100. Bobiller's theorem.

by Virgil Nicula, Sep 7, 2010, 2:55 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=365492
Quote:
Bobiller's theorem. Let $ABC$ be a triangle. For a point $P$ denote $\left|\begin{array}{cc} Q\in BC\ ,\ PA\perp PQ\\\ R\in CA\ ,\ PB\perp PR\\\ S\in AB\ ,\ PC\perp PS\end{array}\right|$ . Prove that $S\in QR$ .
Proof 1. $\left|\begin{array}{ccc} \frac {QB}{QC}=\frac {PB}{PC}\cdot\frac {\sin\widehat{QPB}}{\sin\widehat{QPC}} & \implies & \frac {QB}{QC}=\frac {PB}{PC}\cdot\frac {\cos\widehat{APB}}{\cos\widehat{APC}}\\\\ \frac {RC}{RA}=\frac {PC}{PA}\cdot\frac {\sin\widehat{RPC}}{\sin\widehat{RPA}} & \implies & \frac {RC}{RA}=\frac {PC}{PA}\cdot\frac {\cos\widehat{BPC}}{\cos\widehat{BPA}}\\\\ \frac {SA}{SB}=\frac {PA}{PB}\cdot\frac {\sin\widehat{SPA}}{\sin\widehat{SPB}} & \implies & \frac {SA}{SB}=\frac {PA}{PB}\cdot\frac {\cos\widehat{CPA}}{\cos\widehat{CPB}}\end{array}\right|\ \bigodot\ \implies$ $\frac {QB}{QC}\cdot\frac {RC}{RA}\cdot\frac {PA}{PB}=1$ $\implies$ $S\in QR$ .

Proof 2. $\left|\begin{array}{c} \overline{QB}=\lambda\cdot \overline{QC} \\\\ \overline{RC}=\mu\cdot \overline{RA}\\\\ \overline{SA}=\theta\cdot \overline{SB}\end{array}\right|$ $\implies$ $\left|\begin{array}{c} \overline{PB}-\lambda\cdot\overline{PC}=(1-\lambda )\cdot\overline{PQ}\\\\ \overline{PC}-\mu\cdot\overline{PA}=(1-\mu )\cdot\overline{PR}\\\\ \overline{PA}-\theta\cdot\overline{PB}=(1-\theta )\cdot\overline{PS}\end{array}\right|$ . Observe that

$\left|\begin{array}{ccccccc} PA\perp PQ & \iff & \overline{PA}\cdot\overline{PQ}=0 & \iff & \overline{PA}\cdot\left(\overline{PB}-\lambda\cdot\overline{PC}\right)=0 & \iff & \overline{PA}\cdot\overline{PB}=\lambda\cdot\overline{PA}\cdot\overline{PC}\\\\ PB\perp PR & \iff & \overline{PB}\cdot\overline{PR}=0 & \iff & \overline{PB}\cdot\left(\overline{PC}-\mu\cdot\overline{PA}\right)=0 & \iff & \overline{PB}\cdot\overline{PC}=\mu\cdot\overline{PB}\cdot\overline{PA}\\\\ PC\perp PS & \iff & \overline{PC}\cdot\overline{PS}=0 & \iff & \overline{PC}\cdot\left(\overline{PA}-\theta\cdot\overline{PB}\right)=0 & \iff & \overline{PC}\cdot\overline{PA}=\theta\cdot\overline{PC}\cdot\overline{PB}\end{array}\right|\ \bigodot$

$\implies$ $\lambda\cdot\mu\cdot\theta =1$ $\implies$ $S\in QR$ .

Proof 3. Apply Stewart's orientated relation to the cevian $AQ$ , $PQ$ in $\triangle ABC$ , $\triangle BPC$ respectively :

$\left|\begin{array}{c} AB^2\cdot\overline{QC}+AQ^2\cdot\overline{CB}+AC^2\cdot\overline {BQ}+\overline{QC}\cdot\overline{CB}\cdot\overline {BQ}=0\\\\ PB^2\cdot\overline{QC}+PQ^2\cdot\overline{CB}+PC^2\cdot\overline {BQ}+\overline{QC}\cdot\overline{CB}\cdot\overline {BQ}=0\end{array}\right|$ . Using relation $AQ^2-PQ^2=PA^2$ obtain by

substruct that $\left(c^2-PB^2\right)\cdot \overline{QC}+PA^2\cdot \overline{CB}+\left(b^2-PC^2\right)\cdot \overline{BQ}=0$ . Using relation $\overline{CB}=\overline {QB}-\overline{QC}$ obtain that

$\left(c^2-PB^2\right)\cdot \overline{QC}+PA^2\cdot\left( \overline {QB}-\overline{QC}\right)+\left(b^2-PC^2\right)\cdot \overline{BQ}=0$ $\implies$ $\lambda =\frac {\overline{QB}}{\overline{QC}}=\frac {PA^2+PB^2-c^2}{PA^2+PC^2-b^2}$ .

So obtain analogously that $\left|\begin{array}{ccccc} \lambda =\frac {\overline{QB}}{\overline{QC}}=\frac {PA^2+PB^2-c^2}{PA^2+PC^2-b^2}\\\\ \mu =\frac {\overline{RC}}{\overline{RA}}=\frac {PB^2+PC^2-a^2}{PB^2+PA^2-c^2}\\\\ \theta =\frac {\overline{SA}}{\overline{SB}}=\frac {PC^2+PA^2-b^2}{PC^2+PB^2-a^2}\end{array}\right|\ \bigodot$ $\implies$ $\lambda\cdot\mu\cdot\theta =1$ $\implies$ $S\in QR$ .
This post has been edited 9 times. Last edited by Virgil Nicula, Nov 26, 2015, 9:42 PM

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  • the visits tho

    by GoogleNebula, Apr 14, 2021, 5:25 AM

  • Bro this blog is ripped

    by samrocksnature, Apr 14, 2021, 5:16 AM

  • Holy- Darn this is good. shame it's inactive now

    by the_mathmagician, Jan 17, 2021, 7:43 PM

  • godly blog. opopop

    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 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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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
Blog Stats
  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404396
  • Total comments: 37
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