Difference between revisions of "1983 AHSME Problems/Problem 22"
Sevenoptimus (talk | contribs) (Added a solution) |
Sevenoptimus (talk | contribs) m (Added box at the bottom) |
||
Line 14: | Line 14: | ||
We must describe geometrically those <math>(a,b)</math> for which the equation <math>x^2+2bx+1=2a(x+b)</math>, i.e. <math>x^2+2(b-a)x+(1-2ab)=0</math>, has no solutions (equivalent to the graphs not intersecting). By considering the discriminant of this quadratic equation, there are no solutions if and only if <math>\left(2(b-a)\right)^2 - 4(1)(1-2ab) < 0 \Rightarrow (b-a)^2 < 1 - 2ab \Rightarrow a^2 - 2ab + b^2 < 1 - 2ab \Rightarrow a^2 + b^2 < 1</math>. Thus <math>S</math> is the unit circle (without its boundary, due to the inequality sign being <math><</math> rather than <math>\leq</math>, but this makes no difference to the area), whose area is <math>\pi (1^2) = \pi</math>, so the answer is <math>\boxed{\textbf{(B)} \ \pi}</math>. | We must describe geometrically those <math>(a,b)</math> for which the equation <math>x^2+2bx+1=2a(x+b)</math>, i.e. <math>x^2+2(b-a)x+(1-2ab)=0</math>, has no solutions (equivalent to the graphs not intersecting). By considering the discriminant of this quadratic equation, there are no solutions if and only if <math>\left(2(b-a)\right)^2 - 4(1)(1-2ab) < 0 \Rightarrow (b-a)^2 < 1 - 2ab \Rightarrow a^2 - 2ab + b^2 < 1 - 2ab \Rightarrow a^2 + b^2 < 1</math>. Thus <math>S</math> is the unit circle (without its boundary, due to the inequality sign being <math><</math> rather than <math>\leq</math>, but this makes no difference to the area), whose area is <math>\pi (1^2) = \pi</math>, so the answer is <math>\boxed{\textbf{(B)} \ \pi}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1983|num-b=21|num-a=23}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 23:59, 19 February 2019
Problem
Consider the two functions and , where the variable and the constants and are real numbers. Each such pair of constants and may be considered as a point in an -plane. Let be the set of such points for which the graphs of and do not intersect (in the -plane). The area of is
Solution
We must describe geometrically those for which the equation , i.e. , has no solutions (equivalent to the graphs not intersecting). By considering the discriminant of this quadratic equation, there are no solutions if and only if . Thus is the unit circle (without its boundary, due to the inequality sign being rather than , but this makes no difference to the area), whose area is , so the answer is .
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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.