Question

Let $O$ be an interior point of triangle $ABC,$ and let $s_1=OA+OB+OC.$ If $s_2=AB+BC+CA,$ then:

for every triangle $s_1\gt\dfrac12s_2,$ and $s_1\leq s_2$

for every triangle $s_1\geq\dfrac12s_2,$ and $s_1\lt s_2$

for every triangle $s_1\gt\dfrac12s_2,$ and $s_1\lt s_2$

for every triangle $s_1\geq\dfrac12s_2,$ and $s_1\leq s_2$

neither $(A)$ nor $(B)$ nor $(C)$ nor $(D)$ applies to every triangle

Answer: C

Concepts:bounding to limit cases triangle inequality

Difficulty rating: 1880

Small Hint:

Add $AB\lt OA+OB,$ $BC\lt OB+OC,$ and $CA\lt OC+OA$

Big Hint:

Extend $AO$ to side $BC$ to prove $OA+OB\lt AC+CB,$ and cycle

Solution:

Adding the three strict triangle inequalities $\begin{gathered} AB\lt OA+OB,\\ BC\lt OB+OC,\\ CA\lt OC+OA \end{gathered}$ gives $s_2\lt2s_1.$

For the upper bound, extend $AO$ to $D$ on $BC.$ Then $AO+OD\lt AC+CD$ and $OB\lt OD+DB,$ so cancellation gives $OA+OB\lt AC+CB.$ Cycling and adding yields $2s_1\lt2s_2.$ Therefore $s_1\gt \frac{s_2}{2}$ and $s_1\lt s_2.$

Thus, the correct answer is C.

1966 AMC 12 Problem 35

Answer: C

Difficulty rating: 1880

Small Hint:

Big Hint:

Solution:

Problem 35 in Other Years