Let $ABC$ be a triangle. A point $D$ lies on line $BC$ and points $E,F$ are taken on $AC,AB$ such that $DE \parallel AB$ and $DF\parallel AC$. Let $G = (AEF) \cap (ABC) \neq A$ and $I = (DEF) \cap BC\neq D$. Let $H$ and $O$ denote the orthocenter and the circumcenter of triangle $DEF$. Prove that $A,O,I$ are collinear if and only if $G,H,I$ are collinear. Proposed by Kaan Bilge
Problem
Source: Turkey Olympic Revenge 2023 P2
Tags: geometry, conditional geometry, Triangle Geometry, olympic revenge
v4913
08.03.2023 00:28
Note that $G$ is the center of the spiral similarity sending $FE$ to $BC$, and by angle chasing $\triangle{IEF} \sim \triangle{ACB}$. Thus, if $A'$ is the reflection of $A$ over $BC$ then this spiral similarity takes $I$ to $A'$ and $(AEF)$ to $(ABC)$. So, $I, O, A$ are collinear iff the predescribed spiral similarity also takes $A$ to the point $P$ where $P, A'$, and the circumcenter of $\triangle{A'BC}$ are collinear. Note that in order for $G$ to be the center of the spiral similarity taking $AI$ to $PA'$, we need $\angle{GAI} = \angle{GPA'} \implies AI \cap A'P \in (ABC)$, which is equivalent to $I = AJ \cap BC$ if $J = A'P \cap (ABC)$. Now note that the orthocenter of $\triangle{DEF}$ is the projection of $D$ onto $(AEF)$, and since $AEDF$ is a parallelogram this is also the $A$-antipode wrt $(AEF)$, so the $G, H, I$ collinearity is equivalent to $GI \perp GA$. Because $\triangle{GAP} \sim \triangle{GIA'}$, this is equivalent to $A'I \perp AP$, or $\angle{IAA'}$ equaling the angle between $AP, BC$. By angle chasing this is equivalent to the angle between $A'P, CH_A$ equaling $\angle{C}$ if $H_A = AA' \cap (ABC)$, but by reflection over $BC$, this is simply the angle between the $A$-diameter of $(ABC)$ and the perpendicular from $B$ to $AB$, which is clearly $\angle{C}$, so the two conditions are equivalent.
Cookierookie
15.03.2023 21:04
First of all notice by angle chasing that $H$ is on $(AEF)$, $\triangle IEF \cong \triangle ACB$ and that the collinearity of $I,H,G$ is equivalent to $\angle IGA = 90^\circ$.
Steiner's Ratio Lemma: Let $A,B,X,Y$ be points on a circle $\omega$. If $Z$ is the intersection of $AB$ and $XY$ then $\frac{ZA}{ZB}=\frac{XA}{XB}\cdot \frac{YA}{YB}$.
$\textbf{Claim:}$ Circles $(IGA)$ and $(IEF)$ are tangent.
$\emph{Proof:}$ By the radical axis theorem, it suffices to prove that the tangent to $(IEF)$ at $I$, $EF$ and $AG$ are concurrent. Let the tangent line to $(IEF)$ at $I$ and $AG$ meet $EF$ at $X$, $Y$ respectively. Applying Steiner's ratio lemma gives us: $$\frac{XE}{XF}=\left(\frac{IE}{IF}\right)^2=\left(\frac{AC}{AB}\right)^2 \text{ and } \ \frac{YE}{YF}=\frac{CE}{BF}\cdot \frac{AE}{AF}$$Also by the theorem of Thales,
$$\frac{AE}{AF}= \frac{\frac{DB}{BC}\cdot AC}{\frac{DC}{BC}\cdot AB}=\frac{DB}{DC}\cdot \frac{AC}{AB}$$$$\frac{CE}{BF}=\frac{\frac{CD}{CB}\cdot AC}{\frac{BD}{BC}\cdot AB}=\frac{CD}{BD}\cdot \frac{AC}{AB}$$Multiplying these two together gives the desired result. $\square$
Let the center of $(IGA)$ be $L$. Now, notice that $\angle IGA = 90^\circ$ is equivalent to the collinearity of $I,L,A$, but by the circle tangency $I,L,O$ are collinear, so the collinearity of $I,L,A$ is equivalent to the collinearity of $I,O,A$ which finishes the problem. $\blacksquare$
teomihai
15.03.2023 21:12
v4913 wrote: Note that $G$ is the center of the spiral similarity sending $FE$ to $BC$, and by angle chasing $\triangle{IEF} \sim \triangle{ACB}$. Thus, if $A'$ is the reflection of $A$ over $BC$ then this spiral similarity takes $I$ to $A'$ and $(AEF)$ to $(ABC)$. So, $I, O, A$ are collinear iff the predescribed spiral similarity also takes $A$ to the point $P$ where $P, A'$, and the circumcenter of $\triangle{A'BC}$ are collinear. Note that in order for $G$ to be the center of the spiral similarity taking $AI$ to $PA'$, we need $\angle{GAI} = \angle{GPA'} \implies AI \cap A'P \in (ABC)$, which is equivalent to $I = AJ \cap BC$ if $J = A'P \cap (ABC)$. Now note that the orthocenter of $\triangle{DEF}$ is the projection of $D$ onto $(AEF)$, and since $AEDF$ is a parallelogram this is also the $A$-antipode wrt $(AEF)$, so the $G, H, I$ collinearity is equivalent to $GI \perp GA$. Because $\triangle{GAP} \sim \triangle{GIA'}$, this is equivalent to $A'I \perp AP$, or $\angle{IAA'}$ equaling the angle between $AP, BC$. By angle chasing this is equivalent to the angle between $A'P, CH_A$ equaling $\angle{C}$ if $H_A = AA' \cap (ABC)$, but by reflection over $BC$, this is simply the angle between the $A$-diameter of $(ABC)$ and the perpendicular from $B$ to $AB$, which is clearly $\angle{C}$, so the two conditions are equivalent. more and this it is elegant!
helenagandalf
28.03.2023 21:01
could someone post the proof to the lemma used by Cookierookie please?? It seems really cool but I haven't seen it before and can't seem to find it anywhere...
a_n
28.03.2023 23:05
helenagandalf wrote: could someone post the proof to the lemma used by Cookierookie please?? It seems really cool but I haven't seen it before and can't seem to find it anywhere...
[url] https://artofproblemsolving.com/community/c6h2357938 [/url]
theorem 1.2 Its just law of sines though