Difference between revisions of "1957 AHSME Problems/Problem 46"
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\textbf{(E)}\ \sqrt{65} </math> | \textbf{(E)}\ \sqrt{65} </math> | ||
− | == Solution == | + | == Solution 1 == |
<asy> | <asy> | ||
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</asy> | </asy> | ||
− | <math>\boxed{\textbf{(E) }\sqrt{65}}</math>. | + | Let the chords intersect the circle at points <math>A,B,C,</math> and <math>D</math> to form simple polygon <math>ABCD</math>. Further, let the chords intersect at point <math>P</math> with <math>AP=6,PC=2,BP=3,</math> and <math>PD=4</math>, as in the diagram. Then, because <math>\overline{AC} \perp \overline{BD}</math>, by the [[Pythagorean Theorem]], <math>AB=3\sqrt5</math> and <math>BC=\sqrt{13}</math>. Because a circle is determined by three coplanar points, the [[circumcircle]] of <math>\triangle ABC</math> will be the circumcircle of <math>ABCD</math>, so the circumdiameter of <math>\triangle ABC</math> will be our desired answer. We know that <math>[\triangle ABC]=\tfrac{(6+2) \cdot 3}2=12</math>. Furthermore, we know that we can express this area as <math>\tfrac{abc}{4R}</math>, where <math>a,b,</math> and <math>c</math> are <math>\triangle ABC</math>'s side lengths and <math>R</math> is its circumradius. Setting this expression equal to <math>12</math>, we can now solve for <math>R</math>: |
+ | \begin{align*} | ||
+ | \frac{abc}{4R} &= 12 \ | ||
+ | \frac{\sqrt{13} \cdot 8 \cdot 3\sqrt{15}}{4R} &= 12 \ | ||
+ | \frac{2 \cdot 3\sqrt{65}}R &= 12 \ | ||
+ | 6\sqrt{65} &= 12R \ | ||
+ | R &= \frac{\sqrt{65}} 2 | ||
+ | \end{align*} | ||
+ | Because the problem asks for the diameter of the circle, our answer is <math>2R=2 \cdot \frac{\sqrt{65}}2=\boxed{\textbf{(E) }\sqrt{65}}</math>. | ||
== See Also == | == See Also == |
Revision as of 11:48, 27 July 2024
Problem
Two perpendicular chords intersect in a circle. The segments of one chord are and ; the segments of the other are and . Then the diameter of the circle is:
Solution 1
Let the chords intersect the circle at points and to form simple polygon . Further, let the chords intersect at point with and , as in the diagram. Then, because , by the Pythagorean Theorem, and . Because a circle is determined by three coplanar points, the circumcircle of will be the circumcircle of , so the circumdiameter of will be our desired answer. We know that . Furthermore, we know that we can express this area as , where and are 's side lengths and is its circumradius. Setting this expression equal to , we can now solve for :
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 45 |
Followed by Problem 47 | |
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