## Geometry Marathon : Season 3

For discussing Olympiad level Geometry Problems
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 $ABC$
A smile is the best way to get through a tough situation, even if it's a fake smile.

Posts: 181
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.
Frankly, my dear, I don't give a damn.

Posts: 181
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 $OH$.
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.
A smile is the best way to get through a tough situation, even if it's a fake smile.

joydip
Posts: 48
Joined: Tue May 17, 2016 11:52 am

### 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 .
The first principle is that you must not fool yourself and you are the easiest person to fool.

joydip
Posts: 48
Joined: Tue May 17, 2016 11:52 am

### 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 $A.$
The first principle is that you must not fool yourself and you are the easiest person to fool.

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm

### Re: Geometry Marathon : Season 3

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.
My Solution:
"Questions we can't answer are far better than answers we can't question"

joydip
Posts: 48
Joined: Tue May 17, 2016 11:52 am

### 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
The first principle is that you must not fool yourself and you are the easiest person to fool.

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
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 $PN.PM=PE^2$