linear algebra
matrix
number theory
greatest common divisor

I understand how to solve these types of problems, but when it is in fraction form I don't understand how to solve the Denominator on the last Step. For my Numerator I have 39. Could someone help for me to understand how to solve the denominator please.

Here is a picture of the problem that I have to solve.

[img]http://i19.photobucket.com/albums/b199/Edgar_Thomas_/4.56.jpg[/img]

i'm not sure i understand where ur having a problem. just solve it as u would solve 2x2 matrix with integers: ad-bc

May I note that you are not SOLVING the determinant. You are EVALUATING the determinant. Thank you.

[quote="sludgethrower"]Solving is assosiated with finding solutions to an equation. Remember a determinant is a number, it doesn't make sense to solve a number. Evaluating is assosiated with finding a numerical equivalent of an expression.[/quote]

and ur answer to this proble would be....? :huh:

mihail911, an evaluation has been given by 7h3.D3m0n.117. It is correct, so there is no need to repeat it. That would be spam. This should be moved to the matrices section.

[quote="7h3.D3m0n.117"]In a 2x2 matrix

$ \begin{bmatrix}a & b\\ c &{ d}\end{bmatrix}$

The determinant is found by $ ad-bc$, as mihail911 said.

Therefore, the solution to your problem will be

[hide]
$ \left(\frac{7}{2}\times-\frac{9}{4}\right)-\left(\frac{8}{7}\times\frac{3}{5}\right)$

$ -\frac{63}{8}-\frac{24}{35}$

$ -\frac{2397}{280}$

Whoa not a pretty answer.
[/hide][/quote]

thanks, I understand how to Evaluate the denominator now. My teacher says that I was suppose to find the LCD.  How did you get such a large number for the numerator? did you find the LCD for the Numerator as well?

[quote="i_like_pie"]@ EDgar - You misunderstood "determinant," which refers to a matrix, as "denominator," which refers to a fraction. If you're still learning about fractions, I don't think you're ready for matrices yet, so don't get confused by them.[/quote]

My mistake mr Pie. I had no real trouble solving fractions in the past. But I can admit that I may have forgotten.

The book says. "A determinant is a real number associated with a matrix."

I understand the formula as we multiply A and D and we also multiply B and C and then

AD-BC

I never really had fractions in the 2X2 matrix and at first I had trouble with the denominater, My Teacher explained that I had to find the LCD. but I throw of by the Numerator

Even when my teacher e-mails me back, she's telling me but I still can't figure it out

Here is what she said

"Edgar,
You did it right except for adding the fractions.  Remember, you
need to find a LCD and rewrite the fractions with that LCD."

I'm not quite sure, but am I suppose to cross multiply?

[quote="EDgar"]"Edgar,
You did it right except for adding the fractions.  Remember, you
need to find a LCD and rewrite the fractions with that LCD."

I'm not quite sure, but am I suppose to cross multiply?[/quote]

Yes, but when you cross multiply, you need to find a LCD in order to solve AD-BC (the subtraction part).

For example, adding the fractions $ \frac{1}{2}$ and $ \frac{1}{3}$

The GCD of 2 and 3 is 6, so you multiply both the fractions by a form of 1 that makes the denominators the same.

$ \frac{1}{2}\cdot\frac{3}{3}\plus{}\frac{1}{3}\cdot\frac{2}{2}$
This becomes:
$ \frac{3}{6}\plus{}\frac{2}{6}$, which equals $ \frac{5}{6}$.

That's how you add fractions...the denominators must first be equal.

Thanks for your help everyone. Honestly I wasn't quite able to get the problem. But its not because of you folks. Its me really, sometimes i'm just a knuckle head like that. I will keep working at it.


[b][color=blue]CONAMAT[/color][/b] is a contest organizated by [b][color=blue]Bertolt Bretch School[/color][/b] - [url=http://www.colegiobertoltbrecht.edu.pe/colegio/index.htm]web site[/url]

 Each year, CONAMAT Contest takes place with a first phase in September along all the Peru,
with 5 sedes in Lima and 5 sedes more in provinces ........
and the final exam in October (This year: Oct. 23)

These are large exams with alternatives...

MORE INFORMATION, and the completed exams: [url=http://www.aduni.com.pe/conamat/conamat.html]Here[/url]
---------------
PD: The match about levels.
        4,5,6 grado = grades of Elementary School

  Secundaria (High School):
        1er Año  = 8th Grade
        2do Año  = 9th Grade
        3er  Año  = 10th Grade
        4to Año   = 11st Grade
        5to Año   = 12nd Grade
----------------------

I chose some Problems from these exams:

      [size=150] [color=blue]CONAMAT 2001 - PERU[/color] [/size]

1. (10th Grade / Q24) Let $f: R \to R$, such that:
\[ f(x^3 + y^3) = (x+y)[f^2(x) - f(x)f(y) + f^2(y)],  \]

where $x,y \in \mathbb{R}$, such that: $f(x) \neq 0$.

Find the value of: \[ \frac{f(2001x)}{2001x}.  \]

2. (12nd Grade / Q12) Let the function $G$, such that:
   (1) $G(x+y+z) = \frac{1}{G(x)} + \frac{1}{G(y)} + \frac{1}{G(z)}$, for all $x,y,z \in R$.\\
   (2) $G(x) > 0$, for all $x \in R$.

   Find $G(2001)$

3. (10th Grade/Q25) Factorizar the following Polynomial:
     $P(a,b,c,x) = (a-b)(x-a)(x-b) + (b-c)(x-b)(x-c) + (c-a)(x-c)(x-a).$

 Good Luck   ;)

 Carlos Bravo
 Lima-PERU

[quote="carlosbr"]
3. (10th Grade/Q25) Factorizar the following Polynomial:
     $P(a,b,c,x) = (a-b)(x-a)(x-c) + (b-c)(x-b)(x-c) + (c-a)(x-c)(x-a).$
[/quote]
$P(a,b,c,x)=(x-c)[(a-b)(x-a)+(b-c)(x-b)+(c-a)(x-a)]$
When $a=b$, $(a-b)(x-a)+(b-c)(x-b)+(c-a)(x-a)=0+(a-c)(x-a)+(c-a)(x-a)=0$. So $(a-b)$ is a factor.
When $b=c$, $(a-b)(x-a)+(b-c)(x-b)+(c-a)(x-a)=(a-b)(x-a)+0+(b-a)(x-a)=0$. So $(b-c)$ is a factor.
After comparing coefficients, $P(a,b,c,x)=(x-c)(a-b)(b-c)$

However, I think you have mistyped your problem. I think the polynomial should be $P(a,b,c,x) = (a-b)(x-a)(x-b) + (b-c)(x-b)(x-c) + (c-a)(x-c)(x-a)$ which is more symmetric. In this case $P(a,b,c,x)=(a-b)(b-c)(a-c)$.

Your right  .......leepakhin!!!  :P

However!!..

Does any body have any ideas about these problem's solutions??

Try these!!!

Carlos Bravo :)
Lima -PERU


I was reading through my textbook and there was one thing that was confusing me about double angles. I know that for the half-angle, the sign ($ +$ or $ -$) is dependent on which quadrant ${ \frac {x}{2}}$ is in. The textbook says nothing at all about any restrictions of the double angle. So my question is, are there any rules about using the double angle formula?

For example, in ${ sin \frac {x}{2}} = \pm{\sqrt {\frac {1 - cos x}{2}}}$, the sign choice depends on the quadrant of $ \frac {x}{2}$ and not $ x$. In the double angle formula, is the sign dependent on the quadrant of $ 2x$ or $ x$?

Edit: Okay, thanks a lot guys. I edited the title because I have another question.

The double angle formulas are simpler, and do not require you to determine the sign.

$ \sin{2x} \equal{} 2\sin{x}\cos{x}$ and $ \cos{2x} \equal{} \cos^2{x} \minus{} \sin^2{x}$ for all $ x$, no matter what quadrant it's in.

Notice also that the $ \pm$ in the addition/subtraction formulas is determined by which one you're doing (addition or subtraction).

Sometimes you can use some more convenient formulas for $ \cos2x$

$ \cos2x\equal{}2\cos^2\minus{}1\equal{}1\minus{}2\sin^2x$

Thanks. Also, I was able to find the triple angle formulas for sine, cosine, and tangent and derive the power reduction formulas for powers of 3 for sine and cosine.

My problem is now that I can't find the triple angle formula in a form that makes it easy to find the reduction for $ \sin{^{4}x}$ and $ \cos{^{4}x}$. Or are the reduction formulas for fourth powers derived in a different way? I got $ \sin{4x}\equal{}\sin{(2x\plus{}2x)}\equal{}\sin{2x}\cos{2x}\plus{}\cos{2x}\sin{2x}$. I think the problem is that I can't find a way to subsitute for $ \cos{2x}$ so that there actually is a $ \sin{^{4}x}$ term in the formula.

Um...maybe this will work...

$ \sin4x \equal{} 2\sin2x\cos2x$
$ \equal{} 2(2\sin x\cos x)(1 \minus{} 2\sin^2x)$
$ \equal{} 4\sin x\cos x \minus{} 8\sin^3x\cos x$


[quote="mathpk"]Let $ \Gamma$ be the circumcircle of a triangle $ ABC$. A circle passing through points $ A$ and $ C$ meets the sides $ BC$ and $ BA$ at $ D$ and $ E$, respectively. The lines $ AD$ and $ CE$ meet $ \Gamma$ again at $ G$ and $ H$, respectively. The tangent lines of $ \Gamma$ at $ A$ and $ C$ meet the line $ DE$ at $ L$ and $ M$, respectively. Prove that the lines $ LH$ and $ MG$ meet at $ \Gamma$.

See all problems from APMO 2008 here,
http://www.mathlinks.ro/viewtopic.php?p=1073977#1073977[/quote]
If $ LH$ cut the circle at a X (different from H ) then $ B,X,E,D$ is cyclic .
It mean that $ LH,MG$ cut on at a point on the circle.

Dear Mathlinkers,
a want only to detail the interesting last solution :
1 let U be the meetpoint of BX and AC, Ta, Tc the resp. tangent to the circumcircle of ABC at A, C,
and X the second meetpoint of LH with the circumcircle of ABC.
2. ULE is the Pascal's line of BXHCA Ta B
3. U, L, E and D are collinear
4. By a converse of Monge (or d'Alembert)'s theorem, B, X, E and D are concyclic
5. By a converse of Pascal's theorem applied to BXGAC Tc B, we are done...
Sincerely
Jean-Louis

[quote="mathpk"]Let $ \Gamma$ be the circumcircle of a triangle $ ABC$. A circle passing through points $ A$ and $ C$ meets the sides $ BC$ and $ BA$ at $ D$ and $ E$, respectively. The lines $ AD$ and $ CE$ meet $ \Gamma$ again at $ G$ and $ H$, respectively. The tangent lines of $ \Gamma$ at $ A$ and $ C$ meet the line $ DE$ at $ L$ and $ M$, respectively. Prove that the lines $ LH$ and $ MG$ meet at $ \Gamma$.

See all problems from APMO 2008 here,
http://www.mathlinks.ro/viewtopic.php?p=1073977#1073977[/quote]
Let $ LH\cap (O) \equal{} \{K\}$
We have $ \angle LKA \equal{} \angle HCA \equal{} \angle LDA$ then $ ALKD$ is cyclic.
Let $ ED\cap AC \equal{} \{R\}, RB\cap (O) \equal{} \{K'\}$
We have $ RK'.RB \equal{} RA.RC \equal{} RE.RD$ so $ K'EDB$ is cyclic.
$ \Rightarrow \angle K'DE \equal{} \angle K'BE \equal{} \angle K'AL$ so $ ALK'D$ is cyclic
Therefore $ K'\equiv K$ we get $ KEDB$ is cyclic.
$ \Rightarrow \angle AKC \equal{} \angle EBD \equal{} \angle EKD$
But $ \angle KDE \equal{} \angle KAL \equal{} \angle KCA$ then $ \Delta KED$~$ \Delta KAC$
$ \Rightarrow \angle KED \equal{} \angle KAC \equal{} \angle KCM \Rightarrow KECM$ is cyclic.
Let $ MG\cap (O) \equal{} \{K''\}$ then similarly, we obtain $ K''ECM$ is cyclic so $ K''\equiv K$->QED

Let $LH\cap {\Gamma}=K$, the tangent lines of $\Gamma$ at $A$ and $C$ be $t_1$ and $t_2$, $KG\cap {t_2}=M'$ and $BK\cap AC=P$, and the line $ELDM$ be $l$.
Pascal's Theorem implies ${t_1}\cap KH=L$, $AB\cap HC=E$, $BK\cap CA=P$ are collinear i.e. $P$ lies on $l$.
Pascal's Theorem implies ${t_2}\cap KG=M'$, $CB\cap GA=D$, ${BK\cap AC}=P$ are collinear i.e. $M'$ lies on $l$ i.e. $t_2, KG, l$ concurrent.
Since ${t_2}\cap l=M$, $M,G,K$ are collinear i.e. $LH\cap MG=K$ i.e. on ${\Gamma}$.

[quote="epitomy01"]The fact that $ BCDE$ is concyclic is superfluous (it may make the problem easier, but it is not necessary for the problem to be true).
Let $ N_{1}$ be the point on where $ LH$ intersects the circle (apart from $ H$). By using Pascal, with $ A' \equal{} A$ being a degenerate point, we obtain that the following points are collinear: $ AB \cap HC, AA' \cap HN, CA' \cap BN$ - this means $ DE, AC$ and $ BN_{1}$ are concurrent (since $ E \equal{} AB \cap HC, L \equal{} AA' \cap HN$), so if $ X$ is where $ AC, DE$ meet, then $ N_{1}$ is where $ XB$ intersects the circle again. If we similarly define $ N_{2}$ to be the point where $ MG$ intersects the circle, we will obtain, by similarly applying Pascal, that $ N_{2}$ is also where $ BX$ intersects the circle - this means $ N_{1} \equal{} N_{2}$, which solves our problem.
In the original problem (with $ BCDE$ concylic), this gives us the cute result, that $ N$ is the Miquel point of $ BCDE$.[/quote]

[hide=Solution]
Lemma 1: $MC = MD$.
Proof: $\angle MDC = 180-\angle CDE = \angle EAC = \angle BAC$ and $\angle MCD = \angle MCB = \angle CAB$(because $MC$ is a tangent), so $MC = MD$.

Now, let $K$ be the second intersection point of $(BED)$ with $(ABC)$.

Lemma 2: $(KGD)$ is tangent to $MD$ at $D$.

Proof: $\angle GKD = \angle BKG - \angle BKD = 180 - \angle BAG -(180 - \angle BED) = 180- \angle EAD - \angle DEA = \angle ADE = \angle MDG$, so $MD$ is indeed tangent to $(KGD)$.

Now, note that line $KG$ is the radical axis of $(KGD)$ and $(ABC)$. However, since $MC$ is tangent to $(ABC)$ and $MD$ is tangent to $(KGD)$ and $MD = MC$, $M$ must also lie on the radical axis of $(ABC)$ and $(KGD)$ so $M, G, K$ are collinear. Analogously, we can derive that $L, H, K$ are collinear, so lines $LH$ and $MG$ intersect at $K$, which by definition, lies on $(ABC)$.[/hide]

[quote=v_Enhance]Drop the condition that $AEDC$ is cyclic (this is the hardest part).
[/quote]
Where does the motivation for this come from?

[quote=TAN768092100853]Excuse me,how does the solution of v_Enhance works and what does this transform preserves?[/quote]

When the problem statement is purely intersections, or a circle is involved in the manner it is here, we can take a projective transformation. Essentially, this let's us change our configuration into something easier to work with (think random quadrilateral ----> square) while preserving the properties that are important to what we are working with. In this manner, we can prove something about a more complicated figure, by assuming it's a simpler figure.

Proof:

Let $K'$ be the intersection of $GM$ with the circumcircle. By Pascal's on $(AGK'BCC)$, we see that $D$, $M$, and the intersection of $AC$ and $K'B$ are collinear. Similarly, if $K''$ is the intersection of $LH$ with the circumcircle, $L$, $E$, and the intersection of $AC$ and $BK''$ is collinear. Since $L, E, D, M$ lie on one line, we must have $K'=K''$, so the intersection of $LH$ and $GM$ lies on the circumcircle.

bary-bash practice:
let $A(1,0,0),B(0,1,0),C(0,0,1)$
then circle through $A$ and $C$ has equation: $a^2yz+b^2xz+c ^2xy=(x+y+z)vx$
thus we get: $E(\frac{v}{c^2},1-\frac{v}{c^2},0)$ and $D(0,1\frac{v}{a^2},\frac{v}{a^2})$
lines $AD$ and $CE$ have equations : $\frac{y}{z}=\frac{a^2}{v}-1$ and $\frac{y}{x}=\frac{c^2}{v}-1$
intersecting with $(ABC)$ we get : $G(1,\frac{c^2}{v}-1,\frac{(v-c^2)c^2}{a^2c^+v(b^2-a^2)}=(1,\frac{c^2}{v}-1,A)$
and $H(\frac{a^2(v-a^2}{a^2c^2+v(b^2-c^2},\frac{a^2}{v}-1,1)=(B,\frac{a^2}{v}-1,1)$
$DE$ has equation $x(v-c^2)+yv+z(v-a^2)=0$ and intersecting with the tangents ,we get : $L(\frac{b^2v+c^2(a^2-v}{c^2(v-c^2},-\frac{b^2}{c^2},1)=(C,-\frac{b^2}{c^2},1)$
and $M(1,-\frac{b^2}{a^2},\frac{a^2(v-c^2)-b^2v}{a^2(a^2-v)})=(1,-\frac{b^2}{a^2},D)$
we calculate equations of $LH$ and $MG$ and it remains to verify that the following system has a solution:which can be easily verified

$$x(\frac{b^2}{a^2}D+A(\frac{c^2}{v}-1)+(A-D)y-z(\frac{c^2}{v}+\frac{b^2}{a^2}-1)=0$$
$$x(\frac{a^2}{v}+\frac{b^2}{c^2}-1)+(C-B)y-z(\frac{b^2}{c^2}B+C(\frac{a^2}{v}-1)=0$$
$$a^2yz+b^2xz+c^2xy$$

[hide=Solution?]
Let $\overline{LH}\cap\Gamma=K$. It suffices to show that the intersection of $\overline{KG}$ and the tangent to $\Gamma$ at $C$ intersect on $\overline{ED}$. However this is obvious after Pascal's Theorem on $BKHCAA$ and $KGACCB$.

A solution without Pascal.

By Reim's theorem, $DE \parallel GH$. Let $LH$ meet $(ABC)$ in $P$. Then $\angle APL = \frac{1}{2} \cdot \widehat{HA} = \angle AED$, so $LPEA$ is cyclic. $DE$ is antiparallel to $AC$ in triangle $ABC$, so if $P'$ is the Miquel point of $ADEC$ and $O$ is the circumcenter of $ADEC$, we have $L \in (AOE)$, so $P \in (AOE)$, so $P \equiv P'$. Thus by symmetry, we have $LH$ and $MG$ meeting on the Miquel point of $ADEC$, which is obviously on $(ABC)$.

Super-fast solution by Pascal's theorem!

Let $(BDE)$ meet $\Gamma$ again at $P$. So, $BP, AC$ and $DE$ concur (at the radical center). By Pascal's theorem on $PHCAAB$, $P,H$ and $L$ are collinear. Similarly, $P,G$ and $M$ are collinear. Hence, we are done. $\square$

A simple angle-chasing solution: Let $\odot (BDE) \cap \Gamma = X$. Note that $\angle HAB = \angle HCB =\angle ECD = \angle EAD = \angle BAG \Rightarrow \overarc{HB}=\overarc{BG}$

Also, $\angle BXA = \angle BCA = \angle BED =180^{\circ}-\angle BXD \Rightarrow BX$ is the external angle bisector of $\angle AXD$.
This means that $XA$ and $XD$ are isogonal w.r.t. $\angle GXH$.

Now, Let $GM \cap \Gamma = X'$. Notice that $\angle MDC = \angle EDB =\angle BAC=\angle MCD \Rightarrow MC=MD$.

But, $MG \cdot MX'=MC^2=MD^2 \Rightarrow MD$ is tangent to $\odot (GDX')$ $\Rightarrow \angle GX'D = \angle MDG = \angle ADE = \angle ACH = \angle AXH =\angle GXD \Rightarrow G,X,X',D$ are concyclic $\Rightarrow D$ lies on $\Gamma$, which is not true.

Thus, We get that $X'=X$ $\Rightarrow M,G,X$ are collinear. Similarly, $L,H,X$ are collinear. Hence, $LH$ and $MG$ meet on $\Gamma$.

[quote=GorgonMathDota][quote=v_Enhance]Drop the condition that $AEDC$ is cyclic (this is the hardest part).

Now let $P = \overline{AD} \cap \overline{CE}$.  Take a projective transformation which preserves $(ABC)$ and sends $\overline{BP} \cap \overline{AC}$ to the center of the circle.

Then it's plain by Ceva that $\overline{ED} \parallel \overline{AC}$, so $ALCM$ is a rectangle. Thus $ALHE$ and $CMGD$ are concyclic, and then straight angle chasing. :D[/quote]

Could anyone explain this projective transformation method?[/quote]

Read evan's geometry book, p. 184

[quote=mathpk]Let $ \Gamma$ be the circumcircle of a triangle $ ABC$. A circle passing through points $ A$ and $ C$ meets the sides $ BC$ and $ BA$ at $ D$ and $ E$, respectively. The lines $ AD$ and $ CE$ meet $ \Gamma$ again at $ G$ and $ H$, respectively. The tangent lines of $ \Gamma$ at $ A$ and $ C$ meet the line $ DE$ at $ L$ and $ M$, respectively. Prove that the lines $ LH$ and $ MG$ meet at $ \Gamma$.[/quote]
$  $\\
[b]Solution:[/b] Firstly define the following points : $T = LH \cap \Gamma$ ; $ T^{\prime} = MG \cap \Gamma$ ; $ K = AC \cap BT$ and $ K_{1} = AC \cap BT^{\prime}$ \\
$  $\\
We proceed with the following lemmas.\\
$  $\\
Lemma 1 : $K \in \overline{DE}$\\
Proof:  Observe that by Pascal's theorem on hexagon $\text{ AABTHC}$, we have that:\\
$AA \cap TH (=L)$ ; $ AB \cap HC(=E)$ ; $ BT \cap AC (=K)$\\
are collinear $ K \in \overline{DE}$.\\
$  $\\
Lemma 2 : $K_{1} \in \overline{DE}$\\
Proof: Now, by Pascal's theorem on $\text{CCAG}T^{\prime}B$:\\
$ CC \cap GT^{\prime} (=M)$ ; $ CA \cap BT^{\prime}(=K_{1})$ ; $ AG \cap BC (=D)$\\
are collinear $\implies K_{1} \in DE $\\
$ $\\
Now both these lemmas imply that  $K = AC \cap BT \cap DE \implies AC \cap DE = K_{1}$. This is possible iff $T \equiv T^{\prime} \implies  LH$ and $MG$ meet in $\Gamma$.\\
$ $\\
[hide = PS] Why did they give in the condition of concyclicity?? [/hide]

[quote=Drunken_Master] By Pascal's theorem on $PHCAAB$, $P,H$ and $L$ are collinear. Similarly, $P,G$ and $M$ are collinear. Hence, we are done. $\square$[/quote]

Can anyone explain, how?

Trollol :rotfl:
[Hide=Pollution]
APMO 2008 #3

 Let $K=DE \cap AC$ . Let $BK \cap (ABC) = X$ . Now,$$\text{Pascal on } XHCAAB \implies E,K,XH \cap AA \text{ are collinear} \implies X,H,L \text{ are collinear} $$
Similarly $X,G,M $ are collinear. Done :D

Let $DE \cap AC = Y$ and let the second intersection of $BY$ with the circumcircle be called $X$.

[b]Claim: [/b]$X,M,G$ are collinear.

[i]Proof: [/i]We use Pascal on $CCBXGA$. This tells us that the intersection of the tangent at $C$ and $XG$, $D$, and $Y$ are collinear. But note that $M$ is the intersection of the tangent at $C$ and $DT$, so $X,M,G$ are collinear, as desired.

Similarly, $L,X,H$ are collinear, and we are done.

[hide="Remarks"]Personally, I felt that this was quite a hard problem, despite its deceptively short write-up. Since half the points in this problem had to do with a single circle, and there were tangents, Pascal's Theorem seemed like a good idea. However, in order to apply Pascal, the condition of the problem had to be modified so that three points were collinear, which motivated the addition of point $Y$ after some trial and error.[/hide]

Solved with [b]@blastoor[/b]
It is easy to see that:
$$ \angle LAB = \angle ACB = \angle BEC = \angle LEA $$
This proves that $LA = LE$. Similarly we can obtain that $MD = MC$. Assume that $LH \cap \odot(ABC) = X$. From PoP follows that:
$$ LA^ 2 = LH \cdot LX = LE^2 $$
This implies that $LE$ is tangent to $\odot(HXE)$. This implies that $\angle LEX = \angle LHE \implies \angle XED = \angle XHE $. It is easy to see that $\angle XBD = 180 -\angle XHE$. Consequently quadrilateral $BXED$ is cyclic.
Now if we assume that $MG \cap \odot(ABC) = X'$, then in similar manner we can obtain that $BDEX'$ is cyclic. This implies that $X = X'$, therefore $LH$ and $MG$ intersects on $\odot(ABC)$ as desired.

We claim that the condition that $AEDC$ is cyclic is irrelevant, in which case the only Euclidean condition left is the circumcircle. Denote $T = \overline{AD} \cap \overline{CE}$ and send $\overline{BT} \cap \overline{AC}$ to the center of the circumcircle.

Then, observe that $\triangle ABC$ has a right angle at $B$, and $T$ lies on the median from $B$. Thus, $\overline{DE} \parallel \overline{AC}$, which yields $ALMC$ a rectangle. We also find that $ALHE$ and $MGDC$ are cyclic. Denote $X = \overline{LH} \cap \overline{MG}$; we proceed with a directed angle chase.

$$\begin{aligned}
    \measuredangle HXG &= \measuredangle LXM \\ &=  \measuredangle XLM + \measuredangle LMX \\ &=  \measuredangle HLE + \measuredangle DMG \\ &=  \measuredangle HAE + \measuredangle DCG \\ &= \measuredangle HCB + \measuredangle BCG \\ &= \measuredangle HCG,
\end{aligned}$$ as desired.

Let $DE\cap AC=J$. Let $JB\cap HL=K$. As $AA\cap HK,AC\cap BK,HC\cap AB$ are collinear, $K$ must lie on $(ABC)$ by the converse of Pascal. Similarly, $JB\cap MG$ lies on $(ABC)$, so $MG\cap HL$ is the second point at which $JB$ intersects $(ABC)$.

Can we do this projective transformation?

$(AEDC)$ is cyclic is irrelevant. So we send $AEDC$ to a isosceles trapezoid. So we have $\triangle ABC$ is isosceles and thus by symmetry $LH \cap MG = B$.

Let $F$ be the miquel point of quadrilateral $ACDE$
I claim that it's the intersection point.
set $G'=FM\cap (ABC)$,we have
$\angle{CAD}=\angle{CED}=\angle{CEM}$
$\angle{CAG}=\angle{CFG}=\angle{CFM}$
so it suffices to prove that $FECM$ is cyclic
but this is evident because
$\angle{MCF}=\angle{CAF}=\angle{DEF}=\angle{MEF}$
So $G'=G$ , similar argument for $F,H,L$ and we are done.

Solved with [b]L567[/b].

Let $LH$ and $GM$ meet $\Gamma$ at points $P$ and $Q$. \\
By Pascal on $CCAPHB$ and $AACQGB$ we have that $AP$ and $AQ$ both pass through the same point, namely, the intersection of $CA$ and $ED$ meaning that $P = Q$ and consequently $LH$ and $MG$ meet at $\Gamma$. $\blacksquare$ \\

[quote=mathpk]Let $ \Gamma$ be the circumcircle of a triangle $ ABC$. A circle passing through points $ A$ and $ C$ meets the sides $ BC$ and $ BA$ at $ D$ and $ E$, respectively. The lines $ AD$ and $ CE$ meet $ \Gamma$ again at $ G$ and $ H$, respectively. The tangent lines of $ \Gamma$ at $ A$ and $ C$ meet the line $ DE$ at $ L$ and $ M$, respectively. Prove that the lines $ LH$ and $ MG$ meet at $ \Gamma$.[/quote]

Nice problem.

Let $X \equiv LH \cap \Gamma$ and $Y \equiv MG \cap \Gamma$ distinct from $H$ and $G$ respectively. We prove that $X \equiv Y$.

Using Pascal's Theorem on degenerate hexagon $ABXHCA$, we get that $AB \cap HC \equiv E$, $BX \cap CA \equiv T$ and $XH \cap AA \equiv L$ are collinear. Similarly, if $T' = BY \cap CA$, then using Pascal's Theorem on degenerate hexagon $CBYGAC$, we get that $CB \cap GA \equiv D$, $BY \cap AC \equiv T'$ and $YG \cap CC \equiv M$ are collinear. This means that $T$ and $T'$ both lie on line $DE$ and line $AC$, therefore $T \equiv T'$, therefore $X \equiv Y$ as desired.

Let $K$ be the Miquel point of almost complete quadrilateral $AEBDC$. I contend $K$ lies on $\overline{MG}$ and $\overline{HL}$.
By radical axis theorem on $(BDEK),(ABCK)$ and $(ACDE)$, we obtain that $\overline{KB},\overline{DE}$ and $\overline{AC}$ are concurrent, say at $P$.
By Pascal on $KGACCB$, we get that $\overline{KG}\cap \overline{CC}$, $D=\overline{AG}\cap\overline{BC}$ and $P=\overline{AC}\cap\overline{BK}$ are collinear, we conclude that $K$ lies on $\overline{MG}$, similarly $K$ lies on $\overline{HL}$. $\blacksquare$

Let $\Gamma$ intersect $LH$ and $MG$ at $X_1$ and $X_2,$ respectively. By Pascal on $AABX_1HC,$ $P=\overline{LD}\cap\overline{AC}$ lies on $\overline{BX_1}.$ Similarly, by Pascal on $CCBX_2GA,$ $P$ lies on $\overline{BX_2}.$ Hence, $X_1=X_2.$ $\square$

:whistling:

Rename the problem to be $A$-indexed, rename $G$ and $H$ to $P$ and $Q$, and rename $L$ and $M$ to $R$ and $S$, and let $D$ and $E$ be any points on $\overline{AB}$ and $\overline{AC}$.

Notice that $X=\overline{BC} \cap \overline{DE}$ is outside of $\Gamma$, so there exists a homography fixing $\Gamma$ and taking $X$ to a point at infinity (by a well-known lemma), which means $\overline{BC} \parallel \overline{DE}$. Then, we can angle chase to show that $BEQR$ and $CDPS$ are cyclic, and angle chase to show that both $\overline{QR}$ and $\overline{PS}$ pass through the reflection of $A$ over the perpendicular bisector of $\overline{BC}$.

Apparently solved this in the past but because of my bad memory, here's a different solution than last time.

Let $X = MG \cap \Gamma \neq G$. Note that since $MD^2 = MC^2 = MG.MX$, we have $MD$ is tangent to $(GDX)$. To finish, $\angle BXD = \angle BXG + \angle GXD = \angle EAD + \angle GDM = \angle EAD + \angle ADE = \angle BED$, so $X$ lies on $(BDE)$, similarly $LH \cap \Gamma$ does too, so they're the same point. $\blacksquare$

i heckin love geometry

After being told by a certain "Even-chan" to ignore the fact that $AEDC$ is cyclic, we employ a projective transformation which fixes $\Gamma$ and sends $F$ to the center of $\Gamma$, where $F$ is the point on $\overline{AC}$ such that $\overline{AD},\overline{CE},\overline{BF}$ concur (I hate $B$-indexed geo). Then $\triangle ABC$ is right with $\angle B=90^\circ$, and it is well-known (by Ceva's) that $\overline{DE} \parallel \overline{AC}$. Then $\angle ALD=\angle AHE=\angle ABD=90^\circ$ hence $ALHE$ and $ALBD$ are cyclic. Likewise, $CMGD$ and $CMBE$ are cyclic. Thus, we have
$$\angle BHL=\angle BHE=\angle BCE=\angle BCH=\angle BAH=\angle EAH=\angle ELH=\angle MLH,$$
and likewise $\angle BLH=\angle LMG$. Thus if $\overline{LH} \cap \overline{MG}=T$, then $\triangle BLM$ and $\triangle TML$ are simply reflections of each other over the perpendicular bisector of $\overline{LM}$, hence $T$ lies on $\Gamma$ as well. $\blacksquare$

can we prove this with pappus's hexagon theorem?

Let $F_1 = \overline{LH} \cap \Gamma$ and $F_2 = \overline{GF} \cap \Gamma$. By Pascal on $F_1HCAAB$, $\overline{BF_1} \cap \overline{CA}$ lies on $\overline{EL}$. By similar logic, $\overline{BF_2} \cap \overline{CA}$ lies on $\overline{EM} = \overline{EL}$, hence $F_1=F_2$.

Let $T_1=\overline{LH} \cap \Gamma$ and $T_2=\overline{GM} \cap \Gamma$. By Pascal's theorem on $T_1HCAAB$, $\overline{T_1B}$, $\overline{AC}$, and $\overline{DE}$ concur. Similarly, Pascal on $T_2GACCB$ implies that $\overline{T_2B}$, $\overline{AC}$, and $\overline{DE}$ concur. We perform the following casework. [list]
[*] If $T_1=T_2$, then we are done by phantom point.
[*] If $T_1 \neq T_2$, then it follows that $\overline{AC} \cap \overline{DE}=B$, but this is impossible since $\triangle ABC$ is acute. [/list]
 We are done.

huh?

Let $T = \overline{AC} \cap \overline{DE}$, $X_1 = \overline{LH} \cap \Gamma, X_2 = \overline{MG} \cap \Gamma$. By Pascal's on $AABX_1HC$ and $CCBX_2GA$ we find that $B$ lies on both $TX_1$ and $TX_2$, which is enough to imply $X_1 = X_2$.

Let $F = (BED) \cap \Gamma$. We make the following claim:
[b][color=#f00]Claim:[/color][/b] $\triangle LAE$ and $\triangle MDC$ are isosceles.
[i]Proof:[/i] \[\angle LAE = \angle ACB = \angle C; \phantom{c} \angle LEA = \angle BED = \angle C, \] and a likewise similar proof ensues for $\triangle MDC$. $\square$

Now as $LA = LE$, it would suffice to show $L$ bears the same power with respect to $\Gamma$ as it does to $(HEF)$, as then $L, H, F$ would be collinear due to the existing radical axis.
[b][color=#f00]Claim:[/color][/b] $LE$ is tangent to $(HEF)$.
[i]Proof:[/i]
\begin{align*}
\angle HFE &= \angle AFH - \angle AFE \\
&= \angle ACH - (\angle AFB - \angle EFB) \\
&= \angle ACH + \angle A - \angle C \\
&= \angle C - \angle HCB + \angle A - \angle C \\
&= \angle A - \angle DAE \\
&= \angle A - (\angle A - \angle DEC) \\
&= \angle HEL,
\end{align*}
as desired. $\square$

It therefore similarly follows that $M, G, F$ are collinear and hence both lines $LH$ and $MG$ meet at $F$ on $\Gamma$. $\blacksquare$


Let $ V$ be an $ n$-dimensional inner product space over the field $ \mathbf{R}$ or $ \mathbf{C}$, and let $ L(V,V)$ be the space of linear operators on $ V$. For each $ \alpha,\beta$ in $ V$, let $ T_{\alpha,\beta}$ be the linear operator on $ V$ defined by $ T_{\alpha,\beta}(\gamma)\equal{}<\gamma,\beta>\alpha$. Show that there is a unique inner product on $ L(V,V)$ with the property that $ \parallel{}T_{\alpha,\beta}\parallel{}^2\equal{}\parallel{}\alpha\parallel{}^2 \parallel{}\beta\parallel{}^2$ for all $ \alpha,\beta\in V$. ($ \parallel{}T_{\alpha,\beta}\parallel{}$ is the norm generated by the  inner product on $ L(V,V)$, not an operator norm.)

Question: why is unique?

You have the inner product prescribed on the set of rank-one operators. Any operator is a sum of rank-one operators.

The [b]norm[/b] is prescribed on the set of rank-one operators. Denote $ <\cdot,\cdot>_L$ the inner product on $ L(V,V)$. What will be $ <T_{\alpha,\beta},T_{\gamma,\delta}>_{L}$?


Please tell the truth.
I spend about 4 hrs on this forum and FTW :lol:  but only about 20 mins a week besides that.

I spend about  80 mins a day on AOPS, other than that, I have a 2-hour training session each week with my team. Most of the time on AOPS isn't on MATHCOUNTS, so about 30 mins is. I spend about 5 1/2 hours now that I think about it.

Every monday and wednesday after school about an hour and a half and then about 2 hours at home which equals to about 5 hours.

2 hours aweek with my team and about 2 hours a day on aops, at least, before For The Win was disabled.  Now I do about an hour or so on Aops and an hour elsewhere.

From FTW, about an hour and a half. We have a 1/2 hour session in school, and that's all I do.

2 2-hour sessions in school with the team... 1.5 hours weekly on FTW... 1.5 hours weekly on my own... 7 hours.

I have 1 and a half our training class for mathcounts each monday, 1 hour for school meet on thursday and an average of 3 hours aops per week so thats like 5 and a half hours.


my dad recently gave me this essay contest flyer. the prize is huge but so is the esssay supposed to be. :P i already have 2100 words. :o  Beat that.  :lol:
[url]http://www.umassd.edu/indic/waves/06_wdc_conf/WAVES_Essay_Competition.jpg[/url]
what do you think?

wow...the topic itself is too hard, forget about writing 2000+ words on it...


How many ways are there to divide a set with 6 elements into 3 disjoint subsets?

$\textbf{(A)}\ 90   \qquad\textbf{(B)}\ 105   \qquad\textbf{(C)}\ 120   \qquad\textbf{(D)}\ 180   \qquad\textbf{(E)}\ 243$

Sorry, I edited this post:

The number of ordered triple subsets: $ 3^6$
The number of ordered triple subsets such that at least one of them is empty: $ 3\cdot2^6$
The number of ordered triple subsets such that two of them are empty: 3

Answer: $ \frac16(3^6\minus{}3\cdot2^6\plus{}3)\equal{}90$.

It is the wrong answer. If you click [url=http://www.mathlinks.ro/viewtopic.php?t=111796]Turkey NMO First Round - 1998[/url], you will see the answer key at the end of second post.

[hide="Combinatorial approach"]For the first subset, the possibilities are $ \binom {6}{2}$; for the second and third subsets, the possibilities are $ \binom {4}{2}$. Multiplying, we find that

$ S(6, 3) \equal{} \binom {6}{2} \binom {4}{2} \equal{} \boxed{90}$[/hide]

metafor, it's purely a coincidence that your answer is the same: you counted some combinations (e.g., $ \{\{1, 2\}, \{3, 4\}, \{5, 6\}\}$) six times, but counted others (e.g., $ \{\{1\}, \{2, 3\}, \{4, 5, 6\}\}$) zero times.

The number of elements of subsets can be $ \{1,2,3\}$ or $ \{2,2,2\}$ or $ \{1,1,4\}$. If we choose two sets, the remaining one will automatically will be chosen.
$ {{6}\choose{1}}{{5}\choose{2}} \plus{} \frac {{{6}\choose{1}}{{5}\choose{1}}}{2!} \plus{} \frac {{{6}\choose{2}}{{4}\choose{2}}}{3!} \equal{} 90$ (We divide by $ 2!$ because $ S_1S_2$ and $ S_2S_1$ are same set divisions.)

[b]JBL[/b] — Ha-ha, so it is. Thank you.


How many ordered triples of three prime numbers exist for which the sum of the members of the triple is 24?

One number must be $ 2$, otherwise the result will be odd. $ 24\minus{}2\equal{}22$, leaving for two primes summing to $ 22$.

There's $ 3$ and $ 19$, $ 5$ and $ 17$, and $ 11$ and $ 11$. There are $ 6$ ways to arrange each the first and second cases, and $ 3$ ways to arrange the third, yielding a total of $ 6\plus{}6\plus{}3\equal{}\fbox{15}$ ordered triples.


search
inequalities unsolved
inequalities

let a,b,c positive numbers. prove that ieq
$ \frac{ab}{3a\plus{}4b\plus{}5c}\plus{}\frac{bc}{3b\plus{}4c\plus{}5a}\plus{}\frac{ca}{3c\plus{}4a\plus{}5b}\leq\frac{a\plus{}b\plus{}c}{12}$.

This is a problem from Bulgaria TST 2006. Search on site.


Latex keeps giving me "File `mathdesign.sty' not found. \usepackage" when I try to include the package mathdesign. I have mathdesign installed, I think, because /usr/share/texmf/tex/latex/mathdesign/mathdesign.sty exists. But LaTeX can't find it. Any suggestions?

LaTeX finds files, not by looking in the various directories as that would take too long, but by maintaining a database of where all the files are.

To refresh the database in Linux use the command line and type
texhash
and I hope that cures the problem.

aha, indeed it does cure the problem :). Many many thanks!


Find the number of the positive integers $a$, less than 2003, for which there exists a positive integer $n$, such $3^{2003}| n^{3}+a$.

Just as a tip: in latex, put 3^{2003} so you get:
$3^{2003}|n^{3}+a$

if a is a perfect cube, there is the obvious solution $n+a=3^{2003}$...

could you please explain how you know that are $3^{2002}$ in total?

I am not sure if there are $3^{2002}$ in total.  If we find the cubic residues of mod $3^{2}$, there are only 0,1,-1.  So that means we have a maximum of $3^{2002}$.  But since for example, $(3^{1000})^{3}\equiv (3^{1001})^{3}$ (mod $3^{2003}$), there are less than $3^{2003}$.

 Looking at the cubic residues on my calculator, it seems all that are equal to -1 or 1 mod 9 are cubic residues and if a number is divisible by 3, we divide both the mod and the number by powers of 3 until we can decide whether it is a residue or not.  But that is not very helpful is it.

[quote="ehehheehee"]But since for example, $(3^{1000})^{3}\equiv (3^{1001})^{3}$ (mod $3^{2003}$), there are less than $3^{2003}$. [/quote]

Hmm.  You're right.  Either way, I'm still not sure how to count the cubic residues within a specific bound  :|

[quote="t0rajir0u"]Well, clearly $-a$ is a cubic residue $\bmod 3^{2003}$, and there are $3^{2002}$ of those [i]in total[/i], but I don't know how to count the ones within a range.[/quote]
First I thought to use the same. But it doesn't help, because of the 3^2003 mod and the limitation a<2003.
There's very interesting approach.
I'll give a hint:
[hide]Let:
$(a,3)=1$
and
$3^{s}| n^{3}+a$, but $3^{s+1}\nmid n^{3}+a, s \geq 2$.
That means: $n^{3}+a=3^{s}t,(t,3)=1$.
Now we'll try to construct such number $n_{1}$, that:
$3^{s+1}| n_{1}^{3}+a$.
Now, this is like some puzzle. You know what properties should have something, but you don't see how it looks like. I can't explain the construction of n_1 better than just giving it:
$n_{1}= n+2*3^{s-1}*t$
$n_{1}^{3}+a = (n+2*3^{s-1}*t)^{3}+a =$
$= (n^{3}+a)+3n^{2}2*3^{s-1}*t+3n*4*3^{2s-2}*t^{2}+8*3^{3s-3}*t^{3}=$
$=3^{s}t+2 n^{2}3^{s}+3^{s+1}M=$
$=3^{s}t(1+2n^{2})+3^{s+1}M$
Since (a,3)=1, then (n,3)=1 and:
$1+2n^{2}\equiv 1+2(1) \equiv 0 (\bmod3)$, so:
$n_{1}^{3}+a \equiv 0 (\bmod 3^{s+1}).$
First time I saw it i was very happy - completely amazing approach!
[/hide]


calculus
integration
geometry
vector
function
calculus computations

Single integration gives area, Double integration gives voulme ..........so what does triple integration gives.

Also, are there more dimensions than 3, i know we use vectors of n dimension and some where i saw time was mentioned as 4th dimension.
but plz give some more information on higher dimensions and its uses.

Thank You

That is an iffy question. You can calculate area and volume (though they are special cases) with just a single integral. I think if you wanted to just assign interpretations like that you can say a triple integral is a volume of some subset $ \Omega \subset \mathbb{R}$, or with a mass density function, $ \rho(x,y,z)$ to give the mass of "stuff".
ie. $ \int\int\int \rho(x,y,z) dxdydz$ could be the mass of some volume of compressed air.

Thank you for replying....
You said
[quote="Fermat's Little Turtle"]You can calculate area and volume (though they are special cases) with just a single integral.[/quote]

remember when we calculate volume via single integration...we integrate a special function such that the numerical value of the answer is equal to the volume, but theoritically we are just computing the area under the curve.

Similarly when we calculate area by double integration we put z=1, so the height of the cuboid(or cylindrical) bars becomes 1 when and what ever we get gives us volume with height 1 hence, numerically magnitude wise its area.

I hope u got my point


Let $\cal {C}$ be a circle of radius 1 and $P_{1},P_{2},...,P_{n}$ inside the circle. Prove that there is a semicircle ${\cal S} \subseteq {\cal C}$ s.t. $ MP_{1}+MP_{2}+...+MP_{n} \geq n $ for all $M \in {\cal S}$.
I only hope this is not classical as well.

It's obviously true for $n=1$, and then we can apply the case $n=1$ to the centroid $G$ of the configuration $P_1,\ldots,P_n$, since every point $X$ in the plane satisfies $XP_1+\ldots+XP_n\ge nXG$.

Woo..., you're fast! That's exactly my solution.


function
inequalities
symmetry
algebra proposed
algebra

Find all functions $ f$, defined on the non negative real numbers and taking non negative real numbers such that $ f(x^2)\plus{}f(y)\equal{}f(x^2\plus{}y\plus{}xf(4y))$ for any non negative real numbers $ x,\ y$.

Pretty cool problem.

I don't want to spoil the fun of solving this problem, so it's all hidden.

[hide]First, notice that $ g(x) \equal{} x^2 \plus{} kx$ spans all non-negative reals for non-negative values of $ x$, because $ g(0) \equal{} 0$ and $ g(x)$ goes to infinity as $ x$ increases.

Then $ f$ is non-decreasing: if $ z \equal{} y \plus{} a$, $ a > 0$, then there is $ x > 0$ such that $ x^2 \plus{} 4f(y)\cdot x \equal{} a$. Choosing such $ x$ we have $ f(x^2) \plus{} f(y) \equal{} f(z)$ and since $ f$ can only assume non-negative values, $ f(z) \geq f(y)$.

Suppose that there is $ k > 0$ such that $ f(k) \equal{} 0$. Then, since $ f$ is non-decreasing and, by letting $ x \equal{} y \equal{} 0$, $ f(0) \equal{} 0$, $ f(x) \equal{} 0$ for $ 0 \leq x \leq k$. Letting $ x$ be $ \sqrt x$ and $ y \equal{} k/4$, we obtain $ f(x) \plus{} f(k/4) \equal{} f(x \plus{} k/4)\iff f(x\plus{}k/4) \equal{} f(x)$. This implies that $ f$ is periodic with period $ k/4$. Since $ f([0,k]) \equal{} \{0\}$, $ f(x) \equal{} 0$ for all non-negative $ x$.

It remains to solve the case in which $ f(x) > 0$ for all positive $ x$. In this case, $ f$ is strictly increasing ($ f(z) \geq f(y)$ turns into a strict inequality) and, henceforth, $ f$ is injective. Substituting $ x$ by $ \sqrt x$ we obtain $ f(x) \plus{} f(y) \equal{} f(x \plus{} y \plus{} \sqrt xf(4y))$. By symmetry, $ f(x \plus{} y \plus{} \sqrt xf(4y)) \equal{} f(x \plus{} y \plus{} \sqrt yf(4x)) \iff \frac{f(4x)}{\sqrt x} \equal{} \frac{f(4y)}{\sqrt y}$ This means that $ \frac{f(x)}{\sqrt x}$ is constant, so $ f(x) \equal{} c\cdot \sqrt x$. Direct substitution yields $ c \equal{} 1$ and we are done.[/hide]

[hide]
1) Putting root(a) to be x and na to be y, for n to be non-negative real. Then we get
f(na)/f(a) = root(n)
By putting a = 1, we know that all values except 1 and 0 depends on f(1),
so we can directly manipulate f(x) = cx^(.5) which easily get c = 1 or 0 by substituting back to the original equation and we get f(x) = x^(.5) or 0 for x,yER+, f:R+|->R+
 [/hide][/hide]

Japan-Mathematical_Olympiad_Finals-2009
NO. 5 :   $ f(x^2)\plus{}f(y)\equal{}f(x^2\plus{}y\plus{}xf(4y))$

f(x)=0, or f(x) progressively increase

swap $ x^2$  with $ y$,
$ f(y)\plus{}f(x^2)\equal{}f(y\plus{}x^2\plus{}\sqrt{y}f(4x^2))$
so  $ xf(4y) \equal{}\sqrt{y}f(4x^2)$
$ \dfrac{f(x)}{x} \equal{} c$

[quote="kunny"]Find all functions $ f$ defined on the non negative real numbers and taking non negative real numbers such that
\[ f(x^2) \plus{} f(y) \equal{} f(x^2 \plus{} y \plus{} xf(4y)) \quad (1)\]
for any non negative real numbers $ x$, $ y$.[/quote]

Fix $ y$ and let $ g(x) \equal{} x^2 \plus{} ax$, where $ f(4y) \equal{} a$, which is constant. It's easy to see that $ g(x)$ spans all non-negative real numbers. Hence, $ f(y) \le f(x^2) \plus{} f(y) \equal{} f(y \plus{} g(x))$, which implies that $ f$ is nondecreasing.
Setting $ x \equal{} y \equal{} 0$ yields $ f(0) \equal{} 0$. Assume that there exsits $ y_0 > 0$ such that $ f(y_0) \equal{} 0$. Then we have $ f(x) \equal{} 0$ for $ x \in [0$, $ y_0]$ as a consequence of the monotonicity. Plugging $ y \equal{} y_0/4$ into $ (1)$ we get $ f(x^2) \equal{} f(x^2 \plus{} y_0/4)$. Construct the sequence $ x_n \equal{} \sqrt {x_{n \minus{} 1}^2 \plus{} y_0/4}$ for $ n \ge 1$ with $ x_0 \equal{} 0$. Hence $ f(x_n^2) \equal{} f(x_{n \minus{} 1}^2 \plus{} y_0/4) \equal{} f(0) \equal{} 0$ and together with the monotonicity $ f(x) \equiv 0$ for all non-negative $ x$.
In the sequel, let $ f(x) \ne 0$ for $ x \ne 0$. Then we have $ f(y) < f(x^2) \plus{} f(y) \equal{} f(y \plus{} g(x))$ for $ x$, $ y \ne 0$ and thus $ f$ is strictly increasing.
Plugging $ y \to y^2$ into $ (1)$ we get $ f(x^2) \plus{} f(y^2) \equal{} f(x^2 \plus{} y^2 \plus{} xf(4y^2))$. Switch $ x$ and $ y$ to get $ f(x^2 \plus{} y^2 \plus{} xf(4y^2)) \equal{} f(x^2 \plus{} y^2 \plus{} yf(4x^2))$ and because $ f$ is strictly increasing $ xf(4y^2) \equal{} yf(4x^2)$ or equivalently $ \frac {f(4x^2)}{x} \equal{} \frac {f(4y^2)}{y} \equal{} c$, where $ c$ is some constant. Hence $ f(x) \equal{} \frac {c\sqrt {x}}{2}$. Plugging this into $ (1)$ we get $ c \equal{} 2$ and hence $ f(x) \equal{} \sqrt {x}$ is the second solution.

can anyone explain to me how we can prove that: f is injective

[quote="kunny"]Find all functions $ f$, defined on the non negative real numbers and taking non negative real numbers such that $ f(x^2)+f(y)=f(x^2+y+xf(4y))$ for any non negative real numbers $ x,\ y$.[/quote]

Let $P(x,y): f(x^2)+f(y)=f(x^2+y+xf(4y))$

Now to show function increasing fix $y$ and let's try to show for all $r$ we get $f(y+r)\geq f(y)$
Now so it's enough to show we can choose $x$ such that $x^2+xf(y)-r=0$ now as $r>0$ so this equation must
have one positive root calling that $x$ we're done to show.
Now our function won't be strictly increasing iff there exist some $a$ such that say $f(4a)=0\implies f(a)=0$
Now $P(\sqrt x,a)\implies f(x+a)=f(x)$ now increasing ness property implies $f(x)=0$ for all $x\geq 0$
Now $P(x,y^2),P(y,x^2)\implies f(x^2+y^2+xf(4y^2))=f(y^2+x^2+yf(4x^2))$
As function is strictly increasing so we must have $xf(4y^2)=yf(4x^2)\implies f(x)=c\sqrt x$
Now original equation implies $c=1$.

We claim the only solutions are $0$ and $\sqrt{x}$. Both work. Now we prove they are the only ones.

Firstly, note that $f$ is increasing. Indeed, $x^2+xf(4y)$ spans $[0, +\infty)$ when $x$ varies, and so $y \leq x$ implies $f(y) \leq f(x)$ if we take $x_0$ such that $x_0^2+x_0f(4y)=x-y \geq 0$. Moreover, $x=y=0$ in the given equation implies $f(0)=0$. Now, we split into two cases.

[b]Case 1:[/b] There exists a $u >0$ such that $f(u)=0$. Note that

$f(u+\sqrt{u}) \leq f(u+\sqrt{u}+\sqrt{u}f(4\sqrt{u}))=f(u)+f(u)=0,$

and so $f(u+\sqrt{u})=0$. Define the sequence $(a_n)$ with $a_1=u$ and $a_{n+1}=a_n+\sqrt{a_n}$. Note that $a_n$ is strictly increasing and unbounded: indeed, if it is bounded then it converges to $\ell \in \mathbb{R}$, and so $\ell=\ell+\sqrt{\ell}$, i.e. $\ell=0$, a contradiction.

Hence, for all $x>0$ we can find a term $a_N$ such that $f(a_N)>x$. Since all outputs of $(a_n)$ are zero, we infer that $f(x)=0,$ and so $f$ is identically zero in this case.

[b]Case 2:[/b] $f(x)=0 \Longleftrightarrow x=0$. In this case, slightly modifying the initial argument for proving increasingness, we can prove that $f$ is actually strictly increasing in this case. Hence, by putting $y \rightarrow y^2$, interchanging $x,y$ and using injectivity, we infer that $f(4x^2)/x$ is constant over all $x$, and from here we trivially conclude the second solution of the problem.


The drive propeller of a ship starts from rest and accelerates at 3.5*10^-3 rad/s^2 for 1.8*10^3 s.  For the next 3.60*10^3 s, the propellerrotates at a constant angular speed.  Then it deccelerates at 2.3*10^-3 rad/s^2, until it slows (without reversing direction) to an angular speed of 5.00 rad/s.  Find the total angular displacement of the propeller.


calculus
integration
trigonometry
function
abstract algebra
algebra
polynomial

Evaluate the following definite integral.

\[\int_{0}^{\pi}\frac{\cos nx}{2-\cos x}dx\ (n=0,\ 1,\ 2,\ \cdots)\]

There are a couple of ways for calculating this integral, I will play with this one:
Observe that

$\int_{0}^{\pi}\frac{\cos nx}{2-\cos x}dx=\frac{1}{2}\int_{0}^{2\pi}\frac{\cos nx}{2-\cos x}dx$.
Therefore your integral is the $n^{th}$ Fourier coefficient of the function $f(x)=\frac{1}{2-\cos x}$.

However we have, assuming that I made no computational mistakes, that:

$f(x)=\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}\sum_{n=1}^{\infty}(2-\sqrt{3})^{n}\cos nx$.
Thus,

$\frac{1}{\pi}\int_{0}^{2\pi}\frac{\cos nx}{2-\cos x}dx=\frac{2}{\sqrt{3}}(2-\sqrt{3})^{n}$.
Therefore

$\int_{0}^{\pi}\frac{\cos nx}{2-\cos x}dx=\frac{\pi}{\sqrt{3}}(2-\sqrt{3})^{n}$.

didilica, how did you calculate the fourier coefficient without calculating that integral directly?

I directly calculated. Use the Weierstrass substitution $t = \tan\frac{x}{2}$, then

\begin{eqnarray*}\int_{0}^{\pi}\frac{\cos nx}{2-\cos x}\, dx & = & \frac{1}{2}\int_{-\pi}^{\pi}\frac{\cos nx}{2-\cos x}\, dx\\ & = & \frac{1}{2}\int_{-\pi}^{\pi}\frac{e^{inx}}{2-\cos x}\, dx\\ & = & \int_{-\infty}^{\infty}\frac{1}{3t^{2}+1}\left(\frac{1+it}{1-it}\right)^{n}\, dt\\ & = & 2\pi i \underset{t=\frac{i}{\sqrt{3}}}{\text{Res}}\frac{1}{3t^{2}+1}\left(\frac{1+it}{1-it}\right)^{n}\\ & = & \frac{\pi}{\sqrt{3}}(2-\sqrt{3})^{n}\end{eqnarray*}

I used exactly didilica's method and got exactly the same answer. The starting point was the Poisson kernel formula:

$1+2\sum_{k=1}^{\infty}r^{k}\cos(kx)= \frac{1-r^{2}}{1-2r\cos x+r^{2}}$

Tweak that to make it match the function we have and you get $r=2-\sqrt{3}$ and eventually didi's formula.

One minor note: if $n=0,$ the integral is $\frac{\pi}{\sqrt{3}}.$ That case is slightly different.

[quote="Kent Merryfield"]I should add that replicating didi's calculation was my second attempt. My first attempt ended with the integral being equal to

$\sum_{j=0}^{\infty}\frac{\pi}{2^{2n+4j+1}}\binom{n+2j}{j}$[/quote]

Oh, that was the very result I got in my second attempt using $\frac{1}{2-\cos x}= \sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k}}{2^{2n+1}}\cos(n-2k)x$

For $I_{n}=\int_{0}^{\pi}\frac{\cos nx}{2-\cos x}dx\ (n=0,\ 1,\ 2,\ \cdots)$ you can find the relation of $I_{n},\ I_{n+1},\ I_{n+2}$.

Using $\cos (n+1)x+\cos (n-1)x = 2\cos x \cos nx$, we have

$I_{n+1}+I_{n-1}= 4I_{n}$

Calculating roots of the characteristic polynomial, we have $x^{2}-4x+1 = 0 \quad \Rightarrow \quad x = 2 \pm \sqrt{3}$. So

$I_{n}= A(2-\sqrt{3})^{n}+B(2+\sqrt{3})^{n}$.

But since $I_{n}$ is bounded for all $n$, $B$ must be zero and $I_{n}= I_{0}(2-\sqrt{3})^{n}= \frac{\pi}{\sqrt{3}}(2-\sqrt{3})^{n}$


[quote="tran300491"]If a,b,c>0  and abc=1 then(a+b)(b+c)(c+a)+7 $ \geq$ 5(a+b+c)[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128529
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168544


function
combinatorics unsolved
combinatorics

Calculate $\sum_{k=0}^{2n}(-1)^{k}{ 4n-2k \choose 2n-k}{2k \choose k}$

The answer has such a nice combinatorial form, it suggests a combinatorial proof.  However, I can only find a proof with generating functions:

Consider $G(x) = \sum_{n = 0}^\infty x^{2n}\sum_{k=0}^{2n}(-1)^{k}{ 4n-2k \choose 2n-k}{2k \choose k}$.  Switching the two summations and making the substitution $j= 2n-k$ gives us that $G(x) = \sum_{k \geq 0}\binom{2k}{k}(-1)^{k}x^{k}\sum_{j \geq 0}\binom{2j}{j}x^{j}$.

From here, we just need to know that $\sum_{n \geq 0}\binom{2n}{n}x^{n}= \frac{1}{\sqrt{1-4x}}$ to see that we have $G(x) = \frac{1}{\sqrt{1+4x}}\cdot \frac{1}{\sqrt{1-4x}}= \frac{1}{\sqrt{1-16x^{2}}}$ so that $G(x) = \sum_{n \geq 0}\binom{2n}{n}4^{n}x^{2n}$ and so, equating coefficients, our original expression is just $\binom{2n}{n}4^{n}$.


Let

$A=\begin{pmatrix}1&2\\ 0&2 \end{pmatrix}\ \ \ \ \ ,B=\begin{pmatrix}3&-8\\ 0&-1 \end{pmatrix}$

Find an invertible matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ both are diagonal matrices.

[quote="Kent Merryfield"]All you can really hope for is that $A$ and $B$ have the same eigenvectors - either they do or they don't. Just diagonalize $A$ (easy enough, since we can just read off its eigenvalues) and see if that works for $B$ as well.

In fact, $P=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ works.

Note also that in this case $AB=BA.$ [i]Not[/i] a coincidence.[/quote]

Ok, I've learned this in linear Algebra recently :D  first we have to find the characteristics of matrices $A$ and $B$ :

$\det(xI-A)=\begin{vmatrix}x-1&-2\\ 0&x-2\end{vmatrix}=(x-1)(x-2)$

$\det(xI-B)=\begin{vmatrix}x-3&8\\ 0&x+1\end{vmatrix}=(x-3)(x+1)$

so the eigen values of $A$ are $1$ and $2$ and the eigen vectors related to them are $u_{1}=(1,0),u_{2}=(2,1)$

and the eigen values of $B$ are $3$ and $-1$ and the eigen vectors related to them are $u_{1}=(1,0),u_{2}=(2,1)$

so so $P=\begin{pmatrix}1&2\\ 0&1\end{pmatrix}$
cool :)

sorry but again I can't understand this completely :oops: , Do You mean that if the eigen vectors are not the same or $AB\neq BA$ then we can't find such a matrix $P$...?!![/img]

If $P^{-1}AP$ and $P^{-1}BP$ are both diagonal, $P$ is a matrix of eigenvectors for both $A$ and $B$. The converse is also true, and $A,B$ commute if such a $P$ exists because $P^{-1}AP$ and $P^{-1}BP$ commute.

The nontrivial theorem: If $A$ and $B$ are both diagonalizable and $AB=BA$, $A$ and $B$ are simultaneously diagonalizable (there exists $P$ as in this problem). You don't need that direction here.


I am posted this here because the people that would care will be viewing this forum most likely.

Did anyone here about the Poincar Conjecture being proven?

I saw this on some website:

"Poincar Conjecture Appears Proved, This Time for Real
A series of public lectures given by Grigori Perelman appear to constitute an announcement that work first published in November 2002 has now resulted in a proof of the Poincar conjecture."

Does anyone know if the Poincare Conjecture has any real significance to the world in some way? What type of break-throughs would this proof bring?

I personally don't know much about it.

All I know is that he'll get a million dollars from the [url=http://www.claymath.org]Clay Mathematics Institute[/url].


geometry
circumcircle
incenter
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, 0 the circumcircle of ABC, U the second point of intersection of the parallel to BC through A with 0,
I the incenter, 1 the circle with AI as diameter, P the second point of intersection of 1 and 0,
and V the second point of intersection of PI with 0.
Prove that UV is perpendicular to BC.Sincerely
Jean-Louis

[quote="jayme"]
I the incenter, 1 the circle with AI as diameter, P the second point of intersection of 1 and 0,
[/quote]
What is P ?
$ P \equal{} (AI)\cap (O)$ ????????????
:D

[quote="jayme"]Dear Mr Green,
P is the second point of intersection of the circle 1 and the circle 0.
The first point of intersection is A.
Sincerely
Jean-Louis[/quote]
Dear jayme,
[img]http://i209.photobucket.com/albums/bb172/Ineq/ab.jpg[/img]
Is it your problem's Fig
Sorry if I am wrong

[quote="mr.danh"]Obvious?  :|  (AV is the diamater of circle 0)[/quote]
So I thinks there was a mistake on this Problem


probability
advanced fields
advanced fields theorems

I am studying probability from Stroock (Probability Theory an Analytic view) and I will be interested in finding the solution manual to check some of my solutions.
Any help is greatly appreciated.
Thank you.

I would also be interested in finding a graduate-level textbook with a solution manual.  :maybe:


articles ARML AMC USA(J)MO USAMO geometry circumcircle

http://tools.wikimedia.de/sixdeg/index.jsp

Can you find two wikipedia articles with six or more degrees of separation? Bonus points if either of them aren't stubs.

Most of these have no routes after 10 degrees
http://en.wikipedia.org/wiki/Wikipedia:Orphaned_Articles/A
http://en.wikipedia.org/wiki/Wikipedia:Orphaned_Articles/B
http://en.wikipedia.org/wiki/Wikipedia:Orphaned_Articles/C
etc.

[quote="Ignite168"][quote="Hamster1800"]w00t! I found one!

[url]http://tools.wikimedia.de/sixdeg/index.jsp?from=electroencephalograph&to=w00t[/url]

It goes from electroencephalograph to w00t. :)

That took me a while to find exactly 6 degrees. also, Finish to Start has 7. I know i saw more with more than 6, but exactly 6 is hard to find.

Wow!
[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=electroencephalograph&to=wow]electroencephalograph to wow[/url]
works too.

I love electroencephalograph. If you go from electroencephalograph to sempai, it works too ^_^.[/quote]

It can be more than 6 and that's sorta cheating as electroencephalograph is a redirect to electroencephalography.[/quote]

Fine. Do [url=http://tools.wikimedia.de/sixdeg/index.jsp?from=Electroencephalography&to=AoPS]Electroencephalograph[b]y[/b] to AoPS[/url]

Or, if you prefer, [url=http://tools.wikimedia.de/sixdeg/index.jsp?from=AoPS&to=w00t]AoPS to w00t[/url] <actually that one is a redirect too lol

or, if going from or to a redirect is cheating, use [url=http://tools.wikimedia.de/sixdeg/index.jsp?from=electroencephalography&to=fail]electroencephalography to fail[/url] < err my bad.. I didn't check and fail also is a redirect >_<

Take 6 forums from AoPS and write about them linking one to the other - I don't think they already exist or are linked to something!

:rotfl:  :rotfl:
first try
[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=hi&to=what]tools.wikimedia.de/sixdeg/index.jsp?from=hi&to=what[/url]
hi to what lmao XDXDXD

That's really cool! I remember doing a similar game with MathWorld.

I found one: [url=http://tools.wikimedia.de/sixdeg/index.jsp?from=half+rhyme&to=mountain+dew]Half rhyme to Mountain dew[/url]

look at this

[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=here&to=there]tools.wikimedia.de/sixdeg/index.jsp?from=here&to=there[/url]

6 degrees of separation:
[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=ARML&to=AoPS]ARML to AoPS[/url]
[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=Nintendo&to=AoPS]Nintendo to AoPS[/url]

7 degrees of separation:
[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=USAMO&to=harmonic+sequence]USAMO to harmonic sequence[/url]
[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=Windows&to=AoPS]Windows to AoPS[/url]
[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=inradius&to=circumradius]inradius to circumradius[/url]
[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=orthocenter&to=incenter]orthocenter to incenter[/url]

[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=boring&to=history]this one was slightly amusing[/url] (only 2 degrees of separation)

Seven Degrees:

[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=Molotov_Cocktail&to=AoPS]Molotov Cocktail to AoPS[/url]

It will be more interesting if you put restrictions like:

Find two topics starting with 'Q' the Wikipedia articles on which are exactly separated by six links.

let's see who can find the longest one (as in, longest minimum path).   this one has 8: [url]http://tools.wikimedia.de/sixdeg/index.jsp?from=inside&to=AoPS[/url]

How bout weird to happy?

[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=weird&to=happy]tools.wikimedia.de/sixdeg/index.jsp?from=weird&to=happy[/url]

http://tools.wikimedia.de/sixdeg/index.jsp?from=SOAD&to=AoPS

I know, there's one redirect, but i still think its cool

Also

http://tools.wikimedia.de/sixdeg/index.jsp?from=calvin&to=alvin
couldn't find a route less than 10 degrees.

Here's 6 [url=http://tools.wikimedia.de/sixdeg/index.jsp?from=life&to=AoPS]life to aops[/url]

[quote="frt"]
[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=inradius&to=circumradius]inradius to circumradius[/url]
[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=orthocenter&to=incenter]orthocenter to incenter[/url][/quote]
Wow. I would expect 2. Incenter-> triangle-> circumradius

[url=http://tools.wikimedia.de/sixdeg/index.jsp?from=Harpo+Marx&to=incenter]Harpo Marx to incenter[/url]

this one finds no route in less than 10 [url]http://tools.wikimedia.de/sixdeg/index.jsp?from=Babylon&to=smilie[/url]

I found one with more than 10 degrees of seperation.

http://tools.wikimedia.de/sixdeg/index.jsp?from=G-sharp+minor&to=Navier


geometry
rectangle
combinatorics unsolved
combinatorics

The numbers $1,2,\ldots,100$ are written in the cells of a $10\times 10$ table, each number is written once. In one move, Nazar may interchange numbers in any two cells. Prove that he may get a table where the sum of the numbers in every two adjacent (by side) cells is composite after at most $35$ such moves.
[i]N. Agakhanov[/i]

Unless there's some other definition of composite, I would say it means 'not prime.'

Yes, composite means "not prime", unless I am mistaken.

Our first 25 moves are spent making sure that the left $5 \times 10$ rectangle is either all even or all odd numbers.  (whichever one has at least 25 already there, we just move the rest of that parity in).

So long as the board has this configuration, we are safe for all the adjacent pairs except those crossing the middle line dividing even from odd (as each sum is an even number greater than or equal to 4).  We use our remaining 10 moves to fix up that middle line as follows:

For each odd $x$ along that line, we successively choose a even $y$ such that $x+y$ is a multiple of 3 which is at least 9 and $y$ has not yet been chosen.  We then swap the number currently across from $x$ with $y$.  Since there are at least 16 even $y$ which sum to a multiple of 3 with $x$, there will always be (several) unchosen possibilities, so this can always be done.

This fixes the pairs crossing the middle line while still leaving everything else having an even sum, so we are done.


trigonometry
geometry
incenter
geometric transformation
reflection
geometry unsolved

Let $D$ and $E$ be points on the sides $CA$ and $AB$ of a triangle $ABC$ such that the lines $BD$ and $CE$ are the interior angle bisectors of the angles $ABC$ and $BCA$, respectively. Given that $\angle BDE=24^{\circ}$ and $\angle CED=18^{\circ}$, determine the angles $\angle A$, $\angle B$, $\angle C$ of triangle $ABC$.

The answer is A=96° (this is easy to obtain right from the beginning, the harder thing is getting the values of B and C) and B=60°, C=24° (I think :D ). I'm still working on a solution.

does anybody have a solution?

(with an image for better understanding)

i get the nasty trig system:

$(\sin 18+\frac{C}{2})(\sin 96+\sin B)=(\sin 24+\frac{B}{2})(\sin 96+\sin C)$
$B+C=84$

it is pretty clear that the solution (B,C)=(60,24), all that remains to prove is that there is only one solution, and this should not be so bad...

ohhh man!!!, very nasty, but thanks anyway!!!

anyone has a solution without trig.?

[quote="Altheman"]...it is pretty clear that the solution (B,C)=(60,24), ...[/quote]
That's because you did not try.

$\frac{\angle A}{2}+90^\circ = \angle BIC = 180^\circ-(\angle BDE+\angle CED) = 138^\circ$

Then $\angle A = 96^\circ,$ $\angle B+\angle C = 84^\circ,$ $\angle B,\ \angle C$ are both acute. By the sine theorem for the $\triangle BDE,\ \triangle CED,$

$\frac{\sin \frac{B}{2}}{\sin 24^\circ}= \frac{DE}{BE},\ \ \ \frac{\sin \frac{C}{2}}{\sin 18^\circ}= \frac{DE}{CD},$

$\frac{\sin \frac{B}{2}}{\sin \frac{C}{2}}\cdot \frac{\sin 18^\circ}{\sin 24^\circ}= \frac{CD}{BE}= \frac{ab}{c+a}\cdot \frac{a+b}{ca}= \frac{\sin B}{\sin C}\cdot \frac{\sin 96^\circ+\sin B}{\sin 96^\circ+\sin C}$

$\frac{\cos \frac{C}{2}\sin 18^\circ}{\cos \frac{B}{2}\sin 24^\circ}= \frac{\sin \left(48^\circ+\frac{B}{2}\right) \cos \left(48^\circ-\frac{B}{2}\right)}{\sin \left(48^\circ+\frac{C}{2}\right) \cos \left(48^\circ-\frac{C}{2}\right)}= \frac{\cos \left(42^\circ-\frac{B}{2}\right) \cos \left(48^\circ-\frac{B}{2}\right)}{\cos \left(42^\circ-\frac{C}{2}\right) \cos \left(48^\circ-\frac{C}{2}\right)}$

Using $\frac{\angle B}{2}= 42^\circ-\frac{C}{2},$

$\frac{\sin 18^\circ}{\sin 24^\circ}= \frac{\cos \left(48^\circ-\frac{B}{2}\right)}{\cos \left(48^\circ-\frac{C}{2}\right)}= \frac{\cos \left(\frac{C}{2}+6^\circ\right)}{\cos \left(\frac{C}{2}-48^\circ\right)}$

${\sin \left(\frac{C}{2}-30^\circ\right)+\sin \left(66^\circ-\frac{C}{2}\right) = \sin \left(\frac{C}{2}+30^\circ\right)+\sin \left(18^\circ-\frac{C}{2}\right)}$

$\cos \frac{C}{2}= 2 \sin 30^\circ \cos \frac{C}{2}= 2 \cos \left(42^\circ-\frac{C}{2}\right) \sin 24^\circ = 2 \sin 24^\circ \sin \left(\frac{C}{2}+48^\circ\right) =$

$= 2 \sin 24^\circ \left(\sin \frac{C}{2}\cos 48^\circ+\cos \frac{C}{2}\sin 48^\circ \right) =$

$= (\sin 72^\circ-\sin 24^\circ) \sin \frac{C}{2}-(\cos 72^\circ-\cos 24^\circ) \cos \frac{C}{2}$

$\tan \frac{C}{2}= \frac{1+\cos 72^\circ-\cos 24^\circ}{\sin 72^\circ-\sin 24^\circ}= \frac{1-\cos 72^\circ}{\sin 72^\circ}= \tan 36^\circ$

because

$\frac{1+\cos 72^\circ-\cos 24^\circ}{\sin 72^\circ-\sin 24^\circ}= \frac{1-\cos 72^\circ}{\sin 72^\circ}\ \Longleftrightarrow$

$2\sin 72^\circ \cos 72^\circ = \sin 72^\circ \cos 24^\circ+\sin 24^\circ \cos 72^\circ-\sin 24^\circ\ \Longleftrightarrow$

$\sin 144^\circ = \sin 96^\circ-\sin 24^\circ = 2 \cos 60^\circ \sin 36^\circ = \sin 36^\circ$

As a result, $C = 72^\circ,\ B = 12^\circ.$
___________________________________

Check: Let DE meet BC at F. By Menelaus theorem,

$\frac{DC}{DA}\cdot \frac{EA}{EB}\cdot \frac{FB}{FC}= \frac{a}{c}\cdot \frac{b}{a}\cdot \frac{FC+a}{FC}= 1,\ \ FC = \frac{ab}{c-b}$

$CE = \frac{\sqrt{ab[(a+b)^{2}-c^{2}]}}{a+b}= \frac{\sqrt{2a^{2}b^{2}(1+\cos C)}}{a+b}= \frac{2ab \cos \frac{C}{2}}{a+b}$

$\frac{CE}{FC}= \frac{2ab \cos \frac{C}{2}}{a+b}\cdot \frac{c-b}{ab}= 2 \cos \frac{C}{2}\cdot \frac{\sin C-\sin B}{\sin A+\sin B}=$

$= 2 \cos \frac{C}{2}\cdot \frac{\cos \frac{C+B}{2}\sin \frac{C-B}{2}}{\sin \frac{A+B}{2}\cos \frac{A-B}{2}}= \frac{2 \sin \frac{C-B}{2}\cos \frac{C+B}{2}}{\cos \frac{A-B}{2}}$

Assuming $\angle C = 72^\circ,\ \angle B = 12^\circ,$

$\frac{CE}{FC}= \frac{2 \sin 30^\circ \cos 42^\circ}{\cos 42^\circ}= 1$

The triangle $\triangle CEF$ is then isosceles, $2 \angle CED = \angle CEF+\angle CFE = \angle BCE = \frac{\angle C}{2},$ and $\angle CED = \frac{\angle C}{4}= 18^\circ.$ The other given angle is then $\angle BDE = \angle DBF+\angle FBD = \frac{\angle B}{2}+\angle CED = 24^\circ.$

$I \equiv BD \cap CE$ is the incenter of $\triangle ABC.$

$ \angle BAC \equal{} 2(\angle EID \minus{} 90^{\circ}) \equal{} 2(180^{\circ} \minus{} 24^{\circ} \minus{} 18^{\circ} \minus{} 90^{\circ})  \Longrightarrow  \angle BAC \equal{} 96^{\circ}$

Let $ P,Q$ be the intersections of  the internal bisector of $ \angle BAI$ with $ DE, DB.$ Then

$ \angle BAQ \equal{} \angle QAI \equal{} \angle EDI \equal{} 24^{\circ}$ $\Longrightarrow$ quadrilaterals $ AEQD,APID$ are cyclic

$ \Longrightarrow \angle DEQ \equal{} \angle DAQ \equal{} 72^{\circ} \ , \ \angle QIP \equal{} \angle DAQ \equal{} 72^{\circ} $

$ \angle QEI \equal{} \angle DEQ \minus{} \angle DEI \equal{} 72^{\circ} \minus{} 18^{\circ} \equal{} 54^{\circ} , \ \angle EIP \equal{} \angle QIP \minus{} 42^{\circ} \equal{} 30^{\circ}$

If $R$ is the  the reflection of $ P$ about $ EI,$ we have $ \angle PEI \equal{} \angle REI \equal{} 18^{\circ}.$ Since $ \angle EIP \equal{} 30^{\circ},$ then $ \triangle PRI$ is equilateral $ \Longrightarrow \angle PRI \equal{} \angle RPI \equal{} 60^{\circ}$

On the other hand, $ \angle ESI \equal{} \angle EQI \plus{} \angle QES \equal{} 84^{\circ} \plus{} 36^{\circ} \equal{} 120^{\circ},$ which implies that $ PRSI$ is cyclic $ \Longrightarrow$ $ \angle RPS \equal{} \angle RIS \equal{} 42^{\circ} \minus{} 30^{\circ} \equal{} 12^{\circ}$ $ \Longrightarrow$ $ \angle EPS \equal{} 84^{\circ}.$ This means that $ \triangle QES \cong \triangle PES$  $\Longrightarrow$ $ EP \equal{} EQ$  $\Longrightarrow$ $ \triangle EPQ$ is isosceles with base $ PQ$ $ \Longrightarrow$ $\angle EQP \equal{} 90^{\circ} \minus{} 36^{\circ} \equal{} 54^{\circ}$

Since $ EQIA$ is cyclic, we have $ \angle AIE \equal{} \angle EQP \equal{} 54^{\circ}$

$ \Longrightarrow \angle AIB \equal{} 42^{\circ} \plus{} 30^{\circ} \plus{} 54^{\circ} \equal{} 126^{\circ}$

 $ \angle ACB \equal{} 2(\angle AIB \minus{} 90^{\circ}) \equal{} 252^{\circ} \minus{} 180^{\circ} \equal{} 72^{\circ}  \Longrightarrow \angle ABC \equal{} 12^{\circ}.$


Let $ x,y,z\in [0,1]$ such that $ xyz\equal{}(1\minus{}x)(1\minus{}y)(1\minus{}z)$ . Prove that :

\[ x(1\minus{}z)\plus{}y(1\minus{}x)\plus{}z(1\minus{}y)\ge \frac{3}{4}\]

[quote="Pain rinnegan"]Let $ x,y,z\in [0,1]$ such that $ xyz \equal{} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$ . Prove that :
\[ x(1 \minus{} z) \plus{} y(1 \minus{} x) \plus{} z(1 \minus{} y)\ge \frac {3}{4}\]
[/quote]

$ (xyz)^{2}\equal{}x(1\minus{}x)y(1\minus{}y)z(1\minus{}z)\leq (\frac{x\plus{}1\minus{}x}{2})^{2}(\frac{y\plus{}1\minus{}y}{2})^{2}(\frac{z\plus{}1\minus{}z}{2})^{2}\equal{}\frac{1}{64}$ <--> $ xyz\leq\frac{1}{8}$

$ xyz\equal{}(1\minus{}x)(1\minus{}y)(1\minus{}z)$ <--> $ x\plus{}y\plus{}z\minus{}(xy\plus{}zy\plus{}zx)\equal{}1\minus{}2xyz$ ----->

$ x(1\minus{}z)\plus{}y(1\minus{}x)\plus{}z(1\minus{}y)\equal{}x\plus{}y\plus{}z\minus{}(xy\plus{}zy\plus{}zx)\equal{}1\minus{}2xyz\geq 1\minus{}2*\frac{1}{8}\equal{}1\minus{}\frac{1}{4}\equal{}\frac{3}{4}$

 proved. :)

[quote="Pain rinnegan"]Let $ x,y,z\in [0,1]$ such that $ xyz \equal{} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$ . Prove that :
\[ x(1 \minus{} z) \plus{} y(1 \minus{} x) \plus{} z(1 \minus{} y)\ge \frac {3}{4}\]
[/quote]

<=>
$ S_{\Delta EDF}\le\frac{1}{4}S_{\Delta ABC}$


function
topology
algebra
domain
calculus
calculus computations

Consider a function $ f: M\rightarrow\mathbb{R}$. Its graph is the set:

$ \{(p,y)\in M \times\mathbb{R}: y \equal{} f(p)\}$

a) Prove that if $ f$ is cont., then its graph is closed.

[hide]
I just decided to show that $ M\times\mathbb{R}$ - graph of f  (complement of graph) is open . Is it ok that I didn't use the fact that f is continuous?
[/hide]

b) Prove that if $ f$ is cont., and $ M$ is compact, then its graph is compact.

[hide]
I think this is right. Use the fact that f maps from compact into compact and the fact that the finite direct product of compacts is compact, then it shouldn't be hard to claim that the subset of that compact direct product is also compact.
[/hide]

c) Prove that if the graph of $ f$ is compact, then $ f$ is cont.

[hide]
I'm at a lost for this one. I don't see how the closed and boundedness of the graph could show that $ f$ is continuous.
[/hide]

d) What if the graph is merely closed? Give an example of a discountinuous function $ f: \mathbb{R}\rightarrow\mathbb{R}$ whose graph is closed.

[hide]
No idea.
[/hide]

Any help for c) and d) would be much appreciated. A check of my responses to a) and b) would be extremely helpful.

(a): If you didn't use that $ f$ is continuous, your argument is wrong. Non-continuous functions usually have non-closed graphs.

(b): What sort of subset of that compact set? You certainly don't have a complete argument.

(c): I recommend the contrapositive. Also, this does require some topological properties of the codomain: it should be at least Hausdorff, so that compact sets are closed.

(d): The other parts say what you can't do. Stepping around that, what can still go wrong?

b) I was saying that graph of f is a subset of $ M\times\mathbb{R}$

c) We haven't learned about Hausdorff spaces yet. We're studying topology in metric spaces.

(b): Bzzzt. No points for you.
That's true, but it's irrelevant.

(c): That's OK. All you need is the property that all compact sets in the codomain are closed, which is certainly true in metric spaces- and you've specified that the codomain is $ \mathbb{R}$ anyway.
On the other hand, there are no restrictions at all on the domain; it can be any topological space, or any compact space in the parts that ask for that.

b) I'm still lost with that one, unless $ M$ and $ \mathbb{R}$ are closed, then I could say that $ f$ being continuous maps closed into closed and direct product of closed spaces is closed(??)

c) Might this possibly work (?) : If $ f: M \rightarrow\mathbb{R}$. Let $ K$ be a closed subset in $ \mathbb{R}$ and let $ (p_{n})$ be a sequence in $ f^{ \minus{} 1}(K)$ where $ p_{n}\rightarrow p$ in $ M$. Since $ f$ is not continuous then, $ f(p_{n})$ does not necessarily converge to $ f(p)$....which contradicts closedness of K?

d) So if the graph is merely closed, then there's nothing to say that the graph is bounded or compact by extension. Then would an example of a discontinuous function be a some function with a jump discontinuity or some $ f(x) \equal{}\frac{1}{x^{2} \minus{}1}$?

Any help would be appreciated.

(d) Getting warmer, but $ \frac1{x^2 \minus{} 1}$ is a continuous function. (on its domain, which is the only sense that matters here.)

Please help on on b), c)

d) a dirchlet function?

No. That's entirely the wrong way to look at discontinuities.

As for the rest- I've reached the limit of my cluelessness tolerance for today.

Any help on a,b,c,d is much appreciated, especially if you can stand my cluelessness.

The graph of a Dirichlet function is not closed. Try going back to the idea of $ \frac{1}{x^2\minus{}1}$ but tweak the example. Try to get around the words "on its domain" by doing something about the domain.

$ f(1) \equal{} 0$ and $ f(\minus{}1) \equal{} 0$. Let $ f(x) \equal{} \frac{1}{x^{2}\minus{}1}$ for the rest of the domain?

Clearly not a continuous function. Can you prove that the graph is closed?

To show the graph is closed, this means we would have to suppose we have some sequence $ x_{n} \rightarrow x$ and show that $ f(x_{n}) \rightarrow f(x)$ right? If this is the case, then what I wrote above for a) needs revision.


linear algebra
matrix
real analysis
real analysis unsolved

Show that $Gl_n(R)\cup A$(where $A$ is a non-invertible matrice)is connexe par arcs.

We identify $\mathcal M_n(\mathbb R)$ with $\mathbb R^{n^2}$. $GL_n^+(\mathbb R)$, the set of invertible matrices with positive determinant and $GL_n^-(\mathbb R)$, the set of invertible matrices with negative determinant are open path connected subsets of $\mathbb R^{n^2}$. Moreover, these two sets have the set of invertible matrices as their common border.

It suffices to prove that we can find a path in our set from an element of $GL_n^+(\mathbb R)$ to $A$ (and the same for $GL_n^-(\mathbb R)$, but I'll prove it only for $GL_n^+(\mathbb R)$). For $n\ge 1$, let $M_n$ be an element of $GL_n^+(\mathbb R)$ contained in the ball of radius $\frac 1n$ centered in $A$. Our path from $M_1$ to $A$ will be a composition of paths from $M_n$ to $M_{n+1}$ situated in the ball centered at $A$ of radius $\frac 1n$ respectively, with endpoint $A$.

$E=GL_n(R) \cup A$
take $A=PJ_rQ$, where $P,Q$ are invertible, $J_r$ the matrix with $0$ everywhere and except for $(J_r)_{i,i}=1$, $1 \leq i \leq r$.

It suffices to show that every invertible matrix $M$ is path connected to $A$, with a path in $E=GL_n(R) \cup A$.
Clearly, because $GL_n(R)^+$ and $GL_n(R)^-$ are path connected, it suffices to prove that there is at least one matrix of $GL_n(R)^+$ path connected to $A$, the path with values in $E$, and another in $GL_n(R)^+$.
It is easy: Take $PD_tQ$ where $D_t=diag(1,1,..,1,t,t,..,t)$ to connect $diag(1,1,..,1)$, and $PD_tQ$ where $D_t=diag(1,1,..,1,t,t,..,t,-t)$ to connect $diag(1,1,..,1,-1)$


in how many consecutive zeroes does the product 115*116*117...*201 end?

[hide="solution"]I got 30 zeros

there are 25 numbers divisible by 5

the are 4 numbers divisible by 25

there is one answer divisible by 125

25+4+1=30[/hide]

[hide]How'd u get 25 numbers divisible by 5? 25*5 IS 125, and 201-115 is 86. I got 23; 18 nos divisible by 5, 4 by 25, and 1 by 125; thus the answer is 23 (18+4+1)[/hide]

For some reason it won't let me edit my last post, so I'll post a new one... the reason that 25 zeros doesn't work is that if the difference is 86, then 5* the number of multiiples of 5 should be  <= the difference. For the person that got 18, you forgot the multiples of 25 and 125, which have 2 and 3 fives in their prime factorizations, respectively.

[hide="Spoiler -- Here is a systematic way to solve this problem"]
[hide="Do not peek until you have tried the problem"]The standard way to find the number of zeros at the end of a factorial (say $201!$) is by repeatedly dividing by 5, and rounding down at each step.  [This finds the number of factors of 5 in $201!$.]

\begin{eqnarray*}
\lfloor 201/5 \rfloor &=& 40 \\
\lfloor 40/5 \rfloor &=& \phantom{0}8 \\
\lfloor 8/5 \rfloor &=& \phantom{0}1 \\
\lfloor 1/5 \rfloor &=& \phantom{0}0 \\
\hbox{Total} &=& 49
\end{eqnarray*}

Do the same to find the number of zeros at the end of $114!$.

Take the difference, and you should have the number of zeros at the end of $\frac{201!}{114!}$[/hide][/hide]


It seems much more simple than chess so i figured someone has figured out how to win everytime with either red or black.

Yes, Victor Allis and James Allen (independently) proved that the first player has a winning strategy.  The first move, as expected, is to play in the middle column.  See the following article for some references:
[url]http://mathworld.wolfram.com/Connect-Four.html[/url].

[quote="SirErnest"]The link says it can be done but doesnt tell how.  Does anybody know how?[/quote]

Well, that would be spoiling the game, wouldn't it?  ;)

Well, given the game, I think Id rather know the winning play so that I never have to play it again.


calculus
integration
irrational number
continued fraction
number theory unsolved
number theory

Prove that, given an irrational number $ x$, there exist infinitely many positive integers $ q$ for which, for each $ q$, there exists an integer $ p$ such that $ \left|x\minus{}\frac{p}{q}\right| \le \frac{1}{q^2}$.

[quote="darkdieuguerre"]Prove that, given an irrational number $ x$, there exist infinitely many positive integers $ q$ for which, for each $ q$, there exists an integer $ p$ such that $ \left|x \minus{} \frac {p}{q}\right| \le \frac {1}{q^2}$.[/quote]

Just take the successive truncated continued fractions of $ x$

[quote="darkdieuguerre"]Prove that, given an irrational number $ x$, there exist infinitely many positive integers $ q$ for which, for each $ q$, there exists an integer $ p$ such that $ \left|x \minus{} \frac {p}{q}\right| \le \frac {1}{q^2}$.[/quote]
It is called Dirichlet Approximations Theorem. See here http://mathworld.wolfram.com/DirichletsApproximationTheorem.html

[quote="pco"]Just take the successive truncated continued fractions of $ x$[/quote]

I understand how a proof using continued fractions might work; is there something potentially simpler? It looks a lot like Dirichlet's, except that one must prove that infinitely many integral $ q$ exist such that each $ q$ has an integral $ p$ so that $ \left|qx \minus{} p\right| \le \frac {1}{q}$.

[quote="Allnames"]It is called Dirichlet Approximations Theorem. See here http://mathworld.wolfram.com/DirichletsApproximationTheorem.html[/quote]

Is it some sort of corollary that follows directly from Dirichlet's? I did look at the MathWorld page, but although it stated that there does exist such integers $ p$ and $ q$, it doesn't anything regarding an infinity of such pairs.

[quote="darkdieuguerre"] I understand how a proof using continued fractions might work; is there something potentially simpler? [/quote]

I think that this is a well-known result about continued fraction :

If $ x>0\in\mathbb R\backslash\mathbb Q$ and if it's successive truncated continued fractions are $ \frac{p_n}{q_n}$, then :

$ p_n$ and $ q_n$ are two positive increasing integer sequences and $ \frac{1}{q_n(q_{n\plus{}1}\plus{}q_n)}<\left|x\minus{}\frac{p_n}{q_n}\right|<\frac{1}{q_nq_{n\plus{}1}}<\frac{1}{q_n^2}$


Evaluate $ \sqrt[3]{2 \plus{} \sqrt5} \plus{} \sqrt[3]{2 \minus{} \sqrt5}.$

Let $ x \equal{} \sqrt [3]{2 \plus{} \sqrt5} \plus{} \sqrt [3]{2 \minus{} \sqrt5} \in \mathbb{R}$

[hide="Hint"]$ (a \plus{} b)^3 \equal{} a^3 \plus{} b^3 \plus{} 3ab(a \plus{} b)$[/hide]

[hide="Solution"]$ x^3 \equal{} 4 \plus{} 3 \sqrt [3]{(2 \plus{} \sqrt {5})(2 \minus{} \sqrt {5})} x \equal{} 4 \minus{} 3x \iff x^3 \plus{} 3x \equal{} 4$

$ x^3 \plus{} 3x$ is increasing in $ x$, so there is at most one real solution. By inspection, $ x \equal{} 1$ is this solution.[/hide]

[hide="Alternately"] On a hunch that the single form of the expression can be simplified, it turns out that $ \sqrt[3]{16 \plus{} 8 \sqrt{5}} \equal{} 1 \plus{} \sqrt{5}$.  Hence $ \sqrt[3]{2 \plus{} \sqrt{5}} \equal{} \phi$.

This is not exactly a hunch:  the norm of $ 2 \plus{} \sqrt{5}$ is $ 4 \minus{} 5 \equal{} \minus{}1$, which suggests it could possibly have a "simple" cube root, also of norm $ \minus{}1$. [/hide]


Write down any integer from 0 to 20 inclusive. The teams that wrote down the second-highest integer among the responses received gets that many points.

What a convoluted problem...

Remember, this is HMMT, and quite late in the Guts Round, so the probability that someone writing down 20 just to see someone else get points for 19 is not that high. Or is it?

I would write 18:

Obviously I wouldn't write 20.  Since no one is stupid enough to write 20, I wouldn't write 19 either.  Same reasoning for not writing an 18.  However, out of all the teams, I bet someone would write either a 19 or an 18 just to try their luck.  So 18 is the best bet, or perhaps 17.

Do we know what ended up happening?

Also, even if you calculate some obscure probability with game theory, it's still more than partly luck.  So does the HMMT Guts round affect the entire score for a team?  If this could determine the difference between 1st and 2nd place in the overall contest, then that's pretty...bad.

This is a lot like what we did at ARML. Every team picked a positive integer and the smallest positive integer such that only one team picked it would be the winner (of some random prizes).

Maybe more psychology than statistics....mysmartmouth your reasoning doesn't make any sense  :wink:

Maybe we should try to test this. Get like 20 people to send in integers to a person through pm and then see what the actual number would be

I agree.  I'd be willing to run this, if no one else wants to.

Okay!

If you want to participate, pm me an integer between 0 and 20 inclusive by midnight Eastern Time on Tuesday, October 17th, 2006 (too soon?).  The person(s) with the second-highest integer win a prize!

This is an interesting "problem" (I guess you could call it that), but what would you do if there are ties?  :wink:

If the set of responses is:

20
20
20
19
19
18
17
...

Then the two people who sent in the 19s would win, and g et 19 points each.  (At least, thats what it seems from the wording.)

I have a hypothesis, assuming everybody is smart.

Obviously, no one would pick 20.
Then no one would pick 19 because they know everybody's smart and didn't pick 20.
Then no one would pick 18 because they know everybody's smart and didn't pick 19.
Then no one would pick 17 because they know everybody's smart and didn't pick 18.

So on and so on.

Wouldn't this be a paradox?

Everybody has to give a number though, so the game is guessing what the highest number someone would pick is.

splashD has got a point, and it is genuine... it's a paradox.

the way i see it, it's teams guessing what the other teams would do. obviously, if the teams know their current scores everything changes.

ie: if the leading team is 7 points ahead, they presume that the 2nd team would try to get at least 7 points, if not 8... so, the 1st team goes for 8, as:
either they win and get 8 points, or the 2nd team wins and gets 7 points, which makes them equal, but still in first...

then you could see what the 2nd team would do in order to get to 1st place but not lose 2nd to 3rd... and so on and on...

the important thing is:yo just can't do this problem out of context.

also, nebula asked what if they're tied? well, it must be 2 ways of breaking ties:

1. when teams tie with 2nd largest integer, look at 3rd largest and so on until only one team wins (his own integer, maybe 5th in row but only one team guessed that number)

2. the teams that picked 2nd largest integer split their prize - the integer is divided by the number of tied teams, and rounded down to integer (ie 3 teams tie with 11, each gets 3 points)

Then no one gets any points, since there is no second-highest guess.

I believe the HMMT answer key said something on the lines of "Theory says that everyone will write down the number 1. But it will be interesting to see what actual responses we get."

Well, in that case, I would write down 0, and win :D

Yeah, but nobody else would have any points either, and I would get the pride and joy of winning :rotfl:

Since theory says everyone will write 1, but that's obviously not going to be the case, can we throw theory out altogether, and just take a random guess? (eliminating 0 and 20, of course).

You would have to know more about the other people that are picking. If they aren't very smart, then they might pick something like 19 or 18, but people who put some thought into this would probably pick more along the lines of 10-15.

I will repost this in G&FF when I get back from practice.  Everyone knows that that forum gets much more viewing than this one.  Perhaps then we can get a bigger sample size.

[quote="E^(pi*i)=-1"]Okay, the contest is over.  Here were the results:

17 - mysmartmouth, OHO, paladin8

16 - mustafa

10 - junggi

mustafa wins!  I will pm you your prize.

Not really enough participants to come to any conclusions.[/quote]

Why did so many people get 17? If nobody did 20 or 19, then nobody would be stupid enough to do 18.

Using that logic, everyone would pick the number 1, and the game would be lame.

[quote="mysmartmouth"]Yes, I decided to disregard any logic that took more than 2 steps of thinking.  Thinking ahead more than two steps is trivially cyclic logic in games such as rock-paper-scissors, but I guess we need about 3 or 4 buffer steps here...I really want to see the HMMT results on this one.

EDIT: WHOA!  Totally new theory.  Some Exeter or Jefferson team out there probably felt very confident about their results.  In order to prevent a second-place team from coming from behind and beating them, they would write 20, thus ensuring no one would get points on the question.  This way, they keep the points in the math section, and have more potential points based on their strong suit, math.  (Thus the best guess is 19, if you have one heavily favored, killer team in the field).

Of course, the problem is totally different when it stands alone like it did here on AoPS.[/quote]

Um, what?  Wouldn't writing 20 down [i]ensure[/i] that another will get points (assuming there is at least one other positive response that isn't 20)?

Anyway, I asked my friend who's on the cross country team here who was at that HMMT but he couldn't remember what the winning number was.  He did, however, remember that from another year where the question was along the lines of "the team that chooses the highest number from 0-20 inclusive that no other team chooses gets that many points".  The winner was 20.

Oh, OK sorrrry.  I was looking at it like 20 must have 19 to get points.  (Consecutive integers).


Say a person plays a psuedo-stock-market game , where at each turn he can either lose one point or gain one dollar based on the roll of a regular 6-sided die. If the roll returns either $1$ or $2$, he gains a dollar; otherwise, he loses a dollar. If the game's rules have him beginning with an integer number of dollars between $\$ 1$ and $\$ 50$, and he wins by reaching $\$ 50$, with how many dollars should he begin so that the game is as fair as possible?

[quote="t0rajir0u"]I don't quite understand.  What does the stock market have to do with anything?  Is this a two-player game (that is, are we taking turns rolling and gaining/losing points)?[/quote]

Sorry, I meant that only one player is playing, like an investor putting money in the stock market (i.e. there's a chance that either he will win or lose money).

My guess is 2/3 of 50, which is closest to 33.

[quote="computer_nuke"]So when you say "with how many dollars should he begin so that the game is as fair as possible", you mean so that the probability of winning is closest to the probability of loosing right (P(win)=1/2 and P(loose)=1/2) or whatever will make it closest to that?[/quote]

Yes, that's what I meant.  :)

(And please ignore the typo in the problem statement where I say he loses one "point"; it should be one dollar, but I can't edit my post.)

[quote="Treething"]actually i'm pretty sure it'd be anywhere between 33.01 and 33.99 dollars (inclusive, yeah)

i have an explanation but it's late and insomnia's beginning to fade[/quote]

Can you give the explanation please?

[quote="Treething"]cuz then he's gotta lose 34 times in order to lose and 17 times to win

and omg omg omg 17 * 4 = 34 * 2 = 68[/quote]

Do you have a more rigorious explination, like a summation that computes the expected value (instead of an abstract idea).


Can you tell me the solution of this:
all odd positive numbers can be derived from this formula:
[code]4n+1, n from Z+.[/code]?

[quote="dondigo"]I don't think so.
All odd numbers are $2k+1$, k is a integer, but they don't have to be $4l+1$. For example $7=4\cdot1+3$[/quote]
Thanx for your response but there is not a matter of been or not been true. It is a confession that i must demonstrate.

[quote="Nea"]there is not a matter of been or not been true. It is a confession that i must demonstrate.[/quote]

 :D

Good luck for your demonstration of something false.

:rotfl:

Well.. maybe what she meant was :
show that this formula only gives odd numbers
?

Although it would be way too easy using the fact that any odd number can be written under the form : 2n + 1, so, if you put : n = 2m, you get : 4m +1.. ;)

So prove that all values of $4n+1$ such that $n \in \mathbb{Z}^{+}$ are odd? Notice that $4n$ must always be even, because it contains a factor of 2. When we add 1 to an even number, we get an odd number, so $4n+1$ must be odd.

[quote]Good luck for your demonstration of something false.[/quote]
Hello tµtµ. You quote my words but i think you haven't read well what i've wrote. I've said that it is a confession so it is true(if you don't think that any odd number can be written under the form: $4n+1$ it is your problem). Now, what i've asked from the forum was to prove it.For example you know that you are son of your parents but juridically you need a document to prove it. Hope you've understood me! bye!
[quote]Although it would be way too easy using the fact that any odd number can be written under the form : 2n + 1, so, if you put : n = 2m, you get : 4m +1.. [/quote]
It's true. It is too easy. I've discovered this forum only 2 days ago. So the first think i've done it was to find an easy problem to see if you would tell me the solution of this. Your answer is correct. Now i know i'm in the right forum of maths. Today begins my true activity in this forum. Let's consider this thread closed and now i'm opening a new thread for another problem(not like this one :lol: ) that i haven't solved yet. bye.
oh i've forgot. chess64 thanx man. I appreciate those who helps peoples.

Ok, good :)

Well, in fact, what tµtµ wanted to say is that, you can't get [b]all[/b] the odd numbers with this form.. an infinity, yeah, but, not all of them ;)


How did you obtain these numbers? They match to 1000 digits of precision, using "bc" on UNIX.

bc -l
scale=1000
e(l( e(l(2)/3)-1 )/3)
e(l(1/9)/3) - e(l(2/9)/3) + e(l(4/9)/3)

[quote="nneonneo"]How did you obtain these numbers? They match to 1000 digits of precision, using "bc" on UNIX.[/quote]
Equalities $A(i)=B(i)\; ,\; i\in \{1,2,3,\}.$ are attributed to S.Ramanujan. Thanks for interest !/Alex


[quote="jensen"]find thevalue of  this:
$\sum_{r=0}^{10} i^2P$
($P=\frac{10!}{i!(10-i)!}$)[/quote]

Assuming you meant r=i,

$i^2P=i^2\binom{10}{i}$

This can solved if one knows differentiation. I'm not doing it that way because we are not allowed to use calculus.

[hide]Let $S=\sum_{k=0}^{10} k^2\binom{10}{k}$. This can be rewritten $\sum_{k=0}^{10} 2\binom{k}{2} \binom{10}{k} +k\binom{10}{k}$, where $k<2\Rightarrow \binom{k}{2}=0$.
$\binom{k}{2}\binom{10}{k}=\binom{10}{2}\binom{8}{k-2}$ since each product represents placing k-2 objects out of 10 in one pile and 2 in another pile. (For the first one you take 2 from the first pile.) Then the sum $\sum_{k=2}^{10} 2\binom{k}{2} \binom{10}{k}=2*45\sum_{k=2}^{10} \binom{8}{k-2}=45*2^9$.
To deal with the second part, reverse the summation to get $\sum_{k=0}^{10} k\binom{10}{k}=\sum_{k=0}^{10} (10-k)\binom{10}{10-k}=\sum_{k=0}^{10} (10-k) \binom{10}{k}$. Since it is equal to both the first and third parts, it is equal to half their sum, making the second part $5\sum_{k=0}^{10} \binom{10}{k}=10*2^9$.
Adding both parts, $S=(45+10)2^9=55*2^9=28160$.[/hide]

Your notation is abysmal.  What you actually wanted to write was

Compute

$\sum^{10}_{i=1} i^2\binom{10}{i}$

or

$\sum^{10}_{i=1} i^2\cdot C^{10}_i$

[quote="jensen"]why you can't calcules?
who don't let you?[/quote]
OK, so if you want I give the solution using calculus.

$(1+x)^n=\binom{n}{0} + \binom{n}{1} x + \binom{n}{2} x^2 +\cdot\cdot\cdot + \binom{n}{n} x^n$
Differentiating w.r.t. x
$n(1+x)^{n-1}=\binom{n}{1} + 2\binom{n}{2} x +\cdot\cdot\cdot + n\binom{n}{n} x^{n-1}$
$\Rightarrow nx(1+x)^{n-1}=\binom{n}{1}x + 2\binom{n}{2} x^2 +\cdot\cdot\cdot + n\binom{n}{n} x^{n}$
Again differentiating w.r.t. x
$\Rightarrow n(1+x)^{n-1} + n(n-1)x(1+x)^{n-2} = 1^2\binom{n}{1} + 2^2\binom{n}{2} x +\cdot\cdot\cdot + n^n\binom{n}{n} x^{n-1}$
Put x=1,
$n\cdot2^{n-1}+n(n-1)\cdot c^{n-2}= 1^2\binom{n}{1} + 2^2\binom{n}{2} +\cdot\cdot\cdot + n^n\binom{n}{n}$
Thus adding 1 both sides we get the answer.

Er, i dont think youve got the question, but it is an interesting identity!


Katie is 2 years older than Joel. In 4 years, her age will be 8 less than twice his age. How old is Katie now?

Let Katie's age now be $K$ and Joel's age now be $J$.

We have the equations $K=J+2$ and $K+4=2(J+4)-8$.

$K=8$.


geometry
circumcircle
geometric transformation
homothety
trigonometry
geometry proposed

Let $ \triangle ABC$ be an acute-angled triangle with circumcircle $ (O,R)$ and nine-point circle $ (N).$ Let $ R_a,R_b,R_c$ be the radii of the circles internally tangent to $ (O)$ at $ A,B,C$ and internally tangent to $ (N).$ Let $ r_a,r_b,r_c$ be the radii of the circles internally tangent to $ (O)$ at $ A,B,C$ and externally tangent to $ (N).$ Show the following relations:

1) $ \frac {1}{R_a} \plus{} \frac {1}{R_b} \plus{} \frac {1}{R_c} \equal{} \frac {4}{R}$

2) $\frac {1}{r_a} \plus{} \frac {1}{r_b} \plus{} \frac {1}{r_c} \equal{} \frac {1}{R}(\sec A \sec B \sec C \plus{} 4)$

[b][color=blue]Lemma:[/color][/b] Let $ (R, \varrho)$ be a circle with a chord $ UV.$ Let $ (P, \varrho_1)$ and $ (Q, \varrho_2)$ be circles internally tangent to $ (R)$ at $ U, V$ and externally tangent to each other at $ W.$ Inversion with center $ U$ and power $ \varrho^2$ takes $ (R), (P)$ into 2 parallel lines $ r \parallel p,$ perpendicular to $ UR.$ $ (Q)$ goes to a circle $ (Q')$ tangent to $ r, p$ at $ V', W',$ with radius

$ \varrho_2' = \frac {V'W'}{2} = \frac {1}{2}\left(\frac {\varrho}{2} - \frac {\varrho^2}{2 \varrho_1}\right) = \frac {\varrho(\varrho - \varrho_1)}{4 \varrho_1}$

$ (Q')$ cuts $ UV$ at $ V' \in UV$ and again at $ T'.$ Since $ V'W'$ is diameter of $ (Q'),$ $ W'T' \perp V'T'.$ Power of $ U$ to $ (Q')$ is

$ UV' \cdot UT' = UV' \cdot (UV' + V'T') = UV' \cdot \left(UV' + V'W' \cdot \frac {UV}{2\cdot UR}\right) =$

$ = \frac {\varrho^2}{UV} \cdot \left(\frac {\varrho^2}{UV} + 2 \varrho_2' \cdot \frac {UV}{2 \varrho}\right) = \varrho^2 \cdot \left(\frac {\varrho^2}{UV^2} + \frac {\varrho_2'}{\varrho}\right)$

$ U$ is homothety center of $ (Q), (Q')$ with coefficient $ \frac {\varrho^2}{UV' \cdot UT'},$ hence

$ \boxed{\frac {1}{\varrho_2} = \frac {1}{\varrho_2'} \cdot \frac {UV' \cdot UT'}{\varrho^2} = \frac {1}{\varrho} + \frac {1}{\varrho_2'} \cdot \frac {\varrho^2} {UV^2} = \frac {1}{\varrho} + \frac {4\varrho \varrho_1}{(\varrho - \varrho_1) \cdot UV^2}}$

[b][color=red]Problem:[/color][/b] Let $ H$ be orthocenter and $ D, E, F$ feet of altitudes $ AH, BH, CH.$ Let $ AH, BH, CH$ cut $ (O)$ again at $ X, Y, Z.$ Midpoints $ K, L, M$ of $ AH, BH, CH$ are on 9-point circle $ (N).$ Let $ DN, EN, FN$ cut $ AO, BO, CO$ at $ O_a, O_b, O_c.$ $ AO \parallel KN$ $ \Longrightarrow$ $ \angle ADO_a = \angle KDN = \angle DKN = \angle DAO = \angle DAO_a$ $ \Longrightarrow$ circle $ (O_a)$ with radius $ R_a = O_aD = O_aA$ is internally tangent to both $ (O), (N)$ at $ A, D,$ respectively. Similarly for the circles $ (O_b), (O_c)$ with radii $ R_b, R_c.$ $ A$ is similarity center of $ (O_a), (O)$ $ \Longrightarrow$ $ R_a = R \cdot \frac {AD}{AX}$ and likewise for $ R_b, R_c.$

$ \frac {1}{R_a} + \frac {1}{R_b} + \frac {1}{R_c} = \frac {1}{R} \left(\frac {AX}{AD} + \frac {BY}{BE} + \frac {CZ}{CF}\right) = \frac {1}{R} \left(3 + \frac {HD}{AD} + \frac {HE}{BE} + \frac {HF}{CF}\right)$

$ = \frac {1}{R} \left(3 + \frac {[HBC] + [HCA] + [HAB]}{[ABC]}\right) \Longrightarrow \boxed{\frac {1}{R_a} + \frac {1}{R_b} + \frac {1}{R_c} = \frac {4}{R}}$

Let $ A', B', C'$ be midpoints of $ BC, CA, AB.$ Let $ (P_a), (P_b), (P_c)$ be circles internally tangent to $ (O)$ at $ A, B, C$ and externally tangent to $ (N).$ Obviously, $ (P_a)$ is internally tangent to $ (O_a)$ at $ A,$ etc. Applying the lemma with $ (R), (P), (Q), UV \longrightarrow (O_a), (N), (P_a), AD,$ etc.,

${{ \frac {1}{r_a} = \frac {1}{R_a} + \frac {4R_a R}{(2R_a - R) \cdot AD^2} = \frac {1}{R_a} + \frac {4R}{(2 AD - AX) \cdot AD}} = \frac {1}{R_a} + \frac {4R}{AH \cdot AD}}$

$ \frac {1}{r_a} + \frac {1}{r_b} + \frac {1}{r_c} = \frac {4}{R} + 4R \left(\frac {1}{AH \cdot AD} + \frac {1}{BH \cdot BE} + \frac {1}{CH \cdot CF}\right) =$

$ = \frac {4}{R} + \frac {2R}{[ABC]} \left(\frac {a}{AH} + \frac {b}{BH} + \frac {c}{CH}\right) =$

$ = \frac {4}{R} + \frac {8R^2}{[ABC]} \cdot \frac {[HBC] + [HCA] + [HAB]}{AH \cdot BH \cdot CH}\ \Longrightarrow$

$ \boxed{\frac {1}{r_a} + \frac {1}{r_b} + \frac {1}{r_c} = \frac {4}{R}\left(1 + \frac {2R^3}{AH \cdot BH \cdot CH}\right) = \frac {4}{R}\left(1 + \frac {R^3}{4 \cdot OA' \cdot OB' \cdot OC'}\right)}$

Thanks for your nice solution, this is what I did:

Let $ H_a,H_b,H_c$ the feet of the altitudes and $ P_a,P_b,P_c$ the second intersections of the corresponding altitudes with the circumcircle $(O).$ Consider the circle $ (O_a')$ internally tangent to $ (N)$ at $ H_a$ passing through $ A.$ Negative inversion with center $ H$ and power $ \overline{HH_a} \cdot \overline{HA}$ takes $ (N)$ into $ (O)$ and $ (O_a')$ into itself $\Longrightarrow$ $ (O)$ is internally tangent to $ (O_a'),$ which implies that $ (O_a')$ is identical with $ (O_a,R_a).$ Thus cevians of the circumcenter are diameters of $ (O_a),(O_b),(O_c).$ Let $ D_a,D_b,D_c$ be the feet of the cevians of the circumcenter. Since rays $ AP_a$ and $ AD_a$ are isogonal WRT $ \angle BAC,$ it follows that:

$ AB \cdot AC \equal{} AD_a \cdot AP_a \ \Longrightarrow \ \frac {1}{AD_a} \equal{} \frac {1}{2R_a} \equal{} \frac {BC \cdot AP_a}{BC \cdot CA \cdot AB}$

$\frac{1}{AD_a}  \equal{} \frac {2[ABP_aC]}{BC \cdot CA \cdot AB} = \frac {2[\triangle ABC] \plus{} 2[ \triangle HBC]}{BC \cdot CA \cdot AB }$

The cyclic sum yields

$ \frac {1}{R_a} \plus{} \frac {1}{R_b} \plus{} \frac {1}{R_c} \equal{} \frac {16 [\triangle ABC]}{BC \cdot CA \cdot AB} \equal{} \frac {4}{R}$

Inversion with center $ A$ and power $ {k_a}^2$ equal to the power of $ A$ WRT $ (N)$ takes $ (O_a,R_a)$ and $ (K_a, r_a)$ into two parallel lines $ \tau$ and $ k$  tangent to $ (N).$ If $ d_a$ and $ {d_a}'$ denote the distances from $ A$ to $ k$ and $\tau,$ by inversion properties we have

$ d_a \cdot 2r_a = {d_a}' \cdot 2R_a \equal{} {k_a}^2 \ . $ Since $ R \equal{} d_a \minus{} {d_a}'$ $ \Longrightarrow \ \frac {1}{r_a} \minus{} \frac {1}{R_a} \equal{} \frac {2R}{{k_a}^2}$

The cyclic sum yields

$ \frac {1}{r_a} \plus{} \frac {1}{r_b} \plus{} \frac {1}{r_c} \minus{} \frac {4}{R} \equal{} 2R\left(\frac {1}{{k_a}^2} \plus{} \frac {1}{{k_b}^2} \plus{} \frac {1}{{k_c}^2} \right)$

Powers of $ A,B,C$ with respect to $ (N)$  are given by

$ {k_a}^2 \equal{} [\triangle ABC] \cot A \  \ ,  \ \ {k_b}^2 \equal{} [\triangle ABC] \cot B \ \ , \ \ {k_c}^2 \equal{} [\triangle ABC] \cot C$

$ \frac {1}{r_a} \plus{} \frac {1}{r_b} \plus{} \frac {1}{r_c} \equal{} \frac {2R}{[\triangle ABC]} \cdot (\tan A \plus{} \tan B \plus{} \tan C) \plus{} \frac {4}{R}$

$ \frac {1}{r_a} \plus{} \frac {1}{r_b} \plus{} \frac {1}{r_c} \equal{} \frac {1}{R} \cdot \frac{ \tan A \cdot \tan B \cdot  \tan C}{ \sin A \cdot \sin B \cdot \sin C} \plus{} \frac {4}{R}$

$ \frac {1}{r_a} \plus{} \frac {1}{r_b} \plus{} \frac {1}{r_c} \equal{} \frac {1}{R}(\sec A \sec B \sec C \plus{} 4)$