Problem: Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter.
Point $H$ is the foot of the perpendicular from $A$ to line $BC$,
and points $P$ and $Q$ are the feet of the perpendiculars
from $H$ to the lines $AB$ and $AC$, respectively.

Given that \[ AH^2 = 2AO^2, \]
prove that the points $O$, $P$, and $Q$ are collinear.

Solution: We present two approaches.

\paragraph{First approach (synthetic).}
First, since $AP \cdot AB = AH^2 = AQ \cdot AC$, it follows that $PQCB$ is cyclic.
Consequently, we have $AO \perp PQ$.

\begin{center}
\begin{asy}
size(8cm);
pair A = dir(110);
pair H = A+1.414*dir(-90);
pair U = H+6*dir(0);
pair V = H-6*dir(0);
pair B = OP(U--V, unitcircle);
pair C = IP(U--V, unitcircle);
pair P = foot(H, A, B);
pair Q = foot(H, A, C);

pair D = -A;
draw(A--B--C--cycle, blue);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);

draw(A--H);
draw(P--H--Q);

pair K = circumcenter(A, B, C);
draw(P--Q, blue);

draw(A--D--C, red);
filldraw(circumcircle(P, Q, B), opacity(0.1)+yellow, dashed+orange);
filldraw(circumcircle(Q, C, D), opacity(0.1)+yellow, dashed+orange);

dot("$A$", A, dir(A));
dot("$H$", H, dir(H));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(70));
dot("$D$", D, dir(D));
dot("$K$", K, dir(60));
/* Source generated by TSQ */
\end{asy}
\end{center}

Let $K$ be the foot of $A$ onto $PQ$,
and let $D$ be the point diametrically opposite $A$.
Thus $A$, $K$, $O$, $D$ are collinear.

Since quadrilateral $KQCD$ is cyclic ($\angle QKD = \angle QCD = 90^\circ$),
we have
\[ AK \cdot AD = AQ \cdot AC = AH^2
  \implies AK = \frac{AH^2}{AD} = \frac{AH^2}{2AO} = AO \]
so $K = O$.

\paragraph{Second approach (coordinates), with Joshua Hsieh.}
We impose coordinates with $H$ at the origin
and $A = (0,a)$, $B = (-b,0)$, $C = (c,0)$, for $a,b,c > 0$.

\begin{claim*}
  The circumcenter has coordinates $(\frac{c-b}{2}, \frac a2 - \frac{bc}{2a})$.
\end{claim*}
\begin{proof}
  This is a known lemma but but we reproduce its proof for completeness.
  It uses the following steps:
  \begin{itemize}
    \ii By power of a point, the second intersection
    of line $AH$ with the circumcircle is $(0, -\frac{bc}{a})$.
    \ii Since the orthocenter is the reflection of this point across line $BC$,
    the orthocenter is given exactly by $(0, \frac{bc}{a})$.
    \ii The centroid is is $\frac{\vec A + \vec B + \vec C}{3} = (\frac{c-b}{3}, \frac a3)$.
    \ii Since $\vec H - \vec O = 3( \vec G - \vec O)$
    according to the Euler line, we have $\vec O = \frac 32 \vec G - \frac 12 \vec H$.
    This gives the desired formula. \qedhere
  \end{itemize}
\end{proof}
Note that $HQ = \frac{HA \cdot HC}{AC} = \frac{ac}{\sqrt{a^2+c^2}}$.
If we let $T$ be the foot from $Q$ to $BC$,
then $\triangle HQT \overset{\sim}{+} \triangle AHC$
and so the $x$-coordinate of $Q$ is given by
$HQ \cdot \frac{AH}{AC} = \frac{a^2c}{a^2+c^2}$.
Repeating the analogous calculation for $Q$ and $P$ gives
\begin{align*}
  Q &= \left( \frac{a^2 c}{a^2+c^2}, \frac{ac^2}{a^2+c^2} \right) \\
  P &= \left( -\frac{a^2 b}{a^2+b^2}, \frac{ab^2}{a^2+b^2} \right).
\end{align*}
Then, $O$, $P$, $Q$ are collinear if and only if the
following shoelace determinant vanishes
(with denominators cleared out):
\begin{align*}
  0
  &= \det
  \begin{bmatrix}
    -a^2b & ab^2 & a^2+b^2 \\
    a^2c & ac^2 & a^2+c^2 \\
    a(c-b) & a^2-bc & 2a
  \end{bmatrix}
  = a \det
  \begin{bmatrix}
    -ab & ab^2 & a^2+b^2 \\
    ac & ac^2 & a^2+c^2 \\
    c-b & a^2-bc & 2a
  \end{bmatrix} \\
  &= a \det
  \begin{bmatrix}
    -a(b+c) & a(b^2-c^2) & b^2-c^2 \\
    ac & ac^2 & a^2+c^2 \\
    c-b & a^2-bc & 2a
  \end{bmatrix}
  = a(b+c) \det
  \begin{bmatrix}
    -a & a(b-c) & b-c \\
    ac & ac^2 & a^2+c^2 \\
    c-b & a^2-bc & 2a
  \end{bmatrix} \\
  &= a(b+c) \cdot \big[ -a(a^2c^2-a^4+bc(a^2+c^2))
  + ac(b-c) \left( -a^2-bc \right)
- (b-c)^2 \cdot a^3 \big] \\
  &= a^2(b+c)(a^4-a^2b^2-b^2c^2-c^2a^2).
\end{align*}
On the other hand,
\begin{align*}
  AH^2 &= a^2 \\
  2AO^2 &= 2 \left[ \left( \frac{c-b}{2} \right)^2
      + \left( - \frac{a}{2} - \frac{bc}{2a} \right)^2 \right]
  = \frac{a^2+b^2+c^2 + \frac{b^2c^2}{a^2}}{2} \\
  \implies AH^2 - 2AO^2 &= \half \left( a^2-b^2-c^2-\frac{b^2c^2}{a^2} \right).
\end{align*}
So the conditions are equivalent.