110. German TST 2004 - Exam V , Problem 1

by Virgil Nicula, Sep 11, 2010, 10:25 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=155550&p=874329#p874329
Quote:
The $A$- excircle of a triangle $ABC$ touches the side $BC$ at the point $K$ and the extended side $AB$ at the point $L$. The $B$- excircle
touches the lines $BA$ and $BC$ at the points $M$ and $N$ respectively. Denote $X\in KL\cap MN$. Show that $\widehat{ACX}\equiv\widehat{NCX}$.

Proof I. Suppose w.l.o.g. that $a>b$. Denote $\left\{\begin{array}{c}S_{1}\in MN\cap LK\\\\ S_{2}\in MN\cap CI_{b}\\\\ V\in BA\cap CI_{b}\end{array}\right\|$. Observe that : $\left\{\begin{array}{c}LB=MA=KB=p-c\\\\ KN=KC+CN=(p-b)+(p-a)=c\\\\ LM=LB+BM=(p-c)+p=a+b\end{array}\right\|$

and $\frac{VA}{b}=\frac{VB}{a}=\frac{c}{a-b}$ $\implies$ $VB=\frac{ac}{a-b}$ $\implies$ $VM=VB-MB=\frac{ac}{a-b}-p=\frac{(a+b)(p-a)}{a-b}$.

$1\blacktriangleright$ Apply the Menelaus' theorem to the $\overline{LS_{1}K}/BMN\ :\ \frac{LB}{LM}\cdot\frac{S_{1}M}{S_{1}N}\cdot\frac{KN}{KB}=1$ $\implies$ $\frac{p-c}{a+b}\cdot\frac{S_{1}M}{S_{1}N}\cdot \frac{c}{p-c}=1$ $\implies$ $\boxed{\frac{S_{1}M}{S_{1}N}=\frac{a+b}{c}}\ \ (1)$.

$2\blacktriangleright$ Apply the Menelaus' theorem to $\overline{VS_{2}C}/BMN\ :\ \frac{VM}{VB}\cdot\frac{CB}{CN}\cdot\frac{S_{2}N}{S_{2}M}=1$ $\implies$ $\frac{(a+b)(p-a)}{ac}\cdot\frac{a}{p-a}\cdot\frac{S_{2}N}{S_{2}M}=1$ $\implies$ $\boxed{\frac{S_{2}M}{S_{2}N}=\frac{a+b}{c}}\ \ (2)$.

From $(1)$ , $(2)$ results that $\frac{S_{1}M}{S_{1}N}=\frac{S_{2}M}{S_{2}N}$ $\implies$ $S_{1}\equiv S_{2}\equiv S\in MN$ i.e. the lines $LK$, $CI_{b}$ are concurrently in the point $S\in MN$ for which $\frac{SM}{SN}=\frac{a+b}{c}$.


Lemma. Let $\triangle XYZ$ for which $XY\perp XZ$. For a mobile line $d$ for which $X\in d$ denote the points $\left\{\begin{array}{c}\{U,V\}\subset d\\\\ YU\perp d\ ,\ ZV\perp d\end{array}\right\|$.

Define the point $W$ for which $WU\parallel XZ$ and $WV\parallel XY$. Then $W\in YZ$ and $\frac{WY}{WZ}=\tan^{2}x$, where $x=m(\widehat{UXY})$.


Proof. Indeed, $\left\{\begin{array}{c}W_{1}\in YZ\ ,\ UW_{1}\perp XY\\\\ W_{2}\in YZ\ ,\ VW_{2}\perp XZ\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c}W_{1}Y=YZ\sin^{2}x\\\\ W_{2}Z=YZ\cos^{2}x\end{array}\right\|$ $\implies$ $W_{1}Y+W_{2}Z=YZ$ $\implies$

$W_{1}\equiv W_{2}\equiv W\in YZ$ and $\frac{WY}{WZ}=\frac{YZ\sin^{2}x}{YZ\cos^{2}x}$, i.e. $\frac{WY}{WZ}=\tan^{2}x$.

$\blacksquare$ Otherwise. Denote $\left\{\begin{array}{c}R\in UW_{1}\cap XY\\\\ S\in VW_{2}\cap XZ\end{array}\right\|$ $\implies$ $\triangle UXY\sim\triangle VZX$ $\implies$ $\frac{UY}{UX}=\frac{VX}{VZ}$.

Therefore, $\left\{\begin{array}{c}\frac{W_{1}Y}{W_{1}Z}=\frac{RY}{RX}=\left(\frac{UY}{UX}\right)^{2}\\\\ \frac{W_{2}Y}{W_{2}Z}=\frac{SX}{SZ}=\left(\frac{VX}{VZ}\right)^{2}\end{array}\right\|$ $\implies$ $\frac{W_{1}Y}{W_{1}Z}=\frac{W_{2}Y}{W_{2}Z}$ $\implies$ $W_{1}\equiv W_{2}\equiv W\in YZ$.

Proof II. Apply the above lemma to the right triangle $BI_{a}I_{b}$ and the line $d\equiv BA$.
This post has been edited 15 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:08 AM

Comment

0 Comments

Own problems or extensions/generalizations of some problems which was posted here.

avatar

Virgil Nicula
Archives
+ October 2017
+ September 2017
+ December 2016
+ October 2016
+ February 2016
+ September 2013
+ October 2010
+ September 2010
Shouts
Submit
  • orzzzzzzzzz

    by mathMagicOPS, Jan 9, 2025, 3:40 AM

  • this css is sus

    by ihatemath123, Aug 14, 2024, 1:53 AM

  • 391345 views moment

    by ryanbear, May 9, 2023, 6:10 AM

  • We need virgil nicula to return to aops, this blog is top 10 all time.

    by OlympusHero, Sep 14, 2022, 4:44 AM

  • :omighty: blog

    by tigerzhang, Aug 1, 2021, 12:02 AM

  • Amazing blog.

    by OlympusHero, May 13, 2021, 10:23 PM

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