## Geometry Marathon : Season 3

- Raiyan Jamil
**Posts:**138**Joined:**Fri Mar 29, 2013 3:49 pm

### Re: Geometry Marathon : Season 3

Problem 43:

An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$ such that $M\in\overline{AB}$, $Q\in\overline{AC}$, and $N,P\in\overline{BC}$. Let $S$ be the intersection of $ \overleftrightarrow{MN}$ and $ \overleftrightarrow{PQ}$. Denote by $\ell$ the angle bisector of $\angle MSQ$.

Prove that $\overline{OI}$ is parallel to $\ell$, where $O$ is the circumcenter of triangle $ABC$, and $I$ is the incenter of triangle $ABC$

An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$ such that $M\in\overline{AB}$, $Q\in\overline{AC}$, and $N,P\in\overline{BC}$. Let $S$ be the intersection of $ \overleftrightarrow{MN}$ and $ \overleftrightarrow{PQ}$. Denote by $\ell$ the angle bisector of $\angle MSQ$.

Prove that $\overline{OI}$ is parallel to $\ell$, where $O$ is the circumcenter of triangle $ABC$, and $I$ is the incenter of triangle $ABC$

A smile is the best way to get through a tough situation, even if it's a fake smile.

- ahmedittihad
**Posts:**175**Joined:**Mon Mar 28, 2016 6:21 pm

### Re: Geometry Marathon : Season 3

Define $K=(SPN)\cap (SQM)$ and let $X,Y$ denote the midpoints of $MQ,NP$ respectively. We will show that $XY||IO$, which proves the problem since it's well known in configurations pertaining to $MN=PQ$ that $XY||\ell$.

By spiral similarity and $MN=PQ$, we have $KNM\cong KPQ$. Thus $KM=KQ\Rightarrow K\in \overline {AXI}$. Let $AI\cap (ABC)=L\ne A$. Note that it suffices to show that $\frac {OL}{IL}=\frac {YK}{XK}$, as that will imply $YXK, OIL$ are homothetic and consequently $IO||XY$.

The ratios are easy to chase. Notice that the spiral similarity gives $\frac {YK}{XK}=\frac {PN}{MQ}=\frac {AQ}{MQ}$, while $\frac {OL}{IL}=\frac {OL}{ BL}$. These two ratios are equal due to $AQM\sim OLB$ and we are done.

By spiral similarity and $MN=PQ$, we have $KNM\cong KPQ$. Thus $KM=KQ\Rightarrow K\in \overline {AXI}$. Let $AI\cap (ABC)=L\ne A$. Note that it suffices to show that $\frac {OL}{IL}=\frac {YK}{XK}$, as that will imply $YXK, OIL$ are homothetic and consequently $IO||XY$.

The ratios are easy to chase. Notice that the spiral similarity gives $\frac {YK}{XK}=\frac {PN}{MQ}=\frac {AQ}{MQ}$, while $\frac {OL}{IL}=\frac {OL}{ BL}$. These two ratios are equal due to $AQM\sim OLB$ and we are done.

Frankly, my dear, I don't give a damn.

- ahmedittihad
**Posts:**175**Joined:**Mon Mar 28, 2016 6:21 pm

### Re: Geometry Marathon : Season 3

Problem $44$

Let $\triangle ABC$ be an acute angled triangle satisfying the conditions $AB > BC$ and $AC > BC$. Denote by $O$ and $H$ the circumcentre and orthocentre, respectively, of $\triangle ABC$. Suppose that the circumcircle of the triangle $AHC$ intersects the line $AB$ at $M$ diﬀerent from $A$, and the circumcircle of the triangle $AHB$ intersects the line $AC$ at $N$ diﬀerent from $A$. Prove that the circumcentre of the triangle $MNH$ lies on the line $OH$.

Let $\triangle ABC$ be an acute angled triangle satisfying the conditions $AB > BC$ and $AC > BC$. Denote by $O$ and $H$ the circumcentre and orthocentre, respectively, of $\triangle ABC$. Suppose that the circumcircle of the triangle $AHC$ intersects the line $AB$ at $M$ diﬀerent from $A$, and the circumcircle of the triangle $AHB$ intersects the line $AC$ at $N$ diﬀerent from $A$. Prove that the circumcentre of the triangle $MNH$ lies on the line $OH$.

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

Let $BH,CH$ meet $(O)$ at $X,Y$. We get $CXNH,BYMH$ are similar rhombuses. Let the perpendicular bisectors of $HM,HN$ meet at $K$ and perpendicular bisectors of $BY,CX$ meet at $O$. Let perpendicular bisectors of $BY,CX$ meet $HM,HN$ at $P,Q$ repectively. Its enough to prove that $\frac{HP}{HM/2}= \frac{HQ}{HN/2}$ or, $\frac{HP}{BY/2}= \frac{HQ}{CX/2}$ which follows from similarity.

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

$\text{Problem 45}$

Let $ABC$ be a triangle with orthocentre $H$ and circumcircle $\omega$ centered at $O$. Let $M_a,M_b,M_c$ be the midpoints of $BC,CA,AB$. Lines $AM_a,BM_b,CM_c$ meet $\omega$ again at $P_a,P_b,P_c$. Rays $M_aH,M_bH,M_cH$ intersect $\omega$ at $Q_a,Q_b,Q_c$. Prove that $P_aQ_a,P_bQ_b,P_cQ_c,OH$ are concurrent.

Let $ABC$ be a triangle with orthocentre $H$ and circumcircle $\omega$ centered at $O$. Let $M_a,M_b,M_c$ be the midpoints of $BC,CA,AB$. Lines $AM_a,BM_b,CM_c$ meet $\omega$ again at $P_a,P_b,P_c$. Rays $M_aH,M_bH,M_cH$ intersect $\omega$ at $Q_a,Q_b,Q_c$. Prove that $P_aQ_a,P_bQ_b,P_cQ_c,OH$ 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 45:

Let $OH \cap \omega = \{X,Y\}$ , $OH \cap P_aQ_a=J$ .Let $G$ be the centroid . $A'$ be the antipode of $A$ wrt $\omega$ , then $Q_a ,M_a,A'$ are collinear. Then ,$$(X,Y;J,G)\stackrel{P_a}{=}(X,Y;Q_a,A)\stackrel{A'}{=}(X,Y;H,O)$$ , which is symmetric for $A, B,C$ . So they concur .

Let $OH \cap \omega = \{X,Y\}$ , $OH \cap P_aQ_a=J$ .Let $G$ be the centroid . $A'$ be the antipode of $A$ wrt $\omega$ , then $Q_a ,M_a,A'$ are collinear. Then ,$$(X,Y;J,G)\stackrel{P_a}{=}(X,Y;Q_a,A)\stackrel{A'}{=}(X,Y;H,O)$$ , which is symmetric for $A, B,C$ . So they concur .

The more I learn, the more I realize how much I don't know.

- Albert Einstein

- Albert Einstein

### Re: Geometry Marathon : Season 3

Problem 46:

Given a $ \triangle ABC $ with a point $ P $ lying on the A-bisector of $ \triangle ABC. $ Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and let $ T $ be the point such that $ TB \parallel DF, $ $ TC \parallel DE. $ Prove that the perpendicular from $ T $ to $ EF $ passes through the midpoint of arc $ BC $ in $ \odot (ABC) $ containing $ A. $

Given a $ \triangle ABC $ with a point $ P $ lying on the A-bisector of $ \triangle ABC. $ Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and let $ T $ be the point such that $ TB \parallel DF, $ $ TC \parallel DE. $ Prove that the perpendicular from $ T $ to $ EF $ passes through the midpoint of arc $ BC $ in $ \odot (ABC) $ containing $ A. $

The more I learn, the more I realize how much I don't know.

- Albert Einstein

- Albert Einstein

### Re: Geometry Marathon : Season 3

My Solution:Raiyan Jamil wrote:$\text{Problem 45}$

Let $ABC$ be a triangle with orthocentre $H$ and circumcircle $\omega$ centered at $O$. Let $M_a,M_b,M_c$ be the midpoints of $BC,CA,AB$. Lines $AM_a,BM_b,CM_c$ meet $\omega$ again at $P_a,P_b,P_c$. Rays $M_aH,M_bH,M_cH$ intersect $\omega$ at $Q_a,Q_b,Q_c$. Prove that $P_aQ_a,P_bQ_b,P_cQ_c,OH$ are concurrent.

I like girls and mathematics; both are beautiful.

### Re: Geometry Marathon : Season 3

Solution of Problem 46 :

lets prove a generalization :

"Given $ \triangle ABC $ and a point $ P $. Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and let $ T $ be the point such that $ TB \parallel DF, $ $ TC \parallel DE$ . Let $M$ be the midpoint of $BC$ . Let $ EF \cap BC = K $,$ AP\cap \odot(ABC) = \{A ,S \}$ , $SM \cap \odot(ABC) = \{J ,S \}$ . Prove that $EF \cap JT \in \odot(JMK)$".

Proof:

Let $L \in AK$ such that $LM \|AS$. Let $EF\cap LM=Q.$ Let the line through $C$ parallel to $DE$ meets $LM$ and $EF$ at $T_1$, $R$ respectively . The line through $R$ parallel to $AC$ meets $BC$ and $LM$ at $S$, $N$ respectively .Now the dilation wrt $K$ that takes $D$ to $C$ , takes $C$ to $S$ and $B$ to $M$ .

As, $\dfrac{KC}{KS}=\dfrac{KE}{KR}=\dfrac{KD}{KC}$ and $KB.KC=KD.KM \Rightarrow \dfrac{KB}{KM}=\dfrac{KD}{KC}$ .

So , $-1=(C,B;D,K)=(S,M;C,K)\stackrel{R}{=}(N,M;T_1,Q)$

let us vary $P$ fixing $AS$ .Then when $P$ coincides with $A$ , the points , $Q\equiv L$ and $T_1\equiv\infty$ .So $(M,N;\infty,L)=-1$. So $L$ is the midpoint of $MN$ . Moreover as $N,M,Q$ are symmetric for $B,C$ ,so $T=T_1$

Let $X\in AK$ such that $JX \| AP$. Now , $KD.DM=BD.DC=AD.DS$ .So $AKSM$ is cyclic. As $XJ \| AS$ , so $KMJX$ is also cyclic.

Let $D'$ be the reflection of $D$ wrt $M$ ,$Y$ be the midpoint of $DS$ and $ML \cap \odot(KMJX) = \{M,N' \}$ .

$KM.MD'=BM.MC=SM.MJ$ .So $KSD'J$ is cyclic . As $YM \| SD' \Rightarrow YM$ is tangent to $ \odot(KMJX)$.

So , $(K,J;M,N')\stackrel{M}{=}(D,S;Y,\infty )=-1$

So , $(M,N';\infty,L)\stackrel{X}{=}(M,N';J,K)=-1$ , So $N=N'$

Now , if $KF\cap \odot(KMJX)= \{K,Z \}$ and $JZ \cap ML=T_2$ .Then ,

$(N,M;T_{2},Q)\stackrel{Z}{=}(N,M;J,K)=-1$ , So $T=T_{2}$ and $Z=EF \cap JT \in \odot(JMK)$.

The original problem was created by " Telvcohl " , you can see it here https://artofproblemsolving.com/communi ... 79p8953771

I have no problem to submit. Anybody feel free to take my turn

lets prove a generalization :

"Given $ \triangle ABC $ and a point $ P $. Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and let $ T $ be the point such that $ TB \parallel DF, $ $ TC \parallel DE$ . Let $M$ be the midpoint of $BC$ . Let $ EF \cap BC = K $,$ AP\cap \odot(ABC) = \{A ,S \}$ , $SM \cap \odot(ABC) = \{J ,S \}$ . Prove that $EF \cap JT \in \odot(JMK)$".

Proof:

Let $L \in AK$ such that $LM \|AS$. Let $EF\cap LM=Q.$ Let the line through $C$ parallel to $DE$ meets $LM$ and $EF$ at $T_1$, $R$ respectively . The line through $R$ parallel to $AC$ meets $BC$ and $LM$ at $S$, $N$ respectively .Now the dilation wrt $K$ that takes $D$ to $C$ , takes $C$ to $S$ and $B$ to $M$ .

As, $\dfrac{KC}{KS}=\dfrac{KE}{KR}=\dfrac{KD}{KC}$ and $KB.KC=KD.KM \Rightarrow \dfrac{KB}{KM}=\dfrac{KD}{KC}$ .

So , $-1=(C,B;D,K)=(S,M;C,K)\stackrel{R}{=}(N,M;T_1,Q)$

let us vary $P$ fixing $AS$ .Then when $P$ coincides with $A$ , the points , $Q\equiv L$ and $T_1\equiv\infty$ .So $(M,N;\infty,L)=-1$. So $L$ is the midpoint of $MN$ . Moreover as $N,M,Q$ are symmetric for $B,C$ ,so $T=T_1$

Let $X\in AK$ such that $JX \| AP$. Now , $KD.DM=BD.DC=AD.DS$ .So $AKSM$ is cyclic. As $XJ \| AS$ , so $KMJX$ is also cyclic.

Let $D'$ be the reflection of $D$ wrt $M$ ,$Y$ be the midpoint of $DS$ and $ML \cap \odot(KMJX) = \{M,N' \}$ .

$KM.MD'=BM.MC=SM.MJ$ .So $KSD'J$ is cyclic . As $YM \| SD' \Rightarrow YM$ is tangent to $ \odot(KMJX)$.

So , $(K,J;M,N')\stackrel{M}{=}(D,S;Y,\infty )=-1$

So , $(M,N';\infty,L)\stackrel{X}{=}(M,N';J,K)=-1$ , So $N=N'$

Now , if $KF\cap \odot(KMJX)= \{K,Z \}$ and $JZ \cap ML=T_2$ .Then ,

$(N,M;T_{2},Q)\stackrel{Z}{=}(N,M;J,K)=-1$ , So $T=T_{2}$ and $Z=EF \cap JT \in \odot(JMK)$.

The original problem was created by " Telvcohl " , you can see it here https://artofproblemsolving.com/communi ... 79p8953771

I have no problem to submit. Anybody feel free to take my turn

The more I learn, the more I realize how much I don't know.

- Albert Einstein

- Albert Einstein

### Re: Geometry Marathon : Season 3

Problem 47: Let $ABCD$ be a cyclic quadrilateral. $AB$ intersects $DC$ at $E$. $AD$ intersects $BC$ at $F$. Let $M, N, P$ are midpoints of $BD, AC, EF$ respectively. Prove that $PN.PM=PE^2$

I like girls and mathematics; both are beautiful.