Search found 48 matches

by joydip
Fri Jan 13, 2017 1:10 pm
Forum: Geometry
Topic: IGO 2016 Advanced/3
Replies: 1
Views: 1243

Re: IGO 2016 Advanced/3

A generalization of the problem : Let $P$ be the intersection point of sides $AD$ and $BC$ of a convex qualrilateral $ABCD$. Suppose that $I_1$ and $I_2$ are the incenters of $\triangle PAB $ & $\triangle PDC$, respectively. Let $O$ be the circumcenter of $\triangle PAB$, and $H $ the orthocenter o...
by joydip
Wed Jan 11, 2017 11:13 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52521

Re: Geometry Marathon : Season 3

Solution to problem 15 : Let $H$ be the orthocenter of $\triangle ABC$. $\angle A_2BP= \angle A_2BC -\angle B_1BC=\angle A_1BC -\angle B_1AC=\angle A_1AC -\angle B_2AC=\angle B_2AP $ Now, $\dfrac {BA_2}{BP}=\dfrac {BA_1}{BP}=\dfrac {AB_1}{AP}=\dfrac {AB_2}{AP}$ . So, $\triangle PBA_2 \sim \triangle...
by joydip
Tue Jan 10, 2017 3:06 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52521

Re: Geometry Marathon : Season 3

Solution of problem 13 :

Let $AB \cap PA' = M $ , $AP \cap BC =N$ , $BC \cap PB' =K$ , $AA' \cap PB' = L$

$(M,A ;C',B)=(PA',PA;PC',PB)=(PA,PA';PC,PB')=A(N,A';C,K)=(P,L;B',K)$
$\Rightarrow (A'M,A'A ;A'C',A'B)=(A'P,A'L;A'B',A'k)$
So,$A',B',C'$ are colinear .
by joydip
Tue Jan 10, 2017 2:43 pm
Forum: Geometry
Topic: IGO 2016 Elementary/5
Replies: 1
Views: 1067

Re: IGO 2016 Elementary/5

$\angle DAB +\angle ABD +\angle ADB =180^\circ \Rightarrow 2\angle CBD + \angle ABD + 4\angle CBD +\angle ABD =180^\circ $ So,$\angle CBD +\angle ABD +2\angle CBD = 90^\circ \Rightarrow \angle ABC +\angle DAB =90^\circ $ So, $BC \perp AD$ .$E$ be the reflection of $D$ W.R.T $BC$ .Then $ E \in AD$. $...
by joydip
Sat Jan 07, 2017 10:05 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52521

Re: Geometry Marathon : Season 3

Solution of problem 9: Assume $Q',P_B,P_C$ are collinear. Let a circle $ \omega $, going though $B,C$ intersect $AB,AC$ at $\bar C ,\bar B$ respectively such that $A$ lies in the segment $B\bar C$ . Let $\Gamma $ be the transformation taking $X$ to $\bar X$ such that $X \cup ABC \sim \bar X \cup A\...
by joydip
Sat Jan 07, 2017 9:57 am
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52521

Re: Geometry Marathon : Season 3

Solution to problem 7 : Let $H$ and $O$ be the orthocenter and circumcenter of $\triangle IBC$ respectively.Let $\omega$ be the nine point circle of $\triangle IBC$,$M,N$ be the midpoints of $BC,IH $ respectively.$(I)$ touches $BC$ at $D$.Let $K$ be the reflection of $D$ W.R.T $AI$. Then $A,I,O$ ar...
by joydip
Sat Jan 07, 2017 12:50 am
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52521

Re: Geometry Marathon : Season 3

Solution of problem 6 : Let $I_3$ be the incenter of $\triangle CDE$ . Let $CI_3 \cap I_1I_2=S$ $\triangle FBD \sim \triangle FEA \Rightarrow \triangle FBI_2 \sim \triangle FEI_1 \Rightarrow \triangle FI_2I_1 \sim \triangle FBE$. So,$\angle BI_2I_1+\angle I_2AB=\angle BI_2F+\angle FI_2I_1+\angle I_...
by joydip
Fri Jan 06, 2017 9:42 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52521

Re: Geometry Marathon : Season 3

Solution of problem 4 :

Let $B_1F$ meet (ABC) again at $K$, $KC_1\cap AB = D_1$.Applying pascal's theorem on hexagon $BACC_1KB_1$ we get $ CC_1 \| BB_1 \| FD_1$ . So , $D=D_1$. So $K,D,C_1$ are collinear. Similerly $ K,E,A_1$ are collinear.
by joydip
Wed Jan 04, 2017 11:38 pm
Forum: Geometry
Topic: ISL 2012 G4: angle biscector with circumcenter
Replies: 3
Views: 1730

Re: ISL 2012 G4: angle biscector with circumcenter

Let , $AD$ meet $( ABC )$ again at $L$ .Then $OL \perp BC$ . $M$ be the midpoint of $BC$ , $K$ be the reflection of $L$ W.R.T $ BC$ and $Y'$ be the reflection of $Y$ W.R.T $ OL$ .$ \angle DKL= \angle KLD=\angle XDA=\angle XAD $ gives $\triangle DAX \sim \triangle KLD \Rightarrow \dfrac {KL}{LD}=\df...
by joydip
Thu Dec 22, 2016 10:34 pm
Forum: Geometry
Topic: Nice geo
Replies: 2
Views: 1374

Re: Nice geo

Let $(O)$ be the circle with diameter $AH$.$D$ and $F$ be the projection of $C$ and $B$ on $AB$ and $AC$ respectively. $A,F,P,D,Q \in (O).\angle QAP=90^\circ \Rightarrow Q,O,P$ are collinear. $\angle DAP =\angle PAF \Rightarrow OP$ bisect $DF$ .By Newton's theorem on $ADHF$, $OM$ bisect $DF \Righta...