1957 AHSME Problems/Problem 46

Problem

Two perpendicular chords intersect in a circle. The segments of one chord are $3$ and $4$; the segments of the other are $6$ and $2$. Then the diameter of the circle is:

$\textbf{(A)}\ \sqrt{89}\qquad \textbf{(B)}\ \sqrt{56}\qquad \textbf{(C)}\ \sqrt{61}\qquad \textbf{(D)}\ \sqrt{75}\qquad \textbf{(E)}\ \sqrt{65}$

Solution 1

$[asy] import geometry; point A = (0,0); point B = (6,3); point C = (8,0); point D, P; circle c = circumcircle(A,B,C); // Triangle ABC w/ Circumcircle draw(triangle(A,B,C)); dot(A); label("A",A,W); dot(B); label("B",B,N); dot(C); label("C",C,E); draw(c); // Segment BD, Triangle ADC pair[] d = intersectionpoints(perpendicular(B,line(A,C)),c); D = d[0]; dot(D); label("D",D,S); draw(B--D); draw(triangle(A,D,C)); pair[] p = intersectionpoints(B--D,A--C); P = p[0]; dot(P); label("P",P,SW); // Right angle mark markscalefactor = 0.0577; draw(rightanglemark(A,P,B)); // Length Labels label("3", midpoint(B--P), W); label("4", midpoint(P--D), W); label("6", midpoint(A--P), S); label("2", midpoint(P--C), S); [/asy]$

Let the chords intersect the circle at points $A,B,C,$ and $D$ to form simple polygon $ABCD$. Further, let the chords intersect at point $P$ with $AP=6,PC=2,BP=3,$ and $PD=4$, as in the diagram. Then, because $\overline{AC} \perp \overline{BD}$, by the Pythagorean Theorem, $AB=3\sqrt5$ and $BC=\sqrt{13}$. Because a circle is determined by three coplanar points, the circumcircle of $\triangle ABC$ will be the circumcircle of $ABCD$, so the circumdiameter of $\triangle ABC$ will be our desired answer. We know that $[\triangle ABC]=\tfrac{(6+2) \cdot 3}2=12$. Furthermore, we know that we can express this area as $\tfrac{abc}{4R}$, where $a,b,$ and $c$ are $\triangle ABC$'s side lengths and $R$ is its circumradius. Setting this expression equal to $12$, we can now solve for $R$: \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 $2R=2 \cdot \frac{\sqrt{65}}2=\boxed{\textbf{(E) }\sqrt{65}}$.

Solution 2 (Answer choices, intution)

Because one of the chords decribed in the problem has length $6+2=8$, we know that the diameter, the largest chord in the circle, must have a length $\geq 8$. This fact eliminates options (B) and (C). Furthermore, because the chord of length $8$ is close to bisecting the other perpendicular chord (being only $0.5$ 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 $AC$ is only slightly less than the diameter's length. This fact guides us towards answer $\boxed{\textbf{(E) }\sqrt{65}}$.

See Also

 1957 AHSC (Problems • Answer Key • Resources) Preceded byProblem 45 Followed byProblem 47 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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