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Difference between revisions of "1971 AHSME Problems/Problem 9"

(Created page with "== Problem 9 == An uncrossed belt is fitted without slack around two circular pulleys with radii of <math>14</math> inches and <math>4</math> inches. If the distance between...")
 
(see also box, diagram, solution edits)
 
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== Problem 9 ==
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== Problem ==
  
 
An uncrossed belt is fitted without slack around two circular pulleys with radii of <math>14</math> inches and <math>4</math> inches.  
 
An uncrossed belt is fitted without slack around two circular pulleys with radii of <math>14</math> inches and <math>4</math> inches.  
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==Solution==
 
==Solution==
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(I don't know how to use Asymptote, so if you are a visual learner, I apologize)
 
  
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Let the center of the smaller circle be <math>A</math> and the center of the larger circle be <math>B.</math> Draw perpendicular radii <math>AC</math> and <math>BD.</math> The length <math>CD</math> should equal 24.
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<asy>
  
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From <math>A,</math> draw a perpendicular line <math>AE,</math> where <math>E</math> is the foot of the perpendicular on <math>BD.</math> Line <math>BE = 14 - 4 = 10.</math>
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import geometry;
  
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Points <math>A, E,</math> and <math>B</math> form a right triangle with legs <math>10</math> and <math>24.</math> We are looking for distance <math>AB,</math> which is <math>\sqrt{10^2 + 24^2} = 26</math>
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point C = origin;
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point A = (0,-4);
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point D = (24,0);
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point B = (24,-14);
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point E;
  
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The answer is <math>\textbf{(D)} 26.</math>
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// Defining point E
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pair[] e = intersectionpoints(perpendicular(A,line(B,D)),B--D);
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E = e[0];
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// Circles
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draw(circle(A,length(segment(A,C))));
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draw(circle(B,length(segment(B,D))));
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// Segments
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draw(A--B);
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draw(A--C);
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draw(B--D);
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draw(C--D);
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label("$24$",midpoint(C--D),N);
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draw(A--E);
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// Labelling Points
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dot(A);
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label("A",A,SW);
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dot(B);
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label("B",B,SE);
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dot(C);
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label("C",C,NW);
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dot(D);
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label("D",D,NE);
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dot(E);
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label("E",E,(1,0));
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// Right Angle Marks
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markscalefactor = 0.135;
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draw(rightanglemark(A,C,D));
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draw(rightanglemark(C,D,E));
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draw(rightanglemark(C,A,E));
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draw(rightanglemark(A,E,D));
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draw(rightanglemark(A,E,B));
 +
 
 +
</asy>
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 +
Let the center of the smaller circle be <math>A</math> and the center of the larger circle be <math>B</math>. Draw <math>\overline{AC}</math> and <math>\overline{BD}</math>, as in the diagram. We know that <math>\overline{AC}</math> and <math>\overline{BD}</math> are perpendicular to <math>\overline{CD}</math> because <math>\overline{CD}</math> is [[tangent (geometry)#Tangents to Circles|tangent]] to both circles. From the problem, we know <math>CD=24</math>.
 +
 
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Draw <math>\overline{AE} \perp \overline{BD}</math> with <math>E</math> on <math>\overline{BD}.</math> <math>ABDC</math> is a [[rectangle]], so, knowing the radii of the circles, we see that <math>BE = BD - ED = BD - AC = 14 - 4 = 10.</math>
 +
 
 +
Points <math>A, E,</math> and <math>B</math> form a right triangle with legs <math>10</math> and <math>24.</math> We are looking for <math>AB,</math> which is <math>\sqrt{10^2 + 24^2} = 26</math> by the [[Pythagorean Theorem]].
 +
 
 +
Thus, our answer is <math>\boxed{\textbf{(D) }26}</math>.
  
 
-edited by coolmath34
 
-edited by coolmath34
 +
 +
== See Also ==
 +
{{AHSME 35p box|year=1971|num-b=8|num-a=10}}
 +
{{MAA Notice}}

Latest revision as of 10:49, 1 August 2024

Problem

An uncrossed belt is fitted without slack around two circular pulleys with radii of $14$ inches and $4$ inches. If the distance between the points of contact of the belt with the pulleys is $24$ inches, then the distance between the centers of the pulleys in inches is

$\textbf{(A) }24\qquad \textbf{(B) }2\sqrt{119}\qquad \textbf{(C) }25\qquad \textbf{(D) }26\qquad \textbf{(E) }4\sqrt{35}$

Solution

[asy] import geometry; point C = origin; point A = (0,-4); point D = (24,0); point B = (24,-14); point E; // Defining point E pair[] e = intersectionpoints(perpendicular(A,line(B,D)),B--D); E = e[0]; // Circles draw(circle(A,length(segment(A,C)))); draw(circle(B,length(segment(B,D)))); // Segments draw(A--B); draw(A--C); draw(B--D); draw(C--D); label("$24$",midpoint(C--D),N); draw(A--E); // Labelling Points dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NW); dot(D); label("D",D,NE); dot(E); label("E",E,(1,0)); // Right Angle Marks markscalefactor = 0.135; draw(rightanglemark(A,C,D)); draw(rightanglemark(C,D,E)); draw(rightanglemark(C,A,E)); draw(rightanglemark(A,E,D)); draw(rightanglemark(A,E,B)); [/asy]

Let the center of the smaller circle be $A$ and the center of the larger circle be $B$. Draw $\overline{AC}$ and $\overline{BD}$, as in the diagram. We know that $\overline{AC}$ and $\overline{BD}$ are perpendicular to $\overline{CD}$ because $\overline{CD}$ is tangent to both circles. From the problem, we know $CD=24$.

Draw $\overline{AE} \perp \overline{BD}$ with $E$ on $\overline{BD}.$ $ABDC$ is a rectangle, so, knowing the radii of the circles, we see that $BE = BD - ED = BD - AC = 14 - 4 = 10.$

Points $A, E,$ and $B$ form a right triangle with legs $10$ and $24.$ We are looking for $AB,$ which is $\sqrt{10^2 + 24^2} = 26$ by the Pythagorean Theorem.

Thus, our answer is $\boxed{\textbf{(D) }26}$.

-edited by coolmath34

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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
All AHSME Problems and Solutions

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