# 1964 AHSME Problems/Problem 39

## Problem

The magnitudes of the sides of triangle $ABC$ are $a$, $b$, and $c$, as shown, with $c\le b\le a$. 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: $\textbf{(A) }2a+b\qquad\textbf{(B) }2a+c\qquad\textbf{(C) }2b+c\qquad\textbf{(D) }a+2b\qquad \textbf{(E) }$ $a+b+c$ $[asy] import math; pair A = (0,0), B = (1,3), C = (5,0), P = (1.5,1); pair X = extension(B,C,A,P), Y = extension(A,C,B,P), Z = extension(A,B,C,P); draw(A--B--C--cycle); draw(A--X); draw(B--Y); draw(C--Z); dot(P); dot(A); dot(B); dot(C); label("A",A,dir(210)); label("B",B,dir(90)); label("C",C,dir(-30)); label("A'",X,dir(-100)); label("B'",Y,dir(65)); label("C'",Z,dir(20)); label("P",P,dir(70)); label("a",X,dir(80)); label("b",Y,dir(-90)); label("c",Z,dir(110)); [/asy]$

## Solution

We know that in a $\triangle DEF$, if $\angle D \le \angle E$ then $EF \le DF$, we can use this fact in the different triangles to form inequalities, and then add the inequalities.

In $\triangle ABC$, since $c \le b \le a$, we have $\angle C \le \angle B \le \angle A$ by the above argument.

Now, $\angle AA'C > \angle B \ge \angle C$, hence we have $AC > AA' \implies b > AA'$

And, $\angle BB'C > \angle A \ge \angle C$, hence we have $BC > BB' \implies a > BB'$

And, $\angle CC'B > \angle A \ge \angle B$, hence we have $BC > CC' \implies a > CC'$

Finally, adding all three inequalities, we have $b + a + a > AA' + BB' + CC' \implies AA' + BB' + CC' < \textbf{2a + b}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 