## Search found 48 matches

Fri Jan 13, 2017 1:10 pm
Forum: Geometry
Replies: 1
Views: 1243

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...
Wed Jan 11, 2017 11:13 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52521

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, $\triangle PBA_2 \sim \triangle... Tue Jan 10, 2017 3:06 pm Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 110 Views: 52521 ### Re: Geometry Marathon : Season 3 Solution of problem 13 : Let$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'$are colinear . Tue Jan 10, 2017 2:43 pm Forum: Geometry Topic: IGO 2016 Elementary/5 Replies: 1 Views: 1067 ### Re: IGO 2016 Elementary/5$\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$.$...
Sat Jan 07, 2017 10:05 pm
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 110
Views: 52521

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 $X \cup ABC \sim \bar X \cup A\... Sat Jan 07, 2017 9:57 am Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 110 Views: 52521 ### Re: Geometry Marathon : Season 3 Solution to problem 7 : Let$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$ar... Sat Jan 07, 2017 12:50 am Forum: Geometry Topic: Geometry Marathon : Season 3 Replies: 110 Views: 52521 ### Re: Geometry Marathon : Season 3 Solution of problem 6 : Let$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_...
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.
Wed Jan 04, 2017 11:38 pm
Forum: Geometry
Topic: ISL 2012 G4: angle biscector with circumcenter
Replies: 3
Views: 1730