209. Nice problem, nice proof !

by Virgil Nicula, Jan 16, 2011, 3:15 PM

PP1 (Jayme). Let $ABC$ be a triangle with the orthocenter $H$ , with the circumcircle $w=C(O,R)$ and with the orthic

triangle $A'B'C'$ . The line $AA'$ cut again $w$ in $D$ , the line $DB'$ cut again $w$ in $E$ and $B^*\in BE\cap A'C'$ . Prove that $B^*H\parallel B^{\prime} C^{\prime} $ .

Proof 1 (Luisgeometria). Denote $F \in B'C' \cap AA'$ , $P \in BC \cap B'C'$ and the midpoint $U$ of $B'C'$ . Since cross ratio $(B',C',F,P)$ is harmonicaally it follows that

$\overline{PF} \cdot \overline{PU}=\overline{PB'} \cdot \overline{PC'}=\overline{PB} \cdot \overline{PC}$ $\Longrightarrow$ $B,C,U,F$ are concyclic $\Longrightarrow$ $\angle UBC=\angle B'FC$ . But quadrilateral $FDCB'$ is cyclic on account of

$\angle AB'C'=\angle ABC=\angle ADC,$ thus $\angle EDC=\angle UBC$ $\Longrightarrow$ $E,U,B$ are collinearly. Let the parallel from $H$ to $B'C'$ cut $A'C',AB$ at $B_0,K$ respectively.

Since $H$ is incenter of $\triangle A'B'C',$ then $\triangle C'HB_0$ is $B_0$-isosceles $\Longrightarrow$ $B_0$ is the midpoint of $HK$ $\Longrightarrow$ $BB_0$ is identically to the $B$-median line of $\triangle BB'C'$ $\Longrightarrow$

$EB$ and $A'C'$ intersect at $B_0$ $\Longrightarrow$ $B_0 \equiv B^{*}$ $\Longrightarrow$ $HB^{*} \parallel B'C'$ .


PP2. Consider two points $M$ , $N$ in the interior of the square $ABCD$ so that $M$ belongs to the interior of the angle $\widehat{NAD}$ and $m\left(\widehat{MAN}\right)=$ $m\left(\widehat{MCN}\right)=45^{\circ}$ .

Prove that $[MAD]+[MCN]+[NAB]=[MCD]+[MAN]+[NCB]$ , where denoted $[XYZ]$ - the area of $\triangle XYZ$ .

Proof 1. Let $E$, $F$ be the reflections of $D$ , $B$ in $AM$ , $CN$ . Thus, $\triangle MFN\equiv\triangle MEN$ and $\triangle MCN+\triangle ABN+\triangle ADM=$ $\triangle MCN+\square AMEN=$

$\square ANCM+\triangle MNE$ . Also, $\triangle AMN+\triangle BCN+\triangle CDM=$ $\square ANCM+\triangle MFN$ . Thus, $\triangle MCN+\triangle ABN+\triangle ADM=$ $\triangle AMN+\triangle BCN+\triangle CDM$ .

Proof 2. Let $\left\{\begin{array}{ccc} m\left(\widehat{NAB}\right)=a & ; & m\left(\widehat{NBC}\right)=b\\\\ m\left(\widehat{MCD}\right)=c & ; & m\left(\widehat{MDA}\right)=d\end{array}\right|$ Apply the theorem of Sinus $:\ \left\{\begin{array}{cc} \left|\begin{array}{cccc} \triangle ANC\ : & \frac {NC}{NA} & = & \frac {\sin\left(45^{\circ}-a\right)}{\sin c}\\\\ \triangle BNC\ : & \frac {NB}{NC} & = & \frac{\sin\left(45^{\circ}-c\right)}{\sin b}\\\\ \triangle ANB\ : & \frac {NA}{NB} & = & \frac {\cos b}{\sin a}\end{array}\right| & \bigodot\\\\ \left|\begin{array}{cccc} \triangle AMC\ : & \frac {MC}{MA} & = & \frac {\sin a}{\sin\left(45^{\circ}-c\right)}\\\\ \triangle AMD\ : & \frac {MA}{MD} & = & \frac{\sin d}{\sin\left(45^{\circ}-a\right)}\\\\ \triangle CMD\ : & \frac {MD}{MC} & = & \frac{\sin c}{\cos d}\end{array}\right| & \bigodot \end{array}\right\|$ $\implies$

$\left\{\begin{array}{c} \tan b=\frac {\sin\left(45^{\circ}-a\right)\sin\left(45^{\circ}-c\right)}{\sin a\sin c}\\\\ \tan d=\frac {\sin\left(45^{\circ}-a\right)\sin\left(45^{\circ}-c\right)}{\sin a\sin c}\end{array}\right|\implies$ $\boxed{\tan b=\tan d=\frac {(1-\tan a)(1-\tan c)}{2\tan a\tan c}}$ . Denote $\left\{\begin{array}{c} \tan a=x\\\ \tan b=y\\\ \tan c=z\end{array}\right|$ , where $y=\frac {(1-x)(1-z)}{2xz}$ . Thus,
This post has been edited 13 times. Last edited by Virgil Nicula, Nov 26, 2015, 6:23 PM

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    -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

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