1965 AHSME Problems/Problem 39

Problem

A foreman noticed an inspector checking a $3$"-hole with a $2$"-plug and a $1$"-plug and suggested that two more gauges be inserted to be sure that the fit was snug. If the new gauges are alike, then the diameter, $d$, of each, to the nearest hundredth of an inch, is:

$\textbf{(A)}\ .87 \qquad  \textbf{(B) }\ .86 \qquad  \textbf{(C) }\ .83 \qquad  \textbf{(D) }\ .75 \qquad  \textbf{(E) }\ .71$

Solution

[asy]  import geometry;  point O = (0,0); point A = (-1/2,0); point B = (1,0); point C, D, E;  // 1", 2", and 3" circles draw(circle(O,3/2)); dot(O); label("O", O, S); draw(circle(A,1)); dot(A); label("A", A, SW); draw(circle(B,1/2)); dot(B); label("B", B, SE);  // Other two circles C=(15/14*3/5,15/14*4/5); D=(15/14*3/5,15/14*-4/5); dot(C); label("C",C,NW); draw(circle(C, 3/7)); dot(D); label("D",D,S); draw(circle(D, 3/7));  // Point E pair[] e=intersectionpoints(line(O,C), circle(O, 3/2)); E=e[1]; dot(E); label("E", E, NE);  // Segments draw(A--B--C--A); draw(O--E);  [/asy]

Let the center of the $3$" circle be $O$, that of the $2$" circle be $A$, that of the $1$" circle be $B$, and those of the circles of unknown radius (let their radii have length $r$) be $C$ and $D$, as in the diagram. Also, in this problem, no two circles share a center, so let the circles be named by their corresponding centers (so, the $3$" circle is circle $O$, etc.). Extend $\overline{OC}$ past $C$ to intersect circle $O$ at point $E$. Because circle $O$ has radius $\frac{3}{2}$ and circle $A$ has radius $1$, $OA=\frac{3}{2}-1=\frac{1}{2}$. Likewise, because $B$ has radius $\frac{1}{2}$, $OB=\frac{3}{2}-\frac{1}{2}=1$. Thus, $AB=\frac{3}{2}$. Furthermore, because the line connecting the centers of two tangent circles goes through their point of tangency, $AC=r+1$ and $BC=r+\frac{1}{2}$. Because $OE=\frac{3}{2}$ and $CE=r$, $OC=\frac{3}{2}-r$. With this information, we can now apply Stewart's Theorem to $\triangle ABC$ to solve for $r$: OBOAAB+OC2AB=AC2OB+BC2OA4[(1)(12)(32)+(32r)2(32)]=4[(r+1)2(1)+(r+12)2(12)]3+6(943r+r2)=4(r2+2r+1)+2(r2+r+14)3+27218r+6r2=4r2+8r+4+2r2+2r+123+27124+6r26r2=18r+8r+2r3+134=28r28r=12r=37 Because the question asks for the diameter of the circle, we calculate $2r=\frac{6}{7}\approx \boxed{\textbf{(B) }0.86}$.

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 38
Followed by
Problem 40
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