Difference between revisions of "1971 AHSME Problems/Problem 9"
Coolmath34 (talk | contribs) (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...") |
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− | == Problem | + | == 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== | ||
− | |||
− | + | <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 <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 | -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 inches and inches. If the distance between the points of contact of the belt with the pulleys is inches, then the distance between the centers of the pulleys in inches is
Solution
Let the center of the smaller circle be and the center of the larger circle be . Draw and , as in the diagram. We know that and are perpendicular to because is tangent to both circles. From the problem, we know .
Draw with on is a rectangle, so, knowing the radii of the circles, we see that
Points and form a right triangle with legs and We are looking for which is by the Pythagorean Theorem.
Thus, our answer is .
-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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.