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

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== Solution ==
 
== Solution ==
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<math>\fbox{B}</math>
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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, E;
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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,NW);
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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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// Point E
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pair[] e=intersectionpoints(line(O,C), circle(O, 3/2));
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E=e[1];
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dot(E);
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label("E", E, NE);
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// Segments
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draw(A--B--C--A);
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draw(O--E);
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</asy>
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Let the center of the <math>3</math>" circle be <math>O</math>, that of the <math>2</math>" circle be <math>A</math>, that of the <math>1</math>" circle be <math>B</math>, and those of the circles of unknown radius (let their radii have length <math>r</math>) be <math>C</math> and <math>D</math>, 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 <math>3</math>" circle is circle <math>O</math>, etc.). Extend <math>\overline{OC}</math> past <math>C</math> to intersect circle <math>O</math> at point <math>E</math>. Because circle <math>O</math> has radius <math>\frac{3}{2}</math> and circle <math>A</math> has radius <math>1</math>, <math>OA=\frac{3}{2}-1=\frac{1}{2}</math>. Likewise, because <math>B</math> has radius <math>\frac{1}{2}</math>, <math>OB=\frac{3}{2}-\frac{1}{2}=1</math>. Thus, <math>AB=\frac{3}{2}</math>. Furthermore, because the line connecting the centers of two tangent circles goes through their point of tangency, <math>AC=r+1</math> and <math>BC=r+\frac{1}{2}</math>. Because <math>OE=\frac{3}{2}</math> and <math>CE=r</math>, <math>OC=\frac{3}{2}-r</math>. With this information, we can now apply [[Stewart's Theorem]] to <math>\triangle ABC</math> to solve for <math>r</math>:
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\begin{align*}
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OB*OA*AB+OC^2*AB&=AC^2*OB+BC^2*OA \\
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4[(1)(\frac{1}{2})(\frac{3}{2})+(\frac{3}{2}-r)^2(\frac{3}{2})]&=4[(r+1)^2(1)+(r+\frac{1}{2})^2(\frac{1}{2})] \\
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3+6(\frac{9}{4}-3r+r^2)&=4(r^2+2r+1)+2(r^2+r+\frac{1}{4}) \\
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3+\frac{27}{2}-18r+6r^2&=4r^2+8r+4+2r^2+2r+\frac{1}{2} \\
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3+\frac{27-1}{2}-4+6r^2-6r^2&=18r+8r+2r \\
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3+13-4&=28r \\
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28r&=12 \\
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r&=\frac{3}{7}
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\end{align*}
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Because the question asks for the diameter of the circle, we calculate <math>2r=\frac{6}{7}\approx \boxed{\textbf{(B) }0.86}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 13:11, 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, 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 $r$: \begin{align*} OB*OA*AB+OC^2*AB&=AC^2*OB+BC^2*OA \\ 4[(1)(\frac{1}{2})(\frac{3}{2})+(\frac{3}{2}-r)^2(\frac{3}{2})]&=4[(r+1)^2(1)+(r+\frac{1}{2})^2(\frac{1}{2})] \\ 3+6(\frac{9}{4}-3r+r^2)&=4(r^2+2r+1)+2(r^2+r+\frac{1}{4}) \\ 3+\frac{27}{2}-18r+6r^2&=4r^2+8r+4+2r^2+2r+\frac{1}{2} \\ 3+\frac{27-1}{2}-4+6r^2-6r^2&=18r+8r+2r \\ 3+13-4&=28r \\ 28r&=12 \\ r&=\frac{3}{7} \end{align*} 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 (Problems • Answer Key • Resources)
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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