Many Feuerbach points and similar quads

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Source: Israel 2020 Olympic Revenge

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#1 h Mar 22, 2021, 10:30 PM • 2 Y

Y by RohanQV, Mango247

Original post by shalomrav, but for some reason the mods locked the problem without any solves :noo:

:noo:

Let $ABCD$ be a cyclic quadrilateral inscribed in circle $\Omega$ . Let $F_A$ be the (associated with $\Omega$ ) Feuerbach point of the triangle formed by the tangents to $\Omega$ at $B,C,D$ , that is, the point of tangency of $\Omega$ and the nine-point circle of that triangle. Define $F_B, F_C, F_D$ similarly. Let $A'$ be the intersection of the tangents to $\Omega$ at $A$ and $F_A$ . Define $B', C', D'$ similarly.
Prove that quadrilaterals $ABCD$ and $A'B'C'D'$ are similar

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#2 h Mar 23, 2021, 6:11 PM

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how to bash this easily

This post has been edited 5 times. Last edited by shalomrav, Mar 24, 2021, 6:13 AM

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#3 h Mar 23, 2021, 6:40 PM • 2 Y

Y by amar_04, math_comb01

We use the following key lemma:
Lemma: Let $ABC$ have incenter $I$ , and let the incircle meet $BC$ at $D$ . Let the Feuerbach point be $Fe$ , the reflection of $D$ over $I$ be $D'$ , and the midpoint of $AI$ be $A'$ . Then $A',D',Fe$ are collinear.
Proof: Define $B',C',E',F'$ similarly. Note that $DB'C'\sim DE'F'$ from simple angle chasing, so by spiral similarity, $B'E',C'F'$ meet at the other intersection of $(DE'F')$ and $(DB'C')$ . But $(DE'F')$ is the incircle, and $(DB'C')$ is the nine point circle of $BIC$ . It is known that $Fe$ is the Poncelet point of $ABCI$ , which means it lies on the nine point circle of $BIC$ . Thus, $Fe$ is the intersection of $(DE'F')$ and $(DB'C')$ , so $B',E',Fe$ are collinear. By symmetry, $A',D',Fe$ and $C',F',Fe$ are collinear. $\square$

Note that the above proof also works if we replace the incircle with an excircle. Now we return to the original problem. Let the center of $\Omega$ be $O$ , let the tangents at $A$ and $B$ meet at $T_{AB}$ , and define similarly for the other pairs of points. Let $M_{AB}$ be the midpoint of $OT_{AB}$ , and define similarly for other pairs of points. Let the point diametrically opposite $A$ be $A'$ , and define $B',C',D'$ similarly. Applying our lemma to $T_{CD}T_{AC}T_{AD}$ , we get that $M_{CD},A',F_B$ are collinear. Similarly, $M_{CD},B',F_A$ are collinear. Now applying Pascal to $A'F_BBB'F_AA$ , we get $M_{CD},O,AF_A\cap BF_B$ are collinear. But $A'B'$ is the pole of $AF_A\cap BF_B$ wrt $\Omega$ , so $A'B'\perp OM_{CD}\perp CD$ . Thus, $A'B'\parallel CD$ . Similar statements hold for every other pair of sides. But then $\angle(A'B',B'C')=\angle(CD,AD)=\angle(BC,AB)$ , so $\angle A'B'C'=\angle ABC$ , and since similar identities hold elsewhere, we get $A'B'C'D'$ is similar to $ABCD$ . $\square$

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#5 h Mar 23, 2021, 8:40 PM • 1 Y

Y by Bumblebee60

$\textbf{LEMMA:}$ $ABC$ be a triangle with contact triangle $DEF$ , incenter $I$ and incircle $\odot(I)$ . Let $D^*$ be the $D-$ antipode in $\odot(I)$ . Let $K$ be an arbitary point on $\odot(I)$ and $K^*$ be the $K-$ antipode in $I$ . Let $Fe$ be the Feuerbach point of $\Delta ABC$ and $X$ be the midpoint of $AI$ . Let $XK^*\cap\odot(I)=T$ . Then $KFe\cap DT\in\overline{AI}$ .

Animate $K$ on $\odot(I)$ , the mappings $K\mapsto KFe$ and $K\mapsto K^*\mapsto T\mapsto DT$ are homographies between pencil of lines through $D$ and $Fe$ and $KFe\equiv DT$ when $K\equiv D$ . Thus by degenerate steiner conic theorem we have that $KFe\cap DT$ traces a line. From (Feuerbach point) we have that $XD^*,Fe$ are collinear. Now when $K\equiv D^*$ we have $KFe\cap DT=X$ and when $K\equiv IFe\cap\odot(I)$ we have $KFe\cap DT=I$ . Thus $KFe\cap DT$ traces out the line $AI$ .

Relabel the original problem as follows

$\textbf{RELABELLED PROBLEM:}$ $ABC$ be a triangle with contact triangle $DEF$ and $K\in\odot(I)$ . The tangent at $K$ meets $BC,CA,AB$ at $X,Y,Z$ . Let $\mathcal{F}_{\Delta AYZ}$ be the Feuerbach point of $\Delta AYZ$ . Similaly define $\mathcal{F}_{\Delta CXY}$ , $\mathcal{F}_{\Delta BXZ}$ , $\mathcal{F}_{\Delta ABC}$ . Let $KK\cap YZ=K^*$ . Similarly define $D^*,E^*,F^*$ . Then $\Delta DEF\cup K\stackrel{-}{\sim} \Delta D^*E^*F^*\cup K^*$ .

We now just restrict our attention to lines $DK$ and $E^*F^*$ . Let $P$ be the midpoint of $XI$ . Let $E',F'$ be the $E,F$ antipodes WRT $\odot(I)$ . From (Feuerbach Point) we have $P,E',\mathcal{F}_{\Delta CXY}$ and $P,F',\mathcal{F}_{\Delta BXZ}$ are collinear. Applying our $\textbf{LEMMA}$ we have that $F\mathcal{F}_{\Delta CXY},E\mathcal{F}_{\Delta BXZ},XI$ are concurrent at a point $\mathcal{P}_{E^*F^*}$ . Thus by La Hire we have that $E^*F^*$ is the polar of $\mathcal{P}_{E^*F^*}$ and hence we have $DK\parallel E^*F^*$ . Using the same procedure we can have the analogous parallel lines. Now by a little angle chasing we can get $\Delta DEF\stackrel{-}{\sim}\Delta D^*E^*F^*$ . Repeating the same procedure we get $\Delta EFK\stackrel{-}{\sim}\Delta E^*F^*K^*$ . Thus summing up all we get $\Delta DEF\cup K\stackrel{-}{\sim} \Delta D^*E^*F^*\cup K^*$ . $\quad\blacksquare$

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