Problem: The magnitudes of the sides of triangle ABC are a,b, and c, as shown, with cβ¦bβ¦a. Through interior point P and the vertices A,B,C lines are drawn meeting the opposite sides in Aβ²,Bβ²,Cβ², respectively. Let s=AAβ²+BBβ²+CCβ². Then, for all positions of point P,s is less than:
Answer Choices:
A. 2a+b
B. 2a+c
C. 2b+c
D. a+2b
E. a+b+c
Solution:
The length of each of the lines through P must be less than the length of the larger of the two sides of the triangle adjacent to the line through P. To see this let BB' be fixed and allow point P to take a position very close to Bβ². Then Aβ² takes a position very close to C so that AAβ² is very nearly equal to b, but less than b. Similarly, CC ' remains smaller than a and BBβ² remains smaller than a for all interior positions of P}.
β΄AAβ²+BBβ²+CCβ²<b+a+aβ΄s<2a+b.