Moving the Blog to Wordpress

by Potla, Mar 22, 2012, 4:58 PM

So, guys, this post falls under the category of an Announcement, although I did not make a tag for that. ^_^
With this post, I officially transfer this blog completely to my analogous Wordpress Blog, which is mainly due to the reason that I am having problems with the CSS editing of this AoPS blog.

I appreciate your help for the past years. Please keep on making comments, suggestions, and feedback etc in the Wordpress Blog as well, because else I will miss you people.

Thanks for your coordination till now, and it's not as if I am banning AoPS or anything, it's just that Wordpress has a better interface in terms of editing and customization.

Finally, thanks to Rijul Saini for contributing to the blog, and all the viewers, thanks to whom this blog has come this far, and will continue to go on.

Potla

This post has been edited 2 times. Last edited by Potla, Mar 22, 2012, 8:52 PM

Poles and Polars - Another Useful Tool!

by Potla, Mar 22, 2012, 2:45 PM

As the title suggests, this post is going to deal with the different aspects of poles and polars, which is a really great tool in case of proving problems using cyclic quadrilaterals and a lot of circular reasoning can be exploited by this. It is easier to deal with lines than with circles, so it is better to work with inversion, poles and polars etc - that's why projective geometry is actually better in many aspects compared to plain euclidean geometry!

Anyways, for the time being let us know the definitions of poles and polars. Before continuing further, you have to note that a pole is a point, and a polar is a line (which is the opposite of Kedlaya, but yeah, notations don't matter much.

) In any case, thanks to Rijul Saini for giving me a hand on this post, which we delayed for more than 8 months. At last, this is getting published on the blog, so Cheers, everyone!

Pole of a line
Take the circle $\omega$ with respect to which you're applying the polar map. Now, drop a perpendicular to the given line from the centre $O$ of $\omega$ to the given line. Name that point $P$ . Now, invert $P$ with respect to the circle. We get a point $P'$ which lies on the same line as $O,P.$ Then, $P'$ is the Pole of the given line.

Polar of a point
Take the circle $\omega$ with respect to which you're applying the polar map. Let the centre of the circle be $O.$ Now, invert the given point $P$ with respect to the circle $\omega$ and name the image as $P'.$
Now, Draw the line perpendicular to $OP'$ which passes through the point $P'.$ Then, this line is the Polar of the point $P.$

Now let us look into some theorems and lemmas that we will be using while solving problems.

Theorem 1:(La Hire's theorem)
$(i)$ Every point is the pole of its polar, and every line is the polar of its pole.
$(ii)$ If $P$ lies on the polar of $Q$ then $Q$ lies on the polar of $Q$

Proof:
(i) Again, direct from the definition.
(ii) $\ P$ is a point on the polar of $Q.$ First, extend $OQ$ to the polar of $Q$ and name that point $Q'.$ If $Q'= P$ then we are through. Otherwise, let the inverse of the point $P$ wrt the circle be $P'.$ Now, $OQ \cdot OQ'=OP \cdot OP'=R^2.$ Now, by similar triangles, we have $\angle OP'Q= \angle OQ'P=90^{\circ}$ and so we are through. $\Box$

Theorem 2:
Three points are collinear if and only if their polars are concurrent.
Proof:
I shall prove only the forward direction. The reverse direction is entirely analogous.
Let the three points be $P,Q,R$ and their polars be $p,q,r.$ Now, let the line through $P,Q,R$ be $l,$ and let the pole of $l$ be $L$ . Since, $P,Q,R$ lie on the polar of $L,$ therefore, $L$ must lie on the polar of $P,Q,R$ i.e. lines $p,q,r.$ Thus, the lines $p,q,r$ are concurrent at $L.\Box$

Now let us get to know the poles and polars of some points and lines that are considered to be important.

Polar of a point outside a circle with respect to itself.
Take an arbitrary point $P$ outside a circle. Since the inverse of the point actually lies inside the circle, the polar will be a secant of the circle. Now, let us assume that the inverse point of $P$ w.r.t. $(O,r)$ is $P'.$ The line perpendicular to $\overline{OP'P}$ and passing through $P'$ will intersect the circle in two points which will be symmetric w.r.t $PP'.$ So let one of these two intersections be $Q,$ and since $PO\times P'O=r^2=PQ^2,$ we note that $QOP'\sim POQ,$ so this leads to $\angle OQP=90^{\circ},$ ie:
Theorem 3:
The polar of a point $P$ lying outside the circle is actually the chord of contact of the point $P$ with respect to $(O,r).$

Polar of a point lying inside the circle does not have any special property that can be exploited alike to the previous one. However, we can take any arbitrary two chords passing through $P$ , find their poles and join the two corresponding points to obtain the polar of that point $P.$

Now, let us move on to Theorem 4. The first name was coined by me, and the second one by Rijul Saini. Don't assume that they are the original names of the theorem.
Theorem 4
(The fundamental theorem of poles and polars, or The theorem of two pascals)
Take a quadrilateral $ABCD$ which is cyclic. If $K=AC\cap BD$ then the polar of $K$ with respect to $\odot(ABCD)$ is the line joining $AB\cap CD$ and $AD\cap BC.$
Proof 1.
The second name almost gives it away. This problem, however, was directly quoted and asked to prove in the Turkish TST 1993.
Anyways, using Pascal's theorem in $ACCBDD,$ we obtain $AC\cap BD, CC\cap DD, BC\cap DA$ are collinear. Again using the same in $CAADBB$ we obtain that $CA\cap DB, AA\cap BB, AD\cap BC$ are collinear. Therefore it is noted that the points $AC\cap BD, AD\cap BC, CC\cap DD, AA\cap BB$ are collinear. Denote this line as $l.$
Since $l$ passes through $CC\cap DD$ and $AA\cap BB,$ which is the polar of $AC\cap BD=K,$ we are done. $\Box$

Proof 2.
This uses harmonic divisions.

Let us take $S=AB\cap CD, F=AD\cap BC, E=AC\cap BD.$ If the tangent to $\odot(ABCD)$ at $S$ touches it at $M,N$ respectively such that $M\in\overhat{DC}$, denote $X=MN\cap AD$ and $Y=SE\cap BC, Z=MN\cap CD, T=SE\cap AD.$
It can be shown easily that $(S,X,A,B); (S,Z,C,D)$ and $(S,E,Y,T)$ are harmonic. Then using $(SXAB)=(SZCD)$ we note that $AD,BC,ZX$ concur; and $F,X,Z$ are collinear. Similarly we obtain that $F,X,E$ and $F,Z,E$ are collinear. Since $XY\equiv MN,$ therefore $E,F,M,N$ are collinear.
So $F$ lies on the polar $XZ$ of $S$ w.r.t $\odot(ABCD).$ By symmetry $S$ lies on the polar $YT$ of $F$ w.r.t $\odot(ABCD).$ So the intersection of these two polars will be $E$ , and the polar of this pole will be $FS.$ We are done! $\Box$

There are some useful results that are correlated to harmonic divisions and poles and polars, which I explain in the following few theorems.
Theorem 5 (a).
If $P$ is the pole of a line $AB$ w.r.t $\omega$ then any line $\ell$ through $P$ is cut harmonically by $P,AB,$ and $\omega.$
Proof.
Let $O$ be the centre of $\omega,$ and let $OP\cap \overleftrightarrow{AB}=B, \ \ell\cap \overleftrightarrow{AB}=A, \ PA\cap \omega = F.$
Let $\omega'$ be the circle passing through $P,A,B.$ Then its centre is the midpoint $O'$ of $BP.$ If $\omega\cap \omega' = C,D;$ then $P$ is the pole of $AB.$ So $OP\cdpt OQ=OC^2.$ Then $\omega$ and $\omega'$ are orthogonal circles, which means that $O'D$ is tangent to $\omega.$ Therefore we obtain $O'E\cdot O'F=O'D^2=O'P^2,$ and since $O'$ is the midpoint of $PA,$ from Theorem 1 - Harmonic Divisions, we are done.
$[asy]import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttttff = rgb(0.2,0.2,1); pen ffttqq = rgb(1,0.2,0); draw(circle((0,0),3),ffttqq); draw((-4.97,(+9-1.69*-4.97)/0.62)--(6.6,(+9-1.69*6.6)/0.62),ttttff); draw((-4.97,(+1.31+0.43*-4.97)/3.26)--(6.6,(+1.31+0.43*6.6)/3.26),ttttff); draw((0,(-0+0.62*0)/1.69)--(6.6,(-0+0.62*6.6)/1.69),ttttff); draw(circle((3.32,0.84),1.64),ffttqq); draw((1.69,0.62)--(2.9,-0.75),ttttff); draw((1.69,0.62)--(2.2,2.04),ttttff); dot((0,0),ds); label("$O$", (-0.28,-0.27),S*lsf); dot((1.69,0.62),ds); label("$P$", (1.47,0.4),S*lsf); dot((4.94,1.05),ds); label("$B$", (5.01,1.05),SE*lsf); dot((2.9,0.78),ds); label("$E$", (2.71,0.52),S*lsf); dot((-3,0.01),ds); label("$F$", (-3.22,-0.32),SW*lsf); dot((4.69,1.74),ds); label("$A$", (4.76,1.84),NE*lsf); dot((2.9,-0.75),ds); label("$C$", (2.98,-0.85),S*lsf); dot((2.2,2.04),ds); label("$D$", (2.16,2.19),N*lsf); dot((3.32,0.84),ds); label("$O'$", (3.32,0.8),SE*lsf); clip((-4.97,-3.81)--(-4.97,5.46)--(6.6,5.46)--(6.6,-3.81)--cycle);[/asy]$
$\Box$

Theorem 5 (b).
If two lines $p,q$ are self-conjugate lines and if $p\cap q = T,$ then $p,q$ are harmonic with the two tangents drawn to the circle from $T.$
Proof.
Let the poles of $p,q$ be $P,Q.$ if the two tangents from $T$ meet the circle in $X,Y.$
Since the polar of $P$ passes through $T,$ from La Hire's theorem we see that the polar of $T$ passes through $P,$ and similarly, $Q.$ Then $X,Y,P,Q$ are collinear. Now, from Theorem $5(a)$ we are done.
$[asy] import graph; size(10cm); draw(circle((0,0),3)); dot((1.5,4.45)); dot((1.5,1.9)); dot((6,0)); dot((1.5,2.6)); dot((1.5,-2.6)); draw((6,0)--(1.5,2.6)); draw((6,0)--(1.5,4.45)); draw((6,0)--(1.5,-2.6)); draw((6,0)--(1.5,1.9)); draw((1.5,4.45)--(1.5,-2.6)); label("$p$", (3.8,0.93), S); label("$q$", (4,2), N); label("$T$", (6.2,0), E); label("$X$", (1.5,2.6), NE); label("$Y$", (1.5,-2.6), SE); label("$P$", (1.5, 4.45), W); label("$Q$", (1.5, 1.9), W); [/asy]$
$\Box$
Theorem 6.
Finally, if four points $A,B,C,D$ form a harmonic range then their polars $a,b,c,d$ will create a harmonic pencil.
Proof.
Let $X$ be the pole of $\overleftrightarrow{ABCD},$ and let $Y=OX\cap \overleftrightarrow{ABCD}.$
If we draw four perpendicular lines $XA', XB', XC', XD'$ perpendicular to $OA,OB,OC,OD,$ respectively, then we have an interesting result. The polar of $X$ passes through $A,$ so from La Hire's, the polar of $A$ passes through $X$ , and therefore $XA'$ is the polar of $A$ with respect to our aforementioned circle. It is obvious that the pencils $X(A',B',C',D')$ and $O(A,B,C,D)$ are equiangular pencils, and since $O(A,B,C,D)$ are harmonic, therefore $X(A',B',C',D')$ are also harmonic. $\Box$

Wow, that was enough of theory for a day already! Let us quickly look into some applications of our Dual Pascal and try to find boring solutions to some problems!

$\boxed{E1}.$ Let $XY$ be the diameter of a circle with two points $P,Q$ on its circumference such that $P$ is closer to $X$ than $Q.$ If $PX$ and $QY$ intersect at $S$ outside the circle, and if the tangents at $P,Q$ to the circle meet at $R;$ then show that $RS\perp XY.$
Solution
Let $T=PQ\cap XY.$ We already know that(from Theorem 4), $S$ lies on the polar of $T.$ The polar of $R$ is $\overline{PQT},$ we see that $R$ lies on the polar of $T.$ So $SR$ is the polar of $T$ , and therefore we are done. $\Box$
$[asy] import graph; unitsize(0.9cm); draw(circle((0,0),3)); dot((-2.236,2)); dot((3,0)); dot((-3,0)); dot((1.3, 2.7)); draw((3,0)--(-12.338,0)); dot((-0.8,6)); draw((-0.8,6)--(-3,0)); draw((-0.8,6)--(3,0)); dot((-12.338,0)); draw((-12.338,0)--(1.3,2.7)); dot((-0.8, 3.8)); draw((-0.8,3.8)--(-2.236,2)); draw((-0.8,3.8)--(1.3,2.7)); label("$X$", (-3,0),SW); label("$Y$", (3,0),SE); label("$P$", (-2.236,2),NW); label("$Q$", (1.3,2.7),NE); label("$R$", (-0.8,3.8), N); label("$S$", (-0.8,6), N); label("$T$", (-12.338,0), S); [/asy]$

$\boxed{E2}.$ Let $ABCD$ be a tangential quadrilateral with incentre $I.$ Let the opposite sides of $ABCD$ meet each other at $E,F;$ ie let $E=AB\cap CD, F=AD\cap BC.$ Let its incircle touch the sides $AB,BC,CD,DA$ at $G,H,K,L,$ respectively. If $P=GK\cap HL,$ then show that $OP\perp EF.$
Solution
Note that $EG,EK$ are tangents to the incircle of $ABCD,$ so that $KG$ is the polar of $E.$ Similarly the polar of $F$ is $HL.$ Therefore the pole of the line $EF$ is $KG\cap HL=P,$ and we are done. $\Box$

$\boxed{E3}.$ (China 2006 Western MO) $AB$ is the diameter of a circle with centre $O.$ $C$ is a point on $AB,$ extended. A line through $C$ cuts the circle with centre $O$ at $D,E.$ $OF$ is a diameter of $\odot(BOD)$ which has centre $O_1.$ Let $CF\cap \odot(BOD)=G.$ Prove that $O,E,A,G$ lie on a circle.
Solution
Let $P=AE\cap BD.$ So the polar of $P$ with respect to the circle $(O)$ passes through $BA\cap DE=C$ and $AD\cap BE=J.$ So $PO\perp CJ,$ and it is obvious that $G, O, P$ are collinear.
So by considering the power of $P$ we have $PD\cdot PB=PG\cdot PO=PE\cdot PA;$ and so $E,A,O,G$ are concyclic points. $\Box$

$\boxed{E4}.$ (IMO 1985) A circle with center $O$ passes through the vertices $A$ and $C$ of triangle $ABC$ and intersects segments $AB$ and $BC$ again at distinct points $K$ and $N,$ respectively. The circumcircles of triangles $ABC$ and $KBN$ intersects at exactly two distinct points $B$ and $M.$ Prove that $\angle OMB=90^{\circ}.$
Solution
Note that the spiral similarity centered at $M$ which sends $A$ to $C$ and $K$ to $N$ sends the midpoint of $AK$ (say, $M_1$ ) to that of $CN$ (say, $M_2$ ). So $M$ is the center of unique spiral similarity that sends $A$ to $M_2$
and $C$ to $M_1$ , and thus it follows that $M,M_1,M_2,B$ are concyclic. Again since $\angle OM_2B=\angle OM_1B,$ so $O,M_1,M_2,B$ are concyclic, and $OB$ is the diameter of the common circle. So, we are done. $\Box$

$\boxed{E5}.$ Circles $\omega_1$ and $\omega_2$ meet at points $O$ and $M.$ Circle $\omega,$ centered at $O,$ meets circles $\omega_1$ and $\omega_2$ in four distinct points $A,B,C$ and $D,$ such that $ABCD$ is a convex quadrilateral. Lines $AB$ and $CD$ meet at $N_1.$ Lines $AD$ and $BC$ meet at $N_2.$ Prove that $N_1N_2\perp MO.$
Solution
Actually this is equivalent to our last problem. Note that if $\odot(ADN_1)\cap \odot(BCN_1)=K,$ then from simple angle chasing we can show that $K=\odot(AOCK)\cap \odot(BODK).$ Also it is obvious that $N_1N_2$ passes through $K\equiv M.$ All of this implies that $N_1N_1\perp MO,$ as desired.
$[asy] import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -34.22, xmax = 11.4, ymin = -14.52, ymax = 15.88; /* image dimensions */ pen fftttt = rgb(1,0.2,0.2); pen wwwwff = rgb(0.4,0.4,1); pen zzqqtt = rgb(0.6,0,0.2); /* draw figures */ draw(circle((-2,0), 5.21)); draw(circle((4,0), 6.25)); draw(circle((0,4.81), 7.84), fftttt); draw((7.84,4.94)--(xmin, 0.24*xmin + 3.05), wwwwff); /* ray */ draw((2.5,-2.62)--(xmin, 0.06*xmin-2.77), wwwwff); /* ray */ draw((-7.03,1.35)--(xmax, -0.77*xmax-4.07), wwwwff); /* ray */ draw((7.84,4.94)--(xmin, 1.42*xmin-6.16), wwwwff); /* ray */ draw((xmin, 0*xmin-4.81)--(xmax, 0*xmax-4.81), zzqqtt); /* line */ /* dots and labels */ dot((0,4.81),dotstyle); label("$O$", (-0.21,5.29), NE * labelscalefactor); dot((0,-4.81),dotstyle); label("$M$", (-0.85,-5.98), NE * labelscalefactor); dot((-7.03,1.35),dotstyle); label("$A$", (-8.29,1.54), NE * labelscalefactor); dot((-1.56,-2.87),dotstyle); label("$B$", (-2.46,-4.05), NE * labelscalefactor); dot((2.5,-2.62),dotstyle); label("$C$", (2.93,-3.52), NE * labelscalefactor); dot((7.84,4.94),dotstyle); label("$D$", (8.12,5.36), NE * labelscalefactor); dot((-32.5,-4.81),dotstyle); label("$N_2$", (-32.58,-6.05), NE * labelscalefactor); dot((0.96,-4.81),dotstyle); label("$N_1$", (1.91,-6.05), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

$\boxed{E6}.$ Let $ABC$ be a triangle with incentre $I$ . Reflections of $I$ in $BC,CA, AB$ are $X,Y,Z.$ Prove that $AX,BY,CZ$ are concurrent.
Note: This is a direct consequence of the isogonic theorem. For details, see here.
Solution
I will use a lemma.
Lemma.
If $\triangle DEF$ is the intouch triangle of $\triangle ABC,$ and if $I$ is the incentre of $\triangle ABC,$ and if $\triangle A'B'C'$ is the triangle directly homothetic to $ABC$ such that $A'\in IA, B'\in IB, C'\in IC$ ; then $A'B'C'$ is perspective to $\triangle DEF.$

Now let us come back to our original problem.
Let $DEF$ be, as usual, the intouch triangle of $\triangle ABC.$ Then let $\Omega$ be the incircle of $\triangle ABC.$ Since $XD=DI,$ we see that the polar of $X$ with respect to $\Omega$ is the perpendicular bisector of $\overline{DI}.$ Similarly, the polar of $Y$ w.r.t $\Omega$ is the perpendicular bisector of $\overline{IE}$ and the polar of $Z$ w.r.t $\Omega$ is the perpendicular bisector of $\overline{IF}.$
Let these perpendicular bisectors intersect each other to form a triangle $KLM.$
Denote by $\omega_1, \omega_2, \omega_3;$ the circles $(D,DI), (E,EI),(F,FI).$ Then $K$ is the intersection of $LK$ and $KM.$ Since $LK$ is the radical axis of $\Omega$ and $\omega_3;$ and since $KM$ is the radical axis of $\Omega$ and $\omega_2,$ therefore $K$ is the radical centre of $\omega_2,\omega_3,\Omega.$
Since $AE^2-r^2=AF^2-r^2$ where $r$ is the inradius of $ABC,$ we note that $A$ has the same power with respect to $\omega_2$ and $\omega_3.$ So $A,K,I$ all lie on the radical axis of these two circles, and therefore, collinear.
From here it is also obvious that $AK=KI,\ BL=LI,\ CM=MI$ and so $\triangle ABC$ and $\triangle KLM$ are directly similar and perspective around $I.$ So they are homothetic.
Applying the aforementioned lemma, we note that the intersections $U=LM\cap EF,\ V=MK\cap FD, \ W=KL\cap DE$ are collinear.
Since $AE$ and $AF$ are tangents to $\Omega,$ we again note that $A$ is the pole of $EF$ w.r.t $\Omega.$ Also since the polar of $X$ w.r.t $\Omega$ is $LM,$ it follows from the La Hire's theorem that the pole of $AX$ w.r.t $\Omega$ is $U.$
Due to symmetry we note that $V,\ W$ are respectively the poles of $BY, \ CZ.$ We already have seen that $U,V,W$ are collinear. Therefore it is obvious that $AX,BY,CZ$ meet at the pole of $\overline{VUW}$ w.r.t $\Omega.$ We are done. $\Box$

http://www.artofproblemsolving.com/Forum/download/file.php?id=31328&mode=view

$\boxed{E7}.$ (Polish MO Second Round 2012) Let $ABC$ be a triangle with $\angle A=60^{\circ}$ and $AB\neq AC$ , $I$ -incentre, $O$ -circumcentre. Prove that perpendicular bisector of $AI$ , line $OI$ and line $BC$ have a common point.
Solution
Assume that the incircle touches $BC, CA, AB$ at $D,E,F$ and that $X=AI\cap \odot(DEF).$ Since $\angle FIA=60^{\circ},\angle EIA=60^{\circ},$ therefore $XEI$ and $XFI$ are equilateral. Also $\angle XFA=30^{\circ}=\angle FAX,$ therefore $AX=XE=IX,$ and the incircle of $ABC$ bisects $AI.$
Now, we claim that $OI$ is the perpendicular bisector of $DX.$
Proof of claim.
Let $I'$ be the reflection of $I$ in $BC.$ Then note that we have $\angle BI'C=\angle BIC=120^{\circ},$ therefore $I'$ lies on $\odot(ABC).$ Now we already have $AI=II'=2r, \ I'O= AO,$ which give $OIA\cong OIA',$ leading to the fact that $OI$ bisects $AI'.$ This readily implies that $OI$ bisects $DX.$
$[asy] import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.53, xmax = 14.86, ymin = -6.22, ymax = 13.04; /* image dimensions */ pen qqwuqq = rgb(0,0.39,0); pen zzttqq = rgb(0.6,0.2,0); draw(arc((0,8.98),1.3,-102.59,-42.59)--(0,8.98)--cycle, qqwuqq); draw((0,8.98)--(-2.01,0)--(9.77,0)--cycle, zzttqq); /* draw figures */ draw((0,8.98)--(-2.01,0), zzttqq); draw((-2.01,0)--(9.77,0), zzttqq); draw((9.77,0)--(0,8.98), zzttqq); draw(circle((1.85,3.09), 3.09), linewidth(1.6) + red); draw((0,8.98)--(1.85,3.09)); draw((xmin, 0.31*xmin + 5.75)--(xmax, 0.31*xmax + 5.75)); /* line */ draw((3.88,3.4)--(xmin, 0.15*xmin + 2.81)); /* ray */ draw((9.77,0)--(xmin, 0*xmin + 0)); /* ray */ draw((0.92,6.04)--(1.85,0)); draw(circle((3.88,3.4), 6.8)); draw((1.85,3.09)--(1.85,-3.09)); /* dots and labels */ dot((0,8.98),dotstyle); label("$A$", (0,9.23), NE * labelscalefactor); dot((-2.01,0),dotstyle); label("$B$", (-2.45,-0.9), NE * labelscalefactor); label("$60^\circ$", (-0.15,7.8), NE * labelscalefactor,qqwuqq); dot((9.77,0),dotstyle); label("$C$", (9.58,-0.86), NE * labelscalefactor); dot((1.85,3.09),dotstyle); label("$I$", (2.01,3.34), NE * labelscalefactor); dot((3.88,3.4),dotstyle); label("$O$", (4.04,3.65), NE * labelscalefactor); dot((1.85,0),dotstyle); label("$D$", (2.09,-0.81), NE * labelscalefactor); dot((3.94,5.36),dotstyle); label("$E$", (4.13,5.64), NE * labelscalefactor); dot((-1.17,3.76),dotstyle); label("$F$", (-1.8,4.04), NE * labelscalefactor); dot((0.92,6.04),dotstyle); label("$X$", (0.97,6.24), NE * labelscalefactor); dot((1.85,-3.09),dotstyle); label("$I'$", (2.01,-3.93), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Coming back to our main proof, note that we have $IO$ as the perpendicular bisector of $DX,$ so $OI$ passes through the pole of $DX$ wrt $\odot(DEF).$ Since pole of $DX$ is $XX\cap BC,$ and because $XX$ is the perpendicular bisector of $AI,$ we are done. $\Box$

$\boxed{E8}.$ (2012 Indonesia Round 2 TST 4: P2) Let $\omega$ be a circle with center $O$ , and let $l$ be a line not intersecting $\omega$ . $E$ is a point on $l$ such that $OE$ is perpendicular with $l$ . Let $M$ be an arbitrary point on $M$ different from $E$ . Let $A$ and $B$ be distinct points on the circle $\omega$ such that $MA$ and $MB$ are tangents to $\omega$ . Let $C$ and $D$ be the foot of perpendiculars from $E$ to $MA$ and $MB$ respectively. Let $F$ be the intersection of $CD$ and $OE$ . As $M$ moves, determine the locus of $F$ .
Solution(hatchguy)
Let $E'$ be the inverse of $E$ wrt $\omega.$ I claim $F$ is the midpoint of $EE'$ . Clearly, since $M$ is in the polar of $E'$ then $E'$ is in the polar of $M$ , and therefore $A,E',B$ are collinear. It is easy to see that $M,E,B,O,A$ lie on a circle with diameter $MO$ . From this, it's very natural to think of droping a perpendicular from $E$ to $AB$ . Let $G$ be the foot of this perpendicular. By Simson's line theorem, we have $C,D,F,G$ are collinear. Also, using that $DEGB$ is cyclic we easily get $\angle FGE = \angle DBE = \angle MOE = \angle FEG;$ The last because $MO \parallel EG$ . Hence $FG= FE$ and we get that $F$ is midpoint of $EE'$ since $EGE'$ is a right triangle. We are done.
$[asy] import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.41, xmax = 16.58, ymin = -8.05, ymax = 10.3; /* image dimensions */ pen ffwwww = rgb(1,0.4,0.4); pen ttttff = rgb(0.2,0.2,1); pen zzqqqq = rgb(0.6,0,0); /* draw figures */ draw(circle((0,0), 8), ffwwww); draw((0,0)--(6.69,8.81), ttttff); draw((xmin, -0.76*xmin + 13.9)--(xmax, -0.76*xmax + 13.9), zzqqqq); /* line */ draw((10.93,7.39)--(7.58,-2.55), ttttff); draw((10.45,5.96)--(1.66,7.83), ttttff); draw((10.93,7.39)--(5.1,6.71), ttttff); draw((6.69,8.81)--(6.28,6.85), ttttff); draw((6.69,8.81)--(10.93,7.39), ttttff); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (0.01,-0.71), NE * labelscalefactor); dot((6.69,8.81),dotstyle); label("$E$", (7.02,9.02), NE * labelscalefactor); dot((10.45,5.96),dotstyle); label("$M$", (10.4,5.14), NE * labelscalefactor); dot((1.66,7.83),dotstyle); label("$A$", (1.82,8.07), NE * labelscalefactor); dot((7.58,-2.55),dotstyle); label("$B$", (7.8,-3.02), NE * labelscalefactor); dot((6.28,6.85),dotstyle); label("$C$", (6.52,7.08), NE * labelscalefactor); dot((10.93,7.39),dotstyle); label("$D$", (11.22,7.58), NE * labelscalefactor); dot((5.1,6.71),dotstyle); label("$F$", (5.24,6.13), NE * labelscalefactor); dot((3.5,4.61),dotstyle); label("$E'$", (3.72,4.03), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
$\Box$

$\boxed{E9}.$ Let incircle $\omega$ of $\triangle ABC$ be tangent to $BC , CA , AB$ in $D, E , F$ . A line from $A$ that is parallel to $DE$ meets $DF$ in $K$ and a line from $A$ that is parallel to $DF$ meets $DE$ in $L$ . If $M,N$ be midpoint of $AB , AC$ , show that $KL$ is on $MN.$
Solution
From simple angle chasing, we see that $\angle DEN=\angle KAE=\angle KFE.$ This leads to the fact that $A,F,K,I,L,E$ lie on a circle with $AI$ as diameter. Thus, $IK\perp KA;$ leading to $C,I,K$ being collinear. Let $H$ be the orthocentre of $\triangle BIC.$ Then note that polar of $K$ passes through the pole of $DF$ wrt $\omega,$ which is $B.$ Also since $IK'\perp DE,$ it follows that $K'=BH\cap CI.$ Similarly, $L'=CH\cap BI.$ Therefore, the pole of $KL$ wrt $\omega$ is $H.$ Thus, $HI\perp KL,$ and $HI\perp BC,$ leading to $KL\parallel MN.$
Let $L_1=AL\cap BC,$ then note that we have $BAL_1$ is $B-$ isosceles, and because $BI\perp AL_1,$ therefore $L$ is the midpoint of $L_1A.$ Thence $ML\parallel BC.$ Similarly $NK\parallel BC.$ Using all these, we see that $M,N,K,L$ lie on a line parallel to $BC.$
$[asy] import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.54, xmax = 8.19, ymin = -1.32, ymax = 8.99; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((0,4.91)--(-3.11,0)--(5,0)--cycle, zzttqq); /* draw figures */ draw((0,4.91)--(-3.11,0), zzttqq); draw((-3.11,0)--(5,0), zzttqq); draw((5,0)--(0,4.91), zzttqq); draw(circle((0.35,1.9), 1.9), linewidth(1.6) + red); draw((0.35,0)--(1.68,3.26)); draw((0.35,0)--(-1.26,2.92)); draw((xmin, 2.45*xmin + 4.91)--(xmax, 2.45*xmax + 4.91)); /* line */ draw((xmin, -1.82*xmin + 4.91)--(xmax, -1.82*xmax + 4.91)); /* line */ draw((-1.56,2.46)--(2.5,2.46)); draw(circle((0.17,3.41), 1.51)); /* dots and labels */ dot((0,4.91),dotstyle); label("$A$", (0.2,5.05), NE * labelscalefactor); dot((-3.11,0),dotstyle); label("$B$", (-3.34,-0.42), NE * labelscalefactor); dot((5,0),dotstyle); label("$C$", (5.01,-0.44), NE * labelscalefactor); dot((0.35,0),dotstyle); label("$D$", (0.36,-0.51), NE * labelscalefactor); dot((1.68,3.26),dotstyle); label("$E$", (1.77,3.41), NE * labelscalefactor); dot((-1.26,2.92),dotstyle); label("$F$", (-1.7,3.03), NE * labelscalefactor); dot((-1,2.46),dotstyle); label("$K$", (-0.8,2.62), NE * labelscalefactor); dot((1.35,2.46),dotstyle); label("$L$", (0.84,2.62), NE * labelscalefactor); dot((-1.56,2.46),dotstyle); label("$M$", (-2.1,2.46), NE * labelscalefactor); dot((2.5,2.46),dotstyle); label("$N$", (2.61,2.5), NE * labelscalefactor); dot((0.35,1.9),dotstyle); label("$I$", (0.27,2.09), NE * labelscalefactor); dot((0.35,8.46),dotstyle); label("$H$", (0.24,8.66), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Result:
This problem teaches us that the polar of the orthocentre of $BIC$ wrt $\omega$ bisects $AB$ and $AC.$ This is a very useful fact. Don't ask me why, because even I don't know where it may come in handy.

$\Box$

$\boxed{E10}.$ (Romania MOM 2012)Let $ABC$ be a triangle and let $I$ and $O$ denote its incentre and circumcentre respectively. Let $\omega_A$ be the circle through $B$ and $C$ which is tangent to the incircle of the triangle $ABC$ ; the circles $\omega_B$ and $\omega_C$ are defined similarly. The circles $\omega_B$ and $\omega_C$ meet at a point $A'$ distinct from $A$ ; the points $B'$ and $C'$ are defined similarly. Prove that the lines $AA',BB'$ and $CC'$ are concurrent at a point on the line $IO$ .

Solution(r1234)
Firstly, we'll cite a lemma:
Lemma.
Let $\ell$ be a line and $\Gamma$ be circle.Suppose $P$ is the pole of $\ell$ wrt $\Gamma$ .Now let $\triangle ABC$ be inscribed in $\Gamma$ .Let $\triangle A'B'C'$ be the circumcevian triangle of $\triangle ABC$ wrt $P$ .Then $\ell$ is the perspective axis of $\triangle ABC$ and $\triangle A'B'C'$ .
Proof of lemma.
Let $A_1=BC\cap B'C', B_1=AC\cap A'C', C_1=AB\cap A'B'$ .
Now applying Pascal's theorem on the hexagon $B'C'ABCA'$ we get $A_1,C_1, C'A\cap CA'$ are collinear.
Again $\triangle AC'B$ and $\triangle A'CB'$ are perspective.So $C_1, C'A\cap CA', BC'\cap B'C$ are collinear.Hence we conclude that $A_1,C_1, BC'\cap B'C$ are collinear.But this line is nothing but the polar of $P$ i.e $\ell$ .Hence we conclude that $A_1B_1C_1\equiv \ell.\Box$

Coming back to the main proof,
Clearly $AA'$ is the radical axis of $\omega _b, \omega_c$ , $BB'$ is the radical axis of $\omega_c, \omega_a$ and $CC'$ is the radical axis of $\omega_a, \omega_b$ .So by radical axis theorem $AA', BB', CC'$ are concurrent.
Let $(I)$ be the incircle of $\triangle ABC$ and $\triangle DEF$ is its intouch triangle. $D', E', F'$ are the touch points of $(I)$ with $\omega_a, \omega_b, \omega_c$ respectively.Now let tangents at $D', E', F'$ meets $BC, CA, AB$ at $X,Y,Z$ respectively.Then its easy to show that $XYZ$ is the radical axis of $(I)$ and $\odot ABC$ .So $X$ is the pole of $DD'$ wrt $(I)$ and similar for others.So $DD', EE', FF'$ concur at the pole of $XYZ$ wrt $(I)$ .Now consider the circles $(I), \omega_b, \omega_c$ .Then by radical axis theorem the lines $AA', E'E', F'F'$ are concurrent ,say at $X_1$ .Then $X_1$ is the pole of $E'F'$ wrt $(I)$ .Since $A, X_1, A'$ are collinear, their polars i.e $EF,E'F',\text{Polar of A'}$ are concurrent.So $\triangle DEF$ and the triangle formed by the polars of $A', B', C'$ are perspective wrt the perspective axis of $\triangle DEF, \triangle D'E'F'$ .But according to our lemma this perspective axis is the polar of the perspective point of $\triangle DEF, \triangle D'E'F'$ , i.e the radical axis of $(I), \odot ABC$ .So we conclude that $AA', BB', CC'$ concur at the pole of the radical axis of $(I), \odot ABC$ wrt $(I)$ .Since $IO\perp \text{Radical axis of (I),circumcircle of ABC}$ we conclude that the pole of the radical axis of $(I), \odot ABC$ lies on $IO$ .Hence $AA', BB', CC'$ concur on $IO$ .
$[asy] import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.06, xmax = 6.06, ymin = -1.72, ymax = 6.06; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqzzzz = rgb(0,0.6,0.6); draw((0,5.38)--(-2,0)--(5,0)--cycle, zzttqq); /* draw figures */ draw((0,5.38)--(-2,0), zzttqq); draw((-2,0)--(5,0), zzttqq); draw((5,0)--(0,5.38), zzttqq); draw(circle((0.7,1.88), 1.88), linewidth(1.6) + red); draw(circle((1.5,0.43), 3.53), qqzzzz); draw(circle((2.36,2.56), 3.68), qqzzzz); draw(circle((-0.27,2.42), 2.97), qqzzzz); draw((0,5.38)--(0.33,-0.5)); draw((2.34,3.85)--(-2,0)); draw((-1.31,2.55)--(5,0)); draw((0.19,1.95)--(1.5,1.76)); /* dots and labels */ dot((0,5.38),dotstyle); label("$A$", (-0.13,5.62), NE * labelscalefactor); dot((-2,0),dotstyle); label("$B$", (-2.24,-0.25), NE * labelscalefactor); dot((5,0),dotstyle); label("$C$", (5.1,-0.22), NE * labelscalefactor); dot((0.7,1.88),dotstyle); label("$I$", (0.75,1.98), NE * labelscalefactor); dot((-0.36,3.42),dotstyle); label("$D'$", (-0.38,3.56), NE * labelscalefactor); dot((0.7,0),dotstyle); label("$D$", (0.64,-0.31), NE * labelscalefactor); dot((2.07,3.15),dotstyle); label("$E$", (2.15,3.26), NE * labelscalefactor); dot((-1.06,2.53),dotstyle); label("$F$", (-0.97,2.5), NE * labelscalefactor); dot((-1.08,1.27),dotstyle); label("$E'$", (-1,1.37), NE * labelscalefactor); dot((2.2,0.76),dotstyle); label("$F'$", (2.32,0.55), NE * labelscalefactor); dot((1.5,1.76),dotstyle); label("$O$", (1.57,1.86), NE * labelscalefactor); dot((0.33,-0.5),dotstyle); label("$A'$", (0.24,-0.83), NE * labelscalefactor); dot((2.34,3.85),dotstyle); label("$B'$", (2.41,3.96), NE * labelscalefactor); dot((-1.31,2.55),dotstyle); label("$C'$", (-1.63,2.7), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
$\Box$

As for exercises, I won't give any for this post. I will just write down some references from where you can get plenty of problems. ^_^

References.
1. Cyclic Quadrilaterals - The Big Picture by Yufei Zhao.
2. Power of a Point by Yufei Zhao.
3. Circles by Yufei Zhao.
4. Poles and Polars by Kin Y. Li.
5. Mathscope Topic on Poles and Polars by Hoàng Quốc Khánh.
6. Introduction to the geometry of triangle by Paul Yiu.

This post has been edited 2 times. Last edited by levans, Jan 15, 2016, 4:35 AM

A digression on Calculus... From introductory to advanced?

by Potla, Dec 31, 2011, 6:02 PM

Before anything, wish everyone a happy and prosperous new year full of joy and success!

Surprised, aren't you? Well, though it may be out of your expectations, but I came across a lot of beautiful problems. So I want to introduce a little bit of graphs(is that really necessary?).
Without wasting any more time, let's continue with this hotch-potch discussion on Calculus. Yeah, I know that I am not organised, sorry about that.
I don't know from where to start, so anyway, let me start with the fundamentals of calculus problem-solving. The problems that we(of course, 'we' means that I am going to follow the Indian curriculum of Calculus) usually solve consist of the following concepts: Graphs and Functions, Limits, continuity and differentiability, function-plotting.
So anyway, let me start with some basic concepts of Polynomials.
Polynomials.
These may be defined as functions (that's my way of looking at them) which have terms of integral powers and real or complex coefficients. For the sake of calculus of our level though, we only look at polynomials $P[\mathbb R](x)$ which is generally given by
$P(x)\equiv a_0x^n+a_1x^{n-1}+\cdots+a_n;$ given $a_i\in\mathbb R\forall i=1,2,\cdots n.$
The degree of a polynomial $\deg{P(x)}$ is the highest power of $x$ that is contained in $P.$ If $a_0\neq 0,$ then the degree of $P$ is $n.$
Polynomials of even degree
If $n$ is even, then we can only have two types of graphs possible.
$P(x)=a_0x^n+a_1x^{n-1}+\cdots+a_n=x^n\left(a_0+\frac{a_1}{x}+\cdots+\frac{a_n}{x^n}\right).$
Assume that $x\to\infty,$ so that $\lim_{x\to\infty}P(x)=\lim_{x\to\infty}a_0x^n=\left\{\begin{aligned}&+\infty, a_0>0\\&-\infty, a_0<0.\end{aligned}\right.$
And, $\lim_{x\to-\infty}P(x)=\lim_{x\to-\infty}a_0x^n=\left\{\begin{aligned}&+\infty, a_0>0\\&-\infty, a_0<0.\end{aligned}\right.$
So the graph can be of two different looks:
$[asy] import graph; unitsize(1cm);size(8cm); real f(real x) {return (36-27x/100-13x^2/10000+3x^3/1000000+x^4/100000000);} pair F(real x) {return (x,f(x));} xaxis("$x$"); yaxis("$y$"); draw(graph(f,-600,500,operator ..),red); [/asy]$
$[asy] import graph; unitsize(1cm);size(8cm); real f(real x) {return (-x^4/100000000+200 x^3/100000000+370000 x^2/100000000-86000000 x/100000000-120);} pair F(real x) {return (x,f(x));} xaxis("$x$"); yaxis("$y$"); draw(graph(f,-650,650, operator ..),red); [/asy]$
So, a polynomial with even degree will either rise to the heaven, or fall down to hell on both sides - $x\to\infty$ and $x\to-\infty$ . It's also obvious that it will cut the $x$ axis even number of times.
Polynomials of odd degree
It's easy to deduce that the graphs of polynomials with odd degree will fall to hell at one side, and rise up to heaven on the other. A nice way of recalling is that, if $a_0>0,$ then the graph will rise up towards heaven at $x\to\infty$ and vice versa. Like,
$[asy] import graph; unitsize(1cm);size(8cm); real f(real x) {return (x^3/1000000+x^2/5000-(29 x)/100-30);} pair F(real x) {return (x,f(x));} xaxis("$x$"); yaxis("$y$"); draw(graph(f,-700,600,operator ..),red); [/asy]$
$[asy] import graph; unitsize(1cm);size(8cm); real f(real x) {return (-x^3/1000000-x^2/5000+(29 x)/100-30);} pair F(real x) {return (x,f(x));} xaxis("$x$"); yaxis("$y$"); draw(graph(f,-750,600, operator ..),red); [/asy]$
Roots
The roots of a polynomial are the points where the curve cuts the $x$ -axis. We can use Descartes sign rule to determine the number of positive or negative roots. Let $P(x)$ be a certain polynomial, which has $'n'$ number of sign changes as while proceeding from the lowest to the highest power (ignoring zero coefficients), has a maximum of $n$ positive roots. It may have $n-2, n-4$ etc positive roots too. When this rule is applied to $P(-x),$ we can guess the maximum allowed negative roots.

The derivative
Let $f(x)$ be a function. Then $f'(x_0)$ is the derivative at a point $x_0,$ which is defined as $f'(x_+)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}, f'(x_-)=\lim_{h\to0}\frac{f(x)-f(x-h)}{h}.$ If $f'(x_+)=f'(x_-)=f'(x)$ in an interval $(a,b),$ then the function is said to be differentiable over all points in $(a,b).$ Then $f'(x_0)$ is the slope of the tangent to the curve $f(x)$ at $x=x_0.$ A function is differentiable only if it's continuous, but the reverse is not true.
Again, I do not intend on lecturing on removable and blah blah continuity, so I am assuming prior knowledge without loss of generality. After all, it's tiring to mention everything, lol. So let's get into some serious stuff.

Convexity, concavity, monotonicity.
A function is said to be monotonically, or strictly increasing, or decreasing if and only if $f(a)\geq f(b), f(a)>f(b)$ and its reverses respectively hold for any $a,b$ in a certain interval.
A function f(x) defined on an interval is called convex (or convex downward/concave upward) if the graph of the function lies below the line segment joining any two points of the graph. The formal definition is, a function $f:\mathbb I\to\mathbb R$ is convex if and only if for any two points $x_1,x_2\in\mathbb I$ and $\lambda\in(0,1)$ we have $f(\lambda x_1+(1-\lambda)x_2)\leq \lambda f(x_1)+(1-\lambda)f(x_2).$ In case the equality doesn't occur, the function is called to be strictly convex.
Concavity is just the opposite.
The necessary and sufficient condition for a certain function $f(x)$ to be continuous in an interval $(a,b)$ is that the function must lie above all of its tangents, ie $f(x)\geq f(y)+(x-y)f'(y).$ Concavity: reverse the sign.

This may also be rephrased as $f''(x)\geq 0$ for all $x$ in $\mathbb I.$ We may show that this is another necessary and sufficient condition for convexity.
Jensen's inequality.
Let $f$ be a convex function of one real variable. Let $x_1,\cdots,x_n\in\mathbb R$ and let $a_1,\cdots, a_n\ge 0$ satisfy $a_1+\dots+a_n=1.$ Then
$f(a_1x_1+\cdots+a_n x_n)\le a_1f(x_1)+\cdots+a_n f(x_n).$
For a proof, look here.

Without any more of introduction, let me move on to one of the most important theorems in Calculus, ie Rolle's theorem.
Rolle's Theorem
If $f(x)$ is continuous and differentiable at every point in an interval $[a,b],$ and if $f(a)=f(b),$ then there exists a certain $x_0\in[a,b]$ such that $f'(x_0)=0.$
From graphical considerations, it is obvious that the function must have changed its slope from either positive to negative, or negative to positive at some point $x_0$ in the given interval. It can contradict this at the cost of its differentiability everywhere.

Mean Value Theorem (MVT for short)
If $f(x)$ and $g(x),$ and $f'(x), g'(x)$ are continuous throughout an interval $[a,b],$ and if $f'(x)\neq 0$ everywhere in the given interval, then there will always exist a point $x_0\in[a,b]$ such that
$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(x_0)}{g'(x_0)}.$
Proof.
Consider the function $\phi(x)=\frac{f(b)-f(a)}{g(b)-g(a)}[g(x)-g(a)]-[f(x)-f(a)].$
Since $\phi(a)=\phi(b)=0,$ so applying Rolle's theorem we see that
$\phi'(x)=\frac{f(b)-f(a)}{g(b)-g(a)}g'(x)-f'(x)=0$ for some point $x=x_0.$ Hence done.
Geometric interpretation
For the non-general version of Rolle's theorem for $g(x)=x,$ we can see that the slope of the line joining $(a,f(a))$ and $(b,f(b))$ is $\frac{f(b)-f(a)}{b-a},$ and $f'(x)$ is the slope at $x=x.$ So, obviously there exists a certain $x_0$ in the interval such that $f'(x_0)$ equals the slope of the chord joining the terminal points.
$[asy] import graph; unitsize(1cm);size(8cm); real f(real x) {return (x^3/1000-2x^2/100+2x/10+1);} pair F(real x) {return (x,f(x));} xaxis("$x$"); yaxis("$y$"); draw(graph(f,-10,37,operator ..),red); dot((10,2)); dot((30,16)); label("$A$", (10,2), N); label("$B$", (30,16), NW); draw((10,2)--(30,16)); dot((21.1963,5.776)); draw((10,-2.0606)--(30,11.939), blue); label("$P$", (21.1963,5.776), S); [/asy]$
Extended MVT
Define a constant $R$ such that the equation
$f(b)-f(a)-(b-a)f'(a)-\frac 12(b-a)^2R=0$
Is satisfied. Define a function $F(x)$ as
$F(x)=f(x)-f(a)-(x-a)f;(a)-\frac 12 (x-a)^2R.$
Since $F(a)=F(b)=0$ , therefore $F'(x)=f'(x)-f'(a)-(x-a)R,$ and
$F'(x_1)=f'(x_1)-f'(a)-(x_1-a)R=0$ for some $x_1\in[a,b].$
Since $F'(a)=F'(b)=0,$ therefore $\exists x_2\in[a,b]; \ni F''(x_2)=0,$ and $R=f''(x_2).$ Substituting $R$ , we get
$f(b)=f(a)+(b-a)f'(a)+\frac{(b-a)^2}{2!}f''(x_2) \ \ (x_2\in (a,b)).$
Continuing, we obtain,
$\begin{aligned}f(b)=f(a)+\frac{(b-a)}{1!}f'(a)+\frac{(b-a)^2}{2!}f''(a)+\cdots+\frac{(b-a)^n}{n!}f^{(n)}(a)\\ +\frac{(b-a)^{n+1}}{(n+1)!}f^{(n+1)}(x_1);\end{aligned}$ Where $x_1\in(a,b).$ This expression is known as the extended MVT.

With this, I am wrapping up all my discussions on the theorems that we study in high school (except for the extended MVT, everything is quite fundamental).
And, it also means that I will now move onto some nice problems. Most of the problems that I will discuss in this post will be involving construction of functions and applications of Rolle's theorem, etc. So, let's change the mood from theoretic to a little of applications.
Problems.
1. Given two functions $f$ and $g$ , continuous on $[a, b],$ differentiable on $(a, b)$ , and $f(a)=f(b) = 0.$ Prove that there exists a point $c\in(a, b)$ such that $g'(c)f(c) + f'(c) = 0.$
Solution
Define $h(x)=f(x)e^{g(x)},$ then $h(a)=h(b)=0$ and so, there exists a $c$ in $(a,b)$ such that $h'(c)=0.$ But, note that $h'(c)=e^{g(x)}\left(f'(x)+f(x)g'(x)\right)=0,$ and this leads us to our desired result. $\Box$

2. Let $f$ be a function from reals to reals with at least two roots $a<b$ . Prove that for any real number $k$ there is $c\in (a,b)$ such that $f(c)+kf'(c)=0$ .Solution(Virgil Nicua)
If $k=0$ , then can choose $c\in\{a,b\}$ . Suppose $k\ne 0$ . In this case consider $g:[a,b]\rightarrow \mathbb R$ , Where $g(x)=e^{\frac xk}\cdot f(x)$ . Observe that $g$ satisfies $g(a)=g(b)=0$ and $g'(x)=e^{\frac xk}\cdot\left[f'(x)+\frac 1k\cdot f(x)\right]$ , so by Rolle's theorem we get our desired result. $\Box$

3. Show that for two functions $f,g$ such that $g(x)\neq 0\forall x\in\mathbb R,$ and given $s,r\in\mathbb R,$ we have a certain $c$ between any two roots $a,b$ of $f$ such that
$\frac{f'(x)}{g'(x)}=\frac{s}{r}\cdot\frac{f(x)}{g(x)}.$
Solution
Consider $h(x)=\frac{f(x)^r}{g(x)^s}.$ Indeed, $h(a)=h(b)=0,$ so there exists at least one $c$ between $a,b$ such that
$\frac{\text{d}}{\text{d} \ x}\left(\frac{f(x)^r}{g(x)^s}\right) = \frac{f(x)^{r-1}}{g(x)^{s+1}}(r\cdot g(x)f'(x)-s\cdot f(x)g'(x))=0;$ Which leads to our desired result. $\Box$

4. Solve in $\mathbb{R},$ the following equation:
$2^x-x=1$ .
Solution
Though it's obvious from graphs that $x=0,1$ are the only solutions, why not apply Rolle's? Suppose $f(x)=2^x-x-1$ has at least three distinct roots. By Rolle's theorem, $f'(x)= 2^x \log 2 - 1$ should have at least two distinct roots; however $f'$ is strictly increasing and so this cannot happen. So, $f$ has at most two distinct roots, namely $0$ and $1.$

5. (Amparvardi) Let $f: [a,b] \to \mathbb R$ , $0<a<b$ be a function which is continuous and differentiable in $(a,b)$ . Prove that there exists a real number $c \in (a,b)$ for which
$f'(c)=\frac1{a-c}+\frac1{b-c}+\frac1{a+b}.$
Solution
Let us assume $\displaystyle g(x)=(x-a)(x-b)\cdot e^{f(x)-\frac x{a+b}}.$ Note that $g(a)=g(b)=0,$ and also
$\begin{aligned}g'(x)&={e^{f(x)-\frac x{a+b}}}\cdot\left(2x-a-b-\frac{(x-a)(x-b)}{a+b}+(x-a)(x-b)f'(x)\right)\\&=(x-a)(x-b){e^{f(x)-\frac x{a+b}}}\left(f'(x)+\frac 1{x-a}+\frac 1{x-b}-\frac1{a+b}\right);\end{aligned}$ So from Rolle's theorem, there exists at least one $c\in(a,b)$ such that $g'(c)=0,$ and we are done. $\Box$

Let's see some LMVT problems now.
6. Show that $\frac{x-1}{x} \leq \ln x \leq x-1$ for $x\in(0,\infty).$
Solution
Consider the function $f(x)=\ln x$ , it is continuous in $[1,x]$ and also differentiable in $(1,x).$ Applying LMVT, there must be one $c$ such that $f'(c) =\frac{\ln x}{x-1}$
But $1 \leq c \leq x$ , therefore $\frac{1}{x} \leq \frac{1}{c } \leq 1 \implies \frac{1}{x} \leq \frac{\ln x}{x-1} \leq 1,$ and hence we get our desired inequality.

7. Prove that for any $x\in\left(0,\frac{\pi}{2}\right),$ we have
$\left(\frac x{\sin x}\right)^{\tan x-x}>\left(\frac {\tan x}{x}\right)^{x-\sin x}.$
Solution(Virgil Nicua)
Since $0<\sin x<x<\tan x,$ we can rewrite the given problem into
$\frac{\ln x-\ln \sin x}{x-\sin x}>\frac{\ln\tan x-\ln x}{\tan x-x}.$
Apply LMVT to the function $\ln:(0,\infty)\to\mathbb R$ on the segments $[\sin x, x]$ and $[x,\tan x].$ So there exist $0<\sin x<c<x<d<\tan x$ satisfying $\frac{\ln x-\ln \sin x}{x-\sin x}=\frac 1c$ and $\frac{\ln\tan x-\ln x}{\tan x-x}=\frac 1d.$ So, $0<c<d\implies$
$\frac{\ln x-\ln \sin x}{x-\sin x}>\frac{\ln\tan x-\ln x}{\tan x-x}\ \ \ \forall x\in \left(0,\frac {\pi}{2}\right).\Box$

8. Let $p(x)=ax^2+bx+c$ be a polynomial with real roots. Given that $a$ and $24a+7b+2c$ have the same sign, prove that it's impossible for both roots of $p(x)$ to be in the interval $(3,4)$ .
Solution(marvopnema)
Assume the two roots of $p(x)$ are both real, and situated in $(3,4)$ , hence of the form $3+u, 3+v$ , with $0<u,v<1$ . From Vieta's relations we have $-\dfrac {b} {a} = (3+u) + (3+v) = 6 + (u+v)$ and $\dfrac {c} {a} = (3+u)(3+v) = 9 + 3(u+v) + uv$ .
Then $\dfrac {1} {a} (24a + 7b + 2c) = 24 - 42 - 7(u+v) + 18 + 6(u+v) + 2uv$ $= 2uv - (u+v) \leq 2uv - 2\sqrt{uv}$ $= 2\sqrt{uv}(\sqrt{uv} - 1) < 0$ , hence $a$ and $24a+7b+2c$ are of different signs. $\Box$

9. While $a>0$ and $n\in\mathbb N,$ find the limit
$\lim_{n\to\infty}n^2\left(\sqrt[n]{a}-\sqrt[n+1]{a}\right).$
Solution(hsbatt)
By LMVT,
$\begin{aligned}n^2 \left(a^{\frac{1}{n}} - a^{\frac{1}{n+1}} \right) &= n^2 a^{\frac{1}{t}} \ln a \left( \frac{1}{n}-\frac{1}{n+1} \right)\\&= \frac{n^2 a^{\frac{1}{t}} \ln a }{n(n+1)}\end{aligned}$
For some $t \in (n,n+1).$
Now, let's sandwich the expression between $\frac{n^2 a^{\frac{1}{n}} \ln a }{n(n+1)}$ and $\frac{n^2 a^{\frac{1}{n+1}} \ln a }{n(n+1)}$
Because both the limits of the left and the right bound are $\ln a,$ therefore the function bounded in-between must also have a limit $\ln a$ and $n\to \infty.\Box$

10. Without integrating, show that the sum $S_n=\sum_{k=1}^n\frac 1{k\sqrt k}$ converges.
Solution
Consider $f:[k,k+1]\to\mathbb R,$ where $f(x)=\frac 1{\sqrt x}$ and $k=1,2,\cdots n-1.$ Observe that $f'(x)=-\frac 1{2x\sqrt x}.$ By LMVT, there must exist a certain $c\in[k,k+1]$ such that $f(k+1)-f(k)=f'(c).$ Since $f'$ is increasing, we get
$f'(k)<f(k+1)-f(k)<f'(k+1);$
Or,
$\frac 1{2k\sqrt k}>\frac 1{\sqrt k}-\frac 1{\sqrt{k+1}}>\frac 1{2(k+1)\sqrt{k+1}}.$
Summing up from $1$ to $n,$ we get
$S_n<2\left(1-\frac 1{\sqrt{n+1}}\right)>S_n-1+\frac1{\sqrt{n+1}};$
Or,
$2\left(1-\frac 1{\sqrt{n+1}}\right)<S_n<3\left(1-\frac 1{\sqrt{n+1}}\right).$
So, the sequence is increasing and bounded above and below. So we see that $2-\sqrt 2\leq S_n\leq 3,$ and $S_n$ converges. $\Box$

11. Let $p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ be a monic polynomial of degree $n>2$ , with real coefficients and all its roots real and different from zero. Prove that for all $k=0,1,2,\cdots,n-2$ , at least one of the coefficients $a_k,a_{k+1}$ is different from zero.
Solution
Lemma 1
If all the roots of $p$ are real, then all the roots of $p'$ are real.
* The proof of this lemma is clear, because between two consecutive distinct roots of $p$ there is a root of $p'$ (using Rolle's theorem), and we have that if $r$ is a root of $p$ with multiplicity $m\geq 2$ , then it's a root of $p'$ with multiplicity $m-1$ .
Lemma 2
At least one of $a_1,a_2$ is different from $0$
* Suppose that the (possibly repeated) roots of $p$ are $r_1,r_2,\ldots,r_n$ , and that $a_1=a_2=0$ .
Then from Viete's formulas we get that:
$\begin{cases}a_1=r_1r_2\ldots r_n\left(\frac{1}{r_1}+\ldots\frac{1}{r_n}\right)=0\\ a_2=r_1r_2\ldots r_n\left(\frac{1}{r_1r_2}+\frac{1}{r_1r_3}+\ldots+\frac{1}{r_{n-1}r_n}\right)=0\end{cases}$ As the $r_i$ are nonzero, this immediately implies that $\frac{1}{r_1^2}+\ldots+\frac{1}{r_n^2}=\left(\frac{1}{r_1}+\ldots\frac{1}{r_n}\right)^2-2\left(\frac{1}{r_1r_2}+\frac{1}{r_1r_3}+\ldots+\frac{1}{r_{n-1}r_n}\right)=0$ , a clear contradiction. This concludes the proof of lemma 2. $\Box$
We now prove that the statement is true by strong induction on $n\geq 1$ :
Base cases: $n=1,2$
If $n=1$ then there is nothing to prove, and if $n=2$ then the result is true because $a_0\neq 0$ (because $0$ is not a root).
Induction step: Assume that $n\geq 3$ .
Case 1: $a_1\neq 0$
In this case we have that $p'$ is a polynomial with all roots real and nonzero (using lemma 1), so we may use the induction hypothesis on $\frac{p'(x)}{n}$ (we divide by $n$ just to make the polynomial monic, a minor technicality to allow us to use the hypothesis). As the coefficients of this polynomial are nonzero multiples of the coefficients of $p$ (except $a_0$ ), we conclude that there are no consecutive zero coefficients in $p$ .
Case 2: $a_2\neq 0$
In this case we have that $p''$ is a polynomial with all roots real and nonzero (using lemma 1 twice), so we may use the induction hypotheses on $\frac{p''(x)}{n(n-1)}$ (again, dividing by $n(n-1)$ is just a technicality), because $n-2\geq 1$ . As the coefficients of this polynomial are nonzero multiples of the coefficients of $p$ (except $a_0,a_1$ ), again we conclude that there are no consecutive zero coefficients (using lemma 2 for the coefficients $a_1,a_2$ ).
This concludes the proof of the desired statement. $\Box$

12. (a) Prove that $\sum_{k=0}^{2n-1}{\frac{x^{k}}{k!}$ has only one real root $x_{2n-1}$ .
(b) Prove that $x_{2n-1}$ is decreasing, and $\lim_{x\rightarrow\infty}{x_{2n-1}}=-\infty$ .
Solution to Part (a)
Let $P_{m}(x)=\sum_{k=0}^{m}\frac{x^{k}}{k!}$
The key to proving that $P_{2n-1}$ has exactly one real root is to show that $P_{2n}$ has no real roots.
Note that for $x\ge0,\ P_{m}(x)\ge 1.$ Hence all zeros must occur for negative $x$
By Taylor's Theorem (with Lagrange mean-value form of the remainder),
$e^{x}=P_{2n}(x)+\frac{e^{\xi}x^{2n+1}}{(2n+1)!}$
But for $x<0,\ \frac{e^{\xi}x^{2n+1}}{(2n+1)!}<0$ which implies that $P_{2n}(x)>e^{x}>0.$
Hence, $P_{2n}(x)$ has no zeros. Now consider $P_{2n+1}(x).$ As an odd degree polynomial, it must have at least one zero. However, by Rolle's Theorem, if it has two or more zeros, then its derivative must have at least one zero. But $P_{2n+1}'(x)=P_{2n}(x),$ which has no zeros. Hence, $P_{2n+1}(x)$ has exactly one zero.
Now, given any $B<0,$ we know that $P_{m}(x)$ tends to $e^{x}$ uniformly on $[B,0].$ This implies that there exists an $N,$ depending on $B,$ such that for $m>N,\ P_{m}(x)$ has no zeros in $[B,0].$ This establishes that the sequence of roots, $x_{2n-1},$ must tend to $-\infty$ as $n\to\infty.$
Solution to Part (b)
Let $Q_{k}(x) = \frac{x^{2k}}{(2k)!}+\frac{x^{2k+1}}{(2k+1)!}= \frac{x^{2k}}{(2k+1)!}\cdot \left( x+2k+1 \right)$ .
We have that $P_{2n-1}(-2n-1) = Q_{0}(-2n-1)+Q_{1}(-2n-1)+\ldots+Q_{n-1}(-2n-1)$ .
Note that $Q_{k}(-2n-1) < 0, \, \forall k \in \overline{0,n-1}$ , since $(-2n-1)+2k+1=2(k-n) < 0$ .
Hence, $P_{2n-1}(-2n-1) < 0$ . Since $P_{2n-1}$ is a polynomial of odd degree and has a unique root, we must have that
$-2n-1 < x_{2n-1}.$ Consequently,
$\begin{eqnarray*}P_{2n+1}\left( x_{2n-1}\right) &=& P_{2n-1}\left( x_{2n-1}\right)+Q_{n}\left( x_{2n-1}\right) \\ \ &=& Q_{n}\left( x_{2n-1}\right) \\ \ &=& \frac{x_{2n-1}^{2n}}{(2n+1)!}\cdot \left( x_{2n-1}+2n+1 \right) \\ \ &>& 0 .\end{eqnarray*}$
Since $P_{2n+1}$ is a polynomial of odd degree and has a unique root, we must have that $x_{2n+1}< x_{2n-1}$ , i.e. $\left\{ x_{2n-1}\right\}_{n \geq 1}$ is strictly decreasing. $\Box$

And, I finish the post here only. 12 Problems for this post!
Also, sorry for not being able to normalise. The introduction part is too easy, and the last problem uses Taylor.

This post has been edited 3 times. Last edited by Potla, Jan 1, 2012, 3:57 PM

A nice Functional Equation

by Rijul saini, Mar 18, 2011, 5:40 AM

I solved a cute functional equation today, and was also wondering what I was doing being a contributor here, so that resulted in this.
IMO Short List 2002 A01
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)$
for all real $x,y$

Solution

Harmonic Divisions - A Powerful & Rarely Used Tool!

by Potla, Feb 8, 2011, 5:56 PM

Long since I have ever posted anything on my blog, and more importantly, I have not posted anything on geometry before (except for a similarity trivia), and therefore this is my first extensive attempt on geometry.
As the title explains clearly, I want to post some useful results and solve some problems related to Harmonic Divisions, because I have not found a very good book or ebook explaining the concepts clearly.

HARMONIC DIVISIONS
Though all of us know what a harmonic division is, if four points $A,B,C,D$ are collinear and if they satisfy
$\frac{AC}{CB}=\frac {AD}{DB},$ then the four points $ACBD$ are said to form a harmonic range.
Now let us move onto some theorems.
Theorem 1.
If $(ACBD)$ is a harmonic range and if $O$ is the midpoint of $AB$ then $OB^2=OC\cdot OD.$
Proof.
After componendo and dividendo on $\frac{AC}{CB}=\frac{AD}{DB};$ we get $\frac{AC+CB}{AC-CB}=\frac{AD+DB}{AD-DB};$ leading to
$\frac{2OB}{2OC}=\frac{2OD}{2OB}\iff OB^2=OC\cdot OD.$
The converse of this theorem is also obviously true.
$[asy] dot((0,0)); dot((-5,0)); dot((+5,0)); dot((3,0)); dot((8,0)); draw((-5,0)--(8,0)); label("$A$", (-5,0), S); label("$O$", (0,0), S); label("$C$", (3,0), S); label("$D$", (8,0), S); label("$B$", (5,0), S); [/asy]$
Theorem 2.
If $AX, BY, CZ$ are three concurrent cevians of a triangle $\triangle ABC$ with $X, Y, Z$ lying on $BC, CA, AB$ respectively, then $ZY\cap BC=T\implies (BXCT)$ is a harmonic division.
Proof
Using Menelaus we have,
$\frac{|AZ|}{|BZ|}\cdot\frac{|BT|}{|CT|}\cdot \frac{|CY|}{|AY|}=1;$
And using Ceva we obtain
$\frac{|AZ|}{|BZ|}\cdot\frac{|BX|}{|CX|}\cdot\frac{|CY|}{|AY|}=1;$
Which jointly lead to $\frac{|BX|}{|CX|}=\frac{|BT|}{|CT|}\implies (BXCT)=-1;$ ie the division is harmonic.
$[asy] size(10cm, 8cm); label("$A$", (0,5), N); label("$B$", (-1,0), S); label("$C$", (5,0), S); label("$X$", (3,0), S); label("$Z$", (-0.3, 3.48), NW); label("$Y$", (2.35,2.65), NE); label("$T$", (10.82,0), S); draw((0,5)--(-1,0)); draw((-1,0)--(10.82,0)); draw((5,0)--(0,5)); draw((0,5)--(3,0)); draw((-1,0)--(2.35,2.65)); draw((5,0)--(-0.3,3.48)); draw((-0.3,3.48)--(10.82,0)); [/asy]$
Theorem 3.
If four concurrent lines are such that one transversal cuts them harmonically, then every transversal intersects these four lines to form a harmonic division.
Proof
Assume the lines $OA, OB, OC, OD$ pass through $O$ and cut by a line $d$ in $A,B,C,D$ respectively. Assume that $(ACBD)$ is harmonic, and therefore let another transversal cut these at $A',B',C',D';$ it is sufficient to check that the four points $A', B', C', D'$ form a harmonic division.
Note that using the sine rule we get
$\frac{AC}{CB}=\frac{OA}{OB}\cdot\frac{\sin\angle AOC}{\sin\angle COB};$ and $\frac{AD}{BD}=\frac{OA}{OB}\cdot\frac{\sin\angle AOD}{\sin\angle BOD}.$
Comparing these two and using the fact that $(ACBD)$ is harmonic, we get $\frac{\sin\angle AOC}{\sin\angle COB}=\frac{\sin\angle AOD}{\sin\angle BOD}.$
Now again we can use similar arguments to arrive at $\frac{A'C'}{C'B'}=\frac{A'D'}{B'D'};$ ie $(A'C'B'D')$ is harmonic.
$[asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((-2,0)--(4,0)); draw((2,0)--(10,0)); draw((0,(+12-6*0)/2)--(14.29,(+12-6*14.29)/2)); draw((0,(+24-6*0)/4)--(14.29,(+24-6*14.29)/4)); draw((0,(+60-6*0)/10)--(14.29,(+60-6*14.29)/10)); draw((-2,(+12+6*-2)/2)--(14.29,(+12+6*14.29)/2)); draw((1.33,(+14.29-5.23*1.33)/0.73)--(14.29,(+14.29-5.23*14.29)/0.73)); dot((-2,0),ds); label("$A$", (-2.17,-0.56),S*lsf); dot((4,0),ds); label("$B$", (3.81,-0.47),S*lsf); dot((2,0),ds); label("$C$", (1.67,-0.53),SW*lsf); dot((10,0),ds); label("$D$", (9.83,-0.5),SE*lsf); dot((0,6),ds); label("$O$", (-0.65,6.3),NE*lsf); dot((1.33,9.99),ds); label("$A'$", (1.61,9.93),NE*lsf); dot((2.06,4.76),ds); label("$D'$", (2.2,4.96),NE*lsf); dot((2.39,2.41),ds); label("$B'$", (2.51,2.61),NE*lsf); dot((3.27,-3.8),ds); label("$C'$", (3.53,-3.91),NE*lsf); clip((-5.43,-5.18)--(-5.43,10.64)--(14.29,10.64)--(14.29,-5.18)--cycle);[/asy]$
Note: Such four concurrent lines are called to form a harmonic range or a harmonic pencil, denoted as $O(ACBD)$ or sometimes as $O(AB; CD).$
Theorem 4.
Any two of the following statements implore the third statement to be true, too.
Statement 1: The pencil $O(AB; CD)$ is harmonic.
Statement 2: $OB$ bisects $\angle AOC$ internally.
Statement 3: $OB$ and $OD$ are perpendicular to each other.
Proof.
I will prove that $(1), (3)\implies (2),$ and the rest is left for the reader to prove.
Refer to the diagram. Draw $EF\parallel AO$ such that $E\in OC, F\in OD.$ Now,
$AO\parallel EF\implies \frac{AC}{CB}=\frac{AO}{EB};$ and $AO\parallel BF\implies \frac{AD}{BD}=\frac{AO}{BF}.$
Using $\frac{AC}{CB}=\frac{AD}{BD},$ we get $EB=BF,$ ie $B$ is the midpoint of $EF.$
Now, we already have $\angle OBF=\angle BOA=\frac{\pi}{2}\land EB=BF\implies \triangle OBE\cong\triangle OBF.$ Therefore $OB$ bisects $\angle EOF,$ but since this is perpendicular to $OA,$ therefore $OA$ is the external bisector of $\angle EOF.$
$[asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); pen wwqqff = rgb(0.4,0,1); pen zzzzff = rgb(0.6,0.6,1); pen ffttqq = rgb(1,0.2,0); pen zzwwff = rgb(0.6,0.4,1); pen ffqqww = rgb(1,0,0.4); draw((-0.82,8.97)--(0.21,8.15)--(1.03,9.18)--(0,10)--cycle,wwqqff); draw((-12,0)--(-5,0),wwqqff); draw((0,10)--(-12,0),wwqqff); draw((0,10)--(-5,0),wwqqff); draw((-5,0)--(12.63,0),wwqqff); draw((0,10)--(12.63,0),wwqqff); draw((0,10)--(-7.92,0),wwqqff); pair parametricplot0_cus(real t){ return (2.79*cos(t)+0,2.79*sin(t)+10); } draw(graph(parametricplot0_cus,-2.2406491565943965,-2.0344439357957027)--(0,10)--cycle,ffttqq+linewidth(0.6pt)); pair parametricplot1_cus(real t){ return (2.79*cos(t)+0,2.79*sin(t)+10); } draw(graph(parametricplot1_cus,-2.44685437739309,-2.2406491565943965)--(0,10)--cycle,ffttqq); draw((-7.92,0)--(-13.59,-7.16),zzwwff); draw((3.59,7.16)--(-13.59,-7.16),ffqqww); dot((-12,0),ds); label("$A$", (-12.25,-0.7),NE*lsf); dot((-5,0),ds); label("$B$", (-5.12,-0.75),NE*lsf); dot((0,10),ds); label("$O$", (-0.09,10.39),NE*lsf); dot((-7.92,0),ds); label("$C$", (-8.15,-0.75),NE*lsf); dot((12.63,0),ds); label("$D$", (12.44,-0.84),NE*lsf); dot((-13.59,-7.16),ds); label("$E$", (-13.09,-7.78),NE*lsf); dot((3.59,7.16),ds); label("$F$", (3.78,7.45),NE*lsf); clip((-15,-8.48)--(-15,15.28)--(14.63,15.28)--(14.63,-8.48)--cycle);[/asy]$
Theorem 5.
If $(AB; CD)$ and $(A'B'; C'D')$ are harmonic, and if $AA', CC', BB'$ concur, then the line $DD'$ is also concurrent with these lines.
Proof.
Let $OD'$ meet $BC$ at some point $D'.$ We obtain that $O(A'B'C'D')$ is harmonic, implying $(AB, CD')$ is also harmonic. So we get
$\frac{AC}{CB}=\frac{AD'}{BD'}.$ But also $\frac{AC}{CB}=\frac{AD}{BD};$ so that
$\frac{AD'}{BD'}=\frac{AD}{BD}\implies \frac{AD'}{AD'-BD'}=\frac{AD}{AD-BD}\implies AD=AD',$ so that $D'$ and $D$ are coincident.

Theorem 6. (NEW)
Let $ABCD$ be a convex quadrilateral. If $K=AD\cap BC, M=AC\cap BD, P=AB\cap KM, Q=DC\cap KM,$ then we must have $K,M,P,Q$ to form a harmonic division.
Proof.
Let $R=BQ\cap AK.$ Then from Theorem 2 in $\triangle KDC$ with cevians $KQ,CA,BD$ and transversal $RQB,$ we observe that the division $K,R,D,A$ is harmonic.
So the range $B(K,R,D,A)$ is harmonic and using $KP$ as a transversal, we see that $(KMPQ)=-1.$ We are done.
$[asy] import graph; size(14cm); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen wwwwff = rgb(0.4,0.4,1); pen ffcccc = rgb(1,0.8,0.8); draw((0,4)--(-2,0)--(4,0)--(2.02,3.66)--cycle); draw((0,4)--(-2,0),wwwwff); draw((-2,0)--(4,0),wwwwff); draw((4,0)--(2.02,3.66),wwwwff); draw((2.02,3.66)--(0,4),wwwwff); draw((-2,(+8+4*-2)/2)--(6.39,(+8+4*6.39)/2),wwwwff); draw((-2.92,(-14.64+3.66*-2.92)/-1.98)--(4,(-14.64+3.66*4)/-1.98),wwwwff); draw((-2,0)--(2.02,3.66),wwwwff); draw((4,0)--(0,4),wwwwff); draw((0.88,(+4.05-2.9*0.88)/0.26)--(6.39,(+4.05-2.9*6.39)/0.26),wwwwff); draw((-2.92,(-15.29+3.82*-2.92)/-2.95)--(4,(-15.29+3.82*4)/-2.95),wwwwff); dot((0,4),ds); label("$D$", (-0.23,4.02),W*lsf); dot((-2,0),ds); label("$A$", (-2.17,-0.24),SW*lsf); dot((4,0),ds); label("$B$", (4.02,-0.3),NE*lsf); dot((2.02,3.66),ds); label("$C$", (2.07,3.74),NE*lsf); dot((0.88,5.76),ds); label("$K$", (0.99,5.72),NE*lsf); dot((1.14,2.86),ds); label("$M$", (0.87,2.76),W*lsf); dot((1.05,3.82),ds); label("$Q$", (1.11,3.91),NE*lsf); dot((1.4,0),ds); label("$P$", (1.44,-0.24),SE*lsf); dot((0.36,4.72),ds); label("$R$", (0.3,4.89),N*lsf); clip((-2.92,-1.31)--(-2.92,6.16)--(6.39,6.16)--(6.39,-1.31)--cycle); [/asy]$
$\Box$
Now we have already done some easy theorems which might come in handy here and there, so we need to keep a tab on them. Let's move onto some simple applications.
Note: From now on I might denote $(ACBD)$ as the cross-ratio and $(ACBD)=-1$ obviously denotes the case when $(AB; CD)$ is harmonic.
Note 2: I assume the reader to have some knowledge on Poles and Polars, Inversion, Homothety and Cross ratio. So hope I will not be too ambiguous while using these notations and notions.

$\boxed{E1}.$ If $AD,BE,CF$ are the altitudes of a triangle $\triangle ABC,$ and if $DE, EF$ meet $AB, BC$ at $F', D'$ respectively, then show that $FD$ and $F'D'$ intersect on a point lying on $AC.$
Solution
Let $FD$ meet $AC$ in $E',$ then we have to show that $D',E',F'$ are collinear.
Using Theorem 2, we have that $(BDCD')$ is harmonic, ie $\frac{BD}{DC}=\frac{BD'}{CD'}.$
Similarly we obtain three other relations, and multiplying them out we get
$\frac{BD'}{D'C}\cdot\frac{CE'}{E'A}\cdot\frac{AF'}{F'C}=\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}\stackrel{\text{Ceva}}{=}-1;$
So that using the converse of Menelaus theorem, $D'E'F'$ is a transversal to the triangle $\triangle ABC,$ and therefore a straight line.
Note: Even if I used directed segments in this proof, I think it is best to avoid using them.
$[asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen wwqqff = rgb(0.4,0,1); pen zzttqq = rgb(0.6,0.2,0); pen ffwwww = rgb(1,0.4,0.4); pen qqzzzz = rgb(0,0.6,0.6); draw((0,8)--(-2,0)--(4,0)--cycle); draw((0,8)--(-2,0),zzttqq); draw((-2,0)--(4,0),zzttqq); draw((4,0)--(0,8),zzttqq); draw((4,0)--(-8,0),wwqqff); draw((-8,0)--(7,-6),wwqqff); draw((0,8)--(7,-6),wwqqff); draw((2.8,2.4)--(-8,0),wwqqff); draw((0,8)--(-2.55,-2.18),wwqqff); draw((0,8)--(0,0),ffwwww); draw((-2,0)--(2.8,2.4),ffwwww); draw((4,0)--(-1.65,1.41),ffwwww); draw((-1.65,1.41)--(7,-6),qqzzzz); draw((2.8,2.4)--(-2.55,-2.18),wwqqff); dot((-2,0),ds); label("$B$", (-2.56,-0.45),NE*lsf); dot((4,0),ds); label("$C$", (4.44,-0.36),NE*lsf); dot((0,8),ds); label("$A$", (-0.12,8.28),NE*lsf); dot((0,0),ds); label("$D$", (-0.12,-0.52),NE*lsf); dot((2.8,2.4),ds); label("$E$", (2.78,2.66),NE*lsf); dot((-1.65,1.41),ds); label("$F$", (-1.82,1.58),NW*lsf); dot((-2.55,-2.18),ds); label("$F'$", (-2.9,-2.89),NE*lsf); dot((-8,0),ds); label("$D'$", (-8.33,-0.45),S*lsf); dot((7,-6),ds); label("$E'$", (7.22,-6.01),NE*lsf); clip((-9.6,-6.75)--(-9.6,8.99)--(10.03,8.99)--(10.03,-6.75)--cycle);[/asy]$

$\boxed{E2}.$ The tangent at a point $P$ of a circle cuts a diametre $AB$ at $T$ ; and $PN$ is the perpendicular to $AB,$ the line joining $B$ to the midpoint of $TP$ cuts $PN$ at $Q.$ Prove that $AQ\parallel TP.$
Solution
Let us assume $QA\cap TP=X.$ Note that we have, from the similarity of $\triangle PBN$ and $\triangle PBA;$ that
$\angle BPT=\angle BAP=\angle NPB.$ So using Theorem 4, we obtain that $(ABTN)$ is harmonic. So the pencil $Q(ABTN)$ is also harmonic, leading to (suing $TX$ as a transversal) $(XMTP)$ is harmonic.
But, this will be harmonic if and only if $\frac{XT}{MT}=\frac{XP}{PM}=\frac{XD}{TM}.$ But using $PM=TM,$ we get $XT=XP,$ but $T\neq P.$ Therefore $X=\infty,$ and $AQ\parallel TP.$
$[asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen wwqqff = rgb(0.4,0,1); pen ffwwww = rgb(1,0.4,0.4); pen wwwwff = rgb(0.4,0.4,1); pen qqzzzz = rgb(0,0.6,0.6); draw(circle((0,0),3),ffwwww); draw((-3,0)--(1.82,-3.68),wwwwff); draw((-3,0)--(4.93,0),wwwwff); draw((1.82,2.38)--(1.82,-3.68),qqzzzz); draw((1.82,2.38)--(4.93,0),wwwwff); draw((3.38,1.19)--(1.82,-3.68),wwwwff); draw((1.82,2.38)--(-3,0),wwwwff); draw((3,0)--(1.82,2.38),wwwwff); dot((0,0),ds); label("$O$", (-0.06,-0.3),NE*lsf); dot((1.82,2.38),ds); label("$P$", (1.88,2.47),NE*lsf); dot((-3,0),ds); label("$A$", (-3.2,-0.25),W*lsf); dot((3,0),ds); label("$B$", (3.03,-0.25),NE*lsf); dot((4.93,0),ds); label("$T$", (4.84,-0.3),NE*lsf); dot((1.82,0),ds); label("$N$", (1.86,-0.28),NE*lsf); dot((3.38,1.19),ds); label("$M$", (3.49,1.18),NE*lsf); dot((1.82,-3.68),ds); label("$Q$", (1.91,-3.76),NE*lsf); clip((-3.91,-4.09)--(-3.91,3.75)--(5.87,3.75)--(5.87,-4.09)--cycle); [/asy]$

$\boxed{E4}$ Let $ABC$ be a right angled triangle at $A$ . $D$ is a point on $CB$ . Let $M$ be the midpoint of $AD$ . $CM$ intersects the perpendicular bisector of $AB$ at $E$ . Prove that $BE\parallel DA$ .
Solution
Let $K=AB\cap CE.$ We will show that if $E$ is a point on $CM$ extended such that $BE\parallel DA,$ Then we must have $L$ is the perpendicular from $E$ onto $AB\implies AE=EB.$
Now let $DA$ and $BE$ meet at $\infty,$ then; since $M$ is the midpoint of $AD$ it is forced that $(DMA\infty)$ is harmonic.
Now this also implies that $B(DMA\infty)$ is a harmonic range, so that using $CE$ as a transversal to this range it is implored that $C,M,K,E$ are harmonic.
Now since $\angle CAK = 90^{\circ}$ we must have $\angle MAK=\angle KAE.$ Therefore using $DA\parallel BE$ we have $\angle DAB=\angle LBE,$ and therefore $\triangle LAE\cong\triangle BEL.$
Hence $L$ is the midpoint of $AB.\Box$
$[asy] import graph; size(10cm); draw((0,0)--(4,0)); draw((4,0)--(0,5)); draw((0,0)--(0,5)); draw((0,5)--(2,-2.51)); draw((0,0)--(2,2.5)); draw((0,0)--(2,-2.51)); draw((4,0)--(2,-2.51)); draw((2,0)--(2,-2.51)); label("$A$", (0,0), W); label("$B$", (4,0), S); label("$C$", (0,5), N); label("$D$", (2,2.25), NE); label("$E$", (2,-2.51), E); label("$M$", (1,1.25), W); label("$L$", (2,0), N); [/asy]$
$\boxed{E5}$ In acute triangle $ABC,$ we have points $D$ and $E$ on sides $AC, AB$ respectively satisfying $\angle ADE=\angle ABC.$ Let the angle bisector of $\angle A$ meet $BC$ at $K.\ P$ and $L$ are projections of $K$ and $A$ to $DE,$ respectively, and $Q$ is the midpoint of $AL.$ If the incenter of $\triangle ABC$ lies on the circumcircle of $\triangle ADE,$ prove that $P,\ Q,$ and the incenter of $\triangle ADE$ are collinear.
Solution(motal)
Let $J\equiv AK\cap BD$ , $R$ the incenter of $\triangle ABC$ and $I$ the incenter of $\triangle ADE$ . $(AQL\infty)=-1$ so, considering $P(AQL\infty)$ with transversal $AK$ , the problem is equivalent to proving that $(AIJK)=-1$ . Easy angle chasing leads to $\triangle ERD$ isosceles. Furthermore, $\angle{ARD}=\angle{AED}=\gamma\Longrightarrow RKCD$ concyclic. With similar argument we obtain $BERK$ concyclic. $\angle{EID}+\angle{EKD}=180^{\circ}-\frac{\beta}{2}-\frac{\gamma}{2}+(\frac{\beta}{2}+\frac{\gamma}{2})=180^{\circ}$ . Therefore we have that $IEKD$ is concyclic with diameter $KI$ . In particular we have that both of these facts hold: $\angle{IDK}=90^{\circ}$ and $DI$ is bisector of $\angle{JDA}$ . Hence considering pencil $D(AIJK)$ we can conclude that $(AIJK)=-1$ .
We are done. $\Box$
Sorry I could not attach a figure, I will attach one as soon as possible.

$\boxed{E6}.$ (Desargues) Let $\triangle ABC$ and $\triangle A'B'C'$ be perspective triangles perspective about a point $O.$ If $BC\cap B'C'=X, CA\cap C'A'=Y$ and $AB\cap A'B'=Z,$ then $X,Y,Z$ are collinear.
Solution
Let us denote $OA\cap BC=E, OA\cap B'C'=E'.$ Now we will use a little of cross ratio or projective geometry to prove this fact.
Firstly consider the concurrent points $B,E,C,Y$ and $B',E',C',Y.$ The corresponding lines $BB', EE', CC', YY$ are concurrent at $O.$ Therefore using a simple idea of cross-ratio we obtain that
$O(BY, EC)=O(B'Y, E'C).$ It is also obvious that $O(BY,EC)=A(BY, EC)=A(X,Y,Z,O)$ and $O(B'Y,E'C')=A'(X,Y,Z,O).$ So the points $X,Y,Z$ are collinear since $O$ lies on the line $AA'.$ (Why?)
We are done.

$\boxed{E7}.$ Given that $AB$ is a diameter of a circle and the lines $CD,CB$ are tangents to the circle, prove that $DE = EF,$ where $F$ is the foot of perpendicular from $D$ onto $AB$ and $E=DF\cap CA.$
Solution
Let $U=CD\cap BA$ and $S=BD\cap CA.$
Since $DF\perp AB$ and $BD\perp DA,$ we see that $\angle ADU=\angle ABD=\angle FDA,$ therefore $DA$ bisects $\angle FDU$ and also $AD\perp BD.$ So the range $D(B,F,A,U)$ is harmonic. Using $CA$ as a transversal this leads to the fact that $S,E,A,C$ are harmonic. So $B(S,E,A,C)$ is harmonic, and again, using $DF$ as a transversal to this range we have that $(FED\infty)=-1,$ ie $E$ is the midpoint of $DF.$
These types of midpoint considerations can prove a lot of geometric problems easily, and of course this one is obvious from considering symmedians. Apart from that, harmonic divisions stands out to be a hugely useful and rarely used (in my country, at least) method.

$\boxed{E8}.$ (Proposed in Olympiad Marathon by Fersolve)
Let $O$ be the intersection of the diagonals of convex quadrilateral $ABCD.$ The circumcircles
of $\triangle OAD$ and $\triangle OBC$ meet at $O$ and $M.$ Line $OM$ meets the circumcircles of $\triangle OAB$ and $\triangle OCD$ at $S$ and $T$ respectively. Prove that $M$ is the midpoint of $ST.$ (NEW)
$[asy]import graph; size(14cm); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttff = rgb(0,0.2,1); pen ffcccc = rgb(1,0.8,0.8); pen fftttt = rgb(1,0.2,0.2); draw((0,5)--(-2,0)--(6,0)--(2.31,6.49)--cycle); draw((0,5)--(-2,0),qqttff); draw((-2,0)--(6,0),qqttff); draw((6,0)--(2.31,6.49),qqttff); draw((2.31,6.49)--(0,5),qqttff); draw((-2,0)--(2.31,6.49),qqttff); draw((0,5)--(6,0),qqttff); draw(circle((-1.27,2.61),2.71),fftttt); draw(circle((4.51,3.45),3.76),fftttt); draw((-5.22,(-3.97+2.69*-5.22)/-0.39)--(10.02,(-3.97+2.69*10.02)/-0.39),qqttff); draw(circle((1.24,5.62),1.38),fftttt); draw(circle((2,0.44),4.02),fftttt); dot((0,5),ds); label("$D$", (-0.43,5.18),NW*lsf); dot((-2,0),ds); label("$A$", (-2.34,-0.36),W*lsf); dot((6,0),ds); label("$B$", (6.14,-0.36),NE*lsf); dot((2.31,6.49),ds); label("$C$", (2.4,6.64),NE*lsf); dot((0.85,4.29),ds); label("$O$", (1.18,4.03),NE*lsf); dot((1.24,1.6),ds); label("$M$", (1.37,1.61),NE*lsf); dot((0.49,6.78),ds); label("$T$", (0.56,7.02),NE*lsf); dot((1.99,-3.59),ds); label("$S$", (2.09,-3.45),NE*lsf); clip((-5.22,-4.1)--(-5.22,8.12)--(10.02,8.12)--(10.02,-4.1)--cycle);[/asy]$
Solution
Let us invert everything about $O$ to map points to their primes. By Theorem 6 on the four collinear points $M',T',S',O;$ we have $\dfrac{M'T'}{T'O}=\dfrac{M'S'}{S'O}.$ Also we have the following relations:
$M'T'=\dfrac{MT\times r^2}{OM\cdot OT}, T'O=\dfrac{r^2}{OT}, M'S'=\dfrac{MS\times r^2}{OM\cdot OS}, S'O=\dfrac{r^2}{OS};$ And therefore we obtain $MT=MS$ and so, $M$ is the midpoint of $TS$ as required. $\Box$

I will keep updating this post from time to time because as time goes on, the number of problems you come across also goes on.
I will add some exercises about a month later (after my secondary exams end), so, till then; Rejoice!!
References
[1] Cosmin Pohoata; Harmonic Divisions; 2007 Mathematical Reflections
[2] C. Stanley Ogilvy; (1990) Excursions in Geometry, Dover.
[3] Application 1
[4] Application 2

Last Updated On: 7:00 pm UCT; 03 March, 2011.

This post has been edited 6 times. Last edited by levans, Jan 15, 2016, 3:57 AM

Different maddening numbers

by Potla, Aug 21, 2010, 8:08 PM

The world is simply full of numbers of all kinds. All the clutter on numbers was made by none but man. Fibonacci and Lucas numbers; perfect numbers; Fermat numbers; Multiply-perfect numbers; abundant and super-abundant numbers, practical and impractical (!) numbers, etc. However, there is an interesting correlation between all these junk of numbers. In this entry, I try to let us get an insight into Perfect, Multiply-perfect, Abundant and Super-abundant and Practical numbers.
As usual, all of us have realized from the definition of a perfect number that what a multiply-perfect number is.
Definition and properties of the $\sigma$ function
Let us, for the sake of clarity, define our notations first. We denote $\sigma (n)$ as the sum of the divisors of $n$ , and $\tau (n)$ as the number of the divisors of $n$ . We try to express $\tau(n)$ in terms of the prime divisors of $n$ . Note that, by the fundamental theorem we have that any $n$ can be expressed in the form of (where $p_i$ 's are the primes arranged in order or not in order)
$n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k};$ A simple combinatorial argument leads to the obvious and well-known fact that
$\tau (n)=(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1)$ (including $1$ as a factor).
Now, for a formula of $\sigma(n)$ we see that all the divisors $d_i$ are of the form
$d_i=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k};$ and $0\leq \beta_i\leq \alpha_i$ . Thence the sum, can be factored into (or rather, found out to be) terms in the expansion of $\prod_{i=1}^n(1+p_i+\cdots+p_i^{\alpha_i})=\prod_{i=1}^n\frac{p_i^{\alpha_i+1}-1}{p_i-1}.$
Now it is evident that when $\text{gcd}(m,n)=1$ we have the multiplicity property of this $\sigma$ function, ie $\sigma(mn)=\sigma(m)\sigma(n)$ . The (obvious) proof is left to the reader.
Multiply-Perfect Numbers
Before discussing the general multiply-perfect numbers, we dive into the perfect numbers for a little time. Let us try to list some of the perfect numbers. They are $6,28,496,8128,\cdots$ written in order. The next values can be generated with a computer, but what do we see common in these numbers?
(1) (Obviously) All are even and $\sigma(n)=2n$ ; which makes them perfect.
(2) $6=2\cdot (2^2-1);$ and $3=2^2-1$ is a prime.
$28 = 2^2\cdot (2^3-1);$ and $2^3-1=7$ is a prime.
$496 = 2^4\cdot(2^5-1);$ and $2^5-1=31$ is a prime.
$8128 = 2^6\cdot(2^7-1);$ and $2^7-1=127$ is a prime.
Proceeding in this manner, Euler proved (again, how many theorems did he prove?) that if $n$ is a given number with the property that $2^n-1$ is a prime, that $2^{n-1}\cdot(2^{n}-1)$ is going to be a perfect number.
With this (extremely) brief idea of perfect numbers we are able to give a generalization. We consider a number $n$ such that its $\sigma (n)=kn$ where $k$ is some natural number.
Before getting some information and problems on multiply-perfect numbers, let us try a problem.
$\boxed{E1.}$ If $n$ is a perfect number and $n+1, n-1$ are primes, find all possible values of $n$
Solution.
Obviously considering modulo $6$ we have that $6|n\implies n=6n_1$ for some $n_1\in\mathbb N$ . If so, we consider $n_1>3.$ In this case,
$\sigma(6n_1)>1+2+3+6+n_1+2n_1+3n_1+6n_1>12n_1+12$ , which is way too bizarre for a perfect number $6n_1$ . So $n_1\leq 3$ . So we get $n=6,12,18$ . Indeed, $5,7$ and $11,13$ and $17, 18$ are all primes. $\Box$
Let us denote a multiply perfect number of the order $k$ by $x_k$ whereas $\sigma(x_k)=kx_k$ . We play around with a few multiply-perfect numbers of some orders. Just as in the previous case, we try to list some properties of a $x_k$ number. At the first glance into a $k-$ tuply perfect number like $120$ or $672$ (Verify with a calculator or the prime factorization formulae for $\sigma(n)$ ), we might try to conjecture a few things. Like, all $x_k$ are even. We might try to construct a sequence of these numbers for different $k$ , and conjecture that there does or does not exist such a sequence. These conjectures are needed in mathematics, but we still need a proof of those, that is why they are called conjectures.

Very few has been known about large $k$ 's, so let us start by noting some examples.
$\boxed{E2.}$ Let $n$ be a number such that $3$ does not divide $n$ yet it satisfies $\sigma(n)=3n$ , show that $\sigma(3n)=12n$ , ie $3n$ is a perfect number of order $4$ .
Solution.
It is obvious from the multiplicative property of $\sigma(n)$ that
$\sigma(3n)=3\sigma(n)=12n;$ so that $3n$ is an $x_4-$ type number.
$\boxed{E3.}$ Every number such that $\sigma(n)=5n$ has at least $6$ different prime factors.
Solution.
This, also is obvious on using the prime decomposition of $n$ , which is
$n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ gives us
$\prod_{i=1}^n\frac{p_i^{\alpha_i+1}-1}{p_i^{\alpha_i}(p_i-1)}=5.$ Now using $\frac{p_1^{\alpha_1+1}-1}{p_1^{\alpha_1}(p_1-1)}>\frac{p_1}{p_1-1};$ which is obvious on expanding, we have
$\frac{p_1}{p_1-1}\cdots\frac{p_n}{p_n-1}>5.$ Now with the same old technique, we consider the product upto $5$ terms, which is surely going to be $<5$ (because the statement is modified to contradict this event by origin); therefore we are surely done.
Superabundant Numbers
A number $n$ is known as superabundant if and only if $\dfrac{\sigma(n)}{n}$ exceeds $\dfrac{\sigma(k)}{k}$ for all $k<n$ .
$\boxed{E4}.$ The sequence of numbers $t_n=\dfrac{\sigma(n)}{n}$ does not have any upper bound..
Solution.
If $d_i$ are the divisors of $n$ then we have, $d_i=\dfrac n{d_{n-i}}$ (why?) and summing up on this sequence we get $\sigma(n)$ as,
$\sigma(n)=\sum_{d_i|n}\dfrac 1{d_i}\implies \dfrac{\sigma(n)}{n}=\sum_{d_i|n}\dfrac 1{d_i}.$
Since we have to prove that this has no upper bound, let $n=m!$ for our convenience. Then we have,
$\dfrac{\sigma(n)}{n}=\sum_{d_i|n}\dfrac 1{d_i}=\sum_{d_i|m!}\dfrac1{d_i}\geq 1+\frac 12+\frac 13+\cdots+\frac 1m.$
Then we have $t_n\geq H_m$ where $H$ is the harmonic series. As we know the harmonic series on natural numbers does not have any upper bound as $n$ increases, and therefore $t_n$ is unbounded from above.
Abundant Numbers
The numbers $n$ with the property that $\sigma(n)\geq 2n$ are abundant ones.
$12, 18,20,24\cdots$ are abundant at our first instance, and we can also find out that all abundant numbers $\leq 100$ are even.
We have a property regarding the abundant numbers.
$\boxed{E5}$ Every number $n$ is abundant implies that $mn$ is abundant, $m\ge 2$
Solution to be found by the reader.

We will discuss on some Practical, Highly Composite Numbers and Quasi-perfect, Semi-perfect numbers soon.
Hope you do not get insane with a lot of numbers, we will get some exercises on these too.

(To Be Updated Soon)

This post has been edited 2 times. Last edited by Potla, Oct 22, 2010, 6:46 PM

Colourings

by Potla, Feb 10, 2010, 9:20 AM

INTRODUCTION

As we know, many tough combinatorial problems are based on a very basic step or strategy which can simplify the problem upto a tremendous extent. One of our important strategies in proving combinatorial problems is using colourings; i.e., to use colours to paint the problem and then proving it using a semi-contradictory argument.

Before proving colouring problems we must first come to some important terms that we need to know.

1. Colourings Of Graphs
First of all, I will introduce you briefly to the colourings of graphs. Though it is not very related to problem-solving, still we need to have some basic knowledge on this.

Before that, we need to know what a graph is. I will introduce us to the idea of a graph in brief. Actually; a graph $G$ is defined as a figure, which can be 2-Dimensional (planar) or 3-Dimensional, which must consist of a set $V(G)$ of vertices, a set $E(G)$ of edges. We denote $e$ as an edge with $(x,y)$ as endpoints; and in order for the figure to be a graph we must have a mapping from each sets of vertices to the endpoints.

An edge is called a loop whenever we have $x$ identical with $y$ . A graph is called simple when it has no loops, and no two edges having the same set of endpoints. Not that every polygon can be called a simple graph.We will come to graphs later again.

We say that a graph $G$ is properly coloured if we have no edge having two endpoints of the same colour. We denote the set of colurs with $C;$ and hence for a polygon with an odd number of sides we have $|C| = 2,$ where $|X|$ denotes the cardinality of $X$ or the number of elements in $X.$

We denote $\chi (G) = |C|_{\text{min}},$ i.e. the minimum number of colours required for a proper colouring. If we denote $P_n$ as the polygon with number of sides $= n$ , then we have $\chi \left(P_{2n + 1}\right) = 2.$ This $\chi (G)$ is said to be the chromatic number of a graph $G.$ We also have the Four Colour Theorem, which can be stated equivalently as; if $G$ is planar we must have $ \ka\chi (G)\leq 4.$ We will discuss about this later too, but now let us come to our main focus.

2. COLOURINGS
In case of combinatorial problems, the simpler, the better. Therefore we will show why colurings are simple.

We can cover up a chessboard with $32$ numbers of $2\times 1$ dominoes in finitely many ways. A physicist M.E. Fischer calculated this number to be $12988816\ \lor \ \left(2^4\times 901^2\right).$ But that is not our point of concern. We try to modify the problem a bit. If $2$ diagonally opposite squares are cut off from this chessboard, then in how many ways can we do the covering with $31 \ \ \ 2\times 1$ dominoes?

Though it looks tough, we can explain it with a simple combinatorial argument. Let us consider the number of black and white squares. First of all, we had $64$ squares; $32$ black and $32$ white. If we take out two diagonally opposite squares, then let both the squares be white in colour (WLOG). In that case, we see that in our newly arranged chessboard, we have $30$ white squares and $32$ black squares. If we were to cover it up with $2\times 1$ dominoes, it would have covered $31$ white and $31$ black squares. but we don't have $31$ white squares, we have only $30$ of them left. Therefore the colouring is not possible. $\Box$

Like this, we can prove many problems by just colouring and then contradicting.

$\boxed{E 1.}$ A $10\times 10$ chessboard cannot be covered by $4\times 1$ straight tetrominoes.
Solution
We colour our $10\times 10$ board with $O; G; B; A$ (Orange, green, blue, ash/grey) as follows:

Here each horizontal straight tetromino covers four squares of different colours, while the vertical ones cover four squares of the same colour each. After all horizontal tetominoes have been placed, we are left with $a + 10; a + 10; a; a$ squares of colours $O; G; B; A$ respectively. If such a complete covering would have existed, then we must have had $a; (a + 10)\equiv 0(\mod 4)$ which is not possible. Therefore we are done. $\Box$

$\boxed{E 2.}$ A $a\times b$ rectangle can be covered by $1\times n$ rectangles if and only if $n|a; \lor n|b.$
Solution(from A. Engel-P.S.S)
We have,if $n|a \lor n|b$ then our covering is obvious.
So we consider the case when $n\not | a; \implies n = pn + q \ (0 < q < n).$ Now colour the board accordingly as in our previous diagram using the colours $1,2,\cdots n.$ There are $bp + b$ squares of each of the colours $1,2,\cdots q;$ and $bp$ squares of each of the colours $1,2,\cdots n$ . The $h$ horizontal $1\times n$ tiles each covering one square of each colour, after being placed, we will have $(bp + b - h)$ squares of each of the colours $1,\cdots,q$ and $bp - h$ of each of the colours $r + 1,\cdots, n.$ Thus $n$ must divide their difference.
$\therefore n|[(bp + b - h) - (bp - h)]\implies n|b.$
Note
We can also state this for the space, whereas if an $a\times b\times c$ block can be covered with $n\times 1\times 1$ bricks we have $n|a \lor n|b\lor n|c.$

$\boxed{E 3.}$ An art gallery is the shape of an $n$ -gon. Find the minimum number of watchmen needed to guard the gallery.
Solution
This is known as the Art Gallery problem.
We join non-intersecting diagonals to form disjoint triangles. Now we colour the vertices of each triangle with distinct colours, so that we can make a proper colouring of each of the triangles (recall this term and prove it using induction). Now, the colour used least number of times will be our required number, which is equal to $\left\lfloor \frac n3\right\rfloor.$
This solution or theorem is known as Chvátal's art gallery theorem.
For further research you can visit the wikipedia link I gave above. $\Box$

I will post more problems soon, I have to leave now. I am posting some Exercises and post more within 5-8 hours.

3. Exercises
1. Consider the following figure, where 14 cities are connected by roads.
Find a path passing through each city exactly once.
Note
In case of graphs, a path or walk means a proper way of starting from a vertex of a graph $x_0$ and walking along the edges, to reach a final point $x_k$ is called a path. Thus a path can be written as $\{x_0;e_1;x_1;e_2;\cdots; e_k; x_k\}.$
A path is said to be simple if $x_k \neq x_0$ , and a closed path if $x_0 = x_k.$ I will come back to this topic again in future when discussing about graphs.

2. Every point in the space is coloured with exactly one of the colours Red, Green, or Blue. The sets $\text{R; G; B}$ consist of the lengths of those segments in space with both endpoints red, green, and blue respectively. Show that at least one of these sets contain all non-negative integers.

Experiment with an equilateral triangle
About a month ago, I was researching with colouring problems that are solved using the Pigeonhole principle and can be proved elegantly. However, I have only dealt with planar figures and there colourings only, so I hope to present my little (and flop) experiment that I made about a month ago.
But before that, I want you to get used to some problems on colourings that require the PHP to be proved elegantly.

The PHP states that if $n + 1$ balls are put into $n$ boxed, we must have a box with more than $1$ ball. This can be generalised as follows:
1. If we put $kn + 1$ balls into $k$ boxes, we must have a box with more than $n$ balls.
2. If the average of $n$ positive numbers $a_1; a_2;\cdots;a_n$ is $a;$ then at least one of the numbers is greater than or equal to $a.$ As a consequence we must have at least one of the numbers less than or equal to $a.$
However, our main point of concern is the first and most elementary form of PHP. In order to prove these theorems, we can assume the converse, and show that it leads to a contradiction. The proof is left to the readers.

Now, we might be given problems on colourings like:
$\boxed{E1}$ If the plane is coloured with two colours, then there exists at least one line with its endpoints and its midpoint of the same colour.
Solution
After drawing a diagram it is obvious. Since there are infinitely many points in the plane, we can get two points of the same colour always in the plane.
Denote them by $A(r); B(r)$ where $r$ signifies red and $b$ signifies blue. (say). Now let us see what its midpoint $D$ is. If $D$ is $r$ , we are done. So WLOG assume $D$ to be blue. Now, we produce the line $AB$ to both sides at a length of $AB.$ Now we get a new line $D_1D_2.$ If $D_1\lor D_2 = r$ , we are done since in both cases we get a line $(r) \ D_1AB \ \lor \ D_2BA.$
$[asy] dot((0,0)); dot((2.5,0)); dot((5,0)); dot((-2.5,0)); dot((-5,0)); label("$D(b)$", (0,0), N); label("$D_1(b)$", (-5,0), N); label("$D_2(b)$", (5,0), N); label("$A(r)$", (-2.5,0), N); label("$B(r)$", (2.5,0), N); draw((-5,0)--(5,0)); [/asy]$
So both of $D_i$ are blue and we have a new segment $D_1DD_2$ of our required type. Hence proved. $\Box$

$\boxed{E2}$ If all points of the plane are coloured red or blue, show that there exists an equilateral triangle with three vertices of the same colour.
Solution
We must have a line with its endpoints and its midpoint of the same colour, as according to E1.
$[asy] dot((0,0)); dot((2.5,0)); dot((5,0)); dot((-2.5,0)); dot((-5,0)); dot((0, 4.3)); label("$D(b)$", (0,0), S); label("$A(b)$", (-5,0), W); label("$B(b)$", (5,0), E); label("$C(r)$", (0,4.3), N); draw((-5,0)--(5,0)); draw((-5,0)--(0,4.3)); draw((0,4.3)--(5,0)); [/asy]$
Let $ADB$ be a line segment with $A,D,B$ blue. Draw an equilateral triangle $ABC$ on this, forcing $C$ to be $r$ in colour. Now lets take the midpoints $E,F$ of $AC, AB$ respectively. Then we again have forced $E;F$ to become $r$ (else we get $\triangle EDC$ and $\triangle FDB$ as the required triangles). In this case too we are forced to get $\triangle CEF$ as our required triangle with all vertices red in colour. Hence we are done. $\Box$

$\boxed{E3}$ (My inspiration to this experiment: Indian National MO 2008) If all the lattice points of the plane are coloured with three colours: $b;r;g$ (standing for blue, red, green respectively) show that we have at least one right triangle with three vertices of different colours; given that $(0,0)\in \{R\}\land (0,1)\in\{B\}.$
Solution(Official solution)
Consider the lattice points on the lines $y = 0$ and $y = 1$ other than $(0, 0), (0, 1).$ If any one of them , say $A = (p, 1)$ is coloured green, then we have a right triangle with $(0, 0), (0, 1)$ and $A$ as vertices, all having different colours.
If not, then the lattice points on $y = 1$ and $y = 0$ are all red or blue. Now consider three different cases:
Case 1
Suppose the point $B = (c, 0)$ is blue. Consider a blue point $D = (p, q)$ in the plane. suppose $p\neq 0.$ If it's projection $(p, 0)$ on the axis is red, then $(p, q),(p, 0), (c, 0)$ are the required vertices.
If $(p, 0)$ is blue, then we can consider the vertices $(0, 0), (p, 0), (c, 0)$ are the vertices of the required RAT.
If $p = 0$ then the points $D, (0, 0) , (c, 0)$ form the required RAT.
Case II
A point $D = (c, 1)$ on the line $y = 1$ is red. The logic is similar to case I.
Case III
Suppose all the lattice points on the line $y = 0$ are red and all the points on the line $y = 1$ are blue.
Consider a green point $E = (p, q)$ where $q\neq 0, 1.$
Consider an isosceles RAT $EKM$ with $\angle E = 90^{\circ}$ such that the hypotenuse $KM$ is a part of the x- axis. Let $EM$ intersect $y = 1$ in $L$ . Then $K$ is a red point and $L$ is a blue point. Hence $EKL$ is a desired right triangle. $\Box$
My Solution

So now that we all know how I have been inspired to perform an experiment, why not lets come to the point?

Here; I thought about a small trick. The problem I thought about was to prove that:
There exists a right triangle with three vertices of the same colour, if we have the plane coloured with red and blue.
Solution
According to our above problem, there must exist an equilateral triangle with all its 3 vertices of the same colour. Refer to figure 1. Let the triangle be $\triangle ABC(r)$ and let $D$ be the midpoint of segment $BC.$ But, as we see, we must have $D\in\{B\}$ , because if $D$ were to be red, we could have found a triangle $ADB$ or $ACD$ satisfying our requirements. Now join $A(r)$ and $D(b).$

As in our figure, let $E$ and $F$ be the midpoints of $AC$ and $AB$ segments respectively.Consider right triangle $EBC.$ If $E$ would have been red, we would have done; since $BE$ is also an altitude of equilateral $\triangle ABC.$ So $E; F\in\{B\}$ is forced again.
Now let us join $EF$ ; and let it intersect $AD$ at $O.$ If $O$ were to be blue, we could have found $\triangle DOE$ or $\triangle DOF$ to fulfill our requirements. So again we see that $O$ has to be red in colour.

Now, if $\angle BOC = 90^{\circ},$ we would have been done. But $\triangle ABC$ is equilateral, so why not find out about this angle? Well, we can let the triangle have equal sides of length $a$ units, so that $AO = OD = \frac 12 AD = \frac {\sqrt 3}{4}a.$ Also $DC = \frac a2$ . Note that from $\triangle DOC$ we have
$\tan \measuredangle DOC = \frac {\frac a2}{\frac {\sqrt 3a}{4}} = \frac 2{\sqrt 3};$ So we get $ \angle DOC = \tan^{ - 1}\left(\frac 2{\sqrt 3}}\right),$ therefore we have
$\angle BOC = 2\tan^{ - 1}\left(\frac 2{\sqrt {3}}\right)\approx 98.2132\cdots ^{\circ};$ which is not $90^{\circ}.$

Damn, how can we proceed?
No, never give up hope. We now try to walk a bit farther towards our goal. Let us consider $\triangle AEF,$ as in Figure 2. Note that if any point on $EF$ would have been red, we would have been done. So it is completely forced that all points on $EF$ except $O$ are blue in colour.
What else can we say? we want a right triangle, am I correct? So why not try to make one? Let us take a point $O_1$ on $AO$ which satisfies $\angle FO_1O = \angle EO_1O = 45^{\circ};$ so we must have $O_1\in \{R\}$ . How funny, now we have three points in a row that are red. But have we done yet?
The answer is unfortunately "no". We still have a bit to go, but we can feel it that we are near to our goal.
OK, last try. Refer to Figure 3.
Draw a line $O_1T$ parallel to $FE$ such that $T$ is on $AE$ . Again we have forced $T\in\{B\}$ . Now we can almost finish our proof if proceeded a bit more. Let us draw a right angle $\angle T_1TE$ with $T_1$ on $EF.$ We can show that such a point exists since $\angle O_1PE = 120^{\circ}$ and due to the acuteness property we force $T_1$ to be in the same side of $EF$ with respect to $O_1T.$

Wow, we have proved earlier that $T_1\in\{B\};$ and in this case we see that $\triangle TT_1F$ is our required triangle

; and we are done.

$\Box$

A really nice problem that I made, isn't it? The solution might be lengthy and not elegant, but still, it is an interesting solution, all the same.

(To be updated)

This post has been edited 5 times. Last edited by Potla, Mar 2, 2011, 7:01 PM

My own problems - all in one place

by Potla, Jan 1, 2010, 6:50 PM

$\boxed{1}$
Prove that for all $a, b, c\in\mathbb{R}^{ + }$ , such that $a + b + c = 1$ , we have:
$\frac {a(1 - a)^2}{bc} + \frac {b(1 - b)^2}{ca} + \frac {c(1 - c)^2}{ab}\ge 4$ (Posed Here)

$\boxed{2}$
Prove that for $x, y, z\in \mathbb{R}^{ + }$ such that $x + y + z = 1$ ,
$\frac {1 - x}{2yz} + \frac {1 - y}{2zx} + \frac {1 - z}{2xy}\ge$
$\ge \frac {1}{\sqrt {x^2 + y^2 - xy}} + \frac {1}{\sqrt {y^2 + z^2 - yz}} + \frac {1}{\sqrt {z^2 + x^2 - zx}} >$
$> \frac {1}{1 - x} + \frac {1}{1 - y} + \frac {1}{1 - z}\ge \frac {9}{2}$ (Posted Here)

$\boxed{3}$
Prove that , for all $a, b, c\in\mathbb{R}^{ + }$ such that $a + b + c = 3$ , we have:
$\sqrt {\left( \sum_{cyc}\frac {a^2 + 2c^2}{b + c} \right) \left(\sum_{cyc} \frac {a}{(b + c)^2}\right)}\ge\frac {3\sqrt {3}}{2\sqrt {2}}$ (Posted Here)[/hide]

$\boxed{4}$
For $a, b, c\in \mathbb{R}$ such that $a + b + c\ge1$ , prove that:
$4(a + b + c) + \frac {a^2}{b + c} + \frac {b^2}{c + a} + \frac {c^2}{a + b}\ge\frac {9}{2}$ (Posted Here)

$\boxed{5}$
Prove that, for $a,b,c\ge 0$ satisfying $\sum_{cyc}\frac {1}{1 + a^2} = \frac 12$ we always have:
$\sum_{cyc}\frac1{a^3 + 2}\leq \frac {1}{3}$ (Posted Here; See post # 75)

$\boxed{6}$
For positive reals $a,b,c$ we have:
$\frac {(a + b)(b + c)(c + a)}{8abc}\ge\frac {\sqrt {\prod_{sym}(a^{2} + b)}}{(a + bc)(b + ca)(c + ab)}$ (Posted here)

$\boxed{7}$
For positive reals $x_i$ satisfying $\sum_{i = 1}^n x_i = n$ Find $P_{min}$ given $k\in\mathbb N$ :
$P = \frac {x_{1}^{k}}{x_{n} + x_{1}} + \frac {x_{2}^{k}}{x_{1} + x_{2}} + \cdots + \frac {x_{n}^{k}}{x_{1} + x_{n - 1}}$ (Posted here)

$\boxed{8}$
For positive reals $a,b,c$ prove that :
$\sum\frac {a^{5}}{b^{2}c^{2} + ab^{2}c}\geq\frac {a + b + c}{2}$ (I jumbled up other three inequalities of mine with this one; Posted here)

$\boxed{9}$
For Positive reals $x,y,z$ prove that:
$\sum\frac {x^{5}}{x^{3}yz + yz^{4}}\geq\frac {3}{2}$ (Posted Here)

$\boxed{10}$
For positive reals $a,b,c$ satisfying $a + b + c = 3$ prove that:
$12 - \sum_{cyc}\left[\frac {a}{bc}(13b^{2} - 3a^{2})\right]\geq 6\sum_{cyc}\left[\frac {b^{2} + c}{c}\right]$ (Posted Here)

$\boxed{11}$
Prove that , for $a,b,c > 0$ satisfying $a^{4} + b(b^{2} + 1) + 4c = 7$ :
$\frac {(a + b)^{2}}{b(b + c)^{2}} + \frac {(b + c)^{2}}{c(c + a)^{2}} + \frac {(c + a)^{2}}{a(a + b)^{2}}\geq a + b + c$ (Posted here)

$\boxed{12}$
For $a,b,c > 0; a + b + c = 1$ Prove that:
$\sum_{cyc}\frac {a^3}{(1 - a)^2}\geq \frac 14$ (Posted here; Post #138)

$\boxed{13}$
Prove that for all positive reals $a,b,c$ we always have:
$\frac {a^2 + bc}{a + b} + \frac {b^2 + ca}{b + c} + \frac {c^2 + ab}{c + a}\geq 3\sqrt [3]{abc}$ (Posted Here)

$\boxed{14}$
For $a,b,c > 0$ , show that
$\frac a{5a + 4b + 3c} + \frac b{5b + 4c + 3a} + \frac c{5c + 4a + 3b}\le \frac 19 \sum \frac a{a + b} + \frac {10}{27}$ (Posted here)

$\boxed{15}$
Given $a,b,c$ are the sides of a triangle, show that we have
$\sum_{cyc}\sqrt {\frac {a}{c + a - b}}\ge 3$ (Not posted anywhere yet)

$\boxed{16}$
For all $x,y,z > 0$ we have that
$\frac {x + y + z}{\sqrt {3}}\geq \frac {xy + yz + zx + 2\sqrt {xyz}\left(\sqrt {x} + \sqrt {y} + \sqrt {z}\right)}{\sqrt {x^2 + xy + y^2} + \sqrt {y^2 + yz + z^2} + \sqrt {z^2 + zx + x^2}}$ (Not posted anywhere yet;)
Note

$\boxed{17}$
Given $a,b,c > 0$ such that $ab + bc + ca = abc$ , prove that we have
$\frac 1{a + 3b + 2c} + \frac1{b + 3c + 2a} + \frac 1{c + 3a + 2b}\leq \frac 16$ (Posted Here)

$\boxed{18}$
For all $a,b,c > 0$ , show that
$\sqrt {\frac {a^2 + bc}{b(c + a)}} + \sqrt {\frac {b^2 + ca}{c(b + a)}} + \sqrt {\frac {c^2 + ab}{a(b + c)}}\geq 3$ (Not posted Anywhere Yet)

$\boxed{19}$
(Sayan Mukherjee) For positive reals $a,b,c$ satisfying $a + b + c = 3;$ Prove that we have:
$\displaystyle \sum_{cyc}\frac {a}{bc}\left(a^2 + 7b^2\right) \ge 4\sum_{cyc}\frac {ab}{c} + 2\sum_{cyc}\frac {b^2 + c}{c}.$ (Not posted Anywhere Yet)

$\boxed{20}$
Given positive reals $a,b,c$ ; Prove that we have
$\displaystyle\frac {(a + b + c)^2}{ab + bc + ca}\geq \frac 49 \cdot \frac {(a + b + c)^2}{ab + bc + ca}\cdot \left(\frac b{a + b} + \frac c{b + c} + \frac a{c + a}\right) + 1.$ (Not posted Anywhere Yet)
(Proposed in the contest named War_inequality_Secondary at maths.vn as a part of Team 1)

$\boxed{21}$
For all $a,b,c\geq 0$ , show that we have
$\displaystyle 128(a + b + c)(a - b)(b - c)(c - a)\leq 9(a^2 + b^2 + c^2)^2$ Note
(Not posted Anywhere Yet)
$\boxed{22}$
Given $a,b,c > 0$ , show that we have
$\frac {a^6}{(a + b)(a^4 + a^2b^2 + b^4)} + \frac {b^6}{(b + c)(b^4 + b^2c^2 + c^4)} + \frac {c^6}{(c + a)(c^4 + c^2a^2 + a^4)}\geq \frac {a + b + c}{6}$ (Posted here)
$\boxed{23}$
For positive reals $a,b,c$ show that we have
$\sum_{cyc}\frac 1{(2a^2 + bc)(b + c)^2}\leq \frac 9{4(ab + bc + ca)^2}.$ (Made while proving problem wrongly here)
Remark.
This is actually weaker than a problem by Tran Nguyen Quoc Cuong (mathstarofvn), which is on proving that
$\sum_{cyc}\frac 1{(2a^2 + bc)}\leq \frac {(a + b + c)^2}{(ab + bc + ca)^2};$ Which follows from the method in which I proved the previous inequality.
$\boxed{24}$
For positive reals $x,y,z$ , prove that we always have the following inequality:
$\sum_{cyc}\frac {x(y + z)^2}{2x + y + z} + \frac 92 \left(\frac {x + y + z}{xy + yz + zx}\right)^2\geq 6 + \frac 9{2(xy + yz + zx)}.$ (Not Posted Anywhere Yet)
(To be updated as soon as I make another new problem)

Topic Of Nhocnhoc, Maths.vn

by Potla, Nov 10, 2009, 6:53 PM

Hello all, this is my contribution to mathlinks.ro, which is the full log file of Topic of nhocnhoc or namely nguoivn. The problems have been translated to english and then edited by me. Hope this compilation will help us all.

This entry was quite long at first so I am attaching the (un-revised edition) of the file

* I will try to correct the alignment of the document when I get enough time. Actually I had been preparing this for more than a month

Attachments:

Topic Of Nhocnhoc.pdf (163kb)

Garden

by Potla, Nov 5, 2009, 4:58 AM

Sorry for getting off-mathematics (all my posts until now are related to mathematics) , but do you like this image? It is of my friend's garden. I really like this so posting.......

Attachments:

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first shouts in 2025!
by NicoN9, Apr 25, 2025, 12:56 PM
dang 2 year bump (can the person who next visits PM me? I want to see when the next person visits)
by roribaki, Jan 6, 2024, 9:18 PM
wholsome
by samrocksnature, Aug 24, 2021, 4:16 PM
wholesome
by centslordm, Jul 2, 2021, 2:08 PM
cool blog
by lneis1, Feb 20, 2021, 6:11 AM
Hello.Very nice blog!
by Functional_equation, Aug 29, 2020, 2:20 PM
Great blog!
by freeman66, Jun 3, 2020, 1:25 AM
Hello. Nice blog!
by mathboy282, Apr 10, 2020, 1:39 AM
Masterpiece
by Kamran011, Mar 18, 2020, 6:35 PM
Hello! Why am I looking through random blogs?
by bingo2016, Nov 21, 2019, 2:40 AM
I am late to find this wonderful gem
by RAMUGAUSS, Nov 13, 2019, 7:32 AM
I WILL NECROPOST
I WILL NECROPOST
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by Davis1234567890, Sep 15, 2019, 2:21 AM
This is a nice blog and it is very useful!
by arandomperson123, Jun 7, 2016, 2:32 AM
bump,,,,
by ythomashu, Jan 15, 2016, 12:51 AM
Incredible
by DanielL2000, Jul 24, 2014, 9:34 PM
Are you still in school?
by harekrishna, Feb 8, 2010, 6:58 PM
ANNOUNCEMENT
My Regional Tests are tomorrow (29 th Nov.), and after that I have my school exams, so it seems as if this blog will not be updated for around a week. Sorry for the inconvenience.
Sincerely,
Potla
(creator and owner of this blog )
by Potla, Nov 27, 2009, 7:23 PM
Nice. Keep working
by mathson, Oct 6, 2009, 11:29 AM
Not at all.
by Potla, Sep 4, 2009, 10:35 AM
Wow! You are very smart.
by Maybach, Sep 1, 2009, 12:56 AM
good going
by turikachi, May 16, 2009, 3:40 AM
Because of my exams..... I must do some work now.
by Potla, Apr 9, 2009, 9:51 AM
well, why no activity here?
by Ramchandran, Apr 6, 2009, 1:46 PM
Thanks for all of yours' kind( )coordination, but I cannot update this bulky( : ) blog till my exams are over, so I want the contrib to continue their job (I am making peine a contrib too).......
by Potla, Mar 3, 2009, 12:56 PM
Really nice solutions.......
However Iliked ochas and murgis problem more because of its simplicity.
however mathias DK sol is just awe strucking
by UKO, Feb 14, 2009, 4:47 AM
@ mathias DK really your proof is beautifull

I didnot wanted to be rude .never mind.
by murgi, Feb 14, 2009, 4:12 AM
@murgi: I just wanted to provide a more beatiful proof.
by Mathias_DK, Feb 13, 2009, 11:08 PM
I couldnot understand why such acomplicated proof is necessary?
when there exits such a simple proof.
by murgi, Feb 13, 2009, 5:11 PM
From now on, this is the place of problem solvers.
Each problem for the day will bring 10 points to the first one to give a solution. Better solutions with ideas other than the ones before will also bring 10 points to the user. Meanwhile, incorrect problems will lose 5 points, so be careful while answering.
by Potla, Feb 13, 2009, 10:28 AM