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
![[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]](http://latex.artofproblemsolving.com/8/b/d/8bd59c69fecdc4a648032b82fc852abf7137eb70.png)
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 ![$[\triangle ABC]=\tfrac{(6+2) \cdot 3}2=12$](http://latex.artofproblemsolving.com/c/0/a/c0ae44a7eb875a6d481cee9d62ce7227159f15c2.png)
. 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
| 1957 AHSC (Problems • Answer Key • Resources) | ||
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Followed by Problem 47 | |
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