Geometry Marathon : Season 3
 Raiyan Jamil
 Posts: 138
 Joined: Fri Mar 29, 2013 3:49 pm
Re: Geometry Marathon : Season 3
$\text{Solution of Problem 28 :}$
Let, $T$ be the ninepoint circle of $\bigtriangleup ABC$. Let $l_y$ be the radical axis of $T,(Y)$ and $l_z$ be the radical axis of $T,(Z)$.And let $A_1$=$l_y \cap l_z$. And the nagel point and centroid of $\bigtriangleup DEF$ be $M$ and $P$ respectively.
$\text{Lemma 1:}$ $A_1FDE$ is a parallelogram.
$\text{Proof:}$ Simple angle and length chasing.
$\text{Lemma 2:}$ $H,P,M$ are collinear and $P$ divides $HM$ in the ration $1:2$.
$\text{Proof:}$ It is well known that $H$ is the incentre of $DEF$. And it is also well known that the centroid,incentre,nagel point of a triangle are collinear and the centroid divides line joining incentre and nagel point in the ratio $1:2$. So, the conclusion follows.
$\text{Lemma 3:}$ Let the midpoint of $EF$ be $U$. Then the reflection of $M$ under $U$ lies on the angle bisector of $\angle FDE$ i.e. $AD$.
$\text{Proof:}$ Let the reflection of $D$ under $H$ be $D'$. Applying Menelaus theorem to triangle $DPH$ and the points $D',M,U$ we get that they are collinear. Then applying Menelaus theorem to $DUD'$ and points $H,P,M,$ we get that $D'$ is the reflection of $M$ under $U$ and thus the conclusion follows.
$\text{Lemma 4:}$ $M$ lies on the angle bisector of $\angle FA_1E$.
$\text{Proof:}$ Since the reflection of $M$ under $U$ lies on the angle bisector of $\angle EDF$, applying symmetry to the whole figure gives us the result.
$\text{Lemma 5:}$ The angle bisector of $\angle FA_1E$ is perpendicular to YZ and BC.
$\text{Proof:}$ The angle bisector of $\angle FA_1E$ is parallel to $AD$ because of symmetry and the conclusion thus folllows as $AD$ is perpendicular to $BC$ and $YZ$.
$\text{Lemma 6:}$ $A_1$ lies on the radical axis of $(Y),(Z)$.
$\text{Proof:}$ Using radical axis theorem in circles $(Y),(Z),T$ the conclusion follows.
$\text{Lemma 7:}$ Radical axis of two circles is perpendicular to the line joining their centres.
$\text{Proof:}$ Well known.
Now, at last, by lemmas 4,5,6,7 we get that $A_1M$ is the radical axis of $(Y)$ and $(Z)$. And we come to a conclusion that $M$ is the radical centre of $(X),(Y),(Z)$.
Let, $T$ be the ninepoint circle of $\bigtriangleup ABC$. Let $l_y$ be the radical axis of $T,(Y)$ and $l_z$ be the radical axis of $T,(Z)$.And let $A_1$=$l_y \cap l_z$. And the nagel point and centroid of $\bigtriangleup DEF$ be $M$ and $P$ respectively.
$\text{Lemma 1:}$ $A_1FDE$ is a parallelogram.
$\text{Proof:}$ Simple angle and length chasing.
$\text{Lemma 2:}$ $H,P,M$ are collinear and $P$ divides $HM$ in the ration $1:2$.
$\text{Proof:}$ It is well known that $H$ is the incentre of $DEF$. And it is also well known that the centroid,incentre,nagel point of a triangle are collinear and the centroid divides line joining incentre and nagel point in the ratio $1:2$. So, the conclusion follows.
$\text{Lemma 3:}$ Let the midpoint of $EF$ be $U$. Then the reflection of $M$ under $U$ lies on the angle bisector of $\angle FDE$ i.e. $AD$.
$\text{Proof:}$ Let the reflection of $D$ under $H$ be $D'$. Applying Menelaus theorem to triangle $DPH$ and the points $D',M,U$ we get that they are collinear. Then applying Menelaus theorem to $DUD'$ and points $H,P,M,$ we get that $D'$ is the reflection of $M$ under $U$ and thus the conclusion follows.
$\text{Lemma 4:}$ $M$ lies on the angle bisector of $\angle FA_1E$.
$\text{Proof:}$ Since the reflection of $M$ under $U$ lies on the angle bisector of $\angle EDF$, applying symmetry to the whole figure gives us the result.
$\text{Lemma 5:}$ The angle bisector of $\angle FA_1E$ is perpendicular to YZ and BC.
$\text{Proof:}$ The angle bisector of $\angle FA_1E$ is parallel to $AD$ because of symmetry and the conclusion thus folllows as $AD$ is perpendicular to $BC$ and $YZ$.
$\text{Lemma 6:}$ $A_1$ lies on the radical axis of $(Y),(Z)$.
$\text{Proof:}$ Using radical axis theorem in circles $(Y),(Z),T$ the conclusion follows.
$\text{Lemma 7:}$ Radical axis of two circles is perpendicular to the line joining their centres.
$\text{Proof:}$ Well known.
Now, at last, by lemmas 4,5,6,7 we get that $A_1M$ is the radical axis of $(Y)$ and $(Z)$. And we come to a conclusion that $M$ is the radical centre of $(X),(Y),(Z)$.
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
Another solution to Problem 28, i think this solu is cute sooo...
We can state the problem in a new way:
Let $ABC$ be a triangle and $\omega$ be its circumcircle. Let the external bisectors of $\angle A, \angle B, \angle C$ intersect $\omega$ at $A', B', C'$ respectively. If $X$ and $Y$ are two arbitrary points then we call $(X,Y)$ the circle centered at $X$ with radius $XY$. Then the radical center of $(A',A)$, $(B',B)$, $(C',C)$ is the nagel point of $\triangle ABC$.
Let $(A', A)$ intersect $\omega$ at $A''$. Then trivially we have $AA''\parallel BC$. Similarly we define the points $B'', C''$. Let $AA''\cap CC''= B_0$. Similarly we define $A_0, C_0$. So the radical axis of $(C,C'), (B,B')$ is the line throungh $B_0$ perpendicular to $B'C'$. We call this line $\gamma_A$. We similarly define $\gamma_B, \gamma_C$.
$\text{Lemma 1}$: $\gamma_A$ is the internal angel bisector of $\angle B_0A_0C_0$.
$\text{Proof}$: By some easy angle chasing, we have $\angle (B'C', AB) = \frac{B}{2} + \frac{C}{2}$. The result follows by the fact that $A_0B_0\parallel AB$ and $\gamma_A\perp B'C'$.
Using our lemma gives us that $\gamma_A, \gamma_A, \gamma_C$ meets at the incenter of $\triangle A_0B_0C_0$. We denote this by $N$.
$\text{Lemma 2}$: The incenter of a triangle $\triangle ABC$ is the Nagel point of the median triangle of $\triangle ABC$.
$\text{Proof}$: Let the median triangle be $A'B'C'$. Let $I'$ be the incenter of $\triangle A'B'C'$. Let $(I)$ intersect $BC$ at $D$, $(I')$ intersect $B'C'$ at $D'$. Let $E$ be the reflection of $D$ wrt $I$, and $E'$ be the reflection of $D'$ wrt $I'$. We have $A, E, D'$ collinear. Again by homothety, we have $ED' \parallel IE'$ and $ED'\parallel AE'$. So we have that $A', E', I$ collinear. Using this same arguement for $B'$ and $C'$, we have $I$ is the Nagel point of $A'B'C'$.
Now back to our original proof, we see that in $\triangle A_0B_0C_0$, $\triangle ABC$ is the median triangle and $N$ is the incenter. So we have proved that $N$ is the Nagel point of $\triangle ABC$. (Q.E.D.)
We can state the problem in a new way:
Let $ABC$ be a triangle and $\omega$ be its circumcircle. Let the external bisectors of $\angle A, \angle B, \angle C$ intersect $\omega$ at $A', B', C'$ respectively. If $X$ and $Y$ are two arbitrary points then we call $(X,Y)$ the circle centered at $X$ with radius $XY$. Then the radical center of $(A',A)$, $(B',B)$, $(C',C)$ is the nagel point of $\triangle ABC$.
Let $(A', A)$ intersect $\omega$ at $A''$. Then trivially we have $AA''\parallel BC$. Similarly we define the points $B'', C''$. Let $AA''\cap CC''= B_0$. Similarly we define $A_0, C_0$. So the radical axis of $(C,C'), (B,B')$ is the line throungh $B_0$ perpendicular to $B'C'$. We call this line $\gamma_A$. We similarly define $\gamma_B, \gamma_C$.
$\text{Lemma 1}$: $\gamma_A$ is the internal angel bisector of $\angle B_0A_0C_0$.
$\text{Proof}$: By some easy angle chasing, we have $\angle (B'C', AB) = \frac{B}{2} + \frac{C}{2}$. The result follows by the fact that $A_0B_0\parallel AB$ and $\gamma_A\perp B'C'$.
Using our lemma gives us that $\gamma_A, \gamma_A, \gamma_C$ meets at the incenter of $\triangle A_0B_0C_0$. We denote this by $N$.
$\text{Lemma 2}$: The incenter of a triangle $\triangle ABC$ is the Nagel point of the median triangle of $\triangle ABC$.
$\text{Proof}$: Let the median triangle be $A'B'C'$. Let $I'$ be the incenter of $\triangle A'B'C'$. Let $(I)$ intersect $BC$ at $D$, $(I')$ intersect $B'C'$ at $D'$. Let $E$ be the reflection of $D$ wrt $I$, and $E'$ be the reflection of $D'$ wrt $I'$. We have $A, E, D'$ collinear. Again by homothety, we have $ED' \parallel IE'$ and $ED'\parallel AE'$. So we have that $A', E', I$ collinear. Using this same arguement for $B'$ and $C'$, we have $I$ is the Nagel point of $A'B'C'$.
Now back to our original proof, we see that in $\triangle A_0B_0C_0$, $\triangle ABC$ is the median triangle and $N$ is the incenter. So we have proved that $N$ is the Nagel point of $\triangle ABC$. (Q.E.D.)
Re: Geometry Marathon : Season 3
$\text{Problem 29}$
In acute triangle $ABC$ , segments $AD; BE$ , and $CF$ are its altitudes, and $H$ is its orthocenter. Circle $\omega$, centered at $O$, passes through $A$ and $H$ and intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. The circumcircle of triangle $OPQ$ is tangent to segment $BC$ at $R$. Prove that $\frac{CR}{BR}=\frac{ED}{FD}.$
In acute triangle $ABC$ , segments $AD; BE$ , and $CF$ are its altitudes, and $H$ is its orthocenter. Circle $\omega$, centered at $O$, passes through $A$ and $H$ and intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. The circumcircle of triangle $OPQ$ is tangent to segment $BC$ at $R$. Prove that $\frac{CR}{BR}=\frac{ED}{FD}.$
Re: Geometry Marathon : Season 3
solution to problem 29:
Lemma: O,H,R are collinear and quadrilateral $HRCP$ is cyclic.
proof: Assume that O,H,R are not collinear. Let $OH \cap BC= R_1$. Let $\angle APO= \theta$. Now,
$\angle OR_1C= 180\angle R_1OP\angle OPA\angle ACB= 180\angle HOP\theta\angle ACB$ $= 1802\angle DAC\theta\angle ACB$= $1802(90\angle ACB)\theta\angle ACB= \angle ACB\theta$
We also have,$\angle APH= \angle OPH\angle OPA= \frac{180\angle HOP}{2}\theta= 90\angle HAP\theta=\angle ACB\theta$
So,$\angle HR_1C= \angle HPA$
So, quadrilateral $HR_1CP$ is cyclic...............................................................................(1)
So, $\angle HR_1p=\angle HCP$ or, $\angle OR_1P=\angle HCP$
Now, $\angle ORP= \angle OQP= \frac{180\angle QOP}{2}=90\angle BAC= \angle ACF= \angle PCH$
So we have, $\angle OR_1P= \angle ORP$
So, $R_1 \in \bigodot QOP$
since $\bigodot QOP$ is tangent with $BC$ we have,
$R=R_1$, contradiction.
And also from (1), quadrilateral $HRCP$ is cyclic.
.....................................................End of Lemma.....................................................
Main proof: $\angle QRO= \angle QPO= \angle OQP= \angle ORP$
And $\angle QPH= \angle BAD= \angle FCB= \angle HPR$
So, H is the incentre of $\bigtriangleup QRP$.
So we have, $\angle RQH= \angle HQP= \angle HAC= 90\angle ACB$
So, $\angle RQP= 2(90\angle ACB)$...............................................................................(2)
Now, $\angle PRC= \angle PQR= 2(90\angle ACB)$
So we have, $\angle RPC= 180\angle PRC\angle PCR= \angle C$.................................................(3)
Similarly we can show that $\angle RQB= \angle B$.....................................................................(4)
And we have $\angle QPR= 2\angle QPH= 2\angle QAH= 2(90\angle B)$........................................(5)
Now, $\frac{BR}{CR}=\frac{BR}{RQ}*\frac{RQ}{RP}*\frac{RP}{CR}= \frac{Sin\angle BQR}{Sin\angle QBR}*\frac{Sin\angle QPR}{Sin\angle RQP}*\frac{Sin\angle RCP}{Sin\angle RPC}= \frac{sin\angle 2B}{sinc\angle 2C}= \frac{FD}{ED}$
Q.E.D
Lemma: O,H,R are collinear and quadrilateral $HRCP$ is cyclic.
proof: Assume that O,H,R are not collinear. Let $OH \cap BC= R_1$. Let $\angle APO= \theta$. Now,
$\angle OR_1C= 180\angle R_1OP\angle OPA\angle ACB= 180\angle HOP\theta\angle ACB$ $= 1802\angle DAC\theta\angle ACB$= $1802(90\angle ACB)\theta\angle ACB= \angle ACB\theta$
We also have,$\angle APH= \angle OPH\angle OPA= \frac{180\angle HOP}{2}\theta= 90\angle HAP\theta=\angle ACB\theta$
So,$\angle HR_1C= \angle HPA$
So, quadrilateral $HR_1CP$ is cyclic...............................................................................(1)
So, $\angle HR_1p=\angle HCP$ or, $\angle OR_1P=\angle HCP$
Now, $\angle ORP= \angle OQP= \frac{180\angle QOP}{2}=90\angle BAC= \angle ACF= \angle PCH$
So we have, $\angle OR_1P= \angle ORP$
So, $R_1 \in \bigodot QOP$
since $\bigodot QOP$ is tangent with $BC$ we have,
$R=R_1$, contradiction.
And also from (1), quadrilateral $HRCP$ is cyclic.
.....................................................End of Lemma.....................................................
Main proof: $\angle QRO= \angle QPO= \angle OQP= \angle ORP$
And $\angle QPH= \angle BAD= \angle FCB= \angle HPR$
So, H is the incentre of $\bigtriangleup QRP$.
So we have, $\angle RQH= \angle HQP= \angle HAC= 90\angle ACB$
So, $\angle RQP= 2(90\angle ACB)$...............................................................................(2)
Now, $\angle PRC= \angle PQR= 2(90\angle ACB)$
So we have, $\angle RPC= 180\angle PRC\angle PCR= \angle C$.................................................(3)
Similarly we can show that $\angle RQB= \angle B$.....................................................................(4)
And we have $\angle QPR= 2\angle QPH= 2\angle QAH= 2(90\angle B)$........................................(5)
Now, $\frac{BR}{CR}=\frac{BR}{RQ}*\frac{RQ}{RP}*\frac{RP}{CR}= \frac{Sin\angle BQR}{Sin\angle QBR}*\frac{Sin\angle QPR}{Sin\angle RQP}*\frac{Sin\angle RCP}{Sin\angle RPC}= \frac{sin\angle 2B}{sinc\angle 2C}= \frac{FD}{ED}$
Q.E.D
Re: Geometry Marathon : Season 3
Problem 30:
Let $ABC$ be a triangle. Let $D$ be a point on $AB$ such that the intersection point of perpendicular bisector of AC and the line through $D$ parallel to the angle bisector of $\angle BAC$ is $P$ and it is inside $\bigtriangleup ABC$. Circumcircle of $ABC$ and $ADP$ intersects at $K$ (other than A). $KP$ is parallel to $AC$. Let the intersection of $AK$ and $BC$ be $E$. The circle with diameter $EK$ intersect $\bigodot ABC$ at $S$ (other than $K$). Prove that $AS$ is a symmedian of triangle ABC.
Let $ABC$ be a triangle. Let $D$ be a point on $AB$ such that the intersection point of perpendicular bisector of AC and the line through $D$ parallel to the angle bisector of $\angle BAC$ is $P$ and it is inside $\bigtriangleup ABC$. Circumcircle of $ABC$ and $ADP$ intersects at $K$ (other than A). $KP$ is parallel to $AC$. Let the intersection of $AK$ and $BC$ be $E$. The circle with diameter $EK$ intersect $\bigodot ABC$ at $S$ (other than $K$). Prove that $AS$ is a symmedian of triangle ABC.
 Raiyan Jamil
 Posts: 138
 Joined: Fri Mar 29, 2013 3:49 pm
Re: Geometry Marathon : Season 3
$\text{Solution to Problem 30: }$
Easy angle chase gives that $K$ is the midpoint of $\text{arc BC}$ not containing $A$.Now, Let $M$ be the midpoint of $BC$ and $L$ be the midpoint of $\text{arc BC}$ containing $A$. We get, $S,E,L$ are collinear and $ALME$ is cyclic. So radical axis theorem gives us that $AL,BC,KS$ are concurrent, let at $X$. Now, $LBKC$ is a harmonic quadrilateral. Harmonic pencil at $A$ gives that $X,B,E,C$ is harmonic. Then pencil $L$ gives that $ABSC$ is a harmonic quadrilateral. So $AS$ is the symmedian of $\bigtriangleup ABC$.
Easy angle chase gives that $K$ is the midpoint of $\text{arc BC}$ not containing $A$.Now, Let $M$ be the midpoint of $BC$ and $L$ be the midpoint of $\text{arc BC}$ containing $A$. We get, $S,E,L$ are collinear and $ALME$ is cyclic. So radical axis theorem gives us that $AL,BC,KS$ are concurrent, let at $X$. Now, $LBKC$ is a harmonic quadrilateral. Harmonic pencil at $A$ gives that $X,B,E,C$ is harmonic. Then pencil $L$ gives that $ABSC$ is a harmonic quadrilateral. So $AS$ is the symmedian of $\bigtriangleup ABC$.
Last edited by Raiyan Jamil on Tue Feb 21, 2017 2:57 am, edited 2 times in total.
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
$\text{Problem 31:}$
Let $\bigtriangleup ABC$ be a triangle with incenter $I$. Prove that the $\text{Euler lines}$ of $\bigtriangleup AIB, \bigtriangleup BIC, \bigtriangleup CIA$ and $\bigtriangleup ABC$ are concurrent.
Let $\bigtriangleup ABC$ be a triangle with incenter $I$. Prove that the $\text{Euler lines}$ of $\bigtriangleup AIB, \bigtriangleup BIC, \bigtriangleup CIA$ and $\bigtriangleup ABC$ 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 31:
Let $D,E,F $ be the midpoints of the arc $BC$ (not containg $A$) ,arc $CA$ (not containg $B$),arc $AB$ (not containg $C$) respectively. Let $H_a,H_b,H_c$ be the orthocenters of $\triangle IBC ,\triangle ICA ,\triangle IAB$ respectively .$M ,N ,P$ be the midpoints of $BC,CA,AB $ respectively.
Let us start by proving some lemmas :
Lemma 1:Let the perpendiculars from $D$ to $AC$ & $E$ to $BC$ meet $(ABC)$ at $X , Y$ respectively .Then $DX=EY$.
Proof: Show that $\angle DCX=\angle ECY$ .(Easy angle chasing) .
Lemma 2:Let $D'$ , $E'$ & $F'$ be the projection of $D$ , $E$ & $F$ on $CA$ , $AB$ & $BC$ respectively .Then $DD'+OM=EE'+ON=FF' + OP$.
Proof:Let $S$ be the projection of $E$ on $BC$.Then $EE'OM=ESOM=\dfrac {EY}{2}=\dfrac {DX}{2}=DD'ON \Rightarrow DD'+OM=EE'+ON$. Similarly $EE'+ON=FF' + OP$.
Let $DH_a \cap AH=A_1$. Define $B_1,C_1$ similarly.
Lemma 3: $HA_1=HB_1=HC_1$
Proof:Let $D_1$ be the reflection of $D$ wrt $M$.Then $\angle BD_1C= \angle BDC=180^\circ \angle BAC=2(I80^\circ \angle BIC)=2\angle BH_aC$.But $BD_1=CD_1$ .So $D_1$ is the center of $\triangle BH_aC$.
$\dfrac {AA_1}{H_aI}=\dfrac {AD}{ID}=\dfrac {AD}{DC}=\dfrac {DD'}{DM}=\dfrac {2DD'}{DD_1}=\dfrac {2DD'}{H_aI}$
So, $AA_1=2DD'$ .So ,$HA_1=HA+AA_1=2OM+2DD'=2(DD'+OM).$
So by lemma 2 $HA_1=HB_1=HC_1$.
Solution :Let $OH \cap DH_a=J_1 ,OH \cap EH_b=J_2$.
Then $\dfrac {HJ_1}{J_1O} =\dfrac {HA_1}{OD}=\dfrac {HB_1}{OE}=\dfrac {HJ_2}{J_2O}$
So $J_1=J_2$. Similarly $OH,DH_a,EH_b,FH_c$ are concurrent.
Let $D,E,F $ be the midpoints of the arc $BC$ (not containg $A$) ,arc $CA$ (not containg $B$),arc $AB$ (not containg $C$) respectively. Let $H_a,H_b,H_c$ be the orthocenters of $\triangle IBC ,\triangle ICA ,\triangle IAB$ respectively .$M ,N ,P$ be the midpoints of $BC,CA,AB $ respectively.
Let us start by proving some lemmas :
Lemma 1:Let the perpendiculars from $D$ to $AC$ & $E$ to $BC$ meet $(ABC)$ at $X , Y$ respectively .Then $DX=EY$.
Proof: Show that $\angle DCX=\angle ECY$ .(Easy angle chasing) .
Lemma 2:Let $D'$ , $E'$ & $F'$ be the projection of $D$ , $E$ & $F$ on $CA$ , $AB$ & $BC$ respectively .Then $DD'+OM=EE'+ON=FF' + OP$.
Proof:Let $S$ be the projection of $E$ on $BC$.Then $EE'OM=ESOM=\dfrac {EY}{2}=\dfrac {DX}{2}=DD'ON \Rightarrow DD'+OM=EE'+ON$. Similarly $EE'+ON=FF' + OP$.
Let $DH_a \cap AH=A_1$. Define $B_1,C_1$ similarly.
Lemma 3: $HA_1=HB_1=HC_1$
Proof:Let $D_1$ be the reflection of $D$ wrt $M$.Then $\angle BD_1C= \angle BDC=180^\circ \angle BAC=2(I80^\circ \angle BIC)=2\angle BH_aC$.But $BD_1=CD_1$ .So $D_1$ is the center of $\triangle BH_aC$.
$\dfrac {AA_1}{H_aI}=\dfrac {AD}{ID}=\dfrac {AD}{DC}=\dfrac {DD'}{DM}=\dfrac {2DD'}{DD_1}=\dfrac {2DD'}{H_aI}$
So, $AA_1=2DD'$ .So ,$HA_1=HA+AA_1=2OM+2DD'=2(DD'+OM).$
So by lemma 2 $HA_1=HB_1=HC_1$.
Solution :Let $OH \cap DH_a=J_1 ,OH \cap EH_b=J_2$.
Then $\dfrac {HJ_1}{J_1O} =\dfrac {HA_1}{OD}=\dfrac {HB_1}{OE}=\dfrac {HJ_2}{J_2O}$
So $J_1=J_2$. Similarly $OH,DH_a,EH_b,FH_c$ are concurrent.
Last edited by joydip on Tue Feb 21, 2017 9:15 pm, edited 1 time in total.
The first principle is that you must not fool yourself and you are the easiest person to fool.
Re: Geometry Marathon : Season 3
Problem 32 :
In triangle $ABC$ with incenter $I$ and circumcenter $O$, let $A',B',C'$ be the points of tangency of its circumcircle with its $A,B,C$mixtilinear incircles, respectively. Let $\omega_A$ be the circle through $A'$ that is tangent to $AI$ at $I$, and define $\omega_B, \omega_C$ similarly. Prove that $\omega_A,\omega_B,\omega_C$ have a common point $X$ other than $I$, and that $\angle AXO = \angle OXA'$
In triangle $ABC$ with incenter $I$ and circumcenter $O$, let $A',B',C'$ be the points of tangency of its circumcircle with its $A,B,C$mixtilinear incircles, respectively. Let $\omega_A$ be the circle through $A'$ that is tangent to $AI$ at $I$, and define $\omega_B, \omega_C$ similarly. Prove that $\omega_A,\omega_B,\omega_C$ have a common point $X$ other than $I$, and that $\angle AXO = \angle OXA'$
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
$\text{Solution of Problem 32:}$
Let, $D,E,F$ be the point of tangency of the incircle with $\bigtriangleup ABC$ at $BC,CA,AB$ respectively. $\bigtriangleup M_a,M_b,M_c$ be the midpoints of the arcs $BC,CA,AB$ opposite to $A,B,C$ respectively. Let the perpendicular to $AI$ at $I$ meet $AB,AC,BC$ at $B_1,C_1$ and $X$ respectively. $P$ be the orthocentre of $\bigtriangleup DEF$ and $Q$ be the inverse of $P$ under the incircle of $\bigtriangleup ABC$.
We'll use a few of the lemmas among which some are from the link http://www.mit.edu/~evanchen/handouts/M ... Guessr.pdf
$\text{Lemma 1:}$ $X,A',M_a$ and $N,I,A'$ are collinear.
$\text{Proof:}$ From the Lemma 10,3 in the link.
Now, obviously $w_A$ is the circle $IDA'X$. Let $G$ be the second point of intersection of $w_A$ and the incircle of $\bigtriangleup ABC$. So, $GD$ is the inverse of $w_A$. Obviously $GDAI \Rightarrow DG \perp EF$. So, the inverses of $w_A,w_B,w_C$ are concurrent at the orthocentre of $\bigtriangleup DEF$ i.e. $P$. So, the circles $w_A,w_B,w_C$ are coaxal and their other point of intersection is $Q$ i.e. the common point $X$ is $Q$ as desired. Which also gives that $IDA'XQG$ is cyclic.
$\text{Lemma 2:}$ $O,I,P,Q$ and $A,G,A'$ are collinear.
$\text{Proof:}$ $\bigtriangleup M_aM_bM_c$ and $\bigtriangleup DEF$ are homothetic similar. If $H$ is their homothetic centre, then, $H,I,O$ are concurrent. Again, since $I$ is the orthocentre of $\bigtriangleup M_aM_bM_c$, we get that $P$ also lies on this line. So, this gives that $O,I,P,Q$ are collinear. The latter is just angle chase.
$\text{Lemma 3:}$ $GIAO$.
$\text{Proof:}$ Just angle chase.
Now, $\angle AOQ=\angle GIQ=\angle GA'Q=\angle AA'Q$ which gives that $AQA'O$ is cyclic. And since $AO=OA'$, we get $\angle AXO=\angle OXA'$ as desired.
Let, $D,E,F$ be the point of tangency of the incircle with $\bigtriangleup ABC$ at $BC,CA,AB$ respectively. $\bigtriangleup M_a,M_b,M_c$ be the midpoints of the arcs $BC,CA,AB$ opposite to $A,B,C$ respectively. Let the perpendicular to $AI$ at $I$ meet $AB,AC,BC$ at $B_1,C_1$ and $X$ respectively. $P$ be the orthocentre of $\bigtriangleup DEF$ and $Q$ be the inverse of $P$ under the incircle of $\bigtriangleup ABC$.
We'll use a few of the lemmas among which some are from the link http://www.mit.edu/~evanchen/handouts/M ... Guessr.pdf
$\text{Lemma 1:}$ $X,A',M_a$ and $N,I,A'$ are collinear.
$\text{Proof:}$ From the Lemma 10,3 in the link.
Now, obviously $w_A$ is the circle $IDA'X$. Let $G$ be the second point of intersection of $w_A$ and the incircle of $\bigtriangleup ABC$. So, $GD$ is the inverse of $w_A$. Obviously $GDAI \Rightarrow DG \perp EF$. So, the inverses of $w_A,w_B,w_C$ are concurrent at the orthocentre of $\bigtriangleup DEF$ i.e. $P$. So, the circles $w_A,w_B,w_C$ are coaxal and their other point of intersection is $Q$ i.e. the common point $X$ is $Q$ as desired. Which also gives that $IDA'XQG$ is cyclic.
$\text{Lemma 2:}$ $O,I,P,Q$ and $A,G,A'$ are collinear.
$\text{Proof:}$ $\bigtriangleup M_aM_bM_c$ and $\bigtriangleup DEF$ are homothetic similar. If $H$ is their homothetic centre, then, $H,I,O$ are concurrent. Again, since $I$ is the orthocentre of $\bigtriangleup M_aM_bM_c$, we get that $P$ also lies on this line. So, this gives that $O,I,P,Q$ are collinear. The latter is just angle chase.
$\text{Lemma 3:}$ $GIAO$.
$\text{Proof:}$ Just angle chase.
Now, $\angle AOQ=\angle GIQ=\angle GA'Q=\angle AA'Q$ which gives that $AQA'O$ is cyclic. And since $AO=OA'$, we get $\angle AXO=\angle OXA'$ as desired.
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