Problem 1602

1. Let $R$ be the intersection of two lines $AD$, $BC$ and $S$ be the intersection of two lines $AB$, $CD$ (because $ABCD$ is not a trapezoid then $R$, $S$ are quite specific). Assume that $B$ lies between $A$, $S$ and also lies between $C$, $R$ as in the figure, the other cases can be proved similarly. Let $K$ be the intersection of the circumcircles of triangles $RAB$, $SBC$, then
$$\angle BKR + \angle BKS = \angle BAD + \angle BCD = 180^{\circ} \text{ or } R, K, S \text{ are collinear.}$$
Hence, we have $RK \cdot RS = RB \cdot RC = RA \cdot RD$ and $SK \cdot SR = SB \cdot SA = SC \cdot SD$ so two quadrilaterals $ADSK$ and $CDRK$ are cyclic, this infers $K$ also belongs to two circles $(RCD)$ and $(SDA)$.
Hence, we have
$$\angle AKD = \angle ASD = \angle BSC = \angle BKC \text{ and } \angle ADK = \angle ASK = \angle BSK = \angle BCK$$
so two triangles $KAD$ and $KBC$ are similar. Then $\frac{KA}{KB} = \frac{AD}{BC} = \frac{AE}{BE} = \frac{AM}{BM}$, using the property of angle bisector in the triangle, we can get $KM$ is the interior angle bisector of $\angle AKB$.

On the other hand, we have
$$\angle RNQ = \angle BNE = \angle CBD - \angle BEN = \angle CAD - \angle AEQ = \angle RQN \text{ so } \angle ARB = 2\angle BNM.$$
From this result, we have $\angle BKM = \frac{1}{2}\angle AKB = \frac{1}{2}\angle ARB = \angle BNM$ and quadrilateral $BMNK$ is also cyclic, which leads to $K$ belonging to the circle $(BMN)$.
Similarly, we also have $K$ belongs to circles $(AQM)$, $(CNP)$, $(DPQ)$.

Next, we have to prove that $K$ is the unique common point of these circles. In fact, two circles $(AMP)$, $(BMQ)$ have two common points: $K$, $M$ and two circles $(DNP)$, $(CNP)$ have two common points: $K$, $N$. Therefore, if these four circles have two distinct common points, it leads to $M$, $N$ coinciding, which is a contradiction. So $K$ is the unique common point of these circles $(AQM)$, $(BMN)$, $(CNP)$, $(DPQ)$.

2. From the properties of power of a point, we have
$$RK \cdot RS = RB \cdot RC = RO^2 - R^2, \quad SK \cdot SR = SB \cdot SA = SO^2 - R^2 \text{ then}$$
$$RO^2 - SO^2 = RK \cdot RS - SK \cdot SR = RK^2 - SK^2.$$
It is easy to get $OK \perp RS$ from the above result.
And based on Brocard's theorem in cyclic quadrilateral $ABCD$, point $E$ is the orthocenter of triangle $ORS$, hence $OE \perp RS$.
Therefore, $O$, $E$, $K$ are collinear.
We also have $\angle RKA + \angle SKC = \angle RBA + \angle SBC = 2\angle ADC = \angle AOC$ so $\angle AKC + \angle AOC = 180^\circ$ implies the quadrilateral $AOCK$ is cyclic. Then
$$EO \cdot EK = EA \cdot EC = R^2 - OE^2 \Rightarrow EO(EO + EK) = R^2 \text{ or } OK = \frac{R^2}{EO}.$$
On the other hand, we also have
$$EO \ge \max\{d(O, AC), d(O, BD)\} = \max\left\{\frac{1}{2}\sqrt{4R^2 - AC^2}, \frac{1}{2}\sqrt{4R^2 - BD^2}\right\} = \frac{1}{2}\sqrt{4R^2 - m^2}$$
Therefore, we conclude the inequality $OK \le \frac{2R^2}{\sqrt{4R^2 - m^2}}$ (Q.E.D).