Difference between revisions of "1965 AHSME Problems/Problem 39"

(see also box, statement of answer)
(diagram)
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== Solution ==
 
== Solution ==
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<asy>
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import geometry;
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point O = (0,0);
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point A = (-1/2,0);
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point B = (1,0);
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point C, D;
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// 1", 2", and 3" circles
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draw(circle(O,3/2));
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dot(O);
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label("O", O, S);
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draw(circle(A,1));
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dot(A);
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label("A", A, SW);
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draw(circle(B,1/2));
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dot(B);
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label("B", B, SE);
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// Other two circles
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C=(15/14*3/5,15/14*4/5);
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D=(15/14*3/5,15/14*-4/5);
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dot(C);
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label("C",C,N);
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draw(circle(C, 3/7));
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dot(D);
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label("D",D,S);
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draw(circle(D, 3/7));
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// Segments
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draw(A--B--C--A);
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draw(C--O);
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</asy>
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<math>\fbox{B}</math>
 
<math>\fbox{B}</math>
  

Revision as of 13:33, 20 July 2024

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;  // 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,N); draw(circle(C, 3/7)); dot(D); label("D",D,S); draw(circle(D, 3/7));  // Segments draw(A--B--C--A); draw(C--O);  [/asy]

$\fbox{B}$

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 38
Followed by
Problem 40
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
All AHSME Problems and Solutions

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