1. Points $X$ and $Y$ lie on a circle centered at $O,$ and arc $XY$ is $80^\circ.$ The circle passing through points $O,$ $X,$ and $Y$ is drawn. Find the measure of arc $XOY$ on the smaller circle.

2. Trapezoid $WXYZ$ is inscribed in a circle, with $WX \parallel YZ$. If arc $YZ$ is $30$ degrees, arc $WZ$ is $t^2 + 7t$ degrees, and arc $XY$ is $60 - 4t$ degrees, find arc $WTX$.

3. In cyclic quadrilateral $PQRS, (angle P)/2 = (angle Q)/3, = (angle R)/4

Find the largest angle of quadrilateral $PQRS,$ in degrees.

4. A regular dodecagon $P_1 P_2 P_3 \dotsb P_{12}$ is inscribed in a circle with radius $1.$ Compute

\[(P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2.\](The sum includes all terms of the form $(P_i P_j)^2,$ where $1 \le i < j \le 12.$)

Elen8 Dec 22, 2019

#2**0 **

1. Points $X$ and $Y$ lie on a circle centered at $O,$ and arc $XY$ is $80^\circ.$ The circle passing through points $O,$ $X,$ and $Y$ is drawn. Find the measure of arc $XOY$ on the smaller circle.

2. Trapezoid $WXYZ$ is inscribed in a circle, with $WX \parallel YZ$. If arc $YZ$ is $30$ degrees, arc $WZ$ is $t^2 + 7t$ degrees, and arc $XY$ is $60 - 4t$ degrees, find arc $WTX$.

Omi67 Dec 23, 2019

#3**0 **

3. Since angle S is equal to (angle Q)/3, by the cycic quadrilateral angle condition, the largest angle is Q = 135.

4. There are 12 digaonals that have a length of P_1 P_2, which from the Sine Law, is sin (15 degrees). There are 12 diagonals that have a legnth of P_1 P_3, which from the Sine Law, is sin (30 degrees). We can appy the same reasoning to the other diagonals, which gives us a total sum of

(12 sin 15)^2 + (12 sin 30)^2 + (12 sin 45)^2 + (12 sin 60)^2 + (12 sin 75)^2 + (12 sin 90)^2 + (12 sin 105)^2 + (12 sin 120)^2 + (12 sin 135)^2 + (12 sin 150)^2 + (12 sin 175)^2 = 864.

Since we have double-counted, the answer is 864/2 = 432.

Guest Dec 23, 2019