HMMT February, 2025

Solution:

We use $\angle$ to denote directed angles. Let $Y$ and $Z$ be the feet of the altitudes from $B$ and $C$ to $AC$ and $AB$, respectively. Then $\angle HZF = \angle HXF = 90^\circ$, so $HZFX$ is cyclic. Similarly, $HYEX$ is cyclic. Therefore,

$$\angle BYX = \angle HYX = \angle HEX = \angle FHX = \angle FZX = \angle BZX.$$
Hence, $BZXY$ is cyclic. A symmetric argument shows $C$ lies on this circle as well. It follows that $\angle BXC = \angle BYC = 90^\circ$, as desired.

Solution:

Let $T$ be the foot of altitude from $A$ to $BC$. For any point $X$, let $X'$ denote the image of $X$ under the negative inversion at $H$ with radius $\sqrt{HA \cdot HT}$. Then $B'$ and $C'$ are the feet of the altitudes from $B$ and $C$ to sides $AC$ and $AB$, respectively.

Claim 1. $\angle BX'C = 90^\circ$.

Proof. Because $HX \perp EF$ and $HE \perp HF$, the quadrilateral $HE'X'F'$ is a rectangle. Note that $\angle BE'H = \angle EB'H = 90^\circ$ and $\angle X'E'H = 90^\circ$. Consequently, $X', B$, and $E'$ are collinear. Similarly, $X', C$, and $F'$ are collinear. Then, $\angle BX'C = \angle E'X'F' = 90^\circ$, as desired. $\square$

From the claim, $X'$ lies on the circle with diameter $BC$ (which $B'$ and $C'$ also lie on). Since this circle is invariant under the inversion, $X$ lies on the circle with diameter $BC$ as well, and $\angle BXC = 90^\circ$.

Solution:

We begin by proving the following lemma.

Lemma 2. Let $ABCD$ be a quadrilateral and $P$ be a point such that $\angle APB + \angle CPD = 180^\circ$. Then, the feet of the altitudes from $P$ to each side of $ABCD$ are concyclic.

Proof. Let $P_A, P_B, P_C, P_D$ be the feet of the altitudes from $P$ to $AB$, $BC$, $CD$, and $DA$ respectively. Note that quadrilateral $P_AP P_BB$ is cyclic. By angle chasing,

$$\angle P_D P_A P_B + \angle P_B P_C P_D = \angle P_D P_A P + \angle P P_A P_B + \angle P_B P_C P + \angle P P_C P_D \\ \qquad = \angle P_D A P + \angle P B P_B + \angle P_B C P + \angle P D P_D \\ \qquad = (180^\circ - \angle APD) + (180^\circ - \angle BPC) \\ \qquad = \angle BPA + \angle DPC \\ \qquad = 180^\circ.$$
Therefore, $P_A P_B P_C P_D$ is cyclic as desired.

Let $P_{\infty}$ be the point at infinity on line $AC$. Let $Y$ and $Z$ be the feet of the altitudes from $H$ to $AB$ and $AC$, respectively. Note that $\angle EHF + \angle BHP_{\infty} = 90^\circ + 90^\circ = 180^\circ$. Thus, the feet of the altitudes from $H$ to $EF$, $EB$, $BP_{\infty}$, and $CP_{\infty}$ are concyclic. In other words, $XYZB$ is cyclic. Since $BCYZ$ is a cyclic quadrilateral, we conclude $X$ lies on this circle, giving us that $\angle BXC = 90^\circ$ as desired.

Remark. Here's another way to prove the lemma.

It is well known that, with the provided condition, there is a point $P'$ that is the isogonal conjugate of $P$ with respect to quadrilateral $ABCD$. Let $P_A$, $P_B$, $P_C$, and $P_D$ be the feet of the altitudes from $P$ to $AB$, $BC$, $CD$, and $DA$, respectively, and let $Q$ be the foot of the altitude from $P'$ to $AB$. Because $P$ and $P'$ are isogonal conjugates with respect to the triangle formed by lines $AB$, $BC$, and $CD$, we have $P_A P_B P_C Q$ is cyclic. Similarly, because $P$ and $P'$ are also isogonal conjugate with respect to the triangle formed by lines $DA$, $AB$, and $BC$, we have $P_D P_A P_B Q$ is cyclic. Consequently, $P_A P_B P_C P_D$ is cyclic as desired.