Geometry Marathon : Season 3
 Raiyan Jamil
 Posts:138
 Joined:Fri Mar 29, 2013 3:49 pm
$\text{Problem 33:}$
Let $\bigtriangleup ABC$ be a scalene triangle with circumcenter $O$ and incenter $I$. Let $H, K, L$ be the feet of the altitudes of $\bigtriangleup ABC$ from the vertices $A, B, C$ respectively. Denote by $A_0, B_0, C_0$ the midpoints of these altitudes $AH, BK, CL$ respectively. The incircle of $\bigtriangleup ABC$ touches the sides $BC, CA, AB$ at the points $D, E, F$ respectively. Prove that the four lines $A_0D, B_0E, C_0F$ and $OI$ are concurrent.
Let $\bigtriangleup ABC$ be a scalene triangle with circumcenter $O$ and incenter $I$. Let $H, K, L$ be the feet of the altitudes of $\bigtriangleup ABC$ from the vertices $A, B, C$ respectively. Denote by $A_0, B_0, C_0$ the midpoints of these altitudes $AH, BK, CL$ respectively. The incircle of $\bigtriangleup ABC$ touches the sides $BC, CA, AB$ at the points $D, E, F$ respectively. Prove that the four lines $A_0D, B_0E, C_0F$ and $OI$ are concurrent.
A smile is the best way to get through a tough situation, even if it's a fake smile.
Re: Geometry Marathon : Season 3
Solution of problem 33:
Let $I_a$ denote the excenter opposite to $A$ .The $A$ excircle touches $BC$ at $P$ .Let $A_0D \cap OI = J$ & $AI \cap BC=K$.Let the perpendiculer from $O$ to $BC$ meet $AI$ & $A_0D$ at $M,N$ respectively.
Lemma : $A_0 ,D ,I_a$ are collinear.
Proof :Let $I_aD$ meet $AH$ at $A_1$ . $A$ & $K$ are the internal and external center of similitude of $(I)$ & $(I_a)$.So $1=( A,K;I,I_a)=(H,K;D,P)=(I_aH,I_aA;I_aD,I_aP)$.As $AH \parallel I_aP$ ,so $(H,A;A_1,\infty)=1$.So $A_1$ is the midpoint of $AH \Rightarrow A_0=A_1$
Solution: $M$ is the midpoint of $II_a$. As $MN \parallel ID \Rightarrow MN=\dfrac {ID}{2}=\dfrac {r}{2}.$
So , $\dfrac {JI}{JO}=\dfrac {ID}{ON}=\dfrac {ID}{MN+OM}=\dfrac {r}{\dfrac {r}{2}+R}=\dfrac {2r}{2R+r}$
As the ratio is constant so $A_0D, B_0E, C_0F$ concur in $J$.
Let $I_a$ denote the excenter opposite to $A$ .The $A$ excircle touches $BC$ at $P$ .Let $A_0D \cap OI = J$ & $AI \cap BC=K$.Let the perpendiculer from $O$ to $BC$ meet $AI$ & $A_0D$ at $M,N$ respectively.
Lemma : $A_0 ,D ,I_a$ are collinear.
Proof :Let $I_aD$ meet $AH$ at $A_1$ . $A$ & $K$ are the internal and external center of similitude of $(I)$ & $(I_a)$.So $1=( A,K;I,I_a)=(H,K;D,P)=(I_aH,I_aA;I_aD,I_aP)$.As $AH \parallel I_aP$ ,so $(H,A;A_1,\infty)=1$.So $A_1$ is the midpoint of $AH \Rightarrow A_0=A_1$
Solution: $M$ is the midpoint of $II_a$. As $MN \parallel ID \Rightarrow MN=\dfrac {ID}{2}=\dfrac {r}{2}.$
So , $\dfrac {JI}{JO}=\dfrac {ID}{ON}=\dfrac {ID}{MN+OM}=\dfrac {r}{\dfrac {r}{2}+R}=\dfrac {2r}{2R+r}$
As the ratio is constant so $A_0D, B_0E, C_0F$ concur in $J$.
The first principle is that you must not fool yourself and you are the easiest person to fool.
Re: Geometry Marathon : Season 3
Problem 34:
Let $O$ & $I$ denote the circumcenter & incenter of $\triangle ABC$ respectively.Prove that The reflections of the
$OI$ line in the sides of the intouch triangle of $\triangle ABC$ concur at the Feuerbach point of $\triangle ABC$.
Let $O$ & $I$ denote the circumcenter & incenter of $\triangle ABC$ respectively.Prove that The reflections of the
$OI$ line in the sides of the intouch triangle of $\triangle ABC$ concur at the Feuerbach point of $\triangle ABC$.
The first principle is that you must not fool yourself and you are the easiest person to fool.
 Raiyan Jamil
 Posts:138
 Joined:Fri Mar 29, 2013 3:49 pm
Re: Geometry Marathon : Season 3
$\text{Solution to Problem 34:}$
Let, $\bigtriangleup DEF$ be the intouch triangle. $M$ be the midpoint of $AI$ and $D_1$ be the antipode of $D$ with respect to the incircle. The Feuerbach point be $F_1$. $P,Q$ be the midpoints of the two arcs of the incircle created by the points $E,F$.
$\text{Lemma 1:}$ $F_1,D_1,M$ are collinear.
$\text{Proof:}$ From the Problem 10 of this Geometry Marathon.
$\text{Lemma 2:}$ $AI$ goes through $P,Q$
$\text{Proof:}$ Trivial.
$\text{Lemma 3:}$ $OI$ is the $\text{Euler line}$ of $\bigtriangleup DEF$.
$\text{Proof:}$ From the lemma 2 of the Solution of Problem 32 in this Geometry Marathon.
Now, we use complex numbers. We take the incircle as the unit circle. Let the complex number of $F_1$ be $x$. Then, by using the first two lemmas and intersection formula on the two chords $F_1C$ and $PQ$, we solve for $x$=$\frac{de+ef+fe}{d+e+f}$.
Then, we are only left to show that the reflection of $F_1$ under $DE,EF,FD$ lies on the $\text{Euler line}$ from the lemma 3. Since this is symmetric, it is enough to show for just one. We get the complex number of the reflection of $F_1$ under $EF=e+fef (\frac{d+e+f}{de+ef+fd})=\frac{def+e^2d+f^2d}{de+ef+fd}$. So, we're left to show that $\frac{def+e^2d+f^2d}{(de+ef+fd)(d+e+f)}$ is a real number which can be shown easily. And we're done.
Let, $\bigtriangleup DEF$ be the intouch triangle. $M$ be the midpoint of $AI$ and $D_1$ be the antipode of $D$ with respect to the incircle. The Feuerbach point be $F_1$. $P,Q$ be the midpoints of the two arcs of the incircle created by the points $E,F$.
$\text{Lemma 1:}$ $F_1,D_1,M$ are collinear.
$\text{Proof:}$ From the Problem 10 of this Geometry Marathon.
$\text{Lemma 2:}$ $AI$ goes through $P,Q$
$\text{Proof:}$ Trivial.
$\text{Lemma 3:}$ $OI$ is the $\text{Euler line}$ of $\bigtriangleup DEF$.
$\text{Proof:}$ From the lemma 2 of the Solution of Problem 32 in this Geometry Marathon.
Now, we use complex numbers. We take the incircle as the unit circle. Let the complex number of $F_1$ be $x$. Then, by using the first two lemmas and intersection formula on the two chords $F_1C$ and $PQ$, we solve for $x$=$\frac{de+ef+fe}{d+e+f}$.
Then, we are only left to show that the reflection of $F_1$ under $DE,EF,FD$ lies on the $\text{Euler line}$ from the lemma 3. Since this is symmetric, it is enough to show for just one. We get the complex number of the reflection of $F_1$ under $EF=e+fef (\frac{d+e+f}{de+ef+fd})=\frac{def+e^2d+f^2d}{de+ef+fd}$. So, we're left to show that $\frac{def+e^2d+f^2d}{(de+ef+fd)(d+e+f)}$ is a real number which can be shown easily. And we're done.
A smile is the best way to get through a tough situation, even if it's a fake smile.
 Raiyan Jamil
 Posts:138
 Joined:Fri Mar 29, 2013 3:49 pm
Re: Geometry Marathon : Season 3
$\text{Problem 35:}$
Let $ABC$ be a triangle with incenter $I$ . Let $P$ and $Q$ denote the reflections of $B$ and $C$ across $CI$ and $BI$ respectively. Show that $PQ\perp OI$, where $O$ is the circumcenter of ABC.
Let $ABC$ be a triangle with incenter $I$ . Let $P$ and $Q$ denote the reflections of $B$ and $C$ across $CI$ and $BI$ respectively. Show that $PQ\perp OI$, where $O$ is the circumcenter of ABC.
A smile is the best way to get through a tough situation, even if it's a fake smile.

 Posts:16
 Joined:Wed Aug 10, 2016 1:29 am
Re: Geometry Marathon : Season 3
$\text{Problem 36:}$
Let $ABC$ be a triangle and $O$ be its circumcenter. A point $P$ is on the internal angle bisector of $\angle B$. Let $(P)$ be the circle that touches $BC$ and $BA$ at $X, Y$. Prove that the reflection of $OP$ wrt $XY$ passes throught the midpoint of $BH$.
I request Geodip bro to enlighten us with his beautiful SYNTHETIC solution to Problem 35 and 36.
Let $ABC$ be a triangle and $O$ be its circumcenter. A point $P$ is on the internal angle bisector of $\angle B$. Let $(P)$ be the circle that touches $BC$ and $BA$ at $X, Y$. Prove that the reflection of $OP$ wrt $XY$ passes throught the midpoint of $BH$.
I request Geodip bro to enlighten us with his beautiful SYNTHETIC solution to Problem 35 and 36.
Last edited by M Ahsan Al Mahir on Mon Feb 27, 2017 8:02 pm, edited 1 time in total.
 Raiyan Jamil
 Posts:138
 Joined:Fri Mar 29, 2013 3:49 pm
Re: Geometry Marathon : Season 3
$\text{Solution to Problem 35:}$
Let $D$ be the feet of perpendicular from $I$ on $BC$. And let $R$ be the circumradius of $\bigtriangleup ABC$.
Obviously, $P,Q$ lie on $AC,AB$ respetively such that $BQ=BC=CP$.Also $IP=IB$ and $IQ=IC$. We will use the perpendicular lemma and show that
$IP^2IQ^2=OP^2OQ^2$
or, $IB^2IC^2=OP^2OQ^2$
or, $BD^2CD^2=OP^2OQ^2$
or, $(BD+CD)(BDCD)=OP^2OQ^2$
or, $BC(ABAC)=OP^2OQ^2$.........$(0)$ [Not proved yet]
Now, we use power of point and get,
$OP^2=R^2+AP.CP=R^2+AP.BC$......$(1)$
$OQ^2=R^2+AQ.BQ=R^2+AQ.BC$.......$(2)$
Subtracting $(2)$ from $(1)$, we get,
$OP^2OQ^2=BC.(APAQ)$......$(3)$
So, comparing $(0)$ and $(3)$, we're left to prove that $(ABAC)=(APAQ)$ which can be proved easily by a little length chase. And we're done.
Let $D$ be the feet of perpendicular from $I$ on $BC$. And let $R$ be the circumradius of $\bigtriangleup ABC$.
Obviously, $P,Q$ lie on $AC,AB$ respetively such that $BQ=BC=CP$.Also $IP=IB$ and $IQ=IC$. We will use the perpendicular lemma and show that
$IP^2IQ^2=OP^2OQ^2$
or, $IB^2IC^2=OP^2OQ^2$
or, $BD^2CD^2=OP^2OQ^2$
or, $(BD+CD)(BDCD)=OP^2OQ^2$
or, $BC(ABAC)=OP^2OQ^2$.........$(0)$ [Not proved yet]
Now, we use power of point and get,
$OP^2=R^2+AP.CP=R^2+AP.BC$......$(1)$
$OQ^2=R^2+AQ.BQ=R^2+AQ.BC$.......$(2)$
Subtracting $(2)$ from $(1)$, we get,
$OP^2OQ^2=BC.(APAQ)$......$(3)$
So, comparing $(0)$ and $(3)$, we're left to prove that $(ABAC)=(APAQ)$ which can be proved easily by a little length chase. And we're done.
A smile is the best way to get through a tough situation, even if it's a fake smile.
 ahmedittihad
 Posts:181
 Joined:Mon Mar 28, 2016 6:21 pm
Re: Geometry Marathon : Season 3
Problem $37$
Point $A$ is outside of a given circle $\omega$. Let the tangent from $A$ to $\omega$ meet $\omega$ at $S$,$T$.
$X$,$Y$ are the midpoints of $AT$,$AS$. Let the tangent from $X$ to $\omega$ meet $\omega$ at $R \neq T$ Points $P$,$Q$ are the midpoints of $XT$,$XR$. Let $XY \cap PQ = K$, $SX \cap TK = L$. Prove that the quadrilateral $KRLQ$ is cyclic.
Point $A$ is outside of a given circle $\omega$. Let the tangent from $A$ to $\omega$ meet $\omega$ at $S$,$T$.
$X$,$Y$ are the midpoints of $AT$,$AS$. Let the tangent from $X$ to $\omega$ meet $\omega$ at $R \neq T$ Points $P$,$Q$ are the midpoints of $XT$,$XR$. Let $XY \cap PQ = K$, $SX \cap TK = L$. Prove that the quadrilateral $KRLQ$ is cyclic.
Frankly, my dear, I don't give a damn.
 Raiyan Jamil
 Posts:138
 Joined:Fri Mar 29, 2013 3:49 pm
Re: Geometry Marathon : Season 3
$\text{Solution to Problem 37:}$
Let $KT$ meet $\omega$ at $L'$ and let $XL'$ meet $\omega$ at $S'$. Since by angle chasing we get $TS'KX$ and also $TSKX$, we get that $S'$ coincides with $S$. So, $L'$ coincides with $L$. Also $KPTR$. Now, $\angle TKP=\angle RTK=\angle RSL=\angle LRQ \Rightarrow KRLQ$ is cyclic.
Let $KT$ meet $\omega$ at $L'$ and let $XL'$ meet $\omega$ at $S'$. Since by angle chasing we get $TS'KX$ and also $TSKX$, we get that $S'$ coincides with $S$. So, $L'$ coincides with $L$. Also $KPTR$. Now, $\angle TKP=\angle RTK=\angle RSL=\angle LRQ \Rightarrow KRLQ$ is cyclic.
A smile is the best way to get through a tough situation, even if it's a fake smile.
Re: Geometry Marathon : Season 3
Solution of problem 36:
Let $S$ be the midpoint of arc $AC$ (containing $B$) & $Q$ be the midpoint of arc $AC$ (not containing $B$).$R$ be the reflection of point $P$ wrt $XY$.Now $M$ be the midpoint of $AC$ & $K$ be the orthocenter of $\triangle SAC $.$N$ be the midpoint of $BH$.
$K$ is the reflection of $Q$ wrt $AC$, $SK=2OM =BH$.Now $\angle YBX=\angle ASC$.So $PXBYR \sim QCSAK \Rightarrow \dfrac {BR}{SK}=\dfrac {BP}{SQ} \Rightarrow \dfrac {BR}{BH}=\dfrac {BP}{2BO} \Rightarrow \dfrac {BR}{BN}=\dfrac {BP}{BO}$
$\angle ABH=\angle OBC \Rightarrow \angle NBP=\angle PBO \Rightarrow \triangle NBP \sim \triangle OBP$ .So $\angle NRP=180^\circ \angle RPO \Rightarrow NR $ is the reflection of $OP$ wrt $XY$.
Let $S$ be the midpoint of arc $AC$ (containing $B$) & $Q$ be the midpoint of arc $AC$ (not containing $B$).$R$ be the reflection of point $P$ wrt $XY$.Now $M$ be the midpoint of $AC$ & $K$ be the orthocenter of $\triangle SAC $.$N$ be the midpoint of $BH$.
$K$ is the reflection of $Q$ wrt $AC$, $SK=2OM =BH$.Now $\angle YBX=\angle ASC$.So $PXBYR \sim QCSAK \Rightarrow \dfrac {BR}{SK}=\dfrac {BP}{SQ} \Rightarrow \dfrac {BR}{BH}=\dfrac {BP}{2BO} \Rightarrow \dfrac {BR}{BN}=\dfrac {BP}{BO}$
$\angle ABH=\angle OBC \Rightarrow \angle NBP=\angle PBO \Rightarrow \triangle NBP \sim \triangle OBP$ .So $\angle NRP=180^\circ \angle RPO \Rightarrow NR $ is the reflection of $OP$ wrt $XY$.
The first principle is that you must not fool yourself and you are the easiest person to fool.