$\text{Another Solution:}$Raiyan Jamil wrote:$\text{Problem 3:}$
In Acute angled triangle $ABC$, let $D$ be the point where $A$ angle bisector meets $BC$. The perpendicular from $B$ to $AD$ meets the circumcircle of $ABD$ at $E$. If $O$ is the circumcentre of triangle $ABC$ then prove that $A,E$ and $O$ are collinear.
Geometry Marathon : Season 3
Re: Geometry Marathon : Season 3
"Questions we can't answer are far better than answers we can't question"
Re: Geometry Marathon : Season 3
Problem 5:
Let $I$ be an incenter of $\triangle ABC$. Denote $D, \ S \neq A$ intersections of $AI$ with $BC, \ (ABC)$ respectively. Let $K, \ L$ be incenters of $\triangle DSB, \ \triangle DCS$. Let $P$ be a reflection of $I$ with the respect to $KL$. Prove that $BP \perp CP$.
Let $I$ be an incenter of $\triangle ABC$. Denote $D, \ S \neq A$ intersections of $AI$ with $BC, \ (ABC)$ respectively. Let $K, \ L$ be incenters of $\triangle DSB, \ \triangle DCS$. Let $P$ be a reflection of $I$ with the respect to $KL$. Prove that $BP \perp CP$.
Last edited by tanmoy on Sat Jan 07, 2017 11:15 am, edited 3 times in total.
"Questions we can't answer are far better than answers we can't question"
Re: Geometry Marathon : Season 3
Solution of problem 5:
we know,$SB=SI=SC$
since $K$ is on the angle bisector of $\angle$BSI
$KB=KI$
And since $P$ is the reflection of $I$ across $KL$
$KP=KI$
SO $K$ is the circumcentre of $\triangle$BPI
Similarly $L$ is the circumcentre of $\triangle$IPC
Now,$\angle$BPC= $\angle$BPI+$\angle$IPC= $\frac{1}{2}(\angle$BKI+$\angle$ILC)=$\frac{1}{2}$(1802($\angle$IBK)+ 180 2($\angle$ICL))= 1/2(1802($\frac{1}{4}\angle$A+$\frac{1}{2}\angle$B)+ 180 2($\frac{1}{4}\angle$A+$\frac{1}{2}\angle$C))=90
we know,$SB=SI=SC$
since $K$ is on the angle bisector of $\angle$BSI
$KB=KI$
And since $P$ is the reflection of $I$ across $KL$
$KP=KI$
SO $K$ is the circumcentre of $\triangle$BPI
Similarly $L$ is the circumcentre of $\triangle$IPC
Now,$\angle$BPC= $\angle$BPI+$\angle$IPC= $\frac{1}{2}(\angle$BKI+$\angle$ILC)=$\frac{1}{2}$(1802($\angle$IBK)+ 180 2($\angle$ICL))= 1/2(1802($\frac{1}{4}\angle$A+$\frac{1}{2}\angle$B)+ 180 2($\frac{1}{4}\angle$A+$\frac{1}{2}\angle$C))=90
Last edited by rubab on Sat Jan 07, 2017 12:00 am, edited 2 times in total.
Re: Geometry Marathon : Season 3
Problem 6
In an acute triangle $ABC$ the points $D$,$E$ and $F$ are the feet of the altitudes through A,B,C respectively. The incenters of the triangle $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.
In an acute triangle $ABC$ the points $D$,$E$ and $F$ are the feet of the altitudes through A,B,C respectively. The incenters of the triangle $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.
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_1AB=90^\circ +\dfrac {B}{2}+90^\circB+\dfrac {B}{2} =180^\circ $. So $BI_2I_1A$ is cyclic. Simillerly $BI_2I_3C,CI_3I_1A$ are cyclic .
$\angle I_2I_3S +\angle I_3I_2I_1=\angle CBI_2+\angle I_1AB +\angle I_3CB=\dfrac {A}{2}+\dfrac {B}{2}+\dfrac {C}{2}= 90^\circ \Rightarrow CI_3\perp I_1I_2$,but $CI_3\perp O_1O_2 \Rightarrow I_1I_2 \parallel O_1O_2$
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_1AB=90^\circ +\dfrac {B}{2}+90^\circB+\dfrac {B}{2} =180^\circ $. So $BI_2I_1A$ is cyclic. Simillerly $BI_2I_3C,CI_3I_1A$ are cyclic .
$\angle I_2I_3S +\angle I_3I_2I_1=\angle CBI_2+\angle I_1AB +\angle I_3CB=\dfrac {A}{2}+\dfrac {B}{2}+\dfrac {C}{2}= 90^\circ \Rightarrow CI_3\perp I_1I_2$,but $CI_3\perp O_1O_2 \Rightarrow I_1I_2 \parallel O_1O_2$
The first principle is that you must not fool yourself and you are the easiest person to fool.
Re: Geometry Marathon : Season 3
problem 7:
Let $I$ be the incenter of triangle $ABC$. Prove that the nine point circles of triangle $AIB$, $BIC$ and $CIA$ are concurrent at the feuerbach point of triangle $ABC$.
Let $I$ be the incenter of triangle $ABC$. Prove that the nine point circles of triangle $AIB$, $BIC$ and $CIA$ are concurrent at the feuerbach point of triangle $ABC$.
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 $A,I,O$ are collinear .$IN = \dfrac {IH}{2}=OM \Rightarrow INMO$ is a parallelogram .So, $IO \ MN$.Now,$KD \perp AI \Rightarrow KD \perp MN$.Let $KD$ meet $\omega$ again at $P$.As $MN$ is a diameter of $\omega$,so $MD=MP$
Let$ \psi$ be the circle with center $M$ and radios $MD$.Then the inversion W.R.T $\psi$ sends $\omega$ to $DP$ .This inversion sends the nine point circle of $\triangle ABC$ to the tangent of $(I)$ at $K$ and $(I)$ to $(I)$.So,it sends the feuerbach point($F_e$) to $K$.so,$ K\in DP \Rightarrow F_e\in \omega$,completing the proof.
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 $A,I,O$ are collinear .$IN = \dfrac {IH}{2}=OM \Rightarrow INMO$ is a parallelogram .So, $IO \ MN$.Now,$KD \perp AI \Rightarrow KD \perp MN$.Let $KD$ meet $\omega$ again at $P$.As $MN$ is a diameter of $\omega$,so $MD=MP$
Let$ \psi$ be the circle with center $M$ and radios $MD$.Then the inversion W.R.T $\psi$ sends $\omega$ to $DP$ .This inversion sends the nine point circle of $\triangle ABC$ to the tangent of $(I)$ at $K$ and $(I)$ to $(I)$.So,it sends the feuerbach point($F_e$) to $K$.so,$ K\in DP \Rightarrow F_e\in \omega$,completing the proof.
The first principle is that you must not fool yourself and you are the easiest person to fool.
Re: Geometry Marathon : Season 3
Problem 8:
Given a cyclic quadrilateral $ABCD$ with circumcircle $(O)$. Let $AB \cap CD=E, \ AD \cap BC=F, \ AC \cap BD=G, \ AC \cap EF=P, \ BD \cap EF=Q$. Let $M, \ N$ be midpoints of $AC, \ BD$, respectively and let $MN \cap EF=H$.
(i) Prove that $M, \ N, \ P, \ Q$ are concyclic.
(ii) Let $K$ be the center of the circumcircle of $MNPQ$. $OK \cap EF=L$, $GL$ cuts $(O)$ at two distinct points $S$ and $T$. Prove that $HS, \ HT$ are tangents from $H$ to $(O)$ at $S$ and $T$, respectively.
Given a cyclic quadrilateral $ABCD$ with circumcircle $(O)$. Let $AB \cap CD=E, \ AD \cap BC=F, \ AC \cap BD=G, \ AC \cap EF=P, \ BD \cap EF=Q$. Let $M, \ N$ be midpoints of $AC, \ BD$, respectively and let $MN \cap EF=H$.
(i) Prove that $M, \ N, \ P, \ Q$ are concyclic.
(ii) Let $K$ be the center of the circumcircle of $MNPQ$. $OK \cap EF=L$, $GL$ cuts $(O)$ at two distinct points $S$ and $T$. Prove that $HS, \ HT$ are tangents from $H$ to $(O)$ at $S$ and $T$, respectively.
"Questions we can't answer are far better than answers we can't question"

 Posts: 16
 Joined: Wed Aug 10, 2016 1:29 am
Re: Geometry Marathon : Season 3
$\text{Solution of problem 8:}$
Last edited by M Ahsan Al Mahir on Sun Jan 08, 2017 10:28 pm, edited 2 times in total.
 Thanic Nur Samin
 Posts: 176
 Joined: Sun Dec 01, 2013 11:02 am
Re: Geometry Marathon : Season 3
Problem 9
Let $\{P, P'\}$ and $\{Q,Q'\}$ be two pairs of isogonal conjugates of $\triangle ABC$. Let $\triangle P_AP_BP_C$ be the cevain triangle of $P$ wrt $\triangle ABC$. Define $\triangle Q_AQ_BQ_C$ similarly.
Prove that, $P_B, P_C$ and $Q'$ are collinear if and only if $Q_B,Q_C$ and $P'$ are collinear.
Let $\{P, P'\}$ and $\{Q,Q'\}$ be two pairs of isogonal conjugates of $\triangle ABC$. Let $\triangle P_AP_BP_C$ be the cevain triangle of $P$ wrt $\triangle ABC$. Define $\triangle Q_AQ_BQ_C$ similarly.
Prove that, $P_B, P_C$ and $Q'$ are collinear if and only if $Q_B,Q_C$ and $P'$ are collinear.
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.