New Year's Countdown Day 27: Triangle Tricks

Let $PB = x,$ so that $PA = 27x.$ Using the Law of Cosines on $\triangle APB,$ we have

$\begin{aligned} AB^2 &= PA^2 + PB^2 - 2(PA)(PB) \cos \angle APB \\ &= x^2 + (27x)^2 - 2x(27x) \cos 120^{\circ} \\ &= x^2 + 729^2 + 27x^2 \\ &= 757x^2. \end{aligned}$

Thus, $AB = x\sqrt{757}.$

Let $O$ be the midpoint of $AB.$ From the given information about $\triangle ABC,$ we deduce that it is a 30-60-90 right triangle, so $O$ is also the circumcenter of $\triangle ABC.$ This gives us

$BC = CO = OA = OB = \dfrac{AB}{\sqrt{3}} = x \sqrt{\dfrac{757}{3}}.$

Since $\angle BOA = \angle BPA = 120^{\circ},$ quadrilateral $BPOA$ is a cyclic quadrilateral. By Ptolemy's theorem,

$\begin{aligned} BP \cdot OA + BA \cdot OP &= BO \cdot AP \\ x \left ( x \sqrt{\dfrac{757}{3}} \right ) + x\sqrt{757} \cdot OP &= \left ( x \sqrt{\dfrac{757}{3}} \right )(27x) \\ \dfrac{x}{\sqrt{3}} + OP &= \dfrac{27x}{\sqrt{3}} \\ OP &= \dfrac{26x}{\sqrt{3}}. \end{aligned}$

Now, we focus on $\triangle BCO.$ Since $BC = CO = OB,$ $\triangle BCO$ is equilateral:

We also have $\angle BPO = 150^{\circ},$ since it is supplementary to $\angle BAC = 30^{\circ}.$ To find the length of $PC,$ we rotate $\triangle BPO$ clockwise $60^{\circ}$ about $O$ to form a new triangle:

Since rotations preserve distances, $OP = OP'$ and $BP = B'P = x.$ This means $\triangle POP'$ is isosceles, and $\angle PP'O = \angle P'PO.$ We also have $\angle POP' = 60^{\circ},$ since that is the angle of rotation. Thus,

$\angle PP'O = \angle PP'O = \dfrac{180^{\circ} - \angle POP'}{2} = \dfrac{180^{\circ} - 60^{\circ}}{2} = 60^{\circ},$

and $\triangle POP'$ is equilateral. This gives us $PP' = PO = \dfrac{26x}{\sqrt{3}}.$

Since $\angle B'P'O = \angle BPO = 150^{\circ},$ $\angle B'P'P = 150^{\circ} - 60^{\circ} = 90^{\circ},$ meaning $\triangle B'P'P$ is a right triangle. By the Pythagorean Theorem,

$\begin{aligned} PC^2 &= PP'^2 + B'P'^2 \\ &= \left ( \dfrac{26x}{\sqrt{3}} \right )^2 + x^2 \\ &= \dfrac{676x^2}{3} + x^2 \\ &= \dfrac{679x^2}{3}. \end{aligned}$

Finally, $PC = x \sqrt{\dfrac{679}{3}},$ so $\left ( \dfrac{PC}{PB} \right )^2 = \dfrac{679}{3},$ and $p + q = 679 + 3 = \boxed{682}.$