Problem 1755

First Solution. (By Gabriel Carroll and Tiankai Liu) Because $\angle AOC = 2\beta$, we have $\angle CAO = 90^\circ - \beta$. Note that in right triangle $APC$, $\angle CAP = 90^\circ - \gamma$, so
$$\begin{aligned} \angle PAO &= \angle CAO - \angle CAP \\ &= \gamma - \beta. \end{aligned}$$
By (1), $\sin \angle PAO \ge \frac{1}{2}$.

Let $N$ be the foot of perpendicular from $O$ to segment $AP$. Then $MPNO$ is a rectangle, so
$$\begin{aligned} \frac{MP}{OA} &= \frac{ON}{OA} \\ &= \sin \angle PAO \\ &\ge \frac{1}{2}, \end{aligned}$$
or $2MP \ge OA = OC$. In right triangles $OCM$ and $OPM$, $OC > CM$ and $OP > MP$. Therefore,
$$\begin{aligned} PC - MP &= MC - 2MP \\ &\le MC - OC \\ &< MC - MC \\ &= 0. \end{aligned}$$
We obtain $OP > MP > PC$, as desired.

Second Solution.

Let $K$ and $Q$ be the reflections of $A$ and $P$, respectively, across line $OM$. Then $KQPA$ is a rectangle and $QP = KA$. Note that
$$\begin{aligned} \angle KOA &= \angle BOA - \angle BOK = \angle BOA - \angle COA \\ &= 2\gamma - 2\beta \ge 60^\circ. \end{aligned}$$
Because $K$ and $A$ are reflections of each other across line $OM$, we have $OK = OA$ and thus $K$ lies on $\omega$. Hence, $QP = KA = 2R \sin(\frac{1}{2}\angle KOA) \ge R$. By the Triangle Inequality, we have
$$\begin{aligned} OP + R &= Q + OC > QC \\ &= QP + PC \ge R + PC, \end{aligned}$$
which implies (*).

Third Solution. By (1) and the Product-to-Sum formulas,
$$\begin{aligned} 4 \sin \beta \cos \gamma &= 2[\sin(\gamma + \beta) - \sin(\gamma - \beta)] \\ &= 2 \sin \alpha - 2 \sin(\gamma - \beta) < 2 - 1 = 1. \end{aligned}$$
Applying the Extended Law of Sines gives
$$CP \cdot CB = (AC \cos \gamma)(CB) = (2R \sin \beta \cos \gamma)(2R \sin \alpha) < R^2. \quad (2)$$
We present two proofs of (*) based on (2).
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* We choose a point $J$ on ray $CB$ so that $CJ \cdot CP = R^2$. It follows from this equation and from (2) that $CJ > CB$, so that $\angle OBC > \angle OJC$. Since $OC/CJ = PC/CO$ and $\angle JCO = \angle PCO$, triangles $JCO$ and $OCP$ are similar and $\angle OJC = \angle COP$. It follows that
$$\begin{aligned} \angle COP &= \angle OJC < \angle OBC \\ &= 90^\circ - \angle BAC, \end{aligned}$$
or $\angle CAB + \angle COP < 90^\circ$.
* The power of $P$ with respect to circle $\omega$ is $BP \cdot PC = R^2 - OP^2$. By (2), we obtain
$$\begin{aligned} OP^2 &= R^2 - BP \cdot PC \\ &> PC \cdot CB - BP \cdot PC \\ &= PC^2. \end{aligned}$$
Thus, $OP > PC$, as desired.

Fourth Solution. By the Extended Law of Sines, $AB = 2R \sin \gamma$ and $AC = 2R \sin \beta$. Therefore, by (1) and the Addition and Subtraction formulas, we obtain
$$\begin{aligned} BP - PC &= AB \cos \beta - AC \cos \gamma \\ &= 2R(\sin \gamma \cos \beta - \sin \beta \cos \gamma) \\ &= 2R \sin(\gamma - \beta) \geq R. \end{aligned}$$
By the Triangle inequality,
$$R + OP = BO + OP > BP \geq R + PC,$$
which implies that $OP > CP$, as desired.

Fifth Solution. (By Zhiqiang Zhang, China) By the Extended Law of Sines and the Product-to-Sum formulas, we have
$$\begin{aligned} CP &= AC \cos \gamma = 2R \sin \beta \cos \gamma \\ &= R(\sin(\beta + \gamma) - \sin(\gamma - \beta)) \\ &< R(1 - \sin 30^\circ) = \frac{OC}{2}, \end{aligned}$$
or $2CP < OC$. On the other hand, applying the Law of Cosines in triangle $COP$ yields
$$CP^2 + OC^2 - 2(CP)(OC) \cos \angle OCB = OP^2.$$
Now (*) follows from the inequality
$$OP^2 > CP^2 + OC^2 - OC \cdot OC \cdot 1 = CP^2.$$