Search found 48 matches

by joydip
Thu Feb 16, 2017 9:45 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52527

Re: Geometry Marathon : Season 3

Prooblem 28 :

Let $ABC$ be a triangle with altitudes $AD,BE,CF$. $X,Y,Z$ are midpoints of $BC,CA,AB$. Consider circles $(X,XD),(Y,YE),(Z,ZF)$. Prove that radical center of $(X),(Y),(Z)$ is Nagel point of triangle $DEF$.
by joydip
Thu Feb 16, 2017 1:28 am
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52527

Re: Geometry Marathon : Season 3

Solution of problem 27 : Let $I_a,I_b,I_C$ be the excenters of $ABC$ opposite to $A,B,C$ respectively . $O$ be the center of $(I_aI_bI_C)$. $I_aO \cap BC=A_1$.Let $I_aO \cap (I_aI_bI_C)=A_2 ,I_a$.let $A_2I \cap (I_aI_bI_C)=A_2,A_3$ and $I_aA_3\cap I_bI_c=A_4,$Let points $A_5,A_6 \in I_bI_c$, such t...
by joydip
Tue Feb 14, 2017 9:08 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO 2017 National Round Secondary 9
Replies: 2
Views: 1636

Re: BdMO 2017 National Round Secondary 9

Let $QO$ meet $(DQC)$ again at $O'$, then $\angle DQX = \angle CQY \Rightarrow DO'=O'C$ implying $O=O'$. Then $FD .FC=FQ.FO \Rightarrow Q$ is the invertion of $F$ wrt $(ABCD)$ ,by brocard's theorm $EQ$ is the polar of $F$ wrt $(ABCD) \Rightarrow -1=(EF,EP;ED,EC)=(PF,PE;PD,PC)$. $\angle EPD = \angle...
by joydip
Mon Feb 13, 2017 1:36 am
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52527

Re: Geometry Marathon : Season 3

Solution of problem 22: Lemma : Let $ \omega_1$ & $ \omega_2$ be two circles with center $O_1 $ & $O_2$ respectively. $ \omega_1 \cap \omega_2 = A ,B$.Let the tangents to $ \omega_1$ at $A , B$ meet at $E$.Let $C,D \in \omega_2$ such that $ACBD$ is harmonic .$CE \cap \omega_2 = C,F $ and $FO_2 \cap...
by joydip
Mon Feb 06, 2017 10:18 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52527

Re: Geometry Marathon : Season 3

Problem 22:

Prove that the circumcenter of $\triangle ABC$ and the centroid of the anti pedal $\triangle A'B'C'$ of the symmedian point of $\triangle ABC$ coincide.
by joydip
Mon Feb 06, 2017 10:01 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52527

Re: Geometry Marathon : Season 3

Solution of problem 21: Let the circle with diameter $BC$ intersect $CO,CN,BO,BM$ again at $J,L,K,P$ respectively.Let $H$ denote the orthocenter.$U,V$ be the midpoints of $CK ,BJ$ respectively.$T'$ be the circumcenter of $UDV$. $L \in AB,P \in AC$. $\angle LBK=\angle PBC \Rightarrow LK=PC \Rightarr...
by joydip
Fri Feb 03, 2017 3:25 pm
Forum: Geometry
Topic: When the problem statement beats most horror stories
Replies: 1
Views: 1057

Re: When the problem statement beats most horror stories

Let $ M$ be the midpoint of $BC$ .$D’$ be the antipode of $ D$ wrt $(I)$.N be the reflection of $D$ wrt $M$. Then $D’= FT \cap DS$. By pascals theorem on hexagon $FD’D’DEF$ we get $D’R$ is tangent to $(I) $ .$AD’$ is the polar of $ R$ wrt $(I)$.So $AD’ \perp IR $.So $ P\in AD’$.Again $N \in AD’$ (we...
by joydip
Thu Feb 02, 2017 5:04 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52527

Re: Geometry Marathon : Season 3

Solution of problem 19: Let $M,N,L$ be the midpoints of $BX,XY,YC$ respectively . $\angle AXY=\angle XPB =\angle YPC=\angle AYX$ The spiral similarity centered at $K$ ,taking $BX$ to $YC$ takes $MX$ to $LC$ .So $K \in (AML)$. The spiral similarity centered at $Q$ ,taking $BX$ to $CY$ takes $MX$ to ...
by joydip
Tue Jan 31, 2017 11:48 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52527

Re: Geometry Marathon : Season 3

Another solution of problem 17: Let $AA_1$ meet $(ABC)$ again $M$ and $MA_2$ meet $(ABC)$ again at $A_3$.Define $B_3,C_3$ similarly.Let $J=OI \cap A_2A_3$ . Let $\omega (M,N)$ denote the power of point $M$ w.r.t circle $N$. An inversion with center $M$ and radius $MC$ takes $A_1 $ & $A_2$ to $A$ & ...
by joydip
Tue Jan 31, 2017 1:00 am
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52527

Re: Geometry Marathon : Season 3

Solution of problem 16 : Let $G$ and $G_1$ be the centroids of $\triangle ABC$ & $\triangle I_{a}I_{b}I_{c}$ respectively .$M$ be the midpoint of $NS$ .$I, G , N$ are collinear and $\dfrac {IG}{GN}=\dfrac{HG}{GS}=\dfrac {HI}{SN}=\dfrac {1}{2}$(well known) $\Rightarrow HI \parallel NS$. As $O$ is th...