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.

1964 AMC 12 Problem 39

Answer: A

Difficulty rating: 1850

Small Hint:

Big Hint:

Solution:

Problem 39 in Other Years