Difference between revisions of "1985 AHSME Problems/Problem 10"

(Created page with "==Problem== An arbitrary circle can intersect the graph <math> y=\sin x </math> in <math> \mathrm{(A) } \text{at most }2\text{ points} \qquad \mathrm{(B) }\text{at mos...")
 
(Improved solution and formatting)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
An arbitrary [[circle]] can intersect the [[graph]] <math> y=\sin x </math> in
+
An arbitrary circle can intersect the graph of <math>y = \sin x</math> in
  
<math> \mathrm{(A) } \text{at most }2\text{ points} \qquad \mathrm{(B) }\text{at most }4\text{ points} \qquad \mathrm{(C)   } \text{at most }6\text{ points} \qquad \mathrm{(D) } \text{at most }8\text{ points}\qquad \mathrm{(E)   }\text{more than }16\text{ points}  </math>
+
<math> \mathrm{(A) \ } \text{at most }2\text{ points} \qquad \mathrm{(B) \ }\text{at most }4\text{ points} \qquad \mathrm{(C) } \text{at most }6\text{ points} \qquad \mathrm{(D) \ } \text{at most }8\text{ points}</math>
 +
<math>\mathrm{(E) }\text{more than }16\text{ points}  </math>
  
 
==Solution==
 
==Solution==
Consider a circle with center on the positive y-axis and that passes through the origin. As the radius of this circle becomes arbitrarily large, its shape near the x-axis becomes very similar to that of <math> y=0 </math>, which intersects <math> y=\sin x </math> infinitely many times. It therefore becomes obvious that we can pick a radius large enough so that the circle intersects the sinusoid as many times as we wish, <math> \boxed{\text{E}} </math>.
+
Consider a circle whose center lies on the positive <math>y</math>-axis and which passes through the origin. As the radius of this circle becomes arbitrarily large, its curvature near the <math>x</math>-axis becomes almost flat, and so it can intersect the curve <math>y = \sin x</math> arbitrarily many times (since the <math>x</math>-axis itself intersects the curve infinitely many times). Hence, in particular, we can choose a radius sufficiently large that the circle intersects the curve at <math>\boxed{\text{(E)} \ \text{more than }16 \text{ points}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=9|num-a=11}}
 
{{AHSME box|year=1985|num-b=9|num-a=11}}
 +
{{MAA Notice}}

Latest revision as of 19:35, 19 March 2024

Problem

An arbitrary circle can intersect the graph of $y = \sin x$ in

$\mathrm{(A) \ } \text{at most }2\text{ points} \qquad \mathrm{(B) \ }\text{at most }4\text{ points} \qquad \mathrm{(C) \  } \text{at most }6\text{ points} \qquad \mathrm{(D) \ } \text{at most }8\text{ points}$ $\mathrm{(E) \  }\text{more than }16\text{ points}$

Solution

Consider a circle whose center lies on the positive $y$-axis and which passes through the origin. As the radius of this circle becomes arbitrarily large, its curvature near the $x$-axis becomes almost flat, and so it can intersect the curve $y = \sin x$ arbitrarily many times (since the $x$-axis itself intersects the curve infinitely many times). Hence, in particular, we can choose a radius sufficiently large that the circle intersects the curve at $\boxed{\text{(E)} \ \text{more than }16 \text{ points}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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
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