51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.

by Virgil Nicula, Jul 8, 2010, 8:23 PM

PP1. Fie $\triangle\ ABC$ cu circumcentrul $O$ si ortocentrul $H$ . Notam mijlocul $M$ al segmentului $[AH]$ . Perpendiculara in $M$ pe dreapta $OM$ intersecteaza dreptele $AB$ si $AC$ in punctele $P$ , $Q$ respectiv. Sa se arate ca $MP=MQ$ .

Metoda 1. $\left\|\ \begin{array}{cc} E\in AC\ , & BE\perp AC\\\\ F\in AB\ , & CF\perp AB\end{array}\ \right\|\ \ \wedge\ \ \left\|\ \begin{array}{cc} X\in AC\ , & HX\parallel AB\\\\ Y\in AB\ , & HY\parallel AC\end{array}\ \right\|$ $\Longrightarrow$ $\left\|\ \begin{array}{cc} XA=\frac {HF}{h_c}\cdot b\ ,& XC=\frac {HC}{h_c}\cdot b\\\\ YA=\frac {HE}{h_b}\cdot c\ , & YB=\frac {HB}{h_b}\cdot c\end{array}\ \right\|$ . Observe that $MX=MY$ deoarece $AXHY$ este paralelogram si

$XA\cdot XC=YA\cdot YB$ deoarece $HF\cdot HC=HE\cdot HB$ (patrulaterul $BCEF$ este ciclic) si $bh_b=ch_c=2S$ . Asadar, punctele $X$ si $Y$ au aceeasi putere fata de circumcercul triunghiului $ABC$ ceea ce inseamna

$OX=OY$ , adica $OM\perp XY$ . Insa $OM\perp PQ$ si $M\in XY\cap PQ$ . In concluzie, $P\equiv Y$ , $Q\equiv X$ si $MP=MQ$ .

Metoda 2. Notam punctul $T\in EF$ pentru care $TH\parallel BC$ si $X\in TM\cap AC$ , $Y\in TM\cap AB$ . Deoarece $TH$ este tangenta la cercul de diametru $[AH]$ rezulta dintr-o proprietate cunoscuta faptul ca $MX=MY$ .

Din acest moment rezolvarea continua ca in metoda precedenta.

Metoda 3 (Cosmin Pohoata). Consideram mijlocul $D$ al laturii $[BC]$ si punctele $U\in AB$ , $V\in AC$ astfel incat $H\in UV$ si $DH\perp UV$ . Din proprietatea fluturelui in cercul de diametru $[BC]$ (cu centrul $D$) se obtine $HU=HV$ .

Deoarece $OD=MH$ , $OD\parallel MH$ rezulta $OM\parallel DH$ . Din $OM\perp PQ$ rezulta ca $PQ\parallel UV$ . Deoarece punctul $M$ apartine $A$-medianei din triunghiul $AUV$ si $M\in PQ\parallel UV$ rezulta $MP=MQ$ .


PP2. Fie triunghiul $ABC$ cu cercul inscris $w=C(I,r)$ si cercul circumscris $\rho =C(O,R)$ . Cercul $w$ atinge latura $[BC]$ in punctul $D$ si dreptele $AI$ , $AO$ intalnesc

a doua oara cercul $\rho$ in punctele $M$ , $S$ respectiv. Consideram punctele $X\in DM$ , $Y\in AO$ astfel ca $I\in XY$ . Sa se arate ca $IX=IY\Longleftrightarrow OI\perp XY$ .

Demonstratie. Se observa ca $\triangle DIM\sim\triangle IAS$ deoarece $\widehat{DIM}\equiv\widehat {IAS}$ si $\frac {DI}{IM}=\frac {IA}{AS}$, unde $IA\cdot IM=2Rr$ este puterea punctului $I$ fata de cercul $\rho$ . Asadar

$\widehat{IMD}\equiv\widehat {ASI}$ , adica punctul $R\in SI\cap MD$ apartine cercului $\rho$ . Si acum patrulaterului convex $ARMS$ ii aplicam proprietatea fluturelui : $IX=IY\Longleftrightarrow OI\perp XY$ .

PP3. Fie triunghiul $ABC$ cu ortocentru $H$ si circumcercul $w=C(O)$ . Notam $D\in AH\cap BC$ , $E\in BC$ pentru care $DE=DC$ si intersectia $F\in AB\cap HE$ . Sa se arate ca $DF\perp DO$ .

Demonstratie. Fie simetricul $L$ al lui $H$ fata de $D$ si $M\in FD\cap LC$ . Se stie ca $L\in w$ si patrulaterul $HELC$ este paralelogram. Deci $DF=DM$ si din proprietatea fluturelui la patrulaterului inscriptibil $ABLC$ obtinem $OD\perp FD$ .
This post has been edited 818 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:46 PM

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  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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  • Total entries: 456
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