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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 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 $r$: OB∗OA∗AB+OC2∗AB=AC2∗OB+BC2∗OA4[(1)(12)(32)+(32−r)2(32)]=4[(r+1)2(1)+(r+12)2(12)]3+6(94−3r+r2)=4(r2+2r+1)+2(r2+r+14)3+272−18r+6r2=4r2+8r+4+2r2+2r+123+27−12−4+6r2−6r2=18r+8r+2r3+13−4=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 (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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