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BdMO Online Forum • View topic - Geometry Marathon : Season 3

Geometry Marathon : Season 3

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

Post Number:#91  Unread postby 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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Re: Geometry Marathon : Season 3

Post Number:#92  Unread postby 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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Re: Geometry Marathon : Season 3

Post Number:#93  Unread postby 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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Re: Geometry Marathon : Season 3

Post Number:#94  Unread postby 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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Re: Geometry Marathon : Season 3

Post Number:#95  Unread postby 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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