2017 AMC 10 B

WLOG, let the centroid of be , The centroid of an equiateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times.

Therefore, , so , so since is isosceles and , then by Law of Cosines, . Alternatively, we can use the fact that the circumradius of an equlateral triangle is equal lto . Therefore, the area of the triangle is , so the square of the area of the triangle is .

WLOG, let the centroid of be . Then, one of the verties must be the other curve of the hyperbola. WLOG, let . Then, point must be the refction of across the

line , so let , and ,

where . Because is the centroid, the average of the coordinates of the vertices of the triangle is . So we know that , Multplying by and solving gives us .

So , and . So , and finding the square of the area gives us.

WLOG, let the centroid of be and let point be . It is known that the centroid is equidistant from the three vertices of . Because we have the coordinates of both and , we know that the distance from to any vertice

of is , Therefore，. It follows that from , where and using the formula for the area of a triangle with sine  .

Because and are congruent to ， also have an area of .

Therefore .Squaring that give us the answer of .

WLOG, let the centroid of the triangle be . By symmetry, the other vertexis . The distance between these two points is , so the height of the triangle is , the side length is , and the area is , yielding an answer of .