18th ibmo - argentina 2003/q2

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#1 h Apr 8, 2006, 11:32 PM • 2 Y

Y by Adventure10, Mango247

Let $C$ and $D$ be two points on the semicricle with diameter $AB$ such that $B$ and $C$ are on distinct sides of the line $AD$ . Denote by $M$ , $N$ and $P$ the midpoints of $AC$ , $BD$ and $CD$ respectively. Let $O_A$ and $O_B$ the circumcentres of the triangles $ACP$ and $BDP$ . Show that the lines $O_AO_B$ and $MN$ are parallel.

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#2 h Apr 9, 2006, 2:03 AM • 1 Y

Y by Adventure10

Let O be the midpoint of AB, the center of the semicircle (O) with the points C, D forming a convex quadrilateral ABDC. $OM = OA\ \cos \widehat{AOM}$ and $ON = OB\ \cos \widehat{BON}$ . $ON \parallel AD$ is a midline of the triangle $\triangle ABD$ and $OM \parallel BC$ is a midline of the triangle $\triangle ABC$ . Let $E, F$ be the midpoints of PC, PD. Denote $\alpha = \angle (OO_A, PE) = \angle (OM, PC)$ and $\beta = \angle (OO_B, PF) = \angle (ON, PD)$ . From the trapezoid $OO_AEP$ , $OO_A = \frac{PE}{\cos \alpha}$ and from the trapezoid $OO_BFP$ , $OO_B = \frac{PF}{\cos \beta}$ . Since $OA = OB = \frac{AB}{2}$ and $PE = \frac{PC}{2} = \frac{CD}{4} = \frac{PD}{2} = PF$ , we get

$\frac{OM}{ON} = \frac{\cos \widehat{AOM}}{\cos \widehat{BON}},\ \ \ \frac{OO_A}{OO_B} = \frac{\cos \beta}{\cos \alpha}$

But from the circle (O) and form $AD \parallel ON, BC \parallel OM$ ,

$\alpha = \angle (OM, PC) = \angle (BC, DC) = \angle BCD = \angle BAD = \angle BON$

$\beta = \angle (ON, PD) = \angle (AD, CD) = \angle ADC = \angle ABC = \angle AOM$

Consequently, $\frac{OM}{ON} = \frac{OO_A}{OO_B}$ , which means $O_AO_B \parallel MN$ .

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#3 h Jan 22, 2010, 3:37 PM • 2 Y

Y by Adventure10, Mango247

Let $L$ be the midpoint of $AB$ and $Q \equiv \odot(ACP) \cap \odot(BDP),$ different from $P.$ Clearly $O_AO_B \perp PQ$ and lines $CA,DB$ and $PQ$ concur at the radical center $E$ of $\odot(ACP),$ $\odot(BDP)$ and the circumference with diameter $AB.$ Since $CD$ is antiparalell to $AB$ with respect to $EA,$ $EB,$ then $EP$ and $EL$ are isogonals with respect to $\angle AEB,$ i.e. $EP$ is the E-symmedian of $\triangle EAB.$ Since $EMLN$ is a cyclic quadrilateral on account of $\angle EML = \angle ENL = 90^{\circ},$ we have $\angle MEL = \angle MNL = \angle PEN$ $\Longrightarrow$ $MN \perp PE.$ Hence $O_AO_B \parallel MN.$

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#4 h Jan 22, 2010, 7:20 PM • 2 Y

Y by Adventure10, Mango247

Let be the point $Q\equiv (O_{a})\cap (O_{b}),$ the other than $P,$ where $(O_{a}),\ (O_{b})$ are the circumcircles of the triangles $\bigtriangleup ACP,\ \bigtriangleup BDP,$ respectively.

The line segments $AC,\ PQ,\ BD,$ as the radical axis of the circles $(O),\ (O_{a}),\ (O_{b})$ taken per two of them, are concurrent at one point so be it $R,$ as their radical center, as well.

$\bullet$ Because of now, $PM\parallel AD$ and $PN\parallel BC$ and $AD\perp BD$ and $BC\perp AC,$ we conclude that $P$ is the orthocenter of the triangle $\bigtriangleup RMN$ and so, we have that $RP\perp MN$ $\Longrightarrow$ $PQ\perp MN$ $,(1)$

From $(1)$ and beause of $O_{a}O_{b}\perp PQ$ we conclude that $O_{a}O_{b}\parallel MN$ and the proof of the proposed problem is completed.

Kostas Vittas.

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#5 h Jan 6, 2011, 7:00 PM • 2 Y

Y by Adventure10, Mango247

Take $\{T\}= BD \cap AC$ , then easily $P$ is the orthocenter of $\triangle TNM$ , $PT \perp MN$ , but $P$ and $T$ belong to the radical axis of the 2 circles, hence $TP \perp O_AO_B$ .

Best regards,
sunken rock

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#6 h Jan 9, 2011, 4:41 AM • 2 Y

Y by Adventure10, Mango247

Project all 7 points $A$ , $M$ , $O_A$ , $O$ , $O_B$ , $N$ , and $B$ onto $CD$ , then it's trivial!

Let $h_{X}$ denote the projection of $X$ onto $CD$ . Let $O$ be the midpoint of $AB$ . Then $O$ , $O_A$ , and $M$ are collinear, similarly $O$ , $O_B$ , and $N$ are collinear.
Then since $AO=OB$ , $h_{A}P=h_{B}P$ . Also, $CP=PD$ , so $h_{A}C=h_{B}D$ . Since $M$ and $N$ are the midpoint of $AC$ and $BD$ , $h_{M}P=h_{N}P$ . Also, since $O_A$ and $O_B$ are on the perp bisectors of $CP$ and $PD$ , $h_{O_A}P=h_{O_B}P$ . Thus,
$\frac{MO_A}{O_AO}=\frac{h_{M}h_{O_A}}{h_{O_A}P}= \frac{h_{N}h_{O_B}}{h_{O_B}P}= \frac{NO_B}{O_BO}$
So $MN \parallel O_AO_B$ .

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#7 h Nov 19, 2011, 5:50 AM • 2 Y

Y by Adventure10, Mango247

let AC,BD ntersect at E.circle $O_A,O_B$ intersect at P and T
it suffices to prove that PT is perpendicular to MN
by equal power theorem,E,P,T collinear
then $MP\bot EN,NP\bot EM$
so P is the orthocemter of triangle EMN
hence $EP\bot MN$

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