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 $O'$ $\fbox{\textbf{(B)} remains stationary}$ .

1957 AHSC (Problems • Answer Key • Resources)
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1957 AHSME Problems/Problem 50

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