Prove that there are infinite maps $ f:\mathbb{C}\to\mathbb{C}$ such that
(1)$ f$ is bijection, and
(2) $ f(a\plus{}b) \equal{} f(a)\plus{}f(b)$ and $ f(ab) \equal{} f(a)f(b)$ for all $ a,b\in\mathbb{C}$.

Yes this is a very nice (and may be well-known) one. There's proof base on Zorn lemma. This needs to prove an auxialary result goes as: Any automorphism on subfield of complex field could be extend to an automorphism of $ C$.
With this, just chose subfields of $ C$ to get the result of the problem, or may be as I remember you could start from $ R$ and using Hamel basic(like as proof for existence of Lebesgue non-measurable set)


geometry perimeter incenter inradius ratio

In the adjoining diagram, $BO$ bisects $\angle CBA$, $CO$ bisects $\angle ACB$, and $MN$ is parallel to $BC$. If $AB=12$, $BC=24$, and $AC=18$, then the perimeter of $\triangle AMN$ is

[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, C=(24,0), A=intersectionpoints(Circle(B,12), Circle(C,18))[0], O=incenter(A,B,C), M=intersectionpoint(A--B, O--O+40*dir(180)), N=intersectionpoint(A--C, O--O+40*dir(0));
draw(B--M--O--B--C--O--N--C^^N--A--M);
label("$A$", A, dir(90));
label("$B$", B, dir(O--B));
label("$C$", C, dir(O--C));
label("$M$", M, dir(90)*dir(B--A));
label("$N$", N, dir(90)*dir(A--C));
label("$O$", O, dir(90));[/asy]

$\textbf {(A) } 30 \qquad \textbf {(B) } 33 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 39 \qquad \textbf {(E) } 42$

[hide="Solution"]It's given in the problem that $\angle ABO=\angle CBO$. We also know by alternate interior angles that $\angle MOB=\angle CBO$. Thus, by the transitive property, $\angle ABO=\angle MOB$, and triangle $BOM$ is isosceles. This means that $MO=MB$. By the same argument, triangle $CON$ is isosceles and $NC=NO$. The perimeter of the triangle is now equal to $(AM+MO)+(AN+NO)=AB+AC=30$. [/hide]

[hide="another solution"]note that O is the incenter, drop a perpendicular from O to BC (the length of this is $r$ :the inradius) , drop perpencular from A to BC (say the whole thing is h, the top part is h-r), use the similar with the heights, also since they are similar, the perimeters are in ratio with the sides:
$\frac{AM}{AB}=\frac{P_{AMN}}{P_{ABC}}=\frac{h-r}{h}=1-\frac{r}{h} \implies P_{AMN}=2s-\frac{2rs}{h}$ then use $A=rs$ and $\frac{2A}{h}=b=24$ then we get $P_{AMN}=30\implies \boxed{A}$

i suppose your solution is easier, but this simplifies just the same[/hide]

$D\in BC\cap AO,\ P\equiv 2p=a+b+c=54,\ P'\equiv 2p'=AM+MN+NA\Longrightarrow$
$\frac{P'}{P}=\frac{AO}{AD}=\frac{b+c}{P}\Longrightarrow \boxed {\ P'=b+c\ }\Longrightarrow P'=30$ (answer [b]A[/b]).
[u]Remark.[/u] It is wellknown (or you can prove easily) that $\boxed {\ \frac{AO}{b+c}=\frac{OD}{a}\ }=\frac{AD}{P}$ (the van [b]Aubel's [/b]relation).


Let $x,y,z$ be positive real numbers such that $x+y+z = 1$. Then prove that:
$\sum \sqrt{\frac{x}{(1-y)(1-x)}} > \frac{1-(x^2+y^2+z^2)}{(1-x)(1-y)(1-z)}$

[quote="pleaseread"]What are the limits of summation?[/quote]
That stands for a cyclic sum and is commonly understood here.


function
inequalities
topology
triangle inequality
real analysis
real analysis unsolved

In Exercise 24 of Chapter 7 of Rudin's [i]Principles of Mathematical Analysis[/i], he gives the following as the first step towards proving that every metric space is isometric to a dense subset of a complete metric space:

Let $ X$ be a metric space, with metric $ d$. Fix a point $ a \in X$. Assign to each $ p \in X$ the function $ f_p$ defined by
\[ f_p(x) \equal{} d(x,p)\minus{}d(x,a)\]
Prove that $ |f_p(x)| \leq d(a,p)$ for all $ x \in X$, and that therefore $ F_p \in \mathcal{C}(X)$.

I don't quite understand why $ |f_p(x)| \leq d(a,p)$ allows us to conclude that the function is continuous. For example, it seems to be discontinuous at $ p$ and $ a$ if $ d$ is the discrete metric.

Fix $ a\in X$.  For each $ p\in X$ we associate an element $ f_{p}(x)\in \ell^{\infty}(X)$ by setting $ f_{p}(x)\equal{}d(x,p)\minus{}d(x,a)$.We have that $ \left|f_{p}(x)\right|\equal{}\left|d(x,p)\minus{}d(x,a)\right|\leq\left|d(p,a)\right|$ for all $ x\in X$ (this is a consequence of the triangle inequality).  Note that since there is no dependence on $ x$, we in fact have $ \left\|f_{p}(x)\right\|_{\infty}\leq d(a,p)$.  It remains to show that $ f_{p}(x)$ is an isometry.  From the triangle inequality, $ \left\|f_{p}\minus{}f_{q}\right\|_{\infty}\equal{}\sup_{x\in X}\left|d(p,x)\minus{}d(q,x)\right|\leq d(p,q)$.  Also, $ \left|d(p,x)\minus{}d(q,x)\right|\equal{}d(p,q)$ when $ x\equal{}p$ or $ x\equal{}q$.  This tells us that $ \left\|f_{p}\minus{}f_{q}\right\|_{\infty}\equal{}d(p,q)$, and so $ f_{p}$ is an isometry.

[quote="Phelpedo"]For example, it seems to be discontinuous at $ p$ and $ a$ if $ d$ is the discrete metric.[/quote]  There is no such thing as a discontinuous function mapping from a discrete metric space: always choose $ \delta \equal{} \frac{1}{2}$ regardless of $ \varepsilon$.  The point of continuity is that it says something about the value of the function at "nearby" points -- in the discrete metric space, there are no nearby points, so every function is vacuously continuous.


calculus
integration
trigonometry
calculus computations

How I'd get my Calculus II class to approach that: first make the (normally indicated) trigonometric substitution $ x\equal{}\tan\theta.$ The integral becomes

$ \int_{\minus{}\frac{\pi}2}^{\frac{\pi}2}\cos^{2n}\theta\,d\theta$

I would then tell them to look up or re-derive the appropriate reduction formula for integrals of powers of the cosine. This gives us

$ \int_{\minus{}\frac{\pi}2}^{\frac{\pi}2}\cos^{2n}\theta\,d\theta\equal{}
\frac{2n\minus{}1}{2n}
\int_{\minus{}\frac{\pi}2}^{\frac{\pi}2}\cos^{2n\minus{}2}\theta\,d\theta$

This will lead us to the answer in product form:

$ \frac{2n\minus{}1}{2n}\cdot\frac{2n\minus{}3}{2n\minus{}2}\cdots\frac24\cdot\frac12
\cdot\pi$

This should be the same as Mr. Doe's after he fixes his typo (That's $ 2^{2n},$ not $ 2^{2^n}$.)

Nice work Kent! I just realized I wrote it wrong  :blush: . I have edited it, so now it correct.


geometry
geometric transformation
reflection
trigonometry
angle bisector
power of a point
radical axis

Denote the points $ X \in AE \cap BD$, $ Y \in AE \cap BC$, $ Z \in AE \cap DF$ and $ T \in DF \cap BC$.

In $ \triangle{AEC}$, we observe that the lines $ AF$, $ DE$ and $ BC$ are concurrent, if and only if, the division $ (AYEZ)$ is harmonic.

Since the quadrilateral $ XYTD$ is cyclic, $ \tan{XYB}= \tan{XDZ}$, which is equivalent to
\[ \frac{XB}{XY}=\frac{XZ}{XD}\Longleftrightarrow XB \cdot XD = XY \cdot XZ. \]
Due to the similarity of the triangles $ \triangle{XAB}$ and $ \triangle{XDA}$, we have that $ XA^{2}=XB \cdot XD$, so $ XA^{2}=XY \cdot XZ$, which by using $ XA=XF$, it is equivalent with $ \frac{YA}{YE}=\frac{ZA}{ZE}$, i.e. the division $ (AYEZ)$ is harmonic.

We denote as $K,$ $L,$ the intersection points of $BC,$ from $DF,$ $AE$ respectively and also the points $X\equiv AC\cap BE,$ $Y\equiv AB\cap CE,$ $Z\equiv AE\cap XY.$

From the complete quadrilateral $ABECXY,$ we conclude that the points $L,$ $Z,$ are harmonic conjugates, with respect to the points $A,$ $E.$

Because of $AD\perp AB,$ $DE\perp BE,$ $DK\perp KB,$ we conclude that the pentagon $ABEKD$ is cyclic ( taken as diameter the segment $BD$ ) and so, we have that the segment line $KB,$ is the angle bisector of the angle $\angle AKE,$ because of $AB = BE.$

So, because of $DK\perp KB,$ we conclude that the segment line $DF,$ passes through the point $Z,$ as the external angle bisector of the triangle $\bigtriangleup KAE,$ through vertex $K.$

Because of now, the points $X,$ $Y,$ $Z$ are collinear, based on the [b][size=100]Desarques’s theorem,[/size][/b] we conclude that the triangles $\bigtriangleup BEA,$ $\bigtriangleup CDF,$ are perspective $($ $X\equiv BE%Error. "capCD" is a bad command.
,$ $Y\equiv BA\cap CF,$ $Z\equiv AE\cap DF$ $).$

Hence, the segment lines $AF,$ $DE,$ $BC$ are concurrent at one point and the proof is completed.

Kostas Vittas.

[color=darkblue][b][u]Remark (a short commentary)[/u].[/b] It is a [u]very nice[/u] application  of the following [b]remarkable harmonical division[/b] (see the [u]nice[/u] [b]Pohoatza[/b]'s proof !) :[/color]

[quote][color=darkred][b][u]Lemma[/u].[/b] Let $ w(O)$ , $ w_o$ be two [u]secant[/u] circles in $ \{A,B\}$ so that $ O\in w_o$ . For $ P\in (AB)$ ([u]segment[/u] !) denote $ \{M,N,R\}$

such that $ \{M,N\}\subset w$ , $ R\in w_o$ and $ N$ separates $ P$ , $ R$ . Then the division $ \{\ M\ ,\ N\ ;\ P\ ,\ R\ \}$ is harmonically.

[b][u]Particular case[/u].[/b] Let $ ABC$ be a triangle with the orthocenter $ H$ . Denote $ D\in BC\cap AH$ and the intersections $ N$ , $ S$

between the line $ AH$ and the circle with the diameter $ [BC]$ . Then the division $ \{\ A\ ,\ H\ ,\ N\ ,\ S\ \}$ is harmonically. [/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] $ PO\cdot PR \equal{} PA\cdot PB \equal{} PM\cdot PN\ \implies\ PO\cdot PR \equal{} PM\cdot PN\ \Longleftrightarrow$ the division $ \{\ M\ ,\ N\ ;\ P\ ,\ R\ \}$ is harmonically.[/color]

[color=darkblue]The following interesting problem is a nice consequence of the harmonical division which was mentioned in the above lemma :[/color]

[quote][color=darkred]Let $ w(O)$ , $ w_o$ be two [u]secant[/u] circles so that $ O\in w_o$ . Consider $ A\in w\cap w_o$ . For $ P\in (AB)$ ([u]segment[/u] !) denote $ \{M,N,R\}$ such that

$ \{M,N\}\subset w$ , $ R\in w_o$ and $ N$ separates $ P$ , $ R$ . For $ C\in AB$ ([u]line[/u] !) denote $ \left\|\begin{array}{c} X\in MC\cap RA \\
\ Y\in NC\cap RA\end{array}\right\|$ .Then $ MY\cap NX\cap CP\ne\emptyset$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] From the upper lemma obtain that the division $ \{\ M\ ,\ N\ ;\ P\ ,\ R\ \}$ is harmonically, i.e. $ \frac {\overline {PM}}{\overline {PN}} \equal{} \minus{} \frac {\overline {RM}}{\overline{RN}}$ . Apply the [u][b]Menelaus[/b]' theorem[/u]

to the transversal $ \overline {RXY}$ and the triangle $ CMN\ : \ \frac {\overline{RM}}{\overline{RN}}\cdot \frac {\overline{YN}}{\overline{YC}}\cdot \frac {\overline{XC}}{\overline{XM}} \equal{} \plus{} 1$ $ \implies$ $ \frac {\overline{PM}}{\overline{PN}}\cdot \frac {\overline{YN}}{\overline{YC}}\cdot \frac {\overline{XC}}{\overline{XM}} \equal{} \minus{} 1$ , i.e. $ MY\cap NX\cap CP\ne\emptyset$ .[/color]

[quote="Sergey"][quote="pohoatza"]
Due to the similarity of the triangles $ \triangle{XAB}$ and $ \triangle{XDA}$, we have that $ XA^{2} \equal{} XB \cdot XD$, so $ XA^{2} \equal{} XY \cdot XZ$, which by using $ XA \equal{} XF$, it is equivalent with $ \frac {YA}{YE} \equal{} \frac {ZA}{ZE}$, i.e. the division $ (AYEZ)$ is harmonic.[/quote]
Please could you tell how exactly using $ XA \equal{} XF$ we get $ \frac {YA}{YE} \equal{} \frac {ZA}{ZE}?$  :blush:[/quote]
There is a typo. He means $ XA \equal{} XE,$ instead of $ XA \equal{} XF.$

Kostas Vittas.

Hello!

$ \left\{M\right\}\equal{}BE\cap DF$, $ \left\{N\right\}\equal{}BC\cap DF$, $ \left\{H\right\}\equal{}DE\cap BN$,
$ \left\{S\right\}\equal{}MH\cap BD$, $ \left\{T\right\}\equal{}MH\cap AC$.

$ DE\bot MB$ and $ BN\bot MD\Rightarrow$ the point $ H$ is the orthocenter of the triangle $ \Delta MBD\Rightarrow$
$ \Rightarrow MS\bot BD$ and $ \angle MSN\equal{}\angle MSE$     (1)
$ \Delta ABS\equiv\Delta EBS\Rightarrow\angle ASB\equal{}\angle ESB\Rightarrow\angle AST\equal{}\angle MSE$     (2)
(1), (2) $ \Rightarrow\angle MSN\equal{}\angle AST\Rightarrow$ the points $ A,S,N$ are collinear.

   Let's consider the triangles $ \Delta SDA$ and $ \Delta MEF$.
$ \left\{B\right\}\equal{}SD\cap ME$, $ \left\{C\right\}\equal{}DA\cap EF$, $ \left\{N\right\}\equal{}AS\cap FM$ and the points $ B,C,N$ are collinear.
Based on the Desargues's theorem, we conclude that the triangles $ \Delta SDA$ and $ \Delta MEF$ are
perspective (the lines $ SM,DE$ and $ AF$ are concurrent at the point $ H$).
   Hence, the lines $ AF,DE$ and $ BC$ are concurrent at the point $ H$.

   Best regards, Petrisor Neagoe :)

[quote="PhilAndrew"]Let $ ABC$ be a right triangle with $ A \equal{} 90^{\circ}$ and $ D \in (AC)$. Denote by $ E$ the reflection of $ A$ in the line $ BD$ and $ F$ the intersection point of $ CE$ with the perpendicular in $ D$ to $ BC$. Prove that $ AF, DE$ and $ BC$ are concurrent.[/quote]
Here is another approach:
Let $ T\equiv DF \cap AE$, $ S\equiv AE\cap BC$, $ K\equiv AE\cap BD$, $ H\equiv DT\cap BC$. In order to prove $ AF,DE,BC$ are concurrent, we need to prove $ (A,E,S,T) \equal{} \minus{} 1$. Define $ (BD)$ is the circle with diameter $ BD$, $ (DS)$ is the circle with diameter $ DS$. It is easy to notice that $ \{A,B,D,E,H\}\in (BD)$ and $ \{K,D,H,S\}\in (DS)\Longrightarrow DH$ is the radical axis wrt $ (BD),(DS)$. Thus, $ \mathcal {P}_{T/(BD)} \equal{} \mathcal {P}_{T/(DS)}$
$ \Longrightarrow \overline{TS}.\overline {TK} \equal{} \overline {TE}.\overline {TA}$, but $ K$ is the midpoint of $ AE$, hence $ (AEST) \equal{} \minus{} 1$, which implies to the result of the problem.
Our proof is completed $ \blacksquare$.

[quote="andyciup"]Here is another solution to this nice problem:
Let $DF\cap BC= O$, $AE\cap BD=M$, $AE\cap BC=N$ and assume $AD\geq DC$.
Let $\measuredangle{CBD}=x$, $\measuredangle{DBA}=y$, and $\measuredangle{BCA}=c$. It`s clear that $x+y+c=90^{\circ}$
Because triangle $BAM$ is right-angled, we get $\measuredangle{BAM}=c+x$.
Because the quadrilateral $ABDO$ is cyclic, we have $\measuredangle{OAD}=\measuredangle{OBD}=x$,
thus $\measuredangle{OAE}=y-x$.
In isosceles triangle $ABE$, we have $\measuredangle{BAE}=c+x$, therefore $\measuredangle{ABE}=2y$.
But $\measuredangle{ABN}=x+y$, therefore $\measuredangle{EBO}=y-x=\measuredangle{OAE}$.
This means that quadrilaterel $OABE$ is cyclic, and thus the pentagon $ADOEB$ is cyclic.
This means that $\measuredangle{EDB}=\measuredangle{EAB}=c+x\Longrightarrow \measuredangle{ADE}=2(c+x)\Longrightarrow \measuredangle{EDF}=y-x$.
In triangle $ABE$, $\ds N\in (AE)\Longrightarrow \frac{AN}{NE}=\frac{AB}{BE}\cdot\frac{\sin \measuredangle{ABN}}{\sin\measuredangle{EBN}}\Longrightarrow \frac{AN}{NE}=\frac{\sin(x+y)}{\sin(y-x)}$
In triangle $EDC$, $\ds F\in (EC)\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin\measuredangle{EDF}}{\sin\measuredangle{FDC}}\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin(y-x)}{\sin(x+y)}$
By multiplying the last two relations we obtain $\ds \frac{AN}{NE}\cdot\frac{EF}{FC}\cdot\frac{CD}{DA}=\frac{ED}{DA}=1$,
therefore, by the converse of Ceva`s theorem, the lines $AF, CN, ED$ are concurent, QED.[/quote]
Darn it this was exactly the same proof I just rediscovered a few minutes ago, which I originally wrote down a while ago but threw away the paper.

[quote="PhilAndrew"]Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$. Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$. Prove that $AF, DE$ and $BC$ are concurrent.[/quote]
The inspiration for this solution comes from Cosmin (pohoatza) in https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CCAQFjAA&url=http%3A%2F%2Fdiendantoanhoc.net%2Fforum%2Findex.php%3Fapp%3Dcore%26module%3Dattach%26section%3Dattach%26attach_id%3D16951&ei=dokOVOH4E8S0yASt5YLgBQ&usg=AFQjCNHsxpkOva9b0YPm1Io5LkgocHpOCg&bvm=bv.74649129,d.aWw (Harmonic Divisions and its Applications).

[hide="Solution"][quote="pohoatza"]Denote the points... $ Y \in AE \cap BC$, $ Z \in AE \cap DF$ and $ T \in DF \cap BC$.

In $ \triangle{AEC}$, we observe that the lines $ AF$, $ DE$ and $ BC$ are concurrent, if and only if, the division $ (AYEZ)$ is harmonic.[/quote]
We claim that $ADTEB$ is a cyclic pentagon. Indeed $ADTB$ is cyclic because $\angle BAD + \angle DTB = 90^{\circ} + 90^{\circ} = 180^{\circ}.$ In addition, due to right angles $\angle BAD$ and $\angle DTB$, we see that $\overline{BD}$ is a diameter of $(ADTB).$ Therefore the reflection of $A$ about $\overline{BD}$, i.e. $E$ must also lie on $(ADTB),$ which implies that $ADTEB$ is cyclic, as desired.

Therefore $\angle ETY = \angle ETB = \overarc{EB} = \overarc{AB} = \angle ATB = \angle ATY.$ So line $TY$ is the internal angle bisector of $\angle ATE.$ In addition, $YT \perp ZT.$
It is well known that these last two statements (angle bisector, and perpendicular lines) imply that the division $(AYEZ)$ is harmonic, and therefore $AF, DE$ and $BC$ are concurrent, as desired.  [/hide]

[quote=PhilAndrew]Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$. Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$. Prove that $AF, DE$ and $BC$ are concurrent.[/quote]

[b]My solution. [/b] Let $X$,$Y$, and $Z$ be the intersections of the line $AE$ with the lines $BD$,$BC$, and $DF$, respectively, and let $W$ be the intersection of the lines $DF$ and $BC.$ Because $E$ is the reflection of $A$ in the line $BD$ we have, $\angle DEB=\angle BAD=\angle DWB = 90^\circ$, which means that the pentagon $ADWEB$ is Cyclic. Thus, we have $$\angle YWE=\angle BWE= \angle BAE=\angle BAX=90^\circ - \angle XAD= \angle ADB=\angle AWB=\angle AWY$$
and we know that the lines $YW$ and $WZ$ are perpendicular. Therefore, the cross ratio $(A,E,Y,Z)$ is harmonic bundle, which means that the lines $CB$, $DE$, and $AF$ are concurrent.

[quote=Sergey][quote="pohoatza"]
Due to the similarity of the triangles $ \triangle{XAB}$ and $ \triangle{XDA}$, we have that $ XA^{2} \equal{} XB \cdot XD$, so $ XA^{2} \equal{} XY \cdot XZ$, which by using $ XA \equal{} XF$, it is equivalent with $ \frac {YA}{YE} \equal{} \frac {ZA}{ZE}$, i.e. the division $ (AYEZ)$ is harmonic.[/quote]

Please could you tell how exactly using $ XA\equal{}XF$ we get $ \frac {YA}{YE} \equal{} \frac {ZA}{ZE}?$  :blush:[/quote]

it is newton theorem

Let $DF \cap BC = K $
It is clear that $ADBK $ and $ADBE $ are cyclic which forces $AKEB $ also to be cyclic.
Now ,since $\angle EAB = \angle AEB ,BK $ bisects $\angle AKE $.
Now,by blanchet's theorem in $\triangle CDF $,$AF,DE,BC $ concur. $\blacksquare$

Ig there was a need for a different solution. So Bumped. :D
[quote=PhilAndrew]Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$. Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$. Prove that $AF, DE$ and $BC$ are concurrent.[/quote]

Let $DF\cap BC=T$. So, $\{A,B,E,T,D\}$ all lie on a circle. Now we will use Ratio Lemma and Ceva's Theorem.
$\frac{AK}{KE}=\frac{AB\sin\angle BAC}{BE\sin\angle EBC}$ and $\frac{EF}{FC}=\frac{ED\sin\angle EDF}{CD\sin\angle FDC}$. Now $\angle EDF=\angle EBC$ and $\angle ABC=\angle FDC$.

So now applying Ceva's Theorem we get that

$$\frac{AD}{DC}\cdot\frac{CF}{FE}\cdot\frac{EK}{KA}=\frac{AD}{DC}\cdot\frac{DC\sin\angle FDC}{ED\sin\angle EDF}\cdot\frac{BE\sin\angle EBC}{AB\sin\angle ABC}=\frac{AD}{DC}\cdot\frac{DC}{ED}\cdot{BE}{AB}=1\implies AF,DE,BC\text{ are concurrent.}\blacksquare$$

Let $P=AF\cap BC$, $Q=DF\cap BC$ and $H=AE\cap BC$. See that $Q\in (ABDE)$ and therefore
$$(Q,H,C,P)\overset{A}{=}(G,E,C,F)\overset{Q}{=}(A,E,B,D)=-1.$$
Furthermore, if $P'=DE\cap BC$ and $I=DH\cap (ABD)$
$$(Q,H,C,P')\overset{D}{=}(Q,I,A,E)\overset{H}{=}(B,D,E,A)=-1,$$
and thus $P=P'$, as desired. $\blacksquare$


For every positive integer n, form the number n/s(n), where s(n) is the sum of the digits of n in base 10. Determine the minimum value of n/s(n) in each of the following cases:

(i) 10 :le: n :le: 99
(ii) 100 :le: n :le: 999
(iii) 1000 :le: n :le: 9999
(iv) 10000 :le: n :le: 99999

we can write n as 10a + b for 10 <= n <= 99. Then we are looking for min( (10a + b)/(a+b))  =  (9a/(a+b) + 1) . This expression is minimized when we minimize a and maximize b so that a=1 , b=9 and n/s(n) = 19/10 = 1.9
We can apply the same technique for larger numbers, i.e finding min ((99a+9b)/(a+b+c) +1) when we have a 3 digit number in the form 100a + b + c.

correct, both of you. but bookworm, could you prove that your pattern works? i cant think of any other ways to solve this other than doing it like tetra or you.


function
real analysis
number theory proposed
number theory

Prove that the set of rational-valued, multiplicative arithmetical functions and the set of complex rational-valued, multiplicative arithmetical functions form isomorphic groups with the convolution operation $ f \circ g$ defined by \[{ (f \circ g)(n)= %Error. "displatmath" is a bad command.
\sum_{d|n} f(d)g(\frac nd}).\] (We call a complex number $ \textit{complex rational}$, if its real and imaginary parts are both rational.)

[i]B. Csakany[/i]

We prove this via several isomorphisms.

For any field $K$, the group of $K$-valued multiplicative arithmetic functions (under convolution) is isomorphic, via the Euler product of its Dirichlet series, to the infinite product $(1 + TK[[T]])^\Bbb N$, where the group operation in $1 + TK[[T]] = \{\sum_{n \geq 0} a_n T^n: \forall n, a_n \in K; a_0 = 1\}$ is the product of formal power series.
This isomorphism is given explicitly as follows: if $f$ is a $K$-valued multiplicative arithmetic function, then we map it to $\left(\sum_{n \geq 0} f(p_i^n) T^n\right)_{i \in \Bbb N}$, where $p_i$ denotes the $i$-th prime number.

If $K$ is a field of characteristic $0$, then the group $(1 + TK[[T]], \cdot)$ is isomorphic to $(TK[[T]], +)$ via $\log$ and $\exp$, which are defined as: $\log(1 + F) = \sum_{n \geq 1} \frac{(-1)^{n - 1}} n F^n$ and $\exp(F) = \sum_{n \geq 0} \frac 1{n!} F^n$ for any $F \in TK[[T]]$.
Moreover, it is clear that the group $(TK[[T]], +)$ is isomorphic to $K^\Bbb N$.

Thus we have shown that, when $K$ is a field of characteristic $0$, the group of $K$-valued multiplicative arithmetic functions is isomorphic to the group of $\left(K^\Bbb N\right)^\Bbb N$, which is again isomorphic to $K^\Bbb N$, as there is a bijection between $\Bbb N \times \Bbb N$ and $\Bbb N$.

It only remains to note that $\Bbb Q(i) \simeq \Bbb Q^2$ as additive groups, and hence $\Bbb Q(i)^\Bbb N \simeq \left(\Bbb Q^2\right)^\Bbb N \simeq \Bbb Q^\Bbb N$.


Determine the positive integers $ x,y,z$ so that: $ \frac{x^2}{30}\equal{}\frac{y}{20}\equal{}\frac{3z^6}{40}$ and $ xyz\equal{}2304.$


Any one  here ?

Okay I modified the post I posted last year :) Any one out there? This year (2006) :)

Yeah...I was there last year and I'll be there this year!

Okay - I edited the title again for this year(2007).   Please mark your calendars for this year's Cincinnati competition on Saturday February 24, 2007. It will again be at the University of Cincinnati's Zimmer auditorium. The state competition this year will be March 10 at Columbus State college.

Good luck to all.


I've seen many problems like this and I need to know a good way to solve them.

[quote="4everwise"]In racing over a given distance at uniform speed, Al can beat Bob by 20 meters, Bob can beat Chris be 10 meters, and Al can beat Chris by 28 meters. How many meters is the distance over which they are racing? Please show your work.[/quote]

$d=r_A \cdot t_1$
$d-20=r_B \cdot t_1$
$d=r_B \cdot t_2$
$d-10=r_C \cdot t_2$
$d=r_A \cdot t_1$
$d-28=r_C \cdot t_1$

It looks solvable, but I don't have time right now  :)

[hide="Solution"]If we assume that "$A$ beats $B$ by $x$ meters" means that $B$ was $x$ meters behind $A$ when $A$ crossed the finish line, then :
Chris was 28 meters behind Al when Bob was 8 meters ahead of Chris.  Therefore, in running 20 meters, Bob gains 2 meters on Chris; i.e., Bob travels at $\frac{10}{9}$ Chris's speed.  Therefore, 10 m is $\frac{1}{10}$ of the total length, since Chris has always travelled $\frac{9}{10}$ of the distance Bob has.  Therefore, the total length over which they are racing is 100m     [b]ANS[/b].
Q.E.D.[/hide]

$\boxed{\boxed{\boxed{\boxed{80}}}}$

Let the distance be 100. Then if Al beat Bob by 20 meters, then Bob ran 80 meters when Al ran 100 meters, so Al can run 125 meters in the time that Bob runs 100 meters (although we all know who's really faster :)), and then Bob can beat Chris by 10 meters, so in the time that Bob runs 100 meters, Chris runs 90 meters.

Also, in the time that Chris runs 90 meters, Al runs 125 meteres, which is 35 more meters.

The problem asked for 28 meters, so we do $\frac{28}{35}=\frac4{5}$

$\frac4{5}\cdot100=80$


Οι πραγματικοί αριθμοί x,y είναι τέτοιοι ώστε \[ {x^2} \plus{} xy \plus{} {y^2} \equal{} 1\]
Θέτουμε \[ K \equal{} {x^3}y \plus{} x{y^3}\]. Να αποδείξετε ότι:
α)\[ K \geqslant  \minus{} 2\]
β)Να βρεθεί η μέγιστη τιμη΄του Κ

[hide]α) $ k \equal{} xy(x^2 \plus{} y^2) \equal{} (xy) \minus{} (xy)^2 \geq \minus{} 2 \leftrightarrow (xy \plus{} 1)(xy \minus{} 2) \leq0$ που ισχυει διοτι,
$ 6>1 \equal{} x^2 \plus{} y^2 \plus{} xy \geq3xy$ και $ 1 \plus{} xy \equal{} x^2 \plus{} y^2 \plus{} 2xy \geq0$
β)ειναι $ K \equal{} xy \minus{} (xy)^2 \leq \frac {2}{9}$ ισοτητα για $ x \equal{} y \equal{} \sqrt {\frac {1}{3}}$[/hide]

Νομίζω παλιός Αρχιμήδης Μικρών 4ο θέμα (αν δεν κανω λάθος του 2006)

[hide]Eπειδη κοιτωντας το προηγουμενο ποστ μου πιστευω οτι στο (b) ισως εδωσα την απαντηση λιγο τυποποιημενα την γραφω πιο αναλυτικα.
Πρακτικα λεω πως το Κ παιρνει τις  τιμες του συνολου τιμων της παραβολης $ f(x)\equal{}x\minus{}x^2$ με x στο $ [\minus{}1,\frac{1}{3}]$κατι που προκυπτει αμεσα απο τις ανισοτητες που εγραψα στο (α).
 αρα $ K \leq \frac{2}{9}$ και την ισοτητα να ισχυει για $ (x,y)\equal{}(\sqrt{ \frac{1}{3}},\sqrt{ \frac{1}{3}})$ και αρα αυτο ειναι το μεγιστο που παιρνει (αρκει ενα ζευγος για να δουμε αν ειναι οντως μεγιστο..)[/hide]

[quote="Athinaios"][hide]Eπειδη κοιτωντας το προηγουμενο ποστ μου πιστευω οτι στο (b) ισως εδωσα την απαντηση λιγο τυποποιημενα την γραφω πιο αναλυτικα.
Πρακτικα λεω πως το Κ παιρνει τις  τιμες του συνολου τιμων της παραβολης $ f(x) \equal{} x \minus{} x^2$ με x στο $ [ \minus{} 1,\frac {1}{3}]$κατι που προκυπτει αμεσα απο τις ανισοτητες που εγραψα στο (α).
 αρα $ K \leq \frac {2}{9}$ και την ισοτητα να ισχυει για $ (x,y) \equal{} (\sqrt {\frac {1}{3}},\sqrt {\frac {1}{3}})$ και αρα αυτο ειναι το μεγιστο που παιρνει (αρκει ενα ζευγος για να δουμε αν ειναι οντως μεγιστο..)[/hide][/quote]
Σε ευχαριστλω πολυ για τη λύση είναι αρκετα διαφτιστική. Ναι είναι Αρχιμήσης μικρών 2006


Each dimension of a parallelogram is increased to four times its original size to form a similar parallelogram. If the new parallelogram has an area of 880 square units, what is the number of square units in the area of the original parallelogram?

This means the new parallelogram has an area $ 4^2 \equal{} 16$ times the original. Therefore, our answer is $ 880 \div 16 \equal{} 55$.


probability
combinatorics unsolved
combinatorics

On one of his travels, Bob landed on an island inhabited by equal numbers of two groups of people - the A's who always told the truth and the B's who answered questions truthfully with probability $ \frac{2}{3}$, independently for each question. One afternoon, Bob was lost on the island and on meeting a local inhabitant asked the following two questions: "Is it night or day?" and "Which is the way to the nearest town?"
(i) Find the probability that the answer to the first question is correct.
(ii) Given that the answer to the first question was correct, calculate the conidtional probability that the answer to the second qustion was correct.

Bad with probability, but let's try:

a)$ \frac{5}{6}$
b)$ \frac{3}{4}$


\[ \begin{array}{l}
 n \in N \\
 a_n  \equal{} n^2  \plus{} 20;d_n  \equal{} GCD(a_n ,a_{n \plus{} 1} ) \\ \end{array}
\]
 prove $ d_n |81$

$ HAPPY$ $ NEW$ $ YEAR!$

$ a_n \equal{} n^2 \plus{} 20$ and $ a_{n\plus{}1}\equal{}n^2\plus{}2n\plus{}21$.
Let $ d\equal{}\gcd(a_n,a_{n\plus{}1})$. So: $ d|n^2\plus{}20 , d|n^2 \plus{} 2n \plus{} 21$.
$ \Rightarrow d|(n^2 \plus{} 2n \plus{} 21)\minus{}(d|n^2\plus{}20)$ or $ d|2n\plus{}1$.
$ \Rightarrow d|(2n\plus{}1)^2$ or $ d|4n^2 \plus{} 4n \plus{}1$. But $ d|4(n^2 \plus{} 2n \plus{} 21)$ or $ d|4n^2 \plus{} 8n \plus{} 84$.
So we have: $ d|(4n^2 \plus{} 8n \plus{} 84)\minus{}(4n^2 \plus{} 4n \plus{}1)$ or $ d|4n\plus{}83$.
But $ d|4(2n\plus{}1)$. Finally $ d|4n\plus{}83 \minus{} 4(2n\plus{}1)$ or $ d|81$.

My solution right ?

Your solution is completely correct, but you have made a small mistake in the last part:You wrote: Finally $ d|4n\plus{}83\minus{}4(2n\plus{}1)$
I guess it would be $ d|4n\plus{}83\minus{}2(2n\plus{}1)\Rightarrow d|81$
But anyways, your solution is quite nice & I enjoyed reading it :)

[quote="Potla"]Your solution is completely correct, but you have made a small mistake in the last part:You wrote: Finally $ d|4n \plus{} 83 \minus{} 4(2n \plus{} 1)$
I guess it would be $ d|4n \plus{} 83 \minus{} 2(2n \plus{} 1)\Rightarrow d|81$
But anyways, your solution is quite nice & I enjoyed reading it :)[/quote]
Thank you for your notice. :blush:


The sum of consecutive even integers $ 2\plus{}4\plus{}6\plus{}\cdots\plus{}m \equal{} 2550$. What is the value of $ m$?

The sum of the first n evens is n(n+1)

n(n+1)=2550
n=50

The 50th even is 50*2 = $ \boxed{100}$


algebra
polynomial
abstract algebra
limit
algebra proposed

[color=blue]

   Write the following polynomial as a product of irreductible polynomials in $\mathbb{Z}[X]$

                                               \[ f(X)=X^{2005}-2005X+2004. \]

     Justify your answer. ;)

     [/color]

This problem is from Mathlinks Contest and it was posted by me here.

http://www.mathlinks.ro/Forum/viewtopic.php?t=71727

It was the only i solved in this round and here it is my solution

SOLUTION
During the solution i will use the folllowing lemma

\$ LEMMA
Let a polynomial $a_nx^n+.....+a_0$   with  integer coefficients.
If  there exist
a prime number  $p$  such that  $p$  doesnot divides $a_n$  and $p|a_0a_1....a_n$
and $p^2$  doesnot divide $a_0$ then the plynomial is irreducible over $Z[x]$

\ \$
Now to the solution

\$
From the fact that   the number  $1$  is duble solution and  by Horner we have that
$f(X)=X^{2005}-2005X+2004=(X-1)^2(X^{2003}+2X^{2002}+3X^{2001}+....+2002X^2+2003X+2004)$

We will prove that the polynomial $(X^{2003}+2X^{2002}+3X^{2001}+....+2002X^2+2003X+2004)$  is irreducible
The prime number  $p=7$  satisfays  the conditions of the lemma so the polynomial is irreducible.
So the factorisation is $f(X)=X^{2005}-2005X+2004=(X-1)^2(X^{2003}+2X^{2002}+3X^{2001}+....+2002X^2+2003X+2004)$

Silouanos Brazitikos

Silouan why is your lemma true?
could you post a proof please.

His lemma is in fact Eisenstein's criterion :)
And there's a little typo, it should be : $p|a_0, a_1, ... , a_{n-1}$
And in fact, it even assures that the polynomial is irreducible over $\mathbb{Q}[X]$.

For a proof of this result, see here for example : http://planetmath.org/encyclopedia/EisensteinCriterion.html

[quote="spix"]The roots in $\mathcal l Z$ of the ploynomial $f(X)=x^{2005}-2005X+2004$ should be amongst the divisors of $2004$. Notice that only $1$ is a the root of $f$ and a divisor of $2004$ now $f'(X)=2005X^{2004}-2005$ so $1$ has multiplicity of $2$ so $f(X)=(X-1)^2(X^{2003}+2X^{2002}+3X^{2001}+...+2003X+2004)$ . Hope it`s correct![/quote]

Is there anything wrong in my solution? :roll:

spix you should PROVE that the second polynomial is irreducible.
I dont understand how silouan did it
it is not surely eisenstein's criterion for $7$ because we dont have 7 divides 2,3,.....,2004 and $7^2$ doesn't divide 2004.
so i think mathmanman that there was no mistake in writing his lemma just I would be interested to see a proof of it if there is because I never have seen such a result

I`m sorry but I don`t understand here`show I thought: we want to put $f$ in an irreductible form over $\mathcal l Z[X]$ so it`s form should be something like $f(X)=(X-x_1)(X-x_2).....(X-x_k)g(X)$ where $x_1,x_2,...,x_k$ are integer roots of $f$ and $g\in\mathcal l Z[X]$ is what remains   and evidently $g$ is irreductible. If $f(X)=X^n+a_{n-1}X^{n-1}+...+a_1X+a_0,a_0\neq 0, f\in \mathcal l Z(X)$ then $f(\alpha)=0$ if $\alpha$ is a divisor of $a_0$.So I checked  amongst the divisors of $2004$ and got that $1$ is the only integer root of $f$ more than that  $1$ has multiplicity of $2$. So $f$ should be something like $f(X)=(X-1)^2g(X)$. And I determined $g(X)=X^{2003}+2X^{2002}+3X^{2001}+...+2003X+2004$ and $g$ is irreductible because if it wouldn`t be then $f$ would have another integer root different from $1$(false). Please correct me where I`m wrong!

ZetaX, just have a look at my post ;)
Albanian Eagle, his lemma is definitely Eisenstein criterion..

[quote="silouan"]
Surely :P . My lemma is not Eisenstein's criterion because we cannot find such a prime with this creterion ;)
My lemma is  another  creterion and I find it in an old Greek book. ;)[/quote]

What if we apply your lemma to the original polynomial?

[quote="silouan"]I don't understand. what do you mean?[/quote]

The best way for people to understand your solution is to provide a proof of the lemma you quoted.

Unless I misunderstood something here, I think the lemma is not true.

Take a look of a simple polynomial $x^2 + (p+1)x + p =(x+1)(x+p)$, where $p$ is a prime.

My friend both this and Eisenstein's criterion are only sufficient and by no means a necessary condition. Did you understand?

[quote="enescu"][quote="silouan"]Exactly. What is your opinion about my conjecture ?Thanks[/quote]
$x^4+2x^3+3x^2+4=(x+1)(x^3+x^2-x+4).$[/quote]

Please read my conjecture carefully .The polynomial we want to be irredicuble is not
$x^4+2x^3+3x^2+4$  but  $x^4+2x^3+3x^2+4x+5$   ;)  ;)

[quote="Farenhajt"][quote="ZetaX"]Irreducibility makes no sense for the reals to look at:
by the fundamental theorem of algebra, every real polynomial can be written as product of polynomials of degree $\leq 2$.[/quote]

Dear ZetaX, making mistakes doesn't mean - at least usually - that people don't know their basic stuff. I congratulate you on your assumed infallibility, imperviousness to hasty reading etc etc. but acting as a Patronizing Tutor of Local Significance really isn't necessary - unless, of course, this forum is your whole world?

Enjoy :)[/quote]
I just made a remark... (and hadn't seen your second post before posting)  :(
Sorry for making this impression, but I don't think I'm infalliable etc. (to see some of my nonsense just look some posts above yours).
I just wanted to clarify a more general reason why this fails, not only for you (I also don't think that you don't know your basic stuff), but for all...
But if you dislike this, I will not do again.

[quote="ZetaX"]I just made a remark... (and hadn't seen your second post before posting)  :(
Sorry for making this impression, but I don't think I'm infalliable etc. (to see some of my nonsense just look some posts above yours).
I just wanted to clarify a more general reason why this fails, not only for you (I also don't think that you don't know your basic stuff), but for all...
But if you dislike this, I will not do again.[/quote]

No, the point isn't in NOT doing it (we all need others to scrutinize our thinking), but I thought I disliked the tone... But possibly it was just me overreacting. Sorry.

OK, no more off-topicking. I repeat my last post so it doesn't get overlooked:

[quote="Farenhajt"]Counterexamples (it hasn't been said if all the coefficients must be positive):

$(-2)x^2+(-1)x+(0)=-x(2x+1)$
$(-1)x^2+(0)x+(1)=(1-x)(1+x)$[/quote]

[quote="Farenhajt"]No, the point isn't in NOT doing it (we all need others to scrutinize our thinking), but I thought I disliked the tone... But possibly it was just me overreacting. Sorry.[/quote]
Sorry for offtopicking, but I really want to know what exactly the problem is. Can you please explain¿

Sorry, i forget :blush:  the coefficients are positive

For that conjecture  it is enough to prove that the polynomial(not the same) is irreducible over Zp  for some prime p.Because there is a result
   “Suppose f(x) belongs to Z[X] with degree  greater than or equal to 1.Let g(x) be a polynomial in Zp[X] obtained from f(x)by reducing all the coefficients of  f(x) modulo p.If  g(x) is irreducible over Zp and  degf(x) = deg g(x) ,then f(x) is irreducible over Q”

Not that f(x) is primitive. Every irreducible polynomial over Z is primitive.So there is a great chance for the truth value of your conjecture.

Savithri

[quote="silouan"]Exactly. What is your opinion about my conjecture ?Thanks[/quoteSavithri

It is enough to prove the polynomial is irreducible over Zp for some prime.But that prime?Change that polynomial in terms of a0.Then take p as the smallest prime which is greater than a0+n.After that I cant proceed..

Savithri

[quote="silouan"][quote="Albanian Eagle"]
it is not surely eisenstein's criterion for $7$ because we dont have 7 divides 2,3,.....,2004 and $7^2$ doesn't divide 2004.
[/quote]

Surely :P . My lemma is not Eisenstein's criterion because we cannot find such a prime with this creterion ;)
My lemma is  another  creterion and I find it in an old Greek book. ;)[/quote]

Think about in this way..I dont know this will work.
It is enough to prove f(x) is irrucible over Zp for some prime.But how to get that prime?That was a problem.
For that conjecture it is difficult to get such a prime. But for a particular problem sometimes it may easy.Anyway you all try this way.

[quote="Kuba"]So if we prove that $X^{2003}+2X^{2002}+\ldots+2003X+2004$ is irreducible we are done.

Lemma1. Let $f\in\mathbb R_+[X], f(X)=a_0+a_1X+\ldots+a_nX^n$ so that $a_0\ge a_1\ge \ldots \ge a_n>0$. Then $f$ has only roots of module $\ge1$.

Proof. Suppose $f$ has a root $e, |e|<1$. Then $-a_0=a_1e+a_2e^2+\ldots a_ne^n | (e-1)$,
 $a_0=(a_1-a_0)e+(a_2-a_1)e^2+\ldots+(a_n-a_{n-1})e^{n-1}+a_ne^n\Rightarrow$
 $a_0=|a_0|\le |a_0-a_1||e|+\ldots+|a_{n-1}-a_n||e^n|+|a_n||e^{n+1}|<a_0$ .
 Contradiction. So $f$ has only roots with module $\ge1$.

Lemma2. $f\in \mathbb Z[X]$ is irreducible iff $X^nf(1/X)$ is irreducible, where $n=\deg f$.

Proof. easy.

Now, our goal is to prove that $g(X)=X^{2003}+\ldots+2003X+2004$ is irreducible. We see that $g$ satisfies the conditions of lemma 1.
More we see that $g$ must have at least one root of module $>1$ (because $g(0)=2004>1$).
Now from lemma 2 it is enough to proove that $X^{2003}g(1/X)=h(X)=2004X^{2003}+2003X^{2002}+\ldots+2x+1$ is irreducible. but because $g$
has roots of module $\ge 1$ it follows that $h$ has roots of module $\le1$. Because $g$ has at least one root of module $>1$ results that
$h$ must have at least one with module $<1$. Now if $h(X)=u(X)v(X), u,v\in\mathbb Z[X]$ at least one of $u,v$ must have a root of module $<1$.
Suppose $u$ has one. But $h(0)=1=u(0)v(0)$. So $u(0)=\pm1$ because $u\in\mathbb Z[X]$. This is a contradiction because $u$ has all roots of module $\le1$
and at least one $<1$. So $h$ is irreducible, then so is $g$. So the problem is done.[/quote]
Wrong, why $u(0)=\pm1$ is contradict to  $u$ has all roots of module $\le1$ and at least one $<1$? what about the following:
\[ u=3x^2+4x+1=(3x+1)(x+1) \]

Yeah, that was mentioned earlier in the topic.

Anyway, I'm reviving this thread in the hope of finding out whether that polynomial is irreducible or not.

In the attached paper, Theorem 3.7 states that $ \left( 1 \plus{} x \plus{} \ldots \plus{} x^n \right)^\prime$ is irreducible for all $ n$ such that $ n \plus{} 1$ is squarefree.

That clearly holds for $ n \equal{} 2004$. :| So the problem is finally "solved".

The lemma given can be reformulated as follows
If $ p(x) \equal{} a_0 \plus{} ... \plus{} a_nx^n$ with $ a_0 \geq a_1\geq... \geq a_n>0$ then all the roots $ z$ of $ p$ have $ |z| > 1$ or $ p$ has at least a root $ z$ which is a root of unity of some order different from 1.

With this lemma, and seeing that our polynomial does not have roots which are roots of unity, the problem is solved.

Good luck.


algebra
polynomial
logarithms
algebra unsolved

Let $P(x)$ be the polynomial of degree $4$ and $P(x)$ has four positive real roots.
Prove that the equation:
$\displaystyle \frac {1-4x}{x^2}P(x)+(1- \frac {1-4x}{x^2})P'(x)-P''(x)=0$ has four positive real roots.

This problem is not hard! :D
Try to solve

I've a soln but I want to see another soln :D

Just adding the source: Vietnam TST 1994 if I'm not mistaken :)

I am ready to kill someone...

The equation is equivalent to \[(\ln|P(x)-P'(x)|-\ln x^{4}-x^{-1})'=0.\] Let $0<x_{1}<x_{2}<x_{3}<x_{4}$ are four roots of $P(x)$ and $y_{1}<y_{2}<y_{3}$ are roots of $P'(x)=0$, then \[x_{1}<y_{1}<x_{2}<y_{2}<x_{3}<y_{3}<x_{4}.\] Note that $P(x)-P'(x)=$ has four roots: on $(x_{1},y_{1})$, on $(y_{1},y_{2})$, on $(y_{2},y_{3})$ and on $(y_{3},x_{4})$. It means that $\ln|P-P'|$ walks to $-\infty$ near four positive points, and $-\ln x^{4}-x^{-1}$ walks to $-\infty$ near $x=0$.
All together, it means that $\ln|P(x)-P'(x)|-\ln x^{4}-x^{-1}$ has 4 local maximums, which are roots of the initial equation.


parameterization
probability
expected value
probability and stats

We are observing a light. At time 0, the light flashes on and off instantaneously. After then, the light flashes on and off instantaneously at various points in time. The time between two adjacent flashes is distributed exponentially with parameter $ L$. We observe this process for a total time $ T$. What is the expected number of flashes that will occur within $ \tau$ units of time of another flash?

Exponentially distributed inter-arrival times implies a Poisson process.


A cylindrical tank 20 feet long and 8 feet in diameter is placed so that its axis is horizontal. When the tank is filled to a depth of 2 feet, how many cubic feet of liquid does it contain?

Can someone help? I've seen these problems before and get stuck trying to do these problems. Thanks.

[quote="anyone"][hide]The best thing to do is to draw a diagram, and you can see that when the cylinder is horizontal the base is half a circle of radius 2 ft and the height is 20 ft. So using the formula for volume of a cylinder and accounting for the fact that the base is only half of a circle, we can find that the volume is $(\frac{1}{2})(2^{2}\pi)(20)=(2\pi)(20)=\boxed{40\pi\text{ ft}^{3}}.$
Does that look right? [/hide][/quote]
Sorry, but thats a long way from the right answer. The correct answer was [hide="Dont click if youre still doing the problem."]$\frac{320}{3}\pi-80\sqrt{3}$.[/hide]

[hide]  Note that the height of the cylinder is on the ground.  Say that the cylinder is clear and the water is blue.  Before trying to figure out how much water is in the cylinder, let's first try to figure out the area of the blue portion of the circle (it really helps to imagine the cylinder).

Now, say that the water hits on of the circles of the cylinder at A and B.  Thus, we want to figure out the area of circluar segment contained by AB.  Connect A and B both to the center of the circle (call this O).  We know that OA=OB=4, since the radius is half the diameter.  Now, we know that the AB is 2 feet from the bottom of the circle, so draw radius OM that is perpindicular to AB at N (and hits the circle at M).  We know that MN=2.  ON=4-2=2.  By the Pythogorean Theorem, we can see that $AN=2\sqrt{3}=BM$.

Thus, the area of Triangle OAB is $\frac{AB(ON)}{2}=\frac{(AN+BM)(ON)}{2}=\frac{(4\sqrt{3})(2)}{2}=4\sqrt{3}$.  Now, we can see that $\tan\angle AON=\sqrt{3}\Rightarrow \angle AON=60=\angle BON\Rightarrow \angle AOB=120$.  So, we have that the area of the sector contained by arc AB is $\frac{120}{360}*\pi(4^{2})=\frac{16\pi}{3}$.  Subtracting the area of the triangle from the area of the sector gives us the area of the circular segment.  Thus, the area of the circluar segment is $\frac{16\pi}{3}-4\sqrt{3}$.

Now, we can multiply this area by the height, which is 20, to get the volume to be $(\frac{16\pi}{3}-4\sqrt{3})*20=\boxed{\frac{320\pi}{3}-80\sqrt{3}}$ [/hide]


inequalities
geometry
circumcircle
inradius
trigonometry
IMO Shortlist
geometric inequality

Let triangle $ABC$ be such that its circumradius is $R = 1.$ Let $r$ be the inradius of $ABC$ and let $p$ be the inradius of the orthic triangle $A'B'C'$ of triangle $ABC.$ Prove that \[ p \leq 1 - \frac{1}{3 \cdot (1+r)^2}.  \]

[hide="Similar Problem posted by Pascual2005"]

Let $ABC$ be a triangle with circumradius $R$ and inradius $r$. If $p$ is the inradius of the orthic triangle of triangle $ABC$, show that $\frac{p}{R} \leq 1 - \frac{\left(1+\frac{r}{R}\right)^2}{3}$.

[i]Note.[/i] The orthic triangle of triangle $ABC$ is defined as the triangle whose vertices are the feet of the altitudes of triangle $ABC$.

[b]SOLUTION 1 by mecrazywong:[/b]

$p=2R\cos A\cos B\cos C,1+\frac{r}{R}=1+4\sin A/2\sin B/2\sin C/2=\cos A+\cos B+\cos C$.
Thus, the ineqaulity is equivalent to $6\cos A\cos B\cos C+(\cos A+\cos B+\cos C)^2\le3$. But this is easy since $\cos A+\cos B+\cos C\le3/2,\cos A\cos B\cos C\le1/8$.

[b]SOLUTION 2 by Virgil Nicula:[/b]

I note the inradius $r'$ of a orthic triangle.

Must prove the inequality $\frac{r'}{R}\le 1-\frac 13\left( 1+\frac rR\right)^2.$

From the wellknown relations $r'=2R\cos A\cos B\cos C$

and $\cos A\cos B\cos C\le \frac 18$ results $\frac{r'}{R}\le \frac 14.$

But $\frac 14\le 1-\frac 13\left( 1+\frac rR\right)^2\Longleftrightarrow \frac 13\left( 1+\frac rR\right)^2\le \frac 34\Longleftrightarrow$

$\left(1+\frac rR\right)^2\le \left(\frac 32\right)^2\Longleftrightarrow 1+\frac rR\le \frac 32\Longleftrightarrow \frac rR\le \frac 12\Longleftrightarrow 2r\le R$ (true).

Therefore, $\frac{r'}{R}\le \frac 14\le 1-\frac 13\left( 1+\frac rR\right)^2\Longrightarrow \frac{r'}{R}\le 1-\frac 13\left( 1+\frac rR\right)^2.$

[b]SOLUTION 3 by darij grinberg:[/b]

I know this is not quite an ML reference, but the problem was discussed in Hyacinthos messages #6951, #6978, #6981, #6982, #6985, #6986 (particularly the last message).

[/hide]

Is construction a valid proof? I asked the question [url=https://artofproblemsolving.com/community/c878920h1843440_construction_versus_proof/]here[/url], but it was not made clear to me whether construction is a valid proof or not.

The problem is wrong. The inequality should be $p \leq 1 - \frac{1}{3} (1+r)^2$ but not $p \leq 1 - \frac{1}{3 \cdot (1+r)^2}$.


$x,y,z \in \mathbb{R}^+_0$
Given that \[ \frac{1}{3} \leq xy+yz+xz \leq 3  \]
Find maximum and minimum for $xyz$ and $x+y+z$.

[quote="Jan"]Given that \[ \frac{1}{3} \leq xy+yz+xz \leq 3  \]
Find maximum and minimum for $xyz$ and $x+y+z$.[/quote]
[hide="xyz"]
To find the max. of $xyz$, u use AM-GM inequality.
Set $xy=a$, $yz=b$, and $xz=c$. Then we have $\frac{a+b+c}{3}\leq 1$. And using AM-GM, $1\geq \frac{a+b+c}{3}\geq \sqrt[3]{abc}$. Thus, $xyz\leq 1$.
There is no lower bound for $xyz$. Let $x=1$ and $y=1$. Then our equ. becomes $\frac{1}{3}\leq 1+2z\leq 3$. Now $z=xyz$ can take arbitrarily small number, since there's $1$ added to $2z$.
[/hide]

I'm not sure...

[hide="x+y+z"]
Wlog, suppose $x\geq y \geq z$. By rearrangement, $xy+yz+xz\leq x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$
So $(x+y+z)^2 \geq 3(xy+yz+xz)\geq 1 \implies x+y+z \geq 1$[/hide]

By the way, the question does not state that $x,y,z$ are positive real numbers.
If x, y and z are assumed to be positive numbers, then we can interpret the problem geometrically.

I'm sorry, I posted this when I had to catch my train to school, so I had to hurry...  :(
Indeed luimichael, the original question says that $x,y,z \in \mathbb{R}^+_0$
Anyway, I saw a solution using AM-GM, so kimby already assumed that they are positive.
I edited it

I'm very interested in your geometrical approach to the problem, luimichael...

for xyz: $xyz=\sqrt[3/2]{xyz}^{3/2}\le\left(\frac{xy+yz+zx}3\right)^{3/2}\le1^{3/2}=1$, just 1 am-gm does it.

$x+y+z$ should have no maximum value because you can choose $x$ to be any arbitrarily large value and still find some small values for $y$ and $z$ such that the original inequality is satisfied. For example, let $y=z=\frac{1}{x}$. Then the inequality becomes $\frac{1}{3} \leq 2 + \frac{1}{x^2} \leq 3$. As $x$ gets larger, the inequality is still valid, but $x+y+z$ approaches infinity.

So, $2(xy+yz+zx)$ represents the total surface area of a cuboid with sides x, y and z.
While $xyz$ represents the volume of the cuboid.
By the isoperimetric property, if the total suface area of the cuboid is given, the maximum possible volume of the cuboid will be attained when the cuboid is a cube.
$2(xy+yz+zx)\leq 6 \implies$ the maximum value of $xyz=1$, when $x=y=z=1$.

Indeed. You can also extend it to the other side of course: $4(x+y+z)$ is the sum of all edges. (and $x+y+z$ is minimal exactly when $4(x+y+z)$ is, of course, being in case of a cube)

is this assuming x,y,z are positive reals..


What is the sum of the reciprocals of the natural-number
factors of 6?

The factors are 1,2,3, and 6.

So, our desired sum is

1/1+1/2+1/3+1/6

$ \boxed{2}$

Note that the sum of the reciprocals of the factors of any perfect number is 2. Try it and see for 28, etc.

A perfect number is a number that is the sum of its (improper) divisors. So since $6 = 1 + 2 + 3$ we get that $6$ is a perfect number. Since $1 + 2 + 4 + 7 + 14 = 28$ we find that $28$ is a perfect number. In general, all even perfect numbers are of the form $2^{p - 1}(2^p - 1)$, where $2^p - 1$ must be a prime number, and all numbers of that form are perfect numbers. No odd perfect numbers are known.

I notice a lot of these reciprocal (I'm not sure I spelled that correctly) problems are 2.

One question: How do you get an avatar???

Go to your User Control Panel, and click on the "Change Avatar" over on the left.

Where it says "submit file", submit a file.

Yes. 1/6+1/3+1/2+1=1/6+2/6+3/6+6/6=12/6=2. I don't know why my old self didn't post any solutions. :(


geometry
geometric transformation
reflection
power of a point

[img]http://img23.imageshack.us/img23/5614/ej103.png[/img]

In the figure it shows three semicircles (please imagine they are  :P ). Knowing $ \angle BDA \equal{} 90$ and $ BD \equal{} 1$, calculate the area of the region in grey.

I don't think there's enough information.
You could slide $ BD$ around and let $ AD$ and $ DC$ change freely, couldn't you? I mean, the length of $ BD$ would change, but you could scale the diagram up/down to get $ BD\equal{}1$.

It actually is enough (namely the region's area is $ \frac {\pi}{4} [(AD \plus{} CD)^2 \minus{} AD^2 \minus{} CD^2] \equal{} \frac {\pi}{2} \cdot AD \cdot CD$, which is just $ \pi / 2$ by power of point) :wink:

==> change that to $ \pi / 4$.

Hm. Assumably, then, the area doesn't change no matter where $ BD$ is, so we could just as well make $ D$ the midpoint of $ AC$. Then $ AD\equal{}BD\equal{}CD\equal{}1$. The diameters of the two semicircles are both $ 1$, so their areas are $ \frac{\pi}{4}$, which means the area of the shaded region is $ \pi\minus{}\frac{2\pi}{4}\equal{}\boxed{\frac{\pi}{2}}$.

Cool. ;)

sorry for the ignorance but after you got $ \frac{\pi}{2} \ast AD \ast CD$ how does that make it $ \frac{\pi}{2}$ by power of a point?

Reflect the semicircle over its diameter, and you have $ AD \cdot CD \equal{} BD^2$, which is just 1 because $ BD\equal{}1$.

Sorry just curious... Wouldnt the region's area be $ \frac{\pi}{8}[(AD \plus{} CD)^2  \minus{} AD^2 \minus{} CD^2]$ since they are semicircles? Making the answer $ \frac{\pi}{4}$ instead of $ \frac{\pi}{2}$ ?

Yeah now that I think about it I believe it's $ \frac{\pi}{4}$ instead of $ \frac{\pi}{2}$ but both answers are in the answer choices...


logarithms
geometry
3D geometry
sphere
calculus
derivative
complex analysis

How does one prove formally that if the complement w.r.t. to the riemann sphere of a region $ \Omega \subset \mathbb{C}$ doesn't have a connected component containing both $ 0$ and $ \infty$, then there is no continuous logarithm on $ \Omega$?

Ok, figured out an approach, but I can only make it work if the region is finitely multiply connected. Can someone please help me fix it for the general case? The problem in the proof below is if it's not finitely multiply connected then $ B$ is not nec. closed, so I don't have my least distance $ d$ between $ A$ and $ B$.

In this case, the complement can be written as a finite union of connected components, all closed in the riemann sphere. We can hence write the complement as a disjoint union of closed sets $ A\cup B$, $ 0 \in A$ and $ \infty \in B$. So $ A$ is bounded. Also, there is a least distance $ d > 0$ between $ A$ and $ B$. Write $ \mathbb{C}$ as the union of a net of squares with sides $ < d/2$ so that $ 0$ lies at the center of a square. We consider the boundaries of the squares as positively oriented arcs.

We introduce the cycle $ \gamma$ which is the (formal) sum of all boundaries belonging to squares which have non-empty intersection with $ A$. It's clear that the index of $ \gamma$ w.r.t. to 0 is $ 1$ and that $ \gamma$ does not meet $ B$ in this representation. Certain sides of the squares occur twice in $ \gamma$, being common to two adjacent squares. Any side that has a point in common with $ K$ is of this type, so the side is run through twice in opposite directions. Hence $ \gamma$ has a representation in $ \Omega$ (and it's clear that there is a representation as a formal sum of closed arcs).

Now, however if there was a logarithm in $ \Omega$ it would by elementary calculation have derivative $ 1/z$ and hence the index of $ \gamma$ w.r.t to $ 0$ would be $ 0$.

General case fixed by fedja here: http://www.mathlinks.ro/Forum/viewtopic.php?p=1081726#1081726

(I couldn't edit my post but I wanted to add it for completeness).


[hide="Solution to part 1"]Because $ \frac {2r \plus{} 1}{r^{2}(r \plus{} 1)^{2}} \equal{} \frac {1}{r^2} \minus{} \frac {1}{(r \plus{} 1)^2}$, we see that $ \sum_{r \equal{} 1}^{n}\frac {2r \plus{} 1}{r^{2}(r \plus{} 1)^{2}} \equal{} \sum_{r \equal{} 1}^{n}\left(\frac {1}{r^2} \minus{} \frac {1}{(r \plus{} 1)^2}\right)$, a telescoping sum.  Thus, the desired sum is equal to $ \frac {1}{1^2} \minus{} \frac {1}{(n \plus{} 1)^2} \equal{} \boxed{\frac {n^2 \plus{} 2n}{n^2 \plus{} 2n \plus{} 1}}$[/hide]


Putnam
number theory proposed
number theory

Write the number 1997 as the sum of positive integers for which the product is maximal, and prove there's no better solution.

I think it is a very very old well-known problem which already appeared in the Putnam competition and at IMO.

Pierre.

Is it? doens't really surprise me.

Although I wonder then why they picked it as a question for the national competition :?

just checking .. it is [tex]2 \times 3^{665}[/tex] right?

yes. or at least that's what I had. But my proof was long and complicated...

i thought the proof was quite standard in that it was a smoothing approach.

it did come out for the IMO so it is weird that it appeared for a national competition.  free marks for some i suppose.

dunno, this was in 1997, long before I participated :)

But anyway it's kind of unfair competition like that...


Eight singers take part in a festival. The organizer wants to plan a number of concerts with four singers performing in each. The number of concerts in which a pair of singers performs together is the same for every pair. Determine the minimum number of concerts.

[hide]
There are 7+6+....+1=7*8/2=28 pairs. When there is a performance, there are 6 total pairs whose "perform-together" count increases by 1. Therefore, the LCM of 6 and 28 divided by 6 is the number of concerts needed, which is 84/6=14, where each pair of performers has 3 concerts together.

If we pick 2 of the singers, and without loss of generality call them singers 1 and 2, then we can have them perform with 3,4, 5,6 and 7,8. With these 3 performances, 1 and 2 have performed 3 times together, while the other pairs of performers have exactly 1 performance between all of them. Now pick any pair from {(3,4), (5,6), (7,8)}. That pair already has 1 performance, and having them perform with another 2 pairs that have not already performed together will give them the needed 3 performances. This can be done with any pairs.
[/hide]


If a,b,c positive and abc=1 proove that
  \[ \sum \frac{2}{a^5(b^3+c^3)}\geq 3  \]

why doesn't anybody reply???Is it so stupid...  :blush:


geometry
circumcircle
incenter
Triangle
IMO Shortlist
angle bisector

Given a triangle $ABC$. The points $A$, $B$, $C$ divide the circumcircle $\Omega$ of the triangle $ABC$ into three arcs $BC$, $CA$, $AB$. Let $X$ be a variable point on the arc $AB$, and let $O_{1}$ and $O_{2}$ be the incenters of the triangles $CAX$ and $CBX$. Prove that the circumcircle of the triangle $XO_{1}O_{2}$ intersects the circle $\Omega$ in a fixed point.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Let $M,N$ be the midpoints of the arcs $CA,CB$ respectively. We have $MO_{1}=MC,\ NO_{2}=NC$, so $\frac{MO_{1}}{NO_{2}}=\frac{MC}{NC}$, which is constant (doesn't depend on $X$). Let $P$ be a point on the arc $AB$ s.t. $\frac{MP}{NP}=\frac{MC}{NC}$. Since we alos have $\angle PMO_{1}=\angle PNO_{2}$, it means that $PMO_{1},PNO_{2}$ are similar, so $\angle O_{1}PO_{2}=\angle MPN=\angle MXN=\angle O_{1}XO_{2}$, so $P$ always belongs to $\Omega$.

[color=red][[b]Moderator edit:[/b] This problem has also been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=555 .][/color]

[b]OFFICIAL SOLUTIONS:[/b]

Let $M$ be the midpoint of the arc BC. It is a known result (otherwise easy to prove) that triangles $MO_{2}B$ is isosceles with $MO_{2} = MB$. Therefore $MO_{2} = MB = MC = MO$, where $O$ is the incenter of the triangle $ABC$. In the same way $A,O_{1},O,B$ are on a circle with center $N$, the midpoint of the
arc $AC$. Let $P$ be on $\Omega$ such that $CP||MN$ and $T$ be the second intersection of $OP$ with $\Omega$. Then $MO = MC = NP$ and $MP = NC = NO$, therefore $MPNO$ is a parallelogram. It follows that $S_{MPT} = S_{NPT}$, whence $MP \cdot MT = NP \cdot NT$. Thus $NC \cdot MT = MC \cdot NT$, which shows that $NT : NO_{1} = MT : MO_{2}$. But then $\triangle NO_{1}T \sim \triangle MO_{2}T$ (because $\widehat{O_{1}NT} = \widehat{XNT} = \widehat{XMT} = \widehat{O_{2}MT}$), $\widehat{NTO_{1}} = \widehat{MTO_{2}}, \widehat{O_{1}XO_{2}} = \widehat{MTN} = \widehat{O_{1}TO_{2}}$ therefore the quadrilateral $XO_{1}O_{2}T$ is cyclic. This proves that the circle $(XO_{1}O_{2})$ passes through the fixed point $T$ or every position of $X$.

[b]Another Solution[/b] (Figure 2):

It is easy to prove that if $A,B$ are fixed points and $C$ is a obile point on one of the arcs $\overset{\frown}{AB}$ of a fixed circle then the locus of the incenter of the triangle $ABC$ is an arc with endpoints $A,B$. Therefore $O_{1}$ and $O_{2}$ are on the circles with centers $N$ and $M$ which pass through $A,C$ and $B,C$ respectively. This suggests a transformation of the figure using an inversion of pole $C$. The circle $(ABC)$ becomes the straight line $AB$, the straight line $CO_{1}$ becomes the bisector of the angle $\widehat{CAX}$, the tangent at $C$ to the circle $(ABC)$ becomes the parallel $(p)$ from $C$ to $AB$, the straight line $CN$ becomes the bisector $q$ of the angle $(\widehat{p,CA})$, the circle $(CAO_{1})$ becomes the perpendicular from $A$ on $(q)$, that is the external bisector of the angle $\widehat{BAC}$. Hence $O_{1}$ becomes an excenter for the triangle $CAX$. In the same way $O_{2}$ becomes an excenter for the triangle $BAX$. It follows that $O_{1}X \bot O_{2}X$, therefore the circle $(XO_{1}O_{2})$ becomes the circle of diameter $O_{1}O_{2}$. For $X = A$ we have $O_{1} = A, O_{2} = I_{c}$ (excenter for $\triangle ABC$), therefore the fixed point $T$ must be the projection of $I_{c}$ on $AB$. This implied by the following facts:

- $O_{2}X$ passes through $O^{'}_{1} = $ incenter of $\triangle CAX$; \item $\triangle AO_{1}T \sim \triangle O^{'}_{1}O_{1}O_{2} \Rightarrow \frac{AT}{O^{'}_{1}O_{2}} = \frac{AO_{1}}{O^{'}_{1}O_{1}}$;
- $\triangle CAI_{c} \sim \triangle CO^{'}_{1}O_{2} \Rightarrow \frac{O^{'}_{1}O_{2}}{AI_{c}} = \frac{CO^{'}_{1}}{CA}$; \item $AT = AI_{c} \cdot \frac{AO_{1}}{AC} \cdot \frac{CO^{'}_{1}}{O^{'}_{1}O_{1}} = AI_{c} \cdot \frac{4R \sin \frac{C}{2} \cos \frac{X}{2}}{AC} \cdot \frac{4R \sin \frac{X}{2} \sin \frac{A}{2}}{4R \sin \frac{C}{2}} = AI_{c} \sin \frac{A}{2}$

where $R$ and $A,C,X$ are the usual notations meant in $\triangle ACX$.

[i]Mentioned figures are coming in due course.[/i]

Let $D$ and $E$ be the midpoints of arcs $\widehat{CA}$ and $\widehat{CB}$. If $Y$ denotes the second intersection point of $\Omega$, which is the circumcircle of $\triangle{DEX}$, and the circumcircle of $\triangle{XO_1 O_2}$, then $Y$ is the center of spiral similarity sending segment $DE$ to $O_1 O_2$. This implies that $Y$ is also the center of the spiral similarity sending segment $DO_1$ to $EO_2$. Hence triangles $\triangle{YDO_1}$ and $\triangle{YEO_2}$ are similar which implies that $YD/YE=DO_1/EO_2$. However, since $O_1$ is the incenter of $\triangle{CAX}$ and $D$ is the midpoint of arc $\widehat{CA}$, it follows that $DO_1 = DA = DC$. Similarly, $EO_2 = EB = EC$ and hence $DY/EY=DC/EC$. Since the locus of all points $P$ such that $DP/EP=DC/EC$ is a fixed circle passing through $C$ and not passing through $D$ and $E$, it intersects $\Omega$ again exactly once. Furthermore, $Y$ cannot coincide with $C$ since $\angle{XO_1 C} + \angle{XO_2 C}=270^\circ$, which implies that $XO_1 O_2 C$ is not cyclic. Therefore $Y$ is the unique point satisfying $DY/EY=DC/EC$ and $Y \neq C$ and therefore is fixed.

Dear MLs,

Maybe knowledge of the fact that this fixed point is
the tangency point of mixtilinear  incircle, will provide
some fresh ideas for a solution .

M.T.

Dear Armpist and Mathlinkers,
the synthetic proof can be seen on
http://perso.orange.fr/jl.ayme vol. 4    A new mixtilinear incircle adventure I     p. 41-43.
Sincerely
Jean-Louis

Dear Jean-Louis

In your paper A new mixtilinear incircle adventure I
on p 42   you use   'Un quadrilatere harmonique' property
to prove that the fixed point of intersection of two circles in on
mixtilinear circle tangency point,
but the earlier  proof of this 'harmonique' property on p. 38
is based on the very fact that point P is the mixtilinear tangency point.

I find it a bit confusing.

BTW, who named this point after Longchamps and when?

M.T.

Dear Armpist and Mathlinkers,
yes, I have to read again this part....
Sincerely
Jean-Louis

Let the midpoint of $\widehat{BC}=G$ and $\widehat{AC}=F$, and the intersection of the two circles as $P$. Note that the $F,O_1,X$ and $G,O_2,X$ are collinear respectively.

Then we have $\angle O_1FP=\angle PGO_2$ and $\angle O_2PO_1=\angle O_2XO_1=\angle GBF\rightarrow\angle FPO_1=\angle GPO_2$, thus $\triangle PFO_1\sim\triangle PGO_2$. We note that $GO_2$ and $FO_1$ are constant lengths as $X$ varies since $F$ is always and the circumcenter $BO_2C$ and likewise for $G$, thus we have the lengths a constant (the radius of the circle). Let $FO_1/GO_2=c$ a constant.

But since $PF/PG=FO_1/GO_2=c$, and $P$ lies on the circle we have that $P$ is fixed since the ratio $PF/PG$ is fixed.

Another interesting solution is to apply an inversion centered at C with arbitrary power.

Let the image of $P$ be $P$ for simplicity.

[b]Then our problem can be phrased:[/b]

Let $ABC$ be a triangle, and let $X$ be a variable point on $AB$. Let $O_1, O_2$ be the C-excenters of $ACX$ and $BCX$ respectively.

Prove that the circumcircle of $O_1XO_2$ passes through a fixed point on $AB$.

[b]Proof:[/b]
Let this fixed point be $H$. We in fact prove that $H$ is the projection of $O_C$ onto $AB$, where $O_C$ is the C-excenter of $ABC$.

Since $\angle O_1XO_2 = 90^\circ$ (easy angle chasing), we must prove that $O_1HO_2 = 90^\circ$, which is equivalent to
$O_1H^2+O_2H^2 = O_1X^2 + O_2X^2$.

Let $H_1, H_2$ be the projections of $O_1, O_2$ onto $AB$.

Then $O_1H^2+O_2H^2 = O_1X^2 + O_2X^2$ is equivalent to
$O_1H_1^2+H_1H^2+O_2H_2^2+H_2H^2 = O_1H_1^2+H_1X^2+O_2H_2^2+H_2X^2$
$H_1H^2+H_2H^2 = H_1X^2+H_2X^2$.

So proving that $H_1H = H_2X$ is sufficient. (then $H_2H = H_1X$ would follow, so the equality would be obvious)

But $H_1H = \left\vert AH - AH_1 \right\vert =$
$ \left\vert \frac{AB+BC-AC}{2} - \frac{AX+CX-AC}{2} \right\vert = $
$\left\vert \frac{(AB-AX)+BC-CX}{2} \right\vert =$
$\left\vert \frac{BX+BC-CX}{2} \right\vert = H_2X$ (all equalities are just from simple properties of excircles tangencies).

[quote="SnowEverywhere"] Since the locus of all points $P$ such that $DP/EP=DC/EC$ is a fixed circle passing through $C$ and not passing through $D$ and $E$[/quote]

Can someone provide a proof for why this is true?

EDIT: ok i got it
just let $X$ be the point on $DE$ such that $DX/XE = DC/EC$. Then let $Y$ be the point such that $(DXEY)$ is harmonic. Then $DP/EP = DX/XE$ implies $PX$ is the angle bisector of angle $DPE$. Thus, by a well-known lemma, angle $XPY=90$ and $P$ lies on the circle of diameter $XY$.

Consider $N,M$ the midpoints of arcs $CB, CA$ and $\Omega_1, \Omega_2$ the circles centered at $M,N$ with radii $r_1=MC, r_2=NC$, respectively. Then by continuity, there exists a point $P \in \odot (ABC)$ such that

$ \frac{MP}{PN} = \frac{r_1}{r_2} := r$.

This is true since with $P=B$ the ratio equals $\frac{BM}{r_2}  > r$ and with $P=A$ the ratio equals $\frac{r_1}{AN} < r$.

Then notice $\angle PNX = \angle PMX$ and $\frac{PN}{MO_2}=\frac{PM}{MO_1}$, so $\triangle PNO_2 \equiv \triangle PMO_1$.

Thus, by Miquel's Theorem, $P=\odot (NXM) \cap \odot (XO_1O_2)$. Since $P$ is constant, we're done.

This was pretty tricky but a bit dumb/silly in motivation.

Here is my solution:-

Let $M,N$ be the mid-points of arcs $CA,CB$ and let $T$ be the touch-point of the $C$-mixitlinear incircle with the circumcircle and $E$ be the touch-point of the $C$ excircle with $BC$. Let $M',N'$ be points on line $AB$ such that $CM.CM'=CN.CN'=CB.CA$ and $\angle BCM = \angle ACM'$ and $\angle ACN = \angle BCN'$. (That is, $M',N'$ are the images of $M,N$ in the inversion about $C$ of radius $\sqrt{CA.CB}$ composed with the reflection in the $C$ angle bisector.)

Now, we see that $\angle \frac{A}{2} = \angle ACM'$ and so $AC=AM'$ Similarly, $BC=BN'$

And so we get that $E$ is the mid-point of $M'N'$.

Therefore, $CMTN$ is a harmonic quadrilateral and hence we get that $\frac{TM}{TN}=\frac{MA}{NA}=\frac{MO_1}{NO_2}$ and so we get that $\triangle TO_1M$ is similar to  $\triangle TO_2N$ and hence $T$ is the center of a spiral similarity sending $MO_1$ to $NO_2$ and so points $X,O_1,O_2,T$ are on a circle.

$\blacksquare \text{  } CMTN \text{  is harmonic where M,N are midpoints of arcs CA , CB and T is point of contact of C-mixtilinear incircle with (CAB) .} $

Proof --  Let $TM \cap AC = M' $ ,  $TN \cap CB=N' $  and $TA \cap \omega =A' $ where  $\omega$  is  C-mixtilinear incircle. It is well known that $CM'$ and $CN'$ are tangents to $\omega$ . So , we have $CM'TN'$ is harmonic .But as  $CM'TN'$ and  $CMTN$ are homothetic with center $T$ ,so  $CMTN$ is harmonic .

So , now ,  $\frac{TM}{TN}=\frac{MC}{NC}=\frac{MO_1}{NO_2}$ . This yields $ \triangle TO_1M \sim \triangle TO_2N \implies$ $T$  is unique center of spiral similarity sending $MO_1 \rightarrow NO_2 \implies T \in (XO_1O_2)$.

Let $O_1X \cap \Omega =M_1$ and $O_2X \cap \Omega =M_2$. Also let $T$ be the $C$-mixtillinear intouch point. We claim that $T$ is the desired fixed point. From [url=https://artofproblemsolving.com/community/c5h1230495p11189111]here[/url], this is equivalent to proving that $$\frac{TM_1}{TM_2}=\frac{M_1O_1}{M_2O_2}= \frac{b \cos \frac{A}{2}}{a \cos \frac{B}{2}}=\frac{\sin \frac{B}{2}}{\sin \frac{A}{2}}$$
Now it is well known that $CM_1M_2T$ is harmonic, which gives $$\frac{TM_1}{TM_2}=\frac{CM_1}{CM_2}=\frac{\sin \angle CM_2M_1}{\sin \angle CM_1M_2}=\frac{\sin \frac{B}{2}}{\sin \frac{A}{2}} \text{   } \blacksquare$$

[I](probably a different solution).[/I]

We claim $T$ is the fixed point where $T$ is $C-$ mixtillinear intouch point. Let $M$ and $N$ be points where $M = XO_1 \cap \odot(ABC)$ and $N = XO_2 \cap \odot(ABC)$. By fact 5, $CM = MA$ and $NC = NB$. Let $K$ be the second intersection of $\odot(ABC)$ and $\odot(XO_1O_2)$.

Let $\Psi$ be an inversion with center $X$ and arbitrary radius.

Now $A^*,B^*,M^*,N^*$ are collinear and $O_1$ and $O_2$ are excenters of $\triangle A^*C^*X$ and $\triangle B^*C^*X$ respectively. Now $K^* = O_1^*O_2^* \cap A^*B^*$. Let $Y = O_2^*C^*\cap O_2^*X$. We have \[-1 = (M^*,X;Z,O_1^*) \stackrel{O_2^*}{=} (M^*,N^*;C^*,K^*).\] We need to show that $K^* \equiv T^*$ or \[\frac{N^*C}{M^*C} = \frac{N^*T}{M^*T}\Leftrightarrow \frac{NC}{MC} = \frac{NT}{MT}.\]

[I]Adding a proof to : $CMNT$ is harmonic (because I didn't know it was true till today :P) Note: The $A^*, B^*$, etc. used below are different as the ones used above.

Consider an inversion $\mathcal{I}$ centered at $C$ with radius $\sqrt{CB\cdot CA}$ and reflection over $C-$ internal angle bisector. Now $M^*A^* = A^*C$ and $N^*B^* = B^*C$. Also $\mathcal{I}$ maps $T$ to $C-$ excircle touchpoint of $\triangle A^*B^*C$. Therefore $T^*M^* = T^*N^*$. This gives $\frac{MT}{CM} = \frac{TN}{CN}$ showing the desired result and we are done $\qquad \square$

-----

[I]Also my first G8 :P

Let $M_1,M_2$ denote midpoints of arcs $AC,BC$ respectively, so $X-O_1-M_1,X-O_2-M_2.$
Clearly $\odot (XO_1O_2)$ meet $\Omega$ again at Miquel point $Y$ of $M_1O_1O_2M_2$ and by Trillium theorem $$|YM_1|:|YM_2|=|O_1M_1|:|O_2M_2|=|CM_1|:|CM_2|$$ which is fixed and so defines $Y$ uniquely.

Let $I$ be incenter. Let $E$ be midpoint of arc $AC$, $F$ midpoint of arc $BC.$ We have $AE=O_1E=CE=IE,BF=O_2F=CF=IF.$

Let $G$ on $(ABC)$ such that $CG$ and $EF$ parallel, and $GI$ intersect $(ABC)$ again at $D$ where $I.$ $EI=EC=GF,IF=CF=EG$ implies $EGFI$ parallelogram.

We claim that $\triangle O_1DE\sim \triangle O_2DF.$ Since $\angle DEO_1=\angle DFO_2$ we only need to show \[ED/EO_1=FD/FO_2\]
\[\iff ED/EC=FD/FC\]
\[\iff ED\cdot EG=FP\cdot FG\]
\[\iff [EDG]=[FDG]\]
True by $EGFI$ parallelogram. Thus, our claim is true and so $\angle O_1DO_2=\angle EDF=\angle O_1XO_2.$ Thus, $D$ is the fixed point on $(XO_1O_2)$

Let $(O_1XO_2) \cap (ABC)=T_C$, $M,N$ the midpoints of minor arcs $AC,BC$ respectivily, now by I-E lemma we have that $MC=MO_1=MA$ and $NC=NO_2=NB$ so now note that $M,O_1,X$ and $N, O_2, X$ are colinear so $T_C$ is miquel point of $MNO_1O_2$ so:
$$\frac{MT_C}{T_CN}=\frac{MO_1}{O_2N}=\frac{MC}{CN} \implies -1=(M, N; C, T_C) \implies T_C \; \text{is unique and fixed}$$
Fun fact: If u project through the incenter of $\triangle ABC$ the harmonic and use $\sqrt{ab}$ inversion u get that $T_C$ is in fact the C-mixtilinear intouch point which is very hilarious tbh.

[quote=SL 1999 G8]Given a triangle $ABC$. The points $A$, $B$, $C$ divide the circumcircle $\Omega$ of the triangle $ABC$ into three arcs $BC$, $CA$, $AB$. Let $X$ be a variable point on the arc $CB$, and let $O_{1}$ and $O_{2}$ be the incenters of the triangles $ABX$ and $CAX$. Prove that the circumcircle of the triangle $XO_{1}O_{2}$ intersects the circle $\Omega$ in a fixed point.[/quote]

Nobody thought about complex, lol wutt :rotfl:
Let $(ABC)$ be the unit circle , $A=a^2,B=b^2,C=c^2,X=x^2$
So $O_1=-ab-ax-bx,O_2=-ac+ax-cx$, let $M_C,M_B$ be midpoints of minor arcs $AB,AC$.
Then $M_C=-ab,M_B=-ac$, since $\overline{X-O_1-M_C},\overline{X-O_2-M_B}$ if $(XO_1O_2) \cap (ABC)=T$, we've
$$\triangle TO_1O_2 \sim \triangle TM_BM_C$$
So, by similarity formula,
$$T=\frac{(-ab-ax-bx)(-ac)-(-ac+ax-cx)(-ab)}{-ac-ab-ax-bx-(-ab-ac+ax-cx)}=\frac{a^2(b+c)}{c-2a-b}$$
Which is independent of $x$ , thus $T$ is a fixed point .The End    $\blacksquare$

[b]S[/b]olved with [b]BVKRB[/b]

Let $P$ intersection point and$M,N$ be midpoint of arc $CA,AB$ then we know $\overline {N-O_1-X}$ and $\overline{M-O_2-X}$
From incenter lemma $NO_1=NA=NB$ and $MO_2=MA=MC$

Note $X$ is center of spiral similarity send $O_1O_2 \rightarrow NM$ so that $NO_1 \rightarrow MO_2$

$$\frac{NO_1}{MO_2}=\frac{PN}{PM}=\frac{NA}{MA}$$

$\implies (A,P;M,N)=-1$ and as only one such point satisfied we get $P$ is fixed point.


Euler
induction
modular arithmetic
algebra
binomial theorem
number theory solved
number theory

Prove or disprove : For all $n \in Z^+ , n\geq 3$ , you can find the number $c$ such that : $c^k$ , $k=1,2,..,n-2$ is not congruent to $1$ (mod n)

Just take $c=0$ or any nontrivial divisor of $n$ ;)
Also if $c$ has to be coprime with $n$ and $n$ is not prime, you get by the theorem of Euler-Fermat that $c^{\phi (n)} \equiv 1 \mod n$ and $\phi (n)<n-1$.

More interesting is the following question:
For which $n$ exists a coprime $c$ such that none of the numbers $c,c^2,c^3,...,c^{\phi (n)-1}$ is $\equiv 1 \mod n$¿
There is such a number iff $n=1;2;4;p^k;2p^k$ with some odd prime $p$. It is like your title says called a primitive root.

[b]definitions[/b]:
Let $\phi$ be Eulers totient function.
for coprime $a,n$ we call the smallest positive integer $k$ with $a^k \equiv 1 \mod n$ the order of $a \mod n$ and write $ord_n(a)$.
Then we get very simply that $ord_n(a) | \phi(n)$ for all such $a$.

'[b]known fact[/b]' (the idea for the prove can e.g. be found in post #6 of http://www.mathlinks.ro/Forum/viewtopic.php?t=14688 ):
\[ \sum_{d|n} \phi (n) =n  \]

[u][b]1. part[/b][/u]: such $c$ exists for the mentioned numbers (trivial for $n=1;2;4$, so not under discussion anymore):

[b]Lemma 1[/b]: there exists a primitive root $\mod p$ ($p$ a prime)
prove:
Let $\psi (d)$ denote the number of prime residue classes $\mod p$ with order $d$. Write $m$ instead of $\phi (p) =p-1$.
Then $\psi (d) =0$ for all $d \nmid m$ by the previous said.
Now for $d|m$ the equation $x^d-1 \equiv 0 \mod p$ has at most $d$ solutions (since the residue classes $\mod p$ build up a field). If $a$ is now an element of order exactly $d$, then $a$ is a root of this equation and with it also all powers of $a$, so we get already all solutions by $a,a^2,a^3,...,a^d\equiv 1 \mod p$. But $a^i$ can only has the same order as $a$ if(f) $i$ is coprime with $d$, and there are at most $\phi (d)$ such $i$ that give different residues (since $a^{i+d} \equiv a^i \mod p$). or in other words: $\psi (d) \leq \phi (d)$
But every prime residue class must have an order, so $\sum_{d|m} \psi (d)=m$.
Now cames the crucial point (using the 'known fact'):
\[ m=\sum_{d|m} \psi (d) \leq \sum_{d|m} \phi (d) =m  \]
and by this $\psi (d) = \phi (d)$ for all divisors $d$ of $m$, so especially for $m$ itself. But $\phi (p-1) =\phi (m) >0$, so there is a primitive root.

[b]Lemma 2[/b]: for all odd primes $p$, integers $e,k$ with $k \geq 2$ we have $(1+ep)^{p^{k-2}} \equiv 1+ep^{k-1} \mod p^k$
prove: Simple induction using the binomial theorem.

[b]Lemma 3[/b]: there exists a primitive root $\mod p^k$ ($p$ an odd prime, $k$ an integer $\geq 2$)
prove:
Let be $c$ any primitive root $\mod p$ (exists by lemma 1).
Now at least one of $g=c$ or $g=c+p$ fulfills $g^{p-1} \equiv 1 \mod p$ and $g^{p-1} \not\equiv 1 \mod p^2$.
The first one holds simply be Fermats theorem for both. Assume that $c^{p-1} \equiv 1 \mod p^2$ because otherwise we are done. Then the binomial theorem gives $(g+p)^{p-1} \equiv g^{p-1}+(p-1)g^{p-2}p \equiv 1-g^{p-2}p \not \equiv 1 \mod p^2$.
If $g$ is of the desired type, we can write $g^{p-1} \mod p^2$ as $1+ep$ with $p \nmid e$.
Now $g^{(p-1)p^{k-2}} \equiv (1+ep)^{p^{k-2}} \equiv 1+ep^{k-1} \not\equiv 1 \mod p^k$ after lemma 2.
Now we see that $ord_{p^k}(g) \nmid (p-1)p^{k-2}$, but as said, it divides $\phi (p^k) =(p-1)p^{k-1}$, so $ord_{p^k} (g) =p^{k-1}$ or $=\phi(p^k)$ and the first one is impossible (since $g^p \equiv g \not\equiv 1 \mod p$), so $g$ is a primitive root $\mod p^k$.

[b]Lemma 4[/b]: there exists a primitive root $\mod 2p^k$ ($p$ an odd prime, $k$ an integer $\geq 1$)
prove:
Take simply a primitive root $c \mod p^k$ and add $p^k$ iff $c$ is even. Then $c$ is coprime with $2p^k$ and has an order at least $\phi(p^k)$. But since $\phi (2p^k) = \phi (p^k)$ we don't need more and are done again.

[u]
[b]2. part[/b][/u]: such $c$ doesn't exist for the other numbers (in relation to lemmas 1 and 3 this part is really easy)
If $n$ has a factoring $n=ab$ into coprime numbers and $2| \phi (a), \phi (b)$, we get for all positive $c$ coprime to $n$ and by this also to $a$ and $b$:
\[ c^{\phi(a)} \equiv 1 \mod a \implies c^{\phi(a)\frac{\phi (b)}{2}} \equiv 1 \mod a \\ \& c^{\phi(b)} \equiv 1 \mod b \implies c^{\phi(b)\frac{\phi (a)}{2}} \equiv 1 \mod b  \]
and by the chinese remainder theorem now
\[ c^{\frac{\phi(a)\phi (b)}{2}} \equiv 1 \mod ab=n  \]
But $\frac{\phi(a)\phi (b)}{2} = \frac{\phi(n)}{2} < \phi (n)$, so $c$ is not a primitive root. But $c$ was any integer coprime to $n$, so there do not exist any primitive roots for such $n$. Simple checking gives that by this we already excluded all numbers that we wanted to exclude, except the powers of $2$.
If $n=2^k$, $k\geq 3$, one easily shows also
\[ c^{\frac{\phi(n)}{2}} \equiv 1 \mod n  \]
for all $c$ coprime to $n$.

Oh , I missed a condition : $c^i$ is not congruent to $c^k$ for $i,k = 1,2,..,n-2$
  Anyway , thank you very much , ZetaX

maximal m different means of remainds mod n  $c,c^2,\dots c^m$ is m=t(n). Were t(n) calculate for $n=p_i^{k_i}$ by formula
$t(n)=lcm(t(p_i^{k(i}), t(2^k)=2^{k-1},k=1,2,t(2^k)=2^{k-2},k>2,t(p^k)=p^{k-1}(p-1),p>2$ (p - is prime).

[quote="perfect_radio"]How do you prove that \[ (1+ep)^{p^{k-2}} \equiv 1+ep^{k-1} \pmod{p^k} \, ?  \][/quote]

For $k=2$, we have simply $1+ep \equiv 1+ep \mod p^2$, clearly true.
Now assume that we already proved it for $k-1$, so $(1+ep)^{p^{k-3}} \equiv 1+ep^{k-2} \mod p^{k-1}$.

Whenever $a \equiv b \mod p^{k-1}$ then $a^p \equiv b^p \mod p^k$ follows. The reason is that we can write $a=b+s p^{k-1}$, and then $a^p = \left( b+s p^{k-1} \right)^p = \sum_{n=0}^p \binom{p}{n} b^{p-n} \left(s p^{k-1}\right)^n \equiv b^p \mod p^k$, the last one because the exponent of $p$ in the summands $\neq b^p$ is at least $k$ (counted the one comming from $\binom{p}{n}$, too).

Then $(1+ep)^{p^{k-2}} = \left( (1+ep)^{p^{k-3}} \right)^p \equiv \left(1+ep^{k-2} \right)^p \mod p^k$ by the previous one. Now again expand by the binomial theorem to get the result.


darkquantum wrote:Prove that

nCm+2*[(n-1)Cm]+3*[(n-2)Cm]+...+(n+1-m)*(mCm)=(n+2)C(m+2)

Hint for people who want to work on it further:

[hide]A relatively well-known identity (Mr. Zeitz in TAACOPS calls it the hockey stick identity) that mCm+(m+1)Cm+(m+2)Cm+...+nCm=(n+1)C(m+1).[/hide]

The actual solution:

[hide]Knowing this information, I can partition the LHS of the problem into:

(nCm+(n-1)Cm+(n-2)Cm+...+(n+1-m)Cm)+((n-1)Cm+(n-2)Cm+...+(n+1-m)Cm)+

((n-2)Cm+...+(n+1-m)Cm)+...+mCm.

Using the above identity on each of the expressions in parentheses, I get:

(n+1)C(m+1)+nC(m+1)+(n-1)C(m+1)+...+(m+1)C(m+1)

Using the identity again, I get (n+2)C(m+2)[/hide]

Yes, MysticTerminator, that is correct!

Mr. Zimmermann at Brooklyn Technical High School calls it the "Christmas Stocking Theorem".

Anyway, knowing that theorem makes this olympiad problem seems like AMC 10.

Your proof is induction in disguise.

If you did not know the identity -- as, for example, I didn't -- or couldn't remember the exact form, you would probably figure it out by trying an inductive argument.


modular arithmetic
algebra proposed
algebra

1) $ f(xy)\plus{}f(x)\equal{}f(x)f(y\plus{}1)$
2) $ f(xy)\minus{}f(x)\equal{}f(x)f(y\plus{}1)$

[quote="Edward_Tur"]1) $ f(xy) \plus{} f(x) \equal{} f(x)f(y \plus{} 1)$
2) $ f(xy) \minus{} f(x) \equal{} f(x)f(y \plus{} 1)$[/quote]

what are $ x,y$? integers? reals? :(

$ 1)2f(0) \equal{} f(0)^{2}$  $ \implies$  $ f(0) \equal{} 0$  or $ f(0) \equal{} 2$

if $ f(0) \equal{} 2$  $ \implies$  $ 2f(y \plus{} 1) \equal{} 4$  $ \implies$  $ f(x) \equal{} 2$

if $ f(0) \equal{} 0$  $ \implies$  $ f(x) \equal{} \minus{} f( \minus{} x)$

$ f(xy) \minus{} f(x) \equal{} f(x)f(y \minus{} 1)$

$ f(y) \minus{} f(1) \equal{} f(1)f(y \minus{} 1)$

$ f(x) \equal{} f(x)f(1)$  $ \implies$  $ f(1) \equal{} 1$  $ (f(x)\not \equal{} 0)$

$ \implies$  $ f(y) \equal{} f(y \minus{} 1) \plus{} 1$  $ \implies$  $ f(x) \equal{} cx$  (for $ x\in\mathbb{Q}$)

$ 2)f(xy) \equal{} f(x)[f(y \plus{} 1) \plus{} 1]$

$ f(0) \equal{} f(0).[f(y \plus{} 1) \plus{} 1]$

if $ f(0)\not \equal{} 0$  $ \implies$  $ f(x) \equal{} \minus{} 1$

if $ f(0) \equal{} 0$  $ \implies$  $ f( \minus{} x) \equal{} f(x)$

$ \implies$  $ f(y \plus{} 1) \equal{} f(1 \minus{} y) \equal{} f( \minus{} y \minus{} 1) \equal{} f(y \minus{} 1)$

$ f(y) \minus{} f(1) \equal{} f(1)f(y \plus{} 1)$

if $ y \equal{} 0$ $ \implies$  $ f(1)^{2} \plus{} f(1) \equal{} 0$  $ \implies$  $ f(1) \equal{} 0$ or $ f(1) \equal{} \minus{} 1$

if $ f(1) \equal{} 0$ $ \implies$  $ f(y) \equal{} 0$

if $ f(1) \equal{} \minus{} 1$  $ \implies$  $ f(y) \plus{} f(y \plus{} 1) \plus{} 1 \equal{} 0$

$ f(x \plus{} 2k) \equal{} f(x)$ $ \implies$  if $ x\equiv 0\pmod{2}$  $ \implies$  $ f(x) \equal{} 0$

if $ x\equiv 1\pmod{2}$  $ \implies$  $ f(x) \equal{} 1$


Has anyone here ever heard "In Her Eyes" on Josh Groban's [i]Closer[/i] album? I can't figure out the time signature on it; the rhythm is...interesting. At first I thought it could be 5-8 or 5-4--I had been counting 1, 2, 3, 4, 5 and putting in notes/words in between "beats"--but then I noticed the system of accents in the lyrics and background music was more like a measure of 6-8 and then a measure of 2-4 (1, 2, 3-4). Sorry if my counting is confusing.

If anyone knows this or has the sheet music and can look at it, I'd be very happy to get an answer.

The meter can, and does, change, you know, especially in contemporary music.  :wink:

My woodwind quartet once read a piece where the number of eight notes in each measure went 2 2 3 2 3 4 2 3 4 5 2 3 4 5 6 2 3 4 5 6 7 2 3 4 5 6 7 8. The other movements had equally strange meters, but we gave up after the first movement.

Yeah, I get that meters can and do change. I was just wondering if anyone else had noticed this, or if it was something much simpler and I was just bad at counting beats. :oops:

Ciaccona, my head hurts just looking at that meter.

The meter in the song you're referring to is probably 5/4, subdivided into 4 groups of 3, 3, 2, and 2 eighth notes.  Basically, it alternates between 6/8 and 2/4, like you said, but the larger rhythmic grouping is quadruple in nature.


For the game show Who Wants To Be A Millionaire?, the dollar values of
each question are shown in the following table (where K = 1000).
Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Value 100 200 300 500 1K 2K 4K 8K 16K 32K 64K 125K 250K 500K 1000K
Between which two questions is the percent increase of the value the small-
est?
(A) From 1 to 2 (B) From 2 to 3 (C) From 3 to 4 (D) From 11 to 12 (E) From 14 to 15

-jorian :ninja:

it's definitely b. cuz it increases only 50 percent, while others are 66 and 2/3 or more

[quote="kstan013"][hide][u][b]C[/b][/u] it only increases by 66 2/3 percent. All of the others increase by 100% except for 11 to 12 which increases by about 95%[/hide][/quote]

I don't really blame kstan013 for getting it wrong because the qusetion was kind of hard to read because the numbers werent on top of each other.

yeah, but he didnt see ( i do not think) that 200 to 300 is less of a percent increase in his work.

yes, my aplogies...they were copied directly

whoever chose "B", is correct
-jorian

no its b the percent increase from 200 to 300 is only 50% increase

aww sig deleted

yeah he already said problem be4


1)

n! = 1 x 2 x 3 ....x n for n>=1
if p = 1! + (2x2!) + (3x3!) +.... +(10x10!)

what is the remainder when p+2 is divided by 11!
Please explain your solutions

2) the sum of the 31st and the 38th term of a AP is 333. what is the sum of the 4 consecutive terms starting  from the 33rd term?

Please explain your solutions.

Since this is an arithmetic progression, the sum of the 31st and 38th term=
the sum of the 32 and 37th term=
the sum of the 33 and 36th term=
the sum of the 34 and 35th term=
333.

So the answer is 666.

1. $ p\equal{} 1 \times 1! \plus{} 2 \times 2! \plus{} ...... \plus{} 10 \times 10! \equal{} 11! \minus{}1 \Rightarrow p\plus{}2 \equal{} 11!\plus{}1$ so the remainder when $ 11$ divide $ p\plus{}2$ is $ \boxed{1}$.

2.Let $ t_n$ be the $ n^{th}$ term in the AP and $ a$ is the first term while $ d$ is the common difference.

$ t_{31} \plus{} t_{38} \equal{} (a\plus{}30d) \plus{} (a\plus{}37d) \equal{} 2a\plus{}67d \equal{} 333$

$ t_{33} \plus{} t_{34} \plus{} t_{35} \plus{} t_{36} \equal{} (a\plus{}32d) \plus{} (a\plus{}33d) \plus{} (a\plus{}34d) \plus{} (a\plus{}35d) \equal{} 4a\plus{}134d \equal{} 2(2a\plus{}67d) \equal{} 666$.


function
geometry
geometric transformation
reflection

It was announced this morning that Barack Obama won the Nobel Peace Prize.  See [url=http://nobelprize.org/nobel_prizes/peace/laureates/2009/]here[/url].

Feel free to discuss.

I think it is an excellent choice, for its [color=green]exemplary[/color] value - not so much for the personal achievements of Obama, but towards the prestige of the function of active President of the United States. For so long this position has been scorned during Bush administration (see Michael Moore hoopla), that it is refreshing to see its value reinforced.

The Nobel peace price has a very dubious value these days; I mean, does one really want to be in the same boat with Arafat and Gorbachev? I personally do not. Also, in my opinion it should be given for some real achievements, not just for a couple of good speeches.

Personally, I expected Piedad Cordoba to win the prize - and I am quite a bit surprised with the decision.

Now, I'd like to comment [b]fedja[/b]'s comment above: just like an American wouldn't like to be in the same boat as Arafat and Gorbachev, most of the people in the East wouldn't like to be in the same boat as Kissinger or S. Peres (or even Gorbachev - he's generally not liked in the East either) - so its value is dubious on both sides of the globe.

As I (and I think many people) expected, this has already alienated Obama even more in the US.  Republicans are bashing him, saying that it's based more on speeches than actual policy (which is true), but they're also whining about how Reagan never won it and how it's some sort of leftist favoritism.  Even liberal Democrats are criticizing the choice, arguing that Obama's policies are too close to Bush's (and it seems like one of the main reasons he won is that he is supposedly the anti-Bush).

Frankly, I think it's a bit of a silly choice, but I kind of like it anyway because it's such a poke in the eye of Obama-haters.  They were practically dancing in the streets because Chicago didn't get the Olympics (suck on that, Obama!) and now they are just seething over Nobel.

But seriously, a Peace Prize is a nice idea in theory.  It just doesn't work in practice, unless you have some impartial aliens from the future choosing the winners.

Quite frankly, I would like to know what had Mr Obama done in order to be a recipient of the Nobel Peace Prize? (I'm not an "Obama" hater)

All he had done was that he had given a couple (or probably more) of nice, articulate and eloquent speeches which advocated 'nuclear disarmament' in this world --a proposition which is only theoretically possible. I really can't envisage a world which is devoid of nuclear weapons.

I think, probably after this rather historic judgment regarding this year's Nobel Peace Prize, the objectivity of the decision or judgment regarding the Peace prize might come into question.

Past recipients of the Peace prize had been people like Nelson Mandela, Mother Teresa -- who had done substantial work to propagate peace in this world. A Nobel Peace prize should not be awarded based solely upon a person's persona.

I think it's a good choice, if someone is to be chosen. He's a reasonable person who has managed to achieve a lot of power.

I was kind of surprised by this. I also agree that Obama hasn't physically done much about peace. He gave a few speeches and that was about it. Didn't he promise when he was campaigning that he would like get 60% of U.S. soldiers out of Iraq in the first two months if he is elected?

I completely agree with Gen8. Obama seems to be too idealistic with little practicality.

[quote="eli140"]Didn't he promise when he was campaigning that he would like get 60% of U.S. soldiers out of Iraq in the first two months if he is elected?[/quote]  No, and it's not clear why you would believe something quite this ridiculous when you are sitting in front of a computer on which you have access to the internet.

It seems like a good idea to consult what the Nobel folks actually listed as their reasons for giving him the prize:[quote="Nobel press release"]for his extraordinary efforts to strengthen international diplomacy and cooperation between peoples. The Committee has attached special importance to Obama's vision of and work for a world without nuclear weapons.

Obama has as President created a new climate in international politics. Multilateral diplomacy has regained a central position, with emphasis on the role that the United Nations and other international institutions can play. Dialogue and negotiations are preferred as instruments for resolving even the most difficult international conflicts. The vision of a world free from nuclear arms has powerfully stimulated disarmament and arms control negotiations. Thanks to Obama's initiative, the USA is now playing a more constructive role in meeting the great climatic challenges the world is confronting. Democracy and human rights are to be strengthened.

Only very rarely has a person to the same extent as Obama captured the world's attention and given its people hope for a better future. His diplomacy is founded in the concept that those who are to lead the world must do so on the basis of values and attitudes that are shared by the majority of the world's population.

For 108 years, the Norwegian Nobel Committee has sought to stimulate precisely that international policy and those attitudes for which Obama is now the world's leading spokesman. The Committee endorses Obama's appeal that "Now is the time for all of us to take our share of responsibility for a global response to global challenges."[/quote]  Several of these seem like genuinely legitimate points.  The nuclear issue, for example, really is significant and has already led to some notable success.  (It's worth noting here especially that a bilateral agreement between the US and Russia is much more than just about the US and Russia; China will determine the number of nukes it "needs" largely relative to US and Russian stockpiles, while India will think relative to China and Pakistan will think relative to India.  Ratcheting down the top numbers decreases incentives to build nukes all the way down the line.)

A defense of the prize on other grounds (esp. the US non-intervention in Honduras) is here: http://alterdestiny.blogspot.com/2009/10/some-more-thoughts-on-obamas-nobel.html

Oh, so he promised to bring all troops home from Iraq in 16 months. It says online that the american soldiers in Iraq would come home 1-2 months late if he keeps his promise, but so far he hasnt.

[quote="Kalle"] He's a reasonable person who has managed to achieve a lot of power.[/quote]
Exactly! And that is the problem with how they choose to whom to give the Nobel peace prize nowdays. It reminds me of a famous funny story that goes like that: [i]My neighbor is so nice, so nice! I met him on the street this morning and he just looked at me and said hello while he could easily pull out a gun and shoot me on the spot...[/i].
(compare that with the "non-intervention in Honduras" story Joel refered to).

This is not to say that Obama is a bad man or anything like that, just to say that his Nobel prize is much more due to his powerful position rather than to his achievements, to the bad things he didn't do though he easily could rather than to the good things he did though they were hard to do. His main virtue so far seems to be just the lack of obvious vices.

[quote="Gen8"]
Past recipients of the Peace prize had been people like Nelson Mandela, Mother Teresa -- who had done substantial work to propagate peace in this world. A Nobel Peace prize should not be awarded based solely upon a person's persona.[/quote]

Mother Teresa [url=http://en.wikipedia.org/wiki/The_Missionary_Position_(book)]is actually a significantly more controversial figure than you may think.[/url]

[quote="blahblahblah"] Mother Teresa [url=http://en.wikipedia.org/wiki/The_Missionary_Position_(book)]is actually a significantly more controversial figure than you may think.[/url][/quote]

Wow, what an awesome point! But, most famous people are entangled or embroiled in some controversy or the other. But, my contention was-- just as [b] fedja [/b] had said-- Obama had won the peace prize solely due to his persona and his 'powerful position'. Relatively speaking, Mother Theresa was much more deserving than Mr Obama to win the prize. Barring the controversies surrounding Mother Theresa, I personally feel she had a done a lot for the destitute people of the slums in Calcutta, India.

The points raised by [b] fedja [/b] are highly cogent. Relatively speaking, Obama is more friendly and is dealing far more diplomatically than his predecessor. But, I have to opine that chinese citizen [b] Hu Jia [/b] was far more deserving than Obama to win this year's prize.

For more about Hu Jia, refer the link below:
[url]http://english.ntdtv.com/ntdtv_en/ns_china/2009-10-08/541537788575.html[/url]

Other winners have included Yasser Arafat, Henry Kissinger, and Mikhail Gorbachev, decisions which have generated a lot of controversy.  That's one of the fun things about the Prize: debating who should and shouldn't have won.

I don't believe that Obama deserved the prize, but I don't think we should blame him for it. Some claim that he should have turned it down, but I wonder whether they would have criticized him as impolite or as trying to appear uber-humble had he turned it down.

I don't see why people are saying that all he did was give a couple of speeches. He has accomplished much in terms of international diplomacy, Iraq, Guantanamo, and arms reductions. Sure, this is nowhere near enough to earn a Nobel Peace Prize, but he's done more than just "given a couple of speeches." In fact, I often find, in general, that I hear more from people complaining that others like Obama for his words and not his substance than I actually hear from people who do like Obama for his words and not his substance. It's a great attack tactic, when your opponent happens to be a good speaker. I am in fact very much upset by Obama's policies, yet I view many of his critics as completely unjustified.

Did anybody notice: http://en.wikipedia.org/wiki/2009_Nobel_Peace_Prize#In_Norway. I find it surprising that the older generations were more supportive of Obama, and interesting (though not surprising) that the leftist parties were the most negative about Obama receiving the peace prize. Yes, this is Wikipedia, but both of these claims have sources, so if anyone speaks Norwegian, feel free to check these facts.

As for what will actually happen, it seems right now that the prize will hinder his efforts because so many people (probably including him) believe that it was undeserved.

[b]Fedja[/b], I'm just curious what is negative about Gorbachev? I was always taught that he increased political freedoms in the USSR.

[b]hsiljak[/b], I see what you say about Kissinger, but what is the opposition to Peres? As far as I know, he is known as a leftist dove within Israeli politics.

[quote="calc rulz"]
[b]hsiljak[/b], I see what you say about Kissinger, but what is the opposition to Peres? As far as I know, he is known as a leftist dove within Israeli politics.[/quote]

*disclaimer: the following is a personal analysis of general opinion of S. Peres in one part of the world. It does not reflect my opinion of him*

Although Shimon Peres is (or more precisely, has been) known as a leftist dove in political circles of Israel, he hasn't been popular in the pro-Palestinian circles around the globe. I don't think that reasons for the 'unpopularity' can be completely objective, concerning he's not the 'hardcore fighter' (although he was something like that in the very beginning of his career). Maybe he's simply marked in this part of world as just another Zionist leader, just another in the line of those who want to ghettoize the Palestine, etc (note that his Avoda party was in office during three great wars in the 1960s, so even they have the 'tradition of warfare'). Maybe only Yitzhak Rabin isn't looked at in that manner, probably because of the way he died.

By the way, Avoda and its leaders don't not seem 'left enough' for me. I don't expect them to be like Maki, but the political stands of Avoda don't seem much different from those presented by Kadima (which is current Peres' party, correct?) or even Likud.

Isn't this like giving a rookie QB who has never taken a snap in college ball, but was proclaimed as one of the best high school players ever the Heisman?

Not quite.  As the Nobel Committee said, one of the major reasons for giving him the prize is because he's changing (or attempting to) the way people view the US, diplomacy, and the world.  The Heisman is supposed to go to the best player in college football for the year, not necessarily the one with the most impact (one could argue that a better analogy would be the MVP award in baseball, which is supposed to go to the player who has the biggest impact on his team and league, not necessarily the best numbers).

Some dates to think about:

Date of Inauguration of President Barack Obama: January 20th, 2009
Deadline for sending in nominations for the 2009 Nobel Peace Prize: February 1st, 2009.

He was either nominated before he became president, or less than 11 days after he became president.

Yes, but being nominated is not the same as being awarded -- the committee wouldn't have made its decision until some time in September.  (Note that more than 200 people were officially nominated; "being nominated" happens to lots of people who deserve it less.)

And also remember that there are a huge amount of people who can nominate someone. And iirc, you only need one supporter to get officially put on the nominee list. So even if you think he didn't deserve it, his nomination is not at all a fault of the prize commission.

Actually,the press in India is blaming Obama while he had nothing much to do after he won the prize.I personally feel that he hasn't done enough to win the prize.Again, nuclear disarmament cannot be a reality until the conflicts of the world are sorted out and everything's perfect(which again can only occur atleast after 30 years considering the no. of problems ).So,by awarding the Nobel to Obama,the committee proved why it is one of the most controversial prizes.