Let $ABC$ be a triangle and let $D$ be a point on $BC$ so that $AD$ bisects the angle $\angle BAC$. The common tangents of the circles $(BAD)$, $(CAD)$ meet at the point $A'$. The points $B'$, $C'$ are defined similarly. Show that $A'$, $B'$, $C'$ are collinear.
Problem
Source: 2024 Israel TST Test 3 P1
Tags: geometry, Tangents, Angle Bisectors
AndreiVila
30.01.2024 00:55
If $M_a$ is the midpoint of arc $BAC$, the circle $\omega_a$ of radius $M_aB$ centered in $M_a$ is tangent to both $(ABD)$ and $(ACD)$, implying by Monge that $A'\in BC$. Furthermore, $\frac{A'B}{A'D}=\frac{A'D}{A'C}=\frac{R_{(ABD)}}{R_{(ACD)}}=\frac{BD}{CD}$, thus $\frac{A'B}{A'C}=\frac{DB^2}{DC^2}$, finishing with Menelaus. Interestingly enough, the last equality gives that $A'$ is the intersection of the tangent in $A$ to $(ABC)$ with $BC$, however I couldn't find a satisfactory (synthetic) argument...
IAmTheHazard
30.01.2024 03:27
Apply $\sqrt{bc}$ inversion. Let $E$ be the midpoint of arc $BC$, the image of $D$. $(BAD)$ and $(CAD)$ get sent to $\overline{EB}$ and $\overline{EC}$, and the common tangents get sent to two circles through $A$ tangent to these lines. By symmetry, their second intersection point (the image of $A'$) will simply be the reflection of $A$ over the perpendicular bisector of $\overline{BC}$, so $A'=\overline{AA} \cap \overline{BC}$ Now let $T=\overline{BB} \cap \overline{CC}$. observe that $A'$ is the pole of $\overline{AT}$ ($A$ lying on the polar is obvious, $T$ lies on it by La Hire since it's the pole of $\overline{BC}$). Thus by pole-polar duality this problem is in fact equivalent to the existence of the symmedian point.
khina
30.01.2024 04:55
We claim that $A'$ is the intersection of $AA \cap BC$ (where $AA$ is the tangent to $(ABC)$ at $A$). Indeed observe inversion about $(A', A'A)$ fixes $A$ and $D$, but swaps $B$ and $C$. The claim follows readily (by i.e. Pascal on $AABBCC$).
math_comb01
01.02.2024 18:58
Lol, it is well known by appolonian circle that if $K = AA \cap BC$ then $KA = KD$ $\therefore K$ is the exsimilicenter of the two circles, by pascal on $AABBCC$ we're done.
hellnish
14.03.2024 15:07
Pascal theorem.
Ianis
25.05.2024 20:21
If $A_1$ is the centre of the $A$-Apollonian circle then $\frac{A_1B}{A_1D}=\frac{A_1D}{A_1C}$, so $A_1$ is the centre of the homothety mapping $(BAD)$ to $(CAD)$, because $A_1$ and the centres of both circles lie on the perpendicular bisector of $AD$. It follows that $A_1=A'$. Hence $A'$, $B'$ and $C'$ lie on the perpendicular bisector of the isodynamic points of $ABC$. Basically the same configuration as Ibero 2004 P5.