# 1957 AHSME Problems/Problem 50

## Problem

In circle $O$, $G$ is a moving point on diameter $\overline{AB}$. $\overline{AA'}$ is drawn perpendicular to $\overline{AB}$ and equal to $\overline{AG}$. $\overline{BB'}$ 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 $O'$:

$\textbf{(A)}\ \text{moves on a straight line parallel to }{AB}\qquad \\ \textbf{(B)}\ \text{remains stationary}\qquad\\ \textbf{(C)}\ \text{moves on a straight line perpendicular to }{AB}\qquad\\ \textbf{(D)}\ \text{moves in a small circle intersecting the given circle}\qquad\\ \textbf{(E)}\ \text{follows a path which is neither a circle nor a straight line}$

## Solution

$[asy] import geometry; point A = (0,0); point B = (16,0); point O = midpoint(A--B); point G = (11,0); point A1, B1, O1; // Defining A', B', and O' pair[] a1 = intersectionpoints(circle(A,length(segment(A,G))),perpendicular(A,line(A,B))); A1 = a1[1]; pair[] b1 = intersectionpoints(circle(B, length(segment(B,G))),perpendicular(B,line(A,B))); B1 = b1[1]; O1 = midpoint(A1--B1); // Circle w/ Diameter draw(circle(O,length(segment(A,B))/2)); draw(A--B); // Segments AA', BB', A'B', and OO' draw(A--A1); draw(B--B1); draw(A1--B1); draw(O--O1); // Points w/ Labels dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(A1); label("A'",A1,NW); dot(B1); label("B'",B1,NE); dot(O); label("O",O,S); dot(O1); label("O'",O1,N); dot(G); label("G",G,S); // Right angle marks markscalefactor = 0.11; draw(rightanglemark(A1,A,B)); draw(rightanglemark(B1,B,A)); [/asy]$

Let $AG=AA'=a$ and $BG=BB'=b$. Then, we know that the diameter $d$ of the circle equals $AG+BG=a+b$. Thus, because the circle's diameter does not change, $a+b$ is constant.

Because $A'O'=O'B'$ and $AO=OB$, $\overline{OO'} \parallel \overline{AA'}$. Thus, $\overline{OO'} \perp \overline{AB}$, and so $OO'$ is the distance from $O'$ to $\overline{AB}$.

Let $P$ be some point which moves along $\overline{A'B'}$. Because $A'B'$ is a line segment, as $P$ moves from $A'$ to $B'$, its distance from $\overline{AB}$ will vary linearly with how much it has travelled along $\overline{A'B'}$. Thus, when it is halfway along $\overline{A'B'}$ (in other words, when $P=O'$) its distance from $\overline{AB}$ will be the arithmetic mean of its distance from $\overline{AB}$ at $A'$ (namely, $A'A=a$) and its distance from $\overline{AB}$ at $B'$ (namely, $B'B=b$). Thus, $O'O = \tfrac{a+b}2$.

Because $a+b=d$, a constant, $\tfrac{a+b}2$ is a constant as well. Thus, $OO'$ is the same regardless of the position of $G$. Furthermore, from our work in paragraph 2, we know that $O'$ must lie on the line perpendicular to $\overline{AB}$ through point $O$. Therefore, because $O'$ is a fixed distance from a fixed point on a fixed line, and it will not suddenly "jump across" to the other side of $\overline{AB}$, we can say with confidence that point $O'$ $\fbox{\textbf{(B)} remains stationary}$.