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.
First, let's find the radius of the circle. Since X is the circumcenter of triangle OYZ, we know that XO, YO, and ZO are all radii of the circle. Since the triangle is equilateral, all three sides are equal in length, so all three radii have the same length. Therefore, we can draw three radii of equal length from O to X, Y, and Z, and form an equilateral triangle OYZ inside circle O.
[asy]
pair X,Y,Z,O;
X = (0,0);
Y = dir(60);
Z = dir(120);
O = (0,2*sqrt(3));
draw(circle(O,2));
draw(O--X);
draw(O--Y);
draw(O--Z);
draw(X--Y--Z--cycle);
label("$X$",X,SW);
label("$Y$",Y,NE);
label("$Z$",Z,NW);
label("$O$",O,SE);
[/asy]
The length of each side of triangle OYZ is equal to twice the radius of the circle, so the area of the equilateral triangle is
$$\frac{\sqrt{3}}{4} \cdot (2r)^2 = \frac{3}{2} r^2$$
where $r$ is the radius of the circle.
We can also find the area of the circle using the formula $A=\pi r^2$. Setting this equal to $100\pi$, we get
$$\pi r^2 = 100\pi$$
$$r^2 = 100$$
$$r = 10$$
Substituting $r=10$ into the expression we found for the area of triangle XYZ, we get
$$\frac{3}{2} \cdot 10^2 = \boxed{150}$$
Therefore, the area of triangle XYZ is 150.