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Difference between revisions of "1959 AHSME Problems/Problem 41"

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== Solution ==
 
== Solution ==
  
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Make a line DG which is parallel to FC. We know that GF = BF = 5, since BFE is similar to BGD.
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<asy>
  
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Since ADG is similar to ACB, we know that AG is 10.
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import geometry;
  
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Thus, AF = 10 + 5 = 15, which is answer <math>\boxed{C}</math>.
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point A = (0,4);
 +
point B = (-16,16);
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point C = (16,16);
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point D = (-16,0);
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point E = (16,0);
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point F = (0,16);
 +
 
 +
// The line
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line l = line((-20,0),(20,0));
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draw(l, Arrows);
 +
 
 +
// Circles
 +
draw(circle(A,4));
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dot(A);
 +
label("A",A,(0,-3));
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draw(circle(B,16));
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dot(B);
 +
label("B",B,W);
 +
draw(circle(C,16));
 +
dot(C);
 +
label("C",C,(1,0));
 +
 
 +
//Tangency points
 +
dot(D);
 +
label("D",D,S);
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dot(E);
 +
label("E",E,S);
 +
dot(F);
 +
label("F",F,NE);
 +
 
 +
// Triangle AFB, Segment BD
 +
draw(triangle(A,F,B));
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draw(B--D);
 +
 
 +
// Right angle labels
 +
markscalefactor=0.3;
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draw(rightanglemark(F,B,D));
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draw(rightanglemark(B,D,E));
 +
draw(rightanglemark(B,F,A));
 +
 
 +
</asy>
 +
 
 +
Let the radius of the two large circles be <math>r</math>. Also, let the center of the small circle be <math>A</math> and the centers of the two large circles be <math>B</math> and <math>C</math>, as in the diagram. Further, let the large circles intersect at <math>F</math>, let the circle centered at <math>B</math> be tangent to the line at <math>D</math>, and let the circle centered at <math>C</math> be tangent to the line at <math>E</math>. Because <math>BD=CE=r</math> and <math>\overline{BD},\overline{CE} \perp \overleftrightarrow{DE}</math>, <math>\overline{BD} \parallel \overleftrightarrow{DE}</math>. Because the two large circles are tangent at <math>F</math>, <math>F</math> is on the segment connecting their centers, <math>\overline{BC}</math>. Thus, <math>F</math> is a distance <math>r</math> from <math>\overleftrightarrow{DE}</math>. Because the small circle (with radius <math>4</math>) is tangent to the line, <math>AF=r-4</math>. Also, because the small circle and the circle centered at <math>B</math> are tangent, <math>AB=r+4</math>. Because <math>BF=r</math>, by the [[Pythagorean Theorem]], we have the following equation:
 +
\begin{align*}
 +
BF^2+AF^2 &= AB^2 \\
 +
r^2+(r-4)^2 &= (r+4)^2 \\
 +
r^2+r^2-8r+16 &= r^2+8r+16 \\
 +
r^2-16r &= 0 \\
 +
r(r-16) &= 0
 +
\end{align*}
 +
Because <math>r \neq 0</math>, we are left with <math>r=16</math>. Thus, the radius of the two large circles is <math>\boxed{\textbf{(D) }16}</math>.
 +
 
 +
== See also ==
 +
{{AHSME 50p box|year=1959|num-b=40|num-a=42}}
 +
{{MAA Notice}}

Latest revision as of 19:42, 21 July 2024

Problem

On the same side of a straight line three circles are drawn as follows: a circle with a radius of $4$ inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is: $\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 12$

Solution

[asy] import geometry; point A = (0,4); point B = (-16,16); point C = (16,16); point D = (-16,0); point E = (16,0); point F = (0,16); // The line line l = line((-20,0),(20,0)); draw(l, Arrows); // Circles draw(circle(A,4)); dot(A); label("A",A,(0,-3)); draw(circle(B,16)); dot(B); label("B",B,W); draw(circle(C,16)); dot(C); label("C",C,(1,0)); //Tangency points dot(D); label("D",D,S); dot(E); label("E",E,S); dot(F); label("F",F,NE); // Triangle AFB, Segment BD draw(triangle(A,F,B)); draw(B--D); // Right angle labels markscalefactor=0.3; draw(rightanglemark(F,B,D)); draw(rightanglemark(B,D,E)); draw(rightanglemark(B,F,A)); [/asy]

Let the radius of the two large circles be $r$. Also, let the center of the small circle be $A$ and the centers of the two large circles be $B$ and $C$, as in the diagram. Further, let the large circles intersect at $F$, let the circle centered at $B$ be tangent to the line at $D$, and let the circle centered at $C$ be tangent to the line at $E$. Because $BD=CE=r$ and $\overline{BD},\overline{CE} \perp \overleftrightarrow{DE}$, $\overline{BD} \parallel \overleftrightarrow{DE}$. Because the two large circles are tangent at $F$, $F$ is on the segment connecting their centers, $\overline{BC}$. Thus, $F$ is a distance $r$ from $\overleftrightarrow{DE}$. Because the small circle (with radius $4$) is tangent to the line, $AF=r-4$. Also, because the small circle and the circle centered at $B$ are tangent, $AB=r+4$. Because $BF=r$, by the Pythagorean Theorem, we have the following equation: \begin{align*} BF^2+AF^2 &= AB^2 \\ r^2+(r-4)^2 &= (r+4)^2 \\ r^2+r^2-8r+16 &= r^2+8r+16 \\ r^2-16r &= 0 \\ r(r-16) &= 0 \end{align*} Because $r \neq 0$, we are left with $r=16$. Thus, the radius of the two large circles is $\boxed{\textbf{(D) }16}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 40 		Followed by
Problem 42
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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