In a triangle $ABC$, $I$ is the incenter. $D$ is the reflection of $A$ to $I$. the incircle is tangent to $BC$ at point $E$. $DE$ cuts $IG$ at $P$ ($G$ is centroid). $M$ is the midpoint of $BC$. prove that a) $AP||DM$.(15 points) b) $AP=2DM$. (10 points)
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Tags: geometry, geometric transformation, reflection, incenter, parallelogram, geometry proposed
Luis González
07.08.2010 18:29
Let $\triangle A_0B_0C_0$ be the antimedial triangle of $\triangle ABC.$ Vertices $A_0,B_0,C_0$ against $A,B,C,$ respectively. Nagel point $N_a$ of $\triangle ABC$ is the incenter of $\triangle A_0B_0C_0,$ i.e. $N_a$ lies on $IG$ such that $\overline{GN_a}:\overline{GI}=-2:1.$ Thus, Nagel point $P_0$ of $\triangle A_0B_0C_0$ is the reflection of $N_a$ about $I.$ Nagel point $D'$ of $\triangle A_0BC,$ lying on its $A_0$-Nagel ray $A_0E,$ is then the reflection of $N_a$ about $M,$ since $\triangle ABC$ and $\triangle A_0CB$ are symmetric about $M.$ Because of $A_0D'=D'P_0=AN_a$ and $AN_a \parallel A_0P_0,$ it follows that $AN_aD'P_0$ is a parallelogram $\Longrightarrow$ $D'$ is the reflection of $A$ about $I.$ Consequently, $D \equiv D'$ and $P \equiv P_0.$ From the fact that $AN_aDP$ is a parallelogram, we deduce that $AP \parallel DM$ and $AP=DN_a=2DM.$
ArefS
07.08.2010 19:04
Dear Luis please hide your solutions it is really confusing for me to see the solution before I think on the problem. Best Regards Aref Sadeghi
sunken rock
07.08.2010 21:02
Here the key problem is to prove that $DE\parallel IM$, which is easily provable by metrical means. Best regards sunken rock
vladimir92
08.08.2010 18:54
sunken rock wrote: Here the key problem is to prove that $DE\parallel IM$, which is easily provable by metrical means. Best regards sunken rock If I'm not wrong, I followed the same way as yours. But I think $DE\parallel IM$ isn't all the problem.
W.L.O.G. assume that $AC>AB$. let $AI$ meet $BC$ at $K$. Put $BC=a$, $CA=b$ , $AB=c$ , $BK=x$ and $CK=y$. Then $y=a-x$ but $\frac{x}{y}=\frac{c}{b}=\frac{x}{a-x}$ then $x=\frac{ac}{b+c} \Leftrightarrow \frac{b-c}{a-2x}=\frac{c}{x}$ call this $(*)$. We have $MK=\frac{a}{2}-x$ and $ME=\frac{BC}{2}-BE=\frac{b-c}{2}$, then $(*)$ can be rewritten as: $\frac{ME}{MK}=\frac{BA}{BK}=\frac{IA}{IK} \Longrightarrow \frac{KM}{KE}=\frac{KI}{KD}$ we deduct then that $(MI)\parallel (DE)$.
Now let the parallel to $(MI)$ through $A$ meet $MD$ at $R$ and denote $G_1\equiv(RI)\cap (AM)$. From thales theorem, we have $\frac{G_1M}{G_1A}=\frac{DI}{DA}=\frac{1}{2}$ then $G_1$ trisect $AM$ therefor $G_1=G$. We have $P\in IR$ and $(AR)\parallel (MI) \parallel (DP)$, thus $\frac{IR}{IP}=\frac{IA}{ID}=1$ then $D$ is the reflexion of $P$ across $I$, meaning that $ARDP$ is parallelogram, from which follow that $(MD) \parallel (AP)$.
For the second question, we have $AP=RD$ then $\frac{DM}{AP}=\frac{DM}{DR}=\frac{DI}{DA}=\frac{1}{2}$ we deduce directly that $AP=2DM$.
sunken rock
08.08.2010 23:11
Well, if $\{R\}\equiv IG \cap DM$, by Menelaus in $\triangle ADM$ with the transversal $IGR$ we get $DM=MR \; (\; * \; )$, hence $IM\parallel AR$ and $AR=2\cdot AR$, consequently, once $DE\parallel IM$, with $(*)$ we get $DP=2\cdot IM$, i.e. $AR=DP$ and $AR\parallel PD$, so $APDR$ is parallelogram. Now, to prove that $APDR$ is parallelogram: if $J$ is the reflection of $E$ in $M$, it is the contact of the A-excircle with $BC$, hence the point $K$, diametrically opposite to $E$ in the incircle belongs to $AJ$. From here $IM$ is the midline of the triangle $\triangle EJK$, i.e. $A, K, R, J$ are collinear and, with $IM$ midline in triangles $\triangle ADR$ and $\triangle KEJ$, both of them having their base on $AJ$, we get $DE\parallel AJ$, consequently $APDR$ IS a parallelogram, done. Best regards, sunken rock
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Virgil Nicula
09.08.2010 01:32
Quote: Let $ABC$ be a triangle with incircle $w=C(I)$ and centroid $G$ . Denote $D\in BC\cap w$ , $M\in AG\cap BC$ , reflection $S$ of $A$ w.r.t. $I$ and $P\in SD\cap GI$ . Prove that $AP\parallel MS$ and $AP=2\cdot MS$ . Proof. Denote diameter $[DE]$ of $w$ , Nagel's point $N$ , $D'\in BC\cap AN$ and $S'\in AI\cap NM$ . Since $AG=2\cdot GM$ and $NG=2\cdot GI$ (well-known property) obtain that $G$ is centroid of $\triangle ANS'$ and $MS'=MN$ , $IS'=IS$. In conclusion, $S\in MN$ and $MS=MN$ . From another properties $MD=MD'$ , $E\in AN\cap w$ , obtain that $IM\parallel AN$ . Therefore, $DS\parallel IM\parallel AN$ . In conclusion, $SP\parallel AN$ and $IA=IS$ $\implies$ $APSN$ is parallelogram $\implies$ $AP\parallel MS$ and $AP=2\cdot MS$ .