Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "On the same side of a straight line three circles are drawn as follows: a circle with a radius of <math>4</math> inches is tangent to the line, the other two circles are equal...")
 
(→‎Solution)
 
(4 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 +
== Problem ==
 +
 
On the same side of a straight line three circles are drawn as follows: a circle with a radius of <math>4</math> 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: <math>\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 12</math>
 
On the same side of a straight line three circles are drawn as follows: a circle with a radius of <math>4</math> 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: <math>\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 12</math>
  
−
Solution:
+
== 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));
  
−
Make a line DG which is parallel to FC. We know that GF = BF = 5, since BFE is similar to BGD.
+
</asy>
  
−
Since ADG is similar to ACB, we know that AG is 10.  
+
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>.
  
−
Thus, AF = 10 + 5 = 15, which is answer C.
+
== See also ==
 +
{{AHSME 50p box|year=1959|num-b=40|num-a=42}}
 +
{{MAA Notice}}

Latest revision as of 20:42, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1959_AHSME_Problems/Problem_41&oldid=223356"