My problem Let $C', B'$ be points on $AB, AC$, respectively, such that $BB'=BC=CC'$. Let $X, Y$ be midpoints of $BB'$ and $CC'$, respectively. As $\triangle HPQ$ has fixed shape; by spiral similarity, point $K$ lies on a fixed line. For $P=B$ we get $K=X$ and for $Q=C$ we get $K=Y$. Consequently, $K$ lies on the line segment $XY$. Note that $\angle BMB'=\angle BXC$ and $\angle CMC'=\angle BYC$ where $M$ is the midpoint of $BC$. However, since $\min \{\angle B, \angle C\} \ge 60^{\circ}$ we have $BB'>B'C$ and $CC'>C'B$ so points $X, Y$ lie entirely in the circle $\odot(BC)$. Thus, $K$ lies inside $\odot(BC)$ so $\angle BKC>90^{\circ}$ must hold. $\blacksquare$

Stats- 7 students solved it in the contest.

Define \(D , E\) as the feet of altitude from vertex \(B , C\) respectively. Let \(L\) be the midpoint of segment \(DE\) s.t \(KL \cap AB \equiv X ; KL \cap AC \equiv Y\). We have, \(\Delta HEP \sim \Delta HLK \) \(\Rightarrow \angle HLK = \angle HEP = 90^{\circ}.\) Therefore, line \(XY \perp HL\) and thus \(XY\) is fixed. WLOG, let \(\angle B > \angle C\) then \(Y\) lies between \(D\) and \(C\) and \(X\) lies between \(A\) and \(E\). Draw the circle \(BCDE\) centered at \(M\), midpoint of BC. Case - (I) \(P\) lies between \(A\) and \(E\). \(\Rightarrow K\) lies on \(LY \Rightarrow BKC > 90^{\circ}\) -It lies inside the semicircle. Case - (II) \(P\) lies between \(B\) and \(E\). In this case \(K\) lies on \(LX\) and thus the \(KL\) is largest when \(\angle LHK = \angle EHK\) is maximum, or \(\angle LHK = \angle EHK = \angle BHE = \angle A\) or, \( P \equiv B\) \(\Rightarrow \Delta PQC \equiv \Delta BQC => MK = \dfrac{1}{2} CQ < 1/2 BC => K\) lies inside \(\odot BCDE \Rightarrow \angle BKC > 90^{\circ}\)

Fix $\triangle ABC$ and animate $P$ linearly on $\overline{AB}.$ By spiral lemma, $P \mapsto Q$ is linear, and thus, $K$ varies linearly on some line. It suffices to check two locations of $P.$ Take $P=B$ and $P$ the location when $Q=C.$