calculus
derivative
integration
logarithms
calculus computations

$ \int^1_0 \frac{x^{m\minus{}1}}{\ln x} dx$

I have tried to let that be $ I(m)$ and then differentiate with respect to m and integrate again, and I get $ I(m) \equal{} \ln (m\plus{}1) \plus{} C$ but everything I try tells me C is negative infinity, which isn't correct...

The integral is divergent, because $ \ln x$ behaves like $ x\minus{}1$ as $ x\to 1$.

ahh so there was nothing wrong in my evaluation of c afterall. Thanks.


trigonometry
geometry
IMO Shortlist
symmetry
angle bisector

$ C$ and $ D$ are points on a semicircle. The tangent at $ C$ meets the extended diameter of the semicircle at $ B$, and the tangent at $ D$ meets it at $ A$, so that $ A$ and $ B$ are on opposite sides of the center. The lines $ AC$ and $ BD$ meet at $ E$. $ F$ is the foot of the perpendicular from $ E$ to $ AB$. Show that $ EF$ bisects angle $ CFD$

[quote="Fermat -Euler"]$C$ and $D$ are points on a semicircle. The tangent at $C$ meets the extended diameter of the semicircle at $B$, and the tangent at $D$ meets it at $A$, so that $A$ and $B$ are on opposite sides of the center. The lines $AC$ and $BD$ meet at $E$. $F$ is the foot of the perpendicular from $E$ to $AB$.
Show that $EF$ bisects angle $CFD$[/quote]

Maybe it's another midnight hallucination, but I have some strange feeling of seeing the following solution on ML:

Let the tangents to the semicircle at the points C and D meet each other at a point G. Also, let U be the center of the semicircle. Then, the points A and B, both lying on the diameter of this semicircle, must be collinear with its center U.

Since the semicircle touches the sides AG and BG of the triangle ABG, its center U must lie on the angle bisector of the angle between these sides, i. e. of the angle AGB. In other words, the line GU is the angle bisector of the angle AGB.

Since the line BG is the tangent to the semicircle at the point C, while U is the center of this semicircle, we have $BG\perp UC$; thus, the point C is the foot of the perpendicular from the point U to the line BG. Similarly, the point D is the foot of the perpendicular from the point U to the line AG.

Now, we can apply the result of http://www.mathlinks.ro/Forum/viewtopic.php?t=5460 to the triangle ABG with the point U on its side AB, and obtain that the line GE is an altitude of the triangle ABG if and only if the line GU is an interior angle bisector of the triangle ABG. But the line GU indeed is an interior angle bisector, as we saw above; thus, the line GE is an altitude of triangle ABG. In other words, the point E lies on the altitude of triangle ABG from the vertex G to the side AB. Hence, the point F, being the foot of the perpendicular from the point E to the line AB, must be the foot of this altitude.

Now we can describe our situation as follows: The point F is the foot of the altitude of the triangle ABG from the vertex G, and the point E is a point on this altitude GF. The lines AE and BE intersect the lines BG and AG at the points C and D, respectively.

Now, according to the Blanchet theorem ( http://www.mathlinks.ro/Forum/viewtopic.php?t=20809 post #5 problem [b](a)[/b]), we have < DFG = - < CFG (with directed angles modulo 180°); in other words, the line EF bisects the angle CFD. And the problem is solved.

An alternative approach would be possible by reflecting the whole figure in the line AB and applying some degenerate cases of Brianchon.

  Darij

[quote="mohamadhosein"]it has a solution whit using fermat point just try it. ;)[/quote]

That would be very strange.

  darij

$\blacksquare \ 1^{\circ}.\ Z\in BC\cap AD;\ U\in AB\cap CD;\ P\in AB,\ ZP\perp AB;$

$T\in CD\cap ZP\Longrightarrow$ [b]quadrilaterals[/b] $PCZO,\ PZDO$ [b]are cyclic[/b] $\Longrightarrow$

$\widehat {BPC}\equiv\widehat {BZO}\equiv\widehat {AZO}\equiv \widehat {APD}\Longrightarrow\widehat {BPC}\equiv \widehat {APD}\Longrightarrow$

$(U,C,T,D)-h.d.\Longrightarrow (U,B,P,A)-h.d.$

$\blacksquare \ 2^{\circ}.\ R\in ZE\cap AB,\ U\in AB\cap CD\Longrightarrow (U,B,R,A)-h.d.$

$\blacksquare \ 3^{\circ}.\ (U,B,P,A)-h.d.,\ (U,B,R,A)-h.d.\Longrightarrow$

$P\equiv R\equiv F\ (E\in ZP)\Longrightarrow\widehat {BFC}\equiv \widehat {AFD}\Longrightarrow\overline {\underline {\left| \ \widehat {EFC}\equiv \widehat {EFD}\ \right| }}$

[u]Remark.[/u] I note $h.d.$ - harmonical division.

[hide="An another shorter and nice proof."][color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ABC$ be a triangle for which the projection of the vertex $A$ to the line $BC$ belongs to the side $[BC]$. Denote : the foot $D$ of the $\widehat{BAC}$- bisector; the its points $E\in AC$, $F\in AB$ for which $DE\perp AC$, $DF\perp AB$; the intersection $P\in BE\cap CF$. Prove that $AP\perp BC$. [/color]

[color=darkblue][u][b]Proof.[/u][/b] Denote the intersections $L\in EF\cap BC$ and $R\in BC\cap AP$.

It is well-known that the division $(B,C,R,L)$ is harmonically, i.e. $\frac{LB}{LC}=\frac{RB}{RC}$.

Apply the Menelaus' theorem to the transversal $\overline{EFL}$ and the triangle $ABC$:

$\frac{LB}{LC}\cdot\frac{EC}{EA}\cdot\frac{FA}{FB}=1$. But $AE=AF\implies\frac{LB}{LC}=\frac{FB}{EC}\implies\frac{RB}{RC}=\frac{FB}{EC}$.

From $\{\begin{array}{c}FB=BD\cos B=\frac{ac}{b+c}\cos B\\\\ EC=DC\cos C=\frac{ab}{b+c}\cos C\end{array}$ $\implies\frac{RB}{RC}=\frac{c\cdot \cos B}{b\cdot\cos C}$.

But and for the projection $R'$ of the point $A$ to the line $BC$ exists the same relation

$\frac{R'B}{R'C}=\frac{c\cdot\cos B}{b\cdot\cos C}$, i.e. $R'\equiv R$. Therefore, $AP\perp BC$.[/color][/hide]

Extend $AD$ and $BC$ to meet at $P$,and let $Q$ be the foot of the perpendicular from $P$ to $AB$. Denote $O$ be the center of semicircle.We have:
$\begin{cases}\triangle PAQ\sim\triangle OAD\\ \triangle PBQ\sim\triangle OBC\end{cases}\Longrightarrow\frac{AQ}{AD}=\frac{PQ}{OD}=\frac{PQ}{OC}=\Longrightarrow\frac{AQ}{QB}\cdot\frac{BC}{CP}\cdot\frac{PD}{DA}=1$
$\Longrightarrow AC,\,BD,\,BQ\,\text{are\,concurrent}\Longrightarrow Q\equiv F$
Finally, since the points $O,\,C,\,P,\,D,\,F$ are concyclic, we have: $\angle{DFP}=\angle{DOP}=\angle{POC}=\angle{PFC}$

Let $E$ and $F$ be on the semicircle so that $B,F,E,A$ are collinear, in that order. Let the tangents to the semicircle at $C$ and $D$ intersect at $P$, and let $Q$ be the intersection of $FC$ and $PE$ (possibly the point at infinity.) By Pascal's theorem on $DDECCF$, we get that $P$, $Q$, and $E$ are collinear. $EC \perp QF$ and $FD \perp QE$, so $E$ is the orthocenter of $\triangle QFE$. Hence, $Q$, $E$, and $F$ are collinear, so $P$, $E$, and $F$ are collinear. Since $PF$ is an altitude, Blanchet's theorem tells us that $\angle CFE = \angle DFE$, so our proof is complete.

My previous solution is completely incorrect... sorry about that.

[b]Lemma[/b]: In cyclic quadrilateral $ABCD$, let $AD \cap BC = P$, $AB \cap CD = Q$, $AA \cap BB = R$, $CC \cap DD = S$, and $AC \cap BD = T$. Then $P$, $R$, and $S$ each lie on the polar of $Q$.
[i]Proof:[/i] By Pascal's theorem on $AACBBD$, we find that $P$, $R$, and $S$ are collinear. By Pascal's theorem on $DDBCCA$, we find that $P$, $T$, and $S$ are collinear. The polars of $R$ and $S$ are $AB$ and $CD$. Since $Q$ lies on the polars of $R$ and $S$, $R$ and $S$ lie on the polar of $Q$. Hence, $P$, $R$, $T$, and $S$ all lie on the polar of $Q$. $\blacksquare$

Let $X,Y$ be on the semicircle so that $B,X,Y,A$ are collinear, in that order. Let $CD$ hit $AB$ at $Z$, and let $AC$ and $BD$ intersect the semicircle at $U$ and $V$, respectively. By Pascals' theorem on $CUVDDC$, we find that $U$, $V$, and $Z$ are collinear. By the lemma, the polar of $Z$ passes through $E$.

Let $XC$ hit $YD$ at $Q$, and let $XD$ hit $YC$ at $R$. $YC \perp XQ$ and $XD \perp QY$, so $R$ is the orthocenter of $\triangle QXY$, whence $QR \perp BY$. Furthermore, the polar of $Z$ also passes through $P$, $Q$ and $R$, so by the lemma, $P$, $Q$, $R$, and $E$ are collinear. It follows that $PF$ is an altitude of $\triangle PBA$, so by Blanchet's theorem $\angle CFE = \angle DFE$.

[hide="Solution"] First of all, we establish the following lemma: let $ABCD$ be a tangential quadrilateral, in which circle $\omega$ is inscribed. If $AB\cap CD=X, AB\cap \omega=P, CD\cap\omega=Q$, and $AC\cap BD=O$, then $\frac{BO}{DO}=\frac{BP}{DQ}$.

This follows trivially from Menelaus' Theorem on $\triangle XPQ$ and $\overline{BOD}$. Consequently, by Ceva's Theorem, $AO, DP,$ and $BY$ are concurrent, where $Y=AD\cap \omega$.

Going back to the problem, it is clear that it suffices to prove that $X=CC\cap DD, E,$ and $F$ are collinear. Let $F'$ be the foot of perpendicular from $X$ to $AB$, and let us reflect the whole diagram with respect to $\overline{AB}$.

According to our corollary of the lemma, we have that $BD, AC, XF'$ are concurrent. However, $BD\cap AC=E$, and we must have that $X, E, F'$ are collinear.

Now, $F'\in XE$ and $F'\in AB$, implying that $F=F'$. Hence, $X, E, F$ are collinear, and we are done. $\blacksquare$[/hide]

Let $AB\cap{CD}=X$,$AD\cap{BC}=Y$,$YE\cap{CD}=G$.
Then $(XG;DC)=-1$,so $G$ is on the polar of $X$.As $X$ is on the $Y$'s polar $CD$,we have $YG$ is the polar of X,so $YE\perp{AB}$,we find $Y,E,F$ are collinear.
Since $F(XG;DC)=-1$ and  $EF\perp{AB}$,so $EF$ bisects $\angle{CFD}$.

[quote="Zhero"]
[b]Lemma[/b]: In cyclic quadrilateral $ABCD$, let $AD \cap BC = P$, $AB \cap CD = Q$, $AA \cap BB = R$, $CC \cap DD = S$, and $AC \cap BD = T$. Then $P$, $R$, and $S$ each lie on the polar of $Q$.
[i]Proof:[/i] By Pascal's theorem on $AACBBD$, we find that $P$, $R$, and $S$ are collinear. By Pascal's theorem on $DDBCCA$, we find that $P$, $T$, and $S$ are collinear. The polars of $R$ and $S$ are $AB$ and $CD$. Since $Q$ lies on the polars of $R$ and $S$, $R$ and $S$ lie on the polar of $Q$. Hence, $P$, $R$, $T$, and $S$ all lie on the polar of $Q$. $\blacksquare$
[/quote]
Typo: Pascal on AACBBD gives P, T, R collinear, instead of P, R, S.

So a faster way to finish using this lemma and the fact that Zhero derived about the polar of Z passing through E is that since AB passes through the center of the circle and Z is on AB, then PE is perpendicular to AB, so line PE = line PF = line EF, so P, E, F are collinear, and we can apply Blanchet's Theorem to finish, or cyclic quadrilaterals.

I saw on [b]blackbelt14253[/b]'s blog that he had used Brianchon's theorem on hexagon PCBP'AD where P' is the reflection of P across AB, which is probably a more convenient way of deducing that P, E, F collinear.

:!:  LEMMA: :!:
Suppose we have the triangle ABC and we have h_a
Let h_a intersect BC at H
Let P be a point on h_a
Let BP intersect AC at E
Let CP intersect AB at F
Then:h_a bisects the angle EHF

Solution:
Let BC intersect AD at X
From X draw a perpendicular line to AB
Let's say this line intersects AB at F
It's enough to prove that XF, BD, AC meet at E
We do this by using Seva
Now it's enough to prove that CB/FB=AD/AF
Draw the line OC
XFOC is cyclic --> these angles are equal COB=BXF
Since OCB=XFB=90 and by using the equality above we have CB/BF=OC/XF
Similarly we get AD/AF=OD/XF
Since OC=OD=R
WE ARE DONE :lol:

Could you please describe how you showed that $XOFD$ is cyclic? Thanks. =)

EDIT: Thanks.

Dear Mathlinkers,
this problem was proposed at the 35th OIM (Hongkong) but no selectde by the jury.
Sincerely
Jean-Louis

Let $O$ be the center of the circle. Let the tangents meet at $P$. Let H be the foot of the perpendicular from $P$ to $AB$. The key is that $AC$ and $BD$ intersect on $AH$.

$PAH$ and $OAD$ are similar, so $AH/AD = HP/DO$. $PBH$ and $OBC$ are similar, so $BH/BC = HP/CO$. But $BO = CO$, so $BH/BC = AH/AD$. Also $PC = PD$ (equal tangents), so $(AH/HB)(BC/CP)(PD/DA) = 1$. Hence by Ceva's theorem applied to the triangle $ABC$, the lines $AC$, $BD$ and $PH$ are concurrent. So $H = F$. $\angle PDO = \angle PFO = \angle PCO$, so $D$, $F$ and $O$ lie on the circle diameter $PO$. So $\angle DFP = \angle DOP = \angle COP = \angle CFP$. $\Box$

let $ BC $ and $ DA $ intersect at $ G $, $ O $ is the center of the circle. connect $ GO $, $ CO $, $ DA $.
try to prove $ G $, $ E $, $ F $ are conliear. in $ \Delta BGA $, use ceva theorem, try to prove: $ \frac {GC} {BC} \cdot \frac {BF} {FA} \cdot \frac {DA} {GD}=1 $.
because: $ BF = BG \cdot Sin \angle GBA $, $ FA = GA \cdot Sin \angle GAB $, suppose the radium of circle $ O $ is $ R $, $ BC = \frac {R} {\tan \angle GBA} $, $DA = \frac {R}{\tan \angle GAB} $, $ GB \cdot Sin \angle GBA = GF = GA \cdot Sin \angle GAB $.  so we can find $ G $, $ E $, $ F $ are conliear.

$ \angle GCO = \angle GFO = 90 $, so $ G C F O $ are concylic, so $ \angle CFG = \angle COG $.
$ \angle GFO = \angle GDO = 90 $, so $ G F O D $ are concylic, so $ \angle GFD = \angle GOD $. it is easy to konw $ \angle COG = \angle GOD $. prove is done.

Dear Mathlinkers,
a symmetrization of the figure wrt AB, leads to another proof... without calculation...
Sincerely
Jean-Louis

[quote="omar31415"][hide="My approach using Brianchon"] Let $Q$ be the intersection of $BC$ and $AD$. We reflect the whole diagram by $AB$, and let $C'$, $D'$ and $Q'$ the reflections of $C$, $D$ and $Q$. By symmetry: $BC'$ and $AD'$ are tangents to the circle. Consider the quadrilateral $QBQ'A$, $BQ=BQ'$ and $AQ=AQ'$, so $A$ and $B$ belong to the perpendicular bisector of $QQ'$ and therefore they are perpendicular, let $G$ be their intersection. Applying Brianchon's theorem in the degenerated hexagon $QCBQ'AD$ we conclude that $QQ'$,$BD$ and $AC$ are concurrent, therefore $E$ belongs to $QQ'$ and so $F$ belongs to $QE$ and $F$=$G$.
We apply Brianchon's again in $QBC'Q'AD$ and in $QCBQ'DA$ to conclude that $QQ'$,$CD'$,$AB$ and $C'D$ concur at $F$.
Finally we see that by symmetry triangles $QCF$ and $Q'FC'$ are congruent, thus $\angle QFC=\angle Q'FC'=\angle DFQ$. Therefore $\angle EFC=\angle EFD$ and the problem is solved.[/hide][/quote]

omar31415,your solution is very beautiful!!!

A diagram for posterity:
[asy]
    size(10cm); pointpen = black;
    pair C = Drawing("C", dir(130), dir(130));
    pair D = Drawing("D", dir(60), dir(60));
    pair B = Drawing("B", -1.55572, dir(180));
    pair A = Drawing("A", 2.0000000, dir(0));
    draw(unitcircle);
    pair Q = Drawing(" ", extension(B,C,A,D), dir(0));
    draw(B--Q);
    draw(A--Q);
    draw(A--B);
    pair O = Drawing("  ", 0.0, dir(0));
    pair E = Drawing("   ", extension(A,C,B,D), dir(0));
    pair F = Drawing(" ", extension(Q,E,A,B), dir(0));
    label("$O$", O, SE);
    label("$F$", F, SW);
    label("$E$", E, SE);
    label("$Q$", Q, N);
    draw(Q--F);
    draw(C--A);
    draw(B--D);
    draw(circumcircle(C,Q,D), green+dashed);
    draw(C--F, orange+dashed);
    draw(C--O, red+dashed);
    draw(D--F, orange+dashed);
    draw(D--O, red+dashed);
    draw(D--C, red+dashed);
[/asy]
Let $Q=CB\cap AD$, the circle be $\Omega$, and $O$ the center of $\Omega$.

[b]Lemma:[/b] The points $Q,E,F$ are collinear.

[i]Proof:[/i] Put down Cartesian coordianates.  Let $O=(0,0)$, let $\Omega$ be the unit circle, and let $AB$ be the $x$ axis.  Let $C=(c_1,c_2),D=(d_1,d_2),E=(e_1,e_2),Q=(q_1,q_2)$ where $c_1^2+c_2^2=1,d_1^2+d_2^2=1$.  It suffices to show that $e_1=q_1$.

By similar triangles, $B=(\tfrac{1}{c_1},0)$ and $A=(\tfrac{1}{d_1},0)$.  Then line $CA$ has equation
\[y=\tfrac{c_2}{c_1-\tfrac{1}{d_1}}(x-\tfrac{1}{d_1}).\]Similarly, $DB$ has equation
\[y=\tfrac{d_2}{1-\tfrac{1}{c_1}}(x-\tfrac{1}{c_1}).\]Since $AC\cap BD=E$, we have
\begin{align*}
\tfrac{c_2}{c_1-\tfrac{1}{d_1}}(e_1-\tfrac{1}{d_1})&=\tfrac{d_2}{1-\tfrac{1}{c_1}}(e_1-\tfrac{1}{c_1})
\\ \left(\tfrac{c_2}{c_1-\tfrac{1}{d_1}}-\tfrac{d_2}{d_1-\tfrac{1}{c_1}}\right)e_1&=(\tfrac{c_2}{c_1d_1-1}-\tfrac{d_2}{c_1d_1-1})
\\ e_1&=\tfrac{\tfrac{c_2-d_2}{c_1d_1-1}}{\tfrac{c_2d_1-d_2c_1}{c_1d_1-1}}
\\ &=\tfrac{c_2-d_2}{d_1c_2-c_1d_2}.\end{align*}Now, line $CB$ has equation
\[y=\tfrac{c_2}{c_1-\tfrac{1}{c_1}}(x-\tfrac{1}{c_1})\]and similarly line $AD$ has equation
\[y=\tfrac{d_2}{d_1-\tfrac{1}{d_1}}(x-\tfrac{1}{d_1}).\]Since $AD\cap CB=Q$, we have
\begin{align*}
\left(\tfrac{c_1}{c_1-\tfrac{1}{c_1}}-\tfrac{d_2}{d_1-\tfrac{1}{d_1}}\right)q_1&=(\tfrac{c_2}{c_1^2-1}+\tfrac{d_2}{1-d_1^2})
\\ (\tfrac{c_1c_2}{c_1^2-1}-\tfrac{d_1d_2}{d_1^2-1})q_1&=(\tfrac{c_2}{-c_2^2}+\tfrac{d_2}{d_2^2})
\\ (\tfrac{d_1}{d_2}-\tfrac{c_1}{c_2})q_1&=(\tfrac{1}{d_2}-\tfrac{1}{c_2})
\\ q_1&=\tfrac{\tfrac{c_2-d_2}{c_2d_2}}{\tfrac{d_1c_2-d_2c_1}{c_2d_2}}
\\ &=\tfrac{c_2-d_2}{d_1c_2-d_2c_1}
\\ &=e_1,\end{align*}as desired. $\blacksquare$

Now we claim $FODQC$ is cyclic.  Obviously $ODQC$ is cyclic since $\angle OCQ+\angle ODQ =\tfrac{\pi}{2}+\tfrac{\pi}{2}=\pi.$  Then since
\[\angle OFQ=\angle OFE =\tfrac{\pi}{2}=\angle OCQ=\angle ODQ,\]$F$ lies on the same circle.

Finally, we want
\[\angle CFE=\angle EFD\iff\angle CFQ=\angle QFD\iff\angle QDC=\angle QCD,\]which is obvious by equal tangents. $\square$

By JBMO 2000 we have that $X,E,F$ are collinear where ${X}=AD \cap BC$. It suffices to prove that $XF$ bisects $\angle DFC$. Let $Y=AB \cap CD$ and let $Z=XF \cap CD$. Then $(Y,Z ; D,C)=-1$ and $\angle YFZ=90^{o}$ from this two facts we have that $ZF$ bisects $\angle DFC$ thus $XF$ bisects $\angle DFC$ q.e.d

Since $\angle EFO=\angle EDO=90$ and $\angle EFO=\angle ECF=90$ so we obtain E,F,O,D,c cyclic.$\angle COD=2\angle CDE$ and we have $\angle COE=\angle CDE$ so $EO$ bisects $\angle COD$ and we can obtain by cyclicity that $\angle EFD=\angle EOD=\angle COE=\angle CFE$ as desired.

The Problem can be restated as:
[quote=Equivalent Problem]In $\Delta ABC$, Let $\Delta DEF$ be the orthic triangle WRT $\Delta ABC$. Let tangents at $F,E$ to $\odot (BFEC)$ intersect $BC$ at $X,Y$. Let $XE,FY$ meet at $R$. Let $V$ be the foot from $R$ to $BC$.Prove, $RV$ bisects $\angle FVE$[/quote]
[b]Solution:[/b] Let $H$ be the orthocenter of $\Delta ABC$ and $M$ be the midpoint of $AH$. Then, $M$ lies on $XF, YE$. Let the perpendiculars from $E, F$ to $BC$ intersect $XM,YM$ at $Q,P$. Let $PF,QE$ intersect $\odot (BFEC)$ again at $F', E'$ $\implies$ $FE,BC,E'F'$ concur, suppose say at $T$, then, Apply Pascal on $FFEEE'F'$ $\implies$ $PQ$ passes through $T$. By La Hire's Theorem $\implies$ $XE,FY$ are polars of $P,Q$ WRT $\odot (BFEC)$. Now since, $T,P,Q$ are collinear, their polars are concurrent $\implies$ $XE \cap FY$ $=$ $R$ lies on $AH$ $\implies$ $V \equiv D$ and since $H$ is the incenter of $\Delta DEF$ $\implies$ $RD$ ($RV$) bisects $\angle FDE$ ($\angle FVE$)

[img]https://i.imgur.com/OjBjdQV.png[/img]

So let's define point $P = AD \cap BC$
Let point $Q \in AB$ such that $PQ \perp AB$
Then the pentagon $CQODP$ is cyclic.
Then we have 2 similarity relations $\triangle PAQ \sim \triangle OAD$ and $\triangle PBQ \sim \triangle OBC $
Thus we have the following relation:
$$ \frac{AD}{AQ} = \frac{OD}{PQ} = \frac{OC}{PQ} = \frac{BC}{BQ} $$
Thus we have the following:
$$ \frac{AQ}{QB}\frac{BC}{CP}\frac{PD}{DA} = 1 $$
Because $PD = PC$ and because of the ratios above.
Thus by the reverse Ceva theorem we have that $AC,BD$ and $PQ$ are concurrent cevians,thus $Q \equiv F$
Thus we have that $CFODP$ is cyclic
This angle-chase finishes the problem off:
$$\angle DFP = \angle DOP = \angle POC = \angle PFC$$
Thus we have shown,because $\angle DFP = \angle PFC$, that PF is an angle-bisector of the angle $\angle CFD$

Let $W$ be the intersection of $BC$ and $DA$. Let $W’$ be the foot of the perpendicular from $W$ onto $AB$. Let $O$ be the center of the given semicircle.
Then, since $\triangle WBW’ \sim \triangle BCO$,
$$\frac{OC}{BC} = \frac{WW’}{BW'}$$Since $\triangle WW’A \sim \triangle ODA$,
$$\frac{DA}{OD} = \frac{W'A}{WW'}$$Multiplying, we see that $\frac{BC}{DA} = \frac{W'A}{BW’}$. We also know that $WC = WD$ by power of a point. Therefore, by Ceva’s theorem, we see that $CA,BD$, and $WW’$ are concurrent. Furthermore, we know that $CWDOW’$ is cyclic because $\angle WCO = \angle WDO = \angle WW’O = 90$. Now, we proceed with angle-chasing. Let $\angle WW’D = a$. Then, $\angle WOD = \angle WCD = a$. Then, $\angle DW’O = \angle DCO = \angle OWD = 90-a$. Since $O$ is the center of the semicircle, $\angle CDO = 90-a$ as well, so $\angle WOC = a$.Thus, $WO$ bisects $\angle COD$, which as we showed beforehand is the same as saying that $EF$ bisects $\angle CFD$.

@above I think you misinterpret the problem. Read carefully definition of point $E$.

My solution:
Let $X = AD \cap BC$ and $Y,P$ - reflections of $X,C$ about $AB$.
By Brianchon on $ADXCBY$ we see that $E \in XY$ and so $F \in XY$
By Brianchon on $ADXBPY$ we see that $F \in DP$ which implies the result.

Really similar to post #7. Oh well.

The key observation is that $AD,BC,EF$ concur.

    Let $AD$ and $BC$ intersect at $P$ and let $Q$ be the foot of the altitude from $P$ to $AB.$ Also let the semicircle have center $O.$ Now note
    \[\triangle PAQ\sim \triangle OAD\]
    \[\triangle PBQ\sim \triangle OBC\]
    so $\frac{AQ}{QB}\cdot\frac{BC}{CP}\cdot\frac{PD}{DA}=1.$ Since $AC,BD,PQ$ concur, $Q$ is actually $F,$ and $AC,BD,PF$ concur.

    Now note
    \[\angle OCP=\angle ODP=\angle OFP=90^{\circ},\]
    so $OFCPD$ is cyclic. Thus
    \[\angle COP=\angle DOP\]
    \[\angle CFP=\angle DFP.\]

Let A' be the point on the diameter for semicircle closer to B. It is not hard to see, with a simple diagram (drawing perp. angles are easy), that we can suspect concurrency. If the intersection of AD and BC is G, by similar triangles ADO with AFG and BCO with BFG, we have by Ceva's that they (AD, BC, and EF) concur. <ODG=<OCG=<OFG=90 degrees, so OFCPD is cyclic. We then have <CFG=<CDG=<CA'D, and <DFG=<DCG=<DA'C, where the last two steps follow from the well known tangent angle rule (or the fact that it is an "instantaneous same point" that are angle <DCG=<DCC=DA'C).

Edit: Will attach ggb diagram later


[quote="alex2029"]Demonstrati ca daca $ x,y,z\ge 0$ si $ x \plus{} y \plus{} z \equal{} 1$ , atunci :

$ \sum_{cyc}\frac {x \plus{} 1}{x \plus{} yx}\ge 9$[/quote]
$ \sum\frac{1}{x\plus{}1}\plus{}\sum\frac{1}{x\plus{}xy}\geq 9$
From adreescu we have that:
$ \sum\frac{1}{x\plus{}1}\geq \frac{9}{4}$
And $ \sum\frac{1}{x\plus{}xy}\geq \frac{9}{1\plus{}\sum xy}$
So we have to prove that:
$ \frac{1}{4}\plus{}\frac{1}{1\plus{}\sum xy} \geq 1 \Leftrightarrow
1\geq 3(xy\plus{}yz\plus{}zx)$
Which is obviously true :wink:

[quote=alex2029]Demonstrati ca daca $ x,y,z\ge 0$ si $ x\plus{}y\plus{}z\equal{}1$ , atunci :
$ \sum_{cyc}\frac{x\plus{}1}{x\plus{}yx}\ge 9$[/quote]
By AM-GM,
$ \sum_{cyc}\frac{x\plus{}1}{x\plus{}yx}=\sum_{cyc}\frac{x\plus{}1}{x(y\plus{}1)}\geq\frac{3}{\sqrt[3]{xyz}}\geq 9.$


most AMC B score should come before AIME right? cuz i didn't get my yet

My teacher just got the paper copy for my score today.


For each positive real number $ x$, let $ S(x) \equal{} \{ \lfloor kx \rfloor | \ k \equal{} 1, 2, \ldots \}.$ Let $ x_1, x_2$, and $ x_3$ be three real numbers greater than $ 1$ such that $ \frac{1}{x_1} \plus{} \frac{1}{x_2} \plus{} \frac{1}{x_3} > 1$. Prove that there exist $ i$ and $ j$, with $ 1 \leq i < j \leq 3$, such that $ S(x_i) \cap S(x_j)$ has infinitely many elements.

[hide="Sketchy, unchecked proof"]For any $ N$ there is at least $ \frac {N}{x_1}$ elements in $ S(x_1)$ that are $ \leq N$
All the elements in $ S(x_i)$ are distinct since $ x_i > 1$

The same for $ S(x_2)$ and $ S(x_3)$

Now say $ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \frac {1}{x_3} \equal{} r > 1$

Thus the number of integers represnted $ \leq N$ from all 3 sets is at least $ \frac {N}{x_1} \plus{} \frac {N}{x_2} \plus{} \frac {N}{x_3} \equal{} Nr$

Thus there are at least $ Nr \minus{} N$ numbers represented more than once. But $ Nr \minus{} N$ increases without bound as $ N$ gets large. Thus an infinite number of numbers are represented more than once, which implies that at least one of the intersections of two of our three sets if infinite.[/hide]

EDIT: small hole  :oops: I included $ 0$ in all the sets. The result still hold, talking out the zero element dosnt make an infinite number finite. If you start without the zeroes at the start

[hide]Then the unbounded quantity becomes $ N(r \minus{} 1) \minus{} r$ which is indeed still unbounded.[/hide]

Just as a matter of interest, this can be extended to any number of $ x_i$


When I use something like this:

\begin{description}

\item[(3A-1)]

     \begin{description}

      \item[a]

      \item[b]

      \end{description}
\end{description}

The 'a' is printed on the same line as the '(3A-1)'.  What's the best way to fix this so that the 'a' and 'b' are horizontally aligned.  My current solution is to do \item\textbf{(3A-1)} etc. forcing it to give me a line break.  But I would like a better way....

[quote="joml88"]What's the best way to fix this so that the 'a' and 'b' are horizontally aligned.[/quote]I think you mean you want them vertically aligned underneath each other. To do that add something like:
\setlength{\leftmargini}{1.55cm}
before the first \begin{description}. You may have to play around with the exact length.


Prove that for all positive reals numbers $x,y,z,t$, we have:
$\frac{x}{y^{7}}+\frac{y}{z^{7}}+\frac{z}{t^{7}}+\frac{t}{x^{7}}\ge\frac{1}{x^{6}}+\frac{1}{y^{6}}+\frac{1}{z^{6}}+\frac{1}{t^{6}}$

[hide]Applying Cauchy's inequality[/hide]

[quote="hien"]Applying Cauchy's inequality[/quote]
Using the rearrangement inequality.

[quote="hien"]Yes, this is a the most elemetaty solution to solve this kind of problem, i think[/quote]
 :lol:
Next:
$a,b,c,d \in R+,a^{3}+b^{3}=c^{3}+1,ab>c$
Prove that:$a+b >c+1$


Find such a 5-digit number that is equal to 45 times the product of its digits.

Looks fun, but I really don't want to guess and check. A better way than guess and check please.

[hide]$10000A + 1000B + 100C + 10D + E = 45ABDE$
$2000A + 200B + 20C + 2D + .2E = 9ABDE$

Since the RHS must be an integer, $E$ must be a multiple of 5. The only choices for $E$, then, is either $0$ or $5$.

If we test $5$.

$2000A + 200B + 20C + 2D + 1 = 45ABD$
$400A + 40B + 4C + .5D + .2 = 9ABD$

The RHS must be an intger so $.5D + .2$ must be an integer. This is impossible since $D$ can only be a number from $0 - 9$. The only possible value of $E$, then, is $0$ which makes the product of the digits $0$ so the answer is all five digit numbers that contain a $0$ somewhere in it.[/hide]

Tarquin, $\frac{2}{5} \ne .5$, but your method should work.  (Remember parity)

Since I'm too lazy to work, I wrote a program on the calculator and got 77175 as the only one

First, E cannot be zero, because the product would be zero, so it is 5.  Then A, B,C, and D are all odd because if one was even, it would end in a 0.
so
10000A+1000B+100C+10D+5=45*A*B*C*D*5
2000A+200B+20C+2D+1=45*A*B*C*D
From this we can see
2D=4 mod 5
D=2,7
But D has to be odd so D=7

10000A+1000B+100C+75=45*A*B*C*7*5
400A+40B+4C+3=63*A*B*C
The number ABC75 is divisible by 7, so that means ABC7-10 is also divisible by 7.
1000A+100B+10(C-1)+7 is divisible by 7.
The number ABC7-10 is divisible by 7 so ABC-24 is too.
So I guess then you have to check odd numbers for A B C and see satisfy those

5a - 97h / 5 =17 / k

so therefore 77175 is the answer

My name at Mathpath, a math camp, was 45. Something completely random.


geometry
circumcircle
trigonometry
trig identities
Law of Sines
similar triangles
geometry proposed

I think that you made small mistake Namely you defined point $ P$ and then mentioned about $ AX$, $ BX$ and $ CX$ without definition of $ X$  $ (P\equal{}X)$

We have $ m\angle{MA'C}\equal{}m\angle{MBA'}\equal{}m\angle{A'AC}$, $ m\angle{MCA'}\equal{}m\angle{MAB'}\equal{}m\angle{A'AB}$ (angles directed modulo $ \pi$).  So then, by the Law of Sines,
\[ \frac{MC}{MB}\equal{}\frac{\sin^2 \angle CAA'}{\sin^2 \angle BAA'}\]
Nice.  Now do the same thing for $ N$ and $ P$ and multiply the three equations.  Use Trig Ceva to get that the RHS is $ 1$, then use Menelaus (I kinda ignored directed lengths here but whatever it works if you use directed lengths) to get the three points are collinear.

Yes, sorry ! Thank you, [b]Delegat[/b] ! I corrected.

This problem is coming from the past.

From the similar triangles $ \bigtriangleup MA'B,$ $ \bigtriangleup MA'C$ $ ($ because $ \angle MBA' \equal{} \angle MA'C$ $ ),$ we have that $ \frac {MB}{MA'} \equal{} \frac {MA'}{MC} \equal{} \frac {A'B}{A'C}$ $ \Longrightarrow$ $ \frac {MB}{MC} \equal{} \frac {(A'B)^{2}}{(A'C)^{2}}$ $ ,(1)$

Similarly from the similar triangles $ \bigtriangleup NB'C,$ $ \bigtriangleup NB'A$ we have that $ \frac {NC}{NB'} \equal{} \frac {NB'}{NA} \equal{} \frac {B'C}{B'A}$ $ \Longrightarrow$ $ \frac {NC}{NA} \equal{} \frac {(B'C)^{2}}{(B'A)^{2}}$ $ ,(2)$

and from the similar triangles $ \bigtriangleup PC'A,$ $ \bigtriangleup PC'B$ we have that $ \frac {PA}{PC'} \equal{} \frac {PC'}{PB} \equal{} \frac {C'A}{C'B}$ $ \Longrightarrow$ $ \frac {PA}{PB} \equal{} \frac {(C'A)^{2}}{(C'B)^{2}}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {MB}{MC}\cdot \frac {NC}{NA}\cdot \frac {PA}{PB} \equal{} \left( \frac {A'B}{A'C}\cdot \frac {B'C}{B'A}\cdot \frac {C'A}{C'B}\right)^{2}$ $ ,(4)$

Because of the cyclic hexagon $ AB'CA'BC',$ whose the main diagonals $ AA',$ $ BB',$ $ CC',$ are concurrent at the point $ X,$

we have $ \frac {A'B}{A'C}\cdot \frac {B'C}{B'A}\cdot \frac {C'A}{C'B} \equal{} 1$ $ ,(5)$ as well known ( and easy to prove).

From $ (4),$ $ (5)$ $ \Longrightarrow$ $ \frac {MB}{MC}\cdot \frac {NC}{NA}\cdot \frac {PA}{PB} \equal{} 1$ $ ,(6)$

From $ (6),$ By [b][size=100]Menelaus theorem[/size][/b], we conclude that the points $ M,$ $ N,$ $ P,$ are collinear and the proof is completed.

Kostas Vittas.

An equivalent formulation of this problem is the so-called [b][size=100]Steinbart theorem[/size][/b], as it has already been published by [b][size=100]Darij Grinberg[/size][/b].

http://de.geocities.com/darij_grinberg/SteinbartVar.zip

Kostas Vittas.

Hi, Mr [b]vittasko[/b]!
Your solution is very beautiful! I viewed your web page and i'm interested in it! It's kool! :P
Now, i propose you help me to solve some geometrical problems.
Here
http://www.mathlinks.ro/viewtopic.php?t=184005
And here
http://www.mathlinks.ro/viewtopic.php?t=184006
Thanks for your help! :D
kaka!

[quote="ricardokaka"]Hi, Mr [b]vittasko[/b]!
Your solution is very beautiful! I viewed your web page and i'm interested in it! It's kool! :P
....[/quote]
Of course you mean the [b][size=100]darij grinberg's[/size][/b] web page ( unforunately, I have not any mine ), in which I am very interesting too.

I will try to solve the problems you mentioned and I hope that me, or some other more skilful mathlinkers, will be able to help you.

Kostas Vittas.

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle with the circumcircle $ \mathcal C$. For a point $ X$ define the second intersections $ A'$ , $ B'$ , $ C'$ of the circle $ \mathcal C$ with the lines $ AX$ , $ BX$ , $ CX$ respectively. The tangents in the points $ A'$ , $ B'$ , $ C'$ to the circle $ \mathcal C$ meet the sidelines $ BC$ , $ CA$ , $ AB$ in the points $ M$ , $ N$ , $ P$ respectively. Prove that these last points are collinearly.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $ D\in AX\cap BC$ , $ E\in BX\cap CA$ , $ F\in CX\cap AB$ .

Therefore ([b]Ceva's theorem[/b]), $ \frac {DB}{DC}\cdot\frac {EB}{EA}\cdot\frac {FA}{FB}\equal{}1\ \ (1)$ .

$ \left\{\begin{array}{c}
\frac {DB}{DC}\equal{}\frac {A'B}{A'C}\cdot \frac {AB}{AC}\\\\
\frac {MB}{MC}\equal{}\left(\frac {A'B}{A'C}\right)^2\end{array}\right\|$ $ \implies$ $ \frac {MB}{MC}\equal{}\left(\frac {DB}{DC}\cdot\frac {AC}{AB}\right)^2$ a.s.o.

In conclusion, using the relation $ (1)$ obtain $ \frac {MB}{MC}\cdot\frac {NC}{NA}\cdot\frac {PA}{PB}\equal{}1$ , i.e. $ P\in MN$ .[/color]


Let $a,b,c>0$ which $a+b+c=1$ we need to prove or disprove that
\[ 3(a^3+b^3+c^3)+18abc\geq 1 \]

Doesn't matter, let's say $c \rightarrow 0$.

pluricomplex ? i don't see your web
          :roll:  :roll: http://www.geocities.com/pluricomplex_amath :roll:

i don't beleave i did not see that friends :D . Thanks to all of you!
@capto : it's still there friend  ;)  may be you get the firewall!

I think the following inequality holds for $a+b+c=1$:

\[ 4(a^3+b^3+c^3)+15abc\geq 1  \]

homogenize it and then it is exactly schur with a factor 3 .

[quote="Vasc"]More general is the following:

For $a_i \ge 0$ such that  $a_1+a_2+...+a_n=1$, we have:

$(n-1)^2(a_1^3+a_2^3+...+a_n^3)+n^{n-2}(2n-1)a_1a_2...a_n \ge 1$.[/quote]

nice can anyone prove  this ( Vasc's ineq):wallbash:

[quote="Vasc"]No one can prove this because it is wrong.
Sorry! :P[/quote]
why do you post it if it's wrong :o

Stupid question... probably because he thought it's good?

[quote="Vasc"]But the following is correct (I hope this):

For $a_i \ge 0$ such that  $a_1+a_2+...+a_n=n$, we have:

$\frac{4}{n}(a_1^n+a_2^n+...+a_n^n)+5a_1a_2...a_n \ge 9$.[/quote]

I think it's not true for n=2  :D

[quote="Vasc"][quote="Vasc"]More general is the following:

For $a_i \ge 0$ such that  $a_1+a_2+...+a_n=1$, we have:

$(n-1)^2(a_1^3+a_2^3+...+a_n^3)+n^{n-2}(2n-1)a_1a_2...a_n \ge 1$.[/quote]
I verified again, and I found that it is valid. :lol:[/quote]

then where's the proof? :P

Actually, the best inequalities of this type are:

For $k\le n$ positive integer and $a_i \ge 0$ such that  $a_1+a_2+...+a_n=n$, we have:

$(\frac{n}{n-1})^{k-1}-1 \le \frac{\frac{a_1^k+a_2^k+...+a_n^k}{n}-1}{1-a_1a_2...a_n} \le n^{k-1}-1$.


Josh earns a weekly salary of of $300 plus a 6% commision on sales. Last week his sales totalled
$3500. What was his total pay. I know the answer but i forget the formula. :starwars:

I'll rewrite the problem because it did not turn out right in hobo020892's post

Josh earns a weekly salary of of 300 dollars plus a 6% commision on sales. Last week his sales totalled 3500 dollars. What was his total pay. I know the answer but i forget the formula. :starwars:

[hide]commision = %commision*sale amount
so 300 + 3500(.06) = 300 + 210 = 510. [/hide]


find the next letter:

U, D, T, C, C, S, S, O, N, D, .....

[hide="hint"]
ask djimenez :lol: (wow that was actually a tricky hint)
[/hide]

[hide]O,D,T,C,Q,D,D,D,D,V[/hide]

[hide]Uno, Dos, Tres,...[/hide]

I haven't seen the solution proposed by jmadsen, but given that I am native spanish speaker, the solution is trivial! What do you think about this one

H, T, S, N, T, Y, I, Y, A, Y

or

T, H , T, L, N, C, K, S, K, S

A little more difficoult ones!!

I take Spanish in school, so it made it pretty trivial since the English one is common, too, and I always look for things like that in sequences of letters :P


alex has four bags of candy, weighing 1,2,3, and 4 pounds. to arrange them, he will pick up two bags which are next to one another, compare their weights on a balance scale and put them back, with the heavier one to the left. what is the maximum number of times, he could possibly have to swap two bags, before he hasthe bags in order, from heaviest to lightest?

how would u do this?

It's possible to do it in 6 (14, 13, 12, 24, 23, 34) but I don't know if you can do it in less.

I think there might be a way to do it in 4 if you put 2 on each side at the beginning.


Dar beine ye tedad adam, har do taee ke ba ham doost hastan hich dooste moshtaraki nadaran va har do taee ke ba ham doost nistan daghighan do dooste moshtarak daran,sabet konid tedade doostaye hame ba ham barabare.

rase daraje $max$ ra $v$ dar nazar begirid($deg(v)=k$).agar $N(v)=\{w_{1},w_{2},\dots,w_{k}\}$.be vozooh hichkodam az $w_{i}$ha ba ham mojaver nistand.pas daghighan $2$ hamsaye moshtarak darand.pas $deg(w_{i})\geq\ k$ pas natije mishavad daraje tamame in $w_{i}$ha barabare darajeye $max$ geraf mibashad pas behamin tartib baghie rasha niz daraye darajeye $k$ mibashand pas gerafe hasel montazam ast. :wink:


Write down the greatest number by using four digit 2's without using any mathematical operation sign.

simple problem :)

[hide]
${2}^{{2}^{22}}$

this is that largest I found....it equals $2^{4194304}$ which is too big even for the scientific calculator on the computer.
[/hide]


geometry
trapezoid
projective geometry
geometry proposed

Let $ ABCD$ be a convex quadrilateral. Suppose $ OA<OC$ and $ OD<OB$ , where the point

$ O\in AC\cap BD$. Define the midpoints $ M$ , $ N$ of $ [BD]$ , $ [AC]$ respectively and the intersections

$ E\in MN\cap AB$ , $ F\in DC\cap MN$ , $ P\in CE\cap BF$ , $ L\in OP\cap EF$ . Prove that $ LE\equal{}LF$ .

[quote="Virgil Nicula"]Let $ ABCD$ be a convex quadrilateral. Suppose $ OA < OC$ and $ OD < OB$ , where the point

$ O\in AC\cap BD$. Define the midpoints $ M$ , $ N$ of $ [BD]$ , $ [AC]$ respectively and the intersections

$ E\in MN\cap AB$ , $ F\in DC\cap MN$ , $ P\in CE\cap BF$ , $ L\in OP\cap EF$ . Prove that $ LE \equal{} LF$ .[/quote]
I think it use papus theorem  :roll:

[u]Here is my first vectorial proof[/u] (with many and ugly calculations !). Maybe someone gives a synthetical proof. [b]Thank you.[/b]

[quote="Virgil Nicula"][color=darkred]Let $ ABCD$ be a convex quadrilateral. Suppose $ OA < OC$ and $ OD < OB$ , where the point

$ O\in AC\cap BD$. Define the midpoints $ M$ , $ N$ of $ [BD]$ , $ [AC]$ respectively and the intersections

$ E\in MN\cap AB$ , $ F\in DC\cap MN$ , $ P\in CE\cap BF$ , $ L\in OP\cap EF$ . Prove that $ LE \equal{} LF$ .[/color][/quote]
[color=darkblue][b][u]Proof[/u].[/b] I"ll use the vectorial calculus. Choose the origin $ O$ , i.e. for any $ X$ , $ Y$ we have $ \overrightarrow{OX} \equal{} X$ and $ \overrightarrow{XY} \equal{} Y \minus{} X\ .$

Thus, exist $ m > 1$ , $ n > 1$ so that $ B \equal{} \minus{} mD$ , $ C \equal{} \minus{} nA$ and $ 2M \equal{} B \plus{} D \equal{} (1 \minus{} m)D$ , $ 2N \equal{} A \plus{} C \equal{} (1 \minus{} n)A$ .

$ \blacktriangleright$ Since $ E\in AB\cap MN$ exist $ \alpha$ , $ \beta$ so that

$ E \equal{} (1 \minus{} \alpha )A \plus{} \alpha B \equal{} (1 \minus{} \beta )M \plus{} \beta N$ $ \Longrightarrow$ $ E \equal{} (1 \minus{} \alpha )A \minus{} m\alpha D \equal{} \frac 12(1 \minus{} \beta )(1 \minus{} m)D \plus{} \frac 12\beta (1 \minus{} n)A$ $ \Longleftrightarrow$

$ \left\|\begin{array}{c} 2(1 \minus{} \alpha ) \equal{} \beta (1 \minus{} n) \\
 \\
\minus{} 2m\alpha \equal{} (1 \minus{} \beta )(1 \minus{} m)\end{array}\right\|$ $ \Longleftrightarrow$ $ \Longleftrightarrow$ $ \left\|\begin{array}{c} \beta \equal{} \frac {2(1 \minus{} \alpha )}{1 \minus{} n} \\
 \\
\alpha \equal{} \frac {(m \minus{} 1)(n \plus{} 1)}{2(mn \minus{} 1)} \\
 \\
1 \minus{} \alpha \equal{} \frac {(m \plus{} 1)(n \minus{} 1)}{2(mn \minus{} 1)}\end{array}\right\|$ $ \implies$ $ \boxed {\ E \equal{} \frac {(m \plus{} 1)(n \minus{} 1)}{2(mn \minus{} 1)}A \minus{} \frac {m(m \minus{} 1)(n \plus{} 1)}{2(mn \minus{} 1)}D\ }$ .

$ \blacktriangleright$ Since $ F\in CD\cap MN$ exist $ \alpha$ , $ \beta$ so that

$ F \equal{} (1 \minus{} \alpha )D \plus{} \alpha C \equal{} (1 \minus{} \beta )M \plus{} \beta N$ $ \Longleftrightarrow$ $ F \equal{} (1 \minus{} \alpha )D \minus{} n\alpha A \equal{} \frac 12(1 \minus{} \beta )(1 \minus{} m)D \plus{} \frac 12\beta (1 \minus{} n)A$ $ \Longleftrightarrow$

$ \left\|\begin{array}{c} 2(1 \minus{} \alpha ) \equal{} (1 \minus{} \beta )(1 \minus{} m) \\
 \\
\minus{} 2n\alpha \equal{} \beta (1 \minus{} n)\end{array}\right\|$ $ \Longleftrightarrow$ $ \Longleftrightarrow$ $ \left\|\begin{array}{c} \beta \equal{} \frac {2n\alpha}{n \minus{} 1} \\
 \\
\alpha \equal{} \frac {(m \plus{} 1)(n \minus{} 1)}{2(mn \minus{} 1)} \\
 \\
1 \minus{} \alpha \equal{} \frac {(m \minus{} 1)(n \plus{} 1)}{2(mn \minus{} 1)}\end{array}\right\|$ $ \implies$ $ \boxed {\ F \equal{} \frac {(m \minus{} 1)(n \plus{} 1)}{2(mn \minus{} 1)}D \minus{} \frac {n(n \minus{} 1)(m \plus{} 1)}{2(mn \minus{} 1)}A\ }\ .$

$ \blacktriangleright$ Denote the midpoint $ L$ of the segment $ [EF]$ , i.e. $ 2L \equal{} E \plus{} F$ $ \Longleftrightarrow$ $ \boxed {\ 4(mn \minus{} 1)L \equal{} \minus{} (m \plus{} 1)(n \minus{} 1)^2A \minus{} (n \plus{} 1)(m \minus{} 1)^2D\ }\ .$

$ \blacktriangleright$ Since $ P\in CE\cap BF$ exist $ \alpha$ , $ \beta$ so that $ P \equal{} \alpha E \plus{} (1 \minus{} \alpha )C \equal{} \beta F \plus{} (1 \minus{} \beta )B$ $ \Longrightarrow$

$ 2(mn \minus{} 1)P \equal{} 2(m \minus{} 1)\alpha E \minus{} 2n(mn \minus{} 1)(1 \minus{} \alpha )A \equal{}$ $ 2(mn \minus{} 1)\beta F \minus{} 2m(mn \minus{} 1)(1 \minus{} \beta )D$ $ \Longleftrightarrow$

$ 2(mn \minus{} 1)P \equal{} \alpha \left[(m \plus{} 1)(n \minus{} 1)A \minus{} m(m \minus{} 1)(n \plus{} 1)D\right] \minus{} 2n(mn \minus{} 1)(1 \minus{} \alpha )A \equal{}$ $ \beta \left[(n \plus{} 1)(m \minus{} 1)D \minus{} n(n \minus{} 1)(m \plus{} 1)A\right] \minus{} 2m(mn \minus{} 1)(1 \minus{} \beta )D\ .$

Thus $ \left\|\begin{array}{c} (m \plus{} 1)(n \minus{} 1)\alpha \minus{} 2n(mn \minus{} 1)(1 \minus{} \alpha ) \equal{} \minus{} n(n \minus{} 1)(m \plus{} 1)\beta \\
 \\
\minus{} m(m \minus{} 1)(n \plus{} 1)\alpha \equal{} (n \plus{} 1)(m \minus{} 1)\beta \minus{} 2m(mn \minus{} 1)(1 \minus{} \beta )\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c} (2mn \minus{} m \minus{} 1)(n \plus{} 1)\alpha \plus{} n(n \minus{} 1)(m \plus{} 1)\beta \equal{} 2n(mn \minus{} 1) \\
 \\
m(m \minus{} 1)(n \plus{} 1)\alpha \plus{} (2mn \minus{} n \minus{} 1)(m \plus{} 1)\beta \equal{} 2m(mn \minus{} 1)\end{array}\right\|$ .

$ \Delta \equal{} \left|\begin{array}{cc} (n \plus{} 1)(2mn \minus{} m \minus{} 1) & n(n \minus{} 1)(m \plus{} 1) \\
 \\
m(m \minus{} 1)(n \plus{} 1) & (m \plus{} 1)(2mn \minus{} n \minus{} 1)\end{array}\right| \equal{}$ $ (m \plus{} 1)(n \plus{} 1)\left[\left(3m^2n^2 \minus{} 4mn \plus{} 1\right) \minus{} (m \plus{} n)(mn \minus{} 1)\right] \equal{}$

$ (m \plus{} 1)(n \plus{} 1)\left[(mn \minus{} 1)(3mn \minus{} 1) \minus{} (mn \minus{} 1)(m \plus{} n)\right]$ $ \implies \boxed {\ \Delta \equal{} (m \plus{} 1)(n \plus{} 1)(mn \minus{} 1)(3mn \minus{} m \minus{} n \minus{} 1)\ }$ .

$ \Delta_{\alpha} \equal{} \left|\begin{array}{cc} 2n(mn \minus{} 1) & n(n \minus{} 1)(m \plus{} 1) \\
 \\
2m(mn \minus{} 1) & (m \plus{} 1)(2mn \minus{} n \minus{} 1)\end{array}\right| \equal{}$ $ 2n(mn \minus{} 1)(m \plus{} 1)\cdot\left|\begin{array}{cc} 1 & n \minus{} 1 \\
 \\
m & 2mn \minus{} n \minus{} 1\end{array}\right|$ $ \implies \boxed {\ \Delta_{\alpha} \equal{} 2n(m \plus{} 1)(n \plus{} 1)(mn \minus{} 1)(m \minus{} 1)\ }$ .

$ \Delta_{\beta} \equal{} \left|\begin{array}{cc} (n \plus{} 1)(2mn \minus{} m \minus{} 1) & 2n(mn \minus{} 1) \\
 \\
m(m \minus{} 1)(n \plus{} 1) & 2m(mn \minus{} 1)\end{array}\right| \equal{}$ $ 2m(mn \minus{} 1)(n \plus{} 1)\cdot\left|\begin{array}{cc} 2mn \minus{} m \minus{} 1 & n \\
 \\
m \minus{} 1 & 1\end{array}\right|$ $ \implies \boxed {\ \Delta_{\beta} \equal{} 2m(n \plus{} 1)(m \plus{} 1)(mn \minus{} 1)(n \minus{} 1)\ }$ .

Therefore $ \alpha \equal{} \frac {\Delta_{\alpha}}{\Delta}\implies$ $ \alpha \equal{} \boxed {\ \frac {2n(m \minus{} 1)}{3mn \minus{} m \minus{} n \minus{} 1)}\ }$ , $ \beta \equal{} \frac {\Delta_{\beta}}{\Delta} \equal{} \frac {2m(n \minus{} 1)}{3mn \minus{} m \minus{} n \minus{} 1}$ and $ \boxed {\ 1 \minus{} \alpha \equal{} \frac {(m \plus{} 1)(n \minus{} 1)}{3mn \minus{} m \minus{} n \minus{} 1}\ }$ .

Thus $ P \equal{} \alpha E \plus{} (1 \minus{} \alpha )C \equal{}$ $ \frac {n(m \minus{} 1)}{3mn \minus{} m \minus{} n \minus{} 1}\cdot$ $ \left[\frac {(m \plus{} 1)(n \minus{} 1)}{mn \minus{} 1}A \minus{} \frac {m(m \minus{} 1)(n \plus{} 1)}{mn \minus{} 1}D\right] \minus{} \frac {n(n \minus{} 1)(m \plus{} 1)}{3mn \minus{} m \minus{} n \minus{} 1}A$ $ \implies$

$ (mn \minus{} 1)(3mn \minus{} m \minus{} n \minus{} 1)P \equal{}$ $ n(m \minus{} 1)\left[(m \plus{} 1)(n \minus{} 1)A \minus{} m(m \minus{} 1)(n \plus{} 1)D\right] \minus{}$ $ n(m \plus{} 1)(n \minus{} 1)(mn \minus{} 1)A\implies$

$ (mn \minus{} 1)(3mn \minus{} m \minus{} n \minus{} 1)P \equal{} \equal{}$ $ \minus{} mn(m \plus{} 1)(n \minus{} 1)^2A \minus{} mn(n \plus{} 1)(m \minus{} 1)^2D \equal{} 4mn(mn \minus{} 1)L\implies \boxed {\ L\in OP\ }$ and

$ (3mn \minus{} m \minus{} n \minus{} 1)P \equal{} 4mnL\implies$ $ \frac {\overrightarrow{OL}}{3mn \minus{} m \minus{} n \minus{} 1} \equal{} \frac {\overrightarrow{OP}}{4mn} \equal{} \frac {\overrightarrow {LP}}{(m \plus{} 1)(n \plus{} 1)}$ $ \implies$ $ \boxed {\frac {PL}{PO} \equal{} \frac {(m \plus{} 1)(n \plus{} 1)}{4mn}}\le 1$ for any $ m\ge 1$ and $ n\ge 1$ .[/color]

[b][u]Remark[/u].[/b] $ \frac {LM}{LN}\equal{}\frac {(m\plus{}1)(n\minus{}1)}{(n\plus{}1)(m\minus{}1)}$ and $ R\in \overline {OLP}\cap BC\ \implies\ \frac {RB}{RC}\equal{}\frac {m(m\plus{}1)(n\minus{}1)^2}{n(n\plus{}1)(m\minus{}1)^2}$ .

Because of the lines through the midpoints $ M,\ N$ of $ BD,\ AC$ respectively and parallel to $ AD,\ BC,$ pass through the midpoint $ K$ of the side-segment $ AB$ of the given quadrilateral $ ABCD,$ we will try to solve the follows equivalent problem.

[b][size=100][color=DarkBlue](1) - EQUIVALENT  PROBLEM. – Through the midpoint $ M$ of the side-segment $ AB$ of a given quadrilateral $ ABCD,$ we draw the parallel lines to $ BC,\ AD,$ which intersect $ CD,$ at points $ E,\ F,$ respectively. Prove that the line segment $ PQ,$ where $ P\equiv AC\cap BD$ and $ Q\equiv AE\cap BF,$ passes through the midpoint $ N$ of $ CD.$[/color][/size][/b]

At first, we will prove the below two [b][size=100]Helping propositions[/size][/b], as the particular cases of the equivalent problem.

[b][size=100][color=DarkBlue](2) - HELPING PROPOSITION 1. – A trapezium $ ABCD$ with $ AB\parallel CD$ is given and let $ E,\ F$ be, the points of intersection of its base $ CD,$ from the lines through the midpoint $ M$ of $ AB$ and parallel lines to $ BC,\ AD,$ respectively. Prove that the line segment $ PQ,$ where $ P\equiv AC\cap BD$ and $ Q\equiv AE\cap BF,$ passes through the midpoint $ N$ of $ CD.$[/color][/size][/b]

[b][size=100](3) – PROOF.[/size][/b] – From $ CE\parallel \equal{} MB$ and $ DF\parallel \equal{} MA$ and $ MA \equal{} MB,$ we have that $ NE \equal{} NF$ and then, it is easy to show that the line segments $ AE,\ BF,$ intersect the line connecting the midpoints $ M,\ N$ of $ AB,\ CD$ respectively, at the same point so be it $ Q,$ from $ EF\parallel AB$ and $ \frac {NE}{MA} \equal{} \frac {NF}{MB}.$

So, we conclude that the line segment $ PQ$ as the problem states, passes through the midpoint $ N$ of $ CD$ and the proof of the [b][size=100]Helping proposition 1[/size][/b] is completed.

[b][size=100][color=DarkBlue](4) - HELPING PROPOSITION 2. – A quadrilateral $ ABCD$ is given with $ CM\parallel AD,$ where $ M$ is the midpoint of its side-segment $ AB.$ We draw the line through $ M$ and parallel to $ BC,$ which intersects $ CD$ at the point so be it $ E.$ Prove that the line segment $ PQ,$ where $ P\equiv AC\cap BD$ and $ Q\equiv AE\cap BC,$ passes through the midpoint $ N$ of $ CD.$[/color][/size][/b]

[b][size=100](5) – PROOF.[/size][/b] – We consider the points $ Q,\ N,\ P,$ on the sidelines $ BC,\ CD,\ BD$ respectively, of the triangle $ \bigtriangleup BCD$ and applying the [b][size=100]Menelaos theorem[/size][/b], it is enough to prove that $ \frac {PD}{PB}\cdot \frac {QB}{QC}\cdot \frac {NC}{ND} \equal{} 1$ $ \Longrightarrow$ $ \frac {PD}{PB} \equal{} \frac {QC}{QB}$ $ ,(1)$ because of $ NC \equal{} ND.$

We denote the points $ X\equiv AB\cap CD$ and $ Z\equiv BC\cap AD,$ and $ Y\equiv XZ\cap BD.$

From the complete quadrilateral $ ABCDXZ$ we have that $ \frac {PD}{PB} \equal{} \frac {YD}{YB}$ $ ,(2)$

From $ (1),\ (2),$ it is enough to prove that $ \frac {QC}{QB} \equal{} \frac {YD}{YB}$ $ \Longrightarrow$ $ QY\parallel CD$ $ ,(3)$

$ \bullet$ We consider the point $ D$ inwardly to the triangle $ \bigtriangleup XZB$ and applying the [b][size=100]Ceva theorem[/size][/b], we have that $ \frac {AX}{AB}\cdot \frac {CB}{CZ}\cdot \frac {YZ}{YX} \equal{} 1$ $ \Longrightarrow$ $ \frac {AX}{AB} \equal{} \frac {YX}{YZ}$ $ ,(4)$ because of $ CB \equal{} CZ,$ from $ MA \equal{} MB$ and $ MC\parallel AZ.$

From $ (4)$ $ \Longrightarrow$ $ AY\parallel BZ$ $ \Longrightarrow$ $ KA \equal{} KY$ $ ,(5)$ where $ K\equiv AY\cap CD,$ from the trapezium $ ABZY.$

From $ (5)$ and because of $ EA \equal{} EQ$ $ ($ from $ MA \equal{} MB$ and $ ME\parallel BQ$ $ ),$ we have that $ QC\parallel \equal{} KA$ $ \Longrightarrow$ $ QC\parallel \equal{} YK$ $ \Longrightarrow$ $ QY\parallel CD.$

Hence, we conclude that $ \frac {QC}{QB} \equal{} \frac {YD}{YB} \equal{} \frac {PD}{PB}$ and then, the line segment $ PQ$ as the problem states, passes through the midpoint $ N$ of $ CD$ and the proof of the [b][size=100]Helping proposition 2[/size][/b] is completed.

[b][size=100](6) – PROOF OF THE EQUIVALENT PROBLEM.[/size][/b] – Let $ B'$ be the point on $ BC,$ such that $ AB'\parallel CD$ and we denote as $ F',$ the point of intersection of $ CD,$ from the line through the midpoint $ M'$ of $ AB'$ $ ($ $ M'\equiv ME\cap AB'$ $ )$ and parallel to $ AD.$

Based on the [b][size=100]Helping proposition 1[/size][/b], we have that the line segment $ P'Q',$ where $ P'\equiv AC\cap B'D$ and $ Q'\equiv AE\cap B'F',$ passes through the midpoint $ N$ of $ CD.$

Through the vertex $ C,$ we draw the line parallel to $ AD,$ which intersects the line segment $ ME$ at the point so be it $ M'',$ as the midpoint of $ AB'',$ where $ B''\equiv BC\cap AM''.$

We consider now the quadrilateral $ AB''CD$ with $ M''A \equal{} M''B''$ and $ CM''\parallel AD$ and based on the [b][size=100]Helping proposition 2[/size][/b], we have that the line segment $ P''Q'',$ where $ P''\equiv AC\cap B''D$ and $ Q''\equiv B''C\cap AE,$ passes through the midpoint $ N$ of $ CD.$

$ \bullet$ We consider now the pencil $ D.CB'BB''$ which is intersected from the line segments $ CA,\ CB''$ and so, we have that $ (C,P',P,P'') \equal{} (C,B',B,B'')$ $ ,(1)$

 Applying the [b][size=100]Menelaos theorem[/size][/b] in the triangle $ \bigtriangleup ECQ''$ with transversal $ QFB,$ we have that $ \frac {QQ''}{QE}\cdot \frac {FE}{FC}\cdot \frac {BC}{BQ''} \equal{} 1$ $ \Longrightarrow$ $ \frac {QQ''}{QE} \equal{} \frac {FC}{FE}\cdot \frac {BQ''}{BC}$ $ ,(2)$

But, $ \frac {FC}{FE} \equal{} \frac {MM''}{ME} \equal{} \frac {BB''}{BQ''}$ $ ,(3)$

From $ (2),\ (3)$ $ \Longrightarrow$ $ \frac {QQ''}{QE} \equal{} \frac {BB''}{BC}$ $ ,(4)$

Applying again the [b][size=100]Menelaos theorem[/size][/b] in the triangle $ \bigtriangleup ECQ''$ with transversal $ Q'F'B',$ we have that $ \frac {Q'Q''}{Q'E}\cdot \frac {F'E}{F'C}\cdot \frac {B'C}{B'Q''} \equal{} 1$ $ \Longrightarrow$ $ \frac {Q'Q''}{Q'E} \equal{} \frac {F'C}{F'E}\cdot \frac {B'Q''}{B'C}$ $ ,(5)$

But, $ \frac {F'C}{F'E} \equal{} \frac {M'M''}{M'E} \equal{} \frac {B'B''}{B'Q''}$ $ ,(6)$

From $ (5),\ (6)$ $ \Longrightarrow$ $ \frac {Q'Q''}{Q'E} \equal{} \frac {B'B''}{B'C}$ $ ,(7)$

From $ (4),\ (7)$ $ \Longrightarrow$ $ \frac {QQ''}{QE}\div \frac {Q'Q''}{Q'E} \equal{} \frac {BB''}{BC}\div \frac {B'B''}{B'C}$ $ \Longrightarrow$ $ (Q'',E,Q,Q') \equal{} (B'',C,B,B') \equal{} (C,B'',B',B)$ $ ,(8)$

From $ (8)$ $ \Longrightarrow$ $ (Q'',Q,Q',E) \equal{} (C,B',B,B'')$ $ ,(9)$

From $ (1),\ (9)$ $ \Longrightarrow$ $ (C,P',P,P'') \equal{} (Q'',Q,Q',E) \equal{} (E,Q',Q,Q'')$ $ ,(10)$

Hence, from $ (10)$ and because of the collinearities of the points $ P',\ N,\ Q'$ and $ P'',\ N,\ Q'',$ we conclude that the points $ P,\ N,\ Q,$ are also collinear and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Andreas Varverakis[/size][/b].

Kostas Vittas.

[quote][color=darkred][b][u]An equivalent enunciation[/u].[/b] Let $ ABCD$ be a convex quadrilateral. Suppose $ OA < OC$ and $ OD < OB$ ,

where $ O\in AC\cap BD$. Construct the midpoints $ M$ , $ N$ of $ [BD]$ , $ [AC]$  respectively, the intersections

$ E\in MN\cap AB$ , $ F\in DC\cap MN$ , $ P\in CE\cap BF$ and the midpoint $ L$ of $ EF$ . Prove that $ L\in OP$ .[/color][/quote]

[color=darkblue][b][u]Proof II[/u].[/b] $ \left\|\begin{array}{c} OA \equal{} a\ ,\ OB \equal{} b \\
 \\
OC \equal{} c\ ,\ OD \equal{} d\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c} c > a\ ;\ \frac {OA}{2a} \equal{} \frac {OC}{2c} \equal{} \frac {CA}{2(c \plus{} a)} \equal{} \frac {NA}{c \plus{} a} \equal{} \frac {ON}{c \minus{} a} \\
 \\
b > d\ ;\ \frac {OD}{2d} \equal{} \frac {OB}{2b} \equal{} \frac {BD}{2(b \plus{} d)} \equal{} \frac {MD}{b \plus{} d} \equal{} \frac {OM}{b \minus{} d}\end{array}\right\|\ (1)$ . Apply the [u][b]Menelaos'[/b] theorem[/u] to :

$ \left\|\begin{array}{cccccc} \overline {NME}/AOB\ : & \frac {NO}{NA}\cdot\frac {EA}{EB}\cdot\frac {MB}{MO} \equal{} 1 & \stackrel{(1)}{\implies} & \frac {EA}{EB} \equal{} \frac {c \plus{} a}{c \minus{} a}\cdot\frac {b \minus{} d}{b \plus{} d} & \implies & \frac {EA}{(c \plus{} a)(b \minus{} d)} \equal{} \frac {EB}{(b \plus{} d)(c \minus{} a)} \equal{} \frac {AB}{2(bc \minus{} ad)} \\
 \\
\overline {MNF}/DOC\ : & \frac {MO}{MD}\cdot\frac {FD}{FC}\cdot\frac {NC}{NO} \equal{} 1 & \stackrel{(1)}{\implies} & \frac {FC}{FD} \equal{} \frac {b \minus{} d}{b \plus{} d}\cdot\frac {c \plus{} a}{c \minus{} a} & \implies & \frac {FC}{(c \plus{} a)(b \minus{} d)} \equal{} \frac {FD}{(b \plus{} d)(c \minus{} a)} \equal{} \frac {DC}{2(bc \minus{} ad)}\end{array}\right\|\ (2)$ .

$ \blacktriangleright$ Denote $ \left\|\begin{array}{c} X\in CP\cap DB \\
 \\
Y\in BP\cap AC\end{array}\right\|$ . Apply the [u][b]Menelaos[/b]' theorem[/u] to :

$ \left\|\begin{array}{cccc} \overline {CXE}/AOB\ : & \frac {CO}{CA}\cdot\frac {EA}{EB}\cdot\frac {XB}{XO} \equal{} 1 & \stackrel{(1)\wedge (2)}{\implies} & \frac {XO}{XB} \equal{} \frac {c(b \minus{} d)}{(b \plus{} d)(c \minus{} a)} \\
 \\
\overline {BYF}/DOC\ : & \frac {BO}{BD}\cdot\frac {FD}{FC}\cdot\frac {YC}{YO} \equal{} 1 & \stackrel{(1)\wedge (2)}{\implies} & \frac {YO}{YC} \equal{} \frac {b(c \minus{} a)}{(c \plus{} a)(b \minus{} d)}\end{array}\right\|\ (3)$ . Denote the point $ R\in OP\cap BC$ .

Apply the [u][b]Ceva[/b]'s theorem[/u] in $ \triangle OBC$ for the point $ P\ : \ \ \frac {RB}{RC}\cdot\frac { YC}{YO}\cdot\frac {XO}{XB} \equal{} 1\ \stackrel {(3)}{\implies}\ \boxed{\ \frac {RB}{RC} \equal{} \frac {b(b \plus{} d)(c \minus{} a)^2}{c(c \plus{} a)(b \minus{} d)^2}\ }$ .

$ \blacktriangleright$ [u][b]Menelaos[/b]' theorem[/u] : $ \left\|\begin{array}{cccccc} \overline {CNO}/MDF\ : & \frac {CF}{CD}\cdot\frac {OD}{OM}\cdot\frac {NM}{NF} \equal{} 1 & \stackrel{(1)\wedge (2)}{\implies} & \frac {NM}{NF} \equal{} \frac {bc \minus{} ad}{d(c \plus{} a)} & \implies & \frac {NM}{bc \minus{} ad} \equal{} \frac {NF}{d(c \plus{} a)} \equal{} \frac {MF}{c(b \plus{} d)} \\
 \\
\overline {BMO}/NAE\ : & \frac {BE}{BA}\cdot\frac {OA}{ON}\cdot\frac {MN}{ME} \equal{} 1 & \stackrel{(1)\wedge (2)}{\implies} & \frac {MN}{ME} \equal{} \frac {bc \minus{} ad}{a(b \plus{} d)} & \implies & \frac {NM}{bc \minus{} ad} \equal{} \frac {ME}{a(b \plus{} d)} \equal{} \frac {NE}{b(c \plus{} a)}\end{array}\right\|$ .

Therefore, $ \frac {MN}{bc \minus{} ad} \equal{} \frac {ME}{a(b \plus{} d)} \equal{}$ $ \frac {NE}{b(c \plus{} a)} \equal{} \frac {NF}{d(c \plus{} a)}$ $ \equal{} \frac {EF}{(c \plus{} a)(b \plus{} d)} \equal{} \frac {2\cdot LM}{(b \plus{} d)(c \minus{} a)} \equal{}$ $ \frac {2\cdot LN}{(c \plus{} a)(b \minus{} d)}\implies$

$ \boxed {\ \frac {LM}{LN} \equal{} \frac {(b \plus{} d)(c \minus{} a)}{(c \plus{} a)(b \minus{} d)}\ }\ \ (4)$ . Denote $ S\in OL\cap BC$ and apply an [u]well-known relation[/u] in the triangle $ BOC\ :$

$ \frac {LM}{LN} \equal{} \frac {SB}{SC}\cdot\frac {OM}{OB}\cdot\frac {OC}{ON}\implies$ $ \frac {SB}{SC} \equal{}$ $ \frac {LM}{LN}\cdot \frac {OB}{OM}\cdot\frac {ON}{OC}\ \ \stackrel{(1)\wedge (4)}{ \equal{} }\ \frac {(b \plus{} d)(c \minus{} a)}{(c \plus{} a)(b \minus{} d)}\cdot \frac {b}{b \minus{} d}\cdot\frac {c \minus{} a}{c}$ $ \implies$ $ \boxed {\ \frac {SB}{SC} \equal{} \frac {b(b \plus{} d)(c \minus{} a)^2}{c(c \plus{} a)(b \minus{} d)^2}\ }$ .

In conclusion, $ \frac {RB}{RC} \equal{} \frac {SB}{SC} \equal{} \frac {b(b \plus{} d)(c \minus{} a)^2}{c(c \plus{} a)(b \minus{} d)^2}$ , what means $ R\equiv S$ , i.e. $ L\in OP$ . See[/color] [url=http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv29n1/body/HTML/57?template=CRUX][color=red][b]here[/b][/color][/url] [color=darkblue]an interesting [u][b]Walter Janous[/b]' extension[/u].[/color]

[quote="vittasko"][b][size=100][color=DarkBlue]EQUIVALENT  PROBLEM. – Through the midpoint $ M$ of the side-segment $ AB$ of a given quadrilateral $ ABCD,$ we draw the parallel lines to $ BC,\ AD,$ which intersect $ CD,$ at points $ E,\ F,$ respectively. Prove that the line segment $ PQ,$ where $ P\equiv AC\cap BD$ and $ Q\equiv AE\cap BF,$ passes through the midpoint $ N$ of $ CD.$[/color][/size][/b][/quote]
[b][size=100][color=DarkBlue]EXTENSION OF THE EQUIVALENT PROBLEM. -Through an arbitrary point $ M$ on the side-segment $ AB$ of a given quadrilateral $ ABCD$ (convex or non-convex), we draw the parallel lines to $ BC,\ AD,$ which intersect $ CD$ at points $ E,\ F,$ respectively. Prove that $ \frac {MA}{MB} \equal{} \frac {ND}{NC},$ where $ N\equiv CD\cap PQ$ and $ P\equiv AC\cap BD$ and $ Q\equiv AE\cap BF.$[/color][/size][/b]

Kostas Vittas.

[quote="Agr_94_Math"]DangChienbn,  can you post your proof based on  Pappus Theorem?[/quote]
I am also interested how we can apply the [b][size=100]Pappos theorem[/size][/b] here. Propably, an elementary solution may be arisen from such that approach.

Kostas Vittas.

An elementary proof using Pappus Theorem...
Dedicated to Costas Vittas!


Prove that
$ 1 \minus{} \frac {sin^240}{sin^270} \equal{} \frac {sin^280 \minus{} sin^260}{sin^240}$


Let ABC be a triangle with bisectors AD,BE,CF. Suppose that $ \angle DEF\equal{}90^o$. Calculate all angles of ABC.

You cannot calculate all angles, only one, $ \angle ABC \equal{} 120^\circ.$
See [url]http://www.mathlinks.ro/viewtopic.php?t=114998[/url].

I guess the answer is $ \angle B \equal{} 120$ and $ \angle A,\angle C$ are arbitrary. :wink:


[quote="pianojuggler4"]This problem was brought up in our systems lab (its generalized problem #6 from the AIME I)

A frog is placed at the origin on the numberline, and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of $m$, or to the closest point with a greater integer coordinate that is a multiple of $n$, with $m$ and $n$ relatively prime. A move sequence is a sequence of coordinates which corespond to valid moves, beginning with 0, and ending at $mn$. Express the total number of move sequences possible for the frog in terms of $m$ and $n$.[/quote]
No one has solved it in two weeks, so I'm bumping it.
Do you have a solution for it?


absolute value
combinatorics unsolved
combinatorics

A pair $ (m,n)$ of positive integers is written on the blackboard. In one step one can double any of these numbers and add $ 1$ to another one.

(a) Is it possible to obtain a pair of equal numbers from the pair $ (1,1998)$ using the above operation?

(b) Is it possible to obtain a pair of numbers such that the absolute value of their difference is $ 13579$ from any given pair?

(c) Is it possible to obtain a pair of numbers such that the absolute value of their difference is $ k$, where $ k$ is a fixed positive integer, from any given pair?


function
real analysis
real analysis unsolved


tangents to a circle with center $O$ from an extrnal point $S$ touch the circle at the points $P$ and $Q$. Intersection points of line $SO$ and the circle are points $A$ and $B$, $X$ is a point on small arc $PB$ ann $C$ and $D$ are the intersection points of the lines $QX$ and $PX$ with $SO$. According to the givens prove that $1$/$AC$ + $1$/$AD$ = $2$/$AB$

Using a central projection, project the circle (O) with the interior point C into a circle (C') centered at the projection C' of the point C. Then Q'X' becomes a circle diameter, hence $P'X' \perp P'Q'.$ Since the tangents SP, SQ of the circle (O) are projected into tangents S'P', S'Q' of the circle (C'). $P'Q' \perp S'C' \equiv S'D'.$ It follows that $P'D' \equiv P'X' \parallel S'D',$ which means that the point D' is at infinity. Hence $\frac{A'D'}{B'D'} = 1$ for finite A', B'. Since C' is the center of the projected circle, $\frac{A'C'}{B'C'} = 1$ as well, and the cross ratio of A', B', C', D' is $\frac{A'D'}{B'D'} \cdot \frac{B'C'}{A'C'} = 1.$ But this is preserved in a central projection, hence the cross ratio of the original points A, B, C, D is also $\frac{AD}{BD} \cdot \frac{BC}{AC} = 1.$ Substituting $BD = \frac{AD \cdot BC}{AC}$ into $AD = AB + BD,$

$AD = AB + \frac{AD \cdot BC}{AC}\ \Longrightarrow\ AC \cdot AD = AC \cdot AB + AD \cdot BC$

Adding $AC \cdot AD$ to both sides,

$2 AC \cdot AD = AC \cdot AB + AD \cdot BC + AC \cdot AD = AC \cdot AB + AD \cdot AB$

$\frac{2}{AB} = \frac{1}{AD} + \frac{1}{AC}$

You can see that $\measuredangle{BPD} = \measuredangle{BQX} = \measuredangle{CPB}$. Since $\measuredangle{APB} = 90^0$, it follows that ${ABCD} = -1$. Hence,  according to Descartes' theorem, we obtain the result.


Find the possible values of the digits p and q, given that 'p543q' is a multiple of 36.

$ 9|p543q \Leftrightarrow p+q=6(mod 9) \Leftrightarrow p+q=6$ or $ 15$
$ 4|p543q \Leftrightarrow q=2$ or $ 6$
Combine this two condition we have $ (p, q)=(4,2), (9,6)$ are the solutions.


induction
algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Let $K$ be a finite field, and let $f(x)=\prod_{a\in K}(x-a)+1$. What are the degrees of the irreducible factors of $f$ in $K[x]$?

Clearly, $f$ has no root in $K$. Because finite extensions of $K$ are normal, the splitting field $L$ of $f$ is optained by adjoining an arbitrary root. So, if $g$ is an irreducible factor of $f$ and $a$ a root of $g$, we have $d : = \deg(g) = [K(a) : K] = [L : K]$, which does not depend of $g$. I show $d = char(K)$:

Let $q = p^{k}$ be the order of $K$, s.t. $p = char(K)$. The galois group of $L/K$ is generated by the frobenius-automorphism $x \mapsto x^{q}$, which has order $d$. So, it suffices to show $a^{q^{p}}= a$. It is well-known that $f = x^{q}-x+1$.  Thus, $a^{p^{k}}= a-1$. Induction yields $a^{p^{ik}}={a^{p^{(i-1)k}}-1}$. After another induction, we get $a^{p^{ik}}={a^{p^{(i-j)k}}-j}$. Now we put $i = j = p$ to get the desired result.

[quote="-oo-"]Because finite extensions of $K$ are normal, the splitting field $L$ of $f$ is optained by adjoining an arbitrary root. So, if $g$ is an irreducible factor of $f$ and $a$ a root of $g$, we have $d : = \deg(g) = [K(a) : K] = [L : K]$, which does not depend of $g$.[/quote]

I cannot understand this argument. Why $d$ does not depend on $g$? For example, let $g_{1}$ and $g_{2}$ be irreducible polynomials such that $\deg(g_{1})=2$ and $\deg(g_{2})=3$. Let $f=g_{1}g_{2}$ and let $a$ be a root of $g_{1}$, then $L=K(a)$ is not a splitting field for $f$.

Nevertheless, it seems that it is enough to use your second argument only. Maybe i cannot get something...

The polynomial $f$ is of a very special kind here: if $\alpha$ is one of its roots in an algebraic closure of $K$, then the roots are precisely $\alpha+t$ as $t$ ranges over $K$; hence $K(\alpha)$ [i]is[/i] a splitting field for $f$.

@lofar: Thank you for this remark. In fact I did not have in mind the right definition of normal extensions.

Fortunately, the proof works because of grobber's conclusion :-).


How many three-digit positive integers exist, all of whose digits are 2's and/or 5's?

[hide]Hmmm... 8?

$ 2\cdot 2\cdot 2\equal{}8$

2 choices for the first blank.
2 choices for the second blank.
2 choices for the third blank.

Also: listing:

1. 222
2. 225
3. 252
4. 522
5. 255
6. 525
7. 552
8. 555[/hide]

something is not right  ...

"And/or is widely used in legal and business writing. Its use in general writing to mean “one or the other or both” is acceptable but can appear stilted."

What exactly did the writer of the problem mean by "[b]and/or[/b]" ?
2's and 5's  .... 2's or 5's ???

The writer of the problem meant the numbers can consist of all $ 2$'s, all $ 5$'s, or a combination of $ 2$'s and $ 5$'s.

Thus, as Dojo said, the answer is $ 2^3$ or $ \boxed{8}$.

Since any of the digits may be a 2 or a 5, this leaves us 2 choices for each digit. There are then 2^3 =8 3-digit integers composed of only 2s and/or 5s.Since any of the digits may be a 2 or a 5, this leaves us 2 choices for each digit. There are then 2^3 ={8 3-digit integers composed of only 2s and/or 5s.


inequalities
real analysis
real analysis unsolved

Show that $ \left|{\frac{xy^3}{x^2+y^4}\right|\leq\left|y\right|}$.

Make the right hand side $ \frac{|y|}2$ for a tighter inequality.

$ ab\le\frac12(a^2\plus{}b^2)$ for positive $ a,b.$

Ok thank you.

With your mentioned in equality, the one above is really obvious. But where does ist come from? It reminds me a bit of the AM-GM one but it has nothing to do with it, right? ;-)

It is a specific case of AM-GM, which you can see by writing it as
\[ \sqrt{a^2b^2}\le\frac{a^2\plus{}b^2}2.\]


In two vertices $M,N$ of a cube the number 1 is written, and in all the other six vertices the number 0 is written. A movement is defined by choosing a vertex and adding one unit to the numbers written in the three adiacent vertices (a vertex is adiacent with another vertex if the segment determined by them is a side of the cube).

Prove that there exists a finite set of movements such that in each of the 8 vertices the same number is written if and only if $MN$ is not a diagonal of a face of the cube.

created by Dinu Serbanescu

Did I misunderstand this question? It looks a little easy compared to the rest.. :?

So what I thought:
there are 3 cases:
1) adjacent (to prove it can)
2) face diagonal (to prove it cannot)
3) cube diagonal (to prove it can)

3) is simple: 2 moves, 1 from each vertex
1) is also simple: 2 moves, 1 from each of the opposing side's vertices

2) looks good on a well known Russian problem I think?
Anyway:

Name the planes ABCD and EFGH (A right above E) and say A and H start with 1, the others with 0 (so M=A and N=H here)

Any move will raise 3 elements out of {A,C,F,H} or {B,D,E,G}.
the latter one starts of sum 0 so will have sum = 3k
the first one will thus have sum 3m+2
these sums can never be equal.


quadratics
function
trigonometry
algebra
number theory unsolved
number theory

you can not assume so!!!
it is absolutely wrong!!!
X=23 and n=9

when n is even it is easy to show that n=6 but i dont know for n odd
i'll check it out tonight...

so,
if n is odd substract 25 to each side
and factor it will easily be seen that n=3 is the only solution. :)

[quote="dad"]X^2-2^n=17
where x,n -natural numbers[/quote]

If $n$ is even, everything is easy. Then $n=2k$ for some natural k. We have: $x^2-2^{2k}=(x+2^k)(x-2^k)=17$. Because 17 is prime, we have $x-2^k=1$, so $x=1+2^k$. Returned to original equation, we have: $2^{k+1}=16$, which implies $k=3$ and than $n=6$. So for n even there is only one solution: $(n,x)=(6,9)$.

For n odd we have $x^2-2^{2k+1}=17$ which implies, that x is odd. Let $x=2t+1$. Than we have after division by four: $t^2+t-2^{2k+1}-4=0$. Let us consider last equation as quadratic in $t$. Than its discriminant $D=17+2^{2k+1}$. Necessary is true, that $D=s^2$ which implies $17+2^{2k+1}=s^2$. Than the mentioned quadratic equation has solution positive solution $t=(s-1)/2$ which implies $s$ is greather than $t$. Working back, if we have a solution $(2k+1,s)$, we can always construct a solution $(2k+1,t)$ with $s$ greather than $t$. By infinite descent, let $(2k+1,x)$ be some solution for given value of $k$ with minimal $x$. By what we have proved, there than is a solution $(2k+1,y)$ with $x$ greather than $y$, which is a contradiction. So there are no solution for n odd.

PS.:I hope it is right...:]

Well, let's observe that

$x^2=2^n+17$

the left hand side of the equation is a function which has a "parabople" as graphic [it does not vanish all reals], and the right hand side is a exponential function transladade by 17, which has a enumerable points in the plane. Off course it might has, at most, two solutions (it's visible that the "exponential graphic" [it does not vanish all reals] cuts the parabole at most two times). Sine one of the solutions [if it exists] is negative, and we're dealing with naturals, we should only get one solution, if it exists, and in case, as shown, $(5,3)$ is the solution.

by the way
7^2=2^5+17 is another solution :?

yup =[ off course. I just did my calculations as n varying the same as x, that's not necessarily truth. We can fix x and vay n as much as we want to find the solution and vice-versa.

sorry for the misexplanation...

[quote="conceive"][quote="dad"]X^2-2^n=17
where x,n -natural numbers[/quote]

If $n$ is even, everything is easy. Then $n=2k$ for some natural k. We have: $x^2-2^{2k}=(x+2^k)(x-2^k)=17$. Because 17 is prime, we have $x-2^k=1$, so $x=1+2^k$. Returned to original equation, we have: $2^{k+1}=16$, which implies $k=3$ and than $n=6$. So for n even there is only one solution: $(n,x)=(6,9)$.

For n odd we have $x^2-2^{2k+1}=17$ which implies, that x is odd. Let $x=2t+1$. Than we have after division by four: $t^2+t-2^{2k+1}-4=0$. Let us consider last equation as quadratic in $t$. Than its discriminant $D=17+2^{2k+1}$. Necessary is true, that $D=s^2$ which implies $17+2^{2k+1}=s^2$. Than the mentioned quadratic equation has solution positive solution $t=(s-1)/2$ which implies $s$ is greather than $t$. Working back, if we have a solution $(2k+1,s)$, we can always construct a solution $(2k+1,t)$ with $s$ greather than $t$. By infinite descent, let $(2k+1,x)$ be some solution for given value of $k$ with minimal $x$. By what we have proved, there than is a solution $(2k+1,y)$ with $x$ greather than $y$, which is a contradiction. So there are no solution for n odd.

PS.:I hope it is right..:][/quote]
your solution doesnt work because you start with x=2t+1 and end up with x=2t+1 its always the same solution nothing decreases so as to apply infinit descent. :)

[quote="dad"]X^2-2^n=17
where x,n -natural numbers[/quote]

Anyway, the problem now changes to the following one:

to find all solutions of

$x^2-2^n=17$ for odd n.

Note for dad: a little course about TeX wouldn't hurt you.

moderators: please move it to number theory (and away from solved).


calculus
derivative
logarithms
calculus computations

Can someone help me with the following problem?
(1 divided by (n+1)*(3 to the power of n))

Find the exact sum of the infinite series:

1/((n+1)(3^n))

The key method here: introduce a variable, and look at the power series as a function. In this case, the series is immediately recognizable.
$\sum_{n=0}^\infty \frac{x^n}{n+1}=-\frac{\ln(1-x)}{x}$. Evaluating at $x=\frac13$, the sum is $3(\ln3-\ln2).$

You didn't say where the sum starts, but I'll assume you meant zero:

$\sum_{n=0}^{\infty}\frac1{(n+1)3^n}.$

First idea: replace the geomtric factor by a variable:

$\sum_{n=0}^{\infty}\frac{x^n}{n+1}.$

We're anticipating using derivatives; it would help if the exponent matched the denominator, so

$\frac1x\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}.$

So let $f(x)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}.$

Then $f'(x)= \sum_{n=0}^{\infty}\frac{(n+1)x^n}{n+1}=\sum_{n=0}^{\infty}x^n= \frac1{1-x}.$

Integrate that term by term and use the fact that $f(0)=0$ to get that $f(x)=-\ln(1-x).$ Hence

$\sum_{n=0}^{\infty}\frac{x^n}{n+1} = \frac1xf(x)=-\frac{\ln(1-x)}x$

Finally, $\sum_{n=0}^{\infty}\frac1{(n+1)3^n}= -\frac{\ln\left(1-\frac13\right)}{\frac13}=3\ln\left(\frac32\right).$

jmerry composes his posts quicker than I do.

[quote="Kent Merryfield"]
We're anticipating using derivatives; it would help if the exponent matched the denominator, so

$ \frac1x\sum_{n \equal{} 0}^{\infty}\frac {x^{n \plus{} 1}}{n \plus{} 1}.$

[/quote]

How can you do this? What if x=0?

You're only ever interested in what goes on in a neighborhood of $ x \equal{} \frac{1}{3}$, so introducing potential problems at $ x \equal{} 0$ doesn't matter.


Keystone Bike Shop advertised a sale for $ 40\%$ off any bike. The sale prices were then discounted an additional $ 20\%$. After both discounts, Jerry paid $ \$168$ for his new bike. What was the number of dollars in the original price of Jerry's bike?

From the original price, our new price is $ .6 * .8 \equal{} .48 \equal{} 48\%$.

$ \frac{168}{.48} \equal{} \boxed{350}$ dollars

the original price of the bike can be modeled by this equation (x is the discounted price):

[size=150]$168=8/10 (6/10 x)[/size]
[size=150]$210=6/10 x
[/size]
[size=150]$350=x[/size]

[size=150]ANS: 350[/size]


inequalities
function
algebra unsolved
algebra

Consider all functions $ f: \mathbb N \rightarrow \mathbb N$ such that inequality $ f(m\plus{}n)\geq f(m)\plus{}f(f(n))\minus{}1$ holds for all naturals $ m$ and $ n$. Find all possible vaues of $ f(2007)$

[quote="delegat"]Consider all functions $ f: \mathbb N \rightarrow \mathbb N$ such that inequality $ f(m \plus{} n)\geq f(m) \plus{} f(f(n)) \minus{} 1$ holds for all naturals $ m$ and $ n$. Find all possible vaues of $ f(2007)$[/quote]
[color=green]From the condition we have $ f$ is an increasing function. On the other hand
\[ f(m\plus{}1)\ge f(m)\plus{}f(f(1))\minus{}1\equal{}f(m)\plus{}a
\]
With $ a\equal{}f(f(1))\minus{}1$. Then $ f(m)\ge f(1)\plus{}(m\minus{}1)a$, let $ f(n)\minus{}f(1)\minus{}(n\minus{}1)a\equal{}g(n)$. Thus
\[ na\plus{}g(m\plus{}n)\ge g(m)\plus{}f(1)\plus{}(f(1)\plus{}(n\minus{}1)a\plus{}g(n))a\plus{}g(f(1)\plus{}(n\minus{}1)a\plus{}g(n))
\]
We also have $ na\le (f(1)\plus{}(n\minus{}1)a\plus{}g(n))a$ and then
\[ g(m\plus{}n)\ge g(m)\plus{}g(g(n))\minus{}1
\]
And now, this problem is easy.[/color]


function integration geometry rectangle calculus LaTeX algebra

Vanishing Theorem: Let $f(x)$ be a continuous function in a finite closed interval $[a,b]$. Assume that $f(x) \geq 0$ in the interval and that  $\int_a^b f(x) dx=0$. Then $f(x)$ vanishes, i.e. $f$ is identically zero.

Proof: The key point to the theorem is that a positive function above the x-axis would have zero area under the curve, but is that possible?. So suppose that $f(c)>0$ for some $c \in (a,b)$. Let  $\varepsilon = f(c)/2$ since $f$ is continuous at $c$. Then there exists a  $\delta >0$ such that $|x-c|< \delta$  implies that $f(x)> \varepsilon$. Now the area under the curve over the interval $(a,b)$ covers a rectangle with height  $\varepsilon$  and width $2\cdot \delta$  so in terms of integrals we have

$ \int_a^b f(x) dx \geq  \int_{c- \delta}^{c+ \delta} f(x) dx \geq  \varepsilon \cdot (2\cdot \delta )= \delta f(c)>0$. We have a contradiction on the assumption that the integral is zero!!! Therefore $f(x)=0$ for all $a<x<b$. Since $f$ is continuous, $f(a)=f(b)=0$ as well!!!
End of proof.

This theorem can be extended to higher dimensions (several variables) and perhaps the following theorem is beautiful as well:

Theorem: Let $f(\mathbf{x})$ be a continuous mapping in $\mathbf{D}_0$ such that  $\int \int \int_{\mathbf{D}} f(\mathbf{x}) d\mathbf{x} = 0$ for all subdomains $\mathbf{D}$ a subset of $D_0$. Then $f(\mathbf{x})=0$ in $\mathbf{D}_0$.
Hint for proof: Let $\mathbf{D}$ be an open ball and have its radius $r\rightarrow 0$. :)
Edit 11/21: Corrected the $\LaTeX$ formulas

I think there is a more simple proof.

Take $F(x)=\int_a^xf(x)\;dx$. Then $0\leq F(x)\leq\int_a^bf(x)\;dx=0$, so $F$ is identically zero. Finally, $0=F'(x)=f(x)$.

[quote="Myth"]We don't need continuity in this theorem, but only $ f\in L_{1,loc}(D_0) $.[/quote]
That's a different theorem. If you drop the condition of contiunity, then you cannot conclude that $f$ is zero [i]everywhere[/i], as would be shown by the trivial example of a function that is 1 at one point and zero everywhere else. The theorem Myth is referring to has as its conclusion that $f$ is zero [i]almost everywhere[/i], or, equivalently, that it is the zero element of $L^1.$ But the elements of $L^1$, although often informally referred to as functions, are not functions; they are equivalence classes of functions under the equivalence relation of almost everywhere equality.

For continuous functions, the proof that 3X.lich posted is basically what has to be done, except that he made the too-strong assumption that $f\ge 0.$ Although you need something like that in the proof, you don't have to assume it up front.

Here's another way to put it, letting $f$ be complex-valued. Suppose $f$ is a continuous complex-valued function on $D_0$ such that for any (open) domain $D\subset D_0, \int_Df(x)dx=0.$ Suppose that for some $x_0, f(x_0)\ne0.$ Let $\alpha=\frac{\overline{f(x_0)}}{|f(x_0)|}.$ Then $\alpha f(x_0)=|f(x_0)|.$ By the continuity of $f$ (and hence $Re(\alpha f)$), then there is some neigborhood $B$ of $x_0$ on which $Re(\alpha f(x))>\frac{|f(x_0)|}2.$ Hence $Re\left(\int_B\alpha f(x)dx\right)\ge|B|\frac{|f(x_0)|}2>0$ and thus $\int_B f(x)dx\ne0.$ Since this cannot happen, we conclude that $f$ can never take on a nonzero value.

[quote="Kent Merryfield"][quote="Myth"]We don't need continuity in this theorem, but only $ f\in L_{1,loc}(D_0) $.[/quote]
That's a different theorem. If you drop the condition of contiunity, then you cannot conclude that $f$ is zero [i]everywhere[/i], as would be shown by the trivial example of a function that is 1 at one point and zero everywhere else. The theorem Myth is referring to has as its conclusion that $f$ is zero [i]almost everywhere[/i], or, equivalently, that it is the zero element of $L^1.$ But the elements of $L^1$, although often informally referred to as functions, are not functions; they are equivalence classes of functions under the equivalence relation of almost everywhere equality.[/quote]
It is clear that if I say $f\in L_{1,loc}$ then identity $f=0$ has meaning in $L_1$-sense. But if you take continuous representative of class $0\in L_{1,loc}$ it must be 0 everywhere.

[quote="Myth"][quote="3X.lich"]Theorem: Let $f(x)$ be a continuous mapping in $D_0$ such that  $\int_D f(x)dx=0$ for all subdomains $D$ a subset of $D_0$. Then $f(x)=0$ in $D_0$.[/quote]
We don't need continuity in this theorem, but only $f\in L_{1,loc}(D_0)$.[/quote]
It is possible for the theorem to hold without the continuity assumption in more generalized spaces. I'll check back later.

Kent Merryfield: you are right about the assumption that $f\geq 0$ is too strong. I wanted to present a proof accessible to beginning calculus students that may find several variables hard.


calculus
integration
function
real analysis
real analysis unsolved

By analogy with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1570654]this problem[/url]--it may turn out to be an easier problem.

If $ f: [0,\infty) \to \mathbb{R}$ is a continuous function, let $ f^* (x) \equal{} \int_0^x f(t)dt$.  Does there exist some $ g$ such that $ g, g^*, g^{**},\ldots$ all decay?

Same story as with sums. You just need a fast decaying function $ f$ such that $ \int_0^\infty x^pf(x)\equal{}0$ for all integer $ p\ge 0$. Now you need to use Fourier transform instead of Fourier series but the differences end there.


linear algebra
matrix
linear algebra unsolved

Let $ A,B\in{M_{n}(C)}$  so that $ AB\plus{}BA\equal{}O_{n}$, $ det(A\plus{}B)\equal{}0$.
Prove that $ det(A^3\minus{}B^3)\equal{}0$


calculus
integration
geometry
vector
function
trigonometry
calculus computations

Given a cycloid expressed in a vector function as: [img]http://i32.tinypic.com/vrrm87.gif[/img] for  0 <= t <= 2pi

How would we:

1) find the area enclosed by the Cycloid C, and the x axis?
2) evaluate [img]http://i31.tinypic.com/20a5c2r.gif[/img] if [img]http://i27.tinypic.com/73cvgn.gif[/img]

I'm having a bit of trouble with this one again. On the surface it doesn't seem that difficult but I have been reading all of this stuff on line integrals and Green's Theorem etc.. that i'm totally confused now. Any solutions would be appreciated again.!

[hide="Hint"]The area enclosed by a simple closed positively orientated curve C is given by $ A \equal{}  \minus{} \oint_C y\, dx$[/hide]

The Cycloid C is negatively orientated (it goes clockwise) so drop a minus sign

\[ A\equal{}\minus{}\oint_{\minus{}C} y \, dx\equal{}\int_{0}^{2\pi}(1\minus{}\cos t)^2 \,dt \equal{} \int_{0}^{2\pi}(1\minus{}2\cos t\plus{}\cos^2 t )\,dt\]
\[ \equal{}\left[\frac{3}{2}t\minus{}2\sin t\plus{}\frac{1}{4}\sin 2t \right]_{t\equal{}0}^{2\pi}\equal{}3\pi\]

and the contribution from the line segment along the x-axis is zero.


[quote="nguoivn"]Given $ a, b, c > 0$. Prove that
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} \frac {81}{4}\left(\frac {a^2b}{(2a \plus{} b)^2} \plus{} \frac {b^2c}{(2b \plus{} c)^2} \plus{} \frac {c^2a}{(2c \plus{} a)^2}\right) \ge \frac {13}{4}(a \plus{} b \plus{} c)\]
:)[/quote]
The inequality can be written as
$ 4\sum \frac{a^2}{b}\plus{}81\sum \frac{a^2b}{(2a\plus{}b)^2} \ge 13(a\plus{}b\plus{}c).$
We have
$ \frac{4a^2}{b}\equal{}\frac{4a^2}{b}\plus{}4a\plus{}b\minus{}4a\minus{}b\equal{}\frac{(2a\plus{}b)^2}{b}\minus{}4a\minus{}b.$
Therefore, the above inequality is equivalent to
$ \sum \left[\frac{(2a\plus{}b)^2}{b}\minus{}4a\minus{}b\right]\plus{}81\sum \frac{a^2b}{(2a\plus{}b)^2} \ge 13(a\plus{}b\plus{}c),$
or
$ \sum \left[\frac{(2a\plus{}b)^2}{b}\plus{}\frac{81a^2b}{(2a\plus{}b)^2}\right] \ge 18(a\plus{}b\plus{}c).$
By the AM-GM Inequality, we have
$ \frac{(2a\plus{}b)^2}{b}\plus{}\frac{81a^2b}{(2a\plus{}b)^2} \ge 18a.$
Adding this and its analogous inequalities, the conclusion follows.


[quote="GameBot"]How many times does the digit 9 appear in the list of all integers from 1 to 500?[/quote]

Lets sort this into cases:

[i]1- Digits:[/i] Obviously, $ \boxed{1}$

[i]2 - Digits:[/i] $ \boxed{19}$

[i]3 - Digits:[/i] $ 20$ times during each hundred, and there are $ 4$ hundreds, so $ \boxed{80}$

$ 1 \plus{} 19 \plus{} 80 \equal{}\boxed{\boxed{\boxed{100}}}$

Umm, I don't think so..., there are 18 nines in the 2-digit numbers.
So there are 19 nines in the three-digit numbers, and $ 5 * 19 \equal{} 95$ nines from 1-500...?
Or am I wrong...?

EDIT: oops, nvm...

I think 007math is correct... I would solve it differently though...

We can divide the numbers into two types.
(1) Numbers ending with 9 but not having 9 as a tens digit.

In this case the the tens digit could be 0 - 8 and the hundreds digit 0 - 4. So    we will have 5 * 9 = 45 occurences of 9.

(2) Numbers having 9 as the tens digit.

In this case, we have 11 9's in the ones and tens digit and the hundreds digit could be 0 - 4. So the total would be 5 * 11 = 55.

Thus, considering all cases, there are 45 + 55 = 100 occurences of 9 all in all...

hm. easy way to do it:

we don't care about the hundreds digit, because that'll never be 9. therefore, we look at the two digits. in each hundred, there can be 10 numbers with a 9 in the units place, and 10 unmbers with a 9 in the tens place. even though some will be counted twice (99,199, etc.), we still count them twice because its asking for HOW MANY NINES, not how many NUMBERS. therefore, its $ (10\plus{}10)5\equal{}(20)5\equal{}\boxed{100}$

This is what I got
5*1*9
5*9*1
5*1*1
sum:95.
Does anyone know what I did wrong?

[quote="FrozenFury"]I just got 95 too. Wouldn't 99,199,299,399, and 499 all be counted twice? So we would now have 100 - 5 = $95$[/quote]

The problem is asking for the [b]number of 9s, not the number of numbers that contain a 9.[/b]

Also guys, please hide your solutions.

I'm skeptical of whether or not 100 is the answer.  Wouldn't you have to subtract since 5 from 100 to get 95 as your answer since the numbers 99, 199, 299, 399, and 499 are counted twice?

[quote="ZZmath9"]Apparently you didn't read my last post?

[quote="ZZmath9"]The problem is asking for the [b]number of 9s, not the number of numbers that contain a 9.[/b]

[/quote]

This means that 99, 199, 299, 399, and 499 [b]need to be counted twice[/b] since there are [b]two[/b] 9s in each number.[/quote]
It is very simple.
There are $2$ cases.
[b]1.Putting 9 in units place.[/b]
No. of cases=$5*10*1=50$
[b]2.Putting 9 in tens place.[/b]
No. of cases=$5*1*10=50$
Adding we get $\boxed {100}$

but then wouldn't we count 199,299,399,499 twice??

The easiest approach is to consider how many times 9 can appear in the units place, how many times in the tens place, and how many times in the hundreds place. If we put a 9 in the units place, there are 10 choices for the tens place and 5 choices for the hundreds digit (including 0), for a total of 50 times. Likewise, if we put a 9 in the tens place, there are 10 choices for the units place and 5 choices for the hundreds digit, for a total of 50 times. Since 9 cannot appear in the hundreds digit, there are $50+50=\boxed{100}$ appearances of the digit 9.

$$\bold{W R O N G}$$
It is this[hide=:]100[/hide]

so i did it this way.
how many times is it in 100?
for every 100 it appears 9 times MINUS the nineties
so we have 9 times so far
now we deal with the nineties
10 times BUT you have to count 99 twice being as the number nine is in 99 twice.
so 9+10+1=20
now we have 5 100's
so 20*5=$\boxed{100}$

I made a silly error by also counting the 9's for the 500's as well. A simple way to think about this is to break up the problem into 3 cases :
1. Find # of 9's for 1 digit numbers (1-9) $Ans$ = $ \boxed{1}$ -- Just 9
2. Find # of 9's for 2 digit numbers (10 - 99) $Ans$ = $ \boxed{19}$ -- 19,29,39,49,59,69,79,89,90,91,92,93,94,95,96,97,98,99 ( the 9's gets counted twice in 99)
3. Find # of 9's for 3 digit numbers (100 - 400) strictly  $Ans$ = $ \boxed{80}$
We can continue to list the number of 9's from 100-400 but that might take quite a bit of time!
[b]Case 1[/b] : 9 in tens digit
$4 * 1 * 10 = 40$ , 4 options for hundreds digit (1,2,3,4), 1 option for tens digit (9), and 10 options for ones digit (0,1,2,3,...,9)

[b]Case 2[/b] : 9 is hundreds digit
$4 * 10 * 1 = 40$ , 4 options for hundreds digit (1,2,3,4), 10 options for tens digit (0,1,2,3,...,9), and 1 option for ones digit (9)

Total = 1 + 19 + 40 + 40 = $\boxed{100}$

[hide]Make sure to count the entire tenths place for 90 (90, 91, 92, Etc.)
Also make sure to count 2 9’s for the 99’s (99, 199, 299, etc.)[/hide]

Isn't this question asking how many times $9$ appears, not how many numbers contain it? What is wrong with what I did? [hide=Work]Just count how many times $9$ appears in the units place and in the tens place, since it can't be in the hundreds. In the units place, $9$ appears $10*5 = 50$ times. In the tens place, it appears $1*5 = 5 times$. Therefore, the answer is $50+5 = 55$ times.[/hide]

I did that my first answer. It isn't $10\times5$ because there are more than $10$ ones appearances of $9$ in each hundred.

[hide=Solution]
In each hundred, there are $10$ units digits of $9$ (including one $9$ of $99$) and $10$ tens digits of $9$ (including the other $9$ of $99$). This goes for all $5$ hundreds in $1\text-500$, so we get $(10+10)\times5=20\times5=100$.
[/hide]

[quote=Panther66]Wait... I got 55(incorrect) what is wrong with [hide=this?]We can use cases for how many times it appears in numbers with one digit, numbers with 2 digits, and numbers with 3 digits less the 500 [/hide][/quote]

Probably you did incorrect casework. Look at the solutions above you.

[quote=Panther66]Wait... I got 55(incorrect) what is wrong with [hide=this?]We can use cases for how many times it appears in numbers with one digit, numbers with 2 digits, and numbers with 3 digits less the 500 [/hide][/quote]

You forgot to count 90, 91 92, etc.


topology
superior algebra
superior algebra unsolved

hi can you help me?
show that singular _homology satisfies the dimension axiom.


In a triangle with sides a, b, and c, is there a general way of finding the measures of each angle in the triangle without using trig?

Example: A triangle with sides 5, 6, and 7.

There isn't. Historically, trigonometry was first developed as THE tool for finding the angles of a triangle - that's its basic "practical" purpose. (And the very word "trigonometria" indeed means "the measuring of a triangle" in Greek.)

Not that I know of (unless you're thinking about rational trig with quadrance and spread, which is kind of unconventional right now). So you're stuck with using law of cosines. Here's a link to rational trig in case you're interested anyways [url]http://en.wikipedia.org/wiki/Rational_trigonometry[/url]
(Edit: I guess this still counts as trig  :P )

Ugh...I guess I'm stuck with the Law of Cosines again.


Prove that the families y =Lnx + c and y = (-x^2/2) +d
are orthogonal trajectories of one another.

It is evidently : $\frac{1}{x}\cdot (-x)=-1\ !$