Geometry Marathon : Season 3
 Raiyan Jamil
 Posts: 138
 Joined: Fri Mar 29, 2013 3:49 pm
Re: Geometry Marathon : Season 3
$\text{Problem 23:}$
Triangle $ABC$ is inscribed in circle $\Omega$. The interior angle bisector of angle $A$ intersects side $BC$ and $\Omega$ at $D$ and $L$ (other than $A$), respectively. Let $M$ be the midpoint of side $BC$. The circumcircle of triangle $ADM$ intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. Let $N$ be the midpoint of segment $PQ$, and let $H$ be the foot of the perpendicular from $L$ to line $ND$. Prove that line $ML$ is tangent to the circumcircle of triangle $HMN$.
Triangle $ABC$ is inscribed in circle $\Omega$. The interior angle bisector of angle $A$ intersects side $BC$ and $\Omega$ at $D$ and $L$ (other than $A$), respectively. Let $M$ be the midpoint of side $BC$. The circumcircle of triangle $ADM$ intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. Let $N$ be the midpoint of segment $PQ$, and let $H$ be the foot of the perpendicular from $L$ to line $ND$. Prove that line $ML$ is tangent to the circumcircle of triangle $HMN$.
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
Firstly, we notice that $\angle CMP = \angle QMB = \angle LAC$. So, $LM$ bisects $\angle PMQ$. If $LM \cap \odot HNM = T$, then $T$ is the midpoint of arc $QAP$ of $\odot HMN$. So, $D, N, T$ are collinear. That means $AT$ extarnally bisects $\angle BAC$. That also indicates that $\odot \omega \cap \odot HMN = T$.
Using Spiral Similarity, we see that $T$ sends $N$ to $D$, and $M$ to $L$. So we have $LD \parallel MN$.
As $\angle LHD = \angle LMD$, $\square LHDM$ is cyclic. This implies $\angle LMH =\angle LDH =\angle MND$. This completes our proof.
Using Spiral Similarity, we see that $T$ sends $N$ to $D$, and $M$ to $L$. So we have $LD \parallel MN$.
As $\angle LHD = \angle LMD$, $\square LHDM$ is cyclic. This implies $\angle LMH =\angle LDH =\angle MND$. This completes our proof.

 Posts: 16
 Joined: Wed Aug 10, 2016 1:29 am
Re: Geometry Marathon : Season 3
$\text{Problem no 24}$
Let $ABC$ be a triangle and $m$ a line which intersects $AB, BC$ and $CA$ at $D, E$ and $F$ in a way that $C$ lies between $B$ and $E$. The parallel lines from the points $A, B, C$ to the line $m$ intersects the circumcircle of $ABC$ at the points $A_1, B_1, C_1$. Prove that the lines $A_1E, B_1F$ and $C_1D$ passes through a same point.
Let $ABC$ be a triangle and $m$ a line which intersects $AB, BC$ and $CA$ at $D, E$ and $F$ in a way that $C$ lies between $B$ and $E$. The parallel lines from the points $A, B, C$ to the line $m$ intersects the circumcircle of $ABC$ at the points $A_1, B_1, C_1$. Prove that the lines $A_1E, B_1F$ and $C_1D$ passes through a same point.

 Posts: 16
 Joined: Wed Aug 10, 2016 1:29 am
Re: Geometry Marathon : Season 3
$\text{Problem 25}$
Let $ABC$ be an acute triangle. Consider the equilateral triangle $A'UV$, with $A', U, V$ on $BC, AC, AB$ respectively, and $UV \parallel BC$. The points $B', C'$ are defined similarly. Prove that $AA', BB', CC'$ are concurrent.
Let $ABC$ be an acute triangle. Consider the equilateral triangle $A'UV$, with $A', U, V$ on $BC, AC, AB$ respectively, and $UV \parallel BC$. The points $B', C'$ are defined similarly. Prove that $AA', BB', CC'$ are concurrent.
 Raiyan Jamil
 Posts: 138
 Joined: Fri Mar 29, 2013 3:49 pm
Re: Geometry Marathon : Season 3
$\text{**The Problem 24 has been posted before here as Problem no 4.}$ $\text{So I'll just give the solution of Problem 25**}$
$\text{Solution of Problem 25 :}$
Let, $A_1$ be a point such that $\bigtriangleup A_1BC$ is equilateral where $A_1$ lies on the side of $BC$ oppostie to $A$. We define $B_1,C_1$ similarly. Now, since $\bigtriangleup A'UV$ and $\bigtriangleup A_1BC$ are homothetic similar. So we deduce that $A,A'$ and $A_1$ are collinear. So, it is sufficient to prove that $AA_1,BB_1,CC_1$ are concurrent.
Now, $AB_1,AC_1$ are isogonals with respect to $\angle BAC$. Similarly $BA_1,BC_1$ are isogonals with respect to $\angle ABC$ and $CA_1,CB_1$ are isogonals with respect to $\angle ACB$. So, applying Jacobi's theorem we deduce that $AA_1,BB_1,CC_1$ are concurrent as desired.
$\text{Note:}$ $AA_1,BB_1,CC_1$ are concurrent at the $\text{first Fermat point}$ of $\bigtriangleup ABC$.
$\text{Solution of Problem 25 :}$
Let, $A_1$ be a point such that $\bigtriangleup A_1BC$ is equilateral where $A_1$ lies on the side of $BC$ oppostie to $A$. We define $B_1,C_1$ similarly. Now, since $\bigtriangleup A'UV$ and $\bigtriangleup A_1BC$ are homothetic similar. So we deduce that $A,A'$ and $A_1$ are collinear. So, it is sufficient to prove that $AA_1,BB_1,CC_1$ are concurrent.
Now, $AB_1,AC_1$ are isogonals with respect to $\angle BAC$. Similarly $BA_1,BC_1$ are isogonals with respect to $\angle ABC$ and $CA_1,CB_1$ are isogonals with respect to $\angle ACB$. So, applying Jacobi's theorem we deduce that $AA_1,BB_1,CC_1$ are concurrent as desired.
$\text{Note:}$ $AA_1,BB_1,CC_1$ are concurrent at the $\text{first Fermat point}$ of $\bigtriangleup ABC$.
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
Let $ABC$ be a triangle with $AB=AC$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB$.
Frankly, my dear, I don't give a damn.

 Posts: 16
 Joined: Wed Aug 10, 2016 1:29 am
Re: Geometry Marathon : Season 3
SOLUTION TO PROB 26:
By some trivial angle chasing we have that $BE$ bisects $\angle DBA$. We have $AE\cap BC = M$
Now let $\odot DCB = ω$ intersect $AB$ at $D'$. Using spiral smilarity, we have $\triangle ED'A$ and $\triangle EDF$ are similar. That implies that $FD=AD=CD$. So $CF\perp AF$. We reflect $F$ about $D$ to get $F'$. If $AF$ bisects $\angle BAK$, then $(B,K;F,F')=1=C(B,K;F,F')$. As $CF\perp CF'$, $CF$ will also bisect $\angle KCB$.
Now let us denote $\odot AFC$ by $γ$. Let $γ\cap ω = P$. Let $CF\cap ω = Q$. Our aim is to prove $QP=QB$.
By spiral similarity, we have $\triangle PQB$ is similar to $\triangle PFM$. So our ultimate aim is to prove $\angle PAM=\angle DBA$.
Again buy spiral similarity, we have $\triangle PBM$ is similar to $\triangle PDA$. That implies that $BP =BM$. Now let $DM\cap ω = M'$. It's easy to see that $BM'=BM=BP$, and $BM'$ is tangent to $\odot ABC$ at $B$. Let the tangents at $A$ and $B$ to $\odot ABC$ intersect at $T$. Again let $AT \cap γ = X$. Trivially we have $AX=BM=BM'$. So, $(TB)(TM')=(TA)(TX)$. so that means $T$ lies on the radical axis of $ω$ and $γ$. So $CP$ is the $C$ symmedian of $\triangle ABC$, and as $C, I, P$ is collinear, our claim is proved.
By some trivial angle chasing we have that $BE$ bisects $\angle DBA$. We have $AE\cap BC = M$
Now let $\odot DCB = ω$ intersect $AB$ at $D'$. Using spiral smilarity, we have $\triangle ED'A$ and $\triangle EDF$ are similar. That implies that $FD=AD=CD$. So $CF\perp AF$. We reflect $F$ about $D$ to get $F'$. If $AF$ bisects $\angle BAK$, then $(B,K;F,F')=1=C(B,K;F,F')$. As $CF\perp CF'$, $CF$ will also bisect $\angle KCB$.
Now let us denote $\odot AFC$ by $γ$. Let $γ\cap ω = P$. Let $CF\cap ω = Q$. Our aim is to prove $QP=QB$.
By spiral similarity, we have $\triangle PQB$ is similar to $\triangle PFM$. So our ultimate aim is to prove $\angle PAM=\angle DBA$.
Again buy spiral similarity, we have $\triangle PBM$ is similar to $\triangle PDA$. That implies that $BP =BM$. Now let $DM\cap ω = M'$. It's easy to see that $BM'=BM=BP$, and $BM'$ is tangent to $\odot ABC$ at $B$. Let the tangents at $A$ and $B$ to $\odot ABC$ intersect at $T$. Again let $AT \cap γ = X$. Trivially we have $AX=BM=BM'$. So, $(TB)(TM')=(TA)(TX)$. so that means $T$ lies on the radical axis of $ω$ and $γ$. So $CP$ is the $C$ symmedian of $\triangle ABC$, and as $C, I, P$ is collinear, our claim is proved.
 Raiyan Jamil
 Posts: 138
 Joined: Fri Mar 29, 2013 3:49 pm
Re: Geometry Marathon : Season 3
$\text{Problem 27:}$
Let $ABC$ be a scalene triangle, let $I$ be its incentre , and let $A_1,B_1$ and $C_1$ be the points of contact of the excircles with the sides $BC,CA$ and $AB$ respectively. Prove that the circumcircles of the triangles $AIA_1,BIB_1$ and $CIC_1$ have a common point different from I.
Let $ABC$ be a scalene triangle, let $I$ be its incentre , and let $A_1,B_1$ and $C_1$ be the points of contact of the excircles with the sides $BC,CA$ and $AB$ respectively. Prove that the circumcircles of the triangles $AIA_1,BIB_1$ and $CIC_1$ have a common point different from I.
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 27 :
Let $I_a,I_b,I_C$ be the excenters of $ABC$ opposite to $A,B,C$ respectively . $O$ be the center of $(I_aI_bI_C)$.
$I_aO \cap BC=A_1$.Let $I_aO \cap (I_aI_bI_C)=A_2 ,I_a$.let $A_2I \cap (I_aI_bI_C)=A_2,A_3$ and $I_aA_3\cap I_bI_c=A_4,$Let points $A_5,A_6 \in I_bI_c$, such that $\angle A_5IA_2=\angle A_6A_2I=90^\circ$. Define those points for $B,C$ similerly.
Lemma 1: $A_4,B_4,C_4 $ are collinear.
Proof: $I_aA_3BIC$ , $I_bBCI_c$ , $I_aI_bI_CA_3$ are cyclic. So $A_4 $ lies in the radical axis $BC$ of $(I_aA_3BIC ),(I_bBCI_c)$.So $A_4,B,C$ are collinear.$\triangle ABC,\triangle I_aI_bI_C$ are perspective from a point ,so by Desargues theorm $A_4,B_4,C_4 $ are collinear.
Lemma 2: $A_5,B_5,C_5$ are collinear.
Proof:Let $I_aA_3 \cap I_cC_3=k$ . $I_aA_3BI$ , $I_cC_3BI$ , $I_aA_3I_cC_3$ are cyclic.So $A_4 $ lies in the radical axis $IB$ of $(I_aA_3BI)$ & $(I_cC_3BI)$.So $I,B,K$ are collinear.So $\dfrac {A_4A_5}{A_5I_b}=\dfrac {KI}{II_b}=\dfrac {C_4C_5}{C_5I_b}$ (as $A_4K \parallel A_5I , C_4K \parallel C_5I)$
So $A_4C_4\parallel A_5C_5$,simillerly $A_4B_4\parallel A_5B_5$,completing the proof.
Lemma 3:$A_6,B_6,C_6$ are collinear.
Proof:Let $D,E,F$ be the midpoints of $I_bI_C$ , $I_aI_C$ ,$I_aI_b$ respectively.$I,D,A_2$ are collinear (wellknown).$A_5I \parallel A_2A_6$ .So $A_6 $ is the reflection of $A_5$ wrt $D$.Similar for $B_6,C_6$.So by Menelaus theorem $A_6,B_6,C_6$ are collinear as $A_5,B_5,C_5$ are collinear.
Solution:
$\angle BA_1A_2=\angle BI_cA_2=90^\circ \Rightarrow I_cBA_1A_2$ is cyclic.$AIBI_c$ is cyclic .So ,$I_aI.I_aA= I_aB.I_aI_c=I_aA_1.I_aA_2.$So $AIA_1A_2$ is cyclic.$\angle IAA_6=\angle IA_2A_6=90^\circ \Rightarrow A_6 \in (AIA_1)$.A dilation with center $I$ & ratio $\dfrac {1}{2}$ takes $A_6,B_6,C_6$ to the centers of $(AIA_1),(BIB_1),(CIC_1)$ which must be collinear by lemma 3.So $(AIA_1),(BIB_1),(CIC_1)$ are coaxial , compliting the proof.
Let $I_a,I_b,I_C$ be the excenters of $ABC$ opposite to $A,B,C$ respectively . $O$ be the center of $(I_aI_bI_C)$.
$I_aO \cap BC=A_1$.Let $I_aO \cap (I_aI_bI_C)=A_2 ,I_a$.let $A_2I \cap (I_aI_bI_C)=A_2,A_3$ and $I_aA_3\cap I_bI_c=A_4,$Let points $A_5,A_6 \in I_bI_c$, such that $\angle A_5IA_2=\angle A_6A_2I=90^\circ$. Define those points for $B,C$ similerly.
Lemma 1: $A_4,B_4,C_4 $ are collinear.
Proof: $I_aA_3BIC$ , $I_bBCI_c$ , $I_aI_bI_CA_3$ are cyclic. So $A_4 $ lies in the radical axis $BC$ of $(I_aA_3BIC ),(I_bBCI_c)$.So $A_4,B,C$ are collinear.$\triangle ABC,\triangle I_aI_bI_C$ are perspective from a point ,so by Desargues theorm $A_4,B_4,C_4 $ are collinear.
Lemma 2: $A_5,B_5,C_5$ are collinear.
Proof:Let $I_aA_3 \cap I_cC_3=k$ . $I_aA_3BI$ , $I_cC_3BI$ , $I_aA_3I_cC_3$ are cyclic.So $A_4 $ lies in the radical axis $IB$ of $(I_aA_3BI)$ & $(I_cC_3BI)$.So $I,B,K$ are collinear.So $\dfrac {A_4A_5}{A_5I_b}=\dfrac {KI}{II_b}=\dfrac {C_4C_5}{C_5I_b}$ (as $A_4K \parallel A_5I , C_4K \parallel C_5I)$
So $A_4C_4\parallel A_5C_5$,simillerly $A_4B_4\parallel A_5B_5$,completing the proof.
Lemma 3:$A_6,B_6,C_6$ are collinear.
Proof:Let $D,E,F$ be the midpoints of $I_bI_C$ , $I_aI_C$ ,$I_aI_b$ respectively.$I,D,A_2$ are collinear (wellknown).$A_5I \parallel A_2A_6$ .So $A_6 $ is the reflection of $A_5$ wrt $D$.Similar for $B_6,C_6$.So by Menelaus theorem $A_6,B_6,C_6$ are collinear as $A_5,B_5,C_5$ are collinear.
Solution:
$\angle BA_1A_2=\angle BI_cA_2=90^\circ \Rightarrow I_cBA_1A_2$ is cyclic.$AIBI_c$ is cyclic .So ,$I_aI.I_aA= I_aB.I_aI_c=I_aA_1.I_aA_2.$So $AIA_1A_2$ is cyclic.$\angle IAA_6=\angle IA_2A_6=90^\circ \Rightarrow A_6 \in (AIA_1)$.A dilation with center $I$ & ratio $\dfrac {1}{2}$ takes $A_6,B_6,C_6$ to the centers of $(AIA_1),(BIB_1),(CIC_1)$ which must be collinear by lemma 3.So $(AIA_1),(BIB_1),(CIC_1)$ are coaxial , compliting 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
Prooblem 28 :
Let $ABC$ be a triangle with altitudes $AD,BE,CF$. $X,Y,Z$ are midpoints of $BC,CA,AB$. Consider circles $(X,XD),(Y,YE),(Z,ZF)$. Prove that radical center of $(X),(Y),(Z)$ is Nagel point of triangle $DEF$.
Let $ABC$ be a triangle with altitudes $AD,BE,CF$. $X,Y,Z$ are midpoints of $BC,CA,AB$. Consider circles $(X,XD),(Y,YE),(Z,ZF)$. Prove that radical center of $(X),(Y),(Z)$ is Nagel point of triangle $DEF$.
The first principle is that you must not fool yourself and you are the easiest person to fool.