Oh wow I didn't even know CPhill passed her at all. I haven't been on for like a month or two.
Hint:
2 * Arc LM of the big circle = Arc LM of the small circle
Inscribed angles in a circle that is not part of the center are always half of the arc they intercept at.
Can this help?
https://www.youtube.com/watch?v=QciTMe0rCYI
its 21
Does this help?
https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_28
Hmmm...
I literally just learned this in school!
(Arc XZ + Arc MN) / 2 = Angle XYZ
it is actually 21.
See this elegant proof:
https://www.youtube-nocookie.com/embed/6F6hw_U-AX4?autoplay=1&iv_load_policy=3&loop=1&modestbranding=1&playlist=6F6hw_U-AX4
My Hint:
BDC is an inscribed angle in the circle with endpoints at the diameter. That means BDC is 180/2 - 90 degrees.
Knowing that, can you use similar triangles to figure out CD?
Draw a diagram
There is an equilateral triangle, and any length formed by DG that is at most 1 forms a sector of a circle that has an arc length of 60 degrees.
Find the area of the red part:
+2864 
