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Difference between revisions of "1983 AHSME Problems/Problem 23"

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[[File:pdfresizer.com-pdf-convert-q23.png]]
 
[[File:pdfresizer.com-pdf-convert-q23.png]]
  
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Consider three consecutive circles, as shown in the diagram above; observe that their centres <math>P</math>, <math>Q</math>, and <math>R</math> are collinear by symmetry. Let <math>A</math>, <math>B</math>, and <math>C</math> be the points of tangency, and let <math>PS</math> and <math>QT</math> be segments parallel to the upper tangent (i.e. <math>L_1</math>), as also shown. Since <math>PQ</math> is parallel to <math>QR</math> (the three points are collinear), <math>PS</math> is parallel to <math>QT</math> (as both are parallel to <math>L_1</math>), and <math>SQ</math> is parallel to <math>TR</math> (as both are perpendicular to <math>L_1</math>), we have <math>\triangle PQS \sim \triangle QRT</math>.
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Consider three consecutive circles, as shown in the diagram above; observe that their centres <math>P</math>, <math>Q</math>, and <math>R</math> are collinear by symmetry. Let <math>A</math>, <math>B</math>, and <math>C</math> be the points of tangency, and let <math>PS</math> and <math>QT</math> be segments parallel to the upper tangent (i.e. <math>L_1</math>), as also shown. Since <math>PQ</math> is parallel to <math>QR</math> (the three points are collinear), <math>PS</math> is parallel to <math>QT</math> (as both are parallel to <math>L_1</math>), and <math>SQ</math> is parallel to <math>TR</math> (as both are perpendicular to <math>L_1</math>, due to the tangent being perpendicular to the radius), we have <math>\triangle PQS \sim \triangle QRT</math>.
  
 
Now, if we let <math>x, y</math>, and <math>z</math> be the radii of the three circles (from smallest to largest), then <math>QS = y - x</math> and <math>RT = z - y</math>. Thus, from the similarity that we just proved, <math>\frac{QS}{PQ} = \frac{RT}{QR} \Rightarrow \frac{y-x}{x+y} = \frac{z-y}{y+z}</math> (where e.g. <math>PQ = x + y</math> because of collinearity). This actually simplifies to <math>y^2 = zx</math>, i.e. <math>\frac{y}{x} = \frac{z}{y}</math>, so the ratio of consecutive radii is constant, forming a geometric sequence. In this case, as the first radius is <math>8</math> and, four radii later, the radius is <math>18</math>, this constant ratio is <math>\sqrt[\leftroot{-2}\uproot{2}{4}]{\frac{18}{8}}</math>. Therefore the middle radius is <math>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</math>, which is choice <math>\boxed{\textbf{A}}</math>.
 
Now, if we let <math>x, y</math>, and <math>z</math> be the radii of the three circles (from smallest to largest), then <math>QS = y - x</math> and <math>RT = z - y</math>. Thus, from the similarity that we just proved, <math>\frac{QS}{PQ} = \frac{RT}{QR} \Rightarrow \frac{y-x}{x+y} = \frac{z-y}{y+z}</math> (where e.g. <math>PQ = x + y</math> because of collinearity). This actually simplifies to <math>y^2 = zx</math>, i.e. <math>\frac{y}{x} = \frac{z}{y}</math>, so the ratio of consecutive radii is constant, forming a geometric sequence. In this case, as the first radius is <math>8</math> and, four radii later, the radius is <math>18</math>, this constant ratio is <math>\sqrt[\leftroot{-2}\uproot{2}{4}]{\frac{18}{8}}</math>. Therefore the middle radius is <math>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</math>, which is choice <math>\boxed{\textbf{A}}</math>.

Revision as of 17:00, 27 January 2019

Problem

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]

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

Solution

Pdfresizer.com-pdf-convert-q23.png

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 actually simplifies 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 $\boxed{\textbf{A}}$.

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