1964 AMC 12 Problem 39

Question

The magnitudes of the sides of triangle

ABC

are

a,

b,

c,

as shown, with

c\leq b\leq a.

Through interior point

P

and the vertices

A,

B,

C,

lines are drawn meeting the opposite sides in

A',

B',

C',

respectively. Let

s=AA'+BB'+CC'.

Then, for all positions of point

P,

s

is less than:

2a+b

2a+c

2b+c

a+2b

a+b+c

Answer: A

Concepts:extremal argument bounding to limit cases inequality

Difficulty rating: 1850

Small hint:

A point on a segment is closer to a fixed vertex than the farther endpoint of that segment

Big hint:

Bound

AA'

,

BB'

, and

CC'

separately using their adjacent side lengths

Solution:

For a fixed vertex, squared distance is a convex function along the opposite side, so its maximum occurs at an endpoint. Since each cevian endpoint is interior to its side,

\begin{gathered} AA'\lt\max(AB,AC)=b,\\ BB'\lt\max(BA,BC)=a,\\ CC'\lt\max(CA,CB)=a. \end{gathered}

Adding gives

s\lt2a+b.

Therefore, the correct answer is A.