A Joint Venture To Neuberg Problem

by RSM, Jun 18, 2012, 5:27 AM

Here I will post solution to a problem which I and Potla were trying for more than a month. Last night we solved the problem.

So, let me state the problem at first:-

Given a triangle $ ABC $ and a point $ P $, suppose, $ A',B',C' $ are the reflections of $ P $ on $ BC,CA,AB $. Prove that $ AA',BB',CC' $ are concurrnt iff $ PP^* $ is parallel to the Euler line of $ ABC $ where $ P^* $ is the isogonal conjugate of $ P $ wrt $ ABC $.

The locus of such point $ P $ is the Neuberg Cubic of $ ABC $. But in our proof we will not use any property of cubics. So let's call the locus of points $ P $ which satisfies the above concurrency fact as Neuberg locus(we are avoiding the term cubic).

At first we will prove 3 properties of the Neuberg locus. The properties are as follows:-

Property 1:-

Given a triangle $ ABC $, if a point $ P $ lies on the Neuberg locus of $ ABC $, then isogonal conjugate of $ P $ wrt $ ABC $ also lies on the Neuberg locus of $ ABC $.

Proof:-

Suppose, $ A',B',C' $ are the reflections of $ P $ on $ BC,CA,AB $. Suppose, $ A_1=PA'\cap \odot A'B'C' $. Similarly define $ B_1,C_1 $. Under inversion wrt $ P $ with power $ PA'.PA_1 $, $ A $ goes to the reflection of $ P $ on $ B_1C_1 $ and similar for others. Now note that, if $ A_2,B_2,C_2 $ are the reflections of $ P^* $ on $ BC,CA,AB $, then $ \Delta A_2B_2C_2 $ is homothetic to $ \Delta A_1B_1C_1 $. So if $ A_3,B_3,C_3 $ are the reflections of $ P^* $ on $ B_2C_2, C_2A_2,A_2B_2 $, then $ \odot P^*A_2A_3, \odot P^*B_2B_3, \odot P^*C_2C_3 $ are co-axial. But the center of $ \odot P^*A_2A_3 $ is the intersection point of $ B_2C_2 $ and $ BC $ and similar for others. So $ B_2C_2\cap BC,C_2A_2\cap CA,A_2B_2\cap AB $ are collinear. So by Desargues' Theorem, $ \Delta ABC $ and $ \Delta A_2B_2C_2 $ are perspective. So $ P^* $ lies on the Neuberg locus of $ ABC $.

Property 2:-

Given a triangle $ ABC $ and a point $ P $, prove that $ P $ lies on the Neuberg locus of $ ABC $ iff $ P $ lies on the Neuberg locus of the pedal triangle of $ P $ wrt $ ABC $.

Proof:-

Suppose, $ P^* $ is the isogonal conjugate of $ P $ wrt $ ABC $. $ A',B',C' $ be the reflections of $ P^* $ on $ BC,CA,AB $. Note that, $ P $ is the circumcenter of $ \Delta A'B'C' $. $ P' $ be the isogonal conjugate of $ P^* $ wrt $ A'B'C' $. $ A_1,B_1,C_1 $ be the circumcenters of $ \Delta B'P'C', \Delta C'P'A', \Delta A'P'B' $. By some simple angle-chasing we get that $ PA_1.PA=PA'^2 $. So $ A'A_1 $ is the isogonal conjugate of $ A'A $ wrt $ \angle B'A'C' $ and similar for others. But from property 1, if $ P $ lies on the Neuberg locus of $ \Delta ABC $ iff so does $ P^* $. So $ A'A,B'B,C'C $ concurr. So $ A'A_1,B'B_1,C'C_1 $ concurr. So $ P' $ lies on the Neuberg locus of $ \Delta A_1B_1C_1 $. Now note that, if $ A_2B_2C_2 $ is the pedal triangle of $ P $ wrt $ ABC $ and $ Q $ is the isogonal conjugate of $ P $ wrt $ A_2B_2C_2 $, then the configuration $ A_2B_2C_2Q $ is homothetic to the configuration $ A_1B_1C_1P' $. So $ Q $ lies on the Neuberg locus of $ \Delta A_2B_2C_2 $. So again using property 1, we get that $ P $ lies on the Neuberg locus of $ \Delta A_2B_2C_2 $.

Corollary:-

Given a triangle $ ABC $ and a point $ P $ on its Neuberg Locus, suppose, $ P_a,P_b,P_c $ are the circumcenters of $ \Delta BPC,\Delta CPA,\Delta APB $. Prove that, $ AP_a,BP_b,CP_c $ are concurrent. Call that concurrency point as $ Q $. If $ P^* $ is the isogonal conjugate of $ P $ wrt $ ABC $, then define $ Q^* $ for $ P^* $ similarly. Then $ Q $ and $ Q^* $ are isogonal conjugates wrt $ ABC $.

Property 3:-

Given a triangle $ ABC $ and a point $ P $, prove that $ P $ lies on the Neuberg locus(or circumcircle or line at infinity) of $ ABC $ iff the Euler lines of $ \Delta BPC,\Delta CPA,\Delta APB $ are concurrent.

Proof:-

Let's ignore the case when $ P $ lies on circumcircle and line at infinity.
Let $G_a, G_b, G_c$ be the centroids of the triangles $BPC,CPA,APB.$ and $ O_a,O_b,O_c $ are the their circumcenters. Then we have $G_bG_c\parallel BC$ and $BG_c\cap CG_b=\text{midpoint}(AP).$ So, the homothety that maps $G_bG_c$ to $BC$ maps $O_bO_c\cap G_bG_c\equiv K_1$ to $O_bO_c\cap BC\equiv K_1'.$ Thus, $\frac{G_cK_1}{G_bK_1}=\frac{BK'_1}{CK'_1}.$
Since the triangles $G_aG_bG_c$ and $O_aO_bO_c$ are perspective, so we get $\prod_{cyc}\frac{G_cK_1}{G_bK_1}=-1,$ which leads to $\prod_{cyc}\frac{BK_1'}{CK_1'}=-1,$ so that $ABC$ and $O_aO_bO_c$ are perspective. So using the corollary of Property 2 we get that $ P $ lies on the Neuberg cubic of $ \Delta ABC $.

Before going to the proof of the main problem we will mention some lemmas that we will use in the proof.

Lemma 1:-

Given two triangles $ ABC $ and $ A'B'C' $(not homothetic), consider the set of all triangles(no two of them homothetic), so that both $ ABC $ and $ A'B'C' $ are orthologic to them. Then the locus of the center of orthology of $ ABC $ and this set of triangles lie on a conic passing through $ A,B,C $ or the line at infinity.

Proof:-

On the sides of $ BC,CA,AB $ take points $ C_1,A_1,B_1 $ so that $ \Delta A_1B_1C_1 $ is homothetic to $ \Delta A'B'C' $. Through $ C_1 $ draw parallel to $ AC $ to intersect $ AB $ at $ A_2 $. Cyclically define $ B_2,C_2 $ on $ BC,AC $. Applying the converse of Pascal's theorem on the hexagon $ A_1C_2B_1A_2C_1B_2 $ we get that $ A_1,B_1,C_1,A_2,B_2,C_2 $ lie on a conic(call this conic $ \Gamma(A_1B_1C_1) $). $ A_3=A_2C_1\cap B_2A_1, B_3=B_2A_1\cap B_1C_2, C_3=C_2B_1\cap A_2C_1 $. Clearly, $ ABC $ and $ A_3,B_3,C_3 $ are homothetic. So $ AA_3,BB_3,CC_3 $ concurr. Since $ A_2A_1 $ is the harmonic conjugate of $ AA_3 $ wrt $ AB,AC $, so if we draw parallels to $ A_1A_2,B_1B_2,C_1C_2 $ through $ A,B,C $, then they intersect $ BC,CA,AB $ at three collinear points. So there exists a conic passing through $ A,B,C $ so that the tangent at $ A $ is parallel to $ A_1A_2 $ and similar for $ B,C $. Call this conic $ \Gamma(ABC) $. Take any point $ X $ on $ \Gamma(ABC) $. Through $ A_1 $ draw a line $ l_a $ parallel to $ AX $ and similarly define $ l_b,l_c $. Note that $ (l_a,A_1A_2;A_1C_2,A_1B_2)=(AX,A_1A_2;AC,AB) $ $ =(BX,BA;BC,B_1B_2)=(l_b,B_1A_2;B_1C_2,B_1B_2) $. So $ l_a\cap l_b $ lies on $ \Gamma(A_1B_1C_1) $ and similar for others. So $ l_a,l_b,l_c $ concurr on $ \Gamma(A_1B_1C_1) $. Its easy to prove that its not true if $ X $ doesn't lie on $ \Gamma(ABC) $.

Lemma 2(Sondat's Theorem):-

Given two triangles $ ABC $ and $ A'B'C' $, such that they are orthologic and perspective, prove that the two centers of orthology are collinear with the center of perspectivity.

Proof:-

Suppose, $ AP\perp B'C', BP\perp C'A', CP\perp A'B' $. Similarly define $ P' $ such that $ A'P'\perp BC, B'P'\perp CA, C'P'\perp AB $. Suppose, $ AA',BB',CC' $ concur at some point $ Q $.
Note that, if we define $ \Gamma(ABC) $ wrt $ \Delta A'B'C' $ using Lemma 1, then we get that, its the rectangular hyperbola passing through $ A,B,C,P $ and the orthocenter of $ ABC $. Now note that, $ BPC $ and $ B'P'C' $ are orthologic and $ A,A' $ are the two centers of orthology. Note that, $ \Gamma(ABC) $ passes through the orthocenter of $ \Delta BPC $, but $ \Gamma(BPC) $ wrt $ \Delta B'P'C' $ is the rectangular hyperbola passing through $ B,C,P,A $ and the orthocenter of $ \Delta BPC $. So $ \Gamma(BPC) $ wrt $ \Delta B'P'C' $ is same as $ \Gamma(ABC) $ wrt $ \Delta A'B'C' $. But note that, $ Q $ lies on $ \Gamma(ABC) $. So $ Q $ lies on $ \Gamma(BPC) $ wrt $ B'P'C' $. So $ PQ\parallel P'Q $. So $ P,P',Q $ are collinear.

Lemma 3:-

Given two points $ P,Q $ and their isogonal conjugates wrt $ ABC $ be $ P',Q' $, then suppose, $ X=PQ'\cap QP' $ and $ Y=PQ\cap P'Q' $, then $ Y $ is the isogonal conjugate of $ X $ wrt $ ABC $.

Proof:-

So if the transformation in property 11 takes $ P' $ to $ Q $, then it takes $ Q' $ to $ P $. So $ (P'A,P'B;P'C,P'Q')=(QA,QB;QC,QP) $ $ \implies (P'A,P'B;P'C,P'Y)=(QA,QB;QC,QY) $. So $ A,B,C,Q,Y,P' $ lie on the same conic, which is the isogonal conjugate of $ PQ' $ wrt $ ABC $. So isogonal conjugate of $ Y $ lies on $ PQ' $ and similarly it lies on $ QP' $. So $ X $ is the isogonal conjugate of $ Y $.
Also note that, if two points $ X,Y $ are on $ PQ' $ and $ PQ $, such that $ X,Y $ are isogonal conjugates, then $ X=PQ'\cap QP' $ and $ Y=PQ\cap P'Q' $.

Some Observations:-

1. Given a triangle $ ABC $ and a point $ P $, if the Euler lines of $ BPC,CPA,APB $ are concurrent, then they concur on the Euler line of $ ABC $.

Proof:-

Suppose, $ O_a,O_b,O_c $ are the circumcenters of $ \Delta BPC, \Delta CPA,\Delta APB $. $ G_a,G_b,G_c $ be the centroids of $ \Delta BPC,\Delta CPA,\Delta APB $. Suppose, $ O_aG_a,O_bG_b,O_cG_c $ concur at some point $ X $. Note that, $ \Delta O_aO_bO_c $ and $ \Delta G_aG_bG_c $ are perspective. Also note that, the perpendiculars from $ O_a,O_b,O_c $ to the sides of $ \Delta G_aG_bG_c $ concur at the circumcenter of $ ABC $ and the perpendiculars from $ G_a,G_b,G_c $ to the sides of $ O_aO_bO_c $ concur at the centroid of $ \Delta ABC $. So by Sondat's Theorem $ O_aG_a,O_bG_b,O_cG_c $ concur on the Euler line of $ ABC $.

2. In the corollary of Property 2, we mentioned two points $ Q,Q^* $. Using Sondat's theorem we get that, $ Q $ lies on $ OP $ and $ Q^* $ lies on $ OP^* $. But since $ Q $ and $ Q^* $ are isogonal conjugates, so using Lemma 3, we get that $ Q=OP\cap HP^* $ and $ Q^*=OP^*\cap HP $ where $ H $ is the orthocenter of $ ABC $.

Proof of The Neuberg Problem:-

Given a triangle $ ABC $ and a point $ P $, suppose, $ P^* $ is the isogonal copnjugate of $ P $ wrt $ ABC $. $ P_a,P_b,P_c $ be the circumcenters of $ \Delta BPC,\Delta CPA,\Delta APB $. $ P^*_a,P^*_b,P^*_c $ be the circumcenters of $ \Delta BP^*C,\Delta CP^*A,\Delta AP^*B $. Suppose, $  P $ lies on the Neuberg locus of $ \Delta P_aP_bP_c $. So $ AP_a,BP_b,CP_c $ are concurrent. Clearly, it implies $ AP^*_a,BP^*_b,CP^*_c $ are concurrent. Suppose, $ O $ is the circumcenter of $ \Delta ABC $. Note that, the triangle formed by the circumcenters of $ \Delta OP_bP_c,\Delta OP_cP_a,\Delta OP_aP_b $ is homothetic to $ \Delta ABC $. So if we draw parallels to the Euler lines of $ \Delta OP_bP_c,\Delta OP_cP_a,\Delta OP_aP_b $ through $ A,B,C $ they will meet at some point $ Q $. Similarly define $ Q^* $ for $ P^* $. Now note that, $ P_bP_c $ is anti-parallel to $ P^*_b,P^*_c $ wrt $ \angle P_bOP_c $. So the Euler line of $ \Delta OP_bP_c $ is anti-parallel to the Euler line of $ \Delta OP^*_bP^*_c $ wrt $ \angle P_bOP_c $. Similar for others. So $ Q^* $ is the isogonal conjugate of $ Q $ wrt $ ABC $. Note that, the triangle formed by the centroids of $ \Delta OP_bP_c,\Delta OP_cP_a,\Delta OP_aP_b $ is homothetic to $ \Delta P_aP_bP_c $. So if we draw parallels to $ AQ,BQ,CQ $ through $ P_a,P_b,P_c $, then they will concurr. So using lemma 1 we get that, $ Q $ lies on $ \Gamma(ABC) $ wrt $ \Delta P_aP_bP_c $ which is the rectangular hyperbola passing through $ A,B,C,H,P $ where $ H $ is the orthocenter of $ \Delta ABC $. So $ Q^* $ lies on $ OP^* $. Similarly, $ Q $ lies on $ OP $. So using lemma 3, we get that, $ Q=OP\cap HP^* $ and $ Q^*=OP^*\cap HP $. Now note that, if $ R $ is the concurreny point of $ AP_a,BP_b,CP_c $ and $ R^* $ is the concurrency point of $ AP^*_a,BP^*_b,CP^*_c $, then from Observation 2, we have $ R=OP\cap HP^* $ and $ R^*=OP^*\cap HP $. So $ R\equiv Q $ and $ R^*\equiv Q^* $. From Observation 1, we get that the $ PQ $ is parallel to the Euler line of $ \Delta P_aP_bP_c $. But now we have, $ Q $ lies on $ OP $. $ O $ is the isogonal conjugate of $ P $ wrt $ P_aP_bP_c $. So the line joining $ P $ and isogonal conjugate of $ P $ wrt $ P_aP_bP_c $ is parallel to the Euler line of $ P_aP_bP_c $. So done.

The proof may look long, but the idea is very small. You can easily realize that after reading the proof.

Property 4:-

If $ P $ lies on the Neuberg locus of $ ABC $, then $ A $ lies on the Neuberg Locus of $ BPC $ and similarly.

Proof:-

Note that, we have proved if the Euler lines of $ PBC,PCA,PAB $ are concurrent, then they concurr on the Euler line of $ ABC $. So using property 3, we get that $ A $ lies on the Neuberg cubic of $ PBC $.

Property 5:-

Given a triangle $ ABC $ and a point $ P $, suppose, $ A_1,B_1,C_1 $ are the intersection points of $ AP,BP,CP $ with $ \odot PBC, \odot PCA, \odot PAB $. $ H_a,H_b,H_c $ are the orthocenters of $ \Delta A_1BC, \Delta B_1CA, \Delta C_1AB $. Prove that $ AH_a,BH_b,CH_c $ are concurrent iff $ P $ lies on the Neuberg cubic of $ \Delta ABC $ or the circles with diameter $ BC,CA,AB $ or the line at infinity.

Proof:-

Suppose, $ O_a,O_b,O_c $ are the circumcnters of $ PBC,PCA,PAB $. Note that, $ CO_a,CH_b $ are isogonal conjugates wrt $ \angle ACB $ and similar for others. Suppose, $ A'=BH_c\cap CH_b $. Similarly, define $ B',C' $. Consider the case when $ P $ lies on the Neuberg cubic of $ ABC $. Clearly, $ AO_a,BO_b,CO_c $ concurr at some point $ Q $. $ Q^* $ be the isogonal conjugate of $ Q $ wrt $ ABC $. Note that, $ AA',BB',CC' $ concurr at $ Q^* $. So applying converse of Brianchan's theorem on $ BA'CB'AC' $ we get that, there exists a conic touching $BH_a,H_aC,CH_b, H_bA,AH_c,H_cB $. So applying Brianchan's theorem on $ AH_cBH_aCH_b $ we get that $ AH_a,BH_b,CH_c $ are concurrent.
The other two cases are just special cases which can be verified easily.

Property 6(Concurrency of Brocard axes):-

Given a triangle $ ABC $ and point $ P $, prove that Brocard axes of $ PBC,PCA,PAB $ are concurrent iff $ P $ lies on the Neuberg cubic of $ ABC $ or the circumcircle of $ ABC $ or the line at infinity.

Note that, the last two cases are trivial cases, so we will prove the first case only.

Proof:-

Before proving the problem we will prove two lemmas.

Lemma 1:-

Isotomic line of the Leomine axis of a triangle $ ABC $ is perpendicular to the Euler line of $ ABC $.

Proof:-

Suppose, $ A'B'C' $ is the Lemoine axis of $ ABC $. Consider the circles with diameter $ AA',BB',CC' $. Clearly, the orthocenter of $ ABC $(call it $ H $) lies on their radical axis. Also the circumcenter of $ ABC $(call it $ O $) lies on their radical axis. So they are co-axial with $ OH $ as radical axis. But using Gaussian line theorem we get that the line joining the midpoints of $ AA',BB',CC' $ is parallel to the isotomic line of $ A'B'C' $ wrt $ ABC $. So the isotomic line of $ A'B'C' $ wrt $ ABC $ is perpendicular to $ OH $.

Lemma 2:-

Let $ABC$ be a triangle, and let $A_1B_1C_1$ and $A_2B_2C_2$ be two isotomic lines such that $A_1,A_2\in BC$ and similars. Show that, if $L$ is the point where the line through $A$ parallel to $BC$ meets $A_1B_1C_1$ then we have $ \frac{B_1L}{C_1L}=\frac{A_2B_2}{A_2C_2}$.

Proof:-
Note that, $ \frac {B_1L}{C_1L}=\frac {AB_1}{AC_1}.\frac {AB}{AC} $. And also $ \frac {B_2A_2}{A_2C_2}=\frac {B_2C}{C_2B}.\frac {AB}{AC} $. But since $ B_1,B_2 $ are isotomic points wrt $ A $, so $ AB_1=CB_2 $ and $ AC_1=BC_2 $. SO $ \frac {B_1L}{LC_1}=\frac {B_2A_2}{A_2C_2} $.

Back to Main problem:-

Suppose, $ O_a,H_a,K_a $ are the circumcenter,orthocenter and symmedian point of $ PBC$. Let $ O $ be the circumcenter of $ ABC $. Note that, its enough to prove that, $ \prod_{cyc} (O_aO_b,O_aO_c;O_aP,O_aK_a)=1 $. Suppose, $ P'B'C' $ is the Lemoine axis of $ PBC $. Then note that, $ (O_aO_b,O_aO_c;O_aP,O_aK_a)=(PC,PB;PP',B'C')=\frac {C'P'}{B'P'} $. $ P_1B_1C_1 $ be the isotomic line of $ P'B'C' $. A line through $ P $ parallel to $ BC $ intersects $ B_1C_1 $ at $ U $. Using lemma 2 we get that, $ \frac {B_1U}{C_1U}=\frac {B'P'}{P'C'} $. So $ (PB,PC;PU,B_1C_1)=(PB,PC;PP',B'C') $. But $ (PB,PC;PU,B_1C_1)=(O_aO_c,O_aO_b;O_aO,O_aH_a) $. So $ (O_aO_c,O_aO_b;O_aO,O_aH_a)=(O_aO_c,O_aO_b;O_aP,O_aK_a) $. But since the lines $ O_aH_a $ are concurrent, so $ \prod_{cyclic} (O_aO_c,O_aO_b;O_aO,O_aH_a)=1 $. So we are done.
This post has been edited 8 times. Last edited by levans, Jul 10, 2015, 5:04 PM

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  • nice blog!

    by wardtnt1234, Aug 18, 2024, 10:38 AM

  • Leaving this comment as a bookmark…nice blog!

    by gabrupro, Nov 20, 2023, 5:59 PM

  • bumpity bump

    by samrocksnature, Jan 27, 2023, 7:02 PM

  • Nice Blog!

    by Functional_equation, Aug 8, 2020, 6:13 PM

  • Nice!!!!

    by A-Thought-Of-God, Jul 9, 2020, 11:25 AM

  • Dada tumi kothay, keno math chere dile? :(

    by amar_04, Mar 2, 2020, 8:38 AM

  • Cool blog tho

    by usernameyourself, Jan 12, 2020, 12:31 PM

  • this is a depressingly dead blog

    by cocohearts, Jan 5, 2020, 1:59 AM

  • hello my post is too short so I have to write this unnecessary message

    by Maffeseater, Apr 17, 2019, 6:00 AM

  • Looks too much interesting, but it's really complex. :D

    Nice blog BTW. :)

    by integrated_JRC, Apr 29, 2018, 6:05 PM

  • This blog is a few lightyears above my head :D But unfortunately it's dead dead dead dead dead ...

    by Kayak, Dec 13, 2017, 9:49 AM

  • Great job....

    by Surajit123, Apr 16, 2013, 9:46 AM

  • I use Geogebra

    by RSM, Jun 5, 2012, 6:52 PM

  • can you suggest any tool(software) for digital geometrical drawings?

    by utsab001, Jun 5, 2012, 5:55 PM

  • Very nice blog with geometry, thank you :)!

    by buratinogigle, May 26, 2012, 5:11 AM

  • WOW, this stuff is complex! :o

    by littlejones, Apr 17, 2012, 5:29 AM

  • Most of the things have gone one mile above my head :( :o
    But surely an excellent blog I believe for people with sound knowledge of geometry. :)

    by a123, Apr 14, 2012, 6:28 AM

  • this blog has gave me a brief knowledge about the projective transformation. Awesome as usual! :lol:

    by BBAI, Mar 12, 2012, 4:07 PM

  • Wow.... I see he is targetting us...

    by iarnab_kundu, Mar 8, 2012, 5:51 AM

  • Did the elephatns make the place for the text smaller or not? I hope to find yet time to look more deeply in the texts of this big blog as told lost of times

    by SCP, Feb 19, 2012, 8:26 PM

  • Awesome, as usual. You should become an author of some journal very soon, it will help people.

    by Potla, Feb 13, 2012, 4:07 PM

  • I was really looking for projective transformation....
    great blog Chandan!!!!! :D :lol:

    by r1234, Feb 13, 2012, 11:53 AM

  • The matter's really dense here.

    by Dranzer, Dec 31, 2011, 7:03 AM

  • Hey!

    I started a website, Teen Thinkers, where I'm hoping to get a whole community of intellectually curious people to contribute. I have a math section, but there's not much really going on, would you like to help me vamp up that section?

    Thx pm alexgnow

    by alexgnow, Nov 19, 2011, 3:57 AM

  • :D :lol:
    Awesome blog

    by Carolstar9, Oct 20, 2011, 11:58 PM

  • :D nice chandan

    by sandeepreddy, Sep 6, 2011, 5:22 AM

  • :D
    great blog,chandan.
    U R A GENIUS
    as i told u at the camp

    by abhinavzandubalm, Aug 1, 2011, 2:25 PM

  • Awesome! Nice blog, Kosnita RSM!

    by Potla, Jun 22, 2011, 6:53 PM

  • 1st shout!

    by Goutham, Mar 31, 2011, 7:31 PM

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