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Difference between revisions of "1983 AHSME Problems/Problem 22"

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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>.
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==See Also==
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{{AHSME box|year=1983|num-b=21|num-a=23}}
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{{MAA Notice}}

Latest revision as of 00:59, 20 February 2019

Problem

Consider the two functions $f(x) = x^2+2bx+1$ and $g(x) = 2a(x+b)$, where the variable $x$ and the constants $a$ and $b$ are real numbers. Each such pair of constants $a$ and $b$ may be considered as a point $(a,b)$ in an $ab$-plane. Let $S$ be the set of such points $(a,b)$ for which the graphs of $y = f(x)$ and $y = g(x)$ do not intersect (in the $xy$-plane). The area of $S$ is

$\textbf{(A)} \ 1 \qquad \textbf{(B)} \ \pi \qquad \textbf{(C)} \ 4 \qquad \textbf{(D)} \ 4 \pi \qquad \textbf{(E)} \ \text{infinite}$

Solution

We must describe geometrically those $(a,b)$ for which the equation $x^2+2bx+1=2a(x+b)$, i.e. $x^2+2(b-a)x+(1-2ab)=0$, 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 $\left(2(b-a)\right)^2 - 4(1)(1-2ab) < 0 \Rightarrow (b-a)^2 < 1 - 2ab \Rightarrow a^2 - 2ab + b^2 < 1 - 2ab \Rightarrow a^2 + b^2 < 1$. Thus $S$ is the unit circle (without its boundary, due to the inequality sign being $<$ rather than $\leq$, but this makes no difference to the area), whose area is $\pi (1^2) = \pi$, so the answer is $\boxed{\textbf{(B)} \ \pi}$.

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


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