1957 AHSME Problems/Problem 46
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 : \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 .
Solution 2 (Answer choices, intution)
Because one of the chords decribed in the problem has length , we know that the diameter, the largest chord in the circle, must have a length . This fact eliminates options (B) and (C). Furthermore, because the chord of length is close to bisecting the other perpendicular chord (being only off of its midpoint), it should be rather close to the length of the diameter. Because the circle's tangent lines are perpendicular to the diameter, moving a small distance away from the diameter will not rapidly decrease the length of the chord, so is only slightly less than the diameter's length. This fact guides us towards answer .
See Also
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