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

Latest revision as of 11:49, 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

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

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

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

Latest revision as of 11:49, 1 August 2024

Problem

Solution

See Also

@@ Line 1: / Line 1: @@
-== Problem 9 ==
+== Problem ==
 An uncrossed belt is fitted without slack around two circular pulleys with radii of <math>14</math> inches and <math>4</math> inches.
@@ Line 12: / Line 12: @@
 ==Solution==
-(I don't know how to use Asymptote, so if you are a visual learner, I apologize)
-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.
+<asy>
-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>
+import geometry;
-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>
+point C = origin;
+point A = (0,-4);
+point D = (24,0);
+point B = (24,-14);
+point E;
-The answer is <math>\textbf{(D)} 26.</math>
+// 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 <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>.
+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
+== See Also ==
+{{AHSME 35p box|year=1971|num-b=8|num-a=10}}
+{{MAA Notice}}