This is some very hard geometry
Triangle XYZ is equilateral. Points Y and Z lie on a circle centered at O, such that X is the circumcenter of triangle OYZ, and X lies inside triangle OYZ. If the area of the circle is , then find the area of triangle XYZ.
Let's call the radius of the circle R. Since X is the circumcenter of triangle OYZ, it lies on the perpendicular bisectors of all three sides of the triangle. This means that XY = XZ = R, and YZ = 2R/sqrt(3), which is the diameter of the circle.
Now let's find the area of triangle XYZ. We can use the formula for the area of an equilateral triangle:
Area of XYZ = (sqrt(3)/4) * XY^2
Substituting XY = R, we get:
Area of XYZ = (sqrt(3)/4) * R^2
To find the area of the circle, we use the formula:
Area of circle = pi * R^2
Substituting the given area, we get:
pi * R^2 =
Solving for R, we get:
R = sqrt() / pi
Substituting this value of R into the formula for the area of XYZ, we get:
Area of XYZ = (sqrt(3)/4) * (sqrt() / pi)^2
Area of XYZ = (3 / 4pi) *
Simplifying, we get:
Area of XYZ = (3sqrt(3)) / 4
Therefore, the area of triangle XYZ is (3sqrt(3)) / 4.
