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Difference between revisions of "1969 AHSME Problems/Problem 21"

(→‎Problem)
(Solution to Problem 21)
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== Solution ==
 
== Solution ==
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<math>\fbox{E}</math>
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Note that the first equation represents a circle and the second equation represents a line.  If a line is tangent to a circle, then it only hits at one point, so there will only be one solution to the [[system of equations]].
  
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== See also ==
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In the second equation, <math>y = -x + \sqrt{2m}</math>.  Substitution results in
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<cmath>x^2 + x^2 - 2x\sqrt{2m} + 2m = m</cmath>
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<cmath>2x^2 - 2x\sqrt{2m} + m = 0</cmath>
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In order for the system to have one solution, the [[discriminant]] must equal <math>0</math>.
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<cmath>4(2m) - 4 \cdot 2 \cdot m = 0</cmath>
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<cmath>0 = 0</cmath>
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Thus, <math>m</math> can be a non-negative real number, so the answer is <math>\boxed{\textbf{(E)}}</math>.
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== See Also ==
 
{{AHSME 35p box|year=1969|num-b=20|num-a=22}}   
 
{{AHSME 35p box|year=1969|num-b=20|num-a=22}}   
  
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[[Category: Intermediate Geometry Problems]]
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[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:03, 20 June 2018

Problem

If the graph of $x^2+y^2=m$ is tangent to that of $x+y=\sqrt{2m}$, then:

$\text{(A) m must equal } \tfrac{1}{2}\quad \text{(B) m must equal } \frac{1}{\sqrt{2}}\quad\\ \text{(C) m must equal } \sqrt{2}\quad \text{(D) m must equal } 2\quad\\ \text{(E) m may be any non-negative real number}$

Solution

Note that the first equation represents a circle and the second equation represents a line. If a line is tangent to a circle, then it only hits at one point, so there will only be one solution to the system of equations.

In the second equation, $y = -x + \sqrt{2m}$. Substitution results in \[x^2 + x^2 - 2x\sqrt{2m} + 2m = m\] \[2x^2 - 2x\sqrt{2m} + m = 0\] In order for the system to have one solution, the discriminant must equal $0$. \[4(2m) - 4 \cdot 2 \cdot m = 0\] \[0 = 0\] Thus, $m$ can be a non-negative real number, so the answer is $\boxed{\textbf{(E)}}$.

See Also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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

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