1951 AMC 12 Problem 32

Question

If

\triangle ABC

is inscribed in a semicircle whose diameter is

\overline{AB},

then

\overline{AC}+\overline{BC}

must be:

Equal to

\overline{AB}

Equal to

\overline{AB}\sqrt2

\ge \overline{AB}\sqrt2

\le \overline{AB}\sqrt2

\overline{AB}^{\,2}

The angle subtending the diameter is a right angle

For legs

a,b

, compare

(a+b)^2

with

2(a^2+b^2)

By Thales’ theorem,

AC

and

BC

are the legs of a right triangle with hypotenuse

AB.

Therefore

\begin{aligned} (AC+BC)^2 &\le2(AC^2+BC^2)\\ &=2AB^2. \end{aligned}

Taking positive square roots gives

AC+BC\le AB\sqrt2,

with equality for an isosceles right triangle.

Thus, the correct answer is D.

Problem 32 in other years