272. Outside of triangle. Extension of Fermat's point.

by Virgil Nicula, May 4, 2011, 1:51 PM

Jakobi's theorem. Outside of $\triangle ABC$ construct $\left\{\begin{array}{cc} \triangle BXC\ : & m\left(\widehat{CBX}\right)=y\ ;\ m\left(\widehat{BCX}\right)=z\\\\ \triangle CYA\ : & m\left(\widehat{ACY}\right)=z\ ;\ m\left(\widehat{CAY}\right)=x\\\\ \triangle AZB\ : & m\left(\widehat{BAZ}\right)=x\ ;\ m\left(\widehat{ABZ}\right)=y\end{array}\right\|$ . Prove that $AX\cap BY\cap CZ\ne\emptyset$ .

Proof. Denote $M\in AX\cap BC$ , $N\in BY\cap CA$ , $Z\in CZ\cap AB$ . Observe that $\frac {BX}{\sin z}=\frac {CX}{\sin y}$ . Therefore,

$\frac {MB}{MC}=\frac {[ABX]}{[ACX]}=$ $\frac {BA\cdot BX\cdot \sin\widehat{ABX}}{CA\cdot CX\cdot \sin\widehat{ACX}}=$ $\frac cb\cdot\frac {\sin z}{\sin y}\cdot\frac {\sin (B+y)}{\sin (C+z)}$ $\implies$ $\frac {MB}{MC}=\frac cb\cdot\frac {\sin z}{\sin y}\cdot\frac {\sin (B+y)}{\sin (C+z)}$ . Show analogously

another two relations. In conclusion, $\left\{\begin{array}{c} \frac {MB}{MC}=\frac cb\cdot\frac {\sin z}{\sin y}\cdot\frac {\sin (B+y)}{\sin (C+z)}\\\\ \frac {NC}{NA}=\frac ac\cdot\frac {\sin x}{\sin z}\cdot\frac {\sin (C+z)}{\sin (A+x)}\\\\ \frac {PA}{PB}=\frac ba\cdot\frac {\sin y}{\sin x}\cdot\frac {\sin (A+x)}{\sin (B+y)}\end{array}\right\|\bigodot\implies$ $\frac {MB}{MC}\cdot\frac {NC}{NA}\cdot\frac {PA}{PB}=1\implies$ $AX\cap BY\cap CZ\ne\emptyset$ .


Proposed problem. A triangle $\bigtriangleup ABC$ is given, and let the external angle bisector of the angle $\angle A$ intersect the lines perpendicular to $BC$ and passing

through $B$ and $C$ at the points $D$ and $E$, respectively. Prove that the line segments $BE$, $CD$, $AO$ are concurrent, where $O$ is the circumcenter of $\bigtriangleup ABC$ .


Proof 1. Suppose w.l.o.g. that $B<C$, i.e. $b<c$. Denote the intersection $X\in AO\cap CD$. Observe that $AO\cap BE\cap CD\ne\emptyset$ $\Longleftrightarrow$ $\frac{XD}{XC}=\frac{BD}{CE}$ .

Thus, $m(\widehat{XAD})=m(\widehat{XAB})+m(\widehat{BAD})=(90-C)+\left(90-\frac{A}{2}\right)=180-\left(C+\frac{A}{2}\right)$ , $m(\widehat{XAC})=m(\widehat{OAC})=90-B$

and $\left\{\begin{array}{c}m(\widehat{ADB})=B+\frac{A}{2}\\\ m(\widehat{AEC})=C+\frac{A}{2}\end{array}\right\|$ . Therefore, $\frac{XD}{XC}=$ $\frac{AD}{AC}\cdot\frac{\sin \widehat{XAD}}{\sin\widehat{XAC}}=$ $\frac{BD}{CE}\cdot \frac{AD}{BD}\cdot\frac{CE}{AC}\cdot\frac{\sin\widehat{XAD}}{\sin\widehat{XAC}}=$ $\frac{BD}{CE}\cdot\frac{\sin\widehat{ABD}}{\sin\widehat{BAD}}\cdot\frac{\sin\widehat{CAE}}{\sin\widehat{AEC}}\cdot\frac{\sin\widehat{XAD}}{\sin\widehat{XAC}}$ $\implies$

$\frac{XD}{XC}=$ $\frac{BD}{CE}\cdot\frac{\cos B}{\cos\frac{A}{2}}\cdot\frac{\cos\frac{A}{2}}{\sin\left(C+\frac{A}{2}\right)}\cdot\frac{\sin\left(C+\frac{A}{2}\right)}{\cos B}$ $\implies$ $\frac{XD}{XC}=\frac{BD}{CE}$ $\implies$ $X\in BE\cap CD$ $\implies$ the lines $AO$ , $BE$ , $CD$ are concurrently.

Proof 2. Suppose w.l.o.g. that $B<C$ and denote the intersections $\left\{\begin{array}{c}X\in AO\cap CD\\\ P\in AO\ ,\ BP\parallel DE\end{array}\right\|$. Prove easily that $\widehat{ADB}\equiv \widehat{DAP}$,

i.e. $PA=BD$ . Observe that $\triangle BDP\equiv\triangle ECA$. Therefore, $\frac{XD}{XC}=\frac{DP}{CA}=\frac{BD}{EC}$, i.e. $X\in BE\cap CD$ .


Some reciprocal questions. Let $BCED$ be a trapezoid for which $BD\parallel CE$ and $BD\perp BC$ . Denote $\left\{\begin{array}{c}M\in (DE)\ ,\ N\in (BC)\ ;\ CM\parallel DN\\\ S\in (BC)\ ,\ SM\perp DE\\\ P\in DN\ ,\ BP\parallel DE\end{array}\right\|$ .


and $X\in MP\cap DC$ . Prove the following the chain of the equivalencies : $BD=PM\Longleftrightarrow \widehat{BMS}\equiv\widehat{CMS}\Longleftrightarrow X\in BE\cap CD$ .
This post has been edited 19 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:00 AM

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    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: 404395
  • Total comments: 37
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