1983 AHSME Problems/Problem 23


In the adjoining figure the five circles are tangent to one another consecutively and to the lines $L_1$ and $L_2$. If the radius of the largest circle is $18$ and that of the smallest one is $8$, then the radius of the middle circle is

[asy] size(250);defaultpen(linewidth(0.7)); real alpha=5.797939254, x=71.191836; int i; for(i=0; i<5; i=i+1) { real r=8*(sqrt(6)/2)^i; draw(Circle((x+r)*dir(alpha), r)); x=x+2r; } real x=71.191836+40+20*sqrt(6), r=18; pair A=tangent(origin, (x+r)*dir(alpha), r, 1), B=tangent(origin, (x+r)*dir(alpha), r, 2); pair A1=300*dir(origin--A), B1=300*dir(origin--B); draw(B1--origin--A1); pair X=(69,-5), X1=reflect(origin, (x+r)*dir(alpha))*X, Y=(200,-5), Y1=reflect(origin, (x+r)*dir(alpha))*Y, Z=(130,0), Z1=reflect(origin, (x+r)*dir(alpha))*Z; clip(X--Y--Y1--X1--cycle); label("$L_2$", Z, S); label("$L_1$", Z1, dir(2*alpha)*dir(90));[/asy]

$\textbf{(A)} \ 12 \qquad  \textbf{(B)} \ 12.5 \qquad  \textbf{(C)} \ 13 \qquad  \textbf{(D)} \ 13.5 \qquad  \textbf{(E)} \ 14$



Consider three consecutive circles, as shown in the diagram above; observe that their centres $P$, $Q$, and $R$ are collinear by symmetry. Let $A$, $B$, and $C$ be the points of tangency, and let $PS$ and $QT$ be segments parallel to the upper tangent (i.e. $L_1$), as also shown. Since $PQ$ is parallel to $QR$ (the three points are collinear), $PS$ is parallel to $QT$ (as both are parallel to $L_1$), and $SQ$ is parallel to $TR$ (as both are perpendicular to $L_1$, due to the tangent being perpendicular to the radius), we have $\triangle PQS \sim \triangle QRT$.

Now, if we let $x, y$, and $z$ be the radii of the three circles (from smallest to largest), then $QS = y - x$ and $RT = z - y$. Thus, from the similarity that we just proved, $\frac{QS}{PQ} = \frac{RT}{QR} \Rightarrow \frac{y-x}{x+y} = \frac{z-y}{y+z}$ (where e.g. $PQ = x + y$ because of collinearity). This equation reduces to $y^2 = zx$, i.e. $\frac{y}{x} = \frac{z}{y}$, so the ratio of consecutive radii is constant, forming a geometric sequence. In this case, as the first radius is $8$ and, four radii later, the radius is $18$, this constant ratio is $\sqrt[\leftroot{-2}\uproot{2}{4}]{\frac{18}{8}}$. Therefore the middle radius is $8 \cdot {\left(\sqrt[\leftroot{-2}\uproot{2}{4}]{\frac{18}{8}}\right)}^{2} = 8 \sqrt{\frac{18}{8}} = \sqrt{18 \cdot 8} = \sqrt{144} = 12$, which is choice $\boxed{\textbf{(A)}}$.

See Also

1983 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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