Geometry Marathon : Season 3

For discussing Olympiad level Geometry Problems
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Raiyan Jamil
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Re: Geometry Marathon : Season 3

Unread post by Raiyan Jamil » Tue Aug 08, 2017 12:58 am

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 $ABC$
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ahmedittihad
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Re: Geometry Marathon : Season 3

Unread post by ahmedittihad » Fri Sep 01, 2017 2:00 pm

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.
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ahmedittihad
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Re: Geometry Marathon : Season 3

Unread post by ahmedittihad » Fri Sep 01, 2017 2:15 pm

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 $OH$.
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Raiyan Jamil
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Re: Geometry Marathon : Season 3

Unread post by Raiyan Jamil » Sat Oct 21, 2017 3:58 pm

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.
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Raiyan Jamil
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Re: Geometry Marathon : Season 3

Unread post by Raiyan Jamil » Sat Oct 21, 2017 4:20 pm

$\text{Problem 45}$

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.
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joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Mon Oct 30, 2017 11:15 pm

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 .
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joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Mon Oct 30, 2017 11:54 pm

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. $
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tanmoy
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Re: Geometry Marathon : Season 3

Unread post by tanmoy » Tue Oct 31, 2017 12:01 am

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:
Let $A_1, B_1, C_1$ be the antipodes of $A, B, C$, respectively and let $e$ be the euler line of $\triangle ABC$.

It is well known that $Q_a,H, M_a, A_1$ are collinear. Also, $Q_b, H, M_b, B_1$ are collinear and $Q_c, H, M_c, C_1$ are collinear.

Applying Pascal's Theorem to $AA_1Q_aBB_1Q_b$, we get that $Q_aB \cap Q_bA \in e$.

Again applying Pascal's Theorem to $AP_aQ_aBP_bQ_b$, we get that $P_aQ_a \cap P_bQ_b \in e$.

Similarly, we get that $P_aQ_a \cap P_cQ_c \in e$ and $P_cQ_c \cap P_bQ_b \in e$.

So, ultimately $P_aQ_a \cap P_bQ_b \cap P_cQ_c \in e$
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joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Sat Nov 18, 2017 3:25 pm

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: 282
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

Re: Geometry Marathon : Season 3

Unread post by tanmoy » Mon Nov 20, 2017 12:32 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 $PN.PM=PE^2$
"Questions we can't answer are far better than answers we can't question"

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