# Understand the problem

##### Source of the problem

##### Topic

##### Difficulty Level

##### Comments

# Start with hints

.Note that all points in the plane of satisfy (why?). For any in the interior, Let be the foot of the perpendicular from to and be the foot of the perpendicular from to the XY plane. It is possible to show that (it is because the angle between the plane and the Z axis is ). Hence, . By symmetry, . This relates the two coordinate systems. For a triangle with sides , the square of the area is . For this to be positive, we must have (after simplification) . In cartesian coordinates, this translates to , which is equivalent to . As lies on the plane , this means that . This last equation is that of the interior of a solid sphere. Hence, our desired locus is the intersection of this solid sphere with the plane , which is precisely the interior of the circumcircle of .

# Watch the video (Coming Soon)

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.