117. Some applications of harmonic division/quadrilateral.

by Virgil Nicula, Sep 12, 2010, 10:25 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=2619

PP1.Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the side $BC$ at $A^{\prime}$, and the line $AA^{\prime}$ meets the incircle again at a point $P$ .
$CP$ and $BP$ meet the incircle of triangle $ABC$ again at $N$ and $M$, respectively. Prove that the lines $AA^{\prime}$, $BN$ and $CM$ are concurrent.

An equivalent enunciation. Let $\triangle ABC$ with circumcircle $w$ . Denote $A'\in BB\cap CC$ , $B'\in CC\cap AA$ , $C'\in AA\cap BB$ , i.e. $A'B'C'$
is tangential to $ABC$ and the second intersections $P$ , $M$ , $N$ of $w$ with $AA'$ , $B'P$ , $C'P$ respectively. Prove that $AA'\cap B'N\cap C'M\ne\emptyset$ .

This difficult and very nice problem is an instructive aplication of the harmonical quadrilateral and the its properties. I suggest to see the characterization of the harmonical quadrilateral in the my messages from this topic (<== click).

Proof. I"ll use only the properties of the harmonical quadrilateral $\mathrm{(\ h.q.\ )}$ Define the tangent $XX$ in the point $X\in w$ to the incircle $w\ .$ Denote the intersection $R\in PP\cap BC$ and the points $B'\in AC$ , $C'\in AB$ which belong to the incircle $w$ of the triangle $ABC\ .$ Recall the relations $A'B=BC'$ and $A'C=CB'\ .$

$1\blacktriangleright\ A\in C'C'\cap B'B'$ and $P\in AA'$ $\Longrightarrow$ $PB'A'C'\--\ \mathrm{h.q.}$ $\Longrightarrow$ $R\in B'C'$ and $\frac{PC'}{PB'}=\frac{A'C'}{A'B'}\ \ (1)\ .$

$2\blacktriangleright\left\{\begin{array}{c} B\in C'C'\cap A'A'\ ,\ M\in PB\Longrightarrow PC'MA'\--\ \mathrm{h.q.}\Longrightarrow T\in A'C'\ ,\ PC'\cdot MA'=PA'\cdot MC'=\frac{1}{2}\cdot PM\cdot A'C'\\\\ C\in BB'\cap A'A'\ ,\ N\in PC\Longrightarrow PB'NA'\--\ \mathrm{h.q.}\Longrightarrow S\in A'B'\ ,\ PB'\cdot NA'=PA'\cdot NB'=\frac{1}{2}\cdot PN\cdot A'B'\end{array}\right\|$

$\Longrightarrow$ $\frac{MP}{NP}=\frac{MC'}{A'C'}\cdot \frac{A'B'}{NB'}\ \ (2)\ .$

$3\blacktriangleright\left\{\begin{array}{c}\triangle CB'N\sim\triangle CPB'\Longrightarrow NC\cdot PB'=B'C\cdot B'N\\\\ \triangle BC'M\sim\triangle BPC'\Longrightarrow MB\cdot PC'=C'B\cdot C'M\end{array}\right\|$ $\Longrightarrow \frac{NC}{MB}=\frac{B'C}{C'B}\cdot\frac{B'N}{PB'}\cdot\frac{PC'}{C'M}$ and $(1)$ $\Longrightarrow$ $\frac{NC}{MB}=\frac{B'C}{C'B}\cdot\frac{B'N}{C'M}\cdot\frac{A'C'}{A'B'}\ \ (3)\ .$

Observe that $R\in B'C'\Longrightarrow$ $\frac{RB}{RC}=\frac{A'B}{A'C}\ \ (4)\ .$ From the product of the relations $(2)$ , $(3)$ , $(4)$ obtain

$\frac{RB}{RC}\cdot \frac{NC}{NP}\cdot\frac{MP}{MB}=1\ ,$ i.e. the points $R$ , $M$ , $N$ are collinearly $\Longrightarrow$ the lines $BN$ , $CM$ , $AA'$ are concurrently.


PP2.Let $ ABC$ be a triangle with $c>b$ . Its incircle touches side $ BC$ at point $ E$ . Point $ D$ is the second intersection of the incircle with

segment $ AE$ (different from $ E$) . Point $E\ne F\in $ is taken such that $ CE = CF$ . Let $ G\in [CF\cap BD$ . Show that $ CF = FG$ .

Proof. Assume w.l.o.g. that $ \angle {B} \leq \angle{C}$ (in this case $ F$ will be on the segment $ AE$ , as in the statement). Denote by $ Y$ , $ Z$ the tangency points of the incircle with the sides $ CA$ , and $ AB$ . Let $ T$ be the intersection point of the lines $ YZ$ and $ BC$ . Since $ DZEY$ is an harmonic quadrilateral (because of the concurrency of the tangects at $ Y$ , $ Z$ on the line $ DE$) , the line $ TD$ is tangent to the incircle at $ D$ and thus $ \angle{TED} = \angle{TDE}$. But $ \angle{TED} = \angle{TFE}$ and so the lines $ TD$ and $ CF$ are parallel. Now, since the quadruple $ (B, E, C, T)$ is harmonic, the pencil $ D(B, E, C, T)$ is harmonic and by intersecting it with the line $ CF$ we conclude that $ F$ is the midpoint of $ CG$ (according to the parallelism of $ DT$ and $ CF$) .


PP3. Let $ABC$ be a triangle and for an interior point $P$ denote $\left\{\begin{array}{c} D\in AP\cap BC\\\ E\in BP\cap CA\\\ F\in CP\cap AB\end{array}\right\|$ And $\left\{\begin{array}{c} M\in CF\cap DE\\\\ N\in AM\cap BC\end{array}\right\|$ . Prove that $\frac {NB}{NC}=2\cdot \frac {DB}{DC}$ .

Proof. Is well-known or prove easily that the division $(C,M,P,F)$ is harmonically. Thus, the division $(C,N,D,B)$ which is the intersection between the pencil

$A(C,M,P,F)$ and the sideline $BC$ is also harmonically. From the well-known property of the harmonical divisions obtain the required relation $\frac {NB}{NC}=2\cdot \frac {DB}{DC}$ .

Otherwise. Since $M\in AN\cap DE\cap CP$ in the triangle $ADC$ and $B\in PE\cap DC$ prove easily (Ceva & Menelaus) that the points $N$ and $B$ are harmonical conjugate w.r.t. the points $D$ and $C$ .
This post has been edited 18 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:45 AM

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12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by ryanbear, May 9, 2023, 6:10 AM

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

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

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    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

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