USA IMO

Let $ABQ_1P$ be a rhombus and let segments $AQ_1$ and $BP$ intersect at $Q_2$. Let $X$ be a point on ray $Q_2Q_1$. Because lines $PQ_2$ and $Q_1Q_2$ are perpendicular, $PX$ increases as $X$ moves away from $Q_2$. Note that $AQ_1$ bisects angle $BAP$, that is, $Q$ is on the ray $Q_2Q_1$. Because $\angle ABC > \angle BCA$, $A$ and $B$ are on the same side of line $PQ$. Because $PQ_1 \parallel AB$ and $PQ \perp BC$, $\angle CBA$ is obtuse if and only if the segment $PQ_1$ and $B$ are on different sides of line $PQ$. It follows that $\angle CBA$ is obtuse if and only if $Q_1$ is farther from $Q_2$ on ray $Q_2Q_1$ than $Q$, that is, $PQ < PQ_1 = AB$.

Second Solution. (by Reid Barton) We may assume without loss of generality that $P$ is closer to $A$ than is $Q$. Let $X$ be the intersection of segment $AC$ and ray $QP$, and let $A_1$ be the reflection of $A$ across the line $PQ$. By symmetry, $A_1$ lies on the ray $BX$ and $AA_1 \parallel BC$.

Let $\angle BCA = x$. Then $\angle A_1AC = x = \angle XBC$. Since $\angle ABC = 2x$, $\angle ABA_1 = x$. It follows that
$$\begin{aligned} \angle PXA &= \angle PXB + \angle BXA \\ &= 90^\circ - \angle XBC + \angle BXA \\ &= 90^\circ + \angle BCX = 90^\circ + x. \end{aligned}$$
Since $\angle CAB = 180^\circ - 3x$, it follows that $\angle XAP = \angle CAP = 60^\circ - x$. Hence,
$$\angle APX = 180^\circ - \angle PXA - \angle XAP = 30^\circ.$$
Hence, $\angle APA_1 = 2\angle APX = 60^\circ$. Because $A_1P = AP$, this implies that triangle $PAA_1$ is equilateral and that $AA_1 = AP$. Note also that $\angle BA_1A = \angle A_1BC = \angle ABA_1$. Hence, $AB = AA_1 = AP$. Therefore, $PQ < AB$ if and only if $PQ < AP$. In triangle $APQ$, $PQ < AP$ if and only if $\angle PAQ < \angle AQP$, that is,
$$\angle PAQ < \angle APX - \angle PAQ = 30^\circ - \angle PAQ,$$
or $2\angle PAQ < 30^\circ$. Because $\angle PAQ = 60^\circ - x$, we conclude that $PQ < AB$ if and only if
$$90^\circ < 2x = \angle ABC,$$
as desired.

Third Solution.
As in the first two solutions, let $\angle BCA = x$ and assume that $\angle CAB = 3\angle CAP$. Also, let $\angle PBC = y$. Then $\angle CAB = 180^\circ - 3x$, $\angle PAB = 120^\circ - 2x$, $\angle ABP = 2x - y$, and $\angle BPA = 60^\circ + y$. From the Law of Sines in triangles $ABC$, $ABP$, and $BCP$, we obtain
$$\begin{aligned} \frac{BC}{\sin(180^\circ - 3x)} &= \frac{AB}{\sin x}, \\ \frac{AB}{\sin(60^\circ + y)} &= \frac{BP}{\sin(120^\circ - 2x)}, \\ \frac{PC}{\sin y} &= \frac{BC}{\sin(180^\circ - 2y)}. \end{aligned}$$
Multiplying the above relations and taking into account that $BP = PC$ yields
$$\sin 3x \sin (60^\circ + y) \sin y = \sin x \sin (60^\circ + 2x) \sin 2y,$$
or
$$\frac{\sin 3x}{\sin x} \sin (60^\circ + y) = \frac{\sin 2y}{\sin y} \sin (60^\circ + 2x).$$
By the Triple-angle formulas, $\sin 3x = \sin x(3 - 4\sin^2 x)$. By the Double-angle formulas, $\sin 2y = \sin y \cdot 2\cos y$. It follows that
$$(3 - 4 \sin^2 x) \sin (60^\circ + y) = 2 \cos y \sin (60^\circ + 2x),$$
or, by the Double-angle formulas,
$$(1 + 2 \cos 2x) \sin (60^\circ + y) = 2 \cos y \sin (60^\circ + 2x).$$
Applying the Product-to-Sum formulas gives
$$\begin{aligned} & \sin (60^\circ + y) + \sin (60^\circ + y + 2x) + \sin (60^\circ + y - 2x) \\ &= \sin (60^\circ + 2x + y) + \sin (60^\circ + 2x - y). \end{aligned}$$
We obtain
$$\sin (60^\circ + y) = \sin [60^\circ + (2x - y)] - \sin [60^\circ - (2x - y)],$$
or, using the Difference-to-Product formulas,
$$\sin (60^\circ + y) = 2 \cos 60^\circ \sin(2x - y) = \sin (2x - y).$$
Note that $60^\circ + y > 0$, $2x - y > 0$, and that $(60^\circ + y) + (2x - y) = 60^\circ + 2x < 180^\circ$ (as $3x = 180^\circ - \angle BAC < 180^\circ$). It follows that $60^\circ + y = 2x - y$, i.e., $\angle BPA = \angle ABP$.
Hence, triangle $ABP$ is isosceles and $AP = AB$. It follows that $PQ < AB$ is equivalent to $PQ < AP$. In triangle $APQ$, this is equivalent to
$$60^\circ - x = \angle PAQ < \angle AQP = x - 30^\circ,$$
that is, $\angle B = 2x > 90^\circ$. It follows that $PQ > AB$ if and only if $\angle B$ is obtuse.