Question

In circle $O,$ $G$ is a moving point on diameter $\overline{AB}.$ $\overline{AA'}$ is drawn perpendicular to $\overline{AB}$ and equal to $AG.$ $\overline{BB'}$ is drawn perpendicular to $\overline{AB},$ on the same side of diameter $\overline{AB}$ as $\overline{AA'},$ and equal to $BG.$ Let $O'$ be the midpoint of $\overline{A'B'}.$ Then, as $G$ moves from $A$ to $B,$ point $O':$

moves on a straight line parallel to $AB$

remains stationary

moves on a straight line perpendicular to $AB$

moves in a small circle intersecting the given circle

follows a path which is neither a circle nor a straight line

Answer: B

Concepts:coordinate geometry midpoint invariant

Difficulty rating: 1790

Small Hint:

Place $A=(0,0),$ $B=(L,0),$ and $G=(g,0)$

Big Hint:

Write coordinates for $A'$ and $B'$ , then average them

Solution:

Let $A=(0,0),$ $B=(L,0),$ and $G=(g,0).$ The perpendicular constructions on the same side give $A'=(0,g),\qquad B'=(L,L-g).$ Their midpoint is $\begin{aligned} O' &=\left(\frac L2,\frac{g+L-g}{2}\right)\\ &=\left(\frac L2,\frac L2\right), \end{aligned}$ independent of $g.$ Thus $O'$ remains stationary.

Therefore, the correct answer is B.

1957 AMC 12 Problem 50

Answer: B

Difficulty rating: 1790

Small Hint:

Big Hint:

Solution:

Problem 50 in Other Years