Problem: Let n be the number of points P interior to the region bounded by a circle with radius 1, such that the sum of the squares of the distances from P to the endpoints of a given diameter is 3. Then n is:
Answer Choices:
A. 0
B. 1
C. 2
D. 4
E. infinite
Solution:
Method I. AP2−(r−x)2=OP2−x2,BP2−(r+x)2=OP2−x2. ∴AP2+BP2−2r2−2x2=2OP2−2x2.∴3=2+2OP2.
∴OP=2​1​, that is P may be any point of a circle with radius 2​1​.
Method II. Since AP2+BP2=3<4=AB2, angle APB>90∘ so
that P is interior to the circle. Therefore, there are many points P satisfying the given conditions such that 0<AP≦23​​.
