70. Colinearity with pole/polar - H\in PQ .

by Virgil Nicula, Jul 30, 2010, 2:56 AM

P1. Let $\triangle{ABC}$ be a triangle with orthocenter $H$ . The tangent lines from $A$ to the circle $w_a$ with diameter $[BC]$ touch it on $P$ and $Q$. Prove that $H\in PQ$ .

Proof 1 (power of point - Sunken rock). Denote $DEF$ - orthic triangle of $\triangle ABC$ , where $D\in BA$ , $E\in CA$ , $F\in AB$ and power $p_w(X)$ of $X$ w.r.t. circle $w$ . Observe
that $P$ , $D$ , $Q$ belong to the circle $\rho_a$ with diameter $[AM]$ and $PQ$ is radical axis of $w_a$ , $\rho_a$ . From $p_{w_a}(H)=$ $\overline{HB}\cdot \overline{HE}=$ $\overline{HA}\cdot \overline{HD}=$ $p_{\rho_a}(H)$ obtain that $H\in PQ$ .

Remark. From the pairs of circles $\{w_b, \rho_b\}$ , $\{w_c, \rho_c\}$ denote analogously the pairs of tangent points $\{R,S\}$ , $\{T,U\}$ of the tangents from $B$ , $C$ to $w_b$ , $w_c$ respectively. Thus $HA\cdot HD=HP\cdot HQ$ , $HB\cdot HE=HR\cdot HS$ , $HC\cdot HF=HT\cdot HU$ and $HA\cdot HD=$ $HB\cdot HE=$ $HC\cdot HF=$ $4R^2\cos A\cos B\cos C$ . Hence $HP\cdot HQ=$ $HR\cdot HS=$ $HT\cdot HU$ , i.e. the ponts $P$ , $Q$ , $R$ , $S$ , $T$ , $U$ belong to same circle. $H$ is radical center for $\left\{w_a,w_b,w_c\right\}$ and also for $\left\{w_{\rho_a},w_{\rho_b},w_{\rho_c}\right\}$ . If denote $L\in EF\cap BC$ , then $L\in PQ$ and $LH\perp AM$ , where $M$ is midpoint of $[BC]$ , what is another well-known remarkable property.

Proof 2 (pole/polar). Let the circumcircle $(O)$ of $\triangle ABC$ , $\{A,D_1\}=AH\cap (O)$ , $D\in BC$ for which $AD\perp BC$ and $\{U,V\}=AH\cap w_a$ . Thus, $DU^2=DB\cdot DC=DA\cdot DD_1=DA\cdot DH$ , i.e. $\boxed {\ DU^2=DA\cdot DH\ }$ what is a characterization of $(A,H,U,V)$ - harmonical division, i.e. $H$ is harmonical conjugate of $A$ w.r.t. $\{U,V\}$ . Hence $H$ belongs to polar of $A$ w.r.t. $(O)$ , i.e. $H\in PQ$ .

Proof 3 (metric). Suppose that $AB$ separates $P$ , $C$ . Denote $E\in AC$ so that $BE\perp AC$ and $F\in AB$ so that $CF\perp AB$ . Observe that $\frac {PF}{PB}=\sqrt {\frac {AF}{AB}}$ and $\frac {QE}{QC}=\sqrt {\frac {AE}{AC}}$ . Thus $\frac {PF}{PB}\cdot\frac {QE}{QC}=|\cos A|=\frac {EF}{BC}$ , i.e. $\frac {PF}{PB}\cdot\frac {QE}{QC}=\frac {EF}{BC}$ $\Longleftrightarrow$ $BP\cdot FE\cdot QC=PF\cdot EQ\cdot CB$ $\stackrel{(*)}{\Longleftrightarrow}$ $BE$ , $PQ$ , $FC$ are concurrently $\Longleftrightarrow$ $H\in PQ$ .

Lemma. Let $ABCDEF$ be a cyclic hexagon. Prove that $\boxed {\ AD\cap BE\cap CF\ne\emptyset \Longleftrightarrow AB\cdot CD\cdot EF=BC\cdot DE\cdot FA\ }\ (*)$ .

Proof 4 (inversion). Let $M$ be midpoint of $BC$ . Observe that $P$ , $Q$ , $D$ belong to the circle $\rho_a$ with diameter $AM$ . Apply an inversion with center at $A$ and radius $AP$ . Because $AP^2=AH\cdot AD$ , we obtain that $H$ is the image of $D$ under this inversion, and $P$ , $Q$ remains invariant. For the other hand, $\rho$ is transformed into the line containing $P$ , $Q$ . Because $D\in \rho_a$ we can conclude that $H\in PQ$ .



P2. Let $I$ be the incenter of the non-isosceles triangle $ABC$ and let $A',B',C'$ be the tangency points of the incircle with the sides $BC,CA,AB$ respectively.
The lines $AA'$ and $BB'$ intersect in $P$, the lines $AC$ and $A'C'$ in $M$ and the lines $B'C'$ and $BC$ intersect in $N$. Prove that the lines $IP$ and $MN$ are perpendicular.


Proof Denote $X\in AA'\cap IN,\ Y\in BB'\cap IM$. From the wellknown relations $IN\perp AA',$ $IM\perp BB'$ results $IX\cdot IN=IY\cdot IM=r^2$,
i.e. the line $MN$ is the image of the circumcircle with the diameter $[IP]$ through the inversion with the pole I and the constant $r^2$. Therefore, $IP\perp MN.$
This post has been edited 26 times. Last edited by Virgil Nicula, May 16, 2017, 8:38 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: 404395
  • Total comments: 37
Search Blog
a