\ (X,d)$ is compact, and hence bounded.

Conversely, suppose $(X,\rho)$ is not compact. Then we can find a sequence $(x_n)_{n\ge 1}\subset X$ with no limit point. Define $f(x_n)=\frac 1n,\ \forall n\ge 1$. This is a continuouys function from the closed subset $(x_n)_n$ of $X$ to $(0,\infty)$. By Tietze's Extension theorem (stating that a continuous map from a closed subset of a normal space to $\mathbb R$ can be extended to a continuous map from the whole space to $\mathbb R$; clearly, it makes no difference if we use $(0,\infty)$ instead), we can extend $f$ to a continous function from $(X,\rho)$ to $(0,\infty)$, which we'll also denote by $f$.

$f$ takes arbitrarily small values, but doesn't reach its infimum on $X$, namely $0$. Hence the new metric $d(x,y)=\rho(x,y)+\left|\frac 1{f(x)}-\frac 1{f(y)}\right|$ is well defined. Clearly, it's equivalent to $\rho$ and unbounded, so we're done.

A great solution grobber. Of Course there is no need to Tietze's Extension theorem. We must prove that there exist a function $f: X\longrightarrow\mathbb R$ which is not bounded. For this if there is a sequence of $X$ such as $x_1,x_2,\dots$ which does not have limit point then consider the sets $B_{\epsilon_i}(x_i)$ which are pairwise disjoint and $\epsilon_i\longrightarrow 0$. Then if you consider the map $f$ that for $x\in X\backslash \bigcup_{i=1}^\infty B_{\epsilon_i}(x_i)$ define $f(x)=0$ and for $x\in B_{\epsilon_i}(x_i)$ define $f(x)=\frac{\epsilon_i-d(x,x_i)}{\epsilon_i^2}$. It is easy to check that $f$ is continious.

Can you give me the definition about metric equivalent ?
And please show me why $d$ and $\rho$ are equivalent ?
In my opinion , I think that if $d$ and $\rho$ are equivalent. So $d$ must be bounded because $\rho$ is bounded.
I have bad feeling here.  :?:

There also some similiar pronblems e.g.
$X$ is compact iff every continuous $f: X\longrightarrow\mathbb R$ is bounded or $f(X)$ is compact subset of $\mathbb R$


Or you could apply PM-AM:

$\sqrt{\frac{a^2 + (1-b)^2}{2}} \geq \frac{|a| + |1 - b|}{2}$

Adding the three together gives

$\sqrt{\frac{a^2 + (1-b)^2}{2}} + \sqrt{\frac{b^2 + (1-c)^2}{2}} + \sqrt{\frac{a^2 + (1-c)^2}{2}} \geq \frac{|a| + |1-a| + |b| + |1-b| + |c| + |1-c|}{2} \geq \frac{3}{2} \Rightarrow \sqrt{a^2 + (1-b)^2} + \sqrt{b^2 + (1-c)^2} + \sqrt{a^2 + (1-c)^2} \geq \frac{3\sqrt{2}}{2}$

Note that the middle part of that inequality follows from $|x| + |1-x| \geq |x + 1 - x| = 1$, which is a simple application of the triangle inequality.

[u]A geometrical interpretation in space.[/u]
Let the points $P(a,b,c)$, $A(0,1,c)$, $B(a,0,1)$, $C(1,b,0)$.
Then $\sum \sqrt{a^2+(b-1)^2}=PA+PB+PC\ !$

Yes and you can also use Minkowski's ineq to kill the ineq in one minute

Let $\overrightarrow x=(a,1-b),\ \overrightarrow y=(b,1-c),\ \overrightarrow z=(c,1-a),$ we have $\overrightarrow x+\overrightarrow y+\overrightarrow z=(a+b+c,\ 3-(a+b+c)).$ Thus use triangle inequality for vector,
$|\overrightarrow x|+|\overrightarrow y|+|\overrightarrow z|\geq |\overrightarrow x +\overrightarrow y+ \overrightarrow z|$

$\Longleftrightarrow \sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}$

${\geq \sqrt{(a+b+c)^2+\{3-(a+b+c)\}^2}=\sqrt{2\left(a+b+c-\frac{3}{2}\right)^2+\frac{9}{2}}\geq \frac{3}{\sqrt{2}}}.$


vector
Asymptote
analytic geometry
LaTeX

Can I graph Cartesian planes with just TeXshop?or do i need to download some kind of software?

Thanks!

Do you mean draw a graph in the Cartesian plane? It can be done but you can use [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_(Vector_Graphics_Language)]Asymptote[/url] and then ask questions about it in [url=http://www.artofproblemsolving.com/Forum/index.php?f=519]Asymptote forum[/url] and look for examples there.


Could someone be kind to post Miklos Schweitzer problems
for the years 2003/2004/2005/2006

thank you very much
m.

problems of 2003

Here on the site you only have this: http://www.mathlinks.ro/Forum/post-14975.html


How to solve this question: Let $(\mathbb{R}^2,d)$ be a metric space where
\[ d((x_1,y_1),(x_2,y_2)=|x_1-x_2|+\sqrt{|y_1-y_2|}.  \]
Find a generalized inner metric $d_s$associated by $d$. What is the topology $\mathcal{T}_{d_s}$ determined by $d_s$?

What is "generalized inner metric associated by "? I think we use another term for this metric, so can you write a definition?

I mean it satisfies the standard properties of a metric and it is possible that $d(x,y)=\infty$ for some $x,y$.


function
algebra
functional equation
real analysis
real analysis unsolved

It's been noted that if $f$ is a real valued, measurable function on $\mathbb{R}$ which satisfies the functional equation
\[ f(x+y)=f(x)+f(y)  \]
everywhere, then $f(x)=ax$ for some $a\in\mathbb{R}$.  I finally found a proof.  Does anyone want to take a stab at it?

If  an aditive function is not continuous, it tends to infinity in every point of $\mathbb{R}$, hence $f$ is not measurable. Am I wrong? :?

It is exactly the fact that a aditive function bounded on an interval is liniar. Someone can prove that if $f$ is bounded on an interval and aditive then it has limit 0 in 0. Hence it is continuous in 0, and very easily comes that it is continuous everywhere, and the problem becomes trivial.

To prove that it has limit 0 in 0, assume the contrary and chose an arbitrary sequence which tends to 0 and has a limit $\ell\neq 0$. You will get that $f$ in not bounded on the any interval. Which is false.