Feet to the sides from the A-foot

Today we shall have a look at a simple configuration which is moderately prevalent in modern Olympiads but has not received much attention. Our work shall involve the feet of the perpendiculars from the foot of the $A-$ altitude to the sides $AB$ and $AC$ .

As usual, notation is borrowed from the diagram attached above, with most familiar points carrying their standard labels.

A summary of key results pertaining to this configuration,

First, an easy angle chase yields $BCC_1B_1$ cyclic. Hence, by Reim’s it also follows that $B_1C_1 \parallel EF$ . Moreover, this means $AO \perp B_1C_1$ as well. Also, it is clear that $AB_1DC_1$ is cyclic due to the right angles.
Let and be the feet of the altitudes from and to line respectively. Clearly, and are both cyclic.
- Since $H$ is the incenter of $\triangle DEF$ , it then follows that $B_1D=B_1P$ . Similarly, $C_1D=C_1P$ which implies that $P$ is the reflection of $D$ across line $\overline{B_1C_1}$ .
- As noted before, $EF \parallel B_1C_1$ so it follows that $\overline{B_1C_1}$ bisects segment $DX_a$ .
- Further, the reflection of $A_1$ across $B_1C_1$ must lie on the line through $D$ parallel to $EF$ and hence, it lies on the circle with diameter $AD$ . Thus, $A_1$ is the orthocenter of $\triangle AB_1C_1$ .
Next, consider the inversion centered at with radius . This swaps with and with . Hence, and as well as are fixed under this inversion.
- This implies that $AX=AD=AY$ and thus $A$ is the center of $(XYD)$ . Moreover, since $AD \perp BC$ this circle is tangent to $BC$ at $D$ .
- By the Incenter/Excenter lemma, we then have that $D$ is the incenter of $\triangle XYH_a$ which proves 2023 Azerbaijan JBMO TST Problem 3.
Note that since is concyclic, it is clear that the isogonal conjugate of with respect to lies on the perpendicular bisector of . In particular, this claim generalizes to any antiparallel to .
- It is not hard to see that the isogonal conjugate of the $A-$ antipode $A'$ with respect to $\triangle ABC$ is the point at infinity along $\overline{AD}$ . Hence, $R$ lies on the rectangular circumhyperbola $\mathcal{H}$ of $\triangle ABC$ passing through $A'$ .
- Pascal’s Theorem on conconic hexagon $AA'BRHC$ then implies that $AA' \cap RH$ lies on the line through $C_1$ perpendicular to $AB$ which by our previous observation that $A_1$ is the orthocenter of $\triangle AB_1C_1$ implies that points $R, H$ and $A_1$ are collinear, which is a key claim in several approaches to 2016 USA TSTST Problem 6.
Since is cyclic it is also clear that points and are equidistant from .
- Thus $AO$ is the perpendicular bisector of segment $B'C'$ which implies that quadrilateral $AB'A'C'$ is a kite, and hence harmonic.
Applying Pascal’s Theorem once more on conconic points implies that lies on the line through perpendicular to . Hence, points and are collinear.
- Hence, projecting $(AA';B'C')$ through $R$ gives us $(B,C;T,S)=-1$ and a further projection through $A'$ shows that points $A',T$ and $X_a$ are collinear.
- This immediately shows that points $A,A_1,D,T$ and $X_a$ are concyclic due to the right angles.
Furthermore, the feet of the altitudes from to the sides and must be the reflections of and across and respectively.
- Hence by the Mean Geometry theorem it follows that $E'F' \parallel B_1C_1$ as well, so by Reim’s theorem $BCF'E'$ must also be cyclic.
- Furthermore, since the line through $H$ parallel to $EF$ is the reflection of the line through $H_a$ parallel to $EF$ through line $E'F'$ it follows that the orthocenter of $\triangle AE'F'$ is the intersection of lines $AA'$ and the line through $H$ parallel to $EF$ .
- Returning to a previous claim, this implies that $Z = BF' \cap CE'$ lies on $\mathcal{H}$ . Hence, by Pascal’s Theorem on coconic hexagon $AA'BZHC$ it follows that $HZ \cap AA'$ lies on the line through $F'$ perpendicular to $AB$ .
- Combining this with the previous observation, this implies that $ZH$ is parallel to $E'F'$ , solving 2020 Fall Senior DPC Problem 4.