Through point $ P $ inside triangle $ ABC $, straight lines were drawn, parallel to the sides, until they intersect with the sides. In the three resulting parallelograms, diagonals that do not contain point $ P $, are drawn. Points $ A_1 $, $ B_1 $ and $ C_1 $ are the intersection points of the lines containing these diagonals such that $A_1$ and $A$ are in different sides of line $BC$ and $B_1$ and $C_1$ are similar. Prove that if hexagon $ AC_1BA_1CB_1 $ is inscribed and convex, then point $ P $ is the orthocenter of triangle $ A_1B_1C_1 $.
Problem
Source: 2016 239 S5
Tags: geometry
TheBarioBario
13.12.2020 17:52
First a few famous lemmas: Lemma 1: in triangle $ABC$, point $D$ lies on $BC$. Then : $$\frac{BD}{CD}=\frac{AB}{AC} \cdot \frac{\sin{\widehat{BAD}}}{\sin{\widehat{CAD}}}$$ Lemma 2: given triangle $ABC$ and point $D$ we have: $$\frac{\sin{\widehat{BAD}}}{\sin{\widehat{CAD}}}= \frac{\sin{\widehat{ABD}}}{\sin{\widehat{ACD}}} \cdot \frac{BD}{CD}$$ Lemma 3: if $\alpha +\beta=x+y \neq 180$ and $\frac{\sin{\alpha}}{\sin{\beta}}=\frac{\sin{x}}{\sin{y}}$ then $\alpha=x$ and $\beta=y$ (For proof of 1&2 double count some areas and for 3 use lemma1) We return to the problem at hand: For the naming, start from $B$ on the triangle and walk counterclockwise on the triangle, the six points are respectively $M,N,L,K,Q,R$. 1) regardless of the “inscribedness” of the hexagon we prove that $APA_1$ are collinear. Proof: use lemma 1 in $ARL$ for $P$ and use lemma 2 in $ARL$ for $A_1$ Combined with lemma 3, it is enough to show: $$\frac{RP}{LP} \cdot \frac{AL}{AR} =\frac{RA_1}{LA_1} \cdot \frac{\sin{\widehat{ARA_1}}}{\sin{\widehat{ALA_1}}}$$which in fact(using thales and the law of sines): $$\frac{RA_1}{LA_1} \cdot \frac{\sin{\widehat{ARA_1}}}{\sin{\widehat{ALA_1}}}=\frac{MA_1}{NA_1}\cdot \frac{\sin{\widehat{BRM}}}{\sin{\widehat{CLN}}}= \frac{\sin{\widehat{MNA_1}}}{\sin{\widehat{NMA_1}}}\cdot \frac{\sin{\widehat{BRM}}}{\sin{\widehat{CLN}}}=\frac{BM}{BR} \cdot \frac{CL}{CN}= \frac{RP}{LP} \cdot \frac{AL}{AR}$$ 2) now we know that $AA_1, BB_1, CC_1$ are concurrent in $P$, call the intersection of $AA_1, B_1 C_1$ as $X$ (similarly for $Y,Z$). We know that $AQPK$, $BMPR$, $CLPN$ are parallelograms so $X,Y,Z$ are the middles of $PA, PB, PC$. so $\widehat{YXA_1}=\widehat{YXP}= \widehat{BAP}= \widehat{BB_1 A_1}= \widehat{YB_1 A_1}$ thus $XYA_1 B_1$ is cyclic similarly for $YZB_1 C_1$ and $XZA_1 C_1$. this gives us that $P$ is the orthocenter of $A_1 B_1 C_1$. this concludes the proof. Bonus: Another fun thing that it gives is that $P$ is the incenter of $ABC$