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

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\textbf{(E) }\ .71 </math>
 
\textbf{(E) }\ .71 </math>
  
== Solution ==
+
== Solution 1 ==
  
 
<asy>
 
<asy>
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point A = (-1/2,0);
 
point A = (-1/2,0);
 
point B = (1,0);
 
point B = (1,0);
point C, D;
+
point C, D, E;
  
 
// 1", 2", and 3" circles
 
// 1", 2", and 3" circles
Line 36: Line 36:
 
D=(15/14*3/5,15/14*-4/5);
 
D=(15/14*3/5,15/14*-4/5);
 
dot(C);
 
dot(C);
label("C",C,N);
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label("C",C,NW);
 
draw(circle(C, 3/7));
 
draw(circle(C, 3/7));
 
dot(D);
 
dot(D);
 
label("D",D,S);
 
label("D",D,S);
 
draw(circle(D, 3/7));
 
draw(circle(D, 3/7));
 +
 +
// 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);
  
 
// Segments
 
// Segments
 
draw(A--B--C--A);
 
draw(A--B--C--A);
draw(C--O);
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draw(O--E);
  
 
</asy>
 
</asy>
  
<math>\fbox{B}</math>
+
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>:
 +
\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 \
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3+13-4&=28r \
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28r&=12 \
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r&=\frac{3}{7}
 +
\end{align*}
 +
Because the question asks for the diameter of the circle, we calculate <math>2r=\frac{6}{7}\approx \boxed{\textbf{(B) }0.86}</math>.
 +
 
 +
== Solution 2 (Overkill?) ==
 +
 
 +
This is Descartes' formula for four mutually tangent circles with radii <math>r_1,r_2,r_3,r_4</math>
 +
 
 +
Let <math>k_i</math> denote <math>\frac{1}{r_i}</math>. This is the curvature.
 +
 
 +
Note: A negative radius means that the other circles are internally tangent to it.
 +
 
 +
Then,
 +
 
 +
<math>(k_1+k_2+k_3+k_4)^2=2(k_1^2+k_2^2+k_3^2+k_4^2)</math>
 +
 
 +
First, scale the problem up by two, so the radii are 1",2",3", and d"
 +
 
 +
Next, let <math>k</math> be <math>frac{1}{d}</math>
 +
 
 +
Using the formula, we get:
 +
 
 +
<math>\left(\frac{7}{6}+k\right)=2(\frac{49}{36}+k^2)</math>
 +
 
 +
<math>\frac{49}{36}+\frac{7}{3}k+k^2=\frac{98}{36}+2k^2</math>
 +
 
 +
(You'll see why I didn't simplify <math>\frac{98}{36}</math>)
 +
 
 +
<math>k^2-\frac{7}{3}-\frac{49}{36}=0</math>
 +
 
 +
Using the quadratic formula:
 +
 
 +
<math>k=\frac{\frac{7}{3}\pm\sqrt{\frac{49}{9}-4\cdot1\cdot\frac{49}{36}}}{2}</math>
 +
 
 +
<math>k=\frac{\frac{7}{3}\pm\sqrt{\frac{49}{9}-\frac{49}{9}}}{2}</math>
 +
 
 +
<math>k=\frac{\frac{7}{3}\pm\sqrt{0}}{2}</math>
 +
 
 +
<math>k=\frac{\frac{7}{3}\pm0}{2}</math>
 +
 
 +
<math>k=\frac{\frac{7}{3}}{2}</math>
 +
 
 +
<math>k=\frac{7}{6}</math>
 +
 
 +
So the diameter <math>d=\frac{6}{7}</math>, which is about <math>\boxed{(\textbf{B}).86}</math>
 +
 
 +
~ [[User:Afly|Afly]] ([[User talk:Afly|talk]])
  
 
== See Also ==
 
== See Also ==

Latest revision as of 17:50, 2 January 2025

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 1

[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 $2r=\frac{6}{7}\approx \boxed{\textbf{(B) }0.86}$.

Solution 2 (Overkill?)

This is Descartes' formula for four mutually tangent circles with radii $r_1,r_2,r_3,r_4$

Let $k_i$ denote $\frac{1}{r_i}$. This is the curvature.

Note: A negative radius means that the other circles are internally tangent to it.

Then,

$(k_1+k_2+k_3+k_4)^2=2(k_1^2+k_2^2+k_3^2+k_4^2)$

First, scale the problem up by two, so the radii are 1",2",3", and d"

Next, let $k$ be $frac{1}{d}$

Using the formula, we get:

$\left(\frac{7}{6}+k\right)=2(\frac{49}{36}+k^2)$

$\frac{49}{36}+\frac{7}{3}k+k^2=\frac{98}{36}+2k^2$

(You'll see why I didn't simplify $\frac{98}{36}$)

$k^2-\frac{7}{3}-\frac{49}{36}=0$

Using the quadratic formula:

$k=\frac{\frac{7}{3}\pm\sqrt{\frac{49}{9}-4\cdot1\cdot\frac{49}{36}}}{2}$

$k=\frac{\frac{7}{3}\pm\sqrt{\frac{49}{9}-\frac{49}{9}}}{2}$

$k=\frac{\frac{7}{3}\pm\sqrt{0}}{2}$

$k=\frac{\frac{7}{3}\pm0}{2}$

$k=\frac{\frac{7}{3}}{2}$

$k=\frac{7}{6}$

So the diameter $d=\frac{6}{7}$, which is about $\boxed{(\textbf{B}).86}$

~ Afly (talk)

See Also

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