USA IMO

Let $AB{Q}_{1}P$ be a rhombus and let segments $A{Q}_1$ and $BP$ intersect at $Q_2$. Let $X$ be a point on ray $Q_2{Q}_1$. Because lines $P{Q}_2$ and $Q_1{Q}_2$ are perpendicular, $PX$ increases as $X$ moves away from $Q_2$. Note that $A{Q}_1$ bisects angle $BAP$, that is, $Q$ is on the ray $Q_2{Q}_1$. Because $\angle ABC>\angle BCA$, $A$ and $B$ are on the same side of line $PQ$. Because $P{Q}_1\parallel AB$ and $PQ\perp BC$, $\angle CBA$ is obtuse if and only if the segment $P{Q}_1$ 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 $Q_2{Q}_1$ than $Q$, that is, $PQ<P{Q}_1=AB$.

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 $A{A}_1\parallel BC$.

Let $\angle BCA=x$. Then $\angle A_{1}AC=x=\angle XBC$. Since $\angle ABC=2x$, $\angle AB{A}_1=x$. It follows that $$\begin{array}{rl}\angle PXA& =\angle PXB+\angle BXA\\ & =90^{\circ }-\angle XBC+\angle BXA\\ & =90^{\circ }+\angle BCX=90^{\circ }+x.\end{array}$$ 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, $\angle AP{A}_1=2\angle APX=60^{\circ }$. Because $A_{1}P=AP$, this implies that triangle $PA{A}_1$ is equilateral and that $A{A}_1=AP$. Note also that $\angle B{A}_{1}A=\angle A_{1}BC=\angle AB{A}_1$. Hence, $AB=A{A}_1=AP$. 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 $$\begin{array}{rl}\frac{BC}{\sin (180^{\circ }-3x)}& =\frac{AB}{\sin x},\\ \frac{AB}{\sin (60^{\circ }+y)}& =\frac{BP}{\sin (120^{\circ }-2x)},\\ \frac{PC}{\sin y}& =\frac{BC}{\sin (180^{\circ }-2y)}.\end{array}$$ 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 $$\begin{array}{rl}& \sin (60^{\circ }+y)+\sin (60^{\circ }+y+2x)+\sin (60^{\circ }+y-2x)\\ & =\sin (60^{\circ }+2x+y)+\sin (60^{\circ }+2x-y).\end{array}$$ 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.