Let $ABC$ be a triangle, and let $D, E$ and $F$ be the feet of the altitudes from $A, B$ and $C,$ respectively. A circle $\omega_A$ through $B$ and $C$ crosses the line $EF$ at $X$ and $X'$. Similarly, a circle $\omega_B$ through $C$ and $A$ crosses the line $FD$ at $Y$ and $Y',$ and a circle $\omega_C$ through $A$ and $B$ crosses the line $DE$ at $Z$ and $Z'$. Prove that $X, Y$ and $Z$ are collinear if and only if $X', Y'$ and $Z'$ are collinear. Vlad Robu
Problem
Source: Stars of Mathematics 2020 (senior level)
Tags: geometry, romania
sarjinius
13.12.2021 14:21
Note that $BCEF, CAFD,$ and $ABDE$ are all cyclic due to right angles. Let $\omega_A$ intersect lines $AB$ and $AC$ at points $P$ and $Q$ respectively. Note that $\angle APQ = \angle ACB = \angle AFE$, so $PQ \parallel EF$. From Power of a Point, $BF \cdot FP = XF \cdot X’F$ and $CE \cdot EQ = XE \cdot X’E$, so $\frac{XE \cdot X’E}{XF \cdot X’F} = \frac{CE \cdot EQ}{BF \cdot FP} = \frac{CE}{BF} \cdot \frac{EQ}{FP} = \frac{CE}{BF} \cdot \frac{AE}{AF}$. Similarly, we have $\frac{YF \cdot Y’F}{YD \cdot Y’D} = \frac{AF}{CD} \cdot \frac{BF}{BD}$ and $\frac{ZD \cdot Z’D}{ZE \cdot Z’E} = \frac{BD}{AE} \cdot \frac{CD}{CE}$. Multiplying the three equations, we get $\frac{XE \cdot X’E}{XF \cdot X’F} \cdot \frac{YF \cdot Y’F}{YD \cdot Y’D} \cdot \frac{ZD \cdot Z’D}{ZE \cdot Z’E} = \frac{CE}{BF} \cdot \frac{AE}{AF} \cdot \frac{AF}{CD} \cdot \frac{BF}{BD} \cdot \frac{BD}{AE} \cdot \frac{CD}{CE} = 1 \iff \left(\frac{XE}{XF} \cdot \frac{YF}{YD} \cdot \frac{ZD}{ZE}\right) \cdot \left(\frac{X’E}{X’F} \cdot \frac{Y’F}{Y’D} \cdot \frac{Z’D}{Z’E}\right) = 1$. Thus, $\frac{XE}{XF} \cdot \frac{YF}{YD} \cdot \frac{ZD}{ZE} = 1$ if and only if $\frac{X’E}{X’F} \cdot \frac{Y’F}{Y’D} \cdot \frac{Z’D}{Z’E} = 1$. By Menelaus’s theorem, $X, Y,$ and $Z$ are collinear if and only if $X’, Y’,$ and $Z’$ are collinear.
Thelink_20
20.11.2024 05:20
(I'm assuming $X$ is between $E$ and $X'$ but the other cases are the same, just the orientations that might change.)
Lemma: $\angle FBX=\angle ECX'$ and $\angle EBX=\angle FCX'$ Proof:$ \ \angle FBX=\angle ABC- \angle XBC=180^{\circ}-(180^{\circ}-\angle ABC)-\angle CX'E=\angle ECX'$ and $\angle FBE=\angle ECF\Rightarrow\angle EBX=\angle FCX'$. $_{\blacksquare}$ $\boxed{\lambda =\frac{\sin\angle FBX}{\sin\angle EBX}}\implies\frac{XF}{XE}=\lambda\cdot\frac{BF}{BE} \ \ $ and $\ \ \frac{X'E}{X'F}=\lambda\cdot\frac{CE}{CF}=\frac{XF}{XE}\cdot\frac{BE}{BF}\frac{CE}{CF}$. $X'-Y'-Z'\iff \frac{X'E}{X'F}\cdot\frac{Y'F}{Y'D}\cdot\frac{Z'D}{Z'E}=1\iff \frac{XF}{XE}\cdot\frac{YD}{YF}\cdot\frac{ZE}{ZD}\cdot\bigg{(}\cancel{\frac{BE}{BF}\frac{CE}{CF}\frac{CF}{CD}\frac{AF}{AD}\frac{AD}{AE}\frac{BD}{BE}}\bigg{)}=1\iff X-Y-Z .\ _{\blacksquare}$