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$\text{Problem 33:}$

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.

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 $J$.

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 $\triangle ABC$.

$\text{Solution to Problem 34:}$

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 $EF=e+f-ef (\frac{d+e+f}{de+ef+fd})=\frac{def+e^2d+f^2d}{de+ef+fd}$. 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.

$\text{Problem 35:}$

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.

$\text{Problem 36:}$

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.

$\text{Solution to Problem 35:}$

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

$IP^2-IQ^2=OP^2-OQ^2$
or, $IB^2-IC^2=OP^2-OQ^2$
or, $BD^2-CD^2=OP^2-OQ^2$
or, $(BD+CD)(BD-CD)=OP^2-OQ^2$
or, $BC(AB-AC)=OP^2-OQ^2$.........$(0)$ [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,

$OP^2-OQ^2=BC.(AP-AQ)$......$(3)$

So, comparing $(0)$ and $(3)$, we're left to prove that $(AB-AC)=(AP-AQ)$ which can be proved easily by a little length chase. And we're done.

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.

$\text{Solution to Problem 37:}$

Let $KT$ meet $\omega$ at $L'$ and let $XL'$ meet $\omega$ at $S'$. Since by angle chasing we get $TS'||KX$ and also $TS||KX$, we get that $S'$ coincides with $S$. So, $L'$ coincides with $L$. Also $KP||TR$. Now, $\angle TKP=\angle RTK=\angle RSL=\angle LRQ \Rightarrow KRLQ$ is cyclic.

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 $XY$.