72. Colinearity in a triangle - H , P , M .

by Virgil Nicula, Aug 1, 2010, 3:34 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=358736
Quote:
Let $ABC$ be a triangle with orthocenter $H$ and orthic triangle $IJK$ , where $I\in BC$ , $J\in CA$ , $K\in AB$ . The bisector of $\widehat {BHK}$ meet $AB$ , $AC$ at $D$ , $ E$ respectively. The bisector of $\widehat {BAC}$ meet circumcircle of $ADE$ at point $P$. Let $M$ be midpoint of $[BC]$ . Prove that $P\in HM$ .

Lemma. In any triangle $ABC$ there is a nice identity :

$\boxed {\ (b+c)^2\cos A+a^2\cos B\cos C=bc(1+\cos A)^2\ }\ (*)$ .

Proof. $(b+c)^2\cos A+a^2\cos B\cos C=bc(1+\cos A)^2\ \Longleftrightarrow$

$\left(b^2+c^2\right)\cos A+(a\cdot\cos B)(a\cdot \cos C)=bc\left(1+\cos^2A\right)\ \Longleftrightarrow$

$\left(b^2+c^2\right)\cos A+(c-b\cdot\cos A)(b-c\cdot \cos A)=bc\left(1+\cos^2A\right)\ \ \mathrm{O.K.}$

A synthetical proof.

A metrical proof. Denote $F\in AI\cap BC$ , where $I$ is incenter of $\triangle ABC$ . Prove easily that $AF\perp DE$ . Apply theorem of Sines in $\triangle AHE$ : $\frac {AE}{\sin\left(C+\frac A2\right)}=\frac {AH}{\cos\frac A2}$ $\Longleftrightarrow$ $AE=\frac {2R\cos A\sin\left(C+\frac A2\right)}{\cos\frac A2}=$ $\frac {4R\cos A\sin\left(C+\frac A2\right)\cos\frac A2}{2\cos^2\frac A2}=$ $\frac {2R\cos A(\sin B+\sin C)}{2\cos^2\frac A2}$ $\implies$ $AE=\frac {(b+c)\cos A}{2\cos^2\frac A2}$ $\implies$ $AP=\frac {AE}{\cos\frac A2}$ $\implies$ $AP=\frac {(b+c)\cos A}{2\cos^3\frac A2}$ . Since $AF=\frac {2bc\cos\frac A2}{b+c}$ obtain $\frac {AP}{AF}=\frac {(b+c)\cos A}{2\cos^3\frac A2}\cdot\frac {b+c}{2bc\cos\frac A2}=$ $\frac {(b+c)^2\cos A}{4bc\cos^4\frac A2}=$ $\frac {(b+c)^2\cos A}{bc(1+\cos A)^2}$ $\implies$ $\frac {AP}{(b+c)^2\cos A}=$ $\frac {AF}{bc(1+\cos A)^2}=$ $\frac {PF}{bc(1+\cos A)^2-(b+c)^2\cos A}\stackrel{(*)}{=}\frac {PF}{a^2\cos B\cos C}$ . In conclusion, $\boxed {\frac {PA}{PF}=\frac {(b+c)^2\cos A}{a^2\cos B\cos C}}\ \ (3)$ . Apply Menelaos' theorem to points $H\in AI$ , $M\in IF$ , $P\in FA\ \ :\ \ \frac {MF}{MI}\cdot\frac{HI}{HA}\cdot\frac {PA}{PF}=$ $\frac {\frac {a(b-c)}{2(b+c)}}{\frac {b^2-c^2}{2a}}\cdot\frac {\cos B\cos C}{\cos A}\cdot\frac {(b+c)^2\cos A}{a^2\cos B\cos C}=1$ , i.e. $P\in HM$ .
This post has been edited 8 times. Last edited by Virgil Nicula, Nov 23, 2015, 3:43 PM

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