Difference between revisions of "1969 AHSME Problems/Problem 21"
(Created page with "== Problem == If the graph of <math>x^2+y^2=m</math> is tangent to that of <math>x+y=\sqrt{2m}</math>, then: <math>\text{(A) m must equal } \tfrac{1}{2}\quad \text{(B) m must e...") |
m (→Solution 2) |
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\text{(C) m must equal } \sqrt{2}\quad | \text{(C) m must equal } \sqrt{2}\quad | ||
\text{(D) m must equal } 2\quad\\ | \text{(D) m must equal } 2\quad\\ | ||
| − | \text{(E) m may be | + | \text{(E) m may be any non-negative real number} </math> |
| − | == Solution == | + | == Solution 1 == |
| − | + | 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]]. | |
| − | == See | + | In the second equation, <math>y = -x + \sqrt{2m}</math>. Substitution results in |
| + | <cmath>x^2 + x^2 - 2x\sqrt{2m} + 2m = m</cmath> | ||
| + | <cmath>2x^2 - 2x\sqrt{2m} + m = 0</cmath> | ||
| + | In order for the system to have one solution, the [[discriminant]] must equal <math>0</math>. | ||
| + | <cmath>4(2m) - 4 \cdot 2 \cdot m = 0</cmath> | ||
| + | <cmath>0 = 0</cmath> | ||
| + | Thus, <math>m</math> can be a non-negative real number, so the answer is <math>\boxed{\textbf{(E)}}</math>. | ||
| + | |||
| + | == Solution 2 == | ||
| + | Since <math>x^2+y^2=m</math> has [[radius]] <math>\sqrt{m}</math>, and <math>x+y=\sqrt{2m}</math> is a [[diagonal]] [[line]] in the [[plane]] with [[slope]] <math>-1</math>, if the two [[curves]] are [[tangent]] then <math>x+y=\sqrt{2m}</math> either has to pass through <math>(\frac{\sqrt{2m}}{2}, \frac{\sqrt{2m}}{2})</math> or <math>(-\frac{\sqrt{2m}}{2}, -\frac{\sqrt{2m}}{2})</math>. However, as <math>\sqrt{2m}</math> is [[positive]], the [[line]] can only pass through <math>(\frac{\sqrt{2m}}{2}, \frac{\sqrt{2m}}{2})</math>, which it does for all <math>m</math>. Thus the answer is <math>\boxed{\textbf{(E)}}</math>. | ||
| + | |||
| + | == See Also == | ||
{{AHSME 35p box|year=1969|num-b=20|num-a=22}} | {{AHSME 35p box|year=1969|num-b=20|num-a=22}} | ||
| − | [[Category: | + | [[Category: Introductory Algebra Problems]] |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:37, 13 November 2023
Contents
Problem
If the graph of 
is tangent to that of 
, then:
Solution 1
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, 
. Substitution results in
![\[x^2 + x^2 - 2x\sqrt{2m} + 2m = m\]](http://latex.artofproblemsolving.com/9/8/5/98545a7fcc27b1b528c6f2eae6456f499abaaffa.png)
![\[2x^2 - 2x\sqrt{2m} + m = 0\]](http://latex.artofproblemsolving.com/b/f/c/bfc8ae7e6732394a3956c424a8d7bc5317ca1ecb.png)
In order for the system to have one solution, the discriminant must equal 
.
![\[4(2m) - 4 \cdot 2 \cdot m = 0\]](http://latex.artofproblemsolving.com/6/f/6/6f68a335464c06d887361e1bba2a73f035bee73b.png)
![\[0 = 0\]](http://latex.artofproblemsolving.com/8/1/1/811bb7773c79bdafc86360780f01eb98b793cb10.png)
Thus, 
can be a non-negative real number, so the answer is 
.
Solution 2
Since has radius 
, and is a diagonal line in the plane with slope 
, if the two curves are tangent then either has to pass through 
or 
. However, as 
is positive, the line can only pass through , which it does for all . Thus the answer is .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
