An Inscribed Quadrilateral

Let $O$ be the center of the unit circle. Then $\triangle ACO$ has sides $1$ , $1$ , and $\sqrt{3}$ , and by the law of cosines $\angle AOC = 120°$ . Since an inscribed angle is half of the central angle of the same arc, $\angle ABC = 60°$ . Since opposite angles of a cyclic quadrilateral add up to $180°$ , $\angle ADC = 120°$ .

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Method 1:

Let $x = \angle COD$ . Then $\angle AOD = \angle AOC - \angle COD = 120° - x$ . Since $\angle AOB$ and $\angle AOD$ are central angles that intercept congruent chords, $\angle AOB = \angle AOD = 120° - x$ . Since a full revolution around point $O$ is $360°$ , $\angle BOC$ $=$ $360° - \angle AOB - \angle AOD - \angle COD$ $=$ $360° - (120° - x) - (120° - x) - x$ $=$ $120° + x$ .

Since the chord length $c$ is related to the radius $r$ and central angle $\theta$ by the equation $c = 2r \sin \frac{\theta}{2}$ , $CD = 2 \cdot 1 \cdot \sin \frac{x}{2}$ $=$ $2 \sin \frac{x}{2}$ , $BC = 2 \cdot 1 \cdot \sin \frac{120° + x}{2}$ $=$ $2 \sin (60° + \frac{x}{2})$ , and $AB = AD = 2 \cdot 1 \cdot \sin \frac{120° - x}{2}$ $=$ $2 \sin (60° - \frac{x}{2})$ . Notice that:

$BC$

$= 2 \sin (60° + \frac{x}{2})$

$= 2 \sin 60° \cos \frac{x}{2} + 2 \cos 60° \sin \frac{x}{2}$

$= 2 \sin 60° \cos \frac{x}{2} + \sin \frac{x}{2}$

$= 2 \sin 60° \cos \frac{x}{2} - \sin \frac{x}{2} + 2 \sin \frac{x}{2}$

$= 2 \sin 60° \cos \frac{x}{2} - 2 \cos 60° \sin \frac{x}{2} + 2 \sin \frac{x}{2}$

$= 2 \sin (60° - \frac{x}{2}) + 2 \sin \frac{x}{2}$

$= AD + CD$

Therefore, the perimeter is $P = AB + BC + CD + AD$ $=$ $(AD) + (AD + CD) + CD + AD$ $=$ $3 \cdot AD + 2 \cdot CD$ , which means $m = 3$ and $n = 2$ , and $3 + 2 = \boxed{5}$ .

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Method 2:

Reflect $AB$ and $CB$ in the perpendicular bisector of diagonal $AC$ , so that $BC \cong AD$ , and $AB > CD$ . (As reflection preserves length, this will not change the overall perimeter of quadrilateral $ABCD$ .)

Since $AC$ and $AD$ are congruent opposite sides of a cyclic quadrilateral, $AB$ and $CD$ are parallel. Since consecutive angles of parallel lines are supplementary, $\angle BAD = 180° - \angle ADC$ $=$ $180° - 120° = 60°$ . Similarly, $\angle BCD = 180° - \angle ABC$ $=$ $180° - 60° = 120°$ . Therefore, quadrilateral $ABCD$ is an isosceles trapezoid.

Let $E$ be the intersection of $AB$ and a perpendicular from $C$ , and $F$ be the intersection of $AB$ and a perpendicular from $D$ . Since $AB$ and $CD$ are parallel, $CDFE$ is a rectangle, and $EF = CD$ . From $\triangle BCE$ , $BE = BC \cos 60° = \frac{1}{2}BC = \frac{1}{2}AD$ , and from $\triangle ADF$ , $AF = AD \cos 60° = \frac{1}{2}AD$ . Therefore, $AB = BE + EF + AF$ $=$ $\frac{1}{2}AD + CD + \frac{1}{2}AD$ $=$ $AD + CD$ .

Therefore, the perimeter is $P = AB + BC + CD + AD$ $=$ $(AD + CD) + (AD) + CD + AD$ $=$ $3 \cdot AD + 2 \cdot CD$ , which means $m = 3$ and $n = 2$ , and $3 + 2 = \boxed{5}$ .