1966 AMC 12 Problem 35

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