1972 AHSME Problems/Problem 32

Problem

$[asy] real t=pi/12;real u=8*t; real cu=cos(u);real su=sin(u); draw(unitcircle); draw((cos(-t),sin(-t))--(cos(13*t),sin(13*t))); draw((cu,su)--(cu,-su)); label("A",(cos(13*t),sin(13*t)),W); label("B",(cos(-t),sin(-t)),E); label("C",(cu,su),N); label("D",(cu,-su),S); label("E",(cu,sin(-t)),NE); label("2",((cu-1)/2,sin(-t)),N); label("6",((cu+1)/2,sin(-t)),N); label("3",(cu,(sin(-t)-su)/2),E); //Credit to Zimbalono for the diagram[/asy]$

Chords $AB$ and $CD$ in the circle above intersect at E and are perpendicular to each other. If segments $AE, EB$ , and $ED$ have measures $2, 3$ , and $6$ respectively, then the length of the diameter of the circle is

$\textbf{(A) }4\sqrt{5}\qquad \textbf{(B) }\sqrt{65}\qquad \textbf{(C) }2\sqrt{17}\qquad \textbf{(D) }3\sqrt{7}\qquad \textbf{(E) }6\sqrt{2}$

Solution

Using the chord theorem we can immediately solve for $\overline {CE}$ which will help us. The chord theorem states that if two chords, $\overline {AB}$ and $\overline {CD}$ , intersect at lets say $E$ , then $\overline {AE} \cdot \overline {EB}=\overline {CE} \cdot \overline {ED}$ . Now that we know this we can label length $\overline {CE}$ as $\eta$ . We now have the equation $3\eta=2\cdot6\implies3\eta=12\implies\eta=4$ . We now know the chord lengths $7$ and $8$ . We now label the center of the circle as $O$ . If the circle was on the graph, with $O$ being right on the origin, then make a $x=0$ line cutting through the center. This bisects chord $AB$ at point $M$ . This means $\overline {AM}=4$ and $\overline {EM}=2$ If we now draw a " $y=0$ " line which bisects chord $CD$ at point $N$ making $\overline {CN}=\tfrac{7}{2}$ . If we now move up $\overline {EM}$ to $\overline {ON}$ then we can find the radius( $\overline {CO}$ ) using Pythagorean Theorem.