1961 AMC 12 Problem 38

Question

Triangle

ABC

is inscribed in a semicircle of radius

r

so that its base

AB

coincides with diameter

AB.

Point

C

does not coincide with either

A

or

B.

Let

s=AC+BC.

Then, for all permissible positions of

C:

s^2\le8r^2

s^2=8r^2

s^2\ge8r^2

s^2\le4r^2

s^2=4r^2

The angle subtended by the diameter is a right angle

If the legs are

p

and

q,

compare

(p+q)^2

with

2(p^2+q^2)

By Thales’ theorem,

ABC

is right at

C.

Put

p=AC

and

q=BC.

Then

p^2+q^2=AB^2=4r^2.

Since

(p-q)^2\ge0,

we have

2pq\le p^2+q^2.

Therefore

s^2=(p+q)^2\le2(p^2+q^2)=8r^2.

Thus, the correct answer is A.

Problem 38 in other years