Let $O, I$ be the circumcenter and incenter of triangle $ABC$. The incircle of $\triangle ABC$ touches $BC, CA, AB$ at points $D, E, F$ repsectively. $FD$ meets $CA$ at $P$, $ED$ meets $AB$ at $Q$. $M$ and $N$ are midpoints of $PE$ and $QF$ respectively. Show that $OI \perp MN$.
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Tags: geometry, circumcircle, incenter, trigonometry, power of a point, radical axis, geometry unsolved
oneplusone
28.11.2010 06:13
Note that $AFBQ$ is harmonic, and $N$ is the midpoint of $QF$, so $NF^2=NB\cdot NA$. Similarly $ME^2=MC\cdot MA$. Thus $NI^2-MI^2=NF^2-ME^2=NB\cdot NA-MC\cdot MA=NO^2-MO^2$, thus $OI\perp MN$.
Goutham
28.11.2010 06:22
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=185866&start=0
jgnr
28.11.2010 06:31
oneplusone wrote: Note that $AFBQ$ is harmonic, and $N$ is the midpoint of $QF$, so $NF^2=NB\cdot NA$. This also implies that $N$ lies on the radical axis of the circumcircle and the incircle of $ABC$.
sayantanchakraborty
04.04.2014 11:13
I am just writing the outlines of my steps...(I have done all calculations in my copy and verified).The steps are not so complicated as they seem to be. Let us assume that $b\ge a\ge c$.Then $Q \epsilon [AB)$ and $P \epsilon [CA)$. Apply sine rule in $\triangle{BQD}$ to get $BQ$ and hence $FQ=BQ+(s-b)$.In a similar manner deduce the length of $PE$.Then calculate $BN=FN-FB=\frac{FQ}{2}-(s-b)$.Analogously get $AM$.Note that $OI \perp MN \Leftrightarrow OM^2-ON^2=IM^2-IN^2$Use the cosine rule in $\triangle OBN$ we get $ON^2=R^2+BN^2+2R*BNsinC$.Similarly we get $OM^2=R^2+AM^2+2R*AMsinB$.Thus $OM^2-ON^2=AM^2-BN^2+cBN-bAM$.Again applying the cosine rule in $\triangle{IAM}$ we get $IM^2=IA^2+AM^2+2IA*AMcos\frac{A}{2}$.Similarly $IN^2=IB^2+BN^2+2IB*BNcos\frac{B}{2}$Thus $IM^2-IN^2=(s-a)^2-(s-b)^2+AM^2-BN^2+(b+c-a)AM-(a+c-b)BN$.Now its just nothing but calculations...
jayme
04.04.2014 13:56
Dear Mathlinkers, see also http://perso.orange.fr/jl.ayme vol. 13 Un cercle passant par Fe Sincerely Jean-Louis
navredras
27.06.2015 13:20
oneplusone wrote: Note that $AFBQ$ is harmonic, and $N$ is the midpoint of $QF$, so $NF^2=NB\cdot NA$. Similarly $ME^2=MC\cdot MA$. Thus $NI^2-MI^2=NF^2-ME^2=NB\cdot NA-MC\cdot MA=NO^2-MO^2$, thus $OI\perp MN$. Why is $ NF^2=NB*NA $?
ManuelKahayon
16.03.2018 05:34
'Tis a property of four harmonic points. To see this, you can draw a circle with center N and radius \(NF\) and note that B and A are conjugated with each other wrt to the circle (and are thus inverses of each other with respect to the circle, because the line which contains them also contains the center of this circle).
WizardMath
16.03.2018 07:56
See the proof of Lemma 3.16 here.
AlastorMoody
13.04.2020 04:22
China MO 2007 P4 wrote:
Let $O, I$ be the circumcenter and incenter of triangle $ABC$. The incircle of $\triangle ABC$ touches $BC, CA, AB$ at points $D, E, F$ repsectively. $FD$ meets $CA$ at $P$, $ED$ meets $AB$ at $Q$. $M$ and $N$ are midpoints of $PE$ and $QF$ respectively. Show that $OI \perp MN$.
Since, $AD$ is the $D-$symmedian WRT $\Delta DFE$, thus projecting the harmonic bundle from $(I)$ to $AC$, $AB$ yields
\begin{align*} -1 = (C, A ;E,P) \overset{D}{=} (B, A ; Q ,F) \end{align*}Thus, $A,C$ are Inverse Points under $\odot (M,MP)$ and Similarly, $A,B$ are Inverse Points under $\odot (N,NQ)$. Thus, $ME^2$ $=$ $MA \cdot MC$ and $NF^2$ $=$ $NB \cdot NA$. Hence, $M,N$ have Equal Powers WRT $(I)$ and $(O)$ $\implies$ $\overline{MN}$ is the Radical Axis of $(I)$ and $(O)$ $\implies$ $MN$ $\perp$ $OI$ $\qquad \blacksquare$