Difference between revisions of "1965 AHSME Problems/Problem 39"
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− | + | == Problem == | |
− | + | ||
− | + | A foreman noticed an inspector checking a <math>3</math>"-hole with a <math>2</math>"-plug and a <math>1</math>"-plug and suggested that two more gauges | |
+ | be inserted to be sure that the fit was snug. If the new gauges are alike, then the diameter, <math>d</math>, of each, to the nearest hundredth of an inch, is: | ||
+ | |||
+ | <math>\textbf{(A)}\ .87 \qquad | ||
+ | \textbf{(B) }\ .86 \qquad | ||
+ | \textbf{(C) }\ .83 \qquad | ||
+ | \textbf{(D) }\ .75 \qquad | ||
+ | \textbf{(E) }\ .71 </math> | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point O = (0,0); | ||
+ | point A = (-1/2,0); | ||
+ | point B = (1,0); | ||
+ | point C, D, E; | ||
+ | |||
+ | // 1", 2", and 3" circles | ||
+ | draw(circle(O,3/2)); | ||
+ | dot(O); | ||
+ | label("O", O, S); | ||
+ | draw(circle(A,1)); | ||
+ | dot(A); | ||
+ | label("A", A, SW); | ||
+ | draw(circle(B,1/2)); | ||
+ | dot(B); | ||
+ | label("B", B, SE); | ||
+ | |||
+ | // Other two circles | ||
+ | C=(15/14*3/5,15/14*4/5); | ||
+ | D=(15/14*3/5,15/14*-4/5); | ||
+ | dot(C); | ||
+ | label("C",C,NW); | ||
+ | draw(circle(C, 3/7)); | ||
+ | dot(D); | ||
+ | label("D",D,S); | ||
+ | draw(circle(D, 3/7)); | ||
+ | |||
+ | // Point E | ||
+ | pair[] e=intersectionpoints(line(O,C), circle(O, 3/2)); | ||
+ | E=e[1]; | ||
+ | dot(E); | ||
+ | label("E", E, NE); | ||
+ | |||
+ | // Segments | ||
+ | draw(A--B--C--A); | ||
+ | draw(O--E); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | Let the center of the <math>3</math>" circle be <math>O</math>, that of the <math>2</math>" circle be <math>A</math>, that of the <math>1</math>" circle be <math>B</math>, and those of the circles of unknown radius (let their radii have length <math>r</math>) be <math>C</math> and <math>D</math>, as in the diagram. Also, in this problem, no two circles share a center, so let the circles be named by their corresponding centers (so, the <math>3</math>" circle is circle <math>O</math>, etc.). Extend <math>\overline{OC}</math> past <math>C</math> to intersect circle <math>O</math> at point <math>E</math>. Because circle <math>O</math> has radius <math>\frac{3}{2}</math> and circle <math>A</math> has radius <math>1</math>, <math>OA=\frac{3}{2}-1=\frac{1}{2}</math>. Likewise, because <math>B</math> has radius <math>\frac{1}{2}</math>, <math>OB=\frac{3}{2}-\frac{1}{2}=1</math>. Thus, <math>AB=\frac{3}{2}</math>. Furthermore, because the line connecting the centers of two tangent circles goes through their point of tangency, <math>AC=r+1</math> and <math>BC=r+\frac{1}{2}</math>. Because <math>OE=\frac{3}{2}</math> and <math>CE=r</math>, <math>OC=\frac{3}{2}-r</math>. With this information, we can now apply [[Stewart's Theorem]] to <math>\triangle ABC</math> to solve for <math>r</math>: | ||
+ | \begin{align*} | ||
+ | OB*OA*AB+OC^2*AB&=AC^2*OB+BC^2*OA \ | ||
+ | 4[(1)(\frac{1}{2})(\frac{3}{2})+(\frac{3}{2}-r)^2(\frac{3}{2})]&=4[(r+1)^2(1)+(r+\frac{1}{2})^2(\frac{1}{2})] \ | ||
+ | 3+6(\frac{9}{4}-3r+r^2)&=4(r^2+2r+1)+2(r^2+r+\frac{1}{4}) \ | ||
+ | 3+\frac{27}{2}-18r+6r^2&=4r^2+8r+4+2r^2+2r+\frac{1}{2} \ | ||
+ | 3+\frac{27-1}{2}-4+6r^2-6r^2&=18r+8r+2r \ | ||
+ | 3+13-4&=28r \ | ||
+ | 28r&=12 \ | ||
+ | r&=\frac{3}{7} | ||
+ | \end{align*} | ||
+ | Because the question asks for the diameter of the circle, we calculate <math>2r=\frac{6}{7}\approx \boxed{\textbf{(B) }0.86}</math>. | ||
+ | |||
+ | == See Also == | ||
+ | {{AHSME 40p box|year=1965|num-b=38|num-a=40}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 13:11, 20 July 2024
Problem
A foreman noticed an inspector checking a "-hole with a "-plug and a "-plug and suggested that two more gauges be inserted to be sure that the fit was snug. If the new gauges are alike, then the diameter, , of each, to the nearest hundredth of an inch, is:
Solution
Let the center of the " circle be , that of the " circle be , that of the " circle be , and those of the circles of unknown radius (let their radii have length ) be and , as in the diagram. Also, in this problem, no two circles share a center, so let the circles be named by their corresponding centers (so, the " circle is circle , etc.). Extend past to intersect circle at point . Because circle has radius and circle has radius , . Likewise, because has radius , . Thus, . Furthermore, because the line connecting the centers of two tangent circles goes through their point of tangency, and . Because and , . With this information, we can now apply Stewart's Theorem to to solve for :
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 38 |
Followed by Problem 40 | |
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 | ||
All AHSME Problems and Solutions |
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