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Difference between revisions of "1972 AHSME Problems/Problem 32"
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<math>\boxed{B}</math>
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== Problem ==
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<asy>
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real t=pi/12;real u=8*t;
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real cu=cos(u);real su=sin(u);
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draw(unitcircle);
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draw((cos(-t),sin(-t))--(cos(13*t),sin(13*t)));
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draw((cu,su)--(cu,-su));
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label("A",(cos(13*t),sin(13*t)),W);
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label("B",(cos(-t),sin(-t)),E);
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label("C",(cu,su),N);
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label("D",(cu,-su),S);
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label("E",(cu,sin(-t)),NE);
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label("2",((cu-1)/2,sin(-t)),N);
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label("6",((cu+1)/2,sin(-t)),N);
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label("3",(cu,(sin(-t)-su)/2),E);
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//Credit to Zimbalono for the diagram</asy>
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Chords <math>AB</math> and <math>CD</math> in the circle above intersect at E and are perpendicular to each other.
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If segments <math>AE, EB</math>, and <math>ED</math> have measures <math>2, 3</math>, and <math>6</math> respectively, then the length of the diameter of the circle is
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<math>\textbf{(A) }4\sqrt{5}\qquad
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\textbf{(B) }\sqrt{65}\qquad
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\textbf{(C) }2\sqrt{17}\qquad
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\textbf{(D) }3\sqrt{7}\qquad
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\textbf{(E) }6\sqrt{2}  </math>
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== Solution ==
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Using the chord theorem we can immediately solve for <math>\overline {CE}</math> which will help us. The chord theorem states that if two chords, <math>\overline {AB}</math> and <math>\overline {CD}</math>, intersect at lets say <math>E</math>, then <math>\overline {AE} \cdot \overline {EB}=\overline {CE} \cdot \overline {ED}</math>. Now that we know this we can label length <math>\overline {CE}</math> as <math>\eta</math>. We now have the equation <math>3\eta=2\cdot6\implies3\eta=12\implies\eta=4</math>. We now know the chord lengths <math>7</math> and <math>8</math>. We now label the center of the circle as <math>O</math>. If the circle was on the graph, with <math>O</math> being right on the origin, then make a <math>x=0</math> line cutting through the center. This bisects chord <math>AB</math> at point <math>M</math>. This means <math>\overline {AM}=4</math> and <math>\overline {EM}=2</math> If we now draw a "<math>y=0</math>" line which bisects chord <math>CD</math> at point <math>N</math> making <math>\overline {CN}=\tfrac{7}{2}</math>. If we now move up <math>\overline {EM}</math> to <math>\overline {ON}</math> then we can find the radius(<math>\overline {CO}</math>) using Pythagorean Theorem.
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<math>\sqrt{{\overline {EM}}^2+{\overline {ON}}^2}=r</math>
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<math>\sqrt{2^2+(\tfrac{7}{2})^2}</math>
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<math>\sqrt{4+\frac{49}{4}}</math>
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<math>\sqrt{\frac{65}{4}}</math>
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<math>\frac{\sqrt{65}}{2}=r</math>
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Now that we have <math>r</math>, the diameter is just two times of that, so <math>\boxed{\textbf{(B)}\sqrt{65}}</math>
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[https://artofproblemsolving.com/wiki/index.php/User:Am24 ~AM24]

Latest revision as of 21:33, 23 August 2025

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.

$\sqrt{{\overline {EM}}^2+{\overline {ON}}^2}=r$

$\sqrt{2^2+(\tfrac{7}{2})^2}$

$\sqrt{4+\frac{49}{4}}$

$\sqrt{\frac{65}{4}}$

$\frac{\sqrt{65}}{2}=r$

Now that we have $r$, the diameter is just two times of that, so $\boxed{\textbf{(B)}\sqrt{65}}$

~AM24

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