Problem: Diameter AB of a circle with center 0 is 10 units. C is a point 4 units from A, and on AB. D is a point 4 units from B, and on AB. P is any point on the circle. Then the broken-line path from C to P to D:
Answer Choices:
A. has the same value for all positions of P
B. exceeds 10 units for all positions of P
C. cannot exceed 10 units
D. is the least when CPD is a right triangle
E. is the greatest when P is equidistant from C and D
Solution:
Let P′ be the intersection of AB and a perpendicular from P to AB, and let P′B=x. Then (PP′)2=x(10−x).
CP=x(10−x)+(6−x)2DP=x(10−x)+(4−x)2
CP+DP=36−2x+16+2x
This sum is greatest when 36−2x=16+2x, that is when x=5. Hence, (E).
or
An ellipse through A and B with C and D as foci is the locus of points such that CP′+P′D=10 where P′ is on the ellipse.